A-Level Biology Exam Questions

Page 02 SECTION 1 — 25 marks Attempt ALL questions 1. An experiment was set up to measure the activity of an enzyme using a substrate that produced a coloured product. The absorbance of the coloured product was measured using a colorimeter. Which row in the table describes the variable being measured? discrete continuous qualitative quantitative A B C D 2. The diagram below shows the cell types used in the production of monoclonal antibodies. Cell type 1: Cells isolated from animal treated with antigen Cell type 2: Cancerous cell line Fusion of cell types 1 and 2 using PEG Growth of cells in selective medium Cell type 3: Cells producing monoclonal antibodies Which row in the table identifies cell types 1, 2 and 3? Cell type 1 Cell type 2 Cell type 3 A B lymphocyte myeloma hybridoma B myeloma hybridoma B lymphocyte C hybridoma myeloma B lymphocyte D myeloma B lymphocyte hybridoma

Page 03 3. The diagram below represents a transmembrane protein. Some of the amino acids in the protein have been identified. Membrane Intracellular Extracellular Arg Ala Ser Leu Leu Ser Which row in the table classifies the amino acids shown in this protein? Arginine (Arg) Alanine (Ala) Leucine (Leu) Serine (Ser) A polar hydrophobic hydrophobic polar B hydrophobic polar hydrophobic polar C polar hydrophobic polar hydrophobic D hydrophobic polar polar hydrophobic [Turn over Page 04 Questions 4 and 5 refer to the following information. During muscle contraction, the protein myosin moves along an actin protein filament by the head of the myosin detaching from the actin, swinging forward and rebinding, as shown in the diagram. myosin 36 nm actin [1 nm = 1 × 10–9 m] 4. This reversible conformational change can be brought about by binding of ATP to the myosin head followed by hydrolysis and release of phosphate and ADP. The myosin head is acting as A a kinase B an ATPase C a proteinase D a phosphatase. 5. When the myosin head detaches and swings forward it moves a distance of 36 nanometres (nm). Myosin has been observed to move at a speed of 18 × 103 nm s–1. How many times will the myosin head detach and swing forward in one second? A 50 B 200 C 500 D 2000 6. In animal rod cells rhodopsin absorbs a photon of light initiating the following cell events. 1 nerve impulse is generated 2 sufficient product formation is triggered 3 activation of hundreds of G-protein molecules 4 activation of hundreds of molecules of an enzyme The correct order of events is A 4, 2, 1, 3 B 3, 4, 2, 1 C 4, 3, 1, 2 D 3, 2, 4, 1.

Page 05 7. In multicellular organisms, only target cells respond to a specific signal because A signalling molecules only come into contact with target cells B only target cells have receptor molecules for the signalling molecule C non-target cells do not respond when the signalling molecule binds to its receptor D receptor molecules in non-target cells do not change conformation when the signalling molecule binds. 8. The hormone thyroxine is A hydrophobic and unable to pass through the cell membrane B hydrophilic and unable to pass through the cell membrane C hydrophobic and able to pass through the cell membrane D hydrophilic and able to pass through the cell membrane. 9. Biological molecules move over short distances by diffusion. The time taken for diffusion can be calculated using the equation below. t = time taken (seconds) x = distance travelled by the diffusing molecule (cm) D = diffusion co-efficient (cm2 per second) = x 2 D t 2 Acetylcholine is a neurotransmitter with a diffusion co-efficient of 4 × 10–6 cm2 per second. The gap across the synapse is 5 × 10–6 cm wide. How many seconds would it take acetylcholine to cross the synapse? A 1·250 B 6·250 × 10–6 C 3·125 × 10–6 D 1·600 × 10–6 10. Type 1 diabetes is caused by A excessive production of insulin B loss of insulin receptor function C failure of GLUT4 to respond to insulin binding D insufficient production of insulin. [Turn over

Page 06 11. An enzyme-controlled reaction is taking place in optimum conditions in the presence of a large surplus of substrate. Conditions can be altered by 1 increasing the temperature 2 adding a positive modulator 3 increasing enzyme concentration 4 increasing substrate concentration. Product yield would be increased by A 1 and 2 B 2 and 3 C 2 and 4 D 3 and 4. 12. At which phase of the cell cycle is the retinoblastoma protein phosphorylated allowing progression to the next phase of the cycle? A G1 B S C G2 D M

Page 07 13. The diagram below shows possible outcomes for a cell following DNA damage. Protein X is involved in all three outcomes. DNA Damage Protein X DNA Repair Cell Cycle Arrest Apoptosis Protein X is A Rb B p53 C Cdk D caspase. 14. Two reagents used in testing for the presence of carbohydrates are iodine solution, which turns blue-black in the presence of starch, and Benedict’s solution, which turns brick red in the presence of maltose. In an investigation of the breakdown of starch into maltose by the enzyme amylase, which of the following would be a positive control? A Maltose alone turns Benedict’s solution brick red. B Starch treated with amylase turns Benedict’s solution brick red. C Starch alone tested with Benedict’s solution remains blue. D Starch treated with amylase does not change the colour of iodine solution. [Turn over

Page 08 15. The diagram below shows some phyla in the animal kingdom. Round worms Sea squirts Invertebrates with jointed legs Animal Kingdom X Y Z Which row in the table identifies the phyla X, Y and Z? Phylum X Y Z A Chordata Nematoda Arthropoda B Arthropoda Nematoda Chordata C Nematoda Arthropoda Chordata D Arthropoda Chordata Nematoda 16. Which of the following descriptions of animal behaviour avoids the use of anthropomorphism? A In some primate species, alpha males often bully lower-ranking animals. B In late summer, worker bees like to visit heather flowers. C The grin on the chimpanzee’s face showed that it was amused by the gesture. D The male moth is attracted to the female by the scent molecules that she emits. 17. A population of chafer beetles were damaging the tees and greens of a golf course. Results from a mark and recapture study suggested a population size that was too small to account for the extent of the damage caused. One possible reason for this is that the A white paint used to mark the beetles washed off some of them before the recapture B white paint used to mark the beetles made them more visible to predators than unmarked beetles C total number of beetles in the recaptured sample was less than the number first captured and marked D marked beetles did not have enough time, after release, to spread out and mix with the rest of the population.

Page 09 18. Ellis-van Creveld syndrome is a rare genetic condition. It is much more common in an isolated population in North America, which was founded by a small number of individuals, than in the general population. The most likely explanation for this is A natural selection B sexual selection C random mutation D genetic drift. 19. The frequency of a given allele in a population is a measure of how common that allele is as a proportion of the total number of copies of all alleles at a specific locus. For a locus with one dominant allele (A) and one recessive allele (a), the frequency of the dominant allele (p) and the frequency of the recessive allele (q) can be used to calculate the genetic variation of a population using the equations below. p + q = 1 p = frequency of A allele q = frequency of a allele p2 + 2pq + q2 = 1 p2 = frequency of homozygous (AA) individuals q2 = frequency of homozygous (aa) individuals 2pq = frequency of heterozygous (Aa) individuals If the allele frequency of the recessive allele is 0·7, the proportion of individuals that would be heterozygous is A 0·09 B 0·21 C 0·42 D 0·49. [Turn over

Page 10 20. In the fruit fly Drosophila melanogaster the gene for eye colour is sex-linked. The allele for red eye (R) is dominant to the allele for white eye (r). A cross between two flies produced the offspring shown in the table below. Sex of offspring Number with white eyes Number with red eyes female 23 22 male 21 22 The genotypes of the parents in this cross were A XrXr and XRY B XRXr and XrY C XRXr and XRY D XRXR and XrY. 21. Which row in the table best describes r-selected species? Number of offspring Offspring survival rate Parental care A many low little B few high extensive C many high extensive D few low little

Page 11 22. Shags and cormorants both belong to the genus Phalacrocorax. They look very similar and nest near each other on the same cliffs. The table below shows the main components of each bird’s diet. Prey Percentage composition of diet Shag (Phalacrocorax aristotelis) Cormorant (Phalacrocorax carbo) sand eels 33 0 sprats 49 1 flatfish 1 26 shrimps 2 33 gobies 4 17 other fish 4 18 The data in the table show A competitive exclusion B competition within each species C resource partitioning D the fundamental niche of each species. [Turn over

Page 12 23. A species of parasitic wasp (Nasonia vitripennis) lays its eggs in the larvae of flies where the eggs develop. This species displays a behaviour called “superparasitism” where, following the laying of eggs by one wasp, a second wasp may superparasitise the same host by also laying its eggs. Researchers investigated the effects of superparasitism on the brood size and sex ratio of offspring in this species. Results were compared to a control that had been parasitised only once. Researchers were able to distinguish between the offspring of the first and second wasp. Results are shown in the table below. Offspring Degree of parasitism Superparasitism Single parasitism control Wasp 1 Wasp 2 brood size 18 ± 3 17 ± 4 20 ± 2 percentage of males 7 ± 2 22 ± 4 6 ± 1 The following statements refer to the data in the table. 1 Superparasitism significantly increased the percentage of males produced by both wasp 1 and wasp 2. 2 Superparasitism significantly increased the percentage of males produced by wasp 2 only. 3 Superparasitism had no significant effect on brood size. 4 Superparasitism significantly decreased the brood size produced by wasps 1 and 2. Which of these statements are valid conclusions supported by the data? A 1 and 3 B 1 and 4 C 2 and 3 D 2 and 4

Page 13 24. The statements below describe events that occur following the engulfing of a pathogen by a phagocyte of the mammalian immune system. P long term survival of lymphocytes Q antigen presentation to lymphocytes R antibody production by lymphocytes S clonal selection of B lymphocytes The correct sequence in which these events occur is A Q, R, S, P B R, Q, P, S C S, Q, P, R D Q, S, R, P. 25. Florida scrubjays have evolved a co-operative breeding system in which helper birds assist breeding pairs in raising young. The table below compares the effect of helpers on the breeding success of birds that are either experienced or inexperienced breeders. Breeding experience of breeding pairs Average number of offspring reared Without helpers With helpers inexperienced 1·24 2·20 experienced 1·80 2·38 Helpers increase the average number of offspring reared by inexperienced breeding pairs compared to experienced breeding pairs by A 19% B 23% C 45% D 60%. [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET] Page 14 [BLANK PAGE] do not write on this page Page 15 [BLANK PAGE] do not write on this page Page 16 [BLANK PAGE] do not write on this page AH FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2016 Total marks — 90 SECTION 1 — 25 marks Attempt ALL questions. Instructions for completion of Section 1 are given on Page 02. SECTION 2 — 65 marks Attempt ALL questions. A Supplementary Sheet for Question 1 is enclosed inside the front cover of this question paper. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. You should score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not you may lose all the marks for this paper. X707/77/01 MONDAY, 9 MAY 9:00 AM – 11:30 AM A/HTP Biology Section 1 — Answer Grid and Section 2 Page 02 SECTION 1 — 25 marks The questions for Section 1 are contained in the question paper X707/77/02. Read these and record your answers on the answer grid on Page 03 opposite. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample Question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D Page 03 A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 SECTION 1 — Answer Grid [Turn over Page 04 [BLANK PAGE] do not write on this page Page 05 [Turn over for next question do not write on this page Page 06 MARKS DO NOT WRITE IN THIS MARGIN SECTION 2 — 65 marks Attempt ALL questions It should be noted that question 11 contains a choice. 1. Read through the Supplementary Sheet for Question 1 before attempting this question. (a) Refer to Figure 2 in the Supplementary Sheet for Question 1. (i) Use the data to describe the egg-laying of uninfected mosquitoes. (ii) If the box plots were perfectly symmetrical, mean values for egg-laying would be very close to median values. State what can be deduced about the mean number of eggs laid by infected mosquitoes in relation to the median value. (iii) Describe the effect that Plasmodium infection has on the fecundity of mosquitoes used in the study. (b) Refer to Figure 3 in the Supplementary Sheet for Question 1. (i) The data shows that infection by Plasmodium appears to increase the longevity of female mosquitoes. Explain why the difference between the two groups can be regarded as significant. (ii) Suggest a benefit to the parasite of its vector living longer. 2 1 1 1 1 Page 07 MARKS DO NOT WRITE IN THIS MARGIN 1. (continued) (c) Refer to Figure 4 in the Supplementary Sheet for Question 1. (i) Explain what the lines of best fit indicate about the relationship between longevity and fecundity in both infected and uninfected mosquitoes. (ii) State, with justification, whether or not this data is reliable. [Turn over 2 1 Page 08 MArKS DO NOT WRITE IN THIS MARGIN 2. Scientists have reported that neurons produced in cell culture from human stem cells have the potential to function when grafted into the site of a spinal injury in rats. (a) State why the cell culture medium in which the neurons were cultured should contain serum. (b) Scientists used a haemocytometer to perform a cell count to calculate the number of stem cells that developed into neurons. The diagram below represents a sample from a culture placed in a haemocytometer and viewed under a microscope. The grid is 0·1 mm in depth. Key: 1 mm Neuron Stem cell 1 mm (i) Calculate the number of neurons in 1 cm3 of the culture. Space for calculation neurons (ii) Suggest one disadvantage of cell counts performed using the haemocytometer. 1 1 1 Page 09 MARKS DO NOT WRITE IN THIS MARGIN 2. (continued) (c) Bright field microscopy was used to view the cells grafted into the site of spinal injury. State another type of biological material that can be viewed using bright field microscopy. (d) In studies involving animals, state one way in which harm to the animals can be minimised. [Turn over 1 1 Page 10 MARKS DO NOT WRITE IN THIS MARGIN 3. Multiple sclerosis (MS) is a neurological condition in which the body’s immune system destroys the myelin sheath that surrounds and insulates nerve axons. A clinical study was carried out into the effects of a new drug interferon beta-1b for this condition. A randomised trial, with a negative control group (placebo), was carried out across four different health centres. During the study patients were given one of three treatments: 0·00 mg (placebo), 0·05 mg or 0·25 mg interferon. The patients administered the drug themselves at home. The study measured how effective the drug was by asking patients to record any worsening of symptoms after 2 years of treatment. The study involved 372 patients aged 18-50 years. Fourteen patients dropped out before completing the trial. Patients’ results are shown in Table 1. Table 1 Level of interferon beta-1b in treatment (mg) Proportion of patients reporting no worsening of symptoms after 2 years of treatment (%) 0·00 16 0·05 18 0·25 25 At one health centre 52 patients were MRI scanned every 6 weeks to monitor any new damage to nerve tissue. The results are shown in Table 2. Table 2 Level of interferon beta-1b in treatment (mg) Proportion of patients showing new nerve damage (%) 0·00 29 0·05 no data recorded 0·25 6 (a) Identify the independent variable in this trial. (b) This trial was carried out in vivo. State one advantage of this type of trial. 1 1 Page 11 MARKS DO NOT WRITE IN THIS MARGIN 3. (continued) (c) Explain why a placebo group was included in this trial. (d) Suggest one way in which the results of the trial may not be reliable. (e) Describe an ethical issue that the researchers would need to consider before this trial. (f) Suggest two conclusions that can be drawn from the results of this trial. Conclusion 1 Conclusion 2 [Turn over 1 1 1 2 Page 12 MARKS DO NOT WRITE IN THIS MARGIN 4. Sickle cell anaemia is an inherited blood disorder that reduces the ability of red blood cells to transport oxygen round the body by changing the structure of haemoglobin. In sickle cell anaemia, the primary structure of a haemoglobin subunit is altered; the amino acid glutamic acid is substituted by the amino acid valine. The structures of glutamic acid and valine are shown below. COOH C H H2N C H H C H H C O O COOH C C CH3 H2N H H3C H Glutamic acid Valine R group (a) State the class of amino acids to which valine belongs. (b) Identify one type of secondary structure shown in the haemoglobin molecule in the figure below. Haem Haem Haem Haem 1 1 Page 13 MARKS DO NOT WRITE IN THIS MARGIN 4. (continued) (c) Explain the term cooperativity in relation to oxygen binding to haemoglobin. (d) The graph below shows the oxygen saturation of haemoglobin at different oxygen pressures for an individual with normal haemoglobin and for another individual with sickle cell haemoglobin. Normal haemoglobin Sickle cell haemoglobin 0 10 20 30 40 50 60 70 80 90 100 Oxygen pressure (units) 100 90 80 70 60 50 40 30 20 10 0 Oxygen saturation (%) Key: Use the graph to compare the oxygen saturation of normal and sickle cell haemoglobin as oxygen pressure increases. (e) Molecules of sickle cell haemoglobin clump together preventing access to oxygen binding sites. Suggest why this is a result of the substitution of glutamic acid by valine. 1 2 1 [Turn over Page 14 MARKS DO NOT WRITE IN THIS MARGIN 5. Describe the structure of spindle fibres and explain their role in the movement of chromosomes during cell division. 4 Page 15 MArKS DO NOT WRITE IN THIS MARGIN 6. The sodium potassium pump (Na/KATPase) is a membrane protein found in animal cells. (a) Give one function of sodium potassium pumps. (b) Describe the role of ATP in altering the affinity of the pump for sodium ions (Na+). (c) Digoxin is a chemical that inhibits the sodium potassium pump by binding to the potassium ion (K+) binding site as shown in the diagram below. Outside cell Inside cell K+ and digoxin binding site Cell membrane 2 K+ Na+ binding site 3 Na+ Explain why binding by digoxin prevents further binding of sodium (Na+) ions by the pump. 1 2 2 [Turn over Page 16 MArKS DO NOT WRITE IN THIS MARGIN 7. Binding of antidiuretic hormone (ADH) to its receptor on the plasma membrane of kidney collecting duct cells triggers the recruitment of water channel proteins as shown below. ADH Activated G proteins recruit water channel proteins Extracellular Intracellular Plasma membrane } (a) (i) Name the water channel protein involved in this process. (ii) Name the process by which a response within the cell is triggered by the binding of ADH to its cell surface receptor. 1 1 Page 17 MArKS DO NOT WRITE IN THIS MARGIN 7. (continued) (b) A urine output of greater than 0·05 litres per kg body mass per day is considered diagnostic of diabetes insipidus. The bar chart below shows the urine output over 6 days of a 70 kg individual being investigated for diabetes insipidus. During days 3 and 4 the individual was treated with the drug desmopressin, a synthetic form of ADH. 12 10 8 6 4 2 0 1 2 3 4 5 6 Day Total daily urine output (litres) (i) Use the data to confirm that a diagnosis of diabetes insipidus is correct for this individual. Space for calculation (ii) Give evidence from the graph that supports the conclusion that desmopressin is an effective treatment. (iii) Diabetes insipidus results from failure to recruit water channel proteins to the cell membrane. Identify the cause of recruitment failure in this individual. 1 1 1 [Turn over Page 18 MARKS DO NOT WRITE IN THIS MARGIN 8. The diagram below shows the pairing of homologous chromosomes in a cell undergoing meiosis. chiasma formation (a) Name the type of cell that undergoes meiosis. (b) (i) Explain how the chiasma formation between the paired homologous chromosomes shown in the diagram leads to variation. (ii) Name the process that ensures haploid gametes produced by meiosis contain a mixture of chromosomes of maternal and paternal origin. 1 2 1 Page 19 [Turn over for next question do not write on this page Page 20 MARKS DO NOT WRITE IN THIS MARGIN 9. In 1971, biologists moved five adult pairs of Italian wall lizards (Podarcis sicula) from their small home island of Kopiste to the neighbouring small island of Mrcaru, which did not have a lizard population. On their return in 2005 Mrcaru was found to have a large population of P. sicula (confirmed by genetic analysis) with significantly larger heads and a greater bite force than the lizards from Kopiste. Their digestive systems were also found to contain microorganisms that assist in the breakdown of plant cell walls. The summer diets of the two lizard populations are shown below. Key: plant matter arthropods other Kopiste Mrcaru Island 1·0 0·8 0·6 0·4 0·2 0·0 Proportion of total food mass (a) Describe the most significant change in the summer diet of the lizards on Mrcaru. 1 Page 21 MARKS DO NOT WRITE IN THIS MARGIN 9. (continued) (b) (i) Explain how the information supports the conclusion that the changes to the lizard population on Mrcaru were the result of natural selection. (ii) Evolution of the lizards on Mrcaru occurred very rapidly. State one factor that can increase the rate of evolution. (c) This study involved taking representative samples of the lizard populations on the two islands. State one feature of a representative sample. [Turn over 2 1 1 Page 22 MARKS DO NOT WRITE IN THIS MARGIN 10. The Figures below show male and female capercaillies (Tetrao urogallus) which are found in some Scottish pine forests. Males are much larger and darker than females and the breast feathers of the male have a metallic green sheen. female capercaillie male capercaillie (a) State the term used to indicate the different body forms of males and females belonging to this species. (b) Capercaillies are a lekking species. Males perform displays during which they fan their tails, hold their wings down and make a variety of sounds. These features, which are attractive to females, are thought to serve as honest signals. (i) Explain what is meant by a lekking species. (ii) Explain why this display is often given as an example of sexual selection. (iii) If the display provides honest signals, state the benefit that may be obtained by females receiving these signals. 1 1 1 1 Page 23 MARKS DO NOT WRITE IN THIS MARGIN 10. (continued) (c) Peacocks are the males of another lekking bird species, Pavo cristatus, whose natural habitat is the dense forests of South-East Asia. As well as the visual stimulus of a tail-feather display, peacocks, during mating, can emit a distinctive “hoot”. These hoots are loud enough to be heard by other females, out of sight of the lek, who may be attracted by the calls and provide the dominant males at the lek with additional mating partners. (i) Suggest why auditory stimuli are advantageous to species inhabiting forest ecosystems. (ii) Recent research has found that some peacocks emit hoots in the complete absence of females at the lek. Females are still attracted to the lek by these sounds. Such “solo” hoots have been described as “dishonest signals”. Explain what is meant by a “dishonest signal” in this behaviour. [Turn over for next question 1 1 Page 24 MARKS DO NOT WRITE IN THIS MARGIN 11. Answer either A or B in the space below and on Page 25. A Discuss reproduction under the following headings: (i) costs and benefits of sexual reproduction; (ii) asexual reproduction as a successful reproductive strategy. OR B Discuss endoparasitic infections under the following headings: (i) difficulties involved in their treatment and control; (ii) benefits of improved parasite control to human populations. Labelled diagrams may be used where appropriate. 4 5 7 2 Page 25 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWER FOR QUESTION 11 [END OF QUESTION PAPER] Page 26 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Page 27 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK

Page 28 Acknowledgement Question 4(b) - petarg/shutterstock.com Question 10 – Bildagentur Zoonar GmbH/shutterstock.com Bildagentur Zoonar GmbH/shutterstock.com © National Qualications 2016 AH Supplementary Sheet for Question 1 X707/77/11 Biology Supplementary Sheet MONDAY, 9 MAY 9:00 AM – 11:30 AM A/HTP

Page 02 1. Malaria is caused by unicellular parasites in the genus Plasmodium. Figure 1 shows the life cycle of the parasite with respect to its human and mosquito hosts. Figure 1 Asexual stage in human Sexual stage in female mosquito Plasmodium life cycle Malaria is a well-researched tropical disease of humans, but less is known about the effects of the parasite on its mosquito vector. The parasite Plasmodium relictum causes malaria in birds. A recent study has been carried out to investigate the effects of this parasite on the mosquito Culex pipiens. In particular, two aspects were investigated: fecundity (number of eggs laid) and longevity (measured as survival after egg laying) of the mosquitoes. In Figure 2, box-and-whisker plots show the total egg production by large numbers of uninfected and infected female mosquitoes. Figure 2 250 200 150 100 50 0 Uninfected Infected Female mosquitoes Maximum – highest value Upper quartile – 25% of data higher than this value Median – middle point of dataset. 50% of data higher than this value Lower quartile – 25% of data lower than this value Minimum – lowest value number of eggs laid per female

Page 03 1. (continued) Figure 3 shows mean survival times after egg laying for uninfected and infected female mosquitoes. Figure 3 19 18 17 16 15 Female mosquitoes Uninfected Infected survival after egg laying (days) Fecundity and longevity were measured in the same individual female mosquitoes to see if there was a relationship between the two variables. The lines of best fit for mosquito survival against the number of eggs each female laid were plotted for uninfected females and infected females. This data is shown in Figure 4. Figure 4 30 25 20 15 10 5 survival after egg laying (days) 0 50 100 150 200 250 Number of eggs laid Key: uninfected females infected females uninfected females infected females Page 04 ACKNOWLEDGEMENT Anna Rassadnikova/shutterstock.com agongallud/shutterstock.com

Page 02 SECTION 1 — 20 marks Attempt ALL questions 1. The diagram below shows part of a DNA molecule before and after a mutation. T A T A C G C G A T A T G C G C C G C G A T C G T A T A T A T A G C G C mutation before after The type of mutation shown is A deletion B substitution C insertion D inversion. 2. Which of the following are required in a polymerase chain reaction (PCR)? A DNA polymerase, template strand and primers B RNA polymerase, template strand and primers C DNA polymerase, template strand and ligase D RNA polymerase, ligase and primers 3. Each cycle of a polymerase chain reaction (PCR) takes 5 minutes. If there are 1000 DNA fragments at the start of the reaction, how long will it take for the number of fragments produced by the reaction to be greater than 1 million? A 15 minutes B 35 minutes C 50 minutes D 55 minutes

Page 03 4. The graphs below show possible changes in the body size of a population of barn swallows, Hirudino rusticana, in response to a selection pressure. body size body size body size number of barn swallows in population number of barn swallows in population number of barn swallows in population original population population after selection Graph 1 Graph 2 Graph 3 Which row in the table below matches each graph with the type of selection taking place? Graph 1 2 3 A disruptive directional stabilising B directional disruptive stabilising C stabilising disruptive directional D directional stabilising disruptive [Turn over

Page 04 5. The diagram below represents a phylogenetic tree showing the evolutionary relatedness of several species of cat. Domestic cat Scottish wildcat Puma Lynx Ocelot Caracal Bay cat Lion Jaguar African leopard Tiger Snow leopard Clouded leopard With how many species does the African leopard share a common ancestor in this phylogenetic tree? A 2 only B 5 only C 12 only D 13

Page 05 6. Over millions of years of evolution, mutations occur at a broadly constant rate within a gene. This allows genes to be used as molecular clocks. The diagram below shows how the base sequence in part of a gene changed as two evolutionary lineages diverged from an original base sequence. The base sequence in the gene has changed at a rate of 1 base per 5 million years as shown. Original base sequence 5 million years after diverging 10 million years after diverging Species Y Species X CAATTGATCG CAATTTATCG CGATTTATCG CAATTGCTCG CGGTTTATCG Assuming this rate of mutation continued, by how many bases would this part of the gene differ in Species X compared with Species Y 20 million years after diverging from the original base sequence? A 4 B 8 C 16 D 20 7. In metabolic pathways, the rates of reaction can be affected by the presence of enzyme inhibitors. Which row in the table below is correct? Type of inhibition Inhibitor binds to active site Effect of increasing substrate concentration on inhibition A competitive yes reversed B non competitive yes unaffected C competitive no unaffected D non competitive no reversed 8. Which row in the table below identifies the number of heart chambers and the type of circulatory system in amphibians? Number of heart chambers Type of circulatory system A 3 incomplete double B 4 incomplete double C 3 complete double D 4 complete double [Turn over

Page 06 9. During unexpected periods of drought the South American lungfish, Lepidosiren paradoxa, survives by burying into mud. This type of behaviour is known as A predictive dormancy B daily torpor C aestivation D hibernation. 10. An experiment was set up to investigate the effect of different respiratory substrates on the rate of respiration in yeast. Methylene blue can be used to measure the rate of respiration as it changes from dark blue to colourless when it accepts hydrogen ions. Four test tubes were set up, each containing yeast, methylene blue and one of the respiratory substrates. The table below shows the results of this investigation. Test tube number Respiratory substrate Appearance of the methylene blue after 20 minutes 1 starch dark blue 2 sucrose light blue 3 lactose dark blue 4 glucose colourless Which of the following conclusions is correct? The rate of respiration is A higher with starch than with glucose B lower with sucrose than with lactose C higher with glucose than with lactose D lower with glucose than with sucrose. 11. Stages of aerobic respiration are shown below. 1 Glycolysis 2 Citric acid cycle 3 Electron transfer chain Which stage(s) involve(s) both phosphorylation of intermediates and generation of ATP? A 1 only B 3 only C 1 and 2 only D 1 and 3 only

Page 07 12. Which row in the table below identifies a stage of aerobic respiration, its site and an event which occurs during that stage? Stage Site Event A electron transfer chain inner mitochondrial membrane carbon dioxide is released B electron transfer chain matrix of mitochondrion hydrogen ions combine with oxygen C citric acid cycle inner mitochondrial membrane hydrogen ions combine with oxygen D citric acid cycle matrix of mitochondrion carbon dioxide is released 13. A field trial was set up to investigate the effect of phosphate fertiliser on the yield of the potato cultivar Maris Piper. Potatoes were planted in 5 plots, each of which received a different level of phosphate fertiliser. When they were harvested the yield from each plot was recorded. A list of suggested improvements to this field trial is shown below. 1 Apply equal volumes of water to each plot. 2 Grow the same number of potato plants in each plot. 3 Use 10 plots at each phosphate fertiliser level. 4 Plant different potato cultivars in each plot. Which of the suggestions would improve the validity of the results? A 1 and 2 B 1 and 3 C 2 and 4 D 3 and 4 [Turn over

Page 08 14. Which compound combines with hydrogen during carbon fixation (Calvin cycle)? A Ribulose biphosphate B NADP C Oxygen D 3-phosphoglycerate 15. The following absorption spectra were obtained by testing four different plant extracts. Which extract contains chlorophyll? Blue Green Green 500 Yellow Yellow 600 Red Red 700 Blue 400 Light not absorbed Light absorbed Wavelength (nm) and colour of light A B C D

Page 09 16. The table below shows the biological and economic yields of four different crops. Crop Biological yield (tonnes of dry mass/hectare) Economic yield (tonnes of dry mass/hectare) pea 10 2 rice 15 10 wheat 30 8 potato 30 10 The crop with the highest harvest index is A pea B rice C wheat D potato. 17. The graph below shows the levels of nitrogen and phosphorus applied to crops in an area of Scotland between 1986 and 2006. 130 120 110 100 90 50 45 40 35 30 1986 1990 1994 1998 2002 2006 Nitrogen applied (kg/ha) Phosphorus applied (kg/ha) In which year was there the smallest difference between the levels of nitrogen and phosphorus applied? A 1998 B 2000 C 2002 D 2006 [Turn over

Page 10 18. Which of the following are features of naturally inbreeding crop plants? 1 Susceptible to inbreeding depression 2 Deleterious alleles eliminated by natural selection 3 Self-pollinating A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 19. On returning to their roost after feeding, vampire bats may regurgitate blood to feed an unrelated individual in the same social group. This is an example of A mutualism B altruism C social hierarchy D kin selection. 20. The statements below refer to behaviour sometimes displayed by lions kept in captivity. 1 Repetitive chewing on cage bars 2 Excessive licking of body 3 Continually pacing backwards and forward Which are examples of stereotypy? A 1 only B 1 and 2 only C 2 and 3 only D 1, 2 and 3 [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET] Page 11 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 12 [BLANK PAGE] DO NOT WRITE ON THIS PAGE H FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2016 Total marks — 100 SECTION 1 — 20 marks Attempt ALL questions. Instructions for the completion of Section 1 are given on Page 02. SECTION 2 — 80 marks Attempt ALL questions. Questions 7 and 14 contain a choice. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. You should score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X707/76/01 MONDAY, 9 MAY 9:00 AM – 11:30 AM A/HTP Biology Section 1 — Answer Grid and Section 2 Page 02 SECTION 1 — 20 marks The questions for Section 1 are contained in the question paper X707/76/02. Read these and record your answers on the answer grid on Page 03 opposite. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample Question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D Page 03 A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 SECTION 1 — Answer Grid [Turn over Page 04 [BLANK PAGE] do not write on this page Page 05 [Turn over for next question do not write on this page Page 06 MARKS DO NOT WRITE IN THIS MARGIN SECTION 2 — 80 marks Attempt ALL questions Questions 7 and 14 each contain a choice. 1. The diagram below shows a process involved in the production of a polypeptide in a cell. Ribosome tRNA mRNA A U G C C A C C G G G G G C C C C G C C G G C G A C A molecule X (a) Name molecule X. (b) State one substance, other than ribosomal RNA (rRNA), that makes up the ribosome. (c) Many polypeptides are modified in order to produce functional proteins. Describe one way in which a polypeptide could be modified. (d) In some eukaryotic cells, different mRNA molecules, and therefore different proteins, can be expressed from a single gene. Name and describe the process which results in different mRNA molecules being expressed. Name Description 1 1 1 2 Page 07 MARKS DO NOT WRITE IN THIS MARGIN 2. DNA holds the genetic information in both prokaryotic and eukaryotic cells. (a) (i) Describe one organisational difference between prokaryotic and eukaryotic chromosomal DNA. (ii) Name the substance with which DNA is packaged in eukaryotes. (b) State one location, other than the nucleus, where DNA is found in eukaryotic cells. (c) During DNA replication two new daughter strands are synthesised using the original strands as templates. (i) State why the antiparallel nature of the DNA molecule results in one of the strands being synthesised in short fragments. (ii) Template DNA, enzymes and ATP are necessary for DNA replication. State one other substance required. (d) Explain why cells need to carry out DNA replication. [Turn over 1 1 1 1 1 1 Page 08 MArKS DO NOT WRITE IN THIS MARGIN 3. Stem cells are unspecialised cells which can be found in embryonic and adult tissue. cell division cell differentiation cheek epithelial cells white blood cells red blood cells muscle cells ciliated epithelial cells stem cell nerve cells (a) Explain how the diagram above indicates that the stem cell shown is an embryonic and not a tissue (adult) stem cell. (b) After a stem cell differentiates, only certain genes are expressed. Explain how this results in different cell types. (c) Give one therapeutic use of stem cells. (d) State one ethical issue relating to the use of embryonic stem cells. 1 1 1 1 Page 09 [Turn over for next question do not write on this page Page 10 MARKS DO NOT WRITE IN THIS MARGIN 4. Meristems can be cultured in growth medium to produce new plants. An experiment was carried out to investigate the effects of three different growth media (A, B and C) on the production of shoots by meristems of African violet plants. Five meristems were removed and cultured in each medium for a period of seven weeks. The average number of shoots produced per meristem was recorded at specific times during the investigation. The results are shown in the graph below. 12 11 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 A B C Growth media Time (weeks) Average number of shoots produced per meristem (a) (i) Use values from the graph to describe the average number of shoots produced per meristem over the seven week period in medium A. (ii) Calculate the percentage increase in the average number of shoots produced per meristem between week 1 and week 7 in medium B. Space for calculation % 2 1 Page 11 MARKS DO NOT WRITE IN THIS MARGIN 4. (a) (continued) (iii) Table 1 below shows the number of shoots produced per meristem at three weeks in one of the media. Table 1 Meristem Number of shoots produced per meristem 1 4 2 5 3 7 4 7 5 6 Using information from Table 1 and the graph, state the medium in which these meristems were cultured. Space for calculation Medium (b) Predict which medium would produce plants with the greatest number of shoots after nine weeks growth. Give a reason for your answer. Medium Reason (c) In a further experiment, the average number of roots and average root length at 7 weeks were recorded in each of the media. The results are shown in Table 2 below. Table 2 Medium Average number of roots produced per meristem Average root length (mm) A 12 12 B 11 19 C 12 17 After analysing the results, medium B was used for the commercial production of plants. Use the information in Table 2 to explain why plants cultured in medium B would grow best. 1 2 2 [Turn over Page 12 MARKS DO NOT WRITE IN THIS MARGIN 5. In the North Pacific Ocean there are two different populations of killer whales Orcinus orca. One population feeds mainly on fish while the other feeds mainly on sea mammals. This behavioural barrier has led to considerable genetic variation between these populations. (a) (i) Name the type of speciation which could occur as a result of this barrier. (ii) State the importance of isolation barriers in speciation. (iii) Scientists believe that these two populations are still the same species. Suggest how they could confirm this. (b) Polyploidy can lead to speciation. (i) State what is meant by the term polyploidy. (ii) Describe one example of the importance of polyploidy in evolution. 1 1 1 1 1 Page 13 MARKS DO NOT WRITE IN THIS MARGIN 6. The antibiotic bacitracin is produced by the bacterial species B.subtilis. The graph below shows the growth curve of a population of B.subtilis cultured to produce the antibiotic. Cell number Phase A Phase B Phase C Phase D Time (a) Name Phase A and explain why cells do not divide during this phase. Name Explanation (b) (i) Name the phase in which the bacteria produce the secondary metabolite bacitracin. (ii) Explain why this secondary metabolite gives an ecological advantage to B.subtilis. (c) This growth curve shows viable cell numbers of B.subtilis. Give evidence from the graph to justify this statement. 2 1 1 1 [Turn over Page 14 MARKS DO NOT WRITE IN THIS MARGIN 7. Answer either A or B in the space below. A Describe and compare anabolic and catabolic reactions. OR B Describe and compare metabolism in conformers and regulators. 4 4 Page 15 [Turn over for next question do not write on this page Page 16 MARKS DO NOT WRITE IN THIS MARGIN 8. An investigation was set up to monitor growth of bacteria in compost. The compost was added to a fermenter and the temperature of the compost was recorded over a 20 day period. Samples of the compost were cultured and the numbers of three bacterial species present were recorded. The compost temperatures and the populations of the three species of bacteria are shown in the table below. Time (days) Compost Temperature (°C) Population (millions per gram of compost) Species A Species B Species C 0 21 396·0 0·4 123·0 2 40 4·2 10·2 14·6 4 72 0·1 195·0 0·1 6 53 0 8·5 0 20 32 0 0 0 (a) Calculate how many times greater the population of Species A was compared to Species B at the start of the investigation. Space for calculation (b) Describe the relationship between temperature of the compost and population of Species C over the first four days. 1 1 Page 17 MARKS DO NOT WRITE IN THIS MARGIN 8. (continued) (c) (i) Using information in the table, state which species of bacteria is thermophilic and justify your answer. Species Justification (ii) Describe how thermophilic bacteria are adapted to survive in their environment. (iii) Give an example, other than in compost, of an environment where thermophilic bacteria are adapted to grow successfully. [Turn over 2 1 1 Page 18 MARKS DO NOT WRITE IN THIS MARGIN 9. The diagram below shows how a human gene can be inserted into bacteria to produce human insulin using recombinant DNA technology. bacteria plasmid human insulin gene recombinant plasmids inserted into new bacteria bacteria cultured which produce insulin (a) Name one enzyme used in this process and state its function. Name Function (b) (i) The recombinant plasmid also contains a gene for resistance to the antibiotic, ampicillin. Describe a procedure which would allow only cells containing the recombinant plasmid to be selected. (ii) Plasmids with these antibiotic resistance genes have been passed to other bacterial species by horizontal transfer. Describe the process of horizontal transfer. (c) When culturing the bacteria which produce insulin, sterile conditions are maintained. Explain why this is important. 2 2 1 1 Page 19 [Turn over for next question do not write on this page Page 20 MARKS DO NOT WRITE IN THIS MARGIN 10. An investigation was carried out to compare the rate of metabolism in a species of cricket, Gryllus assimilis, at different temperatures. Five crickets were placed in a sealed flask which was fitted with a carbon dioxide (CO2) sensor as shown in the diagram below. CO2 sensor flask water bath at 5 °C crickets The flask was placed in a water bath at 5 °C and left for 10 minutes. The CO2 produced per minute was then measured. This procedure was repeated at 10, 15, 20 and 30 °C. The results are shown in the table below. Temperature (°C) Rate of CO2 production (units per minute) 5 300 10 500 15 800 20 1200 30 1600 (a) (i) Give a reason why the flask was left for 10 minutes at each temperature before each reading was taken. 1 Page 21 MARKS DO NOT WRITE IN THIS MARGIN 10. (a) (continued) (ii) A control flask should be included in this investigation. Describe the control and explain its purpose in the investigation. Description Explanation (b) Plot a line graph to show the results of the investigation. (Additional graph paper, if required can be found on Page 31). (c) Draw a conclusion from these results. [Turn over 2 2 1 Page 22 MARKS DO NOT WRITE IN THIS MARGIN 11. Colchicine is a chemical used in plant breeding programmes to induce mutations and produce cultivars with improved characteristics. Sesame is an important crop plant grown for its edible seeds and leaves. An investigation was carried out to determine the effects of colchicine concentration on sesame. Sesame seeds were soaked in different concentrations of colchicine solution for 24 hours. Seeds from each concentration were germinated and 50 plants were grown from each concentration. Ninety days later the total leaf area, number of seeds and mass of seeds per plant were recorded. The average results are shown in the table below. Colchicine concentration (m mol l−1) Average total leaf area per plant (cm2) Average number of seeds per plant Average total mass of seeds per plant (g) 0 1500 548 2·8 0·1 2315 532 3·5 0·5 2786 550 4·4 1·0 3500 512 4·7 (a) (i) Identify the independent variable in this investigation. (ii) State an aspect of the investigation which helped to ensure that reliable results were obtained. (b) (i) An important characteristic of food crops is the 1000 seed mass which is the total mass of a sample of 1000 seeds. Calculate the 1000 seed mass for the plants grown from seeds soaked in a colchicine concentration of 0·5 m mol l−1. Space for calculation g 1 1 1 Page 23 MARKS DO NOT WRITE IN THIS MARGIN 11. (b) (continued) (ii) Express, as the simplest whole number ratio, the average total leaf area per plant from seeds soaked in a colchicine concentration of 0 to that at 1·0 m mol l−1. Space for calculation : 0 1·0 m mol l−1 m mol l−1 (c) Explain the relationship between the total leaf area and total mass of seeds. [Turn over 1 2 Page 24 MARKS DO NOT WRITE IN THIS MARGIN 12. Malaria is a disease in humans caused by a parasite which is transmitted from human to human by mosquitoes. The stages of infection in humans are shown in the flow diagram below. A female mosquito carrying parasites in its saliva bites a human to feed on blood which it needs for production of her eggs. Parasites pass into human blood and travel to the liver where they multiply and destroy liver cells. Parasites are released from liver cells into the blood where they enter red blood cells and destroy them. (a) (i) Identify the vector in this parasitic relationship. (ii) Give a reason why only female mosquitoes transmit the malaria parasite. (b) Explain the effect of a parasitic relationship on the host. (c) Two methods used to control the spread of malaria are described below. Method 1 – mosquito repellent applied to the skin. Method 2 – anti-malarial drugs that kill the parasite. Choose one of these methods and explain how it can reduce the number of cases of malaria. Method Explanation 1 1 1 1 Page 25 [Turn over for next question do not write on this page Page 26 MARKS DO NOT WRITE IN THIS MARGIN 13. Freshwater mussels are small animals which live on the beds of lakes and rivers. Zebra mussels are a species of freshwater mussel native to lakes in Russia. They were accidentally introduced by humans into a river in North America in 1991. The populations of zebra mussels and the native unionid mussels were measured over a 12 year period. The results are shown in the graph below. 4000 160 3500 140 3000 120 2500 100 2000 80 1500 60 1000 40 500 20 0 0 zebra mussel population (mussels per m2) unionid mussel population (mussels per m2) 1991 1993 1995 1997 1999 2001 year 2003 (a) (i) State the unionid mussel population in 1993. mussels per m2 (ii) State the zebra mussel population when the unionid mussel population was 50 mussels per m2. mussels per m2 (iii) Calculate the average increase per year in the zebra mussel population between 1991 and 2003. Space for calculation mussels per m2 per year (b) Explain how the graph confirms that zebra mussels are more successful competitors than unionid mussels. 1 1 1 1 Page 27 MARKS DO NOT WRITE IN THIS MARGIN 13. (continued) (c) Using evidence from the graph, explain why zebra mussels are an invasive species. (d) Suggest a reason why the population of zebra mussels may have increased faster in the North American river than in its native habitat. (e) Invasive species have a negative impact on genetic diversity of an ecosystem. State what is meant by genetic diversity. [Turn over for next question 1 1 1 Page 28 MARKS DO NOT WRITE IN THIS MARGIN 14. Answer either A or B in the space below and on Pages 29 and 30. A Write notes on crop protection under the following headings: (i) weeds, pests and diseases; (ii) methods of control. OR B Write notes on social behaviour in animals under the following headings: (i) social hierarchy and cooperative hunting; (ii) social insects. 4 4 4 4 Page 29 SPACE FOR ANSWERS [Turn over Page 30 SPACE FOR ANSWERS [END OF QUESTION PAPER] Page 31 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Additional Graph paper for Question 10 (b) Page 32 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Page 33 ACKNOWLEDGEMENT Section 2 Question 3 - BlueRingMedia/shutterstock.com

3 P52220RA Turn over 5. The marine archaeon Geogemma barossii, for example, can survive at a blistering 121 °C, while other microbes have retained metabolic activity at temperatures as low as −20 °C in Siberian permafrost. The bacterium Deinococcus radiodurans remains viable after exposure to 1,000 times the fatal human radiation dose, and the aquatic archaeon Ferroplasma acidarmanus can withstand extremely acidic water, with pH values as low as 0. Such flashy feats have earned these and other microbes the title “extremophiles”—lovers of extreme conditions. But as remarkable as their metabolic capabilities are, calling them “extreme” is a bit human-centric. Because of our own requirement for oxygen and narrow acceptable ranges of temperature, salinity, pressure, pH, and radiation, the survival of other organisms in a wide range of environments seems extreme to us. But for a microbe that has come to depend on the abundant hydrogen ions of acidic hot springs, an air-conditioned suite at the Ritz is a threatening proposition. The wide variety of biochemical modes of existence reflect billions of years of evolution, adaptation, and niche differentiation rather than a standardized characterization of biological fortitude. 6. For the title of “extremophile” to be broadly meaningful, it must refer to a more objective measure of extremeness—an advantageous capability enacted in response to a common challenge. One such challenge, something that all living organisms must face, is the acquisition of chemical energy to drive cellular reactions. Perhaps the ways in which organisms handle this task could separate the truly industrious from the merely viable. The energy of life 7. Energy is the currency of biology. By harvesting electrons from a stunning range of starting materials, Earth’s organisms produce adenosine triphosphate (ATP), which powers biological reactions. In the case of mammals and most eukaryotes, sugars and other organic molecules are common electron sources, the oxidation of which drives ATP production. Bacteria and archaea can use a range of other chemicals, from sulfide to iron to ammonium. 8. Cells take up these electron-rich molecules and capture their electrons, which jump down an electron transport chain in the mitochondrial or cell membrane. As electrons move along the membrane toward a final electron acceptor, protons are pumped from the cell’s interior to the exterior, setting up a chemical gradient. Finally, protons stream back into the cell, releasing the chemical pressure and generating ATP. With each energy-requiring reaction, from flagella construction to cell division and growth, cells draw upon their ATP bank. Courtesy of Jeffrey Marlow Oceanic extremes This experimental setup on Hydrate Ridge off the coast of Oregon samples microbial metabolism in deep-sea methyl seeps, which host a variety of seemingly strange creatures, including some truly extreme archaea and bacteria that cannot survive without each other.

4 P52220RA 9. This elegant, multistep process is a pervasive feature of life as we know it, but energetic challenges are ever-present. If the electrical potentials of electron donor and acceptor are too closely aligned, for example, it won’t be possible to squeeze much energy from their coupling. The concentrations of the reactants and the speed at which enzymes can mobilize them are also key factors. These two components—the magnitude of energy available from a particular pairing and the rate of such reactions—determine how much energy a cell can produce. 10. The other half of the equation—the cost of living, as it were—is often harder to evaluate. Cataloging the biochemical parts list of a particular cell is one challenge. Individual biosynthetic pathways—the production of lipids from glycerol derivatives, for example—are relatively well characterized under “standard” conditions, but a cell’s ever-changing chemical environment can render baseline calculations inaccurate. Scaled over millions of such reactions, the margin of error may be a substantial proportion of the available energy. And this is just considering the biosynthesis of new cellular material. In most environments, microbes must always be vigilant against biochemical breakdown resulting from environmental stresses, calling on energy reserves to restore old enzymes or patch holes in cell walls. Competition among residents may also demand additional energy expenditure, such as powering flagella to swim around in search of food or producing antibiotic molecules to keep predatory neighbors at bay. 11. If, however, we are able to estimate how much energy is required for survival, and compare that to how much energy is available to be extracted from the environment, we can begin to consider “extreme” organisms in a more objective fashion. Some of the most “exotic” environments actually offer luxurious energetic balances; it’s the microbes with low net energy availability that are the real extremophiles, whether they live an expensive existence in a high-energy environment, or an ascetic life in an energetic desert. Easy living High energy availability, low energy requirements 12. The hot springs of Yellowstone National Park are uniquely beautiful palettes: concentric rings transition from blue in the pools’ centers to green, then yellow, orange, and red at the water’s periphery. The mesmerizing visuals contrast sharply with the damp, sulfurous odors wafting across your nostrils and the stern warnings from signs and rangers to keep your distance. Against this otherworldly backdrop, the 1966 discovery of viable cells living in the ultrahot waters came as a surprise that forced a reconsideration of microbial limits. After all, water temperatures frequently topped out well above the tolerance range of most known organisms. Nearly all of E. coli’s enzymes, for example, unfold and become ineffective at 60 °C. 13. Hot-springs microbes have traditionally been labeled “extremophiles,” yet their energetic bank account is typically well in the black. Like their moderate-temperature relatives that ply the planet’s oceans, thermophilic cyanobacteria gain energy from light-driven reactions that mobilize electrons from water. Along the outer edges of thermal springs, energy-generating light is abundant, and cyanobacteria flourish. Indeed, the vibrant colors we see are the plentiful microbial pigments that coat the limestone surfaces.

5 P52220RA Turn over 14. This is not to say that life at high temperatures is easy. On the contrary, only through a range of sophisticated molecular adaptations, encoded by subtle edits in thermophiles’ genomes, has this forbidding niche become habitable. Protein stability is perhaps the main challenge for life at high temperature. Higher thermal energy causes hyperactive atoms to vibrate with more kinetic energy, threatening the structural integrity of the molecules that perform biochemical reactions. If sulfur-containing cysteine amino acids are positioned strategically within protein structures, disulfide bridges can form interatomic support beams that resist unfolding. Some thermophilic enzymes also have larger hydrophobic cores, away from the proteins’ exposed active sites, which act as additional glue to fight thermal destabilization. Other adaptations, such as simpler protein folds or fewer bound metal ions, further guard against molecular destabilization in the face of thermal stress. 15. Evolving the capability to handle high temperatures may not have been straightforward, and biosynthetic construction costs might have presented some hurdles, but the payoff does seem to have been worth it. By constructing heat-stable enzymes, the cyanobacteria inhabiting hot springs are able to photosynthesize in relative peace, away from the feeding frenzy of predatory microbes or larger creatures in habitats such as the ocean’s surfaces. Thus, while Yellowstone’s hot springs may seem like an extreme environment, not all of the microbes that inhabit them are struggling to survive. An expensive lifestyle High energy availability, high energy requirements 16. One needn’t travel to the bottom of the ocean or into a scalding hot spring to find microbes living at the edge of energetic feasibility. Sometimes, the most remarkable habitats are in your own backyard, beneath well-manicured Kentucky bluegrass or a haphazard array of lawn furniture. “Generic” temperate soils are among the richest microbial milieus on the planet, with each pinch of dirt hosting up to a billion cells, and down there, it’s all-out biochemical warfare. 17. Among the more prominent denizens of this dense microbial metropolis are representatives of the bacterial genus Streptomyces: stringy, rod-shaped organisms that develop centimeter-scale networks branching through the soil. Streptomyces gain energy through heterotrophy, the consumption of organic molecules such as sugars, amino acids, or aromatic compounds. These are energetically juicy molecules, and they’re abundant given the high density of plants in the vicinity, but it’s far from a free lunch. NOAA Okeanos Explorer Program, 2013 ROV Shakedown and Field Trials Methane munchers Off the coast of Virginia, methane bubbles flowing out of the seafloor sediment support a variety of life, including some truly extreme microbes.

6 P52220RA 18. Streptomyces capture organic molecules largely by secreting enzymes into the soil to access and degrade energy-rich polymers before other competitors can get to them. It’s a bold strategy susceptible to freeloader hijacking, with no guarantee that the processed material will find its way back to the same organism that went to the trouble of producing the enzyme. But at scale, the odds become more palatable, and the benefit from degraded organics that find their way to one Streptomyces or another outweighs the inefficiency of the strategy. Building a large network of interconnected cells is the only option that makes this spendthrift approach worthwhile. 19. Oxygen is the highest-potential electron acceptor on the market, and transferring electrons to O2 provides the biggest payoff per electron-donating molecule. This makes the upper, oxygen- perfused layers of soil highly sought-after real estate, but it comes at a price. In a 2005 study, geomicrobiologists Tom McCollom of the University of Colorado, Boulder, and one of us (Amend, then at Washington University in St. Louis) calculated the energetic costs of synthesizing an extensive list of biomolecules, including amino acids, nucleotides, fatty acids, saccharides, and amines, from inorganic precursor molecules. We found that biosynthetic costs were actually higher—by more than an order of magnitude—in oxygen-infused conditions than in anaerobic settings. Part of this discrepancy is due to the fact that many precursors must be reduced from their oxidized state prior to biomolecular construction, but it suggests that the energetic windfall from using oxygen as an electron acceptor may be a necessary copay, not a bankable nest egg. 20. Perhaps Streptomyces’s most impressive adaptation is the genus’s remarkable array of antibiotics, which can attack competitors’ cell walls or protein-synthesis machinery. But these large, sophisticated weapons require a high flux of electron-rich intermediates and the repurposing of cellular supply chains. An analysis led by J. Stefan Rokem of the Hebrew University of Jerusalem showed that antibiotic production represents an enormous drain on biosynthetic pathways, frequently costing more than half of the stocked supply of precursor building blocks (such as pyruvate or acetyl-CoA) that would otherwise be used to construct biomass and generate new cells. But, while expensive, antibiotic synthesis is critical for survival in this crowded environment, minimizing competition for much-needed resources. From an energetic perspective, Streptomyces is an extreme organism, and it’s hiding in plain sight. Just getting by Low energy availability, low energy requirements 21. As Jason is hauled back on deck, packed to the brim with samples from deep-sea methane seeps, Atlantis becomes a hive of activity. We rush the cores of seafloor sediment along with chunks of carbonate rocks to the onboard laboratory, where they’re partitioned and allocated for experiments. Liquid is extracted for geochemical measurements, and a few grams of sediment are frozen for DNA and microscopic analyses. A separate aliquot is scooped into a shiny silver mylar bag, mixed with filtered seawater and isotopically labeled chemicals, and flushed with nitrogen and methane gas. The bag is heat-sealed and set aside, a time capsule to be opened several months later to determine how much of the isotope-labeled substrate has been taken up by the mysterious process of anaerobic methane oxidation.

7 P52220RA Turn over 22. What we ultimately find is confirmation of a bizarre biological partnership operating at the edge of what is energetically possible. As has been found at other methane seep sites around the world, certain types of archaea and bacteria aggregate together in multicellular clumps, with tens, hundreds, or occasionally thousands of microbes linked by mutual energetic necessity. The details of the association are still up for debate, but it appears that archaeal partners oxidize methane and transfer electrons to the bacteria to enable the reduction of sulfate to sulfide, generating energy to power cellular functions. Remarkably, when the energetics numbers are crunched, the archaea come out in the red—their half of the arrangement does not appear to produce enough energy for their own survival. This means that the sulfate-reducing reaction performed by their bacterial partners must supply power to both species. How this mutualism works, especially in an evolutionary framework, is far from certain. 23. The energetic demands for biosynthesis are relatively low given these organisms’ very slow rates of growth, doubling only every few months. Nevertheless, given the difficulty of extracting and sharing energy in methane seep environments, anaerobic methane-oxidizing partnerships deserve the title of extremophiles, as characterized by the energetic framework. Hopefully, future studies will illuminate the nature of this symbiosis and provide insight into how the energetically improbable becomes possible, untangling the intricacies of these and other slow-growing extremophiles. An impossible situation Low energy availability, high energy requirements 24. The final permutation of energetic cost-benefit ratios seems like a non-starter: having higher energy demands than the rates of supply does not make for a sustainable situation. And while growth under such conditions seems impossible (with the notable exception of tightly-coupled metabolisms like those described above), an energy debt need not mean cell death. 25. When the going gets tough, some microbial species, such as the bacterium Bacillus subtilis, initiate a hibernation protocol, shutting down the furnace and turning off the lights before forming a life raft that will hopefully ferry them to greener pastures. The process is called sporulation, and it’s a life-or-death decision not to be made lightly. B. subtilis is commonly found in soil environments susceptible to feast or famine swings in energy availability. When one of these cells senses nutrient stress, it draws on energy stores, activating flagella to search for food, flooding its surroundings with antibiotics to kill off competitors, or desperately importing foreign pieces of DNA in hopes that a novel capability will be the ticket out of a bad situation. If all else fails, it replicates its genetic material and partitions it into a protective capsule that can withstand extreme heat, radiation, chemical stress, desiccation, and energetically untenable conditions. Sensors located on the spore’s outer surface probe the environment for friendlier surroundings and assess the possibility of returning to a more active way of life. Powering up is an extremely energetically demanding undertaking, permitting full resurrection only under ideal circumstances. Thus, while this behavior may be considered extreme in itself, spore formers dodge the true test of their extremophilic nature by waiting out the impossible in a state of metabolic hibernation.

8 P52220RA Rethinking extremes 26. For millennia, microbes have searched for loopholes and experimented with novel molecular machines in an attempt to gain a foothold where others can’t. From backyard soils to seafloor methane seeps, these extremophiles eke out a living, revealing adaptations to the energy equation that may point us toward other organisms awaiting discovery in the headlights of a future robotic spacecraft. 27. Many of the most promising astrobiological targets in our solar system may well possess the baseline requirements for life such as liquid water and key elements, but net energy availability is an unknown. The lakes of Titan, Saturn’s largest moon, have plenty of organic fuel for the taking, but maintaining a viable microenvironment would likely be very difficult given the lack of an appropriate solvent at such frigid temperatures. The Martian subsurface, on the other hand, may lack easily obtainable energy sources, but with relatively few apparent hazards, a low-energy way of life could be feasible. 28. Researchers have yet to fully sample the diversity of bioenergetic regimes on Earth. With continued exploration of the seafloor or the planet’s deep interior, we find novel geochemical cocktails that demand microbial innovation. And so, back on the Atlantis, we load Jason with cleaned collection tubes and send our robotic emissary on yet another journey into the unknown, in search of new microbial responses to the extreme conditions of the deep. The Energy of Life by Jeffrey Marlow and Jan Amend, February 1, 2015. The Scientist Magazine. http://www.the-scientist.com/?articles.view/articleNo/41990/title/The-Energy-of-Life/

Page 02 SECTION 1 — 25 marks Attempt ALL questions 1. Trypan blue is used as a vital stain to identify viable cells when viewed in a haemocytometer. A vital stain A stains all cells B only stains dead cells C only stains living cells D only stains the culture medium. 2. Bovine insulin is a soluble protein with an isoelectric point of pH 5·4. Which of the following graphs represents the level of precipitate formed as the pH of a bovine insulin solution is changed? 9 9 9 9 1 1 1 1 pH pH pH pH mass of precipitate (mg) mass of precipitate (mg) mass of precipitate (mg) mass of precipitate (mg) C A B D

Page 03 3. The figure shows how a scientist used serial dilution followed by plating to check the number of bacteria in a stock culture. 0∙1 cm3 transferred to agar plate followed by incubation for 2 days 9∙0 cm3 119 colonies 1∙0 cm3 1∙0 cm3 1∙0 cm3 1∙0 cm3 Bacterial stock How many bacteria were there in 1 cm3 of the original bacterial stock? A 1·19 × 105 B 1·19 × 106 C 1·19 × 107 D 1·19 × 108 4. Place the following events for a sodium potassium pump into the correct sequence. 1 Phosphorylation 2 Potassium ions released into cell 3 Transporter protein has high affinity for sodium ions inside the cell 4 Dephosphorylation A 2, 1, 4, 3 B 3, 4, 2, 1 C 2, 1, 3, 4 D 3, 1, 4, 2 [Turn over

Page 04 5. The binding of oxygen to haemoglobin is affected by small changes in temperature or pH. Which of the following changes would decrease haemoglobin’s affinity for oxygen? A increased temperature, decreased pH B increased temperature, increased pH C decreased temperature, decreased pH D decreased temperature, increased pH 6. The gene represented in the diagram codes for multiple proteins due to alternative RNA splicing. The coding regions (exons) are labelled 1 to 10. 1 2 3 4 5 6 7 8 9 10 Alternative RNA splicing results in an mRNA that contains exon 2 or 3, exon 4, 5 or 6 and exon 7 or 8. Predict the total number of different proteins that can be produced from this gene. A 3 B 7 C 12 D 15

Page 05 7. In an immunoassay used to detect the presence of an antigen for a disease-causing organism (pathogen), the following samples were tested. 1 A sample from an individual thought to be infected with the pathogen. 2 A sample from an organism known to cause similar symptoms, but unrelated to the pathogen. 3 A sample from the pathogen. 4 A sample of purified water. Which row in the table identifies the purpose of each sample? Positive control Negative control Test assay A 3 4 2 B 3 2 1 C 2 4 3 D 2 3 1 8. Which of the following diagrams represents the sequence of phases involved in the cell cycle? M G2 G2 S G2 S M G2 G1 G1 G1 G1 S M S M B C D A [Turn over

Page 06 9. Which letter in the diagram represents the first stage in cell signalling for a peptide hormone molecule? C D B A extracellular intracellular 10. Red blood cells swell or shrink rapidly in response to changes in the water concentration around them. This is due to one type of water channel protein, Aquaporin 1 (AQP1). Each red blood cell can transport a total of 1 × 1014 water molecules per second and each AQP1 channel can transport 5 × 108 water molecules per second. How many AQP1 channels are present in the membrane of a red blood cell? A 2 × 105 B 5 × 105 C 2 × 106 D 5 × 106 11. Which of the following changes in the rate of the cell cycle could result in a degenerative disease? A A controlled increase B A controlled decrease C An uncontrolled increase D An uncontrolled decrease

Page 07 12. Which of the following diagrams represents the sequence of stages involved in mitosis? Telophase Prophase Prophase Anaphase Anaphase Metaphase Anaphase Prophase Metaphase Anaphase Metaphase Telophase Prophase Telophase Telophase Metaphase C B A D 13. Bacteriophage M13 is a virus that can infect bacteria. When bacteria in the lab are infected with M13, it is essential to use the correct ratio of one virus particle to two bacterial cells. A 0·25 cm3 sample of an E. coli culture having a concentration of 8 × 108 cells per cm3 is placed into a tube. What volume of a bacteriophage stock having a concentration of 2 × 109 bacteriophage per cm3 should be added to the cell sample to give a bacteriophage to bacteria ratio of 1:2? A 0·05 cm3 B 0·2 cm3 C 5 cm3 D 20 cm3 [Turn over

Page 08 14. A student planned to measure the activity of catalase by using an oxygen probe to measure oxygen production. As part of their pilot study, the student used a standard sample with an oxygen concentration of 22·00% to test the accuracy and precision of the probe. Four readings of the standard sample were taken and the results obtained are shown in the table. Reading 1 2 3 4 Oxygen concentration (%) 20·94 20·93 20·93 20·94 The results indicate the measurements taken were A accurate and precise B accurate but not precise C precise but not accurate D neither accurate nor precise. 15. Which of the following formulae would allow relative fitness to be calculated? A frequency of a particular genotype after selection frequency of a particular genotype before selection B frequency of a particular genotype before selection frequency of a particular genotype after selection C number of surviving offspring of a particular genotype number of surviving offspring of other genotypes D number of surviving offspring of other genotypes number of surviving offspring of a particular genotype

Page 09 16. Some invertebrate species can be used as indicator species to assess the level of pollution by organic waste in freshwater systems. Species within the taxa Ephemeroptera (mayflies), Plecoptera (stoneflies) and Trichoptera (caddisflies) are particularly sensitive to pollution. These taxa are referred to collectively as EPT taxa. The presence of these species can be used to assess water pollution levels by calculating an EPT index using the formula shown. The value obtained allows the water quality to be assessed using Table 1. EPT index = total number of EPT taxa present × 100 total number of taxa present Table 1 EPT index Water quality >50% excellent 35–49% good 25–34% moderate <25% poor A study sampled the invertebrate species in a stretch of river and the results are shown in Table 2. Table 2 Taxa Number of individuals Striped stonefly 6 Dragonfly 5 Simulid blackfly 9 Net-spinning caddisfly 6 Bloodworm 8 Flathead mayfly 3 Damselfly 7 The data suggests the water quality of this river is A excellent B good C moderate D poor. [Turn over

Page 10 17. The smaller tea tortrix (Adoxophyes honmai) is a moth that is an economically important pest of tea plants in Japan. A honmai has a number of development stages (larvae) from egg to adult. Viruses that infect crop pests have potential use in controlling pest populations. A study compared the infection and killing of A honmai by two viruses, X and Y. Mean survival time after infection for each larval stage was measured and the results are shown in the table. Uninfected larvae were then released onto tea plants that had either virus X, virus Y or no virus applied to their leaves. The percentage infection of the larvae was measured and the results are shown in the graph. Stage of insect development when infected Mean survival time ± SE (days) Virus X infection Virus Y infection 1st 18·9 ± 0·6 5·8 ± 0·1 2nd 15·2 ± 0·4 7·0 ± 0·1 3rd 12·2 ± 0·2 7·0 ± 0·1 4th 10·1 ± 0·2 8·0 ± 0·1 5th 8·0 ± 0·1 6·9 ± 0·1 15 10 5 0 Virus X Virus Y No virus Treatment Percentage infection of larvae Which of the following conclusions can be drawn from this information? A Virus X has a higher transmission rate and is more virulent than Virus Y. B Virus X has a higher transmission rate and is less virulent than Virus Y. C Virus X has a lower transmission rate and is less virulent than Virus Y. D Virus X has a lower transmission rate and is more virulent than Virus Y.

Page 11 18. Parthenogenesis is most likely to be common in environments with a A warm climate and low parasite density B warm climate and high parasite density C cool climate and low parasite density D cool climate and high parasite density. 19. Asexual reproduction is most likely to be a successful reproductive strategy in A wide, stable niches B narrow, stable niches C wide, unstable niches D narrow, unstable niches. 20. The black grouse male is larger and more brightly coloured than the female and competes with other males at leks. Which of the following pairs of features are characteristic of this species? A Monogamy and sexual dimorphism B Monogamy and reversed sexual dimorphism C Polygamy and sexual dimorphism D Polygamy and reversed sexual dimorphism 21. Which of the following conversions is catalysed by reverse transcriptase? A RNA → DNA B RNA → protein C DNA → RNA D DNA → protein [Turn over

Page 12 22. The figure represents the structure of a Zika virus. Genome Capsid 40 nm 1,000,000 nm = 1 mm Which row in the table describes the structure of a Zika virus? Capsid Genome Diameter (m) A Protein Nucleic acid 40 × 10−6 B Protein Nucleic acid 40 × 10−9 C Nucleic acid Protein 40 × 10−6 D Nucleic acid Protein 40 × 10−9 23. Schistosomiasis in humans is caused by an A ectoparasitic arthropod B endoparasitic amoeba C endoparasitic nematode D endoparasitic platyhelminth. 24. Cholera is a disease which causes diarrhoea and is potentially fatal. It is transmitted through the consumption of food or water contaminated by the bacterium Vibrio cholerae. It often has a higher incidence in refugee camps than in the surrounding countryside. Which of the following measures is not appropriate for reducing the incidence of cholera in refugee camps? A Improved vector control B Increased sanitation C Decreased population density D Reduced costs for cholera vaccines

Page 13 25. The scatterplot shows the results obtained when life expectancy at birth was plotted against age at first reproduction for 24 species of mammals of different sizes. Age at first reproduction Life expectancy Which of the following conclusions can be drawn from the data? A An increase in life expectancy causes an increase in the age of first reproduction. B An increase in the age of first reproduction causes an increase in life expectancy. C Larger animals have longer life expectancy. D Life expectancy and age at first reproduction are correlated. [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET.] Page 14 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 15 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 16 [BLANK PAGE] DO NOT WRITE ON THIS PAGE AH FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2017 Total marks — 90 SECTION 1 —25 marks Attempt ALL questions. Instructions for the completion of Section 1 are given on Page 02. SECTION 2 —65 marks Attempt ALL questions. A Supplementary Sheet for Question 1 is enclosed inside the front cover of this question paper. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. You should score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X707/77/01 TUESDAY, 23 MAY 9:00 AM – 11:30 AM A/HTP Biology Section 1 — Answer Grid and Section 2 Page 02 SECTION 1 — 25 marks The questions for Section 1 are contained in the question paper X707/77/02. Read these and record your answers on the answer grid on Page 03 opposite. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample Question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D Page 03 A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 SECTION 1 — Answer Grid [Turn over Page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 05 [Turn over for next question DO NOT WRITE ON THIS PAGE Page 06 MARKS DO NOT WRITE IN THIS MARGIN SECTION 2 — 65 marks Attempt ALL questions It should be noted that question 11 contains a choice 1. Read through the Supplementary Sheet for Question 1 before attempting this question. (a) Refer to Figure 2 in the Supplementary Sheet for Question 1. Describe the trend shown in Figure 2. (b) Describe the action of caspases in cell destruction. (c) Refer to Figure 3 in the Supplementary Sheet for Question 1. (i) Explain how the data supports the conclusion that the intrinsic pathway for apoptosis is triggered by the extract. (ii) Describe what the three negative values in Figure 3 indicate about the level of apoptosis in these treatments. 1 1 2 1 Page 07 MARKS DO NOT WRITE IN THIS MARGIN 1. (continued) (d) Refer to Figures 4A and 4B in the Supplementary Sheet for Question 1. (i) Calculate the percentage increase in the number of cells with 400 units of DNA in the cells treated with extract compared to the control cells. Space for calculation (ii) It was concluded that the extract initiated cell cycle arrest at a checkpoint after DNA replication but prior to cell division in these cells. Explain how the data in Figures 4A and 4B support this hypothesis. [Turn over 1 2 Page 08 MARKS DO NOT WRITE IN THIS MARGIN 2. The light sensitive layer at the back of the human eye is called the retina. It is able to detect light due to the presence of photoreceptor cells called cones and rods. There are three types of cone cells (blue, green and red) which are sensitive to different wavelengths of light as shown in Figure 1. Different wavelengths of light are perceived as different colours. Figure 1 Wavelength (nm) Absorbance Red cones Green cones Rods Blue cones 50 100 0 400 500 600 700 (a) (i) In cone cells, the light sensitive molecule retinal combines with a membrane protein to form photoreceptor proteins. Name this membrane protein. (ii) One percent of human males do not have functional red cone cells, a colour vision deficiency called protanopia. Affected individuals would perceive a red object reflecting light of a wavelength of 670 nm as black. Use Figure 1 to explain this observation. 1 1 Page 09 MARKS DO NOT WRITE IN THIS MARGIN 2. (continued) (b) Rod cells are more sensitive than cone cells at low light intensities. State how this sensitivity is achieved. (c) Most birds have four types of cone cells. Name the additional wavelength range to which these organisms are sensitive. [Turn over 1 1 Page 10 MARKS DO NOT WRITE IN THIS MARGIN 3. Cortisol is a hydrophobic signalling molecule, produced by the human adrenal gland, that affects a number of different tissues in response to stress. It has a role in increasing blood sugar levels, in suppressing the immune system, and in promoting the metabolism of fats, proteins, and carbohydrates. (a) (i) Cortisol is a steroid hormone. Describe the mechanism by which this type of signalling molecule causes an effect within the target cell. (ii) Suggest a way in which cortisol might have different effects in different tissues. 2 1 Page 11 MARKS DO NOT WRITE IN THIS MARGIN 3. (continued) (b) Addison’s disease is a disorder in which the adrenal glands do not produce sufficient cortisol. One test for this disease is to give the patient an injection of a hormone called ACTH, which stimulates the adrenal gland to release cortisol. In a healthy person, cortisol levels should rise by at least 70 µg per litre after 30 minutes and 110 µg per litre after 60 minutes. The graph shows the results of this test on three patients who were investigated for Addison’s disease. Patient 3 (Has Addison’s disease) Patient 2 Time (minutes) Serum cortisol (µg per litre) Patient 1 (Does not have Addison’s disease) 10 0 20 30 40 50 60 0 50 100 150 200 250 300 350 State the diagnosis that would be appropriate for Patient 2, using data to justify your answer. [Turn over 1 Page 12 MARKS DO NOT WRITE IN THIS MARGIN 4. The diagram shows how two types of enzyme can be involved in controlling the activity of a protein in response to the presence of a signalling molecule within the cell (intracellular signal molecule). Intracellular signalling molecules are often produced as a result of extracellular signals received by cell-surface receptors. Intracellular signal molecule present Intracellular signal molecule removed enzyme X protein kinase target protein active target protein inactive (a) (i) Explain how the action of protein kinase can switch a target protein from inactive to active. (ii) Name the type of enzyme represented by enzyme X. (iii) Explain the importance of the system being able to return the target protein to its inactive state. 1 1 1 Page 13 MARKS DO NOT WRITE IN THIS MARGIN 4. (continued) (b) Protein kinase A (PKA) is an enzyme that is involved in this type of signalling. To test the hypothesis that PKA is found in a variety of cell types, cell extracts were prepared from different cell types and the proteins in the extracts separated by electrophoresis in a gel. The proteins were blotted onto a solid support and an antibody recognising PKA (anti-PKA antibody) was used to detect the presence of PKA. (i) Describe how protein electrophoresis is used to separate proteins. (ii) Explain how the anti-PKA antibody would be used to detect the presence of PKA. [Turn over 1 2 Page 14 MARKS DO NOT WRITE IN THIS MARGIN 5. The water vole (Arvicola amphibius) was common in Scotland but has declined markedly in recent years due to habitat loss and predation by American mink (Neovison vison). A survey was carried out to estimate the population of water voles on a Scottish river system by counting the number of latrines (droppings sites) on a river bank at the water’s edge. Water vole latrines are created as part of a territorial behaviour where a water vole will revisit the same site over and over again to deposit its droppings. Water vole American mink (a) Suggest one reason why counting latrines is an appropriate indirect sampling technique for water voles. (b) The survey team worked on a total of 518 km, using survey sites of 500 m at 5 km intervals along the river system. The number of latrines per kilometre of waterways was counted. Name the type of sampling used. 1 1 Page 15 MARKS DO NOT WRITE IN THIS MARGIN 5. (continued) (c) All surveys were conducted in June, July and August of the same year. There was heavy rainfall for three days in August. Any survey site too deep to walk into was omitted. The remains of one water vole that had been preyed upon were discovered at one survey site. Of the 92 sites sampled, only one site showed any latrines. Identify one aspect of the experimental design that shows: (i) High reliability; (ii) Low reliability. (d) It was concluded that the water vole population on this river had become extremely low due to predation by American mink. Give two reasons why this may not be a valid conclusion. 1 2 [Turn over 1 1 2 Page 16 MARKS DO NOT WRITE IN THIS MARGIN 6. The mechanism of sex determination is not the same for all species. (a) In most mammals the sex of the organism is determined by its genotype. Describe how genetic control determines the phenotype of maleness in mammals. (b) In some reptile species the sex of offspring is environmentally rather than genetically controlled. Describe how an environmental factor can influence the sex ratio of offspring in such species. (c) Diabetes insipidus is a condition characterised by the production of an excessive volume of dilute urine. An hereditary form of the disease in humans is the result of an X-linked mutation in the gene coding for the receptor for the hormone ADH. The mutated allele (Xa) is recessive to the normal allele (XA). (i) Explain why males are more likely to be affected by diabetes insipidus than females. (ii) Explain why carrier females are usually not affected by diabetes insipidus even though they carry a mutated copy of the gene. 1 1 1 2 Page 17 MARKS DO NOT WRITE IN THIS MARGIN 6. (c) (continued) (iii) A man who is not affected by diabetes insipidus has a partner who is a carrier. State the proportion of their sons that would be predicted to have the condition. Space for working % [Turn over 1 Page 18 MARKS DO NOT WRITE IN THIS MARGIN 7. Students observed a group of California sea lions (Zalophus californianus) that were situated on a rocky outcrop off the coast of California. During each observation period, ten sea lions were observed for six minutes each. The sea lions were watched from a distance using binoculars. The checklist was used as a reference when recording the behaviours observed. California sea lion Behaviour Description of behaviour grooming Licking, smoothing self with tongue, scratching observing Sitting up on flippers looking around resting Lying down with some head raising, barking or yawning movement, aggressive Barking, aggressive charging or chasing movement, non-aggressive Moving for better position on rock other Behaviours not specified above (a) State the term used for a behavioural checklist of this kind. (b) As they observed the sea lions the students noted the time at which each new behaviour started. Describe how the data could be used to construct a time budget. 1 1 Page 19 MARKS DO NOT WRITE IN THIS MARGIN 7. (continued) (c) One student noted, “Sometimes the sea lions tried to annoy each other.” Explain why anthropomorphic statements such as this should be avoided in behavioural studies. (d) Observing the sea lions from a distance made distinguishing some details of behaviour difficult. Suggest an improvement to the method, other than direct observation, that would reduce this source of error. [Turn over 1 1 Page 20 MARKS DO NOT WRITE IN THIS MARGIN 8. “So sex exists to keep parasites at bay.” (Lane, 2009) With reference to the Red Queen hypothesis, discuss the importance of sexual reproduction in defence against parasites. 5 Page 21 [Turn over for next question DO NOT WRITE ON THIS PAGE Page 22 MARKS DO NOT WRITE IN THIS MARGIN 9. Acraea encedon is a butterfly found in tropical Africa. Females of this species can be one of two types: either producing broods that are entirely female or producing broods that have males and females in an approximate 1:1 sex ratio. One hypothesis proposed to explain the all-female broods was that bacteria inherited from the mother kill male embryos only. (a) Explain how antibiotics that kill bacteria could be used in a controlled trial to test this hypothesis. (b) Research has shown that the rapid evolution of male-killing bacteria from non-male-killing strains has been enabled by horizontal gene transfer. Explain what is meant by “horizontal gene transfer”. 2 1 Page 23 MARKS DO NOT WRITE IN THIS MARGIN 9. (continued) (c) Some populations of the butterfly are extremely female-biased with over 90% being infected with this bacterium. These females form lekking swarms at landmarks. This has been described as sex-role-reversal. Butterfly populations that are not female-biased do not have lekking swarms. Suggest the purpose of the lekking swarms and explain why they are only found in female-biased populations of the butterfly. (d) Other sex‐role‐reversed mating systems are usually associated with males investing more than females in offspring, but this is not the case with A. encedon. Suggest one way in which males might demonstrate greater reproductive investment than females in a sex-role-reversed mating system. [Turn over 2 1 Page 24 MARKS DO NOT WRITE IN THIS MARGIN 10. The graph shows the number of measles notifications (reported cases) since 1950 and vaccination rates since 1970. Measles notification (× 1000) Key: Vaccine uptake (%) Vaccine uptake (%) Cases (× 1000) Year 50 54 58 62 66 70 74 78 82 86 90 94 98 02 04 100 80 60 40 20 0 900 800 700 600 500 400 300 200 100 0 (a) State the term used to describe the study of data concerned with the outbreak and spread of infectious disease. (b) Describe the pattern of measles notifications prior to the introduction of vaccination. (c) Describe the correlation suggested by the graph. 1 1 1 Page 25 MARKS DO NOT WRITE IN THIS MARGIN 10. (continued) (d) The MMR vaccine was introduced in 1988 to combat the infectious viral diseases of measles, mumps and rubella. In 1998, a report, published in a peer reviewed medical journal, claimed to establish a link between the MMR vaccine and the disorder known as autism. (i) Explain what is meant by the term “peer reviewed”. (ii) Although criticised heavily and eventually discredited, this research led to a considerable reduction in the number of children being vaccinated. In some areas, such as Swansea in South Wales, vaccination rates fell as low as 67·5%. Over the same period, a huge increase in the number of measles notifications occurred in that area. Account for the spread of measles in the Swansea epidemic. (iii) Explain how the events in Swansea confirm that the graph not only shows correlation but also shows causation. (e) The World Health Organisation (WHO) recommends 95% of children should be immunised (vaccinated) against measles in order to protect all. State the term used to describe this WHO threshold. [Turn over for next question 1 1 1 1 Page 26 MARKS DO NOT WRITE IN THIS MARGIN 11. Answer either A or B in the space below and on Page 27. A Discuss the role of amino acid R-groups in: (i) the determination of tertiary structure of proteins; (ii) influencing the location of proteins within cells. OR B Discuss cell membranes under the following headings: (i) the phospholipid bilayer as a selective barrier; (ii) types of transport proteins and their functions. 3 6 2 7 Page 27 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWER FOR QUESTION 11 [END OF QUESTION PAPER] Page 28 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Page 29 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Page 30 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 31 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 32 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Acknowledgement of Copyright Question 5 Ian Schofield/shutterstock.com Stephan Morris/shutterstock.com Question 7 Eric Isselee/shutterstock.com Question 8 Quote is taken from “Life Ascending: The Ten Great Inventions of Evolution” (Page 135) by Nick Lane. ISBN 9781861978486. Published by Profile Books. Reproduced by kind permission of Profile Books. © National Qualications 2017 AH Supplementary Sheet for Question 1 X707/77/11 Biology Supplementary Sheet TUESDAY, 23 MAY 9:00 AM – 11:30 AM A/HTP

Page 02 1. There are two main pathways of programmed cell death (apoptosis): intrinsic (from within the cell) and extrinsic (from outside the cell). Figure 1 summarises some of the main features of the two pathways. Figure 1 Intrinsic Pathway Caspase 3 activation Caspase 8 activation Apoptosis Caspase 9 activation Cell membrane receptor binding p53 activation Extracellular death signal DNA damage Extrinsic Pathway Apoptosis is deregulated in many tumours, resulting in uncontrolled cell division despite the presence of significant DNA damage. One strategy for the discovery of new anti-cancer drugs has been to examine traditional medicinal herbs. A study was carried out to investigate the effect of an extract of the wild ginger plant, Asiasari radix, on the initiation of apoptosis in colon cancer cells. Colon cancer cells were treated with this extract and then assessed for the presence of apoptotic cells. The percentage increase in cells undergoing apoptosis was calculated by comparing the level of apoptosis in treated cells with that in untreated controls as shown in Figure 2. Figure 2 1200 1000 800 600 400 200 0 20 15 10 5 Extract concentration (μg/cm3) Increase in cells undergoing apoptosis (%)

Page 03 1. (continued) To investigate the involvement of caspases in this process of apoptosis, the experiment was repeated using a single dose of extract (10 µg/cm3) but with the addition of a variety of caspase inhibitors (drugs known to prevent the activation of one or more caspases). Inhibitors used included an inhibitor known to prevent activation of all caspases (all ci) and individual inhibitors of caspase 3 (ci 3), caspase 8 (ci 8) and caspase 9 (ci 9). Percentage changes in the number of cells undergoing apoptosis compared to untreated cancer cells are shown in Figure 3. Figure 3 Change in number of cells undergoing apoptosis (%) ci 9 + Extract ci 8 + Extract ci 3 + Extract all ci + Extract Extract Treatment 500 400 300 200 100 0 −100 The distribution of the cancer cells across the different phases of the cell cycle was then investigated by measuring the DNA content of the cells. Cultures of cells were treated with 10 µg/cm3 extract or left untreated for 24 hours as a control and then the DNA content of 10,000 cells was analysed for each cell culture. Results for treated cells are shown in Figure 4A and for control cells in Figure 4B. DNA content is displayed with arbitrary units where 200 units represents the DNA content of a non-dividing diploid cell. 300 300 240 240 180 180 120 120 60 60 0 0 600 600 400 400 200 200 0 0 DNA content (arbitrary units) DNA content (arbitrary units) Number of cells counted Number of cells counted Figure 4A Figure 4B Treated cells Untreated cells Page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

Page 02 SECTION 1 — 20 marks Attempt ALL questions 1. Which of the following diagrams shows the correct structure of DNA? 5' 3' 3' 5' 3' 5' 5' 3' 5' 3' 3' 5' 3' 5' 5' 3' C D A B 2. A section of double stranded DNA was found to have 60 guanine bases and 30 adenine bases. What is the total number of deoxyribose sugars in this section? A 30 B 90 C 180 D 270

Page 03 3. The following terms describe different structures into which DNA can be organised within cells. 1 Linear chromosome 2 Circular chromosome 3 Circular plasmid Which of these terms describe how DNA is organised within photosynthetic plant cells? A 1 only B 2 only C 1 and 2 only D 2 and 3 only 4. Which of the following molecules are required in the replication of the lagging strand of a DNA molecule? A DNA polymerase and ligase only B DNA polymerase and primers only C Ligase and primers only D DNA polymerase, ligase and primers [Turn over

Page 04 5. The diagram shows a molecule of tRNA. Region Z Bond Y Region X A C U Which row in the table identifies Region X, Bond Y and Region Z? Region X Bond Y Region Z A amino acid attachment site hydrogen anticodon B anticodon hydrogen amino acid attachment site C amino acid attachment site peptide anticodon D anticodon peptide amino acid attachment site 6. New species have evolved when two populations have become A isolated by a behavioural barrier B unable to interbreed to produce fertile offspring C very different due to directional selection D very different due to disruptive selection.

Page 05 7. The diagram shows an enzyme, its substrate and a substance which inhibits it. inhibitor enzyme substrate inhibitor binding site The following statements describe some features of enzyme inhibition. 1 The inhibitor binds to the active site. 2 The effect of the inhibitor is reduced by increasing the substrate concentration. 3 The inhibitor is non-competitive. Which of these statements apply to the inhibitor shown in the diagram? A 1 only B 3 only C 1 and 2 only D 2 and 3 only [Turn over

Page 06 8. An investigation was carried out to determine the effect of lead ion concentration on the activity of the enzyme amylase. The results are shown in the table. Lead ion concentration (mol l−1) Amylase activity (% of control) 0·0 (control) 100 0·1 84 0·2 23 0·3 11 0·4 2 0·5 0 A conclusion that can be drawn from these results is that inhibition was A highest at high lead ion concentrations B highest at low lead ion concentrations C lowest at lead ion concentration 0·5 mol l−1 D highest at lead ion concentration 0·1 mol l−1. 9. ATP is recycled to transfer energy within cells. The diagram shows two reactions involving ATP. Reaction 2 ATP ADP + Pi Reaction 1 Which row in the table describes Reaction 1 and Reaction 2? Reaction 1 Reaction 2 A catabolic and energy released anabolic and energy required B anabolic and energy released catabolic and energy required C catabolic and energy required anabolic and energy released D anabolic and energy required catabolic and energy released

Page 07 10. The fungus Aspergillus niger is grown in large fermenters to produce citric acid using starch as a substrate. The graph shows the changes in the citric acid and starch concentrations in a fermenter over 168 hours. 168 144 120 96 72 48 24 0 60 50 40 30 20 10 0 30 25 20 15 10 5 0 Time (hours) Starch concentration (g/100 ml) Citric acid concentration (g/100 ml) The citric acid concentration equals the starch concentration at A 0 hours B 48 hours C 72 hours D 120 hours. [Turn over

Page 08 11. The following list describes changes which take place during temperature regulation in mammals. 1 Vasodilation 2 Vasoconstriction 3 Contraction of hair erector muscles 4 Relaxation of hair erector muscles Which of these changes takes place in response to a decrease in body temperature? A 1 and 3 only B 1 and 4 only C 2 and 3 only D 2 and 4 only 12. The diagram shows a bacterial plasmid with restriction sites for three different restriction endonucleases, Sal1, Eco R1 and Bam H1. restriction site for Bam H1 restriction site for Eco R1 restriction site for Sal1 Bacterial Plasmid Key Which row in the table identifies the number of fragments produced if the plasmid was cut with the combinations of restriction endonucleases shown? Combination Sal1 and Bam H1 Sal1 and Eco R1 A 3 4 B 3 5 C 4 4 D 4 5

Page 09 13. Inbreeding depression is a result of A an increase in heterozygotes B a genetically variable population C crossbreeding for improved characteristics D an accumulation of recessive deleterious alleles. 14. Livestock production generates less food per unit area of land than crop production because A energy is gained between trophic levels of the food chain B livestock production degrades natural resources C energy is lost between trophic levels of the food chain D it is easier to grow crops than raise livestock in difficult habitats. 15. The table shows optimum, maximum and minimum temperatures for the growth of some crop plants. Temperature (°C) Crop Optimum Maximum Minimum Maize 22–26 32–34 20–22 Wheat 20–25 36–38 5–7 Rice 30–33 37–40 18–22 Potato 15–20 28–34 12–14 Soyabean 25–28 37–40 10–14 Which of the following predictions is supported by the evidence in the table? A Maize will grow at lower temperatures than soyabean. B Rice will grow at higher temperatures than soyabean. C Rice will grow in a narrower range of temperatures than maize. D Wheat will grow in a wider range of temperatures than potato. [Turn over

Page 10 16. Triticale is a hybrid cereal species which was produced by crossing Triticum durum (a species of wheat) with Secale cereale (a species of rye) as shown in the diagram. F1 hybrid (sterile) Triticale 2n = 42 Chromosome number artificially induced to double Secale cereale Triticum durum 2n = 28 × What was the diploid (2n) number of chromosomes in Secale cereale? A 7 B 14 C 21 D 28 17. The following statements describe symbiotic relationships between organisms. 1 Mistletoe plants absorb nutrients from apple trees on which they grow. 2 Egyptian Plover birds clean the teeth of Nile crocodiles and feed on the debris they remove. 3 Tapeworms live in the small intestine of pigs and absorb some of their nutrients. Which of these relationships can be described as parasitic? A 2 only B 3 only C 1 and 2 only D 1 and 3 only

Page 11 18. Resveratrol is a substance which may reduce the risk of heart disease. Using recombinant DNA technology, E. coli bacteria have been modified so that they now produce resveratrol when grown in a medium containing coumaric acid. The graph shows concentrations of resveratrol and coumaric acid in the medium over a 30 hour period. 30 25 20 15 10 5 0 220 200 180 160 140 120 100 80 60 40 20 0 110 100 90 80 70 60 50 40 30 20 10 0 Coumaric acid concentration (mg/l) Resveratrol concentration (mg/l) Time (hours) The simplest whole number ratio of the concentration of resveratrol to coumaric acid after 25 hours is A 1 : 2 B 1 : 40 C 10 : 1 D 20 : 1. 19. Which row in the table identifies how the bottleneck effect and habitat corridors may change genetic diversity of a population? Change in genetic diversity Bottleneck effect Habitat corridors A decrease decrease B decrease increase C increase decrease D increase increase [Turn over

Page 12 20. Some species of social insect are of economic importance to humans by providing ecosystem services. Which of the following are examples of ecosystem services? 1 Braconid wasps parasitising hornworms which are a pest of tomatoes. 2 Bumblebees pollinating an orchard of apple trees. 3 Worker termites caring for the queen and her offspring. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET.] H FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2017 Total marks — 100 SECTION 1 — 20 marks Attempt ALL questions. Instructions for the completion of Section 1 are given on Page 02. SECTION 2 — 80 marks Attempt ALL questions. Questions 10 and 15 contain a choice. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. You should score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X707/76/01 TUESDAY, 23 MAY 9:00 AM – 11:30 AM A/HTP Biology Section 1 — Answer Grid and Section 2 Page 02 SECTION 1 — 20 marks The questions for Section 1 are contained in the question paper X707/76/02. Read these and record your answers on the answer grid on Page 03 opposite. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample Question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D Page 03 A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 SECTION 1 — Answer Grid [Turn over Page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 05 [Turn over for next question DO NOT WRITE ON THIS PAGE Page 06 MARKS DO NOT WRITE IN THIS MARGIN SECTION 2 — 80 marks Attempt ALL questions Note that Questions 10 and 15 contain a choice 1. The diagram illustrates steps in the transcription and translation of a gene. Protein B Protein A Alternative splicing Gene DNA Primary mRNA transcript Exon 3 Exon 2 Exon 1 Exon 4 Exon 4 Exon 2 Exon 1 Exon 3 Exon 2 Exon 1 (a) Name the regions always removed from a primary mRNA transcript. (b) Insert numbers in the boxes below to show the three exons in the gene shown above which could be translated to produce a protein which is different from proteins A and B. Exon Exon Exon 1 1 Page 07 MARKS DO NOT WRITE IN THIS MARGIN 1. (continued) (c) Single gene mutations can occur which may affect the structure of the proteins produced. (i) Describe the effect of a nonsense mutation on Protein A and give a reason for your answer. Description Reason (ii) A deletion mutation occurred in Exon 2. Explain why this would have a major effect on the structure of proteins A and B. [Turn over 2 1 Page 08 MARKS DO NOT WRITE IN THIS MARGIN 2. Two heat-tolerant DNA polymerases used in polymerase chain reactions (PCR) are Taq and Pfu. Pfu has “proof reading” activity. It checks that the correct nucleotides are inserted during replication of a target sequence and then corrects any errors. The graph shows the temperatures during a single PCR cycle required to amplify a target sequence using Taq and Pfu. 4·0 3·5 3·0 2·5 2·0 1·5 1·0 0·5 0 100 90 80 70 60 50 40 Taq polymerase Pfu polymerase Key: Time (minutes) Temperature (°C) (a) (i) Calculate the time taken for 16 copies of the target sequence to be made from one DNA fragment using Taq polymerase. Space for calculation minutes 1 Page 09 MARKS DO NOT WRITE IN THIS MARGIN 2. (a) (continued) (ii) Identify the time period during which primers bind to the original DNA fragment. from to minutes. (b) A scientist was planning to amplify DNA using PCR. State which DNA polymerase should be used and describe the advantage of using this polymerase. DNA polymerase Advantage (c) Explain the importance of using heat-tolerant DNA polymerases in PCR. [Turn over 1 1 1 Page 10 MARKS DO NOT WRITE IN THIS MARGIN 3. The herbicide glyphosate is used to control the annual weed charlock (Sinapis arvensis) in cereal fields. An investigation was carried out into the effect of glyphosate on the development of glyphosate resistance in charlock plants in a cereal plot. The charlock plants were treated with glyphosate from 2009 to 2016 and the percentage of glyphosate resistant plants in the plot was recorded every year. The results are shown in the table. Year Charlock plants resistant to glyphosate (%) 2009 10 2010 18 2011 32 2012 42 2013 53 2014 58 2015 66 2016 66 (a) Using values from the table describe the change in glyphosate resistance over the time of investigation. (b) Explain how natural selection resulted in the change in glyphosate resistance. (c) Another investigation was carried out into the development of antibiotic resistance in bacteria. It was observed to be more rapid than the development of glyphosate resistance in charlock. Explain this observation in terms of gene transfer. 2 2 1 Page 11 MARKS DO NOT WRITE IN THIS MARGIN 4. (a) Human muscles contain satellite cells within the muscle tissue. The diagram illustrates the division and differentiation of satellite cells. satellite cell differentiation muscle cells myoblast cell cell division satellite cells (i) Using information from the diagram explain why satellite cells are an example of tissue (adult) and not embryonic stem cells. (ii) State one benefit to the human body of satellite cells differentiating into myoblast cells. (iii) Satellite cells could be used to treat muscle diseases. Give one ethical reason for using satellite cells instead of embryonic stem cells in order to treat such diseases. (b) Give one example of how stem cells are used as model cells in medical research. [Turn over 1 1 1 1 Page 12 MARKS DO NOT WRITE IN THIS MARGIN 5. (a) The phylogenetic tree illustrates the evolutionary relatedness of six groups of animals. 600 500 400 300 200 100 0 Salmon Frogs Eagles Humans Rats Mice Millions of years ago (i) Using information from the phylogenetic tree state when the last common ancestor of salmon and frogs lived. million years ago (ii) Calculate how many million years separate the divergence of eagles and humans from the divergence of rats and mice. Space for calculation million years (iii) Rats are more closely related to humans than they are to frogs. Use evidence from the phylogenetic tree to justify this statement. 1 1 1 Page 13 MARKS DO NOT WRITE IN THIS MARGIN 5. (continued) (b) The graph shows a molecular clock which compares the amino acid sequence of the protein cytochrome c between a range of species. 600 500 400 300 200 100 0 30 20 10 0 Amino acids in cytochrome c which are different (%) Time of last common ancestor (millions of years ago) (i) Cytochrome c is a protein containing 112 amino acids. Calculate the number of amino acids in cytochrome c that are different between two species whose last common ancestor lived 500 million years ago. Space for calculation (ii) Predict the percentage of amino acids in cytochrome c which would be different between two species who shared a common ancestor 550 million years ago. % (c) Using information from the phylogenetic tree and the graph, state the percentage of amino acids in cytochrome c that are different between rats and frogs. % 1 1 1 [Turn over Page 14 MARKS DO NOT WRITE IN THIS MARGIN 6. The diagram shows genetically modified yeast growing in a fermenter in a medium to which the amino acid lysine has been added. air in stirrer filter genetically modified yeast + growth medium + lysine (a) (i) Name the process for which the yeast cells need the amino acid lysine. (ii) The fermenter contains 5·5 litres of growth medium. Calculate the mass of lysine which should be added to the medium to give a concentration of 300 mg/l. Space for calculation mg (iii) The air entering the fermenter passes through a filter to prevent contamination. Explain why it is necessary to prevent contamination of the culture. 1 1 1 Page 15 MARKS DO NOT WRITE IN THIS MARGIN 6. (a) (continued) (iv) The optimum pH for yeast growth is 4·5. Suggest how this pH could be maintained in the fermenter. (b) Some phases of a growth curve of yeast culture are shown. A B Time Yeast cell number Complete the table by selecting growth phase A or B. Name the chosen phase and describe an event which occurs during that phase of growth. Letter Phase of growth Description (c) Describe a safety mechanism used to prevent the survival of genetically modified microorganisms in the external environment. [Turn over 1 2 1 Page 16 MARKS DO NOT WRITE IN THIS MARGIN 7. Sea bass are saltwater fish that can regulate their internal salt concentration. They have specialised cells in their gills with protein pumps in the membrane. These pumps actively transport excess salt from their bodies. (a) The specialised cells have many mitochondria. Explain why this is necessary. (b) Many animal species regulate their body temperature. Explain the importance of regulating body temperature. (c) Compare regulators and conformers in terms of their ecological niches. 2 1 1 Page 17 [Turn over for next question DO NOT WRITE ON THIS PAGE Page 18 MARKS DO NOT WRITE IN THIS MARGIN 8. Deer mice (Peromyscus maniculatus) are small mammals living in a variety of habitats ranging from low to high altitude. An investigation was carried out to compare the haemoglobin from two populations of deer mice living at low and high altitudes. Blood samples were taken from both populations and exposed to different levels of oxygen. The percentage of haemoglobin in the blood samples which had oxygen bound to it was measured. The results are shown in the graph. 120 100 80 60 40 20 100 90 80 70 60 50 40 30 20 10 0 Oxygen level to which blood sample was exposed (units) Haemoglobin with bound oxygen (%) haemoglobin from deer mice living at high altitude haemoglobin from deer mice living at low altitude Key (a) (i) State one variable that should be controlled when exposing the blood samples to oxygen in order for a valid conclusion to be drawn. 1 Page 19 MARKS DO NOT WRITE IN THIS MARGIN 8. (a) (continued) (ii) State the oxygen level at which there is the greatest difference in the percentage of haemoglobin bound to oxygen between the two groups. units (iii) Use information from the graph to explain how the deer mice from the population living at high altitude are adapted to a low oxygen niche. (b) Suggest one physiological adaptation, other than differences in haemoglobin, that deer mice from high altitudes could have to increase the efficiency of oxygen delivery to cells. (c) Describe the structure of a deer mouse heart and explain how this allows efficient delivery of oxygen to cells. Description Explanation [Turn over 1 1 1 2 Page 20 MARKS DO NOT WRITE IN THIS MARGIN 9. Catalase is an enzyme which breaks down hydrogen peroxide into oxygen and water. Paper discs soaked in catalase sink when placed into hydrogen peroxide solution. The discs rise to the surface when oxygen is produced. The time taken for the discs to rise can be used to measure catalase activity. An experiment was set up to investigate the effect of copper sulfate concentration on catalase activity. Six tubes were set up, each containing 10 cm3 of hydrogen peroxide and 5 cm3 of a different concentration of copper sulfate. One paper disc was then placed into each test tube as shown in the diagram. The time taken for each paper disc to rise to the surface was recorded. hydrogen peroxide (10 cm3) + copper sulfate (5 cm3) Paper disc The results are shown in the table. Concentration of copper sulfate solution (mol l−1) Time taken for paper disc to rise (seconds) 0·2 8 0·3 12 0·4 15 0·6 18 0·8 19 1·0 20 (a) (i) Name the independent variable in this experiment. (ii) Describe a suitable control for this experiment. (iii) Suggest how the temperature of the tubes could be kept constant. 1 1 1 Page 21 MARKS DO NOT WRITE IN THIS MARGIN 9. (a) (continued) (iv) Give a feature of the experiment which may make the results unreliable. (b) (i) Draw a line graph using the results in the table. (Additional graph paper, if required, will be found on Page 32.) (ii) Calculate the percentage increase in the time taken for the paper disc to rise when the copper sulfate concentration increased from 0·2 mol l−1 to 1·0 mol l−1. Space for calculation % (c) Draw a conclusion from the results of this experiment. 1 2 1 1 [Turn over Page 22 MARKS DO NOT WRITE IN THIS MARGIN 10. Answer either A or B in the space below. A Write notes on primate behaviour. OR B Write notes on invasive species. 4 4 Page 23 MARKS DO NOT WRITE IN THIS MARGIN 11. During photosynthesis light energy is absorbed by photosynthetic pigments in the chloroplasts. (a) (i) State one fate of the light which is not absorbed by the photosynthetic pigments. (ii) Describe the effect of absorbed light energy on the pigment molecules. (iii) Plants contain several pigments including chlorophyll a, chlorophyll b and carotenoids. Explain the advantage to a plant of having more than one type of photosynthetic pigment. (b) Following photolysis, hydrogen is transferred to the coenzyme NADP. State the source of this hydrogen. (c) Describe the role of the NADPH in the Calvin cycle (carbon fixation). [Turn over 1 1 1 1 1 Page 24 MARKS DO NOT WRITE IN THIS MARGIN 12. Potato plants are attacked by leaf eating caterpillars. Bacillus thuringiensis is a bacterium which can be used to control these pests. The bacteria produce a protein (Bt toxin) which kills these caterpillars. (a) (i) Explain how an attack by leaf eating caterpillars causes a reduction in crop yield. (ii) State an advantage of using this type of biological control rather than using chemicals. (b) Bt toxin does not kill all caterpillars. A study was carried out to investigate the effectiveness of the Bt toxin compared with a modified Bt toxin by exposing different groups of caterpillars to them. The results are shown in the table. Number of caterpillars tested Number of caterpillars surviving Caterpillars killed (%) Toxin tested Bt toxin alone 240 204 Modified Bt toxin alone 300 105 65 Bt toxin and modified Bt toxin used together 210 42 80 (i) Complete the table to show the percentage of caterpillars killed by the Bt toxin alone. Space for calculation 2 1 1 Page 25 MARKS DO NOT WRITE IN THIS MARGIN 12. (b) (continued) (ii) The Bt toxin and modified Bt toxin work by different mechanisms. Use information from the table to justify this statement. [Turn over 1 Page 26 MARKS DO NOT WRITE IN THIS MARGIN 13. Gluten is a protein found in crops that can cause human health problems. Scientists are breeding barley cultivars to produce ultra low gluten levels. A commercially produced barley (Sloop) and a low gluten cultivar (LG) were crossed to produce two different cultivars with ultra low gluten levels (ULG 1 and ULG 2). The gluten content of each cultivar is shown in the table. Barley cultivar Gluten content (mg/g) Sloop 57·0 LG 5·1 ULG 1 1·7 ULG 2 0·004 (a) Calculate how many times greater the gluten content of Sloop is compared to that of ULG 2. Space for calculation times greater (b) The allele for ultra low gluten is recessive. To investigate if the cultivar LG was heterozygous for gluten, it was crossed with the cultivar ULG1 which was homozygous for this recessive allele. offspring ULG1 low gluten cultivar × (i) Name this type of cross. (ii) Describe the expected phenotypes of the offspring if LG was heterozygous. 1 1 1 Page 27 MARKS DO NOT WRITE IN THIS MARGIN 13. (continued) (c) Barley is a naturally inbreeding plant. Explain why inbreeding depression would be unlikely to be a problem when a barley cultivar self-pollinates for many generations. (d) Barley grains contain the enzyme amylase which breaks down starch in the grain to sugar used in brewing beer. Average grain mass, starch content and amylase activity for three barley cultivars are shown in the table. Average mass of a single grain (mg) Starch content of grains (%) Amylase activity (units/mg ) Barley cultivar Sloop 53·6 70 0·6 ULG1 33·5 65 1·0 ULG2 39·2 64 1·4 (i) As well as total mass of all the grains, state the information required in order to calculate the average mass of a single grain. (ii) Select a cultivar from the table that would be best to use in beer production and justify your selection. Cultivar Justification [Turn over 1 1 1 Page 28 MARKS DO NOT WRITE IN THIS MARGIN 14. African wild dogs are carnivores which live in packs and use cooperative hunting. Each wild dog requires an average of 30 000 kJ of energy per day for the pack to survive. The bar chart shows the relationship between pack size and energy gain per wild dog per day. 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 50 000 40 000 30 000 20 000 10 000 0 Pack size (number of wild dogs) Energy gain (kJ/wild dog/day) (a) Using information from the bar chart, state the minimum pack size (i) at which cooperative hunting becomes an advantage; wild dogs (ii) for survival of the pack. wild dogs (b) Suggest why wild dogs in larger packs gain more energy per individual from hunting even though there are more animals to be fed. (c) Most of the wild dogs in a pack are related. Usually only one dominant female has offspring which other members of the pack will feed. Explain why pack members feed offspring which are not their own. 1 1 1 1 Page 29 MARKS DO NOT WRITE IN THIS MARGIN 15. Answer either A or B in the space below and on Pages 30 and 31. A Write notes on the citric acid cycle of cell respiration. OR B Write notes on how animals survive and avoid adverse conditions. 7 7 [Turn over Page 30 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWERS Page 31 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWERS [END OF QUESTION PAPER] Page 32 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK ADDITIONAL GRAPH PAPER FOR QUESTION 9 (b) (i) Page 33 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Page 34 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Page 35 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Page 36 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

2 P52290 Scientific article for use with Question 7 Craving control: how food messes with your mind 1. You’ve just had a hearty lunch, but the doughnuts next to your desk are winking at you. You can’t shake the thought of what the glazed, soft dough would taste like – and know that you won’t be able to get on with your day until you have it. 2. On a basic level our relationship with food is simple – signals between the gut and the brain tell us when we’re hungry, and when we are full. But experience shows us that the drive to eat is much more tangled and irrational. Some of that is down to the reward hit – the feeling of pleasure, mediated by the brain’s reward centre – that we get from eating calorie-dense food like that glistening doughnut. Indeed, the effect of such foods has led some to liken our desire for them to drug addiction. 3. But we now know the gut itself, and also the microbes inside it, manipulate what we crave, painting a much more complex picture of the forces that determine the way we see food. Cravings could even be contagious – literally. When it comes to food, we’re not as in control as we might think. 4. “People think we have much more conscious control over our eating behaviour than we do. There’s a lot going on behind the scenes and it makes it very difficult to exert control on it,” says Tony Goldstone, an endocrinologist at Imperial College London. 5. Even so, knowing about the forces that manipulate the way we think about food opens up new ways to regain control – for instance by retraining the brain or altering our gut flora. Fresh approaches would be more sensible than just expecting people to eat better, says Goldstone: “We don’t just tell asthmatic people to breathe more.” 6. What, when and how much we eat has typically been explained by two systems, one based on hunger and one on reward. The hunger system is mediated by hormones from the gut and from fat cells, which send information to the brain via the gut’s own nervous system about when we last ate and how hungry we should feel. “We can eat very little one day, and a great deal the next, but this system works to ensure that body weight is relatively stable across the years,” says John Menzies, a neurobiologist at the University of Edinburgh, UK. 7. The reward system is more concerned with what type of food we eat. At its heart is the dopamine pathway, which seems to respond most strongly to foods that are high in fat and sugar. This is natural and necessary – it evolved to prompt us to seek out such food, helping us survive. “If we see a high-energy food, it pays to get it while it’s available – a famine may be round the corner,” says Menzies. “However, in our modern environment where food is abundant and cheap, the reward system may work against us, pushing us towards eating sweet and fatty foods even though we already have plentiful energy stores.” 8. The brain even has its own calorie counter that drives our choices without us knowing, according to a recent study. Participants were shown pictures of 50 foods and asked how many calories they thought each contained, and then invited to bid in an auction for a chance to eat the foods. Regardless of their calorie estimations, which were often inaccurate, the individuals were more likely to bid for the foods that were truly the most calorific. MRI scans showed that activity in reward regions of the brain correlated with the true calorific content of foods – the more calories, the greater the reward.

3 P52290 Turn over 9. Although these hunger and reward systems sound very different, there’s a growing awareness of how interconnected they are. Some clues come from genetics. A gene called FTO is strongly linked to weight gain, and one variant of it raises a person’s risk of becoming obese by 70 per cent. A recent study showed that such people have higher than normal levels of the hormone ghrelin, which is released by the gut, telling them they are still hungry after eating, but their reward system works differently too. MRI studies showed that this group’s brains responded differently when they were shown pictures of food: the most pronounced differences being in the reward regions. The reward pathways in the brains of obese people have also been shown to respond less strongly to food – which could be driving them to seek out even more each time. 10. More evidence of the link comes from people who have had gastric bypass surgery – which reduces the capacity of the stomach and makes food pass more quickly into the small intestine. After surgery, not only do people want to eat less, they experience a profound change in what they want to eat, finding they are drawn to much less calorie-dense foods. And brain scans of people before and after gastric bypass surgery showed altered activity in their reward centres. That contrasts with people who have a gastric band inserted. One explanation for these effects is that after a gastric bypass, food reaches the bowel much more quickly, so there’s a faster hormone response, whereas a gastric band has no effect on hormone levels. 11. “These hormones are normally released after a meal to make us feel full, but as we’re discovering, they also have effects on the way the brain works, to regulate the hedonic responses, the pleasure from food,” says Goldstone. “The bypass patient will say, ‘I’m not hungry, and I also don’t want or like the food’. The band patient will say: ‘I’m not hungry, but I could murder the chocolate cake’. ” 12. What if you could recreate these effects without the surgery? Susan Roberts, at Tufts University in Medford, Massachusetts, has designed a diet in which foods look like the kinds of calorie-dense treats people have learned to crave, but with a twist. “We basically confused people’s reward system by giving them foods that had the flavour and appearance of high-calorie foods that are easily digested, but in fact they were lower calorie, slowly digested versions,” she says. For instance, her diet includes a lower-calorie, slowly digested pizza, made with added fibre. 13. In a small trial, she scanned the brains of a group of overweight people before and after putting them on a six-month diet based on these foods. At the end of the study, the scans showed an increase in activity of reward pathways when the participants looked at pictures of healthy, low- calorie foods, compared with a group not eating the diet. Risky rewards 14. “We were effectively retraining their brains,” says Roberts. “You can think of pizza and you start craving pizza because you anticipate that rush of calories. If you eat the food and you fail to get the rush of calories, over time the reward circuitry adapts so it’s no longer expecting a great zoom of carbohydrate coming in,” she says. 15. The added fibre helped recondition cravings by making people feel full, but Roberts says it’s also important that the participants only ate when they were truly hungry, to strengthen the reward they got from the food. And if dieters cheat and tuck into old favourites, it would strengthen the old reward pathways. Roberts is now beginning two larger clinical trials, and has commercialised the diet plan.

4 P52290 16. So we can retrain our brains to desire different foods. 17. While the brain clearly has a huge influence over what we eat, the influence of gut bacteria might be surprisingly large, too, and they can even affect our minds. 18. Joe Alcock at the University of New Mexico in Albuquerque and his colleagues published a review of research on the microbiome and came to an intriguing conclusion – gut microbes don’t just flourish on certain diets, they may also control our food cravings and preferences to serve their own purposes. 19. There are several ways they could do this. Animals’ gut flora has been shown to affect their taste receptors, which changes their food preferences. And many gut microbes can produce proteins that mimic gut hormones. Alcock’s team even thinks that changes in food preferences that people experience after bariatric surgery might be down to changes in gut microbes, not hormones. 20. That means interventions like probiotics, which help to change the composition of the microbiome, might be useful tools in regulating food cravings. And it suggests a varied diet would make it harder for any one type to flourish and exert control. 21. Because the faecal and oral microbiomes of families under the same roof are more similar than people who don’t live together, the idea that food cravings are influenced by gut bacteria also raises the intriguing possibility that through the spread of these microbes, cravings could even be contagious. Of course, this similarity could be because the members of a household have the same diet. But it might also be that gut bacteria are spread person to person. We already know people are much more likely to become obese if they have a friend who is obese, leading some to speculate that the effect is not down to social contagion, but the spread of microbes. 22. More needs to be done to work out how strong all these effects are, but this new appreciation for the hidden forces influencing our perception of food has wide-reaching implications. Goldstone even wonders whether tapping into the connection between the hunger and reward pathways could alter appetites of a different kind. Animal studies have already shown that ghrelin increases intake of alcohol, nicotine and other drugs, while “fullness” hormones reduce intake. 23. He suspects the same is true for humans. “We’ve shown that your nutritional state modifies the way the brain responds not just to food but also to winning money, and to stress,” he says. “That’s because the same reward circuitry is involved. There’s evidence that gut hormones modify not only reward and consumption of food but also any drug of abuse – such as nicotine, cocaine, alcohol,” he says. They are now beginning a large study. 24. At the very least, all this suggests that expecting people to rely purely on willpower to control what they eat, especially if they are obese, is misguided. “There’s a cabal of obesity researchers that have turned up their hands and said the only thing you can do is rely on willpower,” says Roberts, “I don’t think it’s worked for the last 30 years and it’s not going to work next year either. Which is why we’re trying to do it in a different way.”

5 P52290 Turn over U.S. Navy Recruits Gut Microbes to Fight Obesity and Disease 25. In his lab at Rice University synthetic biologist Jeff Tabor is creating a kind of Lilliputian naval academy. The midshipmen are so small they can’t be seen with the naked eye. But they’re part of a vital mission to protect U.S. naval forces from internal enemies, ranging from metabolic disorders to anxiety and depression. 26. In 2014 Tabor received a three-year grant from the U.S. Office of Naval Research (ONR) to genetically modify a harmless species of Escherichia coli bacteria normally found in the human gut. The goal is to create an edible probiotic organism that can hone in on developing disease and stave it off, even before symptoms take hold. He has recently succeeded in engineering E. coli with sensors that can detect the presence of chemicals signaling disease – at least in the mouse gut. 27. His ultimate aim is to design “a precision gut bacterium that manipulates the intestinal environment in humans to keep it healthy,” he says. This involves rewiring the genes of E. coli to transform the cells into predictable and reliable microbial medics loaded with engineered genetic circuits that can sense specific chemical disturbances and fire off a battery of molecules to neutralize them. The cells would live only a short time in the gut, perhaps six hours or so, “just long enough to do their job,” Tabor says. Then they would die naturally or self-destruct. 28. Tabor’s initial target: obesity and related metabolic issues. “We want to use a genetically engineered E. coli cell to sense the chemicals that signal gut disturbances linked with obesity,” Tabor says, “and then deliver beneficial molecules to prevent weight gain.” 29. Tabor’s work represents the fruitful collision of two hot fields: synthetic biology, the engineering of microorganisms to make useful products; and microbiomics, the study of the microbes living on and inside humans and other animals, collectively known as the microbiome. “There’s great potential in this area because there are so many widespread chronic diseases associated with the gut,” says Pamela Silver of the Wyss Institute for Biologically Inspired Engineering at Harvard University, which published a report of the first synthetic engineered gut microbe in 2014. 30. The 100 trillion bacterial cells that reside in our guts play a major role in nearly every aspect of human biology – digesting food, guiding the immune system, even dictating mental health by sending signals to the brain that affect mood, cognition and behavior. It’s not surprising, then, that disruption of these gut microbial communities can lead to disease, including obesity and related problems. 31. Tabor’s project is part of a larger program on the microbiome funded by the ONR to help U.S. naval forces be more robust in the face of stressors – changes in diet or environment, fearful situations, sleep loss or disrupted circadian rhythms from shifting time zones or living in a submarine. “We’re interested in how gut microbiota respond to these stresses,” says Linda Chrisey, program officer in the ONR’s Warfighter Protection and Application Division. “Are they contributing to the host’s response? If so, can we tweak the microbiota to insulate the host from the stress?”

6 P52290 32. Tabor chose to focus on obesity “because we already know a lot about it at the molecular level,” he says, “so it’s a good model to test the concept.” Our microbiota act like a kind of metabolic ‘organ,’ that affects calorie and nutrient absorption, manages energy balance and controls body weight. (Scientists aren’t sure what shapes microbiomic composition. Increasing evidence suggests that it’s determined before birth and has to do with genetics, maternal diet and mode of delivery.) It’s clear that some bacteria make molecules that disrupt the balance within, causing obesity and other disorders. Studies have shown that the gut bacteria of healthy people churn out compounds that strengthen the intestinal wall but those of obese people make compounds that weaken the wall. This allows bacterial molecules to pass into the bloodstream where they do not belong, triggering an immune response. The resulting chronic inflammation is correlated with a laundry list of ailments, from inflammatory bowel disease to mental health disorders, such as anxiety and depression. 33. It’s still early in the game, but Tabor has already isolated several sensors, reengineered them and put them into a single E. coli bacterium. He has fed the modified cells to mice and shown that the sensors have been activated inside the mouse gut, suggesting they have detected the target chemicals. 34. Tabor plans to have a single E. coli bacterium carry up to a dozen sensors so it can detect multiple signals at one time for a more accurate diagnosis. Ultimately, he plans to engineer these cells to produce drugs when and where they’re called for – highly targeted antibiotics designed to bind with and deactivate those bacterial chemicals that might otherwise leak into the blood from the intestine – thereby preventing the changes that lead to obesity, inflammation and associated ills. Delivering these drugs to the exact tissue in the body where they’re needed and nowhere else would both decrease side effects and increase efficacy. 35. However, “these are genetically engineered organisms, so there will be a long debate about them,” Silver says. “We’ll have to weigh the risks versus the potential benefits. But we’re working to develop ways to make these organisms inherently safe. And I think the concern over risks will be neutralized by the benefits, especially for people who suffer from chronic disease.” 36. So far, Tabor has altered only mouse microbiota. But, he says, “it’s hard to imagine a future where we aren’t diagnosing and treating, possibly curing, many diseases in humans by manipulating gut bacteria in this way – diabetes, autoimmune disorders, cancer, neurological disorders,” and, yes, weight issues. 37. In fact, the Navy may find creative ways to deploy these synthetic probiotics not just to avoid obesity and its attendant problems but to quickly shift body weight and metabolism as necessary, Tabor suggests. “Imagine you have a team of marines going from a temperate environment, say, at sea level, to a really cold environment, like up on top of a mountain, in a short period of time. You want them to be able to put on some fat quickly to be more robust in the cold environment.” 38. The solution? A dose of yogurt laced with synthetic probiotics that change warfighters’ metabolism to increase fat for a couple of weeks – and after that another dose to take it off when they return to sea level.

7 P52290 Turn over Can the Bacteria in Your Gut Explain Your Mood? 39. Eighteen vials were rocking back and forth on a squeaky mechanical device the shape of a butcher scale, and Mark Lyte was beside himself with excitement. ‘‘We actually got some fresh yesterday – freshly frozen,’’ Lyte said to a lab technician. Each vial contained a tiny nugget of monkey feces that were collected at the Harlow primate lab near Madison, Wis., the day before and shipped to Lyte’s lab on the Texas Tech University Health Sciences Center campus in Abilene, Tex. 40. Lyte’s interest was not in the feces per se but in the hidden form of life they harbor. The digestive tube of a monkey, like that of all vertebrates, contains vast quantities of what biologists call gut microbiota. The genetic material of these trillions of microbes, as well as others living elsewhere in and on the body, is collectively known as the microbiome. Taken together, these bacteria can weigh as much as six pounds, and they make up a sort of organ whose functions have only begun to reveal themselves to science. Lyte has spent his career trying to prove that gut microbes communicate with the nervous system using some of the same neurochemicals that relay messages in the brain. 41. Inside a closet-size room at his lab that afternoon, Lyte hunched over to inspect the vials, whose samples had been spun down in a centrifuge to a radiant, golden broth. Lyte, 60, spoke fast and emphatically. ‘‘You wouldn’t believe what we’re extracting out of poop,’’ he told me. ‘‘We found that the guys here in the gut make neurochemicals. We didn’t know that. Now, if they make this stuff here, does it have an influence there? Guess what? We make the same stuff. Maybe all this communication has an influence on our behavior.’’ 42. Since 2007, when scientists announced plans for a Human Microbiome Project to catalog the micro-organisms living in our body, the profound appreciation for the influence of such organisms has grown rapidly with each passing year. Bacteria in the gut produce vitamins and break down our food; their presence or absence has been linked to obesity, inflammatory bowel disease and the toxic side effects of prescription drugs. Biologists now believe that much of what makes us human depends on microbial activity. The two million unique bacterial genes found in each human microbiome can make the 23,000 genes in our cells seem paltry, almost negligible, by comparison. ‘‘It has enormous implications for the sense of self,’’ Tom Insel, the director of the National Institute of Mental Health, told me. ‘‘We are, at least from the standpoint of DNA, more microbial than human. That’s a phenomenal insight and one that we have to take seriously when we think about human development.’’ 43. Given the extent to which bacteria are now understood to influence human physiology, it is hardly surprising that scientists have turned their attention to how bacteria might affect the brain. Micro- organisms in our gut secrete a profound number of chemicals, and researchers like Lyte have found that among those chemicals are the same substances used by our neurons to communicate and regulate mood, like dopamine, serotonin and gamma-aminobutyric acid (GABA). These, in turn, appear to play a function in intestinal disorders, which coincide with high levels of major depression and anxiety. Last year, for example, a group in Norway examined feces from 55 people and found certain bacteria were more likely to be associated with depressive patients. 44. At the time of my visit to Lyte’s lab, he was nearly six months into an experiment that he hoped would better establish how certain gut microbes influenced the brain, functioning, in effect, as psychiatric drugs. He was currently compiling a list of the psychoactive compounds found in the feces of infant monkeys. Once that was established, he planned to transfer the microbes found in one newborn monkey’s feces into another’s intestine, so that the recipient would end up with a completely new set of microbes – and, if all went as predicted, change their neurodevelopment. The experiment reflected an intriguing hypothesis. Anxiety, depression and several pediatric disorders, including autism and hyperactivity, have been linked with gastrointestinal abnormalities. Microbial transplants were not invasive brain surgery, and that was the point: changing a patient’s bacteria might be difficult but it still seemed more straightforward than altering his genes.

8 P52290 45. When Lyte began his work on the link between microbes and the brain three decades ago, it was dismissed as a curiosity. By contrast, last September, the National Institute of Mental Health awarded four grants worth up to $1 million each to spur new research on the gut microbiome’s role in mental disorders, affirming the legitimacy of a field that had long struggled to attract serious scientific credibility. Lyte and one of his longtime colleagues, Christopher Coe, at the Harlow primate lab, received one of the four. ‘‘What Mark proposed going back almost 25 years now has come to fruition,’’ Coe told me. ‘‘Now what we’re struggling to do is to figure out the logic of it.’’ It seems plausible, if not yet proved, that we might one day use microbes to diagnose neurodevelopmental disorders, treat mental illnesses and perhaps even fix them in the brain. 46. In 2011, a team of researchers at University College Cork, in Ireland, and McMaster University, in Ontario, published a study in Proceedings of the National Academy of Science that has become one of the best-known experiments linking bacteria in the gut to the brain. Laboratory mice were dropped into tall, cylindrical columns of water in what is known as a forced-swim test, which measures over six minutes how long the mice swim before they realize that they can neither touch the bottom nor climb out, and instead collapse into a forlorn float. Researchers use the amount of time a mouse floats as a way to measure what they call ‘‘behavioral despair.’’ (Antidepressant drugs, like Zoloft and Prozac, were initially tested using this forced-swim test.) 47. For several weeks, the team, led by John Cryan, the neuroscientist who designed the study, fed a small group of healthy rodents a broth infused with Lactobacillus rhamnosus, a common bacterium that is found in humans and also used to ferment milk into probiotic yogurt. Lactobacilli are one of the dominant organisms babies ingest as they pass through the birth canal. Recent studies have shown that mice stressed during pregnancy pass on lowered levels of the bacterium to their pups. This type of bacteria is known to release immense quantities of GABA; as an inhibitory neurotransmitter, GABA calms nervous activity, which explains why the most common anti-anxiety drugs, like Valium and Xanax, work by targeting GABA receptors. 48. Cryan found that the mice that had been fed the bacteria-laden broth kept swimming longer and spent less time in a state of immobilized woe. ‘‘They behaved as if they were on Prozac,’’ he said. ‘‘They were more chilled out and more relaxed.’’ The results suggested that the bacteria were somehow altering the neural chemistry of mice. 49. Until he joined his colleagues at Cork 10 years ago, Cryan thought about microbiology in terms of pathology: the neurological damage created by diseases like syphilis or H.I.V. ‘‘There are certain fields that just don’t seem to interact well,’’ he said. ‘‘Microbiology and neuroscience, as whole disciplines, don’t tend to have had much interaction, largely because the brain is somewhat protected.’’ He was referring to the fact that the brain is anatomically isolated, guarded by a blood- brain barrier that allows nutrients in but keeps out pathogens and inflammation, the immune system’s typical response to germs. Cryan’s study added to the growing evidence that signals from beneficial bacteria nonetheless find a way through the barrier. Somehow – though his 2011 paper could not pinpoint exactly how – micro-organisms in the gut tickle a sensory nerve ending in the fingerlike protrusion lining the intestine and carry that electrical impulse up the vagus nerve and into the deep-brain structures thought to be responsible for elemental emotions like anxiety. Soon after that, Cryan and a co-author, Ted Dinan, published a theory paper in Biological Psychiatry calling these potentially mind-altering microbes ‘‘psychobiotics.’’

9 P52290 Turn over 50. It has long been known that much of our supply of neurochemicals – an estimated 50 percent of the dopamine, for example, and a vast majority of the serotonin – originate in the intestine, where these chemical signals regulate appetite, feelings of fullness and digestion. But only in recent years has mainstream psychiatric research given serious consideration to the role microbes might play in creating those chemicals. Lyte’s own interest in the question dates back to his time as a postdoctoral fellow at the University of Pittsburgh in 1985, when he found himself immersed in an emerging field with an unwieldy name: psychoneuroimmunology, or PNI, for short. The central theory, quite controversial at the time, suggested that stress worsened disease by suppressing our immune system. 51. By 1990, at a lab in Mankato, Minn., Lyte distilled the theory into three words, which he wrote on a chalkboard in his office: Stress o Immune o Disease. In the course of several experiments, he homed in on a paradox. When he dropped an intruder mouse in the cage of an animal that lived alone, the intruder ramped up its immune system – a boost, he suspected, intended to fight off germ-ridden bites or scratches. Surprisingly, though, this did not stop infections. It instead had the opposite effect: stressed animals got sick. Lyte walked up to the board and scratched a line through the word ‘‘Immune.’’ Stress, he suspected, directly affected the bacterial bugs that caused infections. 52. To test how micro-organisms reacted to stress, he filled petri plates with a bovine-serum-based medium and laced the dishes with a strain of bacterium. In some, he dropped norepinephrine, a neurochemical that mammals produce when stressed. The next day, he snapped a Polaroid. The results were visible and obvious: the control plates were nearly barren, but those with the norepinephrine bloomed with bacteria that filigreed in frostlike patterns. Bacteria clearly responded to stress. 53. Then, to see if bacteria could induce stress, Lyte fed white mice a liquid solution of Campylobacter jejuni, a bacterium that can cause food poisoning in humans but generally doesn’t prompt an immune response in mice. To the trained eye, his treated mice were as healthy as the controls. But when he ran them through a plexiglass maze raised several feet above the lab floor, the bacteria- fed mice were less likely to venture out on the high, unprotected ledges of the maze. In human terms, they seemed anxious. Without the bacteria, they walked the narrow, elevated planks. 54. Each of these results was fascinating, but Lyte had a difficult time finding microbiology journals that would publish either. ‘‘It was so anathema to them,’’ he told me. When the mouse study finally appeared in the journal Physiology & Behavior in 1998, it garnered little attention. And yet as Stephen Collins, a gastroenterologist at McMaster University, told me, those first papers contained the seeds of an entire new field of research. ‘‘Mark showed, quite clearly, in elegant studies that are not often cited, that introducing a pathological bacterium into the gut will cause a change in behavior.’’ 55. Lyte went on to show how stressful conditions for newborn cattle worsened deadly E. coli infections. In another experiment, he fed mice lean ground hamburger that appeared to improve memory and learning – a conceptual proof that by changing diet, he could change gut microbes and change behavior. After accumulating nearly a decade’s worth of evidence, in July 2008, he flew to Washington to present his research. He was a finalist for the National Institutes of Health’s Pioneer Award, a $2.5 million grant for so-called blue-sky biomedical research. Finally, it seemed, his time had come. When he got up to speak, Lyte described a dialogue between the bacterial organ and our central nervous system. At the two-minute mark, a prominent scientist in the audience did a spit take. 56. ‘‘Dr. Lyte,’’ he later asked at a question-and-answer session, ‘‘if what you’re saying is right, then why is it when we give antibiotics to patients to kill bacteria, they are not running around crazy on the wards?’’

10 P52290 57. Lyte knew it was a dismissive question. And when he lost out on the grant, it confirmed to him that the scientific community was still unwilling to imagine that any part of our neural circuitry could be influenced by single-celled organisms. Lyte published his theory in Medical Hypotheses, a low- ranking journal that served as a forum for unconventional ideas. The response, predictably, was underwhelming. ‘‘I had people call me crazy,’’ he said. 58. But by 2011 – when he published a second theory paper in Bioessays, proposing that probiotic bacteria could be tailored to treat specific psychological diseases – the scientific community had become much more receptive to the idea. A Canadian team, led by Stephen Collins, had demonstrated that antibiotics could be linked to less cautious behavior in mice, and only a few months before Lyte, Sven Pettersson, a microbiologist at the Karolinska Institute in Stockholm, published a landmark paper in Proceedings of the National Academy of Science that showed that mice raised without microbes spent far more time running around outside than healthy mice in a control group; without the microbes, the mice showed less apparent anxiety and were more daring. In Ireland, Cryan published his forced-swim-test study on psychobiotics. There was now a groundswell of new research. In short order, an implausible idea had become a hypothesis in need of serious validation. 59. Late last year, Sarkis Mazmanian, a microbiologist at the California Institute of Technology, gave a presentation at the Society for Neuroscience, ‘‘Gut Microbes and the Brain: Paradigm Shift in Neuroscience.’’ Someone had inadvertently dropped a question mark from the end, so the speculation appeared to be a definitive statement of fact. But if anyone has a chance of delivering on that promise, it’s Mazmanian, whose research has moved beyond the basic neurochemicals to focus on a broader class of molecules called metabolites: small, equally druglike chemicals that are produced by micro-organisms. Using high-powered computational tools, he also hopes to move beyond the suggestive correlations that have typified psychobiotic research to date, and instead make decisive discoveries about the mechanisms by which microbes affect brain function. 60. Two years ago, Mazmanian published a study in the journal Cell with Elaine Hsiao, then a graduate student [in the lab of Paul Patterson] and now a neuroscientist at Caltech, and others, that made a provocative link between a single molecule and behavior. Their research found that mice exhibiting abnormal communication and repetitive behaviors, like obsessively burying marbles, were mollified when they were given one of two strains of the bacterium Bacteroides fragilis. 61. The study added to a working hypothesis in the field that microbes don’t just affect the permeability of the barrier around the brain but also influence the intestinal lining, which normally prevents certain bacteria from leaking out and others from getting in. When the intestinal barrier was compromised in his model, normally ‘‘beneficial’’ bacteria and the toxins they produce seeped into the bloodstream and raised the possibility they could slip past the blood-brain barrier. As one of his colleagues, Michael Fischbach, a microbiologist at the University of California, San Francisco, said: ‘‘The scientific community has a way of remaining skeptical until every last arrow has been drawn, until the entire picture is colored in. Other scientists drew the pencil outlines, and Sarkis is filling in a lot of the color.’’ 62. Mazmanian knew the results offered only a provisional explanation for why restrictive diets and antibacterial treatments seemed to help some children with autism: altering the microbial composition might be changing the permeability of the intestine. ‘‘The larger concept is, and this is pure speculation: is a disease like autism really a disease of the brain or maybe a disease of the gut or some other aspect of physiology?’’ Mazmanian said. For any disease in which such a link could be proved, he saw a future in drugs derived from these small molecules found inside microbes. (A company he co-founded, Symbiotix Biotherapies, is developing a complex sugar called PSA, which is associated with Bacteroides fragilis, into treatments for intestinal disease and multiple sclerosis.) In his view, the prescriptive solutions probably involve more than increasing our exposure to environmental microbes in soil, dogs or even fermented foods; he believed there were 11 P52290 wholesale failures in the way we shared our microbes and inoculated children with these bacteria. So far, though, the only conclusion he could draw was that disorders once thought to be conditions of the brain might be symptoms of microbial disruptions, and it was the careful defining of these disruptions that promised to be helpful in the coming decades. 63. The list of potential treatments incubating in labs around the world is startling. Several international groups have found that psychobiotics had subtle yet perceptible effects in healthy volunteers in a battery of brain-scanning and psychological tests. Another team in Arizona recently finished an open trial on fecal transplants in children with autism. (Simultaneously, at least two offshore clinics, in Australia and England, began offering fecal microbiota treatments to treat neurological disorders, like multiple sclerosis.) Mazmanian, however, cautions that this research is still in its infancy. ‘‘We’ve reached the stage where there’s a lot of, you know, ‘The microbiome is the cure for everything,’ ’’ he said. ‘‘I have a vested interest if it does. But I’d be shocked if it did.’’ 64. Lyte issues the same caveat. ‘‘People are obviously desperate for solutions,’’ Lyte said when I visited him in Abilene. (He has since moved to Iowa State’s College of Veterinary Medicine.) ‘‘My main fear is the hype is running ahead of the science.’’ He knew that parents emailing him for answers meant they had exhausted every option offered by modern medicine. ‘‘It’s the Wild West out there,’’ he said. ‘‘You can go online and buy any amount of probiotics for any number of conditions now, and my paper is one of those cited. I never said go out and take probiotics.’’ He added, ‘‘We really need a lot more research done before we actually have people trying therapies out.’’ 65. If the idea of psychobiotics had now, in some ways, eclipsed him, it was nevertheless a curious kind of affirmation, even redemption: an old-school microbiologist thrust into the midst of one of the most promising aspects of neuroscience. At the moment, he had a rough map in his head and a freezer full of monkey fecals that might translate, somehow, into telling differences between gregarious or shy monkeys later in life. I asked him if what amounted to a personality transplant still sounded a bit far-fetched. He seemed no closer to unlocking exactly what brain functions could be traced to the same organ that produced feces. ‘‘If you transfer the microbiota from one animal to another, you can transfer the behavior,’’ Lyte said. ‘‘What we’re trying to understand are the mechanisms by which the microbiota can influence the brain and development. If you believe that, are you now out on the precipice? The answer is yes. Do I think it’s the future? I think it’s a long way away.’’ Bibliography Can the Bacteria in Your Gut Explain Your Mood? by Peter Andrey Smith The New York Times Magazine 23 June 2015, www.nytimes.com/2015/06/28/magazine/can-the-bacteria-in-your-gut-explain-your-mood.html Craving Control by Chloe Lambert New Scientist 18 November 2015, www.newscientist.com/article/mg22830483-300-craving-control-how-food-messes-with-your-mind/ U.S. Navy Recruits Gut Microbes to Fight Obesity and Disease by Jennifer Ackerman Scientific American 16 December 2015, www.scientificamerican.com/article/u-s-navy-recruits-gut-microbes-to-fight-obesity-and-disease/ 12 P52290 BLANK PAGE

page 02 SECTION 1 — 25 marks Attempt ALL questions 1. Which row in the table describes properties of proteins that allow them to be separated using the techniques shown? Protein separation technique Centrifugation Gel electrophoresis A density charge B charge density C shape charge D charge shape 2. Which of the pairs of cell types are fused in order to produce monoclonal antibodies? A T lymphocyte and myeloma B B lymphocyte and myeloma C T lymphocyte and hybridoma D B lymphocyte and hybridoma

page 03 3. The three-dimensional structure of a protein is shown. region X Which row in the table describes region X? Type of secondary structure Bonding that stabilises secondary structure A β-sheet peptide B α-helix peptide C β-sheet hydrogen D α-helix hydrogen [Turn over

page 04 4. The diagram shows some interactions between amino acid R-groups in a polypeptide chain. Which letter shows hydrophobic interactions? CH2 CH2 CH2 CH2 S S CH2 CH3 CH3 CH3 CH3 CH CH CH2 CH2 CH2 H3N+ O- CH2 OH C CH2 O O- C O C D B A 5. A student is preparing media for an experiment to investigate the effect of an inhibitor on cell growth. Flask 1 contains a control medium with no inhibitor. The concentration of the stock inhibitor solution used to prepare the final solution is 80%. Flask Inhibitor volume (cm3) Glucose volume (cm3) FBS volume (cm3) Buffer volume (cm3) Final inhibitor concentration (%) 1 0·00 1·00 5·00 19·00 0 2 1·00 5·00 20 What volumes of inhibitor and buffer should be added to flask 2 to give 25 cm3 of medium with an inhibitor concentration of 20%? A 4·75 cm3 inhibitor  +  20·25 cm3 buffer B 4·75 cm3 inhibitor  +  14·25 cm3 buffer C 6·25 cm3 inhibitor  +  18·75 cm3 buffer D 6·25 cm3 inhibitor  +  12·75 cm3 buffer

page 05 6. Transcription of gene Z only occurs when its transcription factor is dephosphorylated. The distribution of the transcription factor together with the activities of a protein kinase and protein phosphatase specific to this transcription factor are shown in the table. Tissue Transcription factor present Protein kinase activity Protein phosphatase activity Muscle – – + Heart + + – Brain + – + Gene Z is transcribed in the A brain only B heart only C muscle and brain only D heart and brain only. 7. The effect of changing the concentration of extracellular potassium ions on the function of sodium potassium pumps was investigated. Starting with an extracellular solution containing no potassium ions, as the concentration of potassium ions is increased the pumps would be expected to A pump out potassium ions at a faster rate B stay in their phosphorylated conformation for longer C pump in sodium ions at a faster rate D hydrolyse ATP at a faster rate. [Turn over

page 06 8. A sample of 106 cells was found to contain 0·4 mg protein. Actin comprises 4·5% of the total protein. 42 g of actin contains 6·02 × 1020 molecules. 1 g = 1000 mg The number of actin molecules per cell is A 2·58 × 108 B 2·58 × 1010 C 2·58 × 1011 D 2·58 × 1014 9. An outcome of the activation of a cell’s thyroid hormone receptors by thyroxine is A decreased production of Na/KATPase B increased production of Na/KATPase C decreased metabolic rate D opening of ligand-gated ion channels.

page 07 10. The flow diagram shows part of the ADH signal transduction pathway in a collecting duct cell. ADH binds to its receptor Receptor activates G-protein G-protein activates protein kinase Which row in the table shows how this pathway would be altered in an individual with diabetes insipidus? Concentration of inactive G-proteins Phosphorylation of proteins A decreased decreased B decreased increased C increased decreased D increased increased 11. Which row in the table describes the states of the proteins p53 and Rb that would increase the rate of cell division? Protein p53 Rb A activated phosphorylated B inhibited phosphorylated C activated dephosphorylated D inhibited dephosphorylated [Turn over

page 08 12. Which diagram shows the correct positions of the cell cycle checkpoints? Checkpoint is represented by G2 S G0 M G1 G2 S G0 M G1 G2 S G0 M G1 G2 S G0 M G1 A B C D 13. In a study of transmission of the rabies virus by vampire bats, the density of bat colonies was estimated using mark and recapture techniques. The total population estimate is given by (MC)/R where the first sample captured is M, the second sample captured is C and the number recaptured is R. One colony was estimated to have a bat population of 440 following the capture of a second sample of 64 bats, of which 8 were marked. The number of bats captured initially, marked and released was A 32 B 55 C 110 D 128 14. Which of the following describes the purpose of a randomised block design? A Controlling for confounding variables B Ranking sample data C Ensuring that sampling is representative D Monitoring the dependent variable

page 09 15. Merlins are small birds of prey which chase and capture skylarks. The effect of skylark song on hunting by merlins was studied. Graph 1 shows the number of successful and unsuccessful attacks on skylarks showing different singing behaviours. Graph 2 shows how the different singing behaviours affected the mean duration of chases by merlins attacking skylarks. Full song Full song Sub-song Sub-song Song behaviour Song behaviour Number of attacks No song No song 0 20 40 60 80 100 120 Mean chase duration (seconds) Graph 1 Graph 2 Successful attack Unsuccessful attack Key 0 10 20 30 40 50 60 Which of the following generalisations about skylark singing behaviour is valid? A No song decreases the number of attacks by merlins and the time they will chase. B No song increases the success of attacks by merlins and the time they will chase. C No song increases the number of attacks by merlins and the time they will chase. D No song decreases the success of attacks by merlins and the time they will chase. [Turn over

page 10 16. Humans and many other primates have opposable thumbs. In the giant panda, a modified wrist bone forms a false ‘thumb’ which is used along with the five digits to manipulate bamboo. The list shows processes related to evolution. 1 Convergent evolution 2 Divergent evolution 3 Natural selection Evolution of the thumbs of primates and the false ‘thumbs’ of giant pandas has involved A 1 only B 2 only C 1 and 3 only D 2 and 3 only. 17. Which statement includes representatives of all three domains of life? A Insects can be vectors for bacterial infections in plants. B Photoreceptor proteins are found in archaea, plants and animals. C Scientists have cloned genes from archaea that can be expressed in Escherichia coli. D Diseases in potato crops can be caused by the transmission of viruses by nematodes.

page 11 18. Haemophilia A is a sex-linked condition that slows blood clotting. The allele for normal clotting (XH) is dominant to the allele for haemophilia A (Xh). The diagram gives information about the inheritance of haemophilia A in one family. unaffected grandmother P affected grandfather Q unaffected grandmother R unaffected grandfather S unaffected mother unaffected father unaffected daughter affected son From the information given, which of the statements is true? A The genotype of grandmother P must be XHXh but the genotype of grandmother R cannot be determined. B The genotype of neither grandmother can be determined. C The genotype of both the unaffected mother and her daughter must be XHXh. D The genotype of neither the mother or her daughter can be determined. [Turn over

page 12 19. Red deer in Scotland have no natural predators. Control of the growth of a population to prevent it from outstripping resources is achieved by annual culling. The number killed annually must be greater than the recruitment (annual population increase due to births). Since birth rates vary, computer models are used to generate three estimates for recruitment based on birth rates of 30%, 35% and 40%. The number of red deer culled annually is recorded in different areas. The table shows cull totals for one year in four areas, along with estimated recruitment at each birth rate. Area Cull total Estimated recruitment based on birth rate 30% 35% 40% North Ross 2654 2151 2557 2963 East Loch Ericht 887 897 1066 1234 Breadalbane 3026 1396 1659 1922 Knoydart 1082 1079 1283 1487 If the true birth rate is 35%, the area(s) in which the cull is sufficient to prevent population growth would be A North Ross, East Loch Ericht, Breadalbane and Knoydart B North Ross, Breadalbane and Knoydart only C North Ross and Breadalbane only D Breadalbane only.

page 13 20. Many species display some characteristics that are typical of r-selection and some that are typical of K-selection. Which of the following species displays only K-selected characteristics? A Leatherback turtles: lay up to nine large clutches of eggs per breeding season; hatchlings receive no parental care; small proportion survives to reach sexual maturity. B Arctic terns: usually lay two eggs per clutch; adults are aggressive in defence of their young; more than 50% of offspring live to 30 years of age. C English oak trees: slow-growing; do not produce seeds until at least 40 years of age; mature trees produce many thousands of seeds annually but only a small proportion germinate. D Common dandelions: readily colonise disturbed ground; grow rapidly; flower several times a year; produce many seeds per flower head. 21. During the ritualised courtship in peafowl, Pavo muticus, the male spreads and shakes his tail feathers to attract a female before stepping back and bowing. This is followed by loud mating calls. This type of fixed action pattern response can be a result of A honest signals B imprinting C male-male rivalry D species-specific sign stimuli. 22. An in vivo study involves observations made in A the natural habitat of an animal B a living cell culture in the laboratory C a living organism D extracts prepared from living tissues. [Turn over

page 14 23. Eggs of the parasitic liver fluke Leucochloridium paradoxum are found on vegetation and can be eaten by marsh snails, Succinea putris. Inside snails, eggs develop into larvae which move to the ends of their tentacles. The tentacles become swollen and brightly coloured, resembling striped caterpillars. Infected snails become more active during daylight when predatory birds mistake the abnormal tentacles for caterpillars and eat them. The larvae within the tentacles complete their life cycle within the birds’ bodies. Eggs are passed out of the birds in faeces. Which items on the list represent part of the extended phenotype of the parasite? 1 Prey selection by birds 2 Modification of snail tentacles 3 Changed activity of snails 4 Feeding method of snails A 1 and 2 only B 1 and 3 only C 2 and 3 only D 3 and 4 only 24. Which row in the table identifies white blood cells capable of long term survival as a part of immunological memory in mammals? Type of white blood cell B lymphocytes T lymphocytes Phagocytes Natural killers A ✓ ✓ B ✓ ✓ C ✓ ✓ D ✓ ✓

page 15 25. The graphs show data derived from a study investigating the effectiveness of the drug praziquantel (PZQ) on Ugandan children with symptoms of schistosomiasis. PZQ dose (mg/kg) 80 (Girls) 80 (Boys) 40 (Girls) 40 (Boys) Individual cure rate 0∙0 0∙0 0∙0 0∙0 1∙0 1∙0 1∙0 1∙0 0∙8 0∙8 0∙8 0∙8 0∙2 0∙2 0∙2 0∙2 0∙4 0∙4 0∙4 0∙4 0∙6 0∙6 0∙6 0∙6 PZQ dose (mg/kg) 80 (Girls) 80 (Boys) 40 (Girls) 40 (Boys) Individual cure rate PZQ dose (mg/kg) 80 (Girls) 80 (Boys) 40 (Girls) 40 (Boys) Individual cure rate PZQ dose (mg/kg) 80 (Girls) 80 (Boys) 40 (Girls) 40 (Boys) Individual cure rate Age 3 years Age 5 years Age 6 years Age 8 years At what age does the data suggest that children would receive most benefit by increasing the dose of the drug? A 3 years B 5 years C 6 years D 8 years [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET]

page 16 ACKNOWLEDGEMENTS Question 3 – Image of Ribbon representation of Cry3Aa toxin structure, adapted from Figure 1, is reprinted by permission of Springer Customer Service Centre GmbH: Springer Nature, Cellular and Molecular Life Sciences, “Signalling versus punching hole: how do Bacillus thuringiensis toxins kill insect midgut cells” by M. Soberón, S. S. Gill, A. Bravo, © 2009. AH FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2018 Total marks — 90 SECTION 1 — 25 marks Attempt ALL questions. Instructions for the completion of Section 1 are given on page 02. SECTION 2 — 65 marks Attempt ALL questions. A supplementary sheet for question 1 is enclosed inside the front cover of this question paper. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. Score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X707/77/01 TUESDAY, 15 MAY 9:00 AM – 11:30 AM A/HTP Biology Section 1 — Answer Grid and Section 2 page 02 SECTION 1 — 25 marks The questions for Section 1 are contained in the question paper X707/77/02. Read these and record your answers on the answer grid on page 03 opposite. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D page 03 A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 SECTION 1 — Answer Grid [Turn over page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 05 [Turn over for next question DO NOT WRITE ON THIS PAGE page 06 MARKS DO NOT WRITE IN THIS MARGIN SECTION 2 — 65 marks Attempt ALL questions Question 11 contains a choice 1. Read through the supplementary sheet for question 1 before attempting this question. (a) Explain why the rate of uptake by GLUT transporters levels off at high glucose concentrations. (b) Refer to Figure 2 in the supplementary sheet for question 1. Figure 2 shows GLUT3 has the lowest KM for glucose. Explain how this supports the conclusion that GLUT3 has the highest affinity for glucose. (c) The rate of glucose transport at a given glucose concentration can be calculated using the formula: × = + max [ ] [ ] M V G V K G V = rate of glucose transport [G] = glucose concentration (mmol per litre) GLUT2 transporters, found mainly in liver and pancreatic cells, have a KM of 17 mmol per litre. At this concentration of glucose the rate of transport by GLUT2 is 0·02 mmol/min. The physiological range of blood glucose concentration in a healthy individual after fasting ranges from approximately 3·9 to 5·5 mmol per litre. (i) Calculate the rate of glucose transport by GLUT2 when the blood glucose concentration is 5·5 mmol per litre. Space for calculation mmol/min 1 1 2 page 07 MARKS DO NOT WRITE IN THIS MARGIN 1. (c) (continued) (ii) Increases in blood glucose concentration lead to increased insulin production by the pancreas. Glucose uptake by GLUT2 is important for this response because as glucose entry via GLUT2 increases the pancreas synthesises more insulin. Suggest why the high KM of GLUT2 is important in this mechanism for sensing glucose concentration. (d) Release of insulin into the bloodstream leads to a rapid increase in the transport of glucose into muscle and fat cells via GLUT4. Explain why this normal response to insulin does not happen in individuals with type 2 diabetes. (e) Refer to Figure 3 in the supplementary sheet for question 1. Describe the trend shown in Figure 3. (f) Blood serum caffeine levels in people who regularly consume caffeine are typically around 6 μmol per litre. Using Figure 3 predict, with justification, whether this level of caffeine consumption would be likely to have a large effect on the transport of glucose by GLUT1. [Turn over 1 2 1 1 page 08 MARKS DO NOT WRITE IN THIS MARGIN 2. (a) The figure represents the four subunits (two α, two β) of a haemoglobin molecule before and after binding with oxygen molecules. haem iron α-globin chain α-globin chain oxygen bound to iron in haem haemoglobin Before oxyhaemoglobin After β-globin chain β-globin chain (i) Name the process whereby binding of oxygen to one subunit of haemoglobin alters the affinity of the remaining subunits. (ii) Tissues with higher metabolic rates require more oxygen. These tissues produce more carbon dioxide, which dissolves in tissue fluids to form carbonic acid. Explain how this increases oxygen delivery at these tissues. (iii) Haem is a non-protein component important in the function of haemoglobin. State the term used to describe such a component. 1 1 1 page 09 MARKS DO NOT WRITE IN THIS MARGIN 2. (continued) (b) During one stage in its life cycle, the human parasite Plasmodium enters a red blood cell. In order to obtain amino acids that it requires, it digests haemoglobin using a mixture of protease enzymes. This releases the product haem, which is toxic to the parasite. Haem is then converted into non-toxic haemozoin by another enzyme called HDP. Proteases Stage 1 HDP Stage 2 Haem (toxic) Haemozoin (non-toxic) Haemoglobin (i) Name the human disease caused by Plasmodium. (ii) Chloroquine is one of a number of drugs used to treat this disease. Suggest how drugs such as chloroquine, that target Stage 2, may provide an effective treatment. [Turn over 1 2 page 10 MARKS DO NOT WRITE IN THIS MARGIN 3. The diagram shows stages in the transmission of a nerve impulse. +30 0 −55 −70 Time Membrane potential (mV) 1 3 4 5 6 2 stimulus applied threshold 1 Membrane potential before nerve impulse initiated 2 Binding of a neurotransmitter to a ligand-gated sodium ion (Na+) channel 3 Voltage gated Na+ channels open 4 Voltage gated Na+ channels become inactivated 5 Voltage gated potassium ion (K+) channels open 6 Membrane potential after nerve impulse has passed (a) (i) State the term that describes the membrane potential at points 1 and 6. (ii) Use the diagram to calculate the change in membrane potential between points 1 and 4. Space for calculation 1 1 page 11 MARKS DO NOT WRITE IN THIS MARGIN 3. (a) (continued) (iii) Use the information in the diagram to explain the importance of K+ channels in nerve transmission. (b) Tetrodotoxin is a poison found in some fish, such as the pufferfish, which has its effect at stage three of the process shown in the diagram. Suggest a possible mechanism for the toxicity of this substance. [Turn over 2 1 page 12 MARKS DO NOT WRITE IN THIS MARGIN 4. The neurotransmitter gamma-aminobutyric acid (GABA) binds to GABAA receptors in nerve cells. GABAA receptors are a family of structurally related transmembrane ion channels. One proposed structure of a GABAA receptor is shown in the diagram. intracellular extracellular phospholipid bilayer region S (a) (i) All GABAA receptors consist of five subunits. Name the level of protein structure describing several connected polypeptide subunits. (ii) Region S has some of the R groups in contact with the bilayer. Predict the class of R groups to which these amino acids belong. 1 1 page 13 MARKS DO NOT WRITE IN THIS MARGIN 4. (continued) (b) It has been suggested that different forms of the GABAA receptor subunit can arise as a result of alternative RNA splicing. Explain how alternative RNA splicing could result in the production of variant forms of GABAA receptor subunits. (c) Suggest what happens to the receptor protein when GABA binds to it. (d) The drug diazepam increases the effect of GABA molecules by binding to a secondary (allosteric) binding site on GABAA receptors. State the term used to describe the effect of diazepam on GABAA receptors. [Turn over 2 1 1 page 14 MARKS DO NOT WRITE IN THIS MARGIN 5. Microtubules are found in all eukaryotic cells. (a) Name the globular protein of which microtubules are composed. (b) Name the structure from which microtubules radiate. (c) The formation and functioning of mitotic spindle fibres depends on the action of specific enzymes such as cytoplasmic dynein. The role of cytoplasmic dynein in mitosis was investigated. Preparations of purified antibodies that inhibit cytoplasmic dynein’s action were injected into cultured mammalian cells at different stages in mitosis. Comparable cells were injected with a buffer solution containing no antibodies. The figure shows a cell undergoing microinjection. The following results were obtained. • Mitosis was blocked in 73% of the cells injected with the antibody at 12 mg/cm3 during prophase. • Injection of buffer alone had no effect on mitosis. • Lower concentrations of antibodies (6 mg/cm3) had no obvious effect on mitosis. • Cells injected with antibody during metaphase or anaphase completed mitosis with no detectable differences compared to cells injected with buffer alone. 1 1 page 15 MARKS DO NOT WRITE IN THIS MARGIN 5. (c) (continued) (i) State the purpose of injecting cells with buffer solution only. (ii) Give a valid conclusion for this experiment. (d) Once mitosis is complete, the cytoplasm separates to give two daughter cells. State the term used to describe this process. [Turn over 1 1 1 page 16 MARKS DO NOT WRITE IN THIS MARGIN 6. The following customer comment was used to promote a product intended to treat cats that suffered from cancer. ‘My cat was diagnosed with bone cancer three years ago. Her leg was amputated, and I was told that she would only live for another six months. I saw advertising for Vivafel and immediately started her on this product. She has been in remission and healthy ever since. I thoroughly recommend this product and the effect it has on cancer in cats.’ (a) (i) What is the conclusion that appears to have been drawn by the cat’s owner? (ii) Apart from being based on one cat, give one reason why this conclusion is invalid. (b) Suggest why the results of the treatment are unlikely to be caused by a placebo effect, in which even a dummy treatment can bring about some improvement. 1 1 1 page 17 MARKS DO NOT WRITE IN THIS MARGIN 6. (continued) (c) Trials to test the effectiveness of the drug Vivafel were set up using living cats. (i) Describe one way to ensure that these trials were ethical. (ii) State an appropriate null hypothesis for these trials. [Turn over 1 1 page 18 MARKS DO NOT WRITE IN THIS MARGIN 7. Where it is impractical to measure every individual in a population, a representative sample of the population must be obtained. Discuss the principles and strategies that should be employed in the collection of representative samples. 4 page 19 [Turn over for next question DO NOT WRITE ON THIS PAGE page 20 MARKS DO NOT WRITE IN THIS MARGIN 8. Giraffes are the tallest terrestrial animals, growing up to 5 m tall. Approximately half of a giraffe’s height is due to its long neck. Modern giraffes have evolved from ancestors with much shorter necks. The figures represent two different hypotheses regarding the evolution of the giraffe’s long neck. Competing browsers hypothesis Necks for sex hypothesis Charles Darwin suggested that the long necks evolved by natural selection: longer necks allowed animals to feed higher up trees with less competition — the ‘competing browsers’ hypothesis. This hypothesis was not thought to be consistent with all the evidence available and a rival hypothesis, ‘necks for sex’, has been put forward. This suggests long necks have evolved as a result of sexual selection through male–male rivalry, where male giraffes fight for access to females by standing side by side and hitting each other with their heads. (a) Use the competing browsers hypothesis to explain how long necks evolved by natural selection. 2 page 21 MARKS DO NOT WRITE IN THIS MARGIN 8. (continued) (b) Long necks evolved around 13 million years ago when much of the African forest was replaced by grassland with a greatly reduced number of trees. Explain how this supports the competing browsers hypothesis. (c) (i) Explain how long necks could have evolved through sexual selection. (ii) A study published in 2013 concluded that there was no sexual dimorphism in neck length in giraffes. Suggest how this finding would cast doubt on the necks for sex hypothesis. [Turn over 1 1 1 page 22 MARKS DO NOT WRITE IN THIS MARGIN 9. New Zealand mud snails, Potamopyrgus antipodarum, are widely distributed in freshwater streams and lakes in New Zealand. Snail populations consist of females that reproduce asexually by parthenogenesis together with females that reproduce sexually by cross fertilisation with males. (a) State two disadvantages of sexual reproduction. 1 2 (b) New Zealand mud snails are commonly infected with parasitic worms of the genus Microphallus. Sexual reproduction is more common in the snails when the prevalence of parasites is high. Explain how this observation supports the Red Queen hypothesis. (c) The New Zealand mud snail has become invasive by spreading beyond its native habitat to colonise areas of Europe and North America. Suggest why invasive populations are found to be composed entirely of parthenogenic females. 2 2 1 page 23 [Turn over for next question DO NOT WRITE ON THIS PAGE page 24 MARKS DO NOT WRITE IN THIS MARGIN 10. The figure shows an Ebola virus, cause of Ebola virus disease (EVD), prevalent in a number of West and Central African countries. The virus is transmitted to people from wild animals and outbreaks may then occur through human to human transmission. lipid membrane polymerase capsid RNA glycoprotein (a) Ebola viruses have a diameter of 8 × 10−2 μm. Give this measurement in nanometres (nm). (1 nm = 10−3 μm) Space for calculation (b) The flow diagram shows some stages in the replication of this virus. New virus particles assembled mRNA is translated into viral proteins Viruses released and cell destroyed Viral RNA replicated Viral RNA is used as a template to synthesise complementary mRNA Virus enters host cell Virus attaches to host cell receptors Explain why the Ebola virus cannot be regarded as a retrovirus. 1 1 page 25 MARKS DO NOT WRITE IN THIS MARGIN 10. (continued) (c) EVD symptoms result from viral disruption of immune cell function such as the loss of lymphocytes by apoptosis. State the cause of cell death during apoptosis. (d) The genes coding for some Ebola virus proteins have a high rate of mutation. Why would this make development of a vaccine more difficult? (e) EVD has a very high mortality rate. Some researchers have suggested that new treatments should not be assessed by clinical trials that use negative control groups. State whether you agree or disagree with this suggestion. Justify your answer. (f) Apart from medical treatments, suggest one measure that could be effective in controlling or preventing outbreaks of EVD in the tropical regions of Africa. [Turn over for next question 1 1 1 1 page 26 MARKS DO NOT WRITE IN THIS MARGIN 11. Answer either A or B in the space below and on page 27. A Discuss the concept of niche under the following headings: (i) realised and fundamental niche; (ii) features of parasite niches. OR B Discuss the formation of variable gametes during meiosis under the following headings: (i) the activity of homologous chromosomes; (ii) meiosis II. 3 6 7 2 page 27 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWER FOR QUESTION 11 [END OF QUESTION PAPER] page 28 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK page 29 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK page 30 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 31 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

page 32 ACKNOWLEDGEMENTS Question 2 (a) – Blamb/Shutterstock.com Question 4 – Image of GABAA receptor by BogHog2 is taken from https://commons.wikimedia. org/wiki/File:NAchR_2BG9.png. Public Domain. Question 5 (c) – koya979/Shutterstock.com Question 8 – AndreAnita/Shutterstock.com Roman Balla/Shutterstock.com Question 10 – Designua/Shutterstock.com © National Qualications 2018 AH Supplementary sheet for question 1 X707/77/11 Biology Supplementary Sheet TUESDAY, 15 MAY 9:00 AM – 11:30 AM B/HTP

page 02 1. The diffusion of glucose across the plasma membrane of mammalian cells is facilitated by a family of related proteins called GLUT transporters. GLUT transporters have a specific binding site for glucose which alternately faces inside and outside the cell. The orientation of the binding site is brought about by a change in conformation as shown in Figure 1. Figure 1 plasma membrane extracellular intracellular glucose transporter glu glu glu = glucose Several studies have measured the changes in rates of glucose uptake by GLUT transporters as the concentration of glucose is increased. Results for four different GLUT transporters are shown in Figure 2. For each type of GLUT transporter, the rate of transport levels off to a maximum value that is termed Vmax. The glucose concentration at which the rate of transport is half Vmax is defined as the KM of the transporter. KM values for four types of GLUT are shown in Figure 2. Figure 2 0 3 6 9 12 15 18 21 0∙00 0∙01 0∙02 0∙03 0∙04 0∙05 Glucose concentration (mmol/l) Glucose uptake (mmol/min) GLUT3 (KM = 1∙4 mmol/l) GLUT4 (KM = 5·0 mmol/l) GLUT1 (KM = 3·0 mmol/l) GLUT2 (KM = 17·0 mmol/l) Key

page 03 1. (continued) Studies have suggested that the chemical caffeine is an inhibitor of glucose transport by GLUT1. Figure 3 shows data obtained from a recent study of the effect of increasing caffeine concentration on the uptake of glucose by GLUT1. The uptake of glucose in these experiments was measured per litre of intracellular fluid. Figure 3 0 0 5 5 10 10 15 15 20 20 Caffeine (mmol/l) Glucose uptake (μmol/l/min) [END OF SUPPLEMENTARY SHEET] page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

page 02 SECTION 1 — 20 marks Attempt ALL questions 1. The following substances are products of fermentation. 1 ATP 2 Lactate 3 Carbon dioxide Which of these are products of fermentation in human muscle cells? A 2 only B 1 and 2 only C 2 and 3 only D 1, 2 and 3 2. The diagram represents a stage of cellular respiration that occurs in a mitochondrion. inner membrane proteins Q P P R water + Which row in the table identifies substances P, Q and R? Substances P Q R A ATP hydrogen ions and electrons oxygen B hydrogen ions and electrons oxygen ATP C oxygen ATP hydrogen ions and electrons D hydrogen ions and electrons ATP oxygen

page 03 3. Part of a metabolic pathway used by cells to produce the amino acid alanine is shown. pyruvate kinase phosphoenol pyruvate pyruvate alanine Alanine is a non-competitive, feedback inhibitor of the enzyme pyruvate kinase. The following statements refer to the metabolic pathway. 1 Pyruvate kinase reduces the activation energy needed to convert phosphoenol pyruvate into pyruvate. 2 Phosphoenol pyruvate is the substrate for pyruvate kinase. 3 Alanine can bind to the active site of pyruvate kinase. Which of these statements are correct? A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 [Turn over

page 04 4. Shrews are small mammals. The graph shows the relationship between body mass and oxygen consumption of shrews at two environmental temperatures. Oxygen consumption (cm³/g of body mass/hour) Body mass (g) Key 10 °C 20 °C 0 4 8 12 16 20 0 4 8 12 16 Which of the following statements about this graph is correct? A Shrews of greater mass consumed less oxygen. B The optimum temperature for oxygen consumption was 10 °C. C As environmental temperature increased oxygen consumption decreased. D At 10 °C a 16 g shrew consumed 6·2 cm3 of oxygen/g of body mass/hour. 5. Yeast cells contain the enzyme catalase which breaks down hydrogen peroxide to produce oxygen. An experiment was carried out into the effect of lead nitrate concentration on the activity of catalase. Six flasks were set up. Each contained 25 cm3 of hydrogen peroxide and 10 cm3 of yeast suspension. 10 cm3 of a different concentration of lead nitrate was then added to each flask. The volume of oxygen produced after 15 minutes was measured. Identify the independent variable in this experiment. A Volume of lead nitrate B Volume of oxygen produced C Activity of catalase D Concentration of lead nitrate

page 05 6. The diagram illustrates the circulatory system of a fish. The arrows indicate the direction of blood flow. capillaries in the body tissues capillaries in the gills heart Which row in the table describes the type of circulatory system of a fish and the blood pressure in the capillaries in the gills and body tissues? Type of circulatory system Blood pressure in the capillaries in the gills Blood pressure in the capillaries in the body tissues A single lower higher B double higher lower C single higher lower D double lower higher [Turn over

page 06 7. In an investigation into fermentation, yeast was grown in a flask of glucose solution for 20 hours at 20 °C. The graph shows the concentrations of ethanol and glucose in the flask over the period of the investigation. Ethanol concentration (grams per litre) Glucose concentration (grams per litre) Time (hours) 0 5 10 15 20 0 1 2 3 4 0 2 4 6 8 What was the glucose concentration when the ethanol concentration was 3·3 grams per litre? A 1·2 grams per litre B 2·2 grams per litre C 2·4 grams per litre D 6·6 grams per litre

page 07 8. The following statements describe stages in the Calvin Cycle (carbon fixation). 1 Carbon dioxide attaches to ribulose bisphosphate (RuBP) producing 3-phosphoglycerate (3PG). 2 3-phosphoglycerate (3PG) forms glyceraldehyde-3-phosphate (G3P). 3 Glyceraldehyde-3-phosphate (G3P) regenerates ribulose bisphosphate (RuBP). Which row in the table identifies the stage which is catalysed by RuBisCO and the stage which requires hydrogen? Catalysed by RuBisCO Requires hydrogen A 1 2 B 1 3 C 3 1 D 3 2 [Turn over

page 08 9. The diagram shows apparatus used in an investigation to measure the rate of photosynthesis in Elodea (pondweed) at different wavelengths of light. Coloured filters were used to change the wavelength of the light. The volume of oxygen collected after 30 minutes was used to measure the rate of photosynthesis. bubbles of oxygen measuring cylinder scale (cm3) coloured filter lamp Elodea Suggested improvements to the investigation are shown. 1 Use a measuring cylinder with a narrower diameter. 2 Repeat the experiment several times and take averages. 3 Use a scale with more divisions. Which of these suggestions would improve the accuracy of the results? A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3

page 09 10. The diagram shows a perennial weed found in agricultural land in Scotland. seeds stem leaf storage organ Which feature of this weed indicates that it should be controlled by a systemic herbicide? A Seeds B Stem C Storage organ D Leaf [Turn over

page 10 11. A field trial was carried out to investigate the effect of mass of phosphate fertiliser applied on the growth of barley. The barley was planted in plots of equal area on a hillside and fertiliser applied as shown in the diagram. 2 g/m2 2 g/m2 2 g/m2 4 g/m2 4 g/m2 4 g/m2 6 g/m2 6 g/m2 6 g/m2 Which of the following procedures would improve the field trial design to take into account higher soil moisture levels at the bottom of the hill? A Increase the range of phosphate fertiliser masses applied. B Randomise the treatment plots. C Increase the number of plots. D Select another hillside. 12. Which of the following is an example of kin selection? A Worker leafcutter ants raising young ants in their colony. B A vampire bat regurgitating blood to feed an unrelated bat. C A dominant lion feeding on a zebra kill before its offspring. D A young orangutan spending a long period in parental care to learn complex social behaviours. 13. An experiment was carried out to investigate the growth rate of pigs. They were put into five groups of eight pigs, each with the same average initial body mass. Each group was fed a diet which contained either 0, 10%, 20%, 30% or 40% faba beans. The pigs were re-weighed each day for 40 days. Which aspect of the experimental design increased reliability of the results? A Five groups of pigs were used. B The pigs were re-weighed each day for 40 days. C Each group had the same average initial body mass. D Each group contained eight pigs.

page 11 14. An investigation was carried out into the social hierarchy in a group of five hens, V, W, X, Y and Z. Hens establish dominance by pecking each other aggressively. The number of pecks given and received was recorded. The results are shown in the table. Number of pecks given by each hen V W X Y Z Number of pecks received by each hen V - - - 10 - W 2 - - 13 - X 6 8 - 7 - Y - - - - - Z 11 10 5 4 - The order of hierarchy from most dominant to least dominant hen is A Z, V, X, W, Y B Y, V, W, X, Z C Z, X, W, V, Y D Y, W, V, X, Z. 15. Each type of human cell has a different structure and function because A only some of their genes are expressed B they contain different genes C some genes are lost during differentiation D some genes are gained during differentiation. [Turn over

page 12 16. The list describes some uses of stem cells. 1 Studying how cells differentiate 2 Researching the development of Parkinson’s disease 3 Producing skin for skin grafts 4 Bone marrow transplants Which of these uses are not therapeutic? A 1 only B 1 and 2 only C 2 and 4 only D 3 and 4 only 17. Which of the following is an example of sexual selection? A Koalas with resistance to disease surviving to reproduce. B Peppered moths with the most effective camouflage avoiding predation. C Plant breeders selecting barley cultivars to cross to improve grain yield. D Female black grouse mating with the male with the best display.

page 13 18. A population of finches became isolated on an island. The graph shows the range of beak sizes within the initial population at the time of isolation and in the population after many generations. Increasing number of finches Key initial population population after many generations Increasing beak size Which row in the table shows the type of selection pressure and the type of speciation which might be expected to occur in this example? Selection pressure Speciation A directional allopatric B directional sympatric C stabilising allopatric D stabilising sympatric [Turn over

page 14 19. Some processes involved in evolution are shown. 1 sexual selection 2 disruptive selection 3 genetic drift Which of these processes involve non-random changes in the frequency of DNA sequences? A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 20. The analysis of DNA sequences from different organisms is used in the production of molecular clocks. This analysis is based on the assumption that over time DNA sequences undergo mutations A randomly B spontaneously C at a varying rate D at a constant rate. [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET.] page 15 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 16 [BLANK PAGE] DO NOT WRITE ON THIS PAGE H FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2018 Total marks — 100 SECTION 1 — 20 marks Attempt ALL questions. Instructions for the completion of Section 1 are given on page 02. SECTION 2 — 80 marks Attempt ALL questions. Questions 11 and 14 contain a choice. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. Score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X707/76/01 TUESDAY, 15 MAY 9:00 AM – 11:30 AM A/PB Biology Section 1 — Answer Grid and Section 2 page 02 SECTION 1 — 20 marks The questions for Section 1 are contained in the question paper X707/76/02. Read these and record your answers on the answer grid on page 03 opposite. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D page 03 A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 SECTION 1 — Answer Grid [Turn over page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 05 [Turn over for next question DO NOT WRITE ON THIS PAGE page 06 MARKS DO NOT WRITE IN THIS MARGIN SECTION 2 — 80 marks Attempt ALL questions Questions 11 and 14 contain a choice 1. The diagram illustrates thermoregulation in mammals following a decrease in body temperature. (a) (i) Name the type of control used in thermoregulation as shown in the diagram. (ii) Name the part of the brain in which thermoreceptors are found. (iii) State how information travels to the effectors in the skin. 1 1 1 thermoreceptors in the brain decrease in body temperature information travels to effectors in the skin corrective mechanism normal body temperature page 07 MARKS DO NOT WRITE IN THIS MARGIN 1. (continued) (b) Effectors in the skin include muscles in the walls of blood vessels. (i) Describe the response of these effectors to a decrease in body temperature. (ii) Explain how this response would help return body temperature to normal. (c) Explain why it is important for a mammal to regulate its body temperature. [Turn over 1 1 1 page 08 MARKS DO NOT WRITE IN THIS MARGIN 2. Daphnia (Daphnia pulex) is a species of water flea that lives in fresh water. An investigation was carried out into the effect of water temperature on the heart rate of one Daphnia. The results are shown in the table. Water temperature (°C) Heart rate (beats per minute) 2 175 7 184 12 194 17 207 22 219 (a) Calculate the average increase in heart rate per °C between 2 °C and 22 °C. Space for calculation beats per minute per °C (b) Daphnia is a conformer. Use evidence from the table to confirm this statement. (c) Explain how an increased water temperature would result in a higher metabolic rate in Daphnia. (d) State the type of response shown by conformers to maintain an optimum metabolic rate. 1 1 2 1 page 09 [Turn over for next question DO NOT WRITE ON THIS PAGE page 10 MARKS DO NOT WRITE IN THIS MARGIN 3. The bacteria Streptomyces is a microorganism found in soil. It produces a secondary metabolite, the antibiotic streptomycin, which kills other microorganisms. Streptomyces live in close association with plant roots. These plants produce soluble carbohydrates which are released into the soil through their roots. (a) (i) Name the growth phase during which streptomycin is produced. (ii) Explain the advantage to Streptomyces of producing an antibiotic such as streptomycin. (b) The relationship between Streptomyces and the plant roots is described as mutualistic. (i) Suggest the benefit to Streptomyces. (ii) Suggest the benefit to the plant. 1 1 1 1 page 11 MARKS DO NOT WRITE IN THIS MARGIN 3. (continued) (c) An investigation was set up to compare the effectiveness of streptomycin with other antibiotics by measuring the survival of bacteria. A species of bacteria was grown in the presence of different concentrations of antibiotics and the percentage which survived was calculated. The results are shown in the table. Antibiotic Concentration of antibiotic (µg/cm3) Survival of bacteria (%) Fusidic acid 10 6 Chloramphenicol 25 42 Erythromycin 5 49 Gentamycin 10 5 Tetracycline 25 35 Streptomycin 10 35 (i) Name one antibiotic with which streptomycin could be validly compared. (ii) Give a conclusion which can be drawn from the results. [Turn over 1 1 page 12 MARKS DO NOT WRITE IN THIS MARGIN 4. The diagram shows information on the breeding and migration of Monarch butterflies (Danaus plexippus). Each generation dies after laying eggs. (a) State one advantage and one disadvantage to the Monarch butterfly of migration to Mexico. Advantage Disadvantage 2 4th generation migrate to North America in the spring and breed to produce a new 1st generation 1st generation breed and produce the 2nd generation in early summer in North America 2nd generation breed and produce the 3rd generation in late summer in North America 4th generation migrate south to Mexico for the winter 3rd generation breed and produce the 4th generation in autumn in North America page 13 MARKS DO NOT WRITE IN THIS MARGIN 4. (continued) (b) The migratory behaviour of the Monarch butterfly from North America to Mexico is innate. Use the information given to justify this statement. (c) Some species of hummingbird also migrate between North America and Mexico. They have high metabolic rates which they reduce while resting each night during the migration period. Name this reduction in metabolic rate. [Turn over 1 1

page 14 5. The diagram shows some features of a plasmid which has been cut open by a restriction endonuclease to allow a gene from a donor chromosome to be inserted. The DNA recognition sites for three restriction endonucleases, E coR1, BamH1, and HindIII, are shown in the table. The arrows indicate where each restriction endonuclease cuts the DNA sequence. ampicillin resistance gene region X G C T T A A C T T A A G Restriction endonuclease DNA sequence recognised BamH1 E coR1 HindIII G G A T C C C C T A G G G A A T T C C T T A A G A A G C T T T T C G A A DO NOT WRITE IN THIS MARGIN page 15 MARKS DO NOT WRITE IN THIS MARGIN 5. (continued) (a) (i) A restriction endonuclease was used to remove a gene from a donor chromosome. Use information from the diagram and the table to identify the restriction endonuclease which would be used to allow the gene to be inserted into the plasmid. Give a reason for your answer. Restriction endonuclease Reason (ii) Name the enzyme which would be used to seal the gene into the plasmid. (iii) A culture of bacterial cells, 20% of which had taken up this modified plasmid, were grown on a nutrient agar plate. The plate was incubated and 250 colonies of this bacteria grew. Predict the number of colonies which would have been expected to grow if the nutrient agar plate had contained the antibiotic ampicillin. Space for calculation (b) Name region X, shown in the diagram, which ensured that the modified plasmid would be passed on to daughter cells. [Turn over 2 1 1 1 page 16 MARKS DO NOT WRITE IN THIS MARGIN 6. An investigation was carried out to monitor the populations of red squirrels (Sciurus vulgaris) and grey squirrels (Sciurus carolinensis) in a 15 km2 wooded area. The average number of breeding pairs of each species was recorded between September 2010 and June 2012. The results are shown in the graph. (a) (i) Use values from the graph to describe the changes in the average number of breeding pairs of grey squirrels from March 2011 to March 2012. (ii) Calculate the total number of breeding pairs of red squirrels in the wooded area in September 2011. Space for calculation breeding pairs (iii) Express, as the simplest whole number ratio, the number of grey squirrels to red squirrels in June 2012. Space for calculation 2 1 1 grey red : Sep 2010 Dec 2010 Mar 2011 Jun 2011 Sep 2011 Dec 2011 Mar 2012 Jun 2012 3·0 2·5 2·0 1·5 1·0 0·5 0·0 Month and Year Average number of breeding pairs of squirrels/km2 Red Squirrel Grey Squirrel Key page 17 MARKS DO NOT WRITE IN THIS MARGIN 6. (continued) (b) Pine martens (Martes martes) live in wooded areas and prey on squirrels. Populations of pine martens, red squirrels and grey squirrels were estimated in two other wooded areas using automatic cameras. These cameras are triggered by the movement of passing animals. The results are shown in the table. Wooded area Estimated number of grey squirrels Estimated number of red squirrels Estimated number of pine martens 1 88 645 45 2 465 112 12 (i) Suggest why the method used to estimate the numbers of pine martens and squirrels may lead to inaccurate results. (ii) Use evidence from the table which could be used to support the following statements. 1 Pine martens are more successful predators of grey squirrels than of red squirrels. 2 Grey squirrels compete more successfully for food than red squirrels. (c) Grey squirrels have spread rapidly and eliminated native red squirrels from much of the UK. State the term used to describe grey squirrels in the UK as a result of this. [Turn over 1 1 1 1 page 18 MARKS DO NOT WRITE IN THIS MARGIN 7. Oil extracted from the seeds of the crop false flax (Camelina sativa) can be used as fuel. An investigation was carried out into the effect of a plant growth regulator paclobutrazol (PBZ) on the photosynthetic pigment content of the leaves and the oil yield from the seeds of false flax. The results are shown in the table. Treatment Average photosynthetic pigment content (mg/g of leaf) Average oil yield (g/plant) Chlorophyll a and b Carotenoids Untreated 3·28 3·02 1·7 Treated with PBZ 3·27 3·98 2·4 Absorption spectra for pigments from the treated and untreated plants were produced and are shown in the graph. (a) (i) Use values from the graph to describe the difference in absorption spectra of the pigments from treated and untreated plants. 1 Wavelength of light (nm) Increasing absorption of light 350 450 550 650 400 500 600 700 pigments from treated plants pigments from untreated plants Key page 19 MARKS DO NOT WRITE IN THIS MARGIN 7. (a) (continued) (ii) Use evidence from the table to explain the difference in the absorption spectra. (iii) State one use that plants make of the light energy absorbed by pigments during photosynthesis. (b) Suggest why seeds from the plants treated with PBZ yield more oil. [Turn over 1 1 2 page 20 MARKS DO NOT WRITE IN THIS MARGIN 8. Salmon can be reared in fish farms where they are sometimes fed small fish such as anchovies. Anchovies feed on animal plankton which feed on plant plankton. The energy contents at each trophic level in this food chain are shown on a log scale in the bar graph. (a) (i) State the energy content fixed in the biomass of salmon. kJ (ii) Plant plankton fix 2% of the solar energy they receive in their biomass. Calculate the total solar energy to which the plant plankton were exposed. Space for calculation kJ 1 1 100,000 10,000 1,000 100 10 1 Plant plankton Animal plankton Anchovies Salmon Organism Energy content fixed in biomass (kJ) page 21 MARKS DO NOT WRITE IN THIS MARGIN 8. (continued) (b) The human population is increasing. (i) State the term used to define the ability of the human population to access food of sufficient quantity. (ii) In terms of energy explain the advantage to the human population of consuming anchovies rather than salmon. [Turn over 1 1 page 22 MARKS DO NOT WRITE IN THIS MARGIN 9. The diagram shows crosses in a breeding programme involving different breeds of sheep. (F2) Commercial lambs Greyface female (F1)  X  Suffolk male (P) Scottish Blackface female  X  Border Leicester male (a) Suggest a reason why breeding programmes such as this include crossbreeding. (b) Explain why Greyface sheep are produced by crossbreeding Scottish Blackfaces with Border Leicesters instead of breeding F1 Greyface sheep together. (c) To produce commercial lambs which show a desired dominant characteristic, Suffolk males homozygous for that characteristic are used. (i) Name the type of cross used to identify if the genotype of the desired characteristic in Suffolk males is homozygous. (ii) Explain the importance of selecting a Suffolk male homozygous for the desired dominant characteristic. 1 1 1 1 page 23 MARKS DO NOT WRITE IN THIS MARGIN 9. (continued) (d) Spider lamb syndrome is a hereditary condition in sheep caused by a recessive deleterious allele which results in limb deformities. State why inbreeding could cause an increase in the number of lambs born with this condition. [Turn over 1 page 24 MARKS DO NOT WRITE IN THIS MARGIN 10. An experiment was carried out to investigate the evolutionary relatedness of four species of fish by comparing proteins extracted from the fish. The more closely related species are, the more proteins they have in common. A sample of muscle tissue from each species of fish was heated in a solution to extract proteins. The protein extracts were analysed by gel electrophoresis which separates proteins according to their mass. A protein standard containing proteins of known masses was also analysed. The results of the gel electrophoresis are shown in the diagram. Each band represents a protein. Mass of protein in protein standard (kDa) Distance travelled (mm) 16 50 20 38 26 30 44 24 66 15 108 10 (a) (i) Identify two variables related to the protein extraction, not already mentioned, which should be kept constant so that a valid conclusion can be drawn. 1 2 (ii) During the preparation the samples were heated. This unfolds the proteins changing their three-dimensional shape. Name one type of bond that could have been broken to cause this change. 2 1 Protein band X Direction of movement of proteins Protein standard Species 1 Species 2 Species 3 Species 4 page 25 MARKS DO NOT WRITE IN THIS MARGIN 10. (continued) (b) (i) Draw a line graph to show the distance travelled by the protein bands in the gel against the mass of protein in the protein standard. (Additional graph paper, if required, can be found on page 33) Mass of protein in protein standard (kDa) (ii) Band X travelled 28 mm. Use the graph to identify the mass of the protein in band X. kDa (iii) Each species of fish contains a protein with a mass of 66 kDa. One amino acid has an average mass of 0·12 kDa. Calculate how many amino acids that would be expected in this protein. Space for calculation (iv) Explain why it was concluded that species 1, 2 and 4 are more closely related to each other than they are to species 3. 2 1 1 1 [Turn over page 26 MARKS DO NOT WRITE IN THIS MARGIN 11. Answer either A or B. A Write notes on components of biodiversity and how these are measured. OR B Write notes on parasitic relationships and transmission of parasites. 4 4 page 27 MARKS DO NOT WRITE IN THIS MARGIN 12. Congenital lactase deficiency in humans is caused by very low activity of the enzyme lactase, resulting in individuals being unable to digest lactose in milk. This is caused by a number of different mutations in the lactase gene. (a) One of the mutations involved causes a frame-shift mutation in the lactase gene. (i) Name a gene mutation which causes a frame-shift. (ii) Describe the effect of this frame-shift mutation on the lactase gene and on the structure of lactase. Effect on lactase gene Effect on structure of lactase (b) (i) Some mutations occur in the sequences which regulate the transcription of the lactase gene. Suggest why this may lead to more lactase enzyme being produced. (ii) Mutations in this gene are more common in Finland than in other parts of Europe. This is thought to be due to a small number of individuals who settled in Finland many generations ago. State the term which describes this change in gene frequency when a small population of individuals breaks away from a larger population. [Turn over 1 2 1 1 page 28 MARKS DO NOT WRITE IN THIS MARGIN 13. Scientists have used chemicals to produce polyploids to try to improve the characteristics of kiwi fruit plants. The table shows some characteristics of fruit from the original diploid plant (2n) and two polyploids produced from it (4n and 6n). Characteristics of fruit Ploidy Average fruit mass (g) Average fruit length (mm) Average fruit diameter (mm) Vitamin C content (units) 2n 96·5 74·4 51·7 123·0 4n 115·8 76·1 60·0 119·0 6n 145·0 86·8 60·0 124·0 (a) (i) State which characteristic is least affected by polyploidy. (ii) Describe the relationship between ploidy and average fruit diameter. (b) Calculate the percentage increase in average fruit mass when the chromosome number is doubled. Space for calculation % 1 2 1 page 29 MARKS DO NOT WRITE IN THIS MARGIN 13. (continued) (c) Describe the event that has occurred in the cells of the kiwi fruit plant that resulted in polyploidy. (d) Polyploids have whole genome duplications. Explain the importance of this in evolution. [Turn over 1 1 page 30 MARKS DO NOT WRITE IN THIS MARGIN 14. Answer either A or B in the space below and on pages 31 and 32. A Write notes on DNA under the following headings. (i) Organisation of DNA in prokaryotic and eukaryotic cells; (ii) The polymerase chain reaction (PCR). OR B Write notes on RNA under the following headings. (i) Structure and functions of different types of RNA; (ii) RNA splicing. 4 5 6 3 page 31 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWERS page 32 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWERS [END OF QUESTION PAPER] page 33 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Additional graph paper for question 10 (b) (i) Mass of protein in protein standard (kDa) page 34 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK page 35 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK page 36 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

page 02 SECTION 1 — 25 marks Attempt ALL questions 1. Primary cell lines have A a limited number of cell divisions and are sourced from tumours B a limited number of cell divisions and are sourced directly from normal animal tissue C an indefinite number of cell divisions and are sourced from tumours D an indefinite number of cell divisions and are sourced directly from normal animal tissue. 2. The proteome is larger than the number of genes in the genome of an organism because A not all genes are expressed as proteins in a particular cell B post translational modifications generate multiple RNAs from a single gene C alternative splicing generates multiple RNAs from a single gene D each mRNA molecule is translated by many ribosomes. 3. A neuron in a squid, running from the brain to the tentacles, is 0·9 m long. A 25 mm length of neuron is depolarised every 0·001 s. The time taken for an impulse to travel the length of this neuron is A 0·000028 s B 0·000036 s C 0·028 s D 0·036 s

page 03 4. The list shows some events that occur in the cell membrane of a neuron during nerve transmission. 1. Binding of neurotransmitter to neuron 2. Closure of voltage-gated ion channels 3. Opening of ligand-gated ion channels 4. Opening of voltage-gated ion channels Which events contribute to the depolarisation of the resting potential of a neuron? A 1 and 2 only B 2 and 3 only C 1, 2 and 3 only D 1, 3 and 4 only 5. Which row in the table describes the expected effects of inhibition of the Na/KATPase? Intracellular Na ion concentration Intracellular K ion concentration Membrane polarity A increase decrease increase B decrease increase increase C increase decrease decrease D decrease increase decrease 6. A decrease in the activity of glucose symport proteins in the cells lining the small intestine could be caused by an increase in the A glucose concentration inside the small intestine B sodium ion concentration in the cells C ATP concentration in the cells D potassium ion concentration in the cells. [Turn over

page 04 7. The surface area to volume ratio of a cell is an important factor affecting transport into cells. The surface area to volume ratio of an E. coli cell is 4·5. A eukaryotic cell has a surface area of 1809 μm2 and a volume of 7235 μm3. Compared to E. coli, the surface area to volume ratio of the eukaryotic cell is approximately A 1·1 times larger B 1·1 times smaller C 18 times larger D 18 times smaller. 8. Alpha helices in proteins are stabilised by A hydrogen bonds B ionic bonds C disulphide bridges D hydrophobic interactions. 9. Which row in the table describes the effects of an increase in temperature on haemoglobin and oxygen delivery to cells? Affinity of haemoglobin for oxygen Oxygen delivery to tissue A increase increase B increase decrease C decrease increase D decrease decrease

page 05 10. The lipid bilayers of cell membranes contain a number of different phospholipids, which are present in different proportions and distributed unevenly across the extracellular and intracellular faces of the bilayer. Figures 1 and 2 show the proportions and distribution of four phospholipids in the cell membrane of human red blood cells. Figure 1 Proportions of four phospholipids in the cell membrane SP PS PE PC PC – phosphatidyl choline SP – sphingomyelin PS – phosphatidyl serine PE – phosphatidyl ethanolamine Figure 2 Distribution of the phospholipids in the two faces of the membrane 100 50 0 50 100 percentage of total extracellular face intracellular face SP PC PE PS The key can be used to identify each of the phospholipids. 1 more than 25 % of total lipid less than 25 % of total lipid go to 2 go to 3 2 found mainly on the extracellular face found mainly on the intracellular face A B 3 found mainly on the extracellular face found mainly on the intracellular face C D Which letter represents sphingomyelin? [Turn over page 06 Questions 11 and 12 refer to the following information. Short-term variations in the time interval between heart beats is termed heart rate variability (HRV). HRV is one measure that can be used to assess health. The box plots give information from one study about the influence of age and gender on HRV. 25–34 35–44 45–54 55–64 65–74 0 20 40 60 80 age (years) 100 heart rate variability (units) females maximum value minimum value upper quartile lower quartile median males Key 11. From the data shown, which of the statements about HRV is correct? A There is a trend towards lower HRV in females B The greatest error in measurements is for ages 25–34 C The mean HRV for females aged 35–44 is lower than 40 units D Gender-related differences in median values are greater than age-related differences 12. Which row in the table describes features of this study? Design of experiment Type of data collected A simple ranked B simple continuous C multifactorial ranked D multifactorial continuous

page 07 13. There are approximately 40 species of birds of paradise in New Guinea, on islands nearby, and in areas of mainland Australia. They are thought to have evolved from a crow-like common ancestor that lived 20 million years ago. The list describes processes that are likely to have contributed to the evolution of the different species. X The food availability on a certain part of one island favoured the survival of male and female individuals with slender curved bills. Y On one island with abundant food choices, females choose mates whose head feathers have elongated plumes. Z Some males and females of a species of crow-like mainland bird were blown by a freak storm to some of the islands. Which row in the table matches processes of evolution with descriptions from the list? Processes of evolution Genetic drift Selection Natural Sexual A X Z Y B Z X Y C Z Y X D Y X Z 14. Which row in the table indicates factors that can all lead to a high rate of evolution? Factor Selection pressure Generation time Gene transfer A high long asexual reproduction B low short horizontal C low long horizontal D high short sexual reproduction [Turn over

page 08 15. Which of the following adaptations can be explained using the Red Queen hypothesis? 1. Acceleration of cheetahs is related to the speed of the antelope they prey on 2. Migration of insectivorous birds is influenced by the availability of insects 3. Resistance of insect larvae to parasitic wasp infection is a response to these wasps 4. Ornamental head plumes of some male birds are attractive to females A 1 and 2 only B 1 and 3 only C 2 and 3 only D 3 and 4 only 16. A student used a colorimetric assay to measure the activity of an enzyme in bananas. An extract was prepared from one banana and used five times to measure the enzyme activity. Three absorbance readings were taken for each of the five assays. The data collected are shown in the table. Absorbance reading at 540 nm Assay 1 2 3 1 0·55 0·56 0·55 2 0·49 0·48 0·50 3 0·56 0·57 0·57 4 0·62 0·63 0·62 5 0·58 0·58 0·59 The student evaluated the data to be reliable and accurate. The student’s evaluation of the data is A correct because the data are reliable and accurate B incorrect because the data are accurate but not reliable C incorrect because the data are reliable but not necessarily accurate D incorrect because the data are not reliable and not necessarily accurate.

page 09 17. Biological fieldwork can sometimes be more dangerous than laboratory work. Which of the following would not generally be involved in a risk assessment for carrying out fieldwork safely? A Identify hazards B Specify control measures C Assess safety training records of participants D Consider hazard severity and likelihood of occurring 18. Some populations of a species may evolve to become more r-selected or K-selected depending on the nature of the habitat they occupy. Which row in the table shows changes in the factors likely to be associated with a population becoming more K-selected? Environmental stability Number of offspring A increase decrease B increase increase C decrease increase D decrease decrease [Turn over

page 10 19. The population of the snail Vertigo antivertigo was investigated at a small site in Wales. Equal numbers of quadrat samples were taken in three areas with different dominant vegetation and the number of individual snails in each was recorded. The results are shown in the chart. 0 10 20 30 40 50 60 iris and rushes long grass short grass dominant vegetation in area quadrats (%) 1–7 individuals absent Key 40 or more individuals 8–39 individuals The information in the chart indicates that this species of snail prefers A short grass to long grass B irises and rushes to long grass C long grass to irises and rushes D short grass to irises and rushes. 20. Biological fieldwork often requires the estimation of population size for a prey species. One method used is mark and recapture. If the method of marking reduced the camouflage coloration of this species, what effect would this be likely to have on the population estimate obtained? A Recapture numbers would be increased and population size would be overestimated B Recapture numbers would be increased and population size would be underestimated C Recapture numbers would be decreased and population size would be overestimated D Recapture numbers would be decreased and population size would be underestimated

page 11 21. In three-spined stickleback fish, males have a distinctive red underside in the breeding season, which is not present in females. Territorial males were presented with model fish, some of which had their undersides painted red and some of which were left unpainted. Males showed an automatic attack response to only the red-painted models. This attack response behaviour is triggered by A sexual dimorphism B sexual selection C an honest signal D a sign stimulus. 22. The list shows three events in meiosis I. Q Independent assortment R Recombination of alleles of linked genes S Pairing of homologous chromosomes The order in which these events occur is A S – R – Q B S – Q – R C Q – R – S D Q – S – R. [Turn over

page 12 23. The graph shows species richness of mollusc populations in areas of fenland. The number of species of mollusc was recorded in several areas as well as the calcium concentration in the fenland water. At each site the concentration of iron was also measured. 0 10 20 30 40 0 100 200 calcium (mg/l) low iron concentration Key high iron concentration number of species 300 400 Which of the following generalisations can be drawn from the graph? A An increase in calcium from 100 to 400 mg/l increases the species richness B High iron concentration leads to the highest species richness C When both calcium and iron concentrations are high the species richness is highest D An increase in calcium up to 150 mg/l increases species richness 24. New patterns of resistance in Plasmodium have increased the challenge experienced in the treatment of malaria. Which of the following strategies is least likely to reduce the challenges in the treatment and control of malaria? A Improved sanitation B Development of new culture methods for Plasmodium C Coordinated vector control D Building of new low-density housing in malarial areas

page 13 25. Which diagram shows the sequence of events in the scientific cycle? form a new hypothesis form a hypothesis form a hypothesis D C B A form a new hypothesis form a hypothesis gather and analyse data gather and analyse data design experiment form a new hypothesis draw conclusions gather and analyse data form a hypothesis design experiment design experiment gather and analyse data draw conclusions draw conclusions draw conclusions design experiment form a new hypothesis [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET.] page 14 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 15 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 16 [BLANK PAGE] DO NOT WRITE ON THIS PAGE AH FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2019 Total marks — 90 SECTION 1 —25 marks Attempt ALL questions. Instructions for the completion of Section 1 are given on page 02. SECTION 2 — 65 marks Attempt ALL questions. A supplementary sheet for question 1 is enclosed inside the front cover of this question paper. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. Score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X707/77/01 TUESDAY, 30 APRIL 1:00 PM – 3:30 PM A/SA Biology Section 1 — Answer grid and Section 2 page 02 SECTION 1 — 25 marks The questions for Section 1 are contained in the question paper X707/77/02. Read these and record your answers on the answer grid on page 03 opposite. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D page 03 SECTION 1 — Answer grid [Turn over A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 05 [Turn over for next question DO NOT WRITE ON THIS PAGE page 06 MARKS DO NOT WRITE IN THIS MARGIN SECTION 2 — 65 marks Attempt ALL questions Question 10 contains a choice 1. Read through the supplementary sheet for question 1 before attempting this question. (a) Refer to Figure 2. Within a species there is usually a positive correlation between size and age, where larger organisms are older. Use Figure 2 to explain if the data support this for veined squid. (b) The researchers used evidence published by other scientists to support their work on ageing squid. Give the term used to describe published summaries of current knowledge and recent findings in a particular field. (c) (i) Explain why many marine organisms use external fertilisation. (ii) Give one cost of this type of fertilisation. 1 1 1 1 page 07 MARKS DO NOT WRITE IN THIS MARGIN 1. (continued) (d) Refer to Figure 3. (i) Explain how Figure 3a suggests that larger males may have a reproductive advantage. (ii) Does the data in Figure 3b also support this conclusion? Justify your answer. (e) Refer to Figure 4. (i) Identify the month that would have the highest breeding activity. (ii) Squid eggs can take 30 days to hatch. Explain how the data for April support this finding. (iii) The squid take over a month to mature from stage 2 to stage 3. Suggest a reason for the unexpected decrease in the proportions of squid at stage 2 caught in May. [Turn over 1 1 1 1 1 page 08 MARKS DO NOT WRITE IN THIS MARGIN 2. Fasciolosis is a disease of cattle caused by the flatworm Fasciola hepatica (F. hepatica). F. hepatica antigens are found both in the blood and the milk of infected cattle and their presence is the basis of an enzyme-linked immunoassay used to identify infected animals. A positive assay is described in the figure. monoclonal antibodies specific to F. hepatica are bound to the assay plate an infected milk sample is added to the plate the plate is washed with a buffer a second monoclonal antibody, specific to F. hepatica and linked to an enzyme, is added the plate is washed with a buffer the colourless enzyme substrate is added and is converted to a coloured product 1. 2. 3. 4. 5. 6. (a) (i) The antibody added at step four is a monoclonal antibody. State the meaning of the term monoclonal. (ii) Monoclonal antibodies can be produced in a laboratory using hybridomas that are prepared by fusing together B lymphocytes and myeloma cells. Name the chemical used to fuse these two cell types together. 1 1 page 09 MARKS DO NOT WRITE IN THIS MARGIN 2. (continued) (b) If the procedure was not carried out correctly a positive result could occur in the absence of F. hepatica antigens; a false positive result. Suggest a possible cause for this false positive result. (c) A pH buffer was used in all reagents and wash solutions. Explain why it is important to control pH in immunoassays. (d) Infection with F. hepatica in cattle results in weight loss and a reduction in milk yield. Suggest a reason for the reduced milk yield in infected cattle. [Turn over 1 2 1 page 10 MARKS DO NOT WRITE IN THIS MARGIN 3. Photoreceptor system proteins are found across the three domains. (a) (i) Name the light sensitive molecule in animals that combines with the protein opsin to form photoreceptors of the eye. (ii) Rod cells contain rhodopsin. Explain why these cells can function in low light intensity. (iii) Describe the role of bacteriorhodopsin in archaea. Figure 1 shows the absorption spectra of rhodopsin in a variety of organisms. Figure 1 rhodopsin absorbance 300 400 wavelength (nm) octopus crab owlfly Key 500 600 (b) Owlflies have rhodopsin that is sensitive to light of 345 nm (UV). Suggest how the photoreceptor proteins of owlflies differ from those of crabs. 1 1 2 1 page 11 MARKS DO NOT WRITE IN THIS MARGIN 3. (continued) Figure 2 shows the depth of penetration in water of the wavelengths in daylight. Figure 2 300 400 350 450 550 500 600 700 650 300 200 100 0 wavelength (nm) depth (m) white light (all colours) (c) Crab species tend to live in shallower coastal waters, whereas octopus species can live in open seas. Explain how the data from Figures 1 and 2 support this statement. [Turn over 2 page 12 MARKS DO NOT WRITE IN THIS MARGIN 4. The CFTR gene encodes a large transmembrane protein (CFTR) with a symmetrical structure. This is composed of two transmembrane regions and two ATP-binding regions. (a) Name the class of R-groups that predominate in the transmembrane regions of proteins. (b) The protein  CFTR is involved in the regulation of water content of extracellular mucus in the lungs and digestive system. The figure represents the action of CFTR. It regulates the passage of chloride ions (Cl−) across membranes of epithelial (lining) cells. In order for this ion channel to open, the protein must bind two ATP molecules, as well as a phosphate group. The increased concentration of Cl− outside the cell draws water out of the epithelial cells into the mucus, maintaining its fluidity. P Cl− extracellular plasma membrane CFTR intracellular ATP ATP phosphorylation ATP binding both required for channel opening } (i) What name is given to a molecule that binds to a protein? (ii) Suggest how the binding of ATP results in the opening of the chloride ion channel. 1 1 1 page 13 MARKS DO NOT WRITE IN THIS MARGIN 4. (continued) (c) Cholera is a disease caused by the bacterium Vibrio cholerae. It causes severe watery diarrhoea, which can lead to dehydration and even death. The bacterium produces cholera toxin which interferes with the control of the CFTR protein channel by constantly activating a kinase enzyme. (i) Describe the reaction catalysed by a kinase enzyme. (ii) Explain how the production of cholera toxin by Vibrio cholerae can lead to more water being drawn out of the epithelial cells. [Turn over 1 2 page 14 MARKS DO NOT WRITE IN THIS MARGIN 5. It is thought that the level of fruit and vegetable consumption could influence the risk of some diseases in humans. One large study has examined the relationship between the level of fruit and vegetable consumption and the risk of mortality. This was a meta-analysis, which is a type of observational study that combines and summarises data from several previous similar studies. (a) Meta-analyses provide good evidence and are widely used in epidemiological studies. State what is meant by the term ‘epidemiology’. A search of previously published research papers was used to identify studies for inclusion in the meta-analysis. However, not all possible studies were included. In most of the studies used, fruit and vegetable consumption was assessed using questionnaires. (b) Suggest why the use of questionnaires to collect information about fruit and vegetable consumption may have led to some errors in the data. (c) One possible criticism of meta-analyses is that the people carrying them out select the previous studies to include, which could result in selection bias. (i) Suggest one reason why an individual study might be considered invalid and therefore excluded from the meta-analysis. (ii) State the effect that selection bias would have on the sample obtained. 1 1 1 1 page 15 MARKS DO NOT WRITE IN THIS MARGIN 5. (continued) The results from this meta-analysis are shown in the graph. The risk of mortality was expressed as a hazard ratio, with a hazard ratio of 1·0 indicating no effect on the risk of mortality. 0 2 4 6 8 10 0∙6 0∙7 0∙8 0∙9 hazard ratio of all causes of mortality fruit and vegetable consumption (servings/day) 95 % confidence intervals 1∙0 (d) The graph shows the 95 % confidence intervals for the data. The confidence interval for six servings per day is wider than that for two servings. What does this indicate about these two sets of data? (e) This study is consistent with the hypothesis that higher fruit and vegetable consumption is associated with a reduced risk of mortality over time. Explain why this observational study can only suggest a possible link between fruit and vegetable consumption and mortality risk over time. 1 1 [Turn over page 16 MARKS DO NOT WRITE IN THIS MARGIN 6. Behaviour is an animal’s response to internal and external stimuli. Discuss how animal behaviour can be measured and recorded. 4 page 17 [Turn over for next question DO NOT WRITE ON THIS PAGE page 18 MARKS DO NOT WRITE IN THIS MARGIN 7. The figure shows the life cycle of the parasitic hairworm, Paragordius varius, found in lakes throughout North and South America. flying insect eg mosquito cricket (definitive host) adult hairworms in cricket cricket enters water aquatic insect larva develops into adult aquatic insect larva (intermediate host) hairworm larva hairworm eggs adult hairworms in water terrestrial environment aquatic environment (a) Hairworm eggs develop into larvae that hatch and move slowly through water at the bottom of the lake, where they may be ingested by aquatic insect larvae. The insects act as intermediate hosts, transporting the parasites into the terrestrial environment when the aquatic insect larvae become flying adults. If the flying insects are eaten by crickets, the parasite larvae penetrate the gut of the cricket to enter the body cavity. (i) Explain why the hairworm and the cricket can be described as symbionts. (ii) Explain why the cricket, and not the aquatic insect larva, is the definitive host. 1 1 page 19 MARKS DO NOT WRITE IN THIS MARGIN 7. (continued) (b) In the presence of water, the hairworm exits the cricket’s body and commences the next stage of its life-cycle. Researchers have shown that the hairworm adults produce ‘mind-controlling’ chemicals, which cause their normally terrestrial cricket hosts to jump into bodies of water. (i) Suggest how this modification of host behaviour benefits the parasite. (ii) Give another example of a host behaviour that can be altered as part of the extended phenotype of a parasite. (c) Adult hairworms are free-living in aquatic environments where they mate. Males die almost immediately after mating so only mate once. Females die shortly after laying eggs fertilised by a single male. State the term used to describe this type of reproductive strategy. (d) Hairworms are closely related to members of the phylum Nematoda. Name the phylum to which both their hosts belong. [Turn over 1 1 1 1 page 20 MARKS DO NOT WRITE IN THIS MARGIN 8. The 1918 ‘Spanish flu’ pandemic was caused by the influenza virus and is thought to have killed approximately 50 million people. (a) (i) Humans have many defences that prevent viruses from entering the body including physical barriers and chemical secretions. Name one of the non-specific immune responses that may result if the virus is able to bypass these defences. (ii) Annual flu vaccinations are designed to give immunity to some strains of influenza for vulnerable individuals. The vaccines typically contain inactivated virus particles that have been purified to leave only two of the proteins that are found on the surface of the virus coat. Suggest why a new flu vaccine needs to be developed annually. 1 2 page 21 MARKS DO NOT WRITE IN THIS MARGIN 8. (continued) (b) Scientists have predicted that another flu pandemic is highly likely in the future. They are particularly concerned that this might arise from the H5N1 strain of influenza virus, which is common in wild bird species, and is able to pass from birds to humans. (i) What term is used to describe the level of harm caused to the host species by a virus? (ii) The case fatality rate (CFR) is the proportion of infected individuals who die as a result of a disease. Spanish flu had an estimated CFR of 2·5 % but the H5N1 flu strain has a CFR of 55 %. If Spanish flu had been caused by the H5N1 strain, and it had infected a similar number of people, what would have been the resulting number of deaths? Space for calculation [Turn over 1 1 page 22 MARKS DO NOT WRITE IN THIS MARGIN 9. The sandperch, Parapercis cylindrica, is a fish that lives on the sea floor. Each female defends a territory in which it feeds and reproduces. Dependent on the population density of the species, a single male may defend a group of up to 10 neighbouring females. (a) All sandperch begin life as hermaphrodites and mature into females. What is meant by the term hermaphrodite? (b) There is a strict size-based dominance hierarchy in the social groups; the mean size of males is greater than that of females. State the term used to describe this difference in mean size. (c) If the male dies, the largest female will undergo a period of growth and will then change sex. State one other cause of sex change in organisms. (d) A study was conducted to measure the growth of female fish undergoing sex-change. Social groups were created in a laboratory by placing groups of similar sized fish in tanks with identical environmental conditions. Two treatments were set up, each with 10 groups. • Treatment 1 with 2 females and 1 male in each group • Treatment 2 with 4 females and 1 male in each group On day 5, the male was removed from half of each of the treatment groups to create experimental groups. On day 35 the sex and growth of each fish was determined. The structure of the groups is shown in the table. Treatment 1 (10 groups) Treatment 2 (10 groups) Control (5 groups) Experimental (5 groups) Control (5 groups) Experimental (5 groups) Group structure Day 1 2 females 1 male 2 females 1 male 4 females 1 male 4 females 1 male Day 5 2 females 1 male 2 females 4 females 1 male 4 females Day 35 2 females 1 male 1 female 1 male 4 females 1 male 3 females 1 male 1 1 1 page 23 MARKS DO NOT WRITE IN THIS MARGIN [Turn over for next question 9. (d) (continued) At day 35 the growth of fish in each experimental group was compared to its control group. This was used to calculate a standardised value that took into account growth in the control group. The figure shows the mean standardised growth values for the experimental groups in treatments 1 and 2. Error bars represent standard error of the mean. 0 0∙005 0∙010 0∙015 0∙020 0∙025 0∙030 treatment 1 treatment 2 Key sex-changing females (now males) non-sex-changing females mean standardised growth value (i) Suggest why it was necessary to have separate control groups for treatments 1 and 2. (ii) Give one conclusion that could be drawn about the effect of treatment 2 on the growth of sex-changing females compared to treatment 1. (iii) Suggest an advantage of greater size in sex-changed females. 1 1 1 page 24 MARKS DO NOT WRITE IN THIS MARGIN 10. Answer either A or B in the space below and on page 25. A Discuss signalling between cells under the following headings. (i) Hydrophilic signalling molecules and signal transduction (ii) Insulin signalling and diabetes OR B Discuss control of the cell cycle under the following headings. (i) Phases of the cell cycle and the importance of cell cycle checkpoints (ii) The role of cyclins and cyclin-dependent kinases 6 4 6 4 page 25 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWER FOR QUESTION 10 [END OF QUESTION PAPER] page 26 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK page 27 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK page 28 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Acknowledgement of copyright Question 8 Everett Historical/shutterstock.com © National Qualications 2019 AH Supplementary sheet for question 1 X707/77/11 Biology Supplementary sheet TUESDAY, 30 APRIL 1:00 PM – 3:30 PM A/SA

page 02 1. The veined squid, Loligo forbesii, is a mollusc species found off the west coast of Scotland. This species undergoes a daily vertical migration in the water, coming up to the surface to feed at night, and diving to deeper depths during the day. Figure 1 shows the anatomical structure of a squid. Figure 1 tail fins eyes mantle head arms and tentacles The relationship between size and age of squid was investigated. The size was determined by measuring the length of the mantle. As the squid migrate up and down in the water, the changes in pressure affect the formation of crystals in the squid balance organs, known as statoliths. These crystals show up as rings that can be counted to show the age of the squid in days. Researchers collected data from squid caught off the west coast of Scotland over a number of weeks. The researchers measured mantle length and counted statolith growth rings for 29 squid. Figure 2 shows age plotted against mantle length. Figure 2 100 120 140 160 180 200 220 240 260 280 300 100 150 200 250 300 mantle length (mm) age (days) 350 400 450

page 03 1. (continued) Many marine organisms use external fertilisation. This process is slightly unusual in squid as male squid produce capsules of male gametes called spermatophores, which are transferred to the females during mating. One individual female squid can lay thousands of eggs. The researchers analysed squid caught by commercial fishing boats. Figures 3a and 3b show the relationship between mantle length of male squid and the length and number of spermatophores produced in mature male squid. Figure 3 0 10 20 30 mean spermatophore length (mm) Figure 3a 0 400 800 1200 number of spermatophores Figure 3b 100 200 300 400 mantle length (mm) mantle length (mm) 500 600 100 200 300 400 500 600 The sexual maturity of individual squid was estimated using a scale of 1–3, where stage 1 was the most immature, and stage 3 the most mature. Squid in stage 3 reproduce actively. Figure 4 shows the proportion of squid at each maturity stage, caught over the course of a year by commercial fishing boats. Figure 4 0 0·1 0·2 0·3 0·4 0·5 0·6 0·7 0·8 0·9 1·0 September October November December January February March April May June July August proportion of squid at each maturity stage stage 3 stage 2 stage 1 month squid maturity stage Key [END OF SUPPLEMENTARY SHEET] page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

page 02 Total marks — 25 Attempt ALL questions 1. Embryonic stem cells can be used therapeutically to differentiate and replace any damaged or diseased tissue. This is because they are A multipotent and specific genes are switched on B pluripotent and all genes are switched on C multipotent and specific genes can be switched on D pluripotent and all genes can be switched on. 2. The table shows the anticodon sequence on tRNA molecules and the specific amino acid each carries. Anticodon sequence Amino acid CCA proline AAG lysine UAU tyrosine AUU isoleucine Part of the amino acid sequence of a protein is shown in the diagram. isoleucine tyrosine lysine lysine proline Triplets of DNA nucleotides that code for the sequence of amino acids in this protein are A CCA UUC UUC AUA UAA B GGT TTC TTC ATA TAA C CCA AAG AAG TAT ATT D CCA AAG AAG UAU AUU.

page 03 3. The graph shows how temperature changes during repeated cycles of a polymerase chain reaction (PCR). 10 9 8 7 6 5 temperature (°C) time (minutes) 4 3 2 1 0 100 90 80 70 60 50 If there were 500 molecules of DNA at the start, predict how many copies there will be after 20 minutes. A 16 000 B 8000 C 2500 D 2000 4. DNA from a mother, child and four men (A, B, C and D) in a paternity suit was analysed. The DNA was amplified using PCR and separated by gel electrophoresis. From the results shown in the diagram, identify the likely father of the child. D C B A Mother Child [Turn over

page 04 5. The table shows three examples of gene transfer. Plasmid transfer by bacteria Sexual reproduction by mice Asexual reproduction by bacteria 1 3 2 Which of these examples illustrate horizontal gene transfer? A 2 only B 3 only C 1 and 2 only D 1 and 3 only 6. Sickle cell anaemia is a human condition that affects haemoglobin, reducing the blood’s ability to carry oxygen. It is caused by a mutation that changes one adenine base to thymine. This results in one amino acid in haemoglobin being changed. The mutation that causes sickle cell anaemia is A deletion B insertion C substitution D translocation.

page 05 7. Subspecies of the wren (Troglodytes troglodytes) have evolved in different island areas in Scotland. The graphs show averages of body length, wing length and body mass for wrens from the islands of St Kilda, Shetland and Fair Isle. 60 50 40 30 20 10 0 12 10 8 6 4 2 0 12 10 8 6 4 0 2 average body length (cm) average body mass (g) average wing length (mm) sub-species sub-species sub-species Fair Isle Fair Isle Fair Isle Shetland Shetland Shetland St Kilda St Kilda St Kilda Which of the following conclusions can be drawn from this data? A Shetland and Fair Isle wrens have a greater average wing length and body mass than St Kilda wrens B St Kilda and Fair Isle wrens have a greater average body length and wing length than Shetland wrens C St Kilda wrens are smaller in each characteristic than the other two subspecies D Fair Isle wrens are larger in each characteristic than the other two subspecies [Turn over

page 06 8. Goldenrod gall flies (Eurosta solidaginis) lay eggs on the stems of several plant species. The newly hatched larvae then burrow into the plant, causing the growth of a mass of plant tissue (gall) around them. The larvae live and feed inside the gall. The size of gall is affected by two different selection pressures. Larvae in smaller galls are more likely to be predated by wasps, while larvae in larger galls are more likely to be predated by birds. Which diagram represents the type of selection affecting gall size? B gall size gall size gall size gall size D C A population after selection number of galls number of galls number of galls number of galls original population Key

page 07 9. The diagram shows the divergence of lineages in the evolution of some primates. 0 10 20 30 40 50 60 70 80 galago lemur marmoset mouse lemur titi monkey squirrel monkey Primates vervet monkey macaque baboon orangutan gorilla chimpanzee human time (millions of years before present) Which row in the table identifies the time that the last common ancestor of vervet monkeys and humans existed, and the number of other species that shared this common ancestor? Time (millions of years before present) Number of other species that shared this common ancestor A 30 5 B 30 11 C 40 8 D 75 11 [Turn over

page 08 10. The diagram represents the conversion of starch into maltose. maltose starch Which row in the table identifies the type of reaction shown in the diagram and whether it requires or releases energy? Type of reaction Energy required or released A catabolic required B catabolic released C anabolic required D anabolic released

page 09 11. The effect of temperature on the metabolic rate of a snail was investigated using the respirometer shown. glass tube solution to absorb carbon dioxide scale (cm3) 0∙4 0∙0 0∙2 tap rubber stopper coloured liquid wire mesh platform snail The experiment was carried out at 10 °C, 15 °C, 20 °C and 25 °C. At each temperature, the tap was left open for 15 minutes then closed and readings were taken from the scale every 2 minutes. Identify how the dependent variable was measured. A Temperature change B Heat production C Oxygen consumption D Carbon dioxide production 12. Which row in the table identifies the temperature monitoring centre, and the location of an effector in thermoregulation in mammals? Temperature monitoring centre Location of effector A hypothalamus nerves B skin hypothalamus C hypothalamus skin D nerves hypothalamus [Turn over

page 10 13. The Gila monster (Heloderma suspectum) is a species of lizard that lives in North America. Its internal temperature is dependent on the environmental temperature. Which row in the table identifies the metabolic cost to and the range of ecological niches of the Gila monster? Metabolic cost Range of ecological niches A low narrow B high narrow C low wide D high wide 14. Which statement describes an event that occurs during the lag phase of microbial growth? A Secondary metabolites are produced B Certain enzymes are induced C Most rapid growth occurs due to plentiful nutrients D The culture medium becomes depleted of nutrients 15. Using recombinant DNA technology, a culture of the bacteria species Micrococcus luteus was genetically modified with a plasmid containing a gene for a human protein. The protein was synthesised by the genetically modified bacteria, but it failed to fold correctly. Which of the following changes to this procedure may lead to a correctly folded protein being produced? A Use a different species of bacteria B Use yeast cells rather than bacteria C Insert an artificial chromosome instead of a plasmid D Insert a regulatory sequence into the plasmid

page 11 16. Bacterial cells were exposed to disinfectant for 14 minutes. Every 2 minutes a sample was taken and the number of viable cells counted. The results are shown in the graph. 0 2 4 6 8 10 12 14 time (minutes) number of viable cells 1 000 000 100 000 10 000 1000 100 10 1 Calculate the percentage decrease in viable cells after being exposed to disinfectant for 6 minutes. A 99·60 B 99·87 C 996 000·00 D 998 700·00 [Turn over

page 12 17. A field trial was carried out to investigate the effect of applying different masses of phosphate fertiliser on grain yield of a wheat cultivar. The wheat was grown in three fields. A mass of 50, 100 or 150 kg of phosphate fertiliser per hectare was applied. How could the design of the field trial be improved to minimise the effects of bias? A Repeat the trial using a wider range of masses of phosphate fertiliser B Increase the number of replicates at each phosphate fertiliser application C Carry out another field trial exactly the same using a different wheat cultivar D Divide each field and randomly apply different masses of phosphate fertiliser 18. Fungi were cultured in a fermenter to produce an enzyme that was released into the culture medium. The graph shows the effect of oxygen concentration on the dry mass of the fungus and the concentration of enzyme in the medium after 48 hours growth. 4 3 2 1 Key fungus enzyme concentration of enzyme (units per litre) dry mass of fungus (units per litre) oxygen concentration (units) 0 30 20 10 0 60 40 20 0 The concentration of enzyme when the dry mass of fungus was 36 units per litre was A 18 units per litre B 28 units per litre C 48 units per litre D 56 units per litre.

page 13 19. Breeding programmes can be used to increase starch levels in barley plants. This involves breeding closely related plants with high starch levels for several generations. This is carried out in order to A increase the frequency of homozygous deleterious alleles B reduce the frequency of homozygous deleterious alleles C increase the frequency of heterozygotes D reduce the frequency of heterozygotes. 20. The bar chart shows the percentage decrease in yield of four crops as a result of the effects of weeds, fungi and insect pests. Key weeds fungi insect pests decrease in yield (%) crop D C B A 40 35 30 25 20 15 10 5 0 Predict which crop will show the greatest increase in yield if herbicides and insecticides are applied. [Turn over

page 14 21. The diagram shows the breeding of two parental pig breeds to produce the Modern Duroc pig. Modern Duroc Brown Duroc × Jersey Red The Modern Duroc pig has increased quality and quantity of meat compared to both parental breeds. The development of the Modern Duroc pig breed is an example of A inbreeding B cross breeding C natural selection D genetic modification. 22. A study into animal welfare in a zoo was carried out by observing the behaviour of some of the animals in their enclosures. The behaviours observed are shown in the table. Animal Behaviour Lion Continuously pacing back and forward Sun bear Feeding young Spider monkey Subordinate monkey grooming the dominant male White rhino Repeatedly chewing bars of enclosure Spotted hyena Sleeping in the sun Which two animals show signs of poor welfare? A Spotted hyena and lion B White rhino and spider monkey C Lion and white rhino D Sun bear and spider monkey

page 15 23. Which of the following is not a behaviour of a worker honeybee? A Producing eggs B Defending the hive C Gathering pollen D Feeding offspring 24. The statements describe examples of behaviour used by animals as protection from predators. 1. A hedgehog rolls into a ball so that its spines protect it from a fox. 2. Mackerel swim in a large shoal making it difficult for a predator to single one out. 3. An individual meerkat watches for predators while the other meerkats feed. Which are examples of social defence? A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 25. The list describes some aspects of biodiversity. 1. The number and frequency of different alleles in a species. 2. The proportion of each species in an ecosystem. 3. The number of different species in an ecosystem. Which are components of species diversity? A 1 only B 3 only C 1 and 2 only D 2 and 3 only [END OF QUESTION PAPER] page 16 SPACE FOR ROUGH WORK page 17 SPACE FOR ROUGH WORK page 18 [BLANK PAGE] DO NOT WRITE ON THIS PAGE page 19 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

page 20 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Acknowledgement of copyright Question 5 Natsmith1/shutterstock.com H FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2019 Instructions for the completion of Paper 1 are given on page 02. Record your answers on the answer grid on page 03. You may use a calculator. Use blue or black ink. Before leaving the examination room you must give your answer booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X807/76/02 TUESDAY, 30 APRIL 09:00 AM – 09:40 AM A/SA Biology Paper 1 — Multiple choice Answer booklet page 02 Paper 1 — 25 marks The questions for Paper 1 are contained in the question paper X807/76/12. Read these and record your answers on the answer grid on page 03. Use blue or black ink. Do NOT use gel pens or pencil. 1. The answer to each question is either A, B, C or D. Decide what your answer is, then fill in the appropriate bubble (see sample question below). 2. There is only one correct answer to each question. 3. Any rough working should be done on the additional space for answers and rough work at the end of this booklet. Sample question The thigh bone is called the A humerus B femur C tibia D fibula. The correct answer is B — femur. The answer B bubble has been clearly filled in (see below). A B C D Changing an answer If you decide to change your answer, cancel your first answer by putting a cross through it (see below) and fill in the answer you want. The answer below has been changed to D. A B C D If you then decide to change back to an answer you have already scored out, put a tick (3) to the right of the answer you want, as shown below: A B C D or A B C D page 03 Paper 1 — Answer grid A B C D 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Biology page 04 [BLANK PAGE] DO NOT WRITE ON THIS PAGE H FOR OFFICIAL USE Fill in these boxes and read what is printed below. Number of seat Town © Mark Full name of centre Forename(s) Surname Scottish candidate number Date of birth Year Day Month National Qualications 2019 Total marks — 95 Attempt ALL questions. You may use a calculator. Questions 3 and 16 contain a choice. Write your answers clearly in the spaces provided in this booklet. Additional space for answers and rough work is provided at the end of this booklet. If you use this space you must clearly identify the question number you are attempting. Any rough work must be written in this booklet. Score through your rough work when you have written your final copy. Use blue or black ink. Before leaving the examination room you must give this booklet to the Invigilator; if you do not, you may lose all the marks for this paper. X807/76/01 TUESDAY, 30 APRIL 10:10 AM – 12:30 PM B/SA Biology Paper 2 page 02 MARKS DO NOT WRITE IN THIS MARGIN Total marks — 95 Attempt ALL questions Questions 3 and 16 contain a choice. 1. The diagram shows the replication of DNA in a chromosome from a eukaryotic cell. The arrows show the directions of replication. 3′ 5′ primer protein X (a) Name protein X. (b) Explain why primers are necessary for DNA replication. (c) (i) Explain why only the leading strand can be replicated continuously. (ii) Name the enzyme that joins fragments together in the lagging strand. (d) Describe how DNA is organised in prokaryotes. 1 1 1 1 2 page 03 MARKS DO NOT WRITE IN THIS MARGIN 2. Three different mutated bacteria, X, Y and Z were studied. Each had a mutation in a different region of its DNA that is transcribed to rRNA. Protein synthesis was measured in cultures of each mutated bacteria and in a culture of unmutated bacteria. The results are shown in the table. Bacterial culture Protein synthesis (%) Unmutated 100 X 9 Y 15 Z 90 (a) Calculate the simplest whole number ratio of percentage protein synthesis in cultures X, Y and Z. Space for calculation : : (b) Name the other component of ribosomes, apart from rRNA. (c) Using your knowledge of biology, suggest how a change in the sequence of bases in DNA transcribed to rRNA in the mutated cultures resulted in a decrease in protein synthesis. (d) Describe one structural difference between DNA and rRNA. 1 1 2 1 [Turn over page 04 MARKS DO NOT WRITE IN THIS MARGIN 3. Attempt either A or B. Write your answer in the space below. A Write an account of ATP synthesis in the electron transport chain during cellular respiration. OR B Write an account of competitive and feedback inhibition of enzymes. You may use labelled diagrams where appropriate. 4 4 page 05 MARKS DO NOT WRITE IN THIS MARGIN 4. Myoglobin and haemoglobin are oxygen-binding proteins in mammals. The myoglobin molecule is a single globin polypeptide. The haemoglobin molecule is composed of both alpha globin and beta globin polypeptides. (a) Myoglobin consists of a chain of 154 amino acids folded in a specific three-dimensional shape. (i) Name a bond that holds the polypeptide chain in this shape. (ii) The gene for myoglobin contains 1154 bases while its mature mRNA transcript contains 462 bases. Give a reason for this difference. (b) The genes for myoglobin and haemoglobin have evolved from a single ancestral globin gene as shown in the diagram. ancestral globin gene beta globin gene myoglobin alpha globin gene myoglobin gene haemoglobin beta globin alpha globin In the genome of mammals there are multiple copies of alpha and beta globin genes. (i) The multiple copies of alpha and beta globin genes are the result of duplication mutations. Describe how a duplication mutation occurs. (ii) Explain the importance of duplication mutations in evolution. 1 1 1 2 [Turn over page 06 MARKS DO NOT WRITE IN THIS MARGIN 5. The Galapagos are a group of islands 600 to 800 miles off the coast of South America. Less than three million years ago some finches of a single ancestral species reached these islands from South America and bred successfully. The number of islands has increased over time, further isolating groups of finches. (a) Use this information to name the type of isolation barrier involved in the speciation of these finches and explain its role. Name Explanation (b) The table shows the estimated number of islands and finch species in the Galapagos over the last 2·8 million years. Time (millions of years ago) Estimated number of islands in the group Estimated number of finch species present 2·8 4 0 1·0 6 5 0·5 18 9 0·0 18 14 (i) Calculate the average increase in the number of finch species per million years over this entire period. Space for calculation species per million years 2 1 page 07 MARKS DO NOT WRITE IN THIS MARGIN 5. (b) (continued) (ii) Suggest what could have caused the change in the number of finch species over the last 0·5 million years. (c) A molecular clock was used to estimate the dates of the divergence of each species. Give one example of the type of data required to generate a molecular clock. [Turn over 1 1 page 08 MARKS DO NOT WRITE IN THIS MARGIN 6. Cystic fibrosis in humans is caused by mutations in a gene that reduces chloride transport across the cell membrane. An investigation was carried out to determine the effectiveness of two drugs, P and Q, on improving chloride transport in individuals with cystic fibrosis. Individuals with different mutations, A, B, and both A and B, were treated as follows. • No drugs • Drug P alone • Drug Q alone • Drug P and drug Q combined Chloride transport across cell membranes was measured and compared to the chloride transport in a control group with no mutations in the gene. The results are shown in the graph. Treatment drug P no drugs drug Q average chloride transport (% of control) drugs P + Q mutation B A 60 50 40 30 20 10 0 both A + B (a) State the purpose of including a control group in this investigation. 1 page 09 MARKS DO NOT WRITE IN THIS MARGIN 6. (continued) (b) Identify the treatment and mutation for which the drug(s) had no effect. Treatment Mutation (c) State how the graph shows that each treatment was carried out on more than one individual with each mutation. (d) State the term used to describe the selection of drugs to treat individuals based on their genomic sequence. [Turn over 2 1 1 page 10 MARKS DO NOT WRITE IN THIS MARGIN 7. The genome of all organisms contains both protein coding genes and non-coding DNA. The size of the genome varies between different species. The table shows the size of the genome and the number of protein coding genes in several different organisms. Organism Size of genome (base pairs) Number of protein coding genes Eukaryotes yeast 1·2 × 107 6600 fruit fly 1·4 × 108 14 000 human 3·2 × 109 21 000 Prokaryotes V. cholera 4·0 × 106 3900 E. coli 4·6 × 106 4200 (a) (i) Using information from the table, compare the size of genomes of eukaryotes and prokaryotes. (ii) Calculate how many times greater the human genome is compared to the V. cholera genome. Space for calculation times greater (iii) The prokaryote M. tuberculosis has a genome size of 4·4 × 106 base pairs. Predict the number of protein coding genes in M. tuberculosis. 1 1 1 page 11 MARKS DO NOT WRITE IN THIS MARGIN 7. (continued) (b) Protein coding genes are used to produce mRNA, which is translated into protein. The bar graph shows the percentage of the genome that codes for protein in four of the organisms shown in the table. V. cholera E. coli yeast fruit fly organism 100 90 80 70 60 50 40 30 20 10 0 percentage of genome coding for protein Use the information in the table and the graph to calculate the size of the genome coding for protein in a fruit fly. Space for calculation base pairs (c) Give one role of the non-coding DNA in the genome. (d) In eukaryotes, alternative RNA splicing occurs. Explain how this results in different proteins being expressed from a single gene. 1 1 1 [Turn over page 12 MARKS DO NOT WRITE IN THIS MARGIN 8. The bacterium E.coli was cultured in a growth medium containing 0·6 mM glucose and 0·6 mM lactose for 180 minutes. Glucose and lactose concentrations were measured every 20 minutes. The results are shown in the table. Time (minutes) Glucose concentration (mM) Lactose concentration (mM) 0 0·60 0·60 20 0·50 0·60 40 0·38 0·60 60 0·09 0·60 80 0·00 0·55 100 0·00 0·44 120 0·00 0·32 140 0·00 0·15 160 0·00 0·04 180 0·00 0·00 (a) Using values from the table, describe changes in the concentration of lactose over the 180 minutes of the experiment. 2 page 13 MARKS DO NOT WRITE IN THIS MARGIN 8. (continued) (b) E.coli breaks down lactose using the enzyme beta-galactosidase as shown. beta-galactosidase lactose glucose + galactose Beta-galactosidase is produced by E.coli only when lactose is present and glucose is absent. (i) Using information from the table, identify the time when E.coli started producing beta-galactosidase. minutes (ii) Suggest a benefit to E.coli of producing beta-galactosidase only when lactose is present. (c) In terms of activation energy, state how enzymes increase the rates of reactions in living cells. [Turn over 1 1 1 page 14 MARKS DO NOT WRITE IN THIS MARGIN 9. Respiration is a cellular process that produces ATP. The first stage in this process is glycolysis. Some steps of glycolysis are shown in the diagram. Pyruvate NAD ATP NADH ADP ADP Step 2 ATP Intermediates Glucose Step 1 (a) State the exact location of glycolysis. (b) (i) Describe the role of ATP in Step 1. (ii) Explain how glycolysis results in a net gain of ATP. (iii) Describe the role of dehydrogenase enzymes in the conversion of intermediates to pyruvate. 1 1 1 2 page 15 MARKS DO NOT WRITE IN THIS MARGIN 9. (continued) (c) When fermentation occurs in animal cells, pyruvate is converted to lactate as shown. pyruvate + NADH lactate + NAD (i) State the conditions required for fermentation. (ii) Using all the information given, suggest why the conversion of pyruvate to lactate is required for glycolysis to continue. [Turn over 1 1 page 16 MARKS DO NOT WRITE IN THIS MARGIN 10. The fungus Aspergillus niger (A.niger) is used to produce citric acid in fermenters using sugars as substrates. Citric acid is an intermediate in a metabolic pathway as shown. isocitrate citric acid aconitase enzyme enzymes enzymes intermediates sugars (a) The aconitase enzyme requires iron to function. Explain why the growth medium used to produce citric acid should not contain iron. (b) The optimum temperature for citric acid production by A.niger is 30 °C. Explain why less citric acid would be produced if the temperature in the fermenter was reduced. 1 1 page 17 MARKS DO NOT WRITE IN THIS MARGIN 10. (continued) (c) A.niger was grown in a fermenter over an 8 day period and the pH was measured every 2 days. The results are shown in the graph. 8 7 time (days) 6 6 5 0 4 4 2 3 pH (i) Predict the pH at 10 days. (ii) Suggest a reason for the change in the pH observed. (iii) State a variable, other than temperature or pH, that should be monitored and controlled in this process. (iv) Explain why it was necessary to sterilise the fermenter before A.niger was added. [Turn over 1 1 1 2 page 18 MARKS DO NOT WRITE IN THIS MARGIN 11. Djungarian hamsters (Phodopus sungorus) are small mammals with high metabolic rates. Under certain conditions they will enter torpor. An experiment was carried out to investigate the effect of daily food intake on torpor frequency. Four groups of six hamsters were fed different masses of food each day and the number of times they entered torpor in one week was recorded. The results are shown in the table. Group Mass of food eaten each day (g) Average torpor frequency (number of times entering torpor in the week) 1 2·5 42 2 3·0 25 3 3·5 15 4 4·0 7 (a) State an advantage to the hamsters of entering torpor. (b) Suggest a measurement that could have been taken to determine whether the hamsters had entered torpor. (c) The average time spent in each period of torpor was 120 minutes. Calculate the average time spent in torpor per day by a hamster with a daily food intake of 2·5 g. Space for calculation minutes 1 1 1 page 19 MARKS DO NOT WRITE IN THIS MARGIN 11. (continued) (d) State how the design of the experiment ensured that the results were reliable. (e) Suggest a factor, other than daily food intake, which could affect torpor frequency. (f) Daily torpor is a way in which animals survive adverse conditions. Give one way in which animals avoid adverse conditions. [Turn over 1 1 1 page 20 MARKS DO NOT WRITE IN THIS MARGIN 12. An investigation was carried out into the effect of increasing time of exposure to UV light on the survival of wild type (WT) and mutant (M) yeast cells. Each type of yeast cell was grown in separate liquid media at 30 °C for 24 hours, diluted and plated onto separate agar plates. They were then exposed to a UV light source for between 0 and 30 seconds in a darkened room. The plates were incubated for four days at 20 °C and the number of yeast colonies that had grown was counted. Each colony grew from a single cell. The results are shown in the table. Time of exposure to UV light (seconds) Number of yeast colonies WT M 0 360 400 5 210 120 10 90 25 15 45 10 20 20 0 30 10 0 (a) (i) State an independent variable in this experiment. (ii) Suggest why exposure to UV light was carried out in a darkened room. 1 1 page 21 MARKS DO NOT WRITE IN THIS MARGIN 12. (continued) (b) On the grid, draw a line graph using the results in the table for both WT yeast and M yeast. (Additional graph paper, if required, can be found on page 31) (c) Draw one conclusion from the results of this investigation. (d) Sunscreen lotions can protect cells from UV damage. Suggest how the investigation could be modified to test the effectiveness of a sunscreen lotion using M yeast as model cells. [Turn over 3 1 2 page 22 MARKS DO NOT WRITE IN THIS MARGIN 13. Apples and plums are grown in North America as food crops. Brown stink bugs (Halyomorpha halys) feed on apples and plums reducing fruit yield. Insecticides are often sprayed onto fruit crops to help control these pests. (a) State how the use of insecticides can be harmful to the environment. (b) Wheel bugs (Arilus cristatus) are a species of insect native to North America that prey on many different insects, including brown stink bugs. Wheel bugs are used along with insecticides to reduce the number of brown stink bugs. (i) Explain why this method of control would require the use of less insecticide. (ii) Name the method of control that involves using both insecticides and wheel bugs. (c) If brown stink bugs spread to the UK, introducing the non-native wheel bugs from North America could be used as a method of control. If this method of control was used, wheel bugs could become an invasive species. (i) Describe evidence that could suggest the wheel bugs had become an invasive species. (ii) Give one reason why invasive species are more successful in their new habitat. 1 1 1 2 1 page 23 [Turn over for next question DO NOT WRITE ON THIS PAGE page 24 MARKS DO NOT WRITE IN THIS MARGIN 14. A tick is a small invertebrate. Some ticks carry the bacteria Borrelia burgdorferi (B.burgdorferi) in their gut after biting infected animals. If these ticks bite humans to feed on blood, the bacteria may enter the human bloodstream and cause Lyme disease. (a) Use the information given to suggest why ticks can be described as (i) vectors (ii) parasites. 1 1 page 25 MARKS DO NOT WRITE IN THIS MARGIN 14. (continued) (b) The tables show the average global temperatures between 1981 and 2010, and the number of cases of Lyme disease in the UK between 1999 and 2009. Table 1 Table 2 Year Average global temperature (°C) Year Number of cases of Lyme disease in the UK 1981–1990 14·12 1999 200 1991–2000 14·26 2004 515 2001–2010 14·47 2009 870 (i) Calculate the percentage increase in cases of Lyme disease in the UK between 1999 and 2009. Space for calculation % (ii) It was concluded from the information in Table 1 and Table 2 that the increase in the number of cases of Lyme disease in the UK was caused by an increase in temperature. Suggest why this conclusion may not be valid. (c) Two methods used to reduce the number of cases of Lyme disease in humans were suggested. Method 1 Use pesticide to kill ticks Method 2 Treat infected animals with antibiotics Explain how each method could reduce the number of cases of Lyme disease in humans. Method 1 Method 2 1 1 2 [Turn over page 26 MARKS DO NOT WRITE IN THIS MARGIN 15. White-faced capuchin monkeys (Cebus capucinus) are primates that live in large social groups in tree tops in South America. Their predators include humans, birds of prey and snakes. If a capuchin sees a predator it gives an alarm call to warn others in the group. Capuchins give a different alarm call for each predator they encounter. Scientists recorded these alarm calls from adult monkeys. The results are shown in the table. Predator Total number of encounters Percentage of encounters when only one monkey called Percentage of encounters when more than one monkey called Birds of prey 155 60 40 Humans 12 67 33 Snakes 65 15 85 (a) (i) Calculate the number of encounters with birds of prey when only one monkey called. Space for calculation (ii) Using the information given, suggest why the highest number of encounters were with birds of prey. (b) Explain why giving alarm calls can be described as altruistic behaviour. 1 1 2 page 27 MARKS DO NOT WRITE IN THIS MARGIN 15. (continued) (c) Primates, such as capuchin monkeys, have a long period of parental care. Explain why the scientists only recorded the calls from adult monkeys. (d) Primates often form alliances with others and carry out appeasement behaviour within their group. State the advantages of these behaviours. Forming alliances Appeasement [Turn over for next question 1 2 page 28 MARKS DO NOT WRITE IN THIS MARGIN 16. Attempt either A or B. Write your answer in the space below and on pages 29 and 30. A Write notes on photosynthesis under the following headings. (i) Use of energy absorbed by photosynthetic pigments (ii) Carbon fixation OR B Write notes on the effects of the following on biodiversity. (i) The bottleneck effect (ii) Habitat fragmentation and habitat corridors You may use labelled diagrams where appropriate. 3 4 2 5 page 29 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWERS [Turn over page 30 MARKS DO NOT WRITE IN THIS MARGIN SPACE FOR ANSWERS [END OF QUESTION PAPER] page 31 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK ADDITIONAL GRAPH PAPER FOR QUESTION 12 (b) page 32 MARKS DO NOT WRITE IN THIS MARGIN ADDITIONAL SPACE FOR ANSWERS AND ROUGH WORK Acknowledgement of copyright Question 14 nechaevkon/shutterstock.com

4 2 The diagram shows the ultrastructure of an animal cell. U T S R Q P Magnification ×9000 (a) (i) The structure labelled R on the diagram represents a (1) A chloroplast B lysosome C mitochondrion D ribosome (ii) The structure labelled Q on the diagram represents the (1) A cell surface membrane B nuclear envelope C rough endoplasmic reticulum D smooth endoplasmic reticulum (iii) Calculate the actual diameter of the structure labelled U. (2) Answer .............................................................. 5 Turn over (iv) Describe how the structures labelled P, S and T are involved in the production and secretion of molecules from this cell. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Which row in the table is correct for structures found in animal cells and prokaryotic cells? (1) Nucleolus Plasmid Ribosome Animal cell Prokaryotic cell Animal cell Prokaryotic cell Animal cell Prokaryotic cell A  ×     B ×    ×  C  × ×    D    ×  × 6 (c) Explain why it is possible to see the detailed structure of a prokaryotic cell with an electron microscope but not with a light microscope. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 10 marks)

7 Turn over 3 Adult female cabbage aphids (Brevicoryne brassicae) undergo either asexual reproduction or sexual reproduction depending on the time of year. The photograph shows some cabbage aphids on a cabbage leaf. Source: © Denis Crawford/Alamy Stock Photo Some of the boxes in the diagram show the number of chromosomes present in a single cell from a particular stage of each of the alternative methods of reproduction. 16 16 Asexual reproduction in spring and summer Sexual reproduction in autumn and winter Growth Growth Sperm Nymph Juvenile Fertilisation Adult female Adult male Eggs (a) (i) Complete the empty boxes in the diagram to show the number of chromosomes present in a single cell at that stage of reproduction. (2) (ii) Label the arrows on the diagram with the letter M to show where meiosis takes place. . (1) 8 (b) Explain why a juvenile aphid is genetically identical to its parent. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) (i) Complete the table with a  or × to compare the events during mitosis and meiosis. The first row in the table has been completed for you. (3) Event Mitosis Meiosis I Meiosis II chromosomes shorten and thicken   × homologous chromosomes pair together crossing over can cause genetic variation homologous chromosomes separate sister chromatids separate 9 Turn over (ii) Which of the following increases the number of different alleles in a population? (1) A crossing over B gene mutation C independent assortment of chromosomes during nuclear division D random fusion of gametes (Total for Question 3 = 10 marks)

10 4 Enzymes are involved in a wide range of metabolic reactions. (a) The diagram represents the structure of a maltose molecule. H HO H H OH H 6CH2OH OH H O   4 5 3 2 1 H OH H OH H 6CH2OH OH H O   4 5 3 2 1 H O Draw a diagram to show the hydrolysis of maltose. (3) 11 Turn over (b) Catalase is an enzyme found in potato cells. It catalyses the breakdown of hydrogen peroxide. hydrogen peroxide water + oxygen catalase In an investigation, cylinders of potato were cut with a cork borer. The cylinders were then sliced into discs with the same thickness and put into a small beaker containing 50 cm3 of hydrogen peroxide. The mass of the beaker and its contents was recorded over a period of 15 minutes. The graph shows the results of the experiment. 0 100 99 98 97 96 95 Mass of contents of beaker as percentage of initial mass Time / minutes 5 10 15 Explain the relationship shown in the graph. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 12 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) Explain how a gene mutation can prevent the production of catalase in potato cells. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 10 marks)

13 Turn over 5 The diagram illustrates part of a flowering plant when pollination has just taken place. pollen grain stigma style ovary (a) Describe the role of enzymes in the growth of the pollen tube. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 14 (b) The styles of some plants secrete RNAase enzymes when pollen from the same flower germinates on the stigma. Explain how these RNAase enzymes affect the growth of the pollen tube. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) Explain the role of double fertilisation in flowering plants. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 10 marks)

15 Turn over 6 Male fertility can be determined by a number of different factors. (a) Describe the process of spermatogenesis. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The mean volume of the semen produced by a male ejaculation is 3.4 cm3. This contains a mean concentration of 17 000 sperm mm–3. Calculate the mean total number of sperm in a single ejaculation. Give your answer in standard form. (2) Answer .............................................................. 16 (c) The distance from the point of ejaculation in the vagina to the upper end of the fallopian tube (where fertilisation takes place) is 19 cm. Some sperm travel this distance in three hours. Calculate the mean speed of these sperm in cm min–1. (1) Answer .............................................................. (d) Male fertility can be estimated using scales that take into account the number of sperm, their mobility and the percentage with a normal ‘head’. To calculate male fertility using these scales:  draw a straight line between the observed number of sperm (scale A) and the percentage of sperm motile after 2 hours (scale C)  from the intersection of this line with scale B, draw another straight line to scale E (the percentage of normal-headed sperm)  the point where this second line crosses scale D (the fertility index), provides a relative assessment of fertility. }         A over 80 over 88 over 55 80 65 84 50 50 55 80 35 10 45 78 30 5 34 25 1 25 20 0.5 15 15 0.1 B C D E under 10 under 70 under 10 fertile fertility index observed number of sperm / millions cm–3 motile sperm after 2 hours / % normal-headed sperm / % 45 25 76 possibly fertile infertile 17 Turn over (i) Estimate the fertility of a man who produces a semen sample with:  25 million sperm per cm3  35% of which are motile after 2 hours and  84% of which have normal heads. (2) Answer .............................................................. (ii) Deduce why fertilisation is unlikely to occur if there are 5 million spermatozoa per cm3. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 10 marks)

18 7 Streptococcus pneumoniae is a bacterium that causes pneumonia. (a) When Streptococcus pneumoniae is grown on a culture plate, the bacterial cells produce:  either smooth, shiny colonies (S‑strain)  or rough colonies (R‑strain). The S‑strain bacteria produce a smooth capsule but the R‑strain bacteria do not. (i) Explain why the R‑strain bacteria are not able to produce the capsule present in the S‑strain bacteria. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) In an investigation, mice were injected with either the S‑strain or the R‑strain. The results are shown in the table. Strain of Streptococcus pneumoniae injected into mice Injected mice S‑strain died R‑strain lived The S‑strain bacteria were then killed by heating them. Mice were then injected with either heat killed S‑strain bacteria or a mixture of heat killed S‑strain bacteria and live R‑strain bacteria. The results are shown in the table. Strain of Streptococcus pneumoniae injected into mice Injected mice S‑strain (heat killed) lived S‑strain (heat killed) and R‑strain (live) died 19 Turn over Analyse the data to explain why the mice died when they were injected with a mixture of heat killed S‑strain and live R‑strain bacteria. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) In another investigation, heat killed S‑strain bacteria were treated with one type of enzyme:  protein digesting enzymes (proteases)  RNA‑digesting enzymes (RNAases)  DNA digesting enzymes (DNAases)  lipid digesting enzymes (lipases). Each type of enzyme treated S‑strain bacteria was then mixed with live R‑strain bacteria and injected into mice. Complete the table to predict whether the mouse lives or dies. Write dead OR alive in the spaces in the table. (3) Enzyme used to treat the heat killed S‑strain bacteria Mouse injected with live R‑strain and enzyme treated S‑strain [dead or alive] protease RNAase dead DNAase lipase 20 (c) Another investigation used radioactively labelled phages. Phages are viruses that consist of DNA contained within a protein head. DNA Protein Phages inject their DNA into a host bacterial cell, but leave the protein heads (called phage ghosts) on the outside of the infected cell. Two different types of phage were developed by growing the phages in cultures of bacteria in two separate media. Medium (A) contained radioactive phosphorus (32P). Medium (B) contained radioactive sulfur (35S). Type A phage had DNA that included the 32P. Type B phage had proteins that included the 35S. (i) State why radioactive carbon, nitrogen or oxygen were not used in this experiment. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 21 Turn over (ii) Separate cultures of the bacteria E. coli were infected with either type A or type B phage. After infection the empty phage heads (phage ghosts) were separated from the bacterial cells. Each was tested for radioactivity. Complete the table to show where radioactivity was present (Y) or absent (N). (1) Phage type (label) Bacterial cells Phage ghosts A (32P) B (35S) (Total for Question 7 = 10 marks)

22 8 The graph shows the results of an experiment investigating the effect of temperature on the hydrolysis of starch using the enzyme amylase. 0 1 2 3 4 5 6 7 10 8 6 4 2 0 Concentration of maltose produced mmol dm–3 Time / hours 55 °C 65 °C (a) Calculate the initial rate of reaction at 55 °C in mmol dm–3 min–1. (2) Answer .............................................................. 23 Turn over (b) Explain the difference between the two curves from 4 and 6 hours. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) In this experiment:  two beakers were prepared containing 1cm3 of amylase and 100 cm3 of starch solution at room temperature  these beakers were then placed in the water baths at two different temperatures, 55 °C and 65 °C  every 30 minutes a sample of the mixture was extracted and the concentration of maltose was measured. (i) Analyse the information to explain why the initial rate of this reaction was faster at 65 °C than it was at 55 °C. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 24 *(d) Evaluate the method used in this investigation and suggest improvements. 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(Total for Question 8 = 13 marks) TOTAL FOR PAPER = 80 MARKS 26 BLANK PAGE 27 BLANK PAGE 28 BLANK PAGE

3 Turn over 2 The table shows two differences between blood plasma and tissue fluid. Blood plasma Tissue fluid Has a higher protein content Has a lower protein content Contains more dissolved oxygen Contains less dissolved oxygen (a) State how tissue fluid is formed. (1) ...................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Explain the differences between blood plasma and tissue fluids shown in the table. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) Some of the tissue fluid is not reabsorbed into the blood capillary. Describe what happens to the tissue fluid that is not reabsorbed into the blood capillary. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 5 marks)

4 3 Mosquitoes transmit a number of diseases, including malaria. The photograph shows a mosquito, Anopheles gambiae. This mosquito transmits malaria. Source: © Sinclair Stammers / Science photo library (a) (i) State the genus of this mosquito. (1) ...................................................................................................................................................................................................................................................................................... (ii) Mosquitoes belong to the class Insecta. Some of the insects in this class are subdivided into a group called Diptera. What is the name of the classification group that includes Diptera? . (1) A Family B Genus C Order D Phylum (iii) Which of the following organisms causes the symptoms of malaria? . (1) A Anopheles gambiae B Plasmodium falciparum C Puccinia graminis D Salmonella enterica 5 Turn over (b) From 2010 to 2015 there was a 29% decrease in the number of deaths from malaria. This was due to prevention and control measures. (i) Explain one implication of a named method of controlling malaria. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) In 2015, the number of deaths from malaria was 429 000. Calculate the number of deaths from malaria in 2010. (2) Answer .............................................................. (c) In 2015, the number of cases of malaria was 211 million. This increased to 216 million cases in 2016. Give the change in the number of cases of malaria in standard form. (1) Answer .............................................................. 6 (d) Scientists are researching new methods of controlling malaria. One group used a procedure called gene drive. This procedure resulted in female mosquitoes that could not bite or could not lay eggs. One of the scientists said: ‘It will still be at least 5 to 10 years before we consider testing any mosquitoes with gene drive, but now we have some encouraging proof that we’re on the right path.’ Explain why this group are sharing the results of their research. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 10 marks)

7 Turn over 4 The photograph shows a blood smear from a healthy person, as seen using a light microscope. K J Source: © biophoto associates/science photo library (a) (i) Which row of the table identifies the cells labelled J and K? (1) Cell J Cell K A eosinophil lymphocyte B lymphocyte neutrophil C monocyte eosinophil D neutrophil monocyte (ii) Which is the approximate ratio of erythrocytes to leucocytes in this blood smear? (1) A 1 : 50 B 3 : 50 C 50 : 1 D 50 : 3 8 (b) Chronic myelogenous leukaemia (CML) is a blood disorder. The photograph shows a blood smear from a person with CML, as seen using a light microscope. Explain why a blood smear can be used to diagnose CML. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 9 Turn over (c) The diagram shows some stages in the production of blood cells from bone marrow stem cells. Bone marrow stem cell Common lymphoid progenitor cell Common myeloid progenitor cell Megakaryocyte Myeloblast Mast cell Erythroblast Basophil Neutrophil Eosinophil Monocyte Small lymphocyte Natural killer cell B lymphocyte T lymphocyte Explain how a bone marrow stem cell differentiates into either a common myeloid progenitor cell or a common lymphoid progenitor cell. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 8 marks)

10 5 The rubella virus stimulates the production of antibodies in humans but usually causes only mild infections. However, infection of pregnant women can cause serious problems for the developing fetus. *(a) Antibodies can be divided into five classes. The table gives some information about these classes of antibody. Information Class of antibody IgM IgA IgG IgD IgE Shape Number of antigen binding sites 10 4 2 2 2 Can cross the placenta no no yes no no Other details secreted into the blood secreted into mucus, tears, saliva, colostrum1 secreted into the blood attached to the surface of B cells involved in allergy and parasitic infections 1 Colostrum is the first type of milk produced by the mother following birth. The graph shows the levels of three of these classes of antibody produced in response to infection with the rubella virus. Levels of antibody 0 1 2 3 4 IgM IgG IgA Time after infection/months 11 Turn over Analyse the information in the table and the graph to assess the role of these five classes of antibody in the immune response to the rubella virus. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 12 (b) Vaccination against rubella has helped to reduce the incidence of infection. Explain the importance of vaccinating as many people as possible against rubella. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 8 marks)

13 Turn over 6 Norovirus is a single-stranded RNA, non-enveloped virus. Norovirus is the most common cause of gastroenteritis. (a) The table shows some features used to classify viruses. Which box in each row shows how these viruses are classified? (3) Virus Classification of viruses DNA enveloped DNA non-enveloped RNA enveloped RNA non-enveloped Ebola λ (lambda) phage Tobacco mosaic 14 (b) One study into the causes of gastroenteritis involved 162 patients. This study found that 24.7% of these gastroenteritis cases were caused by viruses. The pie chart shows the proportion of the different types of virus identified as causing these gastroenteritis cases. astrovirus rotavirus A norovirus GI norovirus GII sapovirus rotavirus B both norovirus GI and GII Estimate the total number of gastroenteritis cases caused by noroviruses. (3) Answer .............................................................. 15 Turn over (c) Norovirus is usually spread through the faecal-oral route. The diagram shows the pathway that norovirus takes through the body. 1  The norovirus is transmitted in contaminated food or water. 2  The norovirus passes through the stomach and into the small intestine. 3  Some noroviruses are taken up by the cells of the small intestine. 4  Some noroviruses pass out of the body in the faeces. This process only takes a few hours. norovirus 1 2 3 4 Describe what happens inside the cells of the small intestine that have taken up the norovirus. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16 (d) Noroviruses can infect an animal as individual viruses or as a cluster of viruses inside a lipid sphere. A cluster of viruses inside a lipid sphere is called a stealth sphere. The diagram shows a stealth sphere. lipid sphere enclosing noroviruses cluster of noroviruses (i) Animals infected with stealth spheres develop gastroenteritis very quickly and with severe symptoms. Animals infected with individual noroviruses develop gastroenteritis more slowly and with less severe symptoms. Analyse the information to explain these findings. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 17 Turn over (ii) Scientists are hoping that new treatments for norovirus infections can be developed to target the stealth sphere. Analyse the information to explain how targeting the stealth sphere could be used to treat these infections. 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(Total for Question 6 = 14 marks)

18 7 Lipids and carbohydrates are used as respiratory substrates. The respiration of lipids generates more ATP than the respiration of carbohydrates. (a) The table shows bonds that may be found in carbohydrates and lipids. Which box in each row shows whether the bond may be found in these molecules? (3) Bond Molecule that bond may be found in carbohydrate only lipid only both carbohydrate and lipid neither carbohydrate nor lipid covalent ester hydrogen (b) The diagram shows some steps in the respiration of carbohydrates and lipids. Glycogen Starch Glucose Pyruvate Acetyl coenzyme A Fatty acids Lipid Krebs cycle (i) Where in a cell does the conversion of pyruvate to acetyl coenzyme A take place? (1) A cytoplasm B matrix C stroma D tonoplast 19 Turn over (ii) Explain why lipids can be respired only in aerobic conditions. 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(c) The electron transport chain uses hydrogen ions to generate ATP. Explain why the respiration of lipids generates more ATP than the respiration of carbohydrates. 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(d) The respiratory quotient (RQ) can indicate which substrate is being used in respiration. The formula used to calculate RQ is RQ volume of carbon dioxide produced volume of oxygen used = (i) State why the RQ for carbohydrate is 1.0. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 20 (ii) The table shows the formula for calculating the volume of oxygen used and the volume of carbon dioxide produced in the respiration of lipids. The number of carbon atoms in the lipid is represented by n. Volume of oxygen used / a.u. Volume of carbon dioxide produced / a.u. 4.5n + 3.75 3n + 6 Calculate the RQ for a lipid containing 21 carbon atoms. (2) Answer .............................................................. (e) Explain why an insect has an RQ of 1.0 at rest and an RQ of 0.7 during flight. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 15 marks)

21 Turn over 8 Skin flora include bacteria that live on the skin. Different parts of the body have different types of bacteria present. Skin flora are beneficial as they inhibit the growth of pathogenic bacteria on the skin. Skin flora compete for nutrients and space or secrete toxic chemicals. (a) The following are types of bacteria found in the female genital tract: • Escherichia coli • Klebsiella species • Lactobacillus species • Staphylococcus species. (i) Describe how these types of bacteria present in the female genital tract could be identified. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22 (ii) The photograph shows Lactobacilli on a skin cell from the female genital tract. skin cell from the female genital tract chains of Lactobacilli { Source: © Janice Carr Content Providers (s): CDC/Dr. Mike Miller Lactobacilli metabolise glycogen into lactic acid. Explain how Lactobacilli inhibit the growth of pathogenic bacteria in the female genital tract. 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(b) Sperm cells are released into the female genital tract in seminal fluid. Seminal fluid contains fructose instead of glucose. The diagram shows the structure of fructose. O H OH H CH20H H OH CH2OH OH 23 Turn over (i) Compare and contrast the structure of fructose with the structure of glucose. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain the advantages of the presence of fructose in seminal fluid. 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(Total for Question 8 = 13 marks) 24 BLANK PAGE

25 Turn over 9 The photograph shows a wombat, an animal that lives in dry parts of Australia. Source: © Blue Gum Pictures/Alamy Stock Photo Wombats are herbivores, feeding on grasses, leaves and bark. Wombats also dig in the soil for roots. They have behavioural and physiological adaptations to survive periods of severe drought. The effect of drought on the health of wild wombats was investigated. Changes in body mass, body condition and the chemical constituents of stomach contents and faeces were recorded. These changes were related to the changes in the quality and quantity of the food of wombats. (a) Explain why drought affects the quality and quantity of the food of wombats. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26 (b) The table shows some of the measurements made in this investigation. Measurement Before drought 10 months into drought 14 months into drought 2 months after rain Body condition index ………………………… 0.86 0.76 0.85 Body mass / kg 26.4 25.1 21.2 25.2 Organic matter in stomach / g kg-1 dry matter 946 732 704 822 Mineral ion content in stomach low high high low Moisture content of faeces / g kg-1 wet mass 815 603 522 748 (i) The body condition was expressed as a body condition index (BCI). The BCI can be calculated using the formula BCI = body mass of wombat mass-for-length of wombat (Y) The mass-for-length (Y) of a wombat can be calculated using the formula Y = (body length × 0.73) – 34.5 A wombat before the drought had a body length of 86.4 cm and a mass of 26.4 kg. Calculate the BCI of this wombat. (2) Answer .............................................................. 27 *(ii) Analyse the data to explain the results of this investigation. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 9 = 13 marks) TOTAL FOR PAPER = 90 MARKS 28 BLANK PAGE

5 Turn over 2 Acetylcholine is a neurotransmitter produced by the parasympathetic nervous system. (a) Describe the process by which acetylcholine is released from a synapse. 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(b) Some chemical pesticides, such as diisopropyl fluorophosphate (DFP), affect the activity of synapses. This pesticide acts by binding irreversibly to the R group of an amino acid in the active site of acetylcholinesterase. (i) Explain why DFP inhibits the activity of acetylcholinesterase. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 6 (ii) Explain why a person poisoned with DFP has a slower heart rate. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 9 marks)

7 Turn over 3 (a) The diagram shows a cell from the proximal convoluted tubule of the nephron. Explain how the features shown in the diagram enable this cell to carry out its function. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... ....................................................................................................................................................................................................................................................................................

8 (b) (i) Which of the following statements about facilitated diffusion are correct? (1) 1. substances move through membrane proteins 2. substances move against a concentration gradient 3. substances move down a concentration gradient 4. substances move between phospholipids A 1 and 2 B 1 and 3 C 2 and 3 D 2 and 4 (ii) The collecting duct is also part of the nephron. Which of the following are the effects of increased release of antidiuretic hormone (ADH)? (1) A decreased permeability of the collecting duct to water, producing more concentrated urine B decreased permeability of the collecting duct to water, producing more dilute urine C increased permeability of the collecting duct to water, producing more concentrated urine D increased permeability of the collecting duct to water, producing more dilute urine (Total for Question 3 = 6 marks)

9 Turn over 4 In cats, the gene for fur length is not located on the X chromosome. The allele for short fur (F) is dominant to the allele for long fur (f). The pedigree diagram shows the inheritance of fur length in a family of cats. 1 2 3 4 5 6 7 9 10 11 12 13 8 male, short fur female, short fur male, long fur female, long fur Key (a) Explain how one piece of evidence from the diagram shows that the allele for short fur is dominant to the allele for long fur. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) A gene for fur colour in cats is located on the X chromosome. The alleles for black fur (XB) and orange fur (Xo) are codominant. Tortoiseshell cats have areas of black and orange fur. (i) Give all the possible genotypes of a male, black cat with short fur. (1) .................................................................................................................................................................................................................................................................................... 10 (ii) Explain why all tortoiseshell cats are female. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) In a genetic cross, a tortoiseshell cat with long fur was mated with an orange male cat, heterozygous for fur length. Deduce the probability of producing a tortoiseshell cat with long fur from this cross. Use a genetic diagram to support your answer. (4) (Total for Question 4 = 9 marks) 11 Turn over BLANK PAGE

12 5 Palm plants and cocoa plants are used to produce lipids for the food industry. Palm oil is produced from palm plants. Cocoa butter is produced from cocoa plants. (a) Palm oil and cocoa butter are made of triglycerides composed of different fatty acids. The diagram shows a triglyceride. C O C R1 H O H C O C R1 O H C O C R2 + 3H2O O H C O C R3 H Complete the diagram to show the products of hydrolysis of this triglyceride. (2) (b) The fatty acid compositions of palm oil and cocoa butter are shown in the table. Fatty acid Chemical formula of fatty acid Percentage of fatty acid found in palm oil and cocoa butter (%) Palm oil Cocoa butter palmitic acid C16H32O2 45 27 oleic acid C18H34O2 40 31 linoleic acid C18H32O2 10 3 myristic acid C14H28O2 1 1 stearic acid C18H36O2 4 38 (i) How many of the named fatty acids shown in the table are saturated? (1) A 1 B 2 C 3 D 4 13 Turn over (ii) The percentage solid fat content (% SFC) is a measure of how much of a lipid is solid at different temperatures. The graph shows the effect of temperature on the % SFC for palm oil and for cocoa butter. 0 10 20 30 40 50 60 70 80 90 100 10 15 20 25 30 35 Temperature / ° C palm oil cocoa butter Percentage solid fat content (% SFC) Comment on the effect of temperature on the % SFC of palm oil and of cocoa butter. . (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 14 (c) Some forests in Borneo have been cleared to grow these plants. Orangutans used to live in forests throughout this island but now they are found in small areas of forest. The map shows the distribution of orangutans in parts of Borneo. N <100 individuals 100–500 individuals 501–1500 individuals >1500 individuals Key (i) Explain why in areas with low populations of orangutans, many of the orangutans have deformities and health problems. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... . 15 Turn over (ii) The Malaysian government is paying local people to create and maintain ‘land-bridges’ of forest to link areas of forest together. Explain why these payments will help to develop palm oil production and support conservation of orangutans. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) State how treaties such as CITES help conserve orangutans. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 13 marks) 16 BLANK PAGE

17 Turn over 6 The speed of a nerve impulse along a neurone depends on the diameter and the presence or absence of myelin. (a) The graph shows an action potential in a neurone. +50 +30 +10 –10 –30 –50 –70 –90 0.5 1.0 1.5 2.0 2.5 Membrane potential / mV Time / ms When are voltage-gated sodium channels open? (1) A 0.25 ms B 0.75 ms C 1.00 ms D 1.25 ms 18 (b) The graph shows the relationship between the diameter of the neurone and the speed of nerve impulses along myelinated and non-myelinated motor neurones. 10.0 8.0 6.0 4.0 2.0 0.0 –2.0 speed of nerve impulse / m s–1 motor neurone diameter / μm 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 myelinated motor neurone non-myelinated motor neurone 19 Turn over (i) Compare and contrast the effect of diameter on the speed of an impulse in myelinated and non-myelinated motor neurones. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Determine the speed of a nerve impulse along a myelinated motor neurone with a diameter of 5 μm. The equation for a linear relationship is y = mx + c (3) Answer .............................................................. 20 (iii) Multiple sclerosis (MS) is a disease that results in the loss of myelin around motor neurones. Explain why the speed of nerve impulses along axons is slower in people with MS. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 11 marks) 21 Turn over BLANK PAGE

22 7 The primary productivity of plants depends on their ability to synthesise carbohydrates and transport solutes to cells. (a) The diagram shows the movement of water across a root. root hair epidermal cell Casparian strip xylem vessel water What is the name of the water movement pathway shown in the diagram? (1) A apoplastic pathway B diffusion pathway C osmotic pathway D symplastic pathway (b) Explain how the mass-flow hypothesis accounts for the movement of sugars from leaves to roots. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 23 Turn over *(c) Mistletoe is a plant that is a parasite on trees. Mistletoe has green leaves. Mistletoe also has a modified root that grows into the xylem of a tree. The diagram shows mistletoe growing on an apple tree branch. mistletoe apple tree branch The effect of mistletoe on the growth of apple trees was investigated. Every year for 20 years, scientists selected apple trees infected with mistletoe and measured the: • mean net primary productivity (NPP) of the apple trees • transpiration rate of the apple trees • transpiration rate of the mistletoe branches on the apple trees They also determined the mineral content of the apple tree leaves after 20 years. A control group of apple trees not infected with mistletoe was included in this investigation. 24 The graphs and table show the results of the investigation. 0 5 10 15 20 25 30 35 1995 2000 2005 2010 2015 Year NPP / kg dry mass of tree per year NPP of apple trees without mistletoe NPP of apple trees with mistletoe 1995 2000 2005 2010 2015 Year 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Transpiration rate / dm3 water per plant per year mistletoe apple tree without mistletoe apple tree with mistletoe Apple tree leaves from Nitrogen content / mg per g of leaf Calcium content / mg per g of leaf trees infected with mistletoe 8.50 6.45 trees not infected with mistletoe 10.60 8.90 25 Turn over Analyse the data to explain the effect of mistletoe on the NPP of apple trees. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 7 = 11 marks)

26 8 The graph shows the oxygen dissociation curves of human haemoglobin and human myoglobin. 12 0 10 20 30 40 50 60 70 80 90 100 0 2 4 6 8 10 Percentage saturation of haemoglobin with oxygen (%) Partial pressure of oxygen (pO2) / kPa haemoglobin myoglobin (a) (i) Calculate the volume of oxygen bound to haemoglobin at a partial pressure of oxygen of 6 kPa in 100 cm3 blood. The concentration of haemoglobin is 15 g per dm3 blood. When fully saturated, there is 1.36 cm3 oxygen per g of haemoglobin. (3) Answer .............................................................. 27 Turn over (ii) Explain why the oxygen dissociation curves for haemoglobin and myoglobin are different. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 28 (b) In an investigation, the oxygen dissociation of haemoglobin from Tibetan people and European people was compared. The Tibetan people were living at high altitude. The European people were living at low altitude. The graph shows the results of this investigation. 0 10 20 30 40 50 60 70 80 90 100 0 2 4 6 8 10 12 Percentage saturation of haemoglobin with oxygen (%) Partial pressure of oxygen (pO2) / kPa haemoglobin from Tibetans, at low partial pressure of CO2 haemoglobin from Europeans, at low partial pressure of CO2 haemoglobin from Europeans, at high partial pressure of CO2 haemoglobin from Tibetans, at high partial pressure of CO2 Comment on the oxygen dissociation curves of the Tibetan people. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 29 Turn over (c) Tibetan populations have a recessive allele that helps them to survive at high altitude. In a village in Tibet, 1950 people out of a population of 2500 people carry two copies of this allele. (i) Calculate the number of heterozygous people in the village. Use the Hardy-Weinberg equation p2 + 2pq+q2 = 1 (3) Answer .............................................................. 30 (ii) This allele is currently found only in human populations in Tibet. The allele has also been found in the fossil bones of an extinct form of modern humans called the Denisovans. Fossils of Denisovans have been found in Tibet. Explain why scientists think that the Denisovans belonged to the same species as modern humans. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 16 marks)

31 Turn over 9 The diagram shows the structure of a rod cell. outer segment inner segment mitochondria nucleus synaptic vesicles Golgi apparatus endoplasmic reticulum (a) (i) How many of the labelled organelles have a double membrane? (1) A 0 B 1 C 2 D 3 (ii) Which of the following statements about the effect of light on rod cells are correct? (1) 1. rhodopsin breaks down into opsin and trans-retinal 2. voltage-gated sodium channels open in the membrane 3. more neurotransmitter is released from the presynaptic terminal A 1 only B 1 and 2 C 1 and 3 D 2 and 3 32 *(b) Graph 1 shows the distribution of photoreceptors across the retina. Graph 2 shows the sensitivity of rod cells and cone cells to light of different wavelengths. 70 0 50 100 150 200 70 60 50 40 30 20 10 0 10 20 30 40 50 60 Cell density / ´103 mm –2 Angle from fovea in degrees Key Rod cells Cone cells Graph 1 Percentage retinal response (%) Wavelength of light / nm 380 450 500 600 650 700 750 550 Blue Cones Rods Green Cones Red Cones Graph 2 33 Turn over Students investigated the ability to detect beams of light of different colours and intensities. A student covered one eye and a bright red light (wavelength 670 nm) was placed at different angles in the field of vision of the other eye. The student stated if they could detect the red light. This was repeated with a dim red light. This was then repeated with a bright green light and a dim green light (wavelength 525 nm). The table shows the results of this investigation. Angle from fovea in degrees Light detected (Yes/No) Bright light Red Dim light Red Bright light Green Dim light Green 60 N N Y Y 30 N N Y Y 15 N N Y Y 10 N N Y Y 5 Y N Y N 0 Y N Y N 5 Y N Y N 10 N N Y Y 15 N N N N 30 N N Y Y 60 N N Y Y Analyse the data to explain the results of this investigation. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 34 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 8 marks) TOTAL FOR PAPER = 90 MARKS 35 BLANK PAGE 36 BLANK PAGE

2 Answer ALL questions. Write your answers in the spaces provided. 1 (a) Describe how oxygen from the air is able to reach insect muscle cells. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) A respirometer was used to compare the rate of aerobic respiration of two species of insect. The diagram shows the respirometer. syringe three-way tap glass tubing small insects gauze glass tube scale coloured liquid soda lime (i) Give the units for a valid comparison of the rate of respiration of these two species of insect. (1) .................................................................................................................................................................................................................................................................................... 3 Turn over (ii) Other factors may affect the results from this respirometer. Describe how one named factor could be controlled when using this respirometer. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Explain why the soda lime in the tube must be replaced when the other species of insect is placed in the respirometer. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 1 = 8 marks)

4 2 The egg of a hen contains albumin, a globular protein. (a) Describe the tertiary structure of a globular protein. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 5 Turn over (b) Pepsin is an enzyme that digests albumin. Albumin can be put into a 50 mm long capillary tube and then made solid by boiling in water for 10 minutes. The diagram shows a 50 mm capillary tube containing solid albumin. capillary tube solid albumin Devise an investigation into the effect of pepsin concentration on the digestion of solid albumin. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 8 marks)

6 3 Water fleas are small animals that live in pond water. A student observed a water flea using a light microscope. The diagram shows the student’s drawing of the water flea. compound eye gut heart The diagram shows an eyepiece graticule and a stage micrometer used by the student to measure the diameter of the heart in this water flea. 50 60 70 0 1 2 3 4 5 6 7 8 9 10 eyepiece graticule stage micrometer (Source: http://biology4alevel.blogspot.com/2014/07/2-cell-structure-microscopy.html) Each small division on the stage micrometer is 0.01 mm. The diameter of the heart was found to be 0.2 mm. (a) Determine the number of eyepiece units that equal the diameter of the water flea heart. (1) Answer .............................................................. 7 Turn over (b) The water flea has a spherical heart with a diameter of 0.2 mm. Cardiac output is the volume of blood pumped out of a heart in one minute. The resting heart rate of a water flea is 200 beats min−1. The volume of a sphere is calculated using the formula 4 3 πr 3 where π = 3.142 (i) Calculate the cardiac output of a water flea at rest using the formula cardiac output = stroke volume × heart rate (3) Answer .............................................................. dm3 min−1 (ii) The cardiac output of a human heart at rest is 5.0 dm3 min−1. Give one reason why the cardiac output of a human heart at rest needs to be greater than the cardiac output of a water flea at rest. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 8 (c) A student investigated the effect of adrenaline and acetylcholine on the heart rate of water fleas using the following method: ● place a water flea in pond water on a microscope slide ● count the number of heart beats in one minute ● repeat with two other water fleas ● repeat this procedure using three water fleas placed in adrenaline solution and three water fleas placed in acetylcholine solution. The table shows the results of this investigation. Treatment Heart rate / beats min−1 Standard deviation (SD) Water flea 1 Water flea 2 Water flea 3 Mean Pond water 195 185 190 190 Adrenaline solution 240 236 238 238 ± 2.0 Acetylcholine solution 116 120 118 118 ± 2.0 (i) Calculate the standard deviation for the mean heart rate in the pond water. Use the formula SD = − − ∑( ) X X n 2 1 (2) Answer .............................................................. 9 Turn over (ii) Plot a graph to show the data for the mean heart rate and SD in each treatment. (2) (iii) Explain why abiotic variables should be controlled in this investigation. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 11 marks)

10 4 A scientist investigated the effect of temperature on the time it takes for blood to clot. The graph shows the results of this investigation. 180 160 140 120 100 80 60 40 20 0 30 32 34 36 38 40 Temperature / °C Clotting time / s (a) Explain how these results show that enzymes are involved in blood clotting. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 11 Turn over (b) Prothrombin is involved in the blood clotting process. The F2 gene codes for the synthesis of prothrombin. This gene is located from base pair 46 719 191 to base pair 46 739 504 on chromosome 11. Determine the number of codons in this gene. (1) Answer .............................................................. (c) A mutation of the F2 gene causes thrombophilia, a condition that results in the production of excess prothrombin. In this gene mutation, guanine is replaced with adenine. (i) Name this type of mutation. (1) .................................................................................................................................................................................................................................................................................... (ii) People without this mutation have a 1 in 1000 risk of producing a blood clot in an artery. The mutation increases this risk by 20 times. State the probability of producing a blood clot for people with this mutation. (1) .................................................................................................................................................................................................................................................................................... (iii) Explain why thrombophilia increases the risk of producing a blood clot in an artery. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 12 (d) A genetic test can be used to find out if a person has thrombophilia. The test involves using a restriction endonuclease to obtain genetic material from white blood cells. This genetic material is then used in the polymerase chain reaction (PCR). (i) State the role of a restriction endonuclease. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Describe the process of PCR. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 13 marks)

13 Turn over 5 Gibberellin stimulates cells in the aleurone layer of cereal grains, such as barley, to produce the enzyme amylase. The diagram shows the location of the aleurone layer in a barley grain. seed coat aleurone layer X (a) The part labelled X contains triploid cells. Name the part labelled X. (1) .................................................................................................................................................................................................................................................................................... (b) The diagram shows a cell from the aleurone layer with some of the structures involved in the production of amylase. GA receptor GA cell wall plasma membrane amylase 14 Gibberellin binds to a protein receptor in the cell surface membrane and this stimulates transcription in the nucleus. (i) Describe the process of transcription in the nucleus of this cell. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Describe the processes occurring after transcription that result in the release of amylase from the cell shown in the diagram. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 15 Turn over (c) Global warming may cause sea levels to rise and salty seawater to flood land used for growing crops. A student read that salt (NaCl) inhibits gibberellin synthesis. This could affect amylase production in cereal grains. Devise an investigation to show the effect of salt concentration on amylase production in cereal grains. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 5 = 13 marks)

16 6 Ultraviolet (UV) radiation damages DNA and is used in microbiology to produce aseptic conditions. The UV radiation changes the bonding found in DNA. This change in bonding creates thymine dimers in which covalent bonds form between adjacent thymine nucleotides. Thymine dimers bend the DNA backbone. This prevents replication of DNA. The diagram shows the change in bonding caused by UV radiation.  thymine dimer A T C C T T G C T A G G A A C G (a) Name the bonds that are lost when thymine dimers are formed. (1) .................................................................................................................................................................................................................................................................................... (b) Explain why preventing the replication of DNA produces aseptic conditions. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 17 Turn over (c) Aseptic conditions are used when growing bacteria in laboratories. Justify two methods, other than UV radiation, for ensuring aseptic conditions. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (d) The effectiveness of UV radiation was investigated, using three species of pathogenic bacteria. The table provides information about the bacteria used in this investigation. Bacteria Gram staining Infected part of body P. aeruginosa negative blood S. aureus positive lung E. coli negative intestine (i) Compare and contrast the structure of the wall of Gram positive bacteria and Gram negative bacteria. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 18 (ii) In this investigation, living bacteria of each species were exposed to UV radiation for four hours. The numbers of living bacteria were counted each hour during the four-hour period. The graph shows the results of this investigation. Key P. aeruginosa S. aureus E. coli 0 1 2 3 4 Time / h Number of bacteria per cm3 108 109 107 106 105 104 103 102 101 0 Calculate the percentage change in numbers of living E. coli after two hours of exposure to UV radiation. (2) Answer .............................................................. % 19 Turn over (iii) A student concluded from this investigation that if food is exposed to UV radiation, it would remove all risk of food poisoning. Criticise the validity of this conclusion. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 14 marks)

20 7 The photograph shows a wood mouse, Apodemus sylvaticus. (Source: © Eric Isselee/Shutterstock) A scientist measured the population size of these mice living in a woodland habitat. The scientist used the capture-mark-recapture method (CMR) to sample the population. This is the CMR method used by the scientist: ● randomly place mammal traps in the woodland for one day ● put a coloured band on a leg of each captured mouse (S1) ● release these marked mice ● after one week place mammal traps in the woodland again for one day ● count the number of mice captured (S2) ● count the number of mice with leg bands in this second sample (L). The table shows some of the results. Number of mice captured S1 S2 200 250 21 Turn over (a) The scientist calculated that the number of mice in the woodland was 1000, using the formula number of mice = S S L 1 2 × Determine the number of mice that had leg bands (L) in the second sample. (1) Answer .............................................................. (b) The diagram shows a section of the trap used by the scientist. food wire trap door board entrance Explain how the design of the trap is effective for capturing mice. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22 (c) Wood mice are small mammals. They cannot maintain their body temperature and will die if left in the trap for a long period of time. Explain why wood mice cannot maintain their body temperature if left in the trap. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 23 Turn over (d) Comment on the claim that the CMR method produces an accurate measure of the population size of mice in the woodland. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 7 = 13 marks)

24 8 The electron transport chain generates ATP. The diagram shows the electron transport chain. carrier (oxidised) carrier (reduced) carrier (oxidised) reduced NAD water carrier (reduced) carrier (oxidised) carrier (reduced) oxidised NAD oxygen ATP ATP ATP (a) State the location of the electron transport chain in a mitochondrion. (1) .................................................................................................................................................................................................................................................................................... (b) Hydrogen cyanide gas is a poison that can kill an animal if it is inhaled. Cyanide is a non-competitive inhibitor of the enzyme cytochrome oxidase. Cytochrome oxidase is the last carrier in the electron transport chain. (i) Explain why hydrogen cyanide gas can kill an animal. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 25 Turn over (ii) A dose of 1.0 mg of hydrogen cyanide per kilogram of body mass is fatal. An animal with a body mass of 10 kg inhaled air containing 0.24 mg dm−3 of hydrogen cyanide. The animal breathed at a rate of 15 dm3 of air per minute. Calculate how many minutes it would take for this animal to obtain a fatal dose. Give your answer to three significant figures. (3) Answer .............................................................. minutes (c) The effect of cyanide on oxygen uptake by mitochondria was investigated using the following method: ● suspend mitochondria in a flask containing a buffer solution ● measure the concentration of oxygen in the flask for the first five minutes ● then add a respiratory substrate and ADP to the flask ● measure the concentration of oxygen in the flask for the next 10 minutes ● then add cyanide solution to the flask ● measure the concentration of oxygen in the flask for the next 10 minutes. The graph shows the results for the first five minutes. Complete the graph to show the results for the next 20 minutes. (2) 25 20 15 10 5 0 Time / minutes Oxygen concentration (Total for Question 8 = 11 marks)

26 9 Biodiversity is monitored by conservationists. Biodiversity can be calculated using the formula D N N n n = − − ∑ ( ) ( ) 1 1 (a) Give the symbol in this formula that represents the total number of organisms of all species. (1) .................................................................................................................................................................................................................................................................................... (b) The photograph shows a hedge. (Source: © 1000 Words/Shutterstock) Many fields in the UK are separated by hedges. Conservationists investigated biodiversity in hedges of different ages. Hedges of different ages were sampled at random and the number of species of woody plants was counted. The graph shows the results of this investigation. 1200 1000 800 600 400 200 0 10 9 8 7 6 5 4 3 2 1 0 Age of hedge / years Mean number of species of woody plants in 30 m of hedge 27 Turn over Analyse the data to explain why conservationists object to the destruction of hedgerows to increase the size of fields. 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(c) State why random sampling was used in this investigation. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 28 (d) The conservationists observed that the plants in older hedges contained more wood than the plants in younger hedges. Wood consists of xylem tissue. (i) Give an advantage for older plants of containing more wood. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Describe a method to show that the stem of an older plant contains more wood. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 29 Turn over (e) A student thought that the age of a plant affects the water potential of its tissue. The student used potatoes of different ages to test this hypothesis. Justify each step of a method the student could use to compare the water potential of potato tissue of different ages. 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(Total for Question 9 = 15 marks)

30 10 The diagram shows the flow of energy in an ecosystem. The numbers represent the energy in kJ m−2 yr−1 in each category. primary consumers 6.2 × 103 producers 8.5 × 104 sunlight 7.0 × 106 (a) (i) Calculate the percentage efficiency of energy transfer from sunlight to the producers. (2) Answer .............................................................. % (ii) Give one reason why the energy transfer from sunlight to the producers is less than 100% efficient. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 31 Turn over (iii) Explain the energy difference between the producers and the primary consumers. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Plant productivity is affected by the abiotic and biotic factors that exist at different latitudes. The diagram shows latitudes of the Earth. Northern hemisphere 0 Equator 30 30 60 60 90 90 30 30 60 60 90 90 North South The table shows the productivity of plants growing on land at different latitudes in the Northern hemisphere. Latitude / degrees Productivity / g m−2 yr−1 0 to 10 (equatorial conditions) 900 to 1500 30 to 40 (warm temperate conditions) 400 to 800 40 to 60 (cool temperate conditions) 200 to 600 60 to 70 (alpine to arctic conditions) 0 to 200 32 A student concluded that primary productivity is determined by abiotic factors that affect the light-dependent and light-independent stages of photosynthesis. Discuss the validity of this conclusion. 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(Total for Question 10 = 14 marks) TOTAL FOR PAPER = 120 MARKS 34 BLANK PAGE 35 BLANK PAGE 36 BLANK PAGE

4 2 The golden snub‑nosed monkey (Rhinopithecus roxellana) is endemic to high mountainous regions of China. This endangered species of monkey feeds on seeds. (a) State what is meant by the term endemic. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The population size of this species has decreased due to changes in its habitat. Climate change is one of the factors affecting the habitat of this species. (i) Which of the following causes global warming? (1) A a decrease in carbon dioxide and methane in the atmosphere that trap infrared radiation B a decrease in carbon dioxide and methane in the atmosphere that trap ultraviolet radiation C an increase in carbon dioxide and methane in the atmosphere that trap infrared radiation D an increase in carbon dioxide and methane in the atmosphere that trap ultraviolet radiation 5 Turn over (ii) Which of the following could provide evidence for climate change? (1) A dendrochronology, peat bog pollen analysis and ice cores B ice cores, classification and peat bog pollen analysis C niche, dendrochronology and classification D niche, dendrochronology and ice cores (c) Golden snub‑nosed monkeys live in areas with very cold winters. The seeds in their diet contain lipids and carbohydrates. Climate change is reducing seed production by plants in their habitat. Discuss the impact of climate change on the monkey population. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks)

6 3 Humans are surrounded by microorganisms in the air, water and food. Some microorganisms are pathogenic. The human body has several barriers to prevent infection by pathogens. (a) Complete the table by putting a tick (ü) in the box if the type of barrier is correct. If the type of barrier is not correct, place a cross (X) in the box. (2) Type of barrier Keratin in the skin Lysozyme in mucus Hydrochloric acid in the stomach Physical Chemical (b) Explain why the presence of microorganisms on the skin and in the gut helps to prevent pathogenic organisms multiplying in the body. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 7 Turn over (c) The human gut contains more than a thousand species of bacteria. Only 30 to 40 of these species are found in the stomach. Explain why there are relatively few species of bacteria in the stomach. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 7 marks)

8 4 Thalassaemia is a recessive genetic disorder that affects the production of haemoglobin. It is caused by a gene mutation. Scientists are developing methods to repair gene mutations such as the one that causes thalassaemia. (a) A gene mutation can be a change in a single base in the (1) A DNA that codes for a different amino acid B DNA that codes for a different monosaccharide C RNA that codes for a different amino acid D RNA that codes for a different monosaccharide (b) One of the most common mutations causing thalassaemia is the substitution of one adenine base with guanine. The diagram shows the location of the mutation in part of the DNA strand coding for four amino acids. substitute base C T G G G C A T A G A A Explain why this mutation affects the function of the haemoglobin molecule. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 9 Turn over (c) The RNA produced during transcription is known as pre‑mRNA. Pre‑mRNA can be modified before being translated on the ribosome. (i) Describe how the pre‑mRNA may be modified before being translated on the ribosome. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain why modification of pre‑mRNA enables one gene to give rise to more than one protein. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 8 marks)

10 5 Reproduction in mammals involves specialised cells called gametes. The production and structure of the gametes can be affected by chemicals in the environment. (a) (i) A sperm cell contains a specialised lysosome called the acrosome. Describe the role of the acrosome in fertilisation. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Following fertilisation a zygote is formed. The zygote then divides by (1) A meiosis to produce diploid cells B meiosis to produce haploid cells C mitosis to produce diploid cells D mitosis to produce haploid cells 11 Turn over (b) Human fertility may be affected by chemicals in the environment such as polychlorinated biphenyls (PCBs). It is thought PCBs affect the functioning of the mitochondria. (i) The diagram shows the structure of a sperm cell. P R Q S © QA International / Science Photo Library Which labelled part of the sperm cell contains mitochondria? (1) A P B Q C R D S 12 (ii) The table shows the effect of different concentrations of PCB on the speed of movement of sperm cells. PCB concentration / mg dm–3 Mean speed of sperm / μm s–1 0 37 1 35 5 29 25 28 Calculate the percentage decrease in the mean speed of sperm in 25 mg dm–3 of PCB compared with the control. (2) Answer .............................................................. (iii) Deduce how PCBs could affect the speed of movement of sperm. 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(Total for Question 5 = 9 marks)

13 Turn over 6 The photograph shows a glacier at the top of a mountain. Glaciers in many areas of the world are retreating (reducing in size). The line on the photograph shows the position of the front edge of the glacier in 1985. Front edge of glacier in 1985 Bare rock is exposed as the glacier retreats. Two hundred years ago, bare rock was exposed after a glacier retreated. A study has been made of the long-term changes in vegetation on that area of rock after the retreat of the glacier. 14 The graphs show the abundance of three species of tree since the bare rock was exposed as the glacier retreated. The abundance of each species was measured every ten years for 200 years. 1500 1200 900 600 300 0 20 40 60 80 100 120 140 160 180 200 Abundance / plants per hectare Time since bare rock exposed / years Alder 1500 1200 900 600 300 0 20 40 60 80 100 120 140 160 180 200 Abundance / plants per hectare Time since bare rock exposed / years Spruce 1500 1200 900 600 300 0 20 40 60 80 100 120 140 160 180 200 Abundance / plants per hectare Time since bare rock exposed / years Hemlock 15 Turn over (a) Comment on the changes in the abundance of these three species over 200 years. 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(b) Explain what happened to the bare rock to allow the growth of these trees. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16 (c) Plants such as lupin are often found in the early stages of glacial retreat. Lupin plants have nodules on their roots containing nitrogen‑fixing bacteria that convert atmospheric nitrogen to ammonium ions. Plants can use ammonium ions as a source of nitrogen. Explain why lupin plants are able to grow in the early stages of glacial retreat. 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(Total for Question 6 =10 marks)

17 Turn over 7 There are two main types of woodland in the UK. One consists of broadleaved deciduous trees, the other contains coniferous trees that have needle‑like leaves. (a) The table shows some of the species of plant growing underneath the trees in a deciduous woodland. Species Number of individuals of each species (n) n(n – 1) Woodrush 2 Holly 8 Bramble 1 Yorkshire fog 9 Sedge 3 (i) Complete the table to show n(n – 1) for each species of plant. (1) (ii) Calculate the diversity index (D) using the data in the table. D = Σ − − N(N 1) n(n 1) (2) Answer .............................................................. 18 (b) Another measure of diversity is species richness. (i) Species richness is the number of (1) A different alleles in one species B different species in one area C individuals of one species in one area D individuals of one species in different habitats (ii) Devise an investigation to compare species richness of the plants growing under the trees in a deciduous woodland with that of a coniferous woodland. 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The table shows the other data recorded in this study. Factor Deciduous woodland Coniferous woodland Mean light intensity 10 cm above soil level / lux 1075 543 Mean pH of soil 7 5 Mean air temperature 10 cm above soil level / °C 18 18 Mean soil moisture content (%) 45 47 Assess the reasons for the differences in the number of species between these two types of woodland. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 13 marks)

20 8 The photograph shows an orangutan. These animals inhabit the forests of Borneo and Sumatra. They are critically endangered, largely due to habitat loss and hunting. Managed breeding programmes in zoos are an important factor in the conservation of the orangutan. (a) Explain how breeding programmes in zoos maintain the genetic diversity of captive populations. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 21 Turn over (b) The orangutan population in Borneo decreased by 60% between 1950 and 2010. This has been linked to loss of habitat. Deforestation has resulted in the loss of natural habitat. The graph shows the area of forest loss in Borneo from 2001 to 2012. 1400 1200 1000 800 600 400 200 0 2000 Year Forest loss / thousand ha 2002 2004 2006 2008 2010 2012 2014 (i) Calculate the mean rate at which forest was lost between 2010 and 2012. (2) Answer .............................................................. 22 *(ii) Land use has changed over time in Borneo, with areas of forest being removed due to logging. The land may then be used for planting oil palms, which are not a suitable habitat for orangutans. The following comments refer to other issues concerning the conservation of orangutans in Borneo. ● Over the past 100 years, 80% of suitable habitat for orangutans has been lost. ● Only 2% of the remaining forest suitable as orangutan habitat is protected. ● From 1999 to 2015, the orangutan population decreased by 50%. ● Captive‑bred orangutans may spread human diseases into wild populations. ● When captive‑bred orangutans are released, their offspring often die young because the females have not learnt parenting behaviour from older individuals. ● Large numbers of orangutans are hunted in Borneo, even though they are a protected animal. It has been stated that habitat protection is more important than captive breeding programmes for the conservation of orangutans in Borneo. Evaluate this statement. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 8 = 11 marks)

24 9 Ebola virus disease (EVD) is a rare and deadly disease most commonly found in Africa. Following a severe outbreak in 2014, in which 11 000 people died, work has been underway to develop a vaccine. The diagram shows the structure of an Ebola virus. RNA strand capsid glycoprotein { (a) Compare and contrast the structure of Ebola virus with that of the human immunodeficiency virus (HIV). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 25 Turn over (b) A vaccine has been developed by genetically modifying a virus that infects cattle. In the genetically modified virus, one of the genes was replaced with a gene for a protein found in the Ebola virus. In a trial of 52 volunteers, 48 developed antibodies against the Ebola virus within 14 days of injection. (i) The type of immunity given by this vaccine is (1) A artificial active immunity B artificial passive immunity C natural active immunity D natural passive immunity (ii) Explain the role of T cells in the immunity to the Ebola virus that develops following the use of this vaccine. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26 (c) The vaccine is still undergoing clinical trials, but was given approval for use in recent Ebola outbreaks. (i) Describe how clinical trials of a vaccine would be conducted. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 27 Turn over (ii) The Ebola vaccine was given to health workers and immediate family of those with the disease. Justify the use of this vaccine, even though the clinical trials had not been completed. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 14 marks)

28 10 Farmers grow wheat plants to produce grain that is used to make flour. Farmers can use modern technology to monitor the mass and the moisture content of the grain in a field. Wheat can be sold at a higher price when it has a lower moisture content. The table shows some data recorded from a field of wheat. Plot number Moisture content of grains (%) Mass of wheat grains / tonnes per Ha 1 0.57 0.759 2 0.58 0.523 3 0.67 4.243 4 0.94 1.954 5 1.64 1.417 6 1.93 1.170 7 2.27 1.321 8 2.53 6.289 9 4.96 3.809 10 10.36 3.404 mean 2.489 (a) (i) Which of the following describes the relationship between the mass of wheat grain and the moisture content of the wheat grain? (1) A inverse correlation B negative correlation C no correlation D positive correlation (ii) Calculate the standard deviation for the mass of wheat using the formula s = Σ − − 2 ( ) 1 x x n (3) Answer .............................................................. 29 Turn over (b) Farmers use fertilisers to increase the growth of crops such as wheat. (i) Fertilisers contain phosphate ion compounds. Plants need phosphate in order to synthesise (1) A both cellulose and phospholipids B both phospholipids and polysaccharides C both polynucleotides and phospholipids D both polynucleotides and polysaccharides (ii) After the grain is harvested, farmers plough the stems (straw) from wheat plants into the soil. This improves the quality of the soil. These stems contain polysaccharides. Explain how microorganisms in the soil break down the stems. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 30 *(c) Biological farming aims to reduce the use of fertilisers to add minerals to the soil. The following is some information on biological farming. ● Biological farming makes use of products containing fungi that can be added to soil to improve soil life. ● Soil life is essential for converting nutrients into a plant‑available form. ● Fungi form a network around the roots of plants to trap water and aid absorption of mineral ions. ● Some fungi provide protection from pathogens. Devise a procedure, using wheat seedlings, to compare the effects of a biological farming product with the use of a chemical fertiliser on the growth of plants. 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(Total for Question 10 = 14 marks) TOTAL FOR PAPER = 100 MARKS 32 BLANK PAGE

4 2 A spirometer can be used to study the performance of an athlete. snabbiology.wordpress.com The trace produced by a spirometer can be used to determine the respiratory minute ventilation and the oxygen consumption of an individual. (a) Describe how a spirometer trace can be used to calculate the respiratory minute ventilation and the oxygen consumption per minute. (4) respiratory minute ventilation .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... oxygen consumption per minute .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 5 Turn over (b) A student compared the spirometer trace for a pair of healthy, genetically identical twins. State two variables that would have to be controlled to make this a valid comparison. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 6 marks)

6 3 Phytochromes and IAA (indole acetic acid) are important substances that bring about growth responses in plants. (a) The diagram shows an oat seedling in part of a field. coleoptile of oat seedling oat seed surface of soil in the field Cells in the tip of the oat coleoptile release IAA. Explain how the IAA affects the growth of the coleoptile. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 7 Turn over (b) Phytochrome is found in two different forms known as PR (or P660) and PFR (or P730). (i) Which row correctly describes the effects of sunlight on phytochrome? (1) Change in form of phytochrome Speed of change A PFR to PR slow B PFR to PR rapid C PR to PFR slow D PR to PFR rapid (ii) Phytochrome can be described as (1) A a form of opsin B a photosensitive pigment C an isomer of retinal D a type of cytochrome (iii) Give one example of a growth response of a plant that is affected by phytochrome. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 7 marks)

8 4 A number of seed banks have been set up around the world. (a) When a seed bank receives new seeds, it processes them in several ways. The processes can include the following stages: • washing the seeds with disinfectant • allowing the seeds to dry. Explain the advantages of these two stages. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The food store of many seeds is an organ called a cotyledon. An organ can be defined as (1) A a group of cells of one type that has a function B a group of several different cell types, each type working together C a group of tissues that are located together that have unrelated functions D part of a system and therefore does not have a function of its own 9 Turn over (c) The processed seeds are stored at minus 20 °C. At intervals, samples of seeds are removed from storage and tested for viability. If at least 75% of the seeds in the sample germinate, the remaining seeds are described as viable. The diagram shows a container with many seeds in it. Sunflower Seeds mass of 1000 seeds = 50 g A sample of 3 g of seeds was removed from this container. These seeds were given optimum conditions for germination. However, only 48 seeds germinated. Determine whether the remaining seeds in the container are viable or not. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 10 (d) Glycolysis occurs during germination. One of the products is adenosine triphosphate (ATP). Which of the following is another product of glycolysis? (1) A carbon dioxide B glucose C oxygen D pyruvate (Total for Question 4 = 8 marks)

11 Turn over 5 An investigation was carried out to study the effect of alcohol concentration on heart rate in Daphnia. (a) Give one reason why some people believe that there are fewer ethical issues when using Daphnia for this investigation rather than using a species of mammal. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) In this investigation, Daphnia were placed into three groups of 10. Each group was placed in alcohol of a different concentration for five minutes. After five minutes, each Daphnia was removed from the alcohol and its heart rate recorded every 15 seconds for one minute. A mean for each concentration was calculated. The results are shown in the table. Alcohol concentration / mol dm−3 Mean Daphnia heart rate / beats per minute 0.00 221 0.17 176 0.34 97 (i) Give one reason why 10 Daphnia were used for each alcohol concentration. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain why the Daphnia were left for five minutes in the alcohol before the heart rate was recorded. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 12 (c) Devise an investigation to find the lowest concentration of alcohol that has an effect on the heart rate of Daphnia. 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(Total for Question 5 = 9 marks)

13 Turn over 6 Wasps are insects that live in groups. One species of wasp (Vespula germanica) has been shown to knock its body repeatedly against a hard surface. This signals the presence and quality of food to other wasps. When threatened by another animal, it may use its stinger to inject a venom to protect itself. bugguide.net (a) Complete the table by giving the name of the type of adaptation. (2) Description of adaptation Type of adaptation shown by the wasp knocking its body to signal food the stinger 14 (b) The stinger injects a venom that contains the enzyme phospholipase. This enzyme hydrolyses phospholipids, releasing fatty acids. Describe how the structure of phospholipase allows it to hydrolyse phospholipids. 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(c) Phospholipase in the venom of wasps can cause allergic reactions. Phospholipase can affect the Golgi apparatus in cells. Phospholipase inhibitors are being investigated as possible drugs to treat the allergic reactions. (i) Give two functions of the Golgi apparatus. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 15 Turn over (ii) A three-phase protocol will be used when developing the phospholipase inhibitor as a new drug. Explain the purpose of each phase of this protocol for a phospholipase inhibitor. 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(Total for Question 6 = 11 marks)

16 7 When scientists visit Antarctica, they need appropriate clothing to help with thermoregulation. www.aa.com (a) An investigation was carried out to study thermoregulation in humans. A woman was wrapped in blankets and her feet were put in hot water for 30 minutes. During this time, the temperature of the skin at the end of one of her fingertips was recorded. The graph shows the results of this investigation. 35 33 31 29 27 25 23 21 19 17 0 0 5 10 15 20 25 30 Time / minutes Fingertip temperature / °C 17 Turn over (i) Explain why there was no change in fingertip temperature between 0 and 3 minutes. 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(ii) Calculate the greatest rate of increase in fingertip temperature. (2) Answer ............................................................................................. 18 (iii) Explain the role of the nervous system in bringing about the increase in temperature of the fingertip as shown in this investigation. 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(b) Sweating is a thermoregulatory mechanism. A student stated that loss of heat when sweating is related to the dipole nature of water molecules. Justify this statement. 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(Total for Question 7 = 12 marks)

19 Turn over 8 There are many reasons why humans may lose muscle mass. Two known causes are hip replacement surgery and some genetically inherited conditions. (a) A muscle fibre is a specialised body cell. Explain how the structure of a muscle fibre is related to its specialised function. 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(b) Muscles can be scanned using magnetic resonance imaging (MRI) to investigate the loss of muscle mass. It is safer to use MRI than computed tomography (CT) because (1) A CT uses X-rays that can cause mutations in the DNA of muscle fibres B CT uses X-rays that can cause mutations in the protein in the muscle fibres C CT uses magnets that can cause mutations in the DNA of muscle fibres D CT uses magnets that can cause mutations in the protein in the muscle fibres 20 (c) Muscular dystrophy (MD) is a range of genetically inherited conditions in which a loss of muscle mass can occur. Limb-girdle muscular dystrophy (LGMD2A) is a rare type of MD. Only six people per million have this recessive condition. In 2018 there were 66.5 million people in the UK. (i) Calculate the number of people in the UK who are carriers of LGMD2A using the Hardy-Weinberg equation. (3) p2 + 2pq + q2 = 1 Answer ............................................................... (ii) In the human population, the frequency of the allele for LGMD2A may change from one generation to the next. Explain why the frequency of this allele may change from one generation to the next. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 21 Turn over (d) Duchenne muscular dystrophy (DMD) is the most common inherited type of MD. DMD is caused by a recessive allele. The table shows some data on DMD from 1993 to 1999. Year Number of deaths due to DMD per year in England and Wales Median age of death / years males females 1993 75 0 19 1994 37 2 19 1995 41 0 20 1996 53 0 19 1997 43 1 20 1998 51 0 21 1999 51 1 20 Analyse the data to explain why DMD causes more deaths in males than in females. 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(Total for Question 8 = 14 marks) 22 BLANK PAGE

23 Turn over 9 Nicotine is a drug found in the smoke of cigarettes. (a) The flow diagram shows how the presence of nicotine can cause the release of noradrenaline. nicotine binds to acetylcholine receptors on the post‑synaptic membrane of a neurone action potential travels along the axon of the post‑synaptic neurone increased uptake of calcium ions by this neurone this neurone then releases noradrenaline (i) Explain how nicotine causes an action potential in the post-synaptic neurone that releases noradrenaline. 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(ii) State how an increase in calcium ion uptake by the neurone leads to the release of noradrenaline. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 24 (b) The effect of inhaling nicotine on the circulatory system of rats was investigated. In this investigation, three variables were considered: • the concentration of nicotine in blood plasma • the diameter of the lumen of one artery • blood pressure Two groups of rats were treated as shown in the table. Group Number of rats in group Nicotine dose / mg A 6 1.0 B 6 0.1 (i) The concentration of nicotine in the blood plasma of the group A rats was recorded at different times and the means calculated. The means are shown in the table along with the range of data for each mean. Time of sampling / minutes Mean concentration of nicotine in blood plasma / ng cm−3 0 (immediately after inhalation) 35.0 ± 9.3 30 (after inhalation) 24.1 ± 5.6 Determine the maximum rate of decrease in the concentration of nicotine in the blood plasma per minute after being given the nicotine. (2) Answer ............................................................................................. ng cm−3 min−1 25 Turn over (ii) The diameter of the lumen of one artery, in each of the 12 rats, was measured when the rats were resting. The blood pressure of each rat was also measured and the mean blood pressure calculated. The diameter of the lumen of the artery and the blood pressure of each rat were then recorded at intervals, for a total of 30 minutes. The rats inhaled nicotine for the first minute (0.0 to 1.0). The table shows the results for the mean diameter of the lumen of the artery. Time / minutes Mean diameter of lumen / µm Group A (1.0 mg nicotine) Group B (0.1 mg nicotine) 0.0 48 48 0.5 44 44 1.0 49 45 2.0 52 48 5.0 57 52 10.0 55 48 15.0 49 48 30.0 49 48 Determine the effect of nicotine concentration on the percentage change in lumen diameter in the first minute. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26 *(iii) The graph shows the results for the mean blood pressure for the two groups of rats. Nicotine was inhaled for the first minute. 160 150 140 130 120 110 100 0 5 10 15 20 25 30 Time / minutes Mean blood pressure / mmHg Group A Group B It has been stated that: ‘Nicotine gained from smoking cigarettes in humans causes an increase in blood pressure and a decrease in the lumen of arteries.’ Analyse all the data from this investigation using rats to evaluate the validity of this statement. 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28 10 Respiration occurs in all healthy living cells. (a) The diagram shows part of the Krebs cycle and the electron transport chain. two CO2 citrate acetyl CoA succinate reduced NAD reduced FAD electron transport chain (i) Which row shows the number of carbon atoms in citrate and succinate? (1) Number of carbon atoms in citrate succinate A 2 4 B 5 4 C 6 4 D 6 8 (ii) Which of the following is transferred to a molecule of FAD to form reduced FAD? (1) A two oxygen atoms B two hydrogen atoms C one oxygen atom and one hydrogen atom D one water molecule 29 Turn over (iii) Explain the need for reduced NAD to be oxidised in a mitochondrion. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iv) A mutation in the gene that codes for the enzyme succinate dehydrogenase stops the conversion of succinate into citrate. Which row states the change in concentration of citrate and reduced FAD as a result of this mutation? (1) Concentration of citrate Concentration of reduced FAD A decreases decreases B decreases increases C increases decreases D increases increases 30 (b) There is a link between the methylation of certain regions of DNA and the risk of developing cancer. The graph shows the relationship between the level of methylation of these regions of DNA and the risk of developing cancer in different parts of the body by the age of 70. Level of DNA methylation / a.u. Risk of developing cancer 0.13 0.15 0.17 0.19 0.21 0.23 lung pancreas bladder liver breast colon cerebrum (i) Draw a line of best fit on the graph to identify any correlation between the independent variable and the dependent variable. (1) (ii) An investigation studied the effect of age on the mean level of DNA methylation. In this investigation, the null hypothesis was rejected at the 5% significance level. Explain what is meant by the phrase: the null hypothesis was rejected at the 5% significance level for this investigation. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 31 *(iii) Scientists have found that a high level of succinate leads to the inhibition of an enzyme called TET. The function of TET is to remove methyl groups from DNA. Explain how a mutation in the gene for succinate dehydrogenase can increase the risk of developing cancer. 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(Total for Question 10 = 14 marks) TOTAL FOR PAPER = 100 MARKS 32 BLANK PAGE

4 2 Mineral ions are needed for plant growth. A student had a plant that had been kept in the same soil for a long time. The growth of the plant had slowed down and the leaves were pale green. The student decided to add fertiliser to the plant and measure the increase in height of the plant. The student made two predictions: • the leaves would gradually become greener • the plant would not start growing until the leaves were greener. (a) Explain why the student made these predictions. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The student used a stock fertiliser solution at a concentration of 500 µg cm−3. Calculate the volumes of stock fertiliser solution and water that had to be mixed together to produce 100 cm3 of fertiliser solution at a concentration of 50 µg cm−3. (1) Volume of stock fertiliser solution .............................................................. cm3 Volume of water .............................................................. cm3 Turn over 5 (c) When the fertiliser was added to the plant, the height of the plant was 130 cm. Six weeks after the fertiliser was added, the plant had grown to a height of 155 cm. Calculate the mean growth rate for this plant, in centimetres per day, in this six-week period. (1) Answer .............................................................. cm day−1 (Total for Question 2 = 5 marks)

6 3 Milk from cows contains a number of different types of lipid. Lipids are made of glycerol and fatty acids. (a) A triglyceride is one type of lipid. Which row of the table describes a triglyceride? (1) Number of molecules Name of bond joining molecules Glycerol Fatty acid A 1 3 glycosidic B 1 3 ester C 3 1 glycosidic D 3 1 ester (b) Fatty acids are either saturated or unsaturated. The diagram shows the structure of an unsaturated fatty acid. HO O C C C C C C C C C C C C C C C C C C H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H Which row of the table shows the differences between this fatty acid and a saturated fatty acid with the same number of carbons? (1) Number of double bonds in the saturated fatty acid Number of hydrogen atoms in the saturated fatty acid A 0 35 B 0 36 C 1 35 D 1 36 Turn over 7 (c) The table shows the composition of some of the different types of lipid in milk from cows. Type of lipid Composition of lipids in milk from cows (%) monoacylglycerols 0.027 diacylglycerols 0.360 triacyglycerols 97.500 free fatty acids 0.027 phospholipids 0.600 Which row of the table shows the types of lipids that are the median and mode for this set of data? (1) Median Mode A diacylglycerols monoacylglycerols B diacylglycerols triacylglycerols C monoacylglycerols diacylglycerols D monoacylglycerols triacylglycerols (d) Phospholipids have a different structure from other lipids, such as triglycerides. Give one difference in the properties of phospholipids and triglycerides. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 8 (e) Milk from cows also contains a solution of proteins, lactose and mineral ions. The lipids in milk are carried inside milk fat globules (MFG). The diagram shows an MFG. lipids carried inside the MFG outer part of MFG consisting of three layers of phospholipid (i) Complete the diagram to show how nine phospholipid molecules would be arranged in the outer part of an MFG. Use this shape to draw each phospholipid: (2) inside the MFG outside the MFG outer part of MFG consisting of three layers of phospholipids Turn over 9 (ii) Explain the structure of an MFG. 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(Total for Question 3 = 9 marks) 10 BLANK PAGE

Turn over 11 4 Insulin is a protein that is composed of two polypeptide chains, chain A and chain B. The diagram shows the amino acid sequence of these two polypeptide chains and some of the bonds involved in the folding of this protein. Thr Thr Thr Lys Pro Tyr Tyr Tyr Phe Phe Gly Arg Glu Glu Glu Glu Gly Gly Gly Cys Cys Cys Cys Cys Cys Val Val Val Val Leu Leu Leu Leu Leu Leu Tyr Ala His His Ser Ser Ser Gln Gln Gln Asn Phe Asn Ile Ile Asn 1 2 3 4 5 Chain B Chain A Bond C (a) Cysteine (Cys) is one type of amino acid found in insulin. The presence of cysteine in insulin is important in determining the structure of this protein. The R group of cysteine is: H C H S H (i) Draw a diagram to show the structure of the amino acid cysteine. (3) 12 (ii) Name the covalent bond labelled C in the diagram. (1) .................................................................................................................................................................................................................................................................................... (b) The table shows the DNA genetic codes on the anti-sense DNA strand (template strand) for amino acids. Genetic code Amino acid Genetic code Amino acid Genetic code Amino acid Genetic code Amino acid AAA AAG Lys CAA CAG Gln GAA GAG Glu TAC TAT Tyr AAC AAT Asn CAT CAC His GAC GAT Asn TCA TCC TCG TCT Ser ACA ACC ACG ACT Thr CCA CCC CCG CCT Pro GCA GCC GCG GCT Ala TGG Try AGA AGG Arg CGA CGC CGG CGT Arg GGA GGC GGG GGT Gly TGC TGT Cys AGC AGT Ser CTA CTC CTG CTT Leu GTA GTC GTG GTT Val TTA TTG Leu ATA ATC ATT Ile TTC TTT Phe ATG Met The genetic codes TAA, TAG and TGA are stop codons. (i) Complete the diagram to show one base sequence in the antisense DNA strand (template strand) coding for the five amino acids numbered in the diagram of insulin. (1) Amino acid number 1 2 3 4 5 Amino acid abbreviation Asn Cys Tyr Asn Glu Base sequence on the antisense DNA strand TGT Turn over 13 *(ii) Analyse the information to deduce how gene mutations in the DNA base sequence for Cys (amino acid number 2) could affect the structure of insulin. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 11 marks)

14 5 Glandular fever is caused by infection with the Epstein–Barr virus (EBV). The virus infects B lymphocytes and epithelial cells. On infection of these cells, the virus enters the lytic cycle. After the lytic cycle is brought under control by the immune system of the body, latency takes place. (a) The diagram shows the structure of EBV. polyhedral capsid glycoproteins envelope DNA Source from: https://www.hindawi.com/journals/jir/2012/370516/ (i) Which virus has both a polyhedral capsid and an envelope? (1) A Ebola B human immunodeficiency virus C λ (lambda) phage D tobacco mosaic virus (ii) State what the genes in the DNA of EBV code for, other than glycoproteins. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 15 (iii) Describe the role of the glycoproteins in EBV. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Complete the graph to show the shape of the growth curve of EBV as a result of one lytic cycle after infection of a cell. (2) Number of viruses in the cell Time after infection number of viruses on infection 16 (c) An investigation studied a group of 103 children infected with EBV, in one town. The number of children with typical symptoms of EBV infection and the number of children with non-typical symptoms of EBV infection were recorded. The graph shows the results of this investigation. 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Age / years Number of children with EBV infection Key non-typical symptoms typical symptoms (i) There were 47% boys in this group of children. Calculate the number of girls in this group of children. (2) Answer .............................................................. Turn over 17 (ii) Determine the effect of age on the proportion of children with non-typical symptoms of EBV. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Describe how this investigation could be improved. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 13 marks)

18 6 The diagram shows four stages in a human cell cycle. G2 of interphase S phase of interphase G1 of interphase mitosis and cytokinesis (a) The cell cycle time for this cell is 24 hours. Give an estimate for the length of time that this cell spends in each of the four stages of the cell cycle. (2) Stage of cell cycle Estimate for the length of time that this cell spends in each stage / hours mitosis and cytokinesis G1 of interphase S phase of interphase G2 of interphase (b) The percentage of cells in any one stage is proportional to the length of time of that stage. Describe an investigation that could be carried out to confirm the length of time of mitosis and cytokinesis for these human cells. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 19 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) A cell in G1 of interphase contains 68 arbitrary units of DNA. Explain how the DNA content and the structure of the chromosomes of this cell will change during the cell cycle. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 11 marks)

20 7 The photograph shows sweet red peppers growing on a plant. © Janet Horton / Alamy Stock Photo Unripe sweet red peppers contain the enzyme catalase. As these peppers ripen, the activity of catalase decreases. The equation summarises the reaction catalysed by catalase. hydrogen peroxide water + oxygen (a) The diagram shows the structure of catalase. one of four protein subunits one of four haem groups Turn over 21 Explain why the enzyme catalase is described as having a tertiary and quaternary structure. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22 (b) The activity of catalase can be measured by collecting the oxygen produced during the reaction. The diagram shows apparatus that can be used to collect the oxygen produced. oxygen collected by displacing water in a test tube tank of water flask containing reaction mixture (i) When measuring the activity of enzymes, the initial rate of reaction should be measured. Explain why the initial rate of reaction should be measured. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 23 *(ii) Devise an investigation, using this apparatus, to show that as sweet red peppers ripen the activity of catalase decreases. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 11 marks)

24 8 In mammals, gametes are produced by spermatogenesis and oogenesis. (a) The diagram shows some of the stages in spermatogenesis. Stage 1 Stage 2 Stage 3 Stage 4 Spermatogonium Spermatid Development Spermatozoa Meiosis I Primary spermatocyte Mitosis Secondary spermatocyte Meiosis II 2n 2n n n 2n n n n n (i) Explain the significance of mitosis in stage 1 of spermatogenesis. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 25 (ii) Explain the events that take place in stage 2, that result in genetic variation. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Compare and contrast the products of stage 2 and stage 3 in spermatogenesis with the products from these stages in oogenesis. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26 (iv) Explain the importance of the acrosome that develops during stage 4. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 27 (b) The effect of storage time on sperm was investigated. The table shows some results of this investigation. Storage time / hours Percentage of sperm with structural defects (%) Percentage of sperm that could swim (%) 0 8.2 ± 3.9 85.0 ±   5.7 6 9.5 ± 3.1 67.5 ± 11.0 12 18.0 ± 3.9 66.3 ±   7.5 18 21.3 ± 6.1 67.5 ±   2.8 24 19.5 ± 3.1 58.8 ±   8.5 30 26.5 ± 3.1 41.3 ±   7.5 Analyse the data to comment on conclusions that can be made from this investigation. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 15 marks) TOTAL FOR PAPER = 80 MARKS 28 BLANK PAGE

4 2 Water can move through plant cells. (a) The diagram shows the cells in a plant root. epidermal cell root hair Casparian strip xylem vessel (Source from: https://cronodon.com/BioTech/Plant_Transport.html) (i) On the diagram, draw the symplastic pathway that water takes from the soil to the xylem vessel. (2) (ii) The length of the root hair cell in the diagram is 60 mm. The magnification of the diagram is ×100. Calculate the actual length of the root hair cell in micrometres. (2) Answer .............................................................. µm Turn over 5 (b) Explain how the cohesion-tension model accounts for the transport of water from plant roots to leaves. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 8 marks)

6 3 The diagram shows a type of potometer. graduated pipette rubber bung shoot water syringe (Source from: Science and Plants for Schools: www.saps.org.uk) The leaves of the shoot must be kept dry when assembling the potometer under water in a sink. The syringe plunger is pushed down until the water level is raised to zero in the graduated pipette. The water level in the graduated pipette is used to measure water uptake by the shoot. (a) (i) Explain why the apparatus should be assembled under water. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain why the leaves must be kept dry whilst assembling the apparatus. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 7 (b) (i) Explain how this potometer can be used to investigate the effect of light intensity on the rate of water uptake by this shoot. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) This potometer measures the uptake of water by a plant shoot. Give two reasons why the water lost by transpiration is less than the water taken up by the plant shoot. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 10 marks) 8 BLANK PAGE

Turn over 9 4 Substances move into cells by several processes. (a) (i) Which of these processes requires energy from respiration? (1) A diffusion B endocytosis C facilitated diffusion D osmosis (ii) Which of these processes can occur against a concentration gradient? (1) A active transport B diffusion C facilitated diffusion D osmosis (iii) The graph shows the effect of increasing oxygen concentration on the rate of uptake of a substance other than oxygen. Rate of uptake Oxygen concentration Which of these processes is shown by the graph? (1) A active transport B diffusion C facilitated diffusion D osmosis 10 (iv) Which substance can enter a cell by diffusion? (1) A amino acid B carbon dioxide C glucose D protein (b) When plant cells absorb water, the water potential of the cells changes. The water potential of potato cells was investigated. The graph shows the percentage change in mean mass of potato samples after immersion in sucrose solutions of different concentrations. 30 20 10 0 −10 −20 1.2 1.0 0.8 0.6 0.4 0.2 0 X Sucrose concentration / mol dm−3 Percentage change in mean mass of potato samples (%) (i) Draw a suitable line on this graph. (1) (ii) Give the sucrose concentration that has the same water potential as the cells in these potato samples. (1) Answer .............................................................. mol dm−3 Turn over 11 (iii) The water potential (ψ) of plant cells can be calculated using the equation: ψ = P + π Explain the changes in the water potential of cell (ψ), turgor pressure (P) and osmotic potential (π), at point X on the graph. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iv) Explain why it is more difficult to determine the osmotic potential (π) than the water potential (ψ) of plant cells. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 12 marks)

12 5 The structure of a frog heart is different from those of mammals and fish. The diagram shows the structure of a frog heart. The arrows show the direction of blood flow. A B (a) (i) Name the parts of the heart labelled A and B. (2) A................................................................................................................................................................................................................................................................................ B................................................................................................................................................................................................................................................................................. (ii) Compare and contrast the structure of the frog heart with the structure of the mammalian heart. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 13 (b) Amphibians, such as the frog: • absorb oxygen through their lungs and through their moist skin • have a lower metabolic rate than mammals • have a double circulation system. (i) State one difference between the double circulation system of the frog and that of the mammal. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain how the circulatory system of the frog is adapted for the metabolism and gas exchange of the frog. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 14 (c) Describe the sequence of stimulation that occurs in a mammalian heart in one cardiac cycle. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 14 marks)

Turn over 15 6 The effect of gas composition on the breathing rate of locusts was investigated. The method used was: Step 1 A locust was placed in a 20 cm3 clear plastic syringe, as shown in the diagram. The piston was inserted so it gently held the locust in place so that the locust had no room to move. plastic tubing 20 cm3 plastic syringe piston Step 2 The number of pumping movements of the abdomen which occurred in 30 seconds was counted. This was repeated three more times. Step 3 Gases of different compositions were added to the syringe through the plastic tubing. Step 4 For each different composition of gases, the number of pumping movements of the abdomen which occurred in 30 seconds was counted. Step 5 Between each composition of gases used, the piston was moved in and out ten times to replace the exhaled air with laboratory air. Step 6 The locust was then left for 5 minutes before the next composition of gases was added. (a) (i) State the purpose of Step 5. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) State the purpose of Step 6. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16 (b) The table shows the results of the investigation. Gas composition (%) Number of movements in 30 seconds Mean Range SD oxygen carbon dioxide 21 0.1 26 22 31 29 27 9 3.9 16 4.1 41 44 46 49 45 8 3.4 100 0.0 5 12 6 8 8 7 3.1 94 6.0 48 42 46 49 46 7 3.1 88 12.0 44 46 50 53 48 9 83 17.0 54 51 53 49 52 5 2.2 (i) Calculate the standard deviation for the number of movements in 30 seconds when the gas has a composition of 88% oxygen and 12% carbon dioxide. Use the formula s n = − − ∑( ) x x 2 1 (3) Answer .............................................................. Turn over 17 (ii) Explain the relationship between the range and the standard deviation. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) Analyse the data to comment on the effect gas composition has on the breathing rate of this locust. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 11 marks)

18 7 Haemoglobin transports oxygen within the blood. The graph shows the relationship between the partial pressure of oxygen and the saturation of haemoglobin with oxygen at three partial pressures of carbon dioxide. 2 4 6 8 10 Partial pressure of oxygen / kPa PCO2 low PCO2 high PCO2 medium Saturation (%) 100 80 60 40 20 0 0 (a) (i) The curve moves to the right when more carbon dioxide is present. This move to the right is due to the (1) A Bohr effect B chloride shift C dissociation curve D oxygen debt Turn over 19 (ii) Explain why the curves are S-shaped. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Calculate the difference between the percentage change in saturation of haemoglobin with oxygen as the partial pressure of oxygen changes from 8 kPa to 2 kPa for low carbon dioxide compared with high carbon dioxide. (2) Difference .............................................................. % (iv) Analyse the data to explain the advantage to the mammal of the curve moving to the right. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 20 (b) Mammals that live at high altitudes are in an environment with a lower partial pressure of oxygen. Explain the position of the haemoglobin dissociation curve for mammals that live at high altitude. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 10 marks) Turn over 21 BLANK PAGE

22 8 The photograph shows a fruit fly, Drosophila. (Source: © Roblan/Shutterstock) Species of Drosophila live in a range of habitats in different parts of the world. The geographic distribution of these species ranges from regions with low temperatures to regions with high temperatures. The graph shows the effect of temperature on the percentage fertility of male flies of two species of Drosophila, D. simulans and D. melanogaster. 100 90 80 70 60 50 40 30 20 10 0 10 11 12 13 14 15 16 Growth temperature / °C Percentage fertility (%) D. simulans D. melanogaster Turn over 23 (a) The effect of low temperature on fertility is often compared using the temperature that results in 50% fertility. Determine the difference in these temperatures for the two species. (1) Temperature difference .............................................................. (b) Scientists also collected data from the two species about the effect of both high and low temperature on the viability of eggs and fertility of male flies. An egg is viable if it develops into an adult. The table shows the effects of temperature on the viability of eggs and fertility of males, for the two species of Drosophila. Species Lowest temperature / °C Highest temperature / °C at which eggs are viable at which males are fertile at which eggs are viable at which males are fertile D. melanogaster 10 12 32 30 D. simulans 10 11 31 28 (i) Give the relationship between temperature and the viability of eggs and fertility of males. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 24 *(ii) Another group of scientists recorded the courtship song of male D. melanogaster and male D. simulans. The male flies vibrate their wings producing a courtship song to attract females. The courtship song is produced in pulses. The interval between pulses and the duration of each pulse are shown in the table. Species Mean interval between pulses / ms Duration of pulse / s D. melanogaster 34.1 1.6 D. simulans 48.9 3.4 Discuss how the data provides evidence for the geographical distribution and type of speciation of the two species of Drosophila. Use the data from tables for temperature and for courtship song. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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.................................................................................................................................................................................................................................................................................... 25 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 8 = 8 marks) TOTAL FOR PAPER = 80 MARKS 26 BLANK PAGE 27 BLANK PAGE 28 BLANK PAGE

4 2 The diagram shows a blood vessel of a person with cardiovascular disease (CVD). R Q P T plaque S X Y © Universal Images Group North America LLC/Alamy Stock Photo (a) (i) Name the type of blood vessel shown in this diagram. (1) ..................................................................................................................................................................................................................................................................................... (ii) Which labelled part of the blood vessel contains tissue that can contract to maintain blood pressure? (1) A P B Q C S D T Turn over 5 (iii) Which labelled parts of the blood vessel contain collagen? (1) A Q B Q and T C Q, S and T D Q, R, S and T (iv) Which of the following shows the cross-sectional area of the lumen of the blood vessel in the diagram? Use the line XY to help you. (1) A 2.01 cm2 B 2.51 cm2 C 5.03 cm2 D 8.04 cm2 (b) Diet is one factor that affects the development of CVD. Explain how the diet of a person could affect the development of CVD. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 2 = 8 marks)

6 3 Monosaccharides join to form disaccharides and polysaccharides. (a) (i) State the type of reaction where two monosaccharides join to form a disaccharide molecule. (1) ..................................................................................................................................................................................................................................................................................... (ii) Which of the following bonds joins two monosaccharides to form a disaccharide molecule? (1) A ester B glycosidic C hydrogen D peptide (b) The diagram shows a molecule of sucrose. CH2OH OH OH OH H H H O H H OH O CH2OH H CH2OH H OH H O (i) Draw the two monosaccharides that are produced when a molecule of sucrose is broken down. (2) Turn over 7 (ii) Name the two monosaccharides that are produced when sucrose is broken down. (1) 1.................................................................................................................................................................................................................................................................................. 2.................................................................................................................................................................................................................................................................................. (c) Compare and contrast the structure of a disaccharide with glycogen. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 3 = 9 marks)

8 4 Scientists studied how perception of risk differs from the actual risk. People were asked to estimate their risk of developing cardiovascular disease (CVD) during their lifetime. This gave their perceived risk. The scientists used information gathered from questionnaires and various tests to calculate a predicted actual risk of developing CVD, for each person. The table shows how the perceived risk given by each person was compared with the predicted actual risk calculated by the scientists. Underestimation of lifetime risk for CVD Correct estimation of lifetime risk for CVD Overestimation of lifetime risk for CVD Perceived risk < predicted actual risk Perceived risk = predicted actual risk Perceived risk > predicted actual risk It was found that some people’s perception of their lifetime risk of developing CVD was different from the predicted actual risk calculated by the scientists. (a) Explain why the perception of the lifetime risk of developing CVD of some people was different from the predicted actual risk. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 9 (b) Explain how blood tests and obesity indicators can be used to collect data to predict the risk of developing CVD. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 4 = 8 marks)

10 5 Blood type is an example of inherited variation. Blood types A, B, AB and O are determined by a single gene. Blood types are due to the presence or absence of antigens on the cell surface membranes of red blood cells. These antigens are glycoproteins. (a) The diagram shows the structure of a cell surface membrane. R Q P S (i) Which of these labels identifies a glycoprotein? (1) A P B Q C R D S Turn over 11 (ii) The alleles that produce blood type antigens A and B are codominant. A person with the genotype IAIB has blood type AB. The allele producing blood type O is recessive. A couple have been told that the probability of having a child with blood type AB is 0.25 and the probability of blood type O is 0.25. Deduce the genotypes and phenotypes of the parents in the table, by using a genetic diagram. (3) Parent Genotype Phenotype 1 2 12 (b) The production of the blood type antigens involves molecules of mRNA and tRNA. (i) Which row in the table describes the structure of tRNA? (1) Bases Number of strands Type of sugar A A, C, G, U 1 deoxyribose B A, C, G, U 1 ribose C A, C, G, T 2 deoxyribose D A, C, G, T 2 ribose (ii) Mononucleotides all contain a base and a sugar as part of their structure. Name the other component of all mononucleotides. (1) ..................................................................................................................................................................................................................................................................................... (iii) The mononucleotides of mRNA are joined together by RNA polymerase. Which part of the eukaryotic cell is the location for this process? (1) A cytoplasm B mitochondrion C nucleus D rough endoplasmic reticulum Turn over 13 (iv) Describe the role of the tRNA in the production of the protein part of a glycoprotein. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (c) Antigens on the cell membranes of microbes can stimulate endocytosis and exocytosis in white blood cells. These processes are involved in transport through a cell surface membrane. Give two differences between endocytosis and exocytosis. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 5 = 11 marks) 14 BLANK PAGE

Turn over 15 6 The fluid mosaic model of cell membranes was first proposed in 1972. One piece of evidence that was used to support the idea of a phospholipid bilayer was the arrangement of phospholipid molecules on a surface between air and water. (a) (i) Complete the diagram to show the arrangement of four phospholipid molecules at this surface. (2) air water (ii) Which of the molecules found in cell membranes have hydrophilic regions? (1) A channel proteins only B only phospholipids and channel proteins C phospholipids, channel proteins and cholesterol D phospholipids only (b) The vacuoles in beetroot cells contain molecules of betalain, a red pigment. Betalains are large polar molecules. These molecules can leave beetroot cells if the vacuole membrane and the cell surface membrane are damaged. Explain why betalain molecules cannot move through intact cell membranes. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 16 (c) Some chemicals affect the permeability of cell membranes. The effect of detergent on the permeability of beetroot cell membranes was investigated. Sections of beetroot were placed in test tubes containing detergent solution. The colour of the detergent solution at the end of the investigation was measured. The graph shows the results of this investigation. Colour intensity of detergent solution / a.u. 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Concentration of detergent solution (%) 0.5 0.4 0.3 0.2 0.1 0.0 (i) The colour of the 0.5% detergent solution was 0.25 a.u. Plot this result on the graph. (1) *(ii) Two students made predictions about the effect of increasing the concentration of detergent. The sketch graphs show the results they expected. 0.5 0.4 0.3 0.2 0.1 0.0 Colour intensity of detergent solution / a.u. Concentration of detergent solution (%) 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.5 0.4 0.3 0.2 0.1 0.0 Colour intensity of detergent solution / a.u. Concentration of detergent solution (%) 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Graph A Graph B Turn over 17 Explain how to carry out an investigation to test these predictions. (6) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 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(Total for Question 6 = 13 marks)

18 7 Starch changes the colour of iodine solution from yellow-brown to blue-black. The enzyme amylase hydrolyses starch. The initial rate of reaction of starch hydrolysis was investigated. A test tube contained 25 cm3 of starch suspension to which 1 cm3 of 1% amylase solution was added. A sample was immediately taken from this test tube and added to iodine solution. The light absorbance of this mixture was measured. This was repeated every 30 seconds during the investigation. The results are shown in the graph. Light absorbance / a.u. 0 5 10 15 Time / minutes 0.8 0.6 0.4 0.2 0.0 (a) (i) Name a piece of apparatus that would be used to measure the light absorbance of this mixture. (1) ..................................................................................................................................................................................................................................................................................... Turn over 19 (ii) Calculate the initial rate of reaction for this experiment. (3) Answer................................................................................ (iii) Draw a line on the graph to show the expected trend if a 5% amylase solution was used instead. (1) (b) Explain why the light absorbance of the mixture changed over time. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 20 (c) A student placed the same volume of starch suspension into two bags made of Visking tubing. Each bag had a different length and width. The thickness of the Visking tubing for each bag was the same. The membrane of the Visking tubing is partially permeable. The same volume and concentration of amylase solution was added to the starch suspension. Each bag was sealed and placed into test tubes containing the same volume of water, as shown in the diagram. Visking tubing water starch and amylase solution Tube A Tube B The water was tested at regular intervals for the presence of maltose. Turn over 21 Maltose was detected in the water in tube A before it was detected in the water in tube B. Assess the factors that would cause maltose to be detected in tube A before tube B. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 7 = 12 marks)

22 8 The photograph shows a ghost shrimp that lives in fresh water. heart © Nicholas Toh/Alamy Stock Photo The effect of caffeine concentration on the heart rate of ghost shrimps was investigated. A shrimp was placed in a caffeine solution and observed using a microscope. The number of heartbeats in one minute was counted and the heart rate was recorded. This was repeated for other concentrations of caffeine solution. The investigation was repeated using two more ghost shrimps. The table shows the results of this investigation. Concentration of caffeine / mg cm–3 Heart rate / beats min–1 Shrimp 1 Shrimp 2 Shrimp 3 3 272 298 304 7 268 284 298 10 260 270 266 13 152 242 292 (a) (i) State and justify a suitable control for this investigation. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 23 (ii) Variables relating to the caffeine solution and the ghost shrimps should have been controlled. State and justify two variables that should have been controlled in this investigation. (4) Variable relating to the caffeine solution........................................................................................................................................................................ ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Variable relating to the ghost shrimps............................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (b) (i) Calculate the percentage change in the heart rate of shrimp 2 when the caffeine concentration of 10 mg cm–3 was replaced with 13 mg cm–3. Give your answer to one decimal place. (2) Answer ................................................................................ % 24 (ii) Comment on the results of this investigation. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (c) Explain one advantage of using ghost shrimps for this investigation. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 8 = 14 marks) TOTAL FOR PAPER = 80 MARKS Every effort has been made to contact copyright holders to obtain their permission for the use of copyright material. Pearson Education Ltd. will, if notified, be happy to rectify any errors or omissions and include any such rectifications in future editions.

6 2 Plant‑based products provide a sustainable alternative to oil‑based plastics. In a single year, over 7600 million single‑use plastic bags were given to customers by supermarkets in England. The mean mass of each bag was 5.5 g. (a) Calculate the total mass in kilograms of plastic bags given out to customers. Give your answer in standard form. (2) Answer .............................................................. kg (b) A sustainable alternative is a plastic material that contains more than 30% plant‑based polymers. Potato starch is a source of these polymers. (i) Describe the structure of starch. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 7 (ii) Explain why this plastic material is only partially sustainable. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) The tensile strength of plant fibres can be tested using different methods. One method is to attach both ends of a fibre to clamps horizontally and hang increasing numbers of masses until the fibre breaks. A second method is to attach one end of a fibre to a clamp and the other to a forcemeter, pulling on the forcemeter until the fibre breaks. Explain which method will give more accurate results. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 9 marks)

8 3 William Withering tested his digitalis ‘soup’ by an early form of drug testing. Withering reported that his first patient found that the digitalis “made him very sick, the sickness recurring at intervals for several days.” Modern drug testing protocols include four stages. (a) (i) In modern drug testing protocols, the first stage that patients with an illness are involved is (1) A preclinical trials B phase 1 C phase 2 D phase 3 (ii) In modern drug testing protocols, side effects such as nausea will first be identified in healthy volunteers during (1) A preclinical trials B phase 1 C phase 2 D phase 3 Turn over 9 (b) Modern drug testing protocols must ensure that new drugs are effective at treating patients. Explain the role that a placebo and a double blind trial have in producing valid conclusions. (4) Placebo .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Double blind trial .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 10 (c) Drug trials may last 15 years before the drug is licensed. The results of each stage are analysed before the drug trial can progress. The results of a trial of a drug for lowering blood pressure are shown in the table. Mean reduction in blood pressure / mmHg People given the drug People given a placebo 7 ± 12 4 ± 8 Testing of this drug did not proceed beyond this stage. Deduce why the drug trial was stopped at this stage. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 9 marks) Turn over 11 BLANK PAGE

12 4 Stages of cell division in plant tissue can be observed using a light microscope. The image shows the result of a root tip squash. A (Source: AL1323910 – Jose Luis Calvo / Shutterstock / PAL) (a) (i) Which stage of cell division is shown by cell A? (1) A anaphase B interphase C metaphase D prophase (ii) The cells shown in the image are dividing by (1) A meiosis to enable sexual reproduction B meiosis to enable growth C mitosis to enable sexual reproduction D mitosis to enable growth Turn over 13 (iii) Describe how a slide can be prepared from a root tip sample to produce this image. 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(iv) The mitotic index gives the percentage of cells in a sample undergoing mitosis. Calculate the mitotic index for the 30 cells in this image. (2) ..............................................................% 14 (b) Two cultures of cells were used to investigate the cell cycle. The cells in culture A have a functioning gene called P21. The cells in culture B have no functioning P21 gene. Both cultures were treated with gamma radiation that can cause damage to DNA. They were then sampled at intervals of 12 hours and the mitotic index calculated. The results are shown in the graph below. 0 12 24 36 48 60 72 84 96 6 5 4 3 2 1 0 Time after radiation / hours Mitotic index (%) Key Culture A Culture B Deduce the effect of the P21 gene on the cell cycle when cells are treated with gamma radiation. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 11 marks) Turn over 15 BLANK PAGE

16 5 Haemophilia is a condition that results in excessive bleeding due to a lack of blood clotting factors such as factor 8. It is an X‑linked recessive condition and one cause is a mutation to the F8 gene. (a) (i) Complete the diagram, by drawing a line, to show a possible locus for the F8 gene. (1) X chromosome Y chromosome (ii) Explain with the use of a genetic diagram why males with haemophilia cannot pass on this condition to their male offspring. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 17 (b) There are 2931 mutations to the F8 gene that have been identified. Haemophilia can also be caused by a mutation to the F9 gene. There are 1133 mutations to the F9 gene that have been identified. Haemophilia is not classed as a polygenic disorder. (i) Give the meaning of the term polygenic. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) There are 575 identified mutations to the F9 gene that cause severe symptoms of haemophilia. The same percentage of these types of mutation is found in the F8 gene. Calculate the number of mutations to the F8 gene that produce severe symptoms. (2) Answer .............................................................. (Total for Question 5 = 9 marks)

18 6 Stem cells can be used to repair damaged organs. Stem cells from the heart can be used to produce mature heart cells. The diagram below shows one method used to produce mature heart cells. Stem cells from the heart Immature heart cells Mature heart cells Whole heart Chemicals isolated from the heart tissue (a) The whole heart contains cardiac muscle tissue. The whole heart is an organ. Give one difference between a tissue and an organ. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Part of the eye contains stem cells. These cells can be used to repair damaged corneas in the eye. (i) Explain why stem cells from the heart cannot be used to grow cells to repair the cornea. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 19 (ii)  Explain why chemicals from the eye are needed to produce corneal cells from a suitable source of stem cells. 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(c) Human embryos are another source of stem cells used in medical therapies. Describe the decisions that society has to make about the use of these embryonic stem cells. 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(Total for Question 6 = 11 marks)

20 7 A range of minerals is necessary for the formation of healthy plant tissue. (a) Which molecule contains magnesium ions? (1) A amino acid B amylose C chlorophyll D DNA (b) Seedlings of rice plants were grown for two weeks in a complete solution containing all the mineral ions required. After two weeks, the plants were divided into three groups, A, B and C. • The plants in Group A were grown in a complete solution. • The plants in Group B were grown in a complete solution lacking magnesium ions. • The plants in Group C were grown in a complete solution lacking calcium ions. The plants were then grown for nine days. The diagrams show the plants at the end of the growing time. The labels L3 to L6 on the diagrams refer to the leaves in order of growth. L6 was the last leaf to grow. L3 Group A Group B Group C L3 L3 L4 L4 L4 L5 L5 L5 L6 L6 L6 At the end of the growing time, L5 was dead and L6 was yellow in the plants in Group B. Turn over 21 Comment on the effects of deficiencies in magnesium ions and calcium ions on the growth of these rice plants. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22 (c) ‘The growth of seedlings is affected by the concentration of magnesium ions whereas germination of seeds is not.’ Devise an investigation to obtain valid evidence to support this statement. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 10 marks) Turn over 23 BLANK PAGE

24 8 Biodiversity can be reduced by management of habitats. (a) A study of plant species, other than grasses, found in two lawns was completed. Lawn A was cut every 14 days. Lawn B was cut every 7 days. The table shows the number of plants of each species identified. Plant species Number of plants Lawn A Lawn B Daisy 18 5 Bee orchid 1 0 Dandelion 12 0 Clover 4 0 Bird’s-foot trefoil 6 0 Biodiversity can be compared in different habitats using a formula to calculate an index of diversity (D). Σ − = − ( 1) ( 1) N N D n n n = total number of organisms of a particular species N = total number of organisms of all species (i) Calculate the index of diversity for lawn A. (3) Answer .............................................................. Turn over 25 (ii) Herbicides are sprayed on lawns to kill unwanted plants. Explain how the calculated value would change if a herbicide was used on this lawn. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The bee orchid, Ophrys apifera, is pollinated by a species of bee, Eucera longicornis, that is found in parts of the UK. Males of this species of bee mistake the flower for a female bee. The bee orchid can also self‑pollinate. The niche of E. longicornis is that it (1) A feeds on the orchid B is found in the same location as the orchid C looks like the orchid flower D pollinates the orchid 26 *(c) The graph shows the changes in the relative abundance of pollinating wild bee species in the UK. The original sample in 1980 represents a value of 100%. 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 110 100 90 80 70 60 50 Year Relative abundance (%) Explain how the changes in the relative abundance of bee species in the UK may result in the development of a new species of bee orchid. 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(Total for Question 8 = 12 marks) TOTAL FOR PAPER = 80 MARKS 28 BLANK PAGE

4 2 An investigation was carried out to compare the aerobic respiration of different sugars by yeast. A suspension of yeast was used to produce three cultures, each one with a different sugar. A control was set up that had a suspension of yeast cultured with water only. The volume of carbon dioxide produced was measured. The mass of yeast and the concentration of each sugar were controlled. The graph shows the results of this investigation. Volume of carbon dioxide produced / cm3 0 10 20 30 40 50 60 25 20 15 10 5 0 Time / mins maltose glucose water lactose (a) (i) Give one difference between the structure of glucose and the structure of maltose. (1) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 5 (ii) Explain the results for the yeast cultured with glucose and the yeast cultured with maltose. 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(b) Explain the results for the yeast cultured with lactose. 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(Total for Question 2 = 5 marks) 6 BLANK PAGE

Turn over 7 3 The structure of a lipid relates to its role in living organisms. (a) Triglycerides are lipids. (i) The diagram shows part of a glycerol molecule and part of a fatty acid. Complete the diagram to show the ester bond joining the fatty acid to the glycerol molecule. (1) C C O rest of fatty acid molecule rest of glycerol molecule (ii) The table shows the relative atomic mass of some elements. Element Relative atomic mass carbon 12 hydrogen 1 nitrogen 14 oxygen 16 A triglyceride is made from a glycerol molecule and three fatty acids. The molecular mass of the glycerol is 92 and the total molecular mass of the three fatty acids is 362. Which is the molecular mass of this triglyceride? (1) A 400 B 436 C 448 D 454 8 (b) The type of lipid in the diet of a person can affect their risk of developing atherosclerosis. The table shows the percentage content of different lipids in some products used to cook food. Product Source of product Saturated lipids Monounsaturated lipids Polyunsaturated lipids Other lipids Omega-6 Omega-3 Beef fat animal 45 39 3 1 12 Butter animal 52 24 2 1 21 Rapeseed oil plant 7 59 20 9 5 Safflower oil plant 10 15 75 0 0 (i) The pie chart shows the approximate ratio of the lipids in beef fat. monounsaturated saturated polyunsaturated other Complete the pie chart to show the approximate ratio of the lipids in safflower oil. (1) Turn over 9 (ii) Calculate the difference in mass of polyunsaturated lipid in 500 g of safflower oil and 500 g of beef fat. (2) Answer ................................................................ g (iii) Explain why cooking food using plant products rather than animal products could lower the risk of developing atherosclerosis. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (c) Cell membranes contain lipids that have a phosphate group attached to the glycerol, instead of one of the fatty acids. Explain the importance of the phosphate group in these lipids. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 3 = 9 marks)

10 4 Transpiration moves water and mineral ions from the roots to the leaves of plants. (a) Which box in the table shows the mineral ions needed to make an amino acid and which box shows those needed to make DNA? (2) Molecule Mineral ions needed to make the molecule nitrate ions only phosphate ions only both nitrate and phosphate ions neither nitrate nor phosphate ions amino acid DNA (b) Water enters the root hair cells and moves through the root tissues by the apoplastic and symplastic pathways. (i) Root hairs have a large surface area that increases the rate of uptake of water. The surface area of a root hair is rounded up to 3.14 × 105 µm2. Which is the actual surface area of this root hair? (1) A 31 460 µm2 B 314 600 µm2 C 313 900 µm2 D 3 139 000 µm2 Turn over 11 (ii) Which row of the table shows the movement of water through the apoplastic and symplastic pathways? (1) Apoplastic pathway Symplastic pathway A B C D (iii) Water moves through the root tissue due to a difference in water potential between one cell and the next cell. Which row of the table describes this movement? (1) Direction of water movement with respect to water potential Direction of water movement with respect to concentration of solutions A higher to lower concentrated to dilute B higher to lower dilute to concentrated C lower to higher concentrated to dilute D lower to higher dilute to concentrated 12 (c) Explain how transpiration depends on the dipole nature of water. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 4 = 8 marks)

Turn over 13 5 Some antibiotics work by inhibiting the production of ribosomes in bacteria. The structure of ribosomes in bacteria is similar to the structure of ribosomes in mitochondria. (a) (i) Describe the structure of a ribosome. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (ii) Describe the role of ribosomes in protein synthesis. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 14 (b) These antibiotics can destroy cancer cells in humans. The diagram shows the membranes of a mitochondrion from a cancer cell treated with these antibiotics. (i) Give one difference in the membranes of this mitochondrion compared with the membranes of a mitochondrion from an untreated cancer cell. (1) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (ii) Some cancer cells depend on oxidative phosphorylation for ATP production. Explain why the antibiotics that inhibit the production of ribosomes prevent oxidative phosphorylation when used to treat cancer cells. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 15 (iii) Some scientists suggested that inhibiting both glycolysis and mitochondrial respiration may be an effective way of treating cancer cells. Explain why this suggestion may be an effective way of treating cancer cells. 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(Total for Question 5 = 9 marks)

16 6 Macroalgae and microalgae are photosynthetic organisms. Seaweeds are macroalgae that live attached to rock in coastal areas. Some seaweeds are green, some are brown and some are red. The colour of seaweeds depends on the photosynthetic pigments contained in their cells. Green seaweeds contain chlorophyll, brown seaweeds contain chlorophyll and fucoxanthin and red seaweeds contain chlorophyll and phycoerythrin. (a) (i) Explain why these seaweeds have different absorption spectra and action spectra. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 17 (ii) The diagram shows the position that these seaweeds occupy on submerged rock and the depth to which different wavelengths of light penetrate into the water. green seaweed brown seaweed red seaweed water level red light green light blue light submerged rock Explain why the seaweeds occupy different positions on the submerged rock. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 18 (b) Fucoxanthin is found in both macroalgae and microalgae. Fucoxanthin has anti-inflammatory, anti-tumour, anti-diabetes, anti-malarial and anti-obesity activity in humans. The diagram outlines the steps taken to extract fucoxanthin from microalgae. Cells of microalgae grown in a culture in a laboratory Cells washed Fucoxanthin extracted The table shows three methods used for washing the cells and extracting the fucoxanthin. Method Liquid used for washing cells Chemical used to extract fucoxanthin A water ethanol B culture media ethanol C culture media water Turn over 19 The graph shows the mass of fucoxanthin extracted from 1 dm3 of cell culture. 90 80 70 60 50 40 30 20 10 0 Mass of fucoxanthin / mg A B C Method (i) Calculate how many times more fucoxanthin was extracted using method A compared with method B. (1) Answer ................................................................ (ii) Explain why different masses of fucoxanthin were extracted using these three methods. 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(Total for Question 6 = 10 marks)

20 7 Bacteria are becoming increasingly resistant to antibiotics. Controlling the spread of antibiotic resistance is necessary. (a) Many types of bacteria are resistant to penicillin and tetracycline. (i) Which row of the table shows the part of a bacterial cell affected by these antibiotics? (1) Penicillin Tetracycline A cell wall cell wall B cell wall ribosome C ribosome cell wall D ribosome ribosome (ii) Which row of the table describes the action of penicillin and tetracycline? (1) Penicillin Tetracycline A bactericidal bactericidal B bactericidal bacteriostatic C bacteriostatic bactericidal D bacteriostatic bacteriostatic Turn over 21 (b) One method of controlling the spread of antibiotic resistance is to avoid unnecessary prescriptions of antibiotics. The table shows some illnesses and whether a prescription for antibiotics is recommended. Illness Antibiotic prescription Bronchitis may be recommended Ear infection may be recommended Influenza not recommended Fluid in the ear not recommended Strep throat recommended Urinary tract infection recommended Explain why these recommendations could reduce the spread of antibiotic resistance. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 22 (c) In an investigation, E.coli bacteria were isolated from animals reared for food and from some wild animals. The mean percentage of the E. coli isolated from each type of animal that were resistant to different types of antibiotics was determined. The table shows the results of this investigation, where n is the number of each type of animal. Antibiotic Mean percentage of resistant E.coli isolated from animals reared for food (%) Mean percentage of resistant E.coli isolated from wild animals (%) Broiler chicken n = 45 Village chicken n = 45 Cattle n = 12 Bat n = 13 Rodent n = 35 Ampicillin 50.00 8.89 8.33 0.00 8.57 Cephalothin 14.29 8.89 8.33 20.00 14.58 Chloramphenicol 46.43 0.00 0.00 0.00 0.00 Neomycin 50.00 2.22 25.00 0.00 5.71 Tetracycline 95.86 55.56 25.00 0.00 2.86 (i) Calculate the ratio of tetracycline-resistant E.coli in all the chickens to tetracycline-resistant E.coli in cattle. (2) Answer ................................................................ (ii) There was a total of 2 × 108 E.coli isolated from cattle and 6 × 109 E.coli isolated from rodents. Calculate how many more E.coli are resistant to cephalothin in rodents than in cattle. (2) Answer ................................................................ Turn over 23 (iii) Antibiotics can be added to animal feed. It is thought this use of antibiotics has contributed to the spread of antibiotic resistance. Evaluate the extent to which the results of this investigation support this idea. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 7 = 12 marks)

24 8 Infections caused by Chlamydia and human papillomavirus (HPV) are sexually transmitted. The HPV is a non-enveloped DNA virus. Chlamydia infection is caused by the bacterium, Chlamydia trachomatis. (a) Compare and contrast the structure of a bacterial cell with the structure of HPV. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (b) Some people with HPV infections have no symptoms but others develop warts and some types of cancer. Over 99% of cervical cancers are caused by HPV. Girls in the UK are offered an HPV vaccine. (i) Explain how an HPV vaccine protects girls from developing cervical cancer. 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Turn over 25 Turn over (ii) Explain why the HPV vaccine will not protect girls from infection with Chlamydia. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... *(iii) The HPV vaccination programme is being extended to boys. Explain the advantages of offering the HPV vaccine to boys, as well as girls, in schools. 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(Total for Question 8 = 15 marks)

26 9 Weddell seals spend a lot of their time swimming underwater, diving to find food and diving to avoid predators. However, they do have to come to the surface to breathe. (a) The graphs give some information about dives by Weddell seals. Length of dive / mins 40 30 20 10 0 Percentage of dives (%) 0–5 6–10 11–15 16–20 21–25 26–30 31–35 36–40 41–45 46–50 51–55 56–60 60+ Depth of dive / m 25 20 15 10 5 Percentage of dives (%) 0–50 51–100 101–150 151–200 201–250 251–300 301–350 351–400 401–450 451–500 501–550 551–600 0 Turn over 27 Turn over Comment on the conclusions that can be made about the dives of these seals. 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(b) Weddell seals carry higher concentrations of oxygen in their bodies than humans. (i) The haemoglobin of Weddell seals is adapted to carry higher levels of oxygen than the haemoglobin of humans, at each partial pressure of oxygen. Which row of the table describes the haemoglobin of the Weddell seal? (1) Position of oxygen dissociation curve compared with human haemoglobin Affinity for oxygen compared with human haemoglobin A to the left higher B to the left lower C to the right higher D to the right lower 28 *(ii) The table shows how much oxygen is stored in different parts of the body of humans and Weddell seals. Part of body Total oxygen stored / a.u. kg–1 body tissue humans Weddell seals lungs 12 3 blood 10 58 muscle 2 23 body fluids, other than in the blood 2 2 Weddell seals also store erythrocytes in their spleen that are released into the bloodstream during deep dives. Turn over 29 Analyse the data to explain the differences between these two mammals. 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..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 30 (c) An investigation studied the change in the concentration of oxygen in the blood and muscles of a Weddell seal during a dive. The graph shows the results of this investigation and the diagram shows part of the circulation of blood in the seal during the dive. Concentration of oxygen Start of dive 0 End of dive brain lung heart in blood in muscles Time Key oxygenated blood deoxygenated blood Analyse the data to explain these results. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 31 (d) The graph shows the changes in the lactic acid concentration in the muscles and blood of a Weddell seal, after a prolonged dive. Concentration of lactic acid / a.u. muscles Time after the end of the dive / mins 100 80 60 40 20 0 0 10 20 30 40 50 60 blood Explain the importance of these changes after the Weddell seal comes to the surface to breathe. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 9 = 17 marks) TOTAL FOR PAPER = 90 MARKS 32 BLANK PAGE

6 2 (a) The diagram shows a human karyotype. 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 x 22 21 20 19 (Source: © DEPT. OF CLINICAL CYTOGENETICS, ADDENBROOKES HOSPITAL/SCIENCE PHOTO LIBRARY) (i) Which of the following identifies the human with this karyotype? (1) A female with Down’s syndrome B female with Turner’s syndrome C male with Down’s syndrome D male with Turner’s syndrome (ii) Which of the following would cause the karyotype in the diagram? (1) A non-disjunction producing a monosomy B non-disjunction producing a polysomy C translocation producing a monosomy D translocation producing a polysomy Turn over 7 (b) (i) The table shows the chromosome number and chromatid number of some of the cells formed during oogenesis in humans. Cell type Chromosome number Chromatid number ovum 23 23 primary oocyte 46 92 secondary oocyte 23 46 Which of the cells are haploid? (1) A ovum and primary oocyte B ovum and secondary oocyte C primary oocyte and secondary oocyte D ovum alone (ii) Explain how meiosis produces new combinations of alleles in gametes. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks)

8 3 The diagram shows two amino acids found in the primary sequence of collagen. N C C H H O O H H H N C C CH3 H O O H H H glycine alanine (a) (i) Which is the R-group for alanine? (1) A CH3 B COOH C H D NH2 (ii) Draw the products of a condensation reaction between glycine and alanine. (2) Turn over 9 (iii) Describe the structure of collagen. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 10 (b) Ehlers-Danlos syndrome is a human condition caused by a mutation in a collagen gene. This causes changes to the elasticity and strength of arteries. Mice were genetically modified to have this mutation. The elastic constant and force required to break arteries from genetically modified mice and unmodified mice were measured. The elastic constant is calculated by using the formula: elastic constant = force applied increase in length of artery The results are shown in the table. Artery Elastic constant / N mm−1 Force required to break artery / N unmodified mice 0.26 2.0 genetically modified mice 0.12 1.6 Turn over 11 (i) Calculate the force required to produce an increase in length of 0.5 cm for an artery taken from an unmodified mouse. (1) Answer.............................................................. (ii) People with Ehlers-Danlos syndrome are at risk of developing aneurysms. An aneurysm is an area of artery wall that weakens and can burst. Explain why people with Ehlers-Danlos syndrome have a high risk of aneurysms. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 10 marks)

12 4 The photograph shows an insect fossilised in amber resin. (Source: Bjoern Wylezich. 123rf.com/PAL) Scientists have classified another fossilised insect found in amber resin as a new species of mosquito, Priscoculex burmanicus. This mosquito is anatomically very similar to modern species of mosquito. (a) (i) Which of these processes produces new species without geographically separating populations? (1) A allopatric speciation B genetic speciation C stabilising speciation D sympatric speciation (ii) Priscoculex burmanicus became extinct about 100 million years ago. State why it was difficult for scientists to decide if this insect belonged to a different species from modern mosquitoes. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 13 (iii) State how scientists would have reached agreement on classifying this mosquito. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iv) Complete the classification table for Priscoculex burmanicus. (2) Taxon Name Domain Eukarya Kingdom Animalia Arthropoda Diptera Order Culicidae Family Anophelinae Genus Species 14 (b) Mosquito populations are often controlled using pesticides that inhibit the enzyme acetylcholinesterase. Long‑term use of pesticides often leads to the development of resistance in these mosquitoes. The ace-1 gene codes for the enzyme acetylcholinesterase. The ace-1r allele is a recessive form of this gene that gives resistance to pesticides. The changes in the numbers of resistant and non-resistant mosquitoes in an area of farmland where pesticides were used in 2007 was investigated. The use of the pesticides was stopped in 2016. The table shows the results of this investigation. Mosquito type Number of mosquitoes 2005 2015 2018 resistant 125 5250 4120 non-resistant 8755 1250 3950 (i) Calculate the number of heterozygous mosquitoes in 2015. Use the Hardy-Weinberg equation p2 + 2pq + q2 = 1 (3) Answer.............................................................. Turn over 15 (ii) Explain why the changes in the numbers of the resistant and non-resistant mosquitoes, from 2005 to 2018, are related to the function of the ace-1 gene. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 12 marks) 16 BLANK PAGE

Turn over 17 5 (a) (i) Which bonds join the nucleotides in a single strand of DNA? (1) A peptide B disulfide C glycosidic D phosphodiester (ii) Describe how the process of DNA replication produces two identical DNA molecules. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 18 (b) The graph shows the relationship between genome size and mutation rates of organisms. 1010 109 108 107 106 105 104 103 102 10−11 10−10 10−9 10−8 10−7 10−6 10−5 10−4 10−3 10−2 Genome size / number of nucleotides Mutation rate / substitutions per nucleotide per generation higher eukaryotes lower eukaryotes RNA viruses single stranded DNA viruses double stranded DNA viruses bacteria (i) Analyse the graph to determine the approximate mutation rate for tobacco mosaic virus (TMV). (2) Answer.............................................................. Turn over 19 (ii) Comment on the relationship between genome size and mutation rate of the organisms shown in the graph. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 10 marks)

20 6 Acetylcholine is a neurotransmitter released by neurones in the autonomic nervous system. (a) (i) Which of the following substances bind to acetylcholine receptors? (1) 1. cobra venom 2. nicotine 3. lidocaine A 1 and 2 B 1 and 3 C 1, 2 and 3 D 2 and 3 (ii) Which row is correct for acetylcholine? (1) Type of neurone that releases acetylcholine Effect of acetylcholine A parasympathetic neurone increases heart rate B parasympathetic neurone decreases heart rate C sympathetic neurone increases heart rate D sympathetic neurone decreases heart rate Turn over 21 (iii) Describe the events that lead to the release of acetylcholine at a synapse. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22 (b) Capsaicin is a substance found in chilli plants that makes them taste ‘hot.’ Capsaicin binds to temperature receptor proteins called TPV receptors. This stimulates impulses along sensory neurones that send impulses to the brain. The effect of capsaicin on the depolarisation of the membrane of a sensory neurone was investigated. The potential difference across the membrane was measured at two places on the sensory neurone after adding capsaicin. The results for a high concentration and a low concentration of capsaicin are shown in the diagram. Low concentration of capsaicin capsaicin TPV receptor potential difference / mV potential difference / mV receptor cell sensory neurone time time High concentration of capsaicin time potential difference / mV potential difference / mV capsaicin TPV receptor receptor cell sensory neurone time Turn over 23 (i) Explain the effect of capsaicin on the depolarisation of the sensory neurone. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 24 *(ii) Capsaicin affects sweat production. It has been suggested that capsaicin could be used to lower the temperature of people with fevers. The effect of three different doses of capsaicin on sweat production was investigated. Five students took one 10 mg tablet of capsaicin each day, for one week. The mass of sweat produced from a 10 cm2 area of skin was measured one hour after taking each tablet of capsaicin. The mean mass of sweat per square centimetre of skin for the group of students was calculated for each day. This was repeated with different groups of students taking 50 mg of capsaicin and taking 100 mg of capsaicin. Most of the students who were given 100 mg of capsaicin reported a burning sensation and redness of the skin. The results are shown in the graph. Error bars represent standard deviations. Key 10 mg capsaicin 50 mg capsaicin 100 mg capsaicin Time / days 7 6 5 4 3 2 1 0 60 50 40 30 20 10 0 Mean mass of sweat produced / mg cm−2 Turn over 25 Analyse the data to discuss the use of capsaicin to lower the temperature of people with fevers. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 14 marks)

26 7 (a) (i) The diagram shows the transfer of energy through a grassland food chain. All values are given in kJ m−2 yr−1. Decomposers Secondary consumer Producer = 1.4 × 105 Primary consumer 9.1 × 102 3.9 × 102 7.6 × 103 3.1 × 103 4.4 × 104 8.4 × 104 energy loss Calculate the percentage efficiency of the energy transfer from the producer to the primary consumer. (2) Answer.............................................................. Turn over 27 (ii) Explain the differences in the efficiency of energy transfer between the trophic levels in a food chain. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 28 *(b) For many years, hedgerows have been removed to increase field sizes to grow crop plants. Hedgerows provide a habitat for many species of plants and animals. Some animals that live in hedgerows eat pests of crops and some pollinate crop plants. The crop yield, temperature and wind speed were measured across the field starting at the hedge. These factors and the crop yield are shown as a relative percentage compared with the centre of the field in the graph. Relative percentage compared with the centre of the field (%) 30 20 10 0 −10 −20 −30 25 20 15 10 5 Key crop yield temperature wind speed Distance from hedgerow / m Turn over 29 Analyse the information to comment on the value of conserving hedges around fields used for growing crops. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 7 = 11 marks)

30 8 (a) Haemophilia is a sex linked, genetic condition that prevents blood clotting. XH is the allele for blood clotting Xh is the allele for haemophilia The pedigree diagram shows a family that has individuals who have haemophilia. Key male with haemophilia male without haemophilia female with haemophilia female without haemophilia 1 2 3 4 5 7 6 8 9 10 11 (i) What are the genotypes of individuals 1 and 2? (1) A XHY and XHXH B XhY and XHXh C XhY and XHXH D XHY and XHXh Turn over 31 (ii) Use a genetic diagram to determine the probability that the next child of individuals 4 and 5 will be a boy without haemophilia. (3) Answer.............................................................. 32 (b) Kidney dialysis machines are used to remove urea from the blood of patients who are suffering from kidney disease. Anti-clotting drugs are given to patients to reduce the risk of blood clot formation whilst undergoing kidney dialysis. The potential use of two anti-clotting drugs, Drug A and Drug B, was investigated. Blood samples were taken from 20 healthy adults and placed into test tubes. The times taken for the blood samples to clot after addition of 2 mg dm−3 of Drug A were recorded. This was repeated with Drug B and again with a control. The results are shown in the graph. Time taken for blood to clot / s 600 500 400 300 200 100 0 Control Drug B Drug A Drug type mean mean mean Turn over 33 It takes 40 minutes for Drug A to become inactive in a person and 110 minutes for Drug B to become inactive. Analyse the data to discuss the use of Drug A and Drug B to prevent the formation of blood clots. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 9 marks) 34 BLANK PAGE

Turn over 35 9 Auxins and cytokinins are plant growth regulators that control apical dominance and cell elongation. (a) Explain how auxins and cytokinins control the growth of lateral buds. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Auxin is an uncharged molecule when it diffuses through the cell surface membrane. When inside the cytoplasm, auxin develops a negative charge. Auxin leaves cells by facilitated diffusion. Explain how the structure of the cell membrane affects the transport of auxin. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 36 (c) When auxin enters a cell, it causes hydrogen ions to be pumped out of the cell into the cell wall. The effect of auxin on the pH of the cell wall and cell elongation was investigated. Auxin was added to plant cells. The pH of the cell wall and cell elongation were measured over a period of 60 minutes. The results are shown in the graph. 240 200 160 120 80 40 0 Elongation of cell / μm 60 50 40 30 20 10 0 Time / min 6.5 6.0 5.5 5.0 4.5 pH of cell wall Key elongation of cell pH of cell wall (i) Calculate the maximum rate of elongation of the cell. (2) Answer.............................................................. 37 (ii) Expansin is a protein located in cell walls. Expansin disrupts hydrogen bonds and ionic bonds of molecules in the cell wall. Deduce a mechanism by which auxin and expansin cause cells to elongate. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 10 marks) TOTAL FOR PAPER = 90 MARKS 38 BLANK PAGE 39 BLANK PAGE 40 BLANK PAGE

2 Answer ALL questions. Write your answers in the spaces provided. 1 Plant stems contain xylem and phloem, tissues specialised for transport. The photomicrograph shows a longitudinal section of phloem tissue.               X (a) (i) Draw the cell labelled X. (2) Turn over 3 (ii) Explain how phloem tissue is adapted for its function. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) State why stains, such as methylene blue, are sometimes used when preparing a microscope slide. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 1 = 6 marks)

4 2 (a) Compare and contrast the central nervous system and the peripheral nervous system. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) During the First World War, some soldiers received head wounds that damaged one side of the cerebellum. One soldier said ‘ The movements of my left hand are done subconsciously but I have to think out each movement of my right arm.’ (i) State one function of the cerebellum. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) State the part of the brain the soldier used to think about moving his right arm. (1) ...................................................................................................................................................................................................................................................................................... Turn over 5 (iii) Explain why this soldier had problems moving his right arm. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks)

6 3 Anthocyanin is a purple pigment found in a range of vegetable plants. The photograph shows some carrots from a variety called Purple Haze. (a) Describe how you would investigate the effect of temperature on the permeability of the membranes in the root cells of these purple carrots. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 7 (b) The anthocyanins in purple carrots are used as antioxidants in the food industry. Purple Haze carrots with a mass of 750g contain 1265.25 mg of anthocyanin. Calculate the percentage of anthocyanin in these carrots. Give your answer to two decimal places. (2) Answer .............................................................. (c) Purple cauliflower plants also contain anthocyanins. Cauliflowers can be light purple, dark purple or white. The photograph shows a dark purple cauliflower. 8 The colour in cauliflowers is controlled by a single gene with two alleles. Purple colour is controlled by the allele Pr. White colour is controlled by the allele pr. In an investigation, dark purple cauliflowers were crossed with pure bred white cauliflowers. The F1 generation cauliflowers were all light purple in colour. (i) State the genotype of the F1 generation. (1) .................................................................................................................................................................................................................................................................................... (ii) The F1 generation was then self‑fertilised to produce the F2 generation. The numbers observed (O) in the F2 generation were: 31 white cauliflowers, 52 light purple and 20 dark purple cauliflowers. It was concluded that the purple colour in cauliflowers is controlled by a single gene with two codominant alleles. State the expected (E) number of each of the three phenotypes in the F2 generation. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Calculate the value for Chi squared using these results. Use the formula for Chi squared: − = ∑ χ 2 2 ( ) O E E (3) Answer .............................................................. Turn over 9 (iv) The table shows some critical values for Chi squared. Degrees of freedom p value 0.900 0.500 0.100 0.050 0.010 1 0.016 0.455 2.706 3.841 6.635 2 0.211 1.386 4.605 5.991 9.210 3 0.584 2.366 6.251 7.815 11.345 4 1.064 3.357 7.779 9.488 13.277 Explain why the conclusion that colour in cauliflowers is controlled by a single gene with two codominant alleles is likely to be valid. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 16 marks)

10 4 A survey was carried out to compare the biodiversity in different areas of a village. The photograph shows a children’s playing field in this village. (a) Quadrats placed in the playing field showed predominantly grass with a few dandelion species, ribbed plantains, yarrow and clover. No animal species were observed. The biodiversity of this playing field was low. (i) Explain the reasons why the biodiversity of this playing field was low. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 11 (ii) The photograph shows a path in a woodland nature reserve near the playing field in this village. After surveying the playing field, a survey of the woodland reserve was carried out using the following method: • quadrats were placed at various intervals either side of the path • plant species were counted in the quadrats • animal species were counted at regular intervals along the path. The table shows the results of this survey. Species Number Plant Dock 10 Thistle 10 Hawthorn 6 Wild cherry 2 Hazel 5 Yarrow 5 Animal Meadow brown butterfly 2 Shield bug 4 Beetle 20 Honeybee 20 Dragonfly 4 Spider 5 12 Calculate the biodiversity index for this woodland nature reserve. Use the formula Σ − = − ( 1) ( 1) N N D n n (3) Answer .............................................................. (iii) It was suggested that the biodiversity index calculated from this survey was an underestimate. Criticise the method used in the survey of the woodland nature reserve. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 13 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Honeybees are important pollinators of flowers. Surveys have suggested that butterflies, ladybirds, beetles, flies and dragonflies are also important pollinators of flowers. A recent report has suggested that a large increase in honeybees kept in towns and cities could affect biodiversity. Explain how the increase in honeybees in towns and cities could affect biodiversity. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 14 marks)

14 5 The photographs show leaves from six species of plant. Daffodil Frogbit floating on water Sunflower Maize Oak tree Pea Turn over 15 In an investigation into transpiration, a survey was carried out to measure the distribution of stomata on leaves. The table shows the results of this investigation. Plant species Density of stomata in leaf surface / number per mm2 Upper epidermis Lower epidermis Beech tree 0 340 Daffodil 65 68 Frogbit (floating leaf) 89 0 Maize 52 58 Oak tree 0 450 Pea 100 220 Sunflower 85 156 Sycamore tree 0 860 16 (a) Analyse the data and the photographs to explain the distribution of the stomata. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) As part of this investigation, the effect of light on the transpiration rate of seven‑day‑old maize seedlings was measured. The transpiration rate of maize seedlings placed in dim light was measured every 30 minutes for 3.5 hours. Extra illumination was then added and the transpiration rate again measured every 30 minutes for another 3.5 hours. The graph shows the results of this investigation. Turn over 17 7 6 5 4 3 2 1 0 Transpiration rate / a.u. extra illumination added Time / hours dim light bright light 1 2 3 4 5 6 7 (i) Calculate the approximate percentage increase in the transpiration rate due to the extra illumination. (2) Answer .............................................................. (ii) Explain two factors that should have been controlled in this investigation. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 9 marks)

18 6 Acetyl CoA is produced in the link reaction when pyruvate is broken down. The photograph shows a molecule of the enzyme acetyl CoA carboxylase. This enzyme is involved in the conversion of acetyl CoA into fatty acids. The enzyme is activated by citrate. Citrate does not bind at the active site. Citrate is produced in the Krebs cycle. The enzyme is inhibited by long chain fatty acyl CoA molecules. Fatty acyl CoA molecules have a longer carbon chain than acetyl CoA. (a) (i) Analyse the information to explain how the activity of the enzyme acetyl CoA carboxylase is controlled. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 19 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain why inhibitors of this enzyme may be useful in treating obesity in humans. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The enzyme acetyl CoA carboxylase has a different structure in bacteria. The enzyme in bacteria is made of several polypeptide subunits whereas in humans, the enzyme is one large polypeptide. Inhibitors of the bacterial enzyme are being developed to treat bacterial infections. Explain why inhibitors of the bacterial enzyme may be useful when treating bacterial infections. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 9 marks)

20 7 A student investigated the effect of the concentration of sodium chloride solution on blood. The student set up six test tubes, each with a different concentration of sodium chloride solution. Five drops of blood were added to each test tube. The appearance of the solutions was then recorded. Samples of each solution were observed using a light microscope. The table shows the results of this investigation. Concentration of sodium chloride (%) Appearance of solution after blood was added Observation with light microscope 3.0 very cloudy cells seen with shrunken edges 1.0 very cloudy cells seen 0.9 very cloudy cells seen 0.7 slightly cloudy cells seen 0.5 cloudy at first, then went clear no cells seen 0.3 clear no cells seen Turn over 21 (a) Analyse the data to comment on the effect of the concentration of sodium chloride solution on blood. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22 (b) The student then investigated the effect of using glucose solutions and sucrose solutions, instead of sodium chloride solutions. The appearance of the solutions was recorded and samples were observed using a light microscope. When the concentration of glucose solution was below 3%, the solution was clear and no blood cells could be seen. When the concentration of glucose solution was above 3%, the solution was cloudy and blood cells could be seen. When the concentration of sucrose solution was below 6%, the solution was also clear and no blood cells could be seen. When the concentration of sucrose solution was above 6%, the solution was cloudy and blood cells could be seen. Explain the differences in the effects of the solutions of glucose and sucrose. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 23 (c) (i) State the part of the brain responsible for osmoregulation. (1) .................................................................................................................................................................................................................................................................................... (ii) Explain how, on a very hot day, the brain ensures that the water potential of the blood remains constant. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 12 marks)

24 8 Cysteine proteases are enzymes found in fruits such as pineapples. When a protease enzyme is added to fat‑free skimmed milk, the milk turns from cloudy to clear. (a) Devise an investigation to find the optimum temperature for the activity of the enzyme cysteine protease found in pineapples. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 25 (b) Cysteine proteases are also found in the predatory bacterium Bdellovibrio bacteriovorus (BvB). These bacteria are harmless to humans but prey on Gram negative bacteria. (i) Give two differences between Gram negative and Gram positive bacteria. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) The cell of the predatory BvB is 1µm in length and can swim 100 times its length per second. Calculate the swimming speed of this bacterial cell in mm s–1. (2) Answer .............................................................. 26 (iii) The BvB bacteria have been described as ‘living antibiotics’. Explain why the BvB bacteria might be useful in the future for treating bacterial infections. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 12 marks) Turn over 27 BLANK PAGE

28 9 Gilbert’s potoroo is one of the most critically endangered species in the world. Gilbert’s potoroo was thought to be extinct until one was found in 1994. It is thought that there are only 70 individuals. The photograph shows a potoroo and its pouched baby feeding on berries. The map shows the only part of Australia where Gilbert’s potoroo is found. Turn over 29 The diet of Gilbert’s potoroo is about 90% fungi. Analysis has shown that the fungal spores are spread in the faeces of the potoroos. These spores then germinate. (a) Explain the advantage to the ecosystem of fungal spores being spread by potoroos. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) In 1994, there were only 20 Gilbert’s potoroo individuals, 12 of which were in a captive colony. A captive breeding programme was set up using: • four adult females and one young female • one adult male, two young males and one young male still in its mother’s pouch. (i) Explain why allele frequencies may cause problems in this population of potoroos. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 30 (ii) Foxes and feral cats, which are predators of potoroos, were introduced into Australia by settlers. The photograph shows a feral cat with its prey. The Australian government is using poisoned sausages in an attempt to kill 2 million feral cats by 2020. The sausages contain chunks of kangaroo meat combined with chicken fat, herbs, spices and a poison. The content of the sausages was based on flavours irresistible to cats. The poison used is produced by the Swan River pea plant that is found in Western Australia.  Other Australian plants that contain the poison are: Acacia, Wedge Peas and Nemcia. This poison is a natural toxin that biodegrades rapidly and the native wildlife in Australia has evolved natural resistance to it. A group of 160 000 people in Australia have signed a petition against the plan to poison feral cats. Discuss whether the plan to poison feral cats is justified. (9) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 31 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 9 = 14 marks)

32 10 Lanzarote is a volcanic island in the Atlantic Ocean. It lies near the coast of West Africa. The photograph shows lichens growing on volcanic rock in Lanzarote. Lichens are organisms composed of a fungus and algae living together. Lichens colonise bare rock. (a) Explain how colonisation of bare rock by lichens can result in succession over long periods of time. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 33 (b) Devise an investigation to show how the lichen distribution is affected by a named abiotic factor. 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(Total for Question 10 = 10 marks)

34 11 White clover plants provide a rich source of nitrogen for cattle when grown with grass plants in fields. The photograph shows a clover plant growing in a field of grass. A student noticed that the grass plants growing near patches of clover were a darker green than other grass plants. (a) Devise an investigation to show whether the presence of clover plants affects the concentration of the green pigment chlorophyll in grass plants. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 35 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Clover plants have bacteria living in root nodules. These bacteria convert atmospheric nitrogen (N2) to ammonia (NH3) using the enzyme nitrogenase. (i) State one organic molecule containing nitrogen atoms which is found in plant cells. (1) .................................................................................................................................................................................................................................................................................... (ii) Nitrogen is a relatively unreactive gas that forms 80% of the atmosphere. The enzyme nitrogenase catalyses the reaction shown: N2 + 8H+ + 16 ATP → 2NH3 + H2 + 16 ADP + 16 Pi The activation energy for this reaction is very high. Analyse the information to explain how the enzyme nitrogenase is involved in the energy changes required for this reaction. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 11 = 11 marks) TOTAL FOR PAPER = 120 MARKS 36 BLANK PAGE

4 2 Skulls can be used as evidence for human evolution. The only type of human present today is the modern human. In the past, another type of human, called Neanderthal, occupied the same geographical area. The photograph shows the skulls of both types of human. Modern human skull Neanderthal skull © hairymuseummatt/DrMikeBaxter Genetic studies now show that these two types of human have interbred. (a) Only small quantities of DNA can be extracted from the remains of Neanderthal humans. The quantity of extracted DNA is amplified using (1) A DNA profiling B polymerase chain reaction C RNA profiling D translation chain reaction Turn over 5 (b) Devise a procedure, using electrophoresis, to compare the amplified DNA from these two types of human. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (c) Explain how the results of electrophoresis could provide evidence for interbreeding between the two types of human. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks)

6 3 During the civil war in Mozambique, from 1977 to 1992, 90% of the African elephants were killed for the ivory in their tusks. Since 1992, the elephant population in Mozambique has increased. The photograph shows a group of elephants in Mozambique. Elephant without tusks Elephant with tusks © ALEXANDER JOE/Staff/Getty Images A study of an elephant population in Mozambique was carried out in 2017. The elephants studied were all born before 1992. The table shows the number of elephants with and without tusks. Elephants Number of elephants with tusks 98 without tusks 102 Turn over 7 (a) The presence of tusks is controlled by a single gene. The allele for tusks is known to be dominant. The frequency of alleles in a population can be determined using the Hardy‑Weinberg equation. Calculate the frequency of the dominant and recessive alleles in this population. Give your answer to 2 decimal places. (3) Frequency of the dominant allele .............................................................. Frequency of the recessive allele .............................................................. 8 (b) In most African elephant populations, 2% to 4% of elephants do not grow tusks. (i) Explain why the percentage of elephants without tusks in Mozambique will probably increase in the future. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (ii) Describe how the Hardy‑Weinberg equation can be used to provide evidence for changes in the elephant population in Mozambique. 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(Total for Question 3 = 8 marks) Turn over 9 BLANK PAGE

10 4 Fertilisation in humans involves the fusion of a sperm cell with an egg cell. (a) (i) After the sperm cell nucleus fuses with the egg cell nucleus, the single fertilised cell is described as a (1) A diploid gamete B diploid zygote C haploid gamete D haploid zygote (ii) The fertilised cell then divides by (1) A meiosis to produce genetically different cells B meiosis to produce genetically identical cells C mitosis to produce genetically different cells D mitosis to produce genetically identical cells (b) Cell division of the fertilised cell produces a ball of totipotent cells. (i) Give the meaning of the term totipotent cell. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 11 (ii) This ball of cells continues to divide to form the embryo. The cells of the embryo become specialised to form tissues and organs. Describe how cells become specialised. 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(c) Factors such as the diet of the mother during pregnancy can cause epigenetic changes in the embryo. Explain how epigenetic changes affect the development of tissues in the embryo. 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(Total for Question 4 = 10 marks)

12 5 Wildlife conservation can involve keeping animals in zoos as well as protecting habitats. (a) Describe the roles that zoos play in animal conservation. 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(b) Habitats can be destroyed by road building. A link road is planned to connect two major roads. The diagram shows the two possible sites for the link road and two woodland habitats. major road major road woodland P woodland Q possible sites for link road Turn over 13 (i) State what is meant by the term habitat. (1) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (ii) To decide which of the two link roads should be built, studies of the biodiversity of woodlands P and Q need to be carried out. Discuss what information needs to be collected from the two woodlands to make a decision about the site of the link road. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 5 = 8 marks)

14 6 Hereditary spherocytosis is a condition that affects red blood cells. This inherited condition is caused by a gene mutation that affects the shape of the cell surface membrane. (a) Describe the structure of the cell surface membrane. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 15 (b) The red blood cells of individuals with spherocytosis are spherical instead of having a biconcave disc shape. Explain the effect of spherocytosis on the uptake of oxygen by red blood cells. 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(c) Most cases of hereditary spherocytosis are caused by a dominant allele. Use a genetic diagram to determine the probability of a child inheriting this condition if one parent is heterozygous and the other parent does not have the condition. (2) Answer .............................................................. (Total for Question 6 = 8 marks)

16 7 The stems of plants contain tissues involved in transport and in support. (a) The photograph shows a longitudinal section through one plant stem. E H G F Cell image library - image ID 38928 (i) Which labelled structure in the photograph contains lignin? (1) A E B F C G D H (ii) The diagram shows a transverse section through a plant stem. J Which of the labelled structures in the photograph is located in the area labelled J in the diagram? (1) A E B F C G D H Turn over 17 (b) The photograph shows a cross section of part of the stem of a plant. Fibres can be extracted from stems similar to this one and used instead of fibres produced from oil‑based plastics. X L N M John Bebbington FRPS Science and Plants for schools – https://www.saps.org.uk (i) The part of the stem labelled L contains (1) A phloem that transports organic solutes B sclerenchyma fibres that provide support C sieve tubes that transport water D xylem vessels that transport water and mineral ions (ii) The part of the stem labelled M contains (1) A phloem that transports organic solutes B sclerenchyma fibres that transport water C sieve tubes that synthesise organic solutes D xylem vessels that transport water and mineral ions 18 (iii) The structure labelled N is a (1) A part of the phloem that transports water B sclerenchyma fibre that provides support C sieve tube that transports organic solutes D xylem vessel that transports water and mineral ions (iv) The line labelled X on the photograph represents the width of the vascular bundle. The actual width of the vascular bundle is 320 µm. Calculate the magnification of the image. (3) Answer .............................................................. Turn over 19 (c) Fibres can be extracted from the stems of nettle plants and used to make clothing. Explain why the production of fibres from nettles is more sustainable than fibres produced from crude oil. 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(Total for Question 7 = 11 marks)

20 8 Penicillin is an antibiotic. It was discovered in 1928. Since then many antibiotics have been identified and are widely used in the treatment of bacterial infections. (a) State what is meant by the term bacteriostatic antibiotic. (1) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (b) Scientists have recently discovered a new class of antibiotics that bind to ribosomes. (i) Explain why these antibiotics could affect the production of proteins in bacteria. 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Turn over 21 (ii) These new antibiotics attach to a site on the ribosome not affected by any known antibiotics. Deduce why these new antibiotics might be used to treat bacteria that are resistant to other antibiotics. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... *(iii) Scientists have isolated these new antibiotics and tested their effectiveness against bacteria that are resistant to other types of antibiotic. Devise a laboratory procedure to compare the effectiveness of penicillin with one of the new antibiotics. 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(Total for Question 8 = 12 marks) Turn over 23 BLANK PAGE

24 9 The human immunodeficiency virus (HIV), shown in the image, causes acquired immunodeficiency syndrome (AIDS). © Liya Graphics/Shutterstock (a) HIV particles contain (1) A DNA and DNA polymerase B DNA and reverse transcriptase C RNA and DNA polymerase D RNA and reverse transcriptase (b) HIV enters and destroys T helper cells. This can cause AIDS. (i) Describe how HIV particles are able to enter T helper cells. 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Turn over 25 (ii) Explain why the destruction of T helper cells causes the symptoms of AIDS. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 26 (c) A small number of people have been identified who are resistant to HIV. They have a mutation in a gene coding for a protein in the cell membrane. (i) Deduce why this mutation makes these people resistant to HIV infection. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (ii) Stem cell therapy can be used to treat patients infected with HIV. The bone marrow of these patients can be destroyed using radiotherapy. The patients can then be given stem cells from the bone marrow of a donor who has this mutation. Explain why these stem cells may prevent HIV causing AIDS. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 9 = 14 marks) Turn over 27 BLANK PAGE

28 10 Anthropogenic climate change is linked to an increase in carbon dioxide in the atmosphere. (a) The World Meteorological Association recorded carbon dioxide levels of 405 ppm in 2017. This is an increase of 46% since 1817. Calculate the level of carbon dioxide in 1817. (2) .............................................................. ppm Turn over 29 (b) A study has investigated the effect of temperature changes on plants growing in the Arctic. This is an area with cold and short growing seasons. Warming in the Arctic is leading to a change in the community of plants. The mean height of plants in the area was studied in both dry and wet growing seasons over a 30‑year period. The summer temperatures over this period were recorded. The results are shown in the graph. Key Wet seasons Dry seasons 30 25 20 15 10 5 0 0 2 4 6 8 10 12 Mean summer temperature / °C Mean height of plants / cm Explain the effect of these environmental factors on the mean height of the plants. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 30 (c) The effect of temperature on the activity of RUBISCO was studied. The results are shown in the graph. 7 6 5 4 3 2 1 0 60 50 40 30 20 10 0 Temperature / °C Enzyme activity / a.u. Analyse the data to explain why temperature affects the growth of plants. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 31 *(d) The world is divided into climate zones based on temperature and rainfall. Details of some of the climate zones are shown in the table. Climate zone Mean temperature Mean precipitation (rain or snow) / mm year−1 Polar (e.g. Arctic) Warmest month below 10 °C 150 to 250 Temperate Coolest month above 0 °C but below 18 °C Warmest month above 10 °C 500 to 2000 Results from the two studies described in this question can be used to explain how climate affects plant growth. Discuss how anthropogenic climate change can affect the growth of plants in polar and temperate climate zones. (6) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 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(Total for Question 10 = 14 marks) TOTAL FOR PAPER = 100 MARKS Every effort has been made to contact copyright holders to obtain their permission for the use of copyright material. Pearson Education Ltd. will, if notified, be happy to rectify any errors or omissions and include any such rectifications in future editions

4 2 Children are different from their parents due to genetic variation and other factors. Genetic variation can be the result of meiosis and random fertilisation of gametes. (a) Describe how meiosis leads to genetic variation in the gametes produced. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) During random fertilisation, only one sperm cell fuses with the cell surface membrane of an egg cell. Vesicles play an important role in this process. Explain how vesicles are involved in the successful fertilisation of an egg cell by only one sperm. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 5 (c) Variation between children and their parents is affected by linkage of genes. Explain why some genes show linkage and others show sex-linkage. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks)

6 3 Mucopolysaccharides are complex molecules found in the human body. (a) Mucopolysaccharides are large molecules containing unbranched polysaccharides. Describe the structure of an unbranched polysaccharide. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Mucopolysaccharides can be broken down by enzymes. Describe how an enzyme could break down the polysaccharide component of mucopolysaccharides. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 7 (c) Mucopolysaccharidosis type 1 (MPS 1) is a recessive genetic condition. People with MPS 1 cannot break down mucopolysaccharides. The pedigree diagram shows the inheritance of MPS 1 in a family. Key female without condition female with condition male without condition 1 2 3 Determine the probability that person 2 has the same sex and MPS 1 phenotype as person 1. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 8 marks)

8 4 Scientists can use models to explain the need for a circulation system in animals. (a) The shapes in the diagram represent two different animals that live in water. The figures represent the height, width and breadth of the animals. Determine why animal A does not need a circulation system but animal B does. (4) A 1 mm 4 mm 16 mm 4 mm 4 mm 4 mm B .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 9 (b) During the development of the mammalian heart, there is a hole between the left ventricle and the right ventricle. This hole usually becomes sealed before the mammal is born. If it is not sealed, the mammal will become easily tired due to a lack of energy. Explain why a mammal born with a hole between the two ventricles will have these symptoms. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) In the lungs, oxygen moves from the alveoli into red blood cells. (i) How many times does an oxygen molecule cross a cell surface membrane to move from the centre of an alveolus to the centre of a red blood cell? (1) A 2 B 3 C 4 D 5 (ii) Oxygen enters the red blood cell by (1) A active transport B diffusion C facilitated diffusion D osmosis (Total for Question 4 = 9 marks)

10 5 The cardiac cycle is the sequence of events in one heartbeat. In humans, the cardiac cycle can be divided into three stages. (a) In the ventricular systole stage of the cardiac cycle, the (1) A ventricles contract, atrioventricular valves close and semilunar valves open B ventricles contract, atrioventricular valves open and semilunar valves close C ventricles relax, atrioventricular valves close and semilunar valves open D ventricles relax, atrioventricular valves open and semilunar valves close (b) Caffeine is a stimulant found in coffee. The effect of different types of coffee on blood pressure was investigated. Decaffeinated coffee has most of the caffeine removed. Four groups of volunteers had their blood pressure measured before having a drink and again 30 minutes later. The results of this investigation are shown in the table. Drink provided to volunteers Mean systolic blood pressure / mm Hg Before the drink 30 minutes later Water 115 119 Decaffeinated coffee 120 118 Regular coffee 106 118 (i) Calculate the percentage increase in the mean systolic blood pressure after drinking regular coffee. (1) .............................................................. % Turn over 11 (ii) Daphnia can be used to investigate the effects of caffeine on heart rate. Devise a procedure to compare the effects of decaffeinated coffee and regular coffee on the heart rate of Daphnia. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 12 (c) Explain why high blood pressure can increase the risk of developing cardiovascular disease (CVD). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 9 marks) Turn over 13 BLANK PAGE

14 6 Spider silk is a very strong and flexible natural fibre. It is of interest to humans as a possible fibre for protective clothing. Scientists have genetically modified a range of organisms to produce spider silk, including goats and plants such as alfalfa. (a) A gene for spider silk is copied. A gene for resistance to antibiotic A is also copied. The flow diagram shows some of the stages in genetically modifying alfalfa plants to produce spider silk using the copied genes. Stage 1 An enzyme is used to join each copied gene for spider silk to a copied antibiotic resistance marker gene to form pairs of joined genes Stage 2 Plasmids are cut open so that these joined genes may be inserted Stage 3 The plasmids are placed inside Agrobacterium tumefaciens bacteria, which are then exposed to antibiotic A Stage 4 Surviving Agrobacterium tumefaciens bacteria are used to infect cells of the alfalfa plant Stage 5 Infected alfalfa plant cells are grown on agar containing IAA Stage 6 Spider silk is extracted from alfalfa plants Turn over 15 (i) Explain how an enzyme is involved in joining the two different genes together in stage 1. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Which one of the following enzymes can be used to cut open the plasmids in stage 2? (1) A DNA polymerase B RNA ligase C RNA polymerase D restriction endonuclease (iii) Explain why antibiotic A is used in stage 3. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16 (iv) Give reasons why the infected alfalfa plant cells are grown on agar containing IAA in stage 5. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (v) The table shows the mass of spider silk produced from the genetically modified alfalfa in stage 6. It also shows the mass of spider silk produced from the genetically modified goats. Organism Mass of spider silk produced per year Alfalfa 218 kg per acre Goat 10 kg per goat A typical number of goats that can be kept on one acre of land is 12. Calculate the percentage increase in spider silk produced by the alfalfa plants compared with the goats. (3) .............................................................. % Turn over 17 (b) Give two reasons why some people may be concerned about the use of genetically modified alfalfa as a source of spider silk. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 13 marks) 18 BLANK PAGE

Turn over 19 7 Moving a limb involves the interaction of muscles, tendons and ligaments. (a) Muscles can contain fast twitch and slow twitch muscle fibres. (i) Describe two structural differences between fast twitch muscle fibres and slow twitch muscle fibres. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Muscles can be extensors or flexors. Which of the following correctly describes the state of the muscles described? (1) When the arm is bent at the elbow, the flexor muscle is When the lower leg is straightened at the knee, the flexor muscle is A contracted contracted B contracted relaxed C relaxed contracted D relaxed relaxed 20 (b) Tendons and ligaments are important structures in elbow and knee joints. (i) Which of the following identifies the structures that join bones to bones in an elbow joint? (1) A ligaments only B ligaments and tendons C tendons only D neither ligaments nor tendons (ii) One type of joint injury is a torn ligament. This may be treated by adding a piece of tendon to the ligament. This is because after a period of time, the tendon tissue changes and responds in the same way as a ligament. Which of the rows in the table correctly describe the changes in this piece of tendon? (1) Row Piece of tendon shows a change in its The change is 1 genotype an anatomical adaptation 2 genotype a physiological adaptation 3 phenotype an anatomical adaptation 4 phenotype a physiological adaptation A row 1 only B row 3 only C rows 1 and 2 D rows 3 and 4 Turn over 21 *(iii) The photograph shows athletes competing in the modern triathlon. © Gonzalo Arroyo Moreno / Stringer The modern triathlon involves three sports: swimming, cycling and running. An investigation was carried out to compare the level of demand on the body of these three sports during a triathlon. The investigation involved 12 athletes who were all males of the same age. Each athlete carried out the triathlon as shown in the flow diagram. There was no rest period between each sport. 1.5 km swim 40 km cycle 10 km run The heart rate for each athlete was measured as they completed each sport. The mean heart rate for each sport was then calculated and is shown in the table. Sport Mean heart rate / bpm Swimming 163 Cycling 165 Running 159 22 The blood lactate level for each athlete was also measured as they completed each sport. Means for lactate level after each sport were calculated. The results are shown in the graph. Swimming Cycling Running 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 Blood lactate concentration / mmol dm−3 It was concluded that cycling was the most demanding of the three triathlon sports. This was followed by swimming and then running. Evaluate the validity of this conclusion. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 23 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 7 = 11 marks)

24 8 Yeast is a single-celled organism that can respire aerobically. Mitochondria are the sites of aerobic respiration in yeast cells. (a) The diagram shows a mitochondrion. P R Q S (i) Which labelled component in the diagram is the site of the Krebs cycle? (1) A P B Q C R D S (ii) Which labelled component in the diagram is the site of oxidative phosphorylation? (1) A P B Q C R D S (iii) Yeast cells can have many small mitochondria. Calculate the magnification of the diagram if the maximum length of the mitochondrion is 0.5 µm. (2) Answer .............................................................. Turn over 25 (b) (i) Name two molecules needed for aerobic respiration that can move into the mitochondria. (2) 1 ................................................................................................................................................................................................................................................................................ 2 ................................................................................................................................................................................................................................................................................ (ii) The outer mitochondrial membrane is not permeable to hydrogen ions (H+). Explain the importance of this feature of the membrane. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26 (c) It has been stated that if the temperature of yeast is raised by 10 °C, the rate of respiration will double. The diagram shows some apparatus that can be used to measure the rate of respiration in yeast. scale (mm) clip potassium hydroxide solution coloured liquid yeast suspension 200 180 160 140 120 100 80 60 40 20 0 Devise an investigation using this apparatus to determine whether an increase of 10 °C doubles the rate of respiration in yeast. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 14 marks) Turn over 27 BLANK PAGE

28 9 When astronauts are in space, the force of gravity is less than when they are on a planet such as Earth or Mars. Scientists need to investigate how varying gravity affects astronauts. NASA/Science photo library In one investigation, scientists grew human muscle cells in culture. These cells were split into two groups:  Group 1 was taken into space for 10 days  Group 2 remained on Earth for 10 days. Data were collected on protein synthesis and protein breakdown in each group of cells. (a) State why data were collected for Group 2 as well as for Group 1. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 29 (b) The scientists proposed a null hypothesis for the rate of breakdown of protein in the muscle cells of the two groups. The data from this investigation showed that there was no significant difference at the 0.05 probability level. (i) Explain what is meant by no significant difference at the 0.05 probability level for the rate of breakdown of protein. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Some of the myosin in the muscle cells was broken down. Describe how the tertiary structure of myosin is related to its function. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 30 *(c) Data for the rate of protein synthesis were collected for the groups of muscle cells as shown in the table. Group of muscle cells Mean rate of protein synthesis / arbitrary units Group 1 cells during the 10-day space mission 40 Group 1 cells after the 10-day space mission returned to Earth 380 Group 2 cells 190 It has been estimated that astronauts will travel to Mars by 2030. It will take about nine months to travel from Earth to Mars. The force of gravity is less on Mars than it is on Earth, but greater than it is in space. Discuss the potential effects of gravity on the muscles of astronauts when travelling to Mars and then living on Mars. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 31 .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 12 marks)

32 10 The retina of the human eye contains rod cells. These cells detect light energy as photons. The light energy is converted to a nerve impulse that is interpreted by the brain. (a) A current is produced by a rod cell when ions move through the cell surface membrane of a rod cell. The graph shows the effect of increasing light intensity on the current produced by a rod cell. 10 000 1000 100 30 20 10 0 Light intensity / photons Current / pA Turn over 33 Explain the effect of increasing light intensity on the current produced by a rod cell. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 34 (b) (i) The transmission of an impulse between a neurone in the optic nerve and a cell in the brain involves ions and neurotransmitter molecules. Describe how these ions and neurotransmitter molecules are involved in the transmission of an impulse. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 35 (ii) The diagram shows a human brain. T W V U Which label on the diagram identifies the area of the brain where an image is interpreted? (1) A T B U C V D W 36 (c) In one experiment, Hubel and Wiesel kept the right eye of a kitten closed during the critical period. They then studied the effect of this on the visual region of the brain of this kitten. Describe the differences in the visual region of this kitten compared with the visual region of a kitten that had both eyes open during the critical period. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 10 = 12 marks) TOTAL FOR PAPER = 100 MARKS

2 Answer ALL questions. Write your answers in the spaces provided. 1 Plants can respond to and use light. The photograph shows a seedling starting to grow from a germinating seed. (a) Explain why the seedling needs a supply of magnesium ions. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 3 (b) Describe the role of IAA (auxin) in the phototropic response of plants. 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(Total for Question 1 = 6 marks) 4 BLANK PAGE

Turn over 5 2 Acetylcholinesterase is an enzyme involved in regulating the transmission of nerve impulses across some synapses. (a) Acetylcholinesterase is found on the cell surface membranes of neurones and red blood cells. These acetylcholinesterase molecules have different primary structures. In humans, a single gene codes for acetylcholinesterase. (i) Explain how a single gene can give rise to acetylcholinesterase molecules with different primary structures. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain how the acetylcholinesterase gene can be expressed in some tissues but not others. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 6 (b) Alzheimer’s disease is associated with the loss of neurones that produce acetylcholine. It has been suggested that inhibitors of acetylcholinesterase may be useful in the treatment of Alzheimer’s disease. (i) Explain why inhibitors of acetylcholinesterase could be useful in the treatment of Alzheimer’s disease. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 7 (ii) Trials of a new inhibitor were carried out using tissue cultures. The effect of the concentration of the inhibitor on acetylcholinesterase activity and cell viability was measured. Percentage cell viability was measured as the percentage of cells that were not killed by the inhibitor. The graph shows the results for this inhibitor. Acetylcholinesterase activity / a.u. Percentage cell viability (%) Concentration of inhibitor / µmol dm−3 Key Cell viability Acetylcholinesterase activity 4 3 2 1 0 250 200 0 50 150 100 100 75 50 25 0 State and justify a suitable concentration of inhibitor to use in clinical trials. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 11 marks)

8 3 Doctors sometimes prescribe beta-blockers for their patients. Beta-blockers are a type of drug with antihypertensive properties. (a) In one study, the effect of beta-blockers on the heart rate during exercise was investigated. In this study, volunteers were placed randomly into one of four groups shown in the table. Group Dose of beta-blocker A none B very low C low D high The heart rate of each volunteer was recorded during a period of exercise. A mean value was calculated separately for the males and females in each group. The graph shows the results of this study. Mean heart rate during exercise / beats per minute Beta-blocker dose 250 200 150 100 50 0 Very low Low High None Key Males Females (i) Calculate the percentage change in male heart rate caused by increasing the dose of beta-blocker from very low to high. (2) Answer .............................................................. Turn over 9 (ii) Deduce the effect of beta-blockers on the supply of blood to muscle during exercise. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Explain why beta-blockers are prescribed for some people. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 10 (c) Beta-blockers work by blocking the effects of a hormone called adrenaline. Adrenaline is produced by the adrenal glands located on top of each kidney. Adrenaline acts on the heart to cause changes in heart rate. Deduce how adrenaline can cause a change in heart rate. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 12 marks) Turn over 11 BLANK PAGE

12 4 Lipoprotein lipase is an enzyme found in different tissues including the brain. (a) The enzyme lipoprotein lipase is involved in the metabolism of lipids in the brain. The effect of temperature on lipid metabolism in the brain tissue of a species of fish was investigated. The results of this investigation are shown in the table. Temperature / °C Lipoprotein lipase mRNA / a.u. Triglyceride concentration / mmol mg−1 Fatty acid concentration / µmol mg−1 Lipoprotein lipase activity per milligram of tissue / a.u. 5 4.50 0.58 0.84 1.70 17 1.00 0.60 0.69 1.55 30 1.10 0.70 0.64 0.99 Comment on the effect of temperature on lipid metabolism in this species of fish. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 13 (b) Lipoprotein lipase can be extracted from brain tissue to investigate the effect of temperature on the rate of reaction of this enzyme. (i) State what is meant by the term Q10 temperature coefficient. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Devise an investigation to determine Q10 for an extract of lipoprotein lipase enzyme. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 9 marks)

14 5 Forests are important habitats. (a) The effect of cutting down trees on the number of bird species observed in two different forest habitats was investigated. (i) Give two biotic factors, other than cutting down trees, that could affect the number of bird species observed in a forest. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Some of the results of the investigation are shown in the table. Forest Number of bird species in areas of the forest where no trees are cut down Number of bird species in areas of the forest where some trees are cut down A 35 19 B 16 10 Calculate the Chi-squared value (χ2) for forest B using the formula shown. 2 2 ( ) O E χ E − = ∑ (3) Answer .............................................................. Turn over 15 (iii) The table gives some critical values for the Chi-squared test. Probability level Critical value 0.05 3.84 0.01 6.64 0.001 10.83 The Chi-squared value for forest A is 4.74. Deduce the effect of some trees being cut down on the number of species of birds in these two forests. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16 (b) Many forests are exploited by humans. (i) Describe how forests can be managed as a sustainable resource. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain the impact of cutting down trees on climate change. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 13 marks)

Turn over 17 6 The neurones of the central nervous system contain TAU proteins. These proteins help to maintain cell structure. In humans, six different TAU proteins can be produced from a single gene. Parkinson’s disease has been linked to the different forms of the TAU proteins present in neurones. Scientists are studying the effect of these different TAU proteins in animal models. One model used is the fruit fly, Drosophila. (a) Describe how Drosophila flies could be genetically modified to produce one form of the human TAU protein. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 18 (b) In one investigation, the effect of ageing on Drosophila making different human TAU proteins was studied. Flies making different forms of human TAU protein were tested in a climbing test. •  Fifteen one-day-old Drosophila flies were placed at the bottom of a measuring cylinder. •  A video recording was carried out and paused at 10 seconds. The height climbed by each fly was recorded. •  The test was repeated for flies of different ages. mean height climbed time = 0 seconds time = 10 seconds The table shows the results of this investigation. Age of flies / days Mean height climbed at 10 seconds / mm Control (no human TAU protein) 0N3R TAU protein 0N4R TAU protein 1 60 ± 6 52 ± 5 54 ± 5 7 61 ± 4 43 ± 1 50 ± 4 14 53 ± 3 18 ± 2 28 ± 8 21 45 ± 9 12 ± 4 18 ± 5 28 32± 8 10 ± 5 19 ± 1 35 26 ± 9 3 ± 1 11 ± 1 42 15 ± 3 1 ± 1 3 ± 1 Turn over 19 Comment on the effect of TAU proteins on the ability of Drosophila flies to climb. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 20 (c) In another investigation, the effect of these TAU proteins and age on the conduction of nerve impulses along the axon of neurones was studied. The length of time between impulses was measured for Drosophila flies of different ages. The results are shown in the graph. 9 8 7 6 5 4 3 2 1 0 Length of time between impulses / ms control 0N3R 0N4R Key Age of flies / days 10 45 Determine the effect of these TAU proteins on the maximum frequency at which nerve impulses can be conducted along the axon of the neurone. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 11 marks) Turn over 21 BLANK PAGE 22 *7 Humans are complex multicellular organisms. Cell division is involved in: •  the production of gametes •  the development of tissues •  the development of tumours •  the response to infection in humans. The figure summarises some information about the organisation of cells in a human. More than 3.7 × 1013 cells in one human More than 200 different cell types The typical cell cycle for a human cell is about 24 hours. However, some cells can divide every eight minutes and others do not divide at all. Four main tissue types (each with many ‘sub-types’) At least 79 different organs Eleven different organ systems The table summarises some information about human genetics and cancers. Number of protein coding genes Approximately 20 000 Number of different genetic disorders Approximately 2200 Proportion of cancers caused by an inherited genetic defect 5 to 10% Turn over 23 Evaluate the role of cell division in processes affecting human health and disease. (9) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 7 = 9 marks)

Turn over 25 8 The scientific article you have studied is adapted from National Geographic. Use the information from the scientific article and your own knowledge to answer the following questions. (a) Explain how two populations of mosquito could be shown to belong to different species (paragraph 4). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Explain why scientists believe CRISPR could be used to prevent AIDS developing in people infected with HIV (paragraph 7). (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26 (c) Describe how plant biologists could demonstrate the effect of deleting genes that attract pests to crops (paragraph 8). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (d) Explain why the editing of the germ line of a human embryo would result in the changes being inherited by the next generation (paragraph 9). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 27 (e) Explain why an RNA guide can be used to identify the precise nucleotides that Cas9 has been sent to cut (paragraph 12). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (f) Gene drives can be used to ‘force almost any genetic trait through a population’ (paragraph 16). Multiple genetic crosses were carried out between individuals homozygous for a recessive allele and individuals heterozygous for the same gene. Describe how the outcome of these crosses would be affected if a gene drive was used with the recessive allele. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 28 (g) Dengue is a disease caused by a virus carried by Aedes aegypti. Using CRISPR and a gene drive, it may be possible to stop the reproduction of Aedes aegypti mosquitoes (paragraphs 20 and 21). Explain why stopping the reproduction of this species of mosquito might not stop the spread of this disease. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (h) Deduce how PERVs present in pig tissues could infect human cells (paragraph 25). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 29 (i) Explain why some pig genes can cause reactions in the human immune system (paragraph 27). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (j) Explain how a vaccine developed against human plague can provide lifelong immunity to sylvatic plague in the black-footed ferret (paragraph 33). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 30 (k) State what is meant by the term endemic (paragraph 35). (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (l) Deduce the pattern of inheritance for Tay-Sachs disease (paragraph 39). (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 29 marks) TOTAL FOR PAPER = 100 MARKS 31 BLANK PAGE 32 BLANK PAGE

4 2 A zygote is formed when gametes fuse at fertilisation. (a) Explain how meiosis results in genetic variation in the gametes. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Describe how the process of fertilisation results in the formation of a zygote from the gametes in humans. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 5 marks)

Turn over 5 3 Malaria is a serious and sometimes fatal disease. Scientists are constantly looking for new ways of controlling this disease. (a) Which row of the table shows the name of the pathogen that causes malaria, and its classification group? (1) A B C D Name of pathogen Classification group Plasmodium genus Plasmodium species Puccinia genus Puccinia species (b) One group of scientists has genetically modified a fungus to produce a spider toxin that kills mosquitoes. Describe how a fungus could be genetically modified to produce spider toxin. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 6 (c) Another group of scientists has discovered a type of fungus that completely protects mosquitoes from infection by the pathogen that causes malaria. This fungus does not kill the mosquitoes. Explain why this approach is less controversial than the approach used by the scientists who are developing the genetically‑modified fungus. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 7 marks)

Turn over 7 4 Bacteria are the host cells for λ (lambda) phage viruses. (a) Which is a description of a λ phage? (1) A DNA virus with a complex protein capsid B DNA virus with a helical protein capsid C RNA virus with a complex protein capsid D RNA virus with a helical protein capsid (b) The image shows phage viruses attacking a bacterium. bacterium phage (Source: © nobeastsofierce Science/Alamy Stock Photo) The length of this bacterium is 1.7 µm. Calculate the length of the labelled phage. Give your answer in nanometres (nm). (2) Answer .............................................................. nm 8 (c) Viruses can be cultured and a growth curve can be produced. The flow chart shows how this can be done. Grow the host cells of the virus on agar Add the viruses to the host cells View the cells under a microscope and count the number of lysed (burst) cells Repeat counts at regular intervals The graph shows a growth curve for viruses. 6 5 4 3 2 1 0 20 40 60 80 100 120 Time after adding viruses to cells / min Log10 number of lysed cells Turn over 9 (i) Explain why there was a delay before the number of lysed cells started to increase. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Calculate the mean rate of increase in the actual number of lysed cells between 50 minutes and 90 minutes. (2) Answer .............................................................. cells min–1 (iii) A sketch has been made of this growth curve. Time after adding viruses to cells Log10 number of lysed cells Complete this sketch to predict the shape of the growth curve after 120 minutes, assuming there is an excess of host cells. (2) (Total for Question 4 = 10 marks) 10 BLANK PAGE

Turn over 11 5 The light‑dependent stage and the light‑independent stage of photosynthesis both take place in the chloroplast. The rate of photosynthesis is affected by a number of different factors, including carbon dioxide concentration and temperature. (a) The diagram shows a chloroplast. Q R S T U V W X (i) Where does the light‑dependent stage take place? (1) A Q B T C V D X (ii) The light‑dependent stage produces hydrogen ions. Where do these hydrogen ions accumulate? (1) A Q B R C S D W (iii) Where does translation take place? (1) A Q B R C S D U 12 (b) Scientists measured the effect of two different concentrations of carbon dioxide on the rate of photosynthesis at different leaf temperatures, in one species of plant. The results are shown in the graph. 10 15 20 25 30 35 40 50 40 30 20 10 0 Leaf temperature / °C Rate of photosynthesis plants grown in high levels of carbon dioxide plants grown in low levels of carbon dioxide (i) Which units are suitable for measuring the rate of photosynthesis in leaves? (1) A µmol m–1 sec–1 B µmol m–1 sec–2 C µmol m–2 sec–1 D µmol m–2 sec–2 Turn over 13 (ii) Analyse the data to identify three conclusions that can be made from this graph. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Explain the effects of carbon dioxide concentration and temperature on the rate of formation of GALP. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 11 marks)

14 6 The lipid content of the cell membranes of prokaryotic organisms changes in response to changes in the environmental temperature. (a) Phospholipids form a bilayer in the cell membranes of bacteria. Describe the structure of a phospholipid. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The cell membranes of most organisms belonging to the domain Archaea are lipid monolayers. The lipid that forms this monolayer is a bipolar lipid. Explain why a bipolar lipid is a suitable molecule to form the cell membrane. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 15 (c) The percentage of branched‑chain fatty acids in membranes of bacteria that have different optimum growth temperatures was investigated. The graph shows the results of this investigation. 0 20 40 60 80 100 80 60 40 20 0 Percentage of branched‑chain fatty acids (%) Optimum growth temperature of bacteria / °C (i) Calculate the mean percentage of branched‑chain fatty acids in bacteria whose optimum growth temperature is <20 °C. (1) Answer .............................................................. % (ii) Analyse the data to describe two conclusions that can be drawn from this investigation. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16 (d) A change in temperature can affect the permeability and fluidity of the membrane. Explain why it is important that the lipid composition of the membrane of prokaryotic organisms changes if the temperature changes. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 10 marks) Turn over 17 BLANK PAGE

18 7 Macrophages are involved in response to infection. (a) Macrophages engulf bacteria by surrounding the bacteria with pseudopodia. The image shows a macrophage forming pseudopodia around some bacteria. macrophage pseudopodia bacteria (Source: © urfin/Shutterstock) The flow chart shows one theory for the formation of pseudopodia. Step 1 antibody binds to the cell surface of the macrophage Step 2 fibrous proteins in the cytoplasm, close to the antibody binding site, break down to form soluble components Step 3 water flows into the part of the cell where the soluble components are located Step 4 pseudopodia start to form around the bacteria Turn over 19 (i) Which diagram shows one antibody binding to the surface of a macrophage (Step 1)? (1) A B C D (ii) Actin is a fibrous protein. Which row of the table describes the breakdown of actin (Step 2)? (1) A B C D monomer formed process by which bond is broken amino acid condensation amino acid hydrolysis nucleotide condensation nucleotide hydrolysis (iii) Explain why water flows into the part of the cell where the soluble components are located (Step 3). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 20 (iv) Describe the events that take place resulting in T helper cell activation, following the formation of pseudopodia by the macrophages (Step 4). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 21 BLANK PAGE 22 *(b) Phagocytosis by macrophages is affected by a number of factors. Microorganisms with a fluorescent green dye attached to them were used to measure phagocytosis by macrophages. The microorganisms were added to the macrophages and incubated. The macrophages were then washed and the extent to which they glowed green was determined. The extent of the green glow is proportional to the number of microorganisms engulfed by the macrophages. The graphs show the results of investigations into phagocytosis by macrophages. 0 20 40 60 150 000 100 000 50 000 0 1.00 0.20 0.10 0.02 Length of investigation / mins Extent of green glow / a.u. Concentration of microorganisms / mg cm–3 0 20 40 60 100 000 80 000 60 000 40 000 20 000 0 100 000 50 000 25 000 12 500 Length of investigation / mins Extent of green glow / a.u. Number of macrophages per culture Source of macrophages Extent of green glow / a.u. 10 000 8 000 6 000 4 000 2 000 0 bone marrow abdomen lung Key bacteria A bacteria B Turn over 23 0 10 100 1 000 20 000 15 000 10 000 5 000 0 Concentration of ATP synthase inhibitor / a.u. Extent of green glow / a.u. Analyse the data to discuss the factors that affect phagocytosis by macrophages. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 13 marks)

24 8 Hypoxia is an inadequate supply of oxygen to tissues and cells that restricts their function. (a) The normal partial pressures of oxygen vary from tissue to tissue. The table shows the normal partial pressure of oxygen in two tissues. Tissue Partial pressure of oxygen / kPa Pulmonary arterial blood 5.3 Other arterial blood 13.3 (i) Calculate the ratio of the partial pressures of oxygen in these two tissues. Give your answer to one decimal place. (1) Answer .............................................................. (ii) Explain why the partial pressures of oxygen in these two tissues are different. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 25 (b) The body responds to hypoxia by releasing hypoxia‑inducible transcription factors (HIF). Investigations have shown that one effect of HIF is an increase in the rate of glycolysis in the affected cells. (i) Explain how HIF could result in an increase in the rate of glycolysis. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain why cells need to respond to hypoxia with an increase in the rate of glycolysis. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26 (iii) The graph shows the changes in levels of two HIFs, HIF‑1 and HIF‑2, before and during hypoxia. Time Before hypoxia During hypoxia HIF‑1 HIF‑2 Level of HIF Compare and contrast the changes in the levels of HIF‑1 and HIF‑2 during hypoxia. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iv) Explain the changes in levels of HIF‑1 and HIF‑2 during hypoxia. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 13 marks)

Turn over 27 9 Plants and animals are adapted for gas exchange. (a) (i) Which row of the table describes gas exchange in the root of a plant? (1) A B C D Net movement of carbon dioxide Net movement of oxygen in only out only out only in only in and out in and out neither in nor out neither in nor out (ii) Which is the gas exchange surface in the stem of a woody plant? (1) A lenticel B pit C plasmodesmata D stomata 28 (b) Spiracles are small openings in the exoskeletons of insects that allow air to enter the respiratory system. Water can evaporate out of the spiracles when they are open. The insect can close the spiracles to reduce water loss. In an investigation, the water loss from insects in air with different humidities was measured. The insects were kept in air with 80% humidity and then moved into air with a lower humidity. Water loss was then measured. The investigation was repeated in air high in carbon dioxide to keep the spiracles open. The table shows the results of this investigation. Percentage humidity (%) Water loss from insects / mg hr–1 Insects in air Insects in air high in carbon dioxide 0 0.10 0.90 20 0.13 0.68 40 0.15 0.50 60 0.13 0.35 80 0.07 0.07 (i) State how the water loss could have been measured in this investigation. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Calculate the percentage increase in water loss from the insects kept in air at 0% humidity compared with those kept at 80% humidity. Give your answer to two decimal places. (1) Answer .............................................................. % Turn over 29 (iii) Explain why the insects were kept in air with 80% humidity at the start of this investigation. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iv) Explain the results of this investigation. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 30 (c) The gas exchange surfaces of fish are the gills. The diagram shows part of the structure of a gill. direction of flow of water gill filaments lamellae Fish can be divided into two groups: species of fish that are active and species of fish that are inactive. The table shows some information about some species of fish and their gills. Species of fish Mass of fish / g Total number of gill filaments Number of lamellae on each side of filament per mm Surface area of gills / cm2 per g of fish Diffusion distance between water and blood / µm Active species Trachurus 26 1665 39 7.8 2.2 Lucioperca 70 1811 15 18.0 no data Salmo 394 1606 19 2.0 6.4 Katsuwonus 3258 6066 32 13.5 0.6 Thunnus 26600 6480 24 8.9 no data Inactive species Callionymus 39 478 16 2.1 no data Ictalurus 239 no data 10 1.2 no data Opsanus 251 660 11 1.9 5.0 Tinca 268 1764 22 1.8 2.5 Turn over 31 (i) Lamellae are present on both sides of the gill filaments. The mean length of a gill filament is 25 mm. Calculate the total number of lamellae on the gill filaments of Thunnus. Express your answer in standard form. (1) Answer .............................................................. 32 *(ii) Analyse the data to explain the relationships between the activity of these fish and the structure of their gills. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 16 marks) TOTAL FOR PAPER = 90 MARKS

4 2 Plant growth substances and phytochromes are chemicals that affect the growth of plants. (a) (i) Which row of the table shows the concentrations required for the fastest growth of lateral buds? (1) A B C D Auxin concentration Cytokinin concentration high high high low low high low low (ii) Some types of auxin affect the activity of cells in the same way as hormones such as oestrogen affect human cells. These auxins pass through the cell membrane and affect the synthesis of enzymes. Explain how these auxins affect the synthesis of enzymes. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 5 BLANK PAGE 6 (b) Phytochrome is involved in the germination of seeds. The effect of red (R) light and far-red (FR) light on the germination of lettuce seeds was investigated. A scientist exposed five groups of 50 seeds to red light for one minute. The seeds were planted and the mean number that germinated was calculated. Another five groups each of 50 seeds were then exposed to red light for one minute, followed by far-red light for four minutes. These seeds were planted and the mean number that germinated was calculated. The experiment was repeated with different combinations of red light and far‑red light. The results are shown in the table. Sequence of red (R) and far‑red (FR) light Mean number of seeds that germinated Standard deviation (SD) R 45 ± 4 R, FR 5 ± 2 R, FR, R 45 ± 5 R, FR, R, FR 10 ± 4 R, FR, R, FR, R 40 ± 9 Turn over 7 Analyse the data to comment on the effect of red (R) and far-red (FR) light on the germination of lettuce seeds. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 8 marks)

8 3 The diagram shows the structure of an animal cell. S Q R P (a) (i) Which of the following is the name of the structure labelled Q? (1) A centriole B nucleus C rough endoplasmic reticulum D smooth endoplasmic reticulum (ii) These cells were cultured in radioactively-labelled amino acids. In which sequence would the cell structures in the diagram become radioactive during protein synthesis? (1) A Q S R B S P R C P R Q D P S R Turn over 9 (iii) Explain one function of lysosomes. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 10 (b) The diagram shows the development of a zygote into a blastocyst. zona pellucida trophectoderm inner cell mass Day 0 = 1 cell Day 1 = 2 cells Day 2 = 4 cells Day 3 = 8 cells Day 4 = 16 cells Day 5 = 32 cells Day 6 = 64 –100 cells The graph shows how the ratio of the volume of the nucleus to the volume of the cytoplasm of each embryonic cell changes as the blastocyst develops. 0 1 2 3 4 5 6 200 180 160 140 120 100 80 60 40 20 0 Time / days Ratio cell of inner cell mass trophectoderm Turn over 11 (i) The ratio shown in the graph can be calculated using the formula ratio = volume of nucleus total volume of cell – volume of nucleus On day 2, the volume of the cell nucleus was 900 μm3. Calculate the total volume of this cell on day 2. Give your answer to two significant figures. (3) Answer .............................................................. μm3 (ii) Comment on the changes in the ratios as the zygote develops into a blastocyst. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 10 marks) 12 BLANK PAGE

Turn over 13 4 The autonomic nervous system controls the heart rate. (a) The diagram shows a human brain. W X Y Z (i) Which region of the brain controls the heart rate? (1) A W B X C Y D Z (ii) Which one of the following would lead to an increase in heart rate? (1) A release of acetylcholine by the sympathetic nervous system B release of acetylcholine by the parasympathetic nervous system C release of noradrenaline by the sympathetic nervous system D release of noradrenaline by the parasympathetic nervous system 14 (b) The graphs show the pressure changes that occur in a healthy heart and the pressure changes that occur in a heart with an atrioventricular valve that leaks. 16 14 12 10 8 6 4 2 0 0.0 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 Pressure / kPa Time / s Healthy heart 16 14 12 10 8 6 4 2 0 0.0 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 Pressure / kPa Time / s Heart with an atrioventricular valve that leaks Key left ventricle aorta left atrium (i) An atrioventricular valve in the healthy heart has a surface area of 3.5 cm2. Determine the force that is applied to this atrioventricular valve when it closes. Use the formula Pressure in kPa = Force in newtons Area in m2 (3) Answer .............................................................. newtons Turn over 15 (ii) Explain why people with an atrioventricular valve that leaks find exercise difficult. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 9 marks) 16 BLANK PAGE

Turn over 17 5 The diagram shows a section of a human retina. X (a) Which of the following is the name of the cell labelled X? (1) A bipolar neurone B ganglion neurone C motor neurone D optic neurone (b) Explain the role of rhodopsin in the generation of a nerve impulse in cell X when light falls on the retina. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 18 *(c) A scientist investigated the effect of how the length of time a volunteer spent in darkness affected the sensitivity of their retina to light. This is the method used: • the volunteer sat in a room illuminated with white light for twenty minutes • the lights were switched off and the room was placed into total darkness for one minute • an LED emitting light of wavelength 500 nm at low light intensity was switched on • the light intensity was gradually increased until the volunteer was able to see the LED and this minimum light intensity was recorded • this was repeated with increasing lengths of time that the volunteer was kept in total darkness • the investigation was then repeated with an LED that emitted red light with a wavelength of 620 nm. The graph shows the results of this investigation. Time spent in darkness Minimum light intensity at which LED was seen 620 nm light 500 nm light The graph shows the relative sensitivity of rods and three types of cone to light of different wavelengths. 350 400 450 500 550 600 650 700 Wavelength / nm Relative sensitivity / a.u. Blue cones Green cones Rods Red cones Turn over 19 Analyse the graphs to explain the results of this investigation. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 5 = 10 marks)

20 6 The diagram shows a nephron. W X Y Z (a) Which labelled parts of this nephron would normally contain glucose? (1) A W only B W and X C W, X, and Y D W, X, Y, and Z (b) Oedema is an accumulation of tissue fluid. This is due to less tissue fluid being returned to the blood capillaries. (i) Explain how tissue fluid is returned to capillaries. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 21 (ii) Furosemides are drugs that are used to reduce oedema. They are diuretics that increase urine production. Furosemides reduce the active transport of sodium ions by the loop of Henle. Explain how furosemides cause an increase in urine production. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 8 marks)

22 7 The photograph shows a tawny owl. These owls live in areas close to the Arctic Circle. (Source: © LINDA WRIGHT/SCIENCE PHOTO LIBRARY) Scientists investigated the feather colours of a population of tawny owls. Feather colours of the owls were scored on a scale of 1 to 11, according to how grey or brown they were. Grey feathered owls were given a score between 6 and 1, with 1 being the most grey. Brown feathered owls were given a score between 7 and 11, with 11 being the most brown. The graph shows the percentages of owls in the population with each of the colour scores. 1 11 10 9 8 7 6 5 4 3 2 35 30 25 20 15 10 5 0 Colour score Percentage of owls with colour score (%) Grey Brown (a) The scientists concluded that feather colour in owls is controlled by a single genetic locus. The brown allele is dominant to the grey allele. (i) State what is meant by a dominant allele. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 23 (ii) Explain the range of feather colours, shown in the graph, in this population of owls. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) In this population of 120 owls, 84 of them were grey. Calculate the number of owls that are heterozygous for feather colour. Use the Hardy-Weinberg equation p2 + 2pq + q2 = 1 (3) Answer .............................................................. 24 *(b) The scientists investigated the effect of climate change on the frequencies of these two alleles for feather colour in populations of tawny owls. The percentage survival rates of both colours of owl were measured when there were different depths of snow. The mean depth of snow was recorded each year between 1980 and 2007. The percentage of owls that were coloured brown between 1960 and 2007 was also recorded. The graphs show the results. 0 5 10 15 20 25 100 80 60 40 20 0 Snow depth / cm Percentage of owls that survive (%) grey owls brown owls 1980 2010 2005 2000 1995 1990 1985 25 20 15 10 5 0 Year Mean snow depth / cm 1960 2010 2000 1990 1980 1970 50 40 30 20 10 0 Year Percentage of owls that are brown (%) Turn over 25 The scientists concluded that increased greenhouse gases from human activity are resulting in an increase in the frequency of the brown allele. Analyse the data to evaluate this conclusion. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 7 = 12 marks)

26 8 The photograph shows a glacier that is melting. As the glacier gradually melts, it leaves behind bare rock. melting glacier (Source: © CHARLES D. WINTERS/SCIENCE PHOTO LIBRARY) Scientists sampled the soil from areas that had been uncovered for different lengths of time after the glacier ice had melted. The table shows the results from the samples. Length of time since glacier melted / years Soil depth / cm Mass of nitrate in soil / mg m–2 Mass of leaf litter that falls to ground each year / g m–2 yr–1 5 5.2 3.8 1.5 15 6.4 4.1 1.7 40 7.0 5.3 12.8 100 9.2 21.8 277.0 150 10.8 35.4 335.0 200 15.1 53.3 277.0 250 16.2 61.4 261.0 (a) Which of the following should be used to determine if there is an association between length of time since the glacial ice has melted and soil depth? (1) A chi squared B correlation coefficient C standard deviation D t-test Turn over 27 (b) Explain how succession brings about the changes in the soil during this period of time. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 28 (c) The roots of some of the plants growing around the glacier were found to contain an oxygen binding protein called leghaemoglobin. The graph shows the oxygen dissociation curve for leghaemoglobin and the oxygen dissociation for haemoglobin. 0 20 18 16 14 12 10 8 6 4 2 100 80 60 40 20 0 Partial pressure of oxygen / kPa Oxygen saturation (%) haemoglobin leghaemoglobin (i) Leghaemoglobin is a protein that is similar in structure to myoglobin. Both of these proteins have one subunit. Explain how the oxygen dissociation curve shows that the structure of leghaemoglobin is more similar to myoglobin than to haemoglobin. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 29 (ii) Explain how leghaemoglobin enables these plants to grow in waterlogged and compact soils. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 12 marks) 30 BLANK PAGE

Turn over 31 9 In Kenya, an in-situ conservation project has been set up to make paper from elephant dung. This project aims to help conserve elephant populations, conserve biodiversity and help local communities. (a) Paper is produced from cellulose molecules present in elephant dung. The diagram shows part of a cellulose molecule. H O H OH CH2OH H OH H H H O H OH OH CH2OH H H O H X (i) Which of the following is the name of the monomer labelled X on the diagram? (1) A α glucose B β glucose C α ribose D β ribose (ii) Draw one of the products of the hydrolysis of this part of the cellulose molecule. (2) 32 (iii) Explain how the structure of cellulose is adapted for its function in plant cell walls. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Elephant conservation often involves the setting up of protected areas to stop poaching. Local people are banned from entering these areas. This method of elephant conservation sometimes causes conflict with local people, as elephants can trample and eat crops in nearby farmland. Farmers sometimes burn areas of grassland and forest to stop elephants destroying their farmland. The new paper conservation project encourages local people to make paper from elephant dung they collect from the conservation area. The paper they make is sold at a high price. Conservationists measured the effect of the new paper project on the index of diversity of the area over two years. They also measured the effect of a protected conservation area where local people were banned from entering on the index of diversity. The results are shown in the table. Area Index of diversity At start of project After two years Conservation area with paper project 2.8 3.6 Conservation area where local people are banned from entering 2.4 2.2 Turn over 33 (i) State why calculating an index of diversity is a better measure of biodiversity than counting the number of different species. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Give two advantages of in-situ conservation of elephants compared with ex-situ conservation of elephants. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Explain the effect of the new paper project on biodiversity compared with the protected area where local people were banned. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 34 (c) Describe the limitations of CITES in the conservation of organisms such as elephants. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 15 marks) TOTAL FOR PAPER = 90 MARKS 35 BLANK PAGE 36 BLANK PAGE

2 Answer ALL questions. Write your answers in the spaces provided. 1 Nerve impulses are transmitted along neurones. (a) (i) In the time period between two nerve impulses, the potential difference across the membrane is –70 mV. State the term given to this potential difference. (1) .................................................................................................................................................................................................................................................................................... (ii) When the neurone is stimulated, an action potential may occur in the axon, due to the change in permeability of the membrane to ions. The graph shows an action potential. 2.5 2.0 1.5 1.0 0.5 +50 +30 +10 –10 –30 –50 –70 –90 Time / ms Membrane potential / mV Describe what happens in the axon membrane to cause the change in potential difference between 0.55 ms and 0.80 ms. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 3 (b) Scientists investigated the effect of stimulating an axon with different voltages and measured the response. The table shows the results of this investigation. Stimulus voltage / mV Response 25 No action potential 35 No action potential 45 No action potential 55 Action potential 65 Action potential 75 Action potential Analyse the data to describe the conclusions which can be drawn. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 1 = 7 marks)

4 2 A student investigated the number of dividing cells in an onion root tip. The student made a root tip squash and counted the number of cells at each stage of the cell cycle. The table shows the results. Number of cells at each stage of the cell cycle Prophase Metaphase Anaphase Telophase Interphase 4 2 1 1 90 (a) (i) Calculate the mitotic index for these results using the formula Mitotic index = number of actively dividing cells total number of cells × 100 (2) Answer .............................................................. (ii) The mitotic index depends on the distance from the root tip. The graph shows the relationship between the mitotic index and the distance from the root tip. 2.0 1.5 1.0 0.5 0 14 12 10 8 6 4 2 0 Distance from root tip / mm Mitotic index Turn over 5 Explain the relationship shown in the graph. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The student used the following method to prepare the root tip squash. 1. Cut a small piece from the tip of a growing onion root and place it in weak acid. 2. Transfer the root tip to a microscope slide. 3. Add a drop of stain. 4. Use a mounted needle to flatten the piece of root and place a coverslip on top. 5. View using low power objective lens of a microscope. 6. Count the cells at each stage of the cell cycle. Justify two improvements to this method. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 6 marks)

6 3 Single-celled algae can be trapped in gel beads and used to study photosynthesis. The beads are placed in a test tube of hydrogencarbonate indicator. The table shows the colour of the indicator when it contains different concentrations of carbon dioxide. Colour of indicator Relative carbon dioxide concentration yellow highest orange higher than atmospheric air red same as atmospheric air magenta lower than atmospheric air purple lowest A student used the following method to investigate the effect of light intensity on the rate of photosynthesis. 1. Set up five test tubes, each half-filled with red hydrogencarbonate indicator. 2. Add a teaspoon of gel beads containing single-celled algae to each test tube and close with a bung. 3. Place each test tube at a different distance from a lamp in a dark room. 4. Leave the tubes for 30 minutes. 5. Record the colour of the hydrogencarbonate indicator in each tube and the position of the gel beads. The table shows the results. Distance from lamp / cm 5 15 25 35 45 Colour of indicator after 30 minutes purple purple red orange orange Position of gel beads in tube after 30 minutes half‑way up less than half‑way up bottom bottom bottom Turn over 7 (a) (i) Describe two control tubes that should be used in this investigation. 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(ii) Explain the changes in colour of the hydrogencarbonate indicator in this investigation. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 8 (b) The colour of the indicator and the position of the beads can be used to give a quantitative measure of the effect of light intensity. Describe how the method could be modified to give valid, quantitative results. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 8 marks)

Turn over 9 4 Fruit flies, Drosophila melanogaster, are often used in genetics investigations. The diagrams show male and female flies. Female Male Actual size = 3 mm (a) Give one feature, visible in the diagrams, that you could use to distinguish between male and female flies and state how it differs between them. (1) Feature .................................................................................................................................................................................................................................................................................... Difference .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) A scientist investigated the inheritance of several genes in fruit flies. (i) A grey-bodied fly was crossed with a black-bodied fly. All the offspring in the F1 generation were grey-bodied. Two of the grey-bodied flies from the F1 generation were then crossed. Determine the expected ratio of phenotypes of the offspring in the resulting F2 generation by using a genetic diagram. (2) Answer .......................................................................................................................................................... 10 (ii) Normal fruit flies have long wings but some flies have very short wings called vestigial wings. The allele for long wing (L) is dominant to the allele for vestigial wing (l). The gene for wing length is located on chromosome 2. The allele for normal antennae (A) is dominant to the allele for bushy antennae (a). The gene for antennae shape is located on chromosome 3. Two flies which are heterozygous for both characteristics are crossed. Determine the ratio of phenotypes that you would expect in the next generation, using a genetic diagram. 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Turn over 11 (iii) The allele for red eyes (R) is dominant to the allele for white eyes (r). The scientist investigated the inheritance of eye colour. The crosses used were: cross 1 – a red-eyed female fly was crossed with a white-eyed male fly: half of the offspring were red-eyed females and half were red-eyed males cross 2 – a white-eyed female fly was crossed with a red-eyed male fly: half of the offspring were red-eyed females and half were white-eyed males. Explain the results of these crosses. 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(Total for Question 4 = 10 marks)

12 5 (a) The photograph shows a germinating pollen grain as seen through a light microscope. (Source: © CAROLINA BIOLOGICAL SUPPLY COMPANY/SCIENCE PHOTO LIBRARY) The actual length of this pollen tube is 136 µm. Calculate the magnification of this photograph. (2) Answer .............................................................. Turn over 13 (b) (i) The diagram shows a pollen grain that has begun to germinate after landing on the surface of the stigma of a flower. wall of pollen grain B A pollen tube top of stigma cytoplasm Identify the structures labelled A and B. (2) A ............................................................................................................................................................................................................................................................................... B ............................................................................................................................................................................................................................................................................... (ii) Describe the process of double fertilisation in flowering plants. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 14 (c) Devise an investigation to determine the effect of pH on the rate of growth of pollen tubes. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 12 marks)

Turn over 15 6 Fritillaria delavayi is a small plant (height 7 cm), that grows on rocky slopes on mountains in China. The image shows Fritillaria delavayi. bright green leaves bright yellow flower For at least 2 000 years, this plant has been collected and used in Chinese medicine. It is not known to be eaten by animals. In less accessible regions, where few humans go, the plants are bright green with bright yellow flowers. In locations where bulbs are collected in high numbers, most plants have greyish- brown leaves and flowers. Scientists believe that the greyish-brown plants are the same species as the brightly‑coloured plants. (a) Explain why most of the plants in areas where bulbs are collected in high numbers have greyish-brown leaves and flowers. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16 (b) (i) Describe how scientists can use gel electrophoresis to show that these plants belong to the same species. 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(ii) Explain why the features of the brightly-coloured plants enable them to grow successfully in the areas where they are not collected by humans. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 10 marks)

Turn over 17 7 Plants require mineral ions from the soil for healthy growth. (a) (i) Describe how mineral (inorganic) ions are taken up by active transport. 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(ii) Describe the function of a named mineral ion that is vital for the growth of plants. (2) Mineral ions .................................................................................................................................................................................................................................................................................... Function .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 18 (b) The photograph shows Galium verum (Lady’s bedstraw), a plant that grows in meadows, hedges, road verges and sand dunes. (Source: © Alfio Scisetti/Alamy Stock Photo) This plant grows to between 15 cm and 60 cm in height. A student investigated whether Galium verum growing in meadows was taller than Galium verum growing on sand dunes. (i) Give a suitable null hypothesis for this investigation. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 19 (ii) Devise a method that can be used to collect valid results to test this null hypothesis. 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(iii) Explain which statistical test would be most suitable to test this null hypothesis. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 13 marks)

20 8 The activity of enzymes is affected by a number of factors. (a) Explain why increasing the temperature of an enzyme-controlled reaction changes the rate of reaction. 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Turn over 21 (b) A student investigated the activity of the enzyme catalase in potato. This enzyme is found in potato cells. It catalyses the conversion of hydrogen peroxide to oxygen and water. The diagram shows the apparatus used to collect the data. beaker containing water retort stand burette oxygen potato hydrogen peroxide rubber tube (i) Describe how two named variables, other than the potato, can be controlled in order to obtain valid results. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22 (ii) The graph shows the results of the investigation. 300 270 240 210 180 150 120 90 60 30 0 50 45 40 35 30 25 20 15 10 5 0 Time / seconds Volume of oxygen produced / cm3 Determine the initial rate of reaction from this graph. (1) Answer .............................................................. Turn over 23 (iii) Explain why the rate of reaction decreases over time. 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(c) The student used this apparatus to investigate the effect of using increasing numbers of potato cubes on the rate of reaction. The potato was cut into 1 cm3 cubes to be used in five trials. In the first trial, one cube was added to the flask and the volume of oxygen produced in 30 seconds was measured. The hydrogen peroxide was then replaced. This was repeated in the other four trials using 2, 3, 4 and 5 cubes of potato. Explain the results you would expect from this investigation. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 13 marks) 24 BLANK PAGE

Turn over 25 9 Water moves through xylem vessels in a plant during transpiration. Phloem tissue is responsible for transport of organic materials such as sucrose. (a) Compare and contrast the structure of xylem tissue and phloem tissue. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) A student used the potometer shown in the diagram to investigate the rate of water uptake in a leafy shoot. cut shoot reservoir scale air bubble beaker of water capillary tube The student investigated the effect of a combination of environmental conditions on the rate of water uptake. 26 The table shows the results of this investigation. Conditions Distance moved by the bubble in 5 minutes / cm Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Mean Still air, in light 2.4 2.2 2.2 2.2 2.1 2.2 Moving air, in light 6.5 6.7 6.4 6.2 6.3 6.4 Still air, in dark 0.9 0.4 0.0 0.0 0.0 0.3 Moving air, in dark 1.7 0.8 0.0 0.0 0.0 0.5 (i) The internal diameter of the capillary tubing is 0.3 mm. The volume of a cylinder is calculated using the formula πr2h Calculate the mean rate of water uptake for the shoot in moving air, in the light. Give your answer in mm3 min–1 to two significant figures. (3) Answer .............................................................. mm3 min–1 Turn over 27 (ii) Analyse the data to explain the results of this investigation. 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(iii) Describe how this investigation could be modified to make a valid comparison of water loss from the upper and lower surfaces of the leaves. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 13 marks) 28 BLANK PAGE

Turn over 29 10 Grassland is an important habitat for grazing animals such as cattle. (a) Fields used for grazing cattle must have a high gross primary productivity. State what is meant by the term gross primary productivity and include the units in which it is measured. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Cattle graze on the plants on grassland. Describe what happens to the energy in the plants that are not eaten by cattle. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 30 *(c) When grazing animals such as cattle, sheep and horses are kept in fields, their faeces are left on the ground, and are known as dung. The dung is eaten by dung beetles, which in turn are eaten by other animals. Dung beetles bury dung up to one metre underground in tunnels they have made. This removes the dung from the surface of the field. A cow can produce over eight tonnes of dung per year. There are around 60 species of dung beetle in the UK. In 2016, 50% of these were identified as being scarce or threatened. The photograph shows a dung beetle. (Source: © Simon Webster/Alamy Stock Photo) Scientists believe factors causing the decline of dung beetles include: • farmers using more land to grow crops, and using more land for building and development • keeping some grazing animals in sheds over the winter, rather than them spending the winter in fields • fewer habitats for dung beetles, leading to populations of dung beetles becoming isolated and preventing outbreeding • grazing animals routinely being treated with anti-parasitic drugs that can kill dung beetles. Turn over 31 Some scientists have stated that farming practices need to change to conserve the species of dung beetle, and to maintain biodiversity and the stability of ecosystems. Discuss the validity of this statement. 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(Total for Question 10 = 13 marks)

Turn over 33 11 Yeast, Saccharomyces cerevisiae, is a single-celled organism used in brewing and baking. The photograph shows yeast cells seen using a light microscope. (Source: © Kateryna Kon/Shutterstock) 34 A student counted the number of yeast cells, at 90-minute intervals, growing in a liquid culture at 30°C. The graph shows the results. 12 11 10 9 8 7 6 5 4 3 2 1 Time / hours Number of yeast cells × 102 per cm3 0 200 100 0 400 300 Turn over 35 (a) (i) Calculate the percentage increase in the number of yeast cells from 2 hours to 6 hours. (2) Answer .............................................................. % (ii) Predict the number of cells per cm3 of the culture at 12 hours. Assume that the culture continues to grow at the same rate. (1) Answer .............................................................. × 102 per cm3 (b) Describe two precautions that would prevent the yeast culture becoming contaminated with bacteria. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 36 (c) The student counted the yeast cells using a haemocytometer. This is a slide with chambers containing a known volume of liquid. The yeast cells are counted when viewed using a light microscope. Any yeast cells totally within the counting square should be counted, along with those which overlap the top or left-hand side of the square. (i) The volume of liquid in the counting square shown is 0.004 mm3 . (Source: © Joseph Elsbernd/Flickr) Calculate the number of yeast cells per mm3 . Give your answer in standard form. (2) Answer .............................................................. per mm3 (ii) Identify two causes of inaccuracy in this method of counting cells. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 37 (d) The table shows the number of cells counted using a haemocytometer. Time / hours Number of cells 1 2 3 4 5 Mean Standard deviation (SD) 0.0 1 2 2 1 2 1.6 0.5 1.5 2 3 3 3 2 2.6 0.5 3.0 4 9 9 3 7 6.4 4.5 13 10 12 9 16 12.0 2.7 6.0 43 31 18 14 15 24.2 12.5 (i) The student calculated the standard deviations using the following formula. SD x x n 2 1 Calculate the standard deviation for the sample at 3 hours. (3) Answer .............................................................. 38 (ii) The student concluded that the yeast population doubles every 90 minutes. Comment on the validity of this conclusion. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 11 = 15 marks) TOTAL FOR PAPER = 120 MARKS 39 BLANK PAGE 40 BLANK PAGE

4   2 Anthropogenic activities are contributing to global warming. Scientists from the IPCC (Intergovernmental Panel on Climate Change) say that changes to diet can have a major impact on greenhouse gas emissions. The method of farming affects the level of impact. The Paris Agreement of 2015 committed countries to reduce carbon emissions. (a) How many of the following statements about global warming are correct? (1) • global warming is caused by a reduction in greenhouse gases • deforestation may contribute to global warming • global warming is leading to the melting of sea ice • global warming may lead to a change in rainfall patterns • global warming is affecting only the ice caps A 2 B 3 C 4 D 5 Turn over 5   (b) The table shows the mass of greenhouse gas released in the production of one serving of a variety of sources of protein. Source of protein Mass of greenhouse gas released in the production of one serving of the protein / kg Beef 7.0 Lamb 5.0 Chicken 2.5 Cheese 2.0 Tofu (from soya beans) 1.0 Nuts 0.5 (i) Explain why a diet based on plant protein produces lower greenhouse gas emissions than a diet based on animal protein. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 6   (ii) Beef cattle are traditionally reared by grazing on natural pastures (grassland). In recent years, large areas of rainforest have been cleared to produce beef. Explain why the farming of beef cattle on deforested land produces more greenhouse gas emissions than from those reared on natural pastures. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks) Turn over 7   BLANK PAGE

8   3 Peat bogs are formed over millions of years from the remains of plants and animals. Waterlogging and acidic conditions prevent the decomposition of plants and animals in peat bogs. The photograph shows peat being cut from a peat bog. (Source: © Reimar/Shutterstock) (a) (i) State how the age of the layers in a peat bog can be determined. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Explain why the conditions in peat bogs prevent decomposition. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 9   (b) In the 1980s, there was a scheme to plant trees in areas of peat bog. This dries out the bog and allows it to decompose, releasing the carbon dioxide locked in the peat bog. Conservationists are cutting down trees and raising the water level to recreate the peat bog. It is estimated that the peat bogs will recover in 15 years. The photograph shows sphagnum moss that grows on the surface of peat bogs and helps their formation. (Source: © Jan Holm/Shutterstock) Devise a procedure to measure the rate of recovery of the peat bog. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 8 marks)

10   4 Malaria is a disease that kills more than 400 000 people every year. It is caused by Plasmodium, a single-celled organism that lives in the blood. Plasmodium is transmitted by mosquitoes. The diagram shows a Plasmodium organism. (a) State and justify two structures shown in the diagram that indicate that Plasmodium is a eukaryotic organism. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 11   (b) Malaria can be controlled by killing the mosquitoes that transmit the disease. Scientists are genetically modifying Metarhizium pingshaense, a fungus that infects mosquitoes. The genetically modified (GM) fungus contains a gene from a species of spider. This gene codes for a protein that kills mosquitoes. (i) The GM fungus transcribes and translates the gene for this protein. Describe the primary structure of a protein. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) The primary structure of this protein is then converted into a tertiary structure and modified by organelles in the cell. Describe the role of the organelles involved in these processes. 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(Total for Question 4 = 9 marks)

12   5 Porphyria is a life-threatening genetic disease. It is caused by a mutation in the gene coding for an enzyme involved in the production of haem. In people with porphyria, haem cannot be produced. (a) Explain why this mutation may prevent the enzyme involved in the production of haem from functioning correctly. 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Turn over 13   (b) Porphyria is caused by a dominant allele and may not develop until later in life. The pedigree diagram shows a family in which some individuals have porphyria. Key Affected male Affected female Unaffected male Unaffected female A B C (i) State the genotype and phenotype of person C. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Person A and Person B have one son with porphyria. What was the probability of them having a child that was male and had porphyria? (1) A 0.00 B 0.25 C 0.50 D 1.00 14   (c) A new technique known as gene silencing has been developed to treat this disease. Molecules known as small interfering RNA (siRNA) combine with mRNA to prevent translation. Deduce how siRNA may be used to prevent the development of porphyria. 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(Total for Question 5 = 10 marks) Turn over 15   6 Leigh syndrome is a disease of the nervous system that affects the brain. It is caused by a mutation in the DNA present inside mitochondria. (a) Draw and label a mitochondrion. (2) (b) Mitochondrial diseases such as Leigh syndrome are passed on by the mother during fertilisation. Explain why a fertilised egg cell will contain only maternal mitochondria. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16   (c) Using IVF, it is possible to produce an embryo that does not have the mitochondrial disease. The nucleus is removed from a donor egg cell from another individual. The nucleus from a fertilised egg from a mother with the mutation is then placed in the egg cell from the donor. (i) Explain why the resulting embryo does not develop mitochondrial disease. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Comment on the inheritance of parental characteristics in offspring produced in this way. 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(Total for Question 6 = 9 marks) Turn over 17   7 Measles is a contagious disease that can be controlled by vaccination. Not all parents choose to have their children vaccinated. The table shows the number of cases of measles and the percentage of children vaccinated in the UK between 2012 and 2017. Year Number of cases of measles Percentage of children vaccinated (%) 2012 1564 91.2 2013 1855 92.3 2014 135 92.7 2015 71 92.3 2016 556 91.9 2017 216 91.6 (a) (i) Calculate the percentage change in the number of cases of measles from 2013 to 2014. Give your answer to three significant figures. (2) ................................................................ % 18   (ii) One conclusion from the data is that it takes time for an increase in vaccination rate to reduce the number of cases of measles in children. Explain why this is a valid conclusion. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 19   (b) The incidence of measles has risen by 300% worldwide in recent years. A study has found that some B memory cells and antibodies are destroyed by the measles virus. (i) Antibodies are released into the blood when (1) A B cells are activated to become killer cells B B cells are activated to become plasma cells C macrophages are activated to become B cells D plasma cells are activated to become memory cells. (ii) Vaccinations are carried out against many serious diseases. Vaccination leads to the production of antibodies and memory cells. Which kind of immunity is provided by this type of vaccination? (1) A active artificial B active natural C passive artificial D passive natural (iii) Explain why, following a measles infection, it may be advisable for children to repeat other vaccinations they have had. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 11 marks)

20   8 Tuberculosis (TB) is a disease that affects the lungs. It is caused by the bacterium Mycobacterium tuberculosis. When these bacteria enter the body an immune response is triggered. (a) The initial response when bacteria enter the body is a (1) A non-specific response and bacteria are destroyed by antibodies B non-specific response and bacteria are destroyed by phagocytes C specific response and bacteria are destroyed by antibodies D specific response and bacteria are destroyed by phagocytes. (b) M. tuberculosis bacteria can remain dormant in the body after infection. Explain why these dormant bacteria are not destroyed by the immune system. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 21   (c) The Human Immunodeficiency Virus (HIV) affects the immune system. People infected with HIV are more likely to develop TB. A study investigated the relationship between TB infections and HIV status. In 2008, the number of people in the world estimated to have TB was 9.4 million. The table shows the percentage of these people who were either HIV positive or HIV negative. The estimated number of deaths due to TB is also shown in the table. HIV status Percentage of people with TB in 2008 (%) Estimated number of patients with TB who died in 2008 Percentage of deaths due to TB (%) HIV positive 15 521 700 HIV negative 85 1 278 400 16 (i) Calculate the percentage of TB patients infected with HIV who died of TB. (2) ................................................................ % (ii) Describe the effect of HIV on the number of deaths from TB. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22   *(iii) There are increased numbers of deaths from TB as a result of HIV infection. The table shows data for deaths worldwide from HIV and TB in 2016. Type of infection Number of deaths TB only 1 300 000 HIV only 826 000 TB and HIV 374 000 Treatment of HIV with anti-viral drugs has reduced the incidence of TB associated with HIV infection. The replication of HIV reduces the T-helper cell count. These anti-viral drugs reduce the replication of the virus in the body. The graph shows the effect of this on the incidence of TB. 500 400 300 200 100 0 Number of T-helper cells mm−3 25 20 15 10 5 0 TB incidence / a.u. Turn over 23   Explain the effect of using anti-viral drugs to treat HIV on the number of deaths from TB. 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(Total for Question 8 = 14 marks) 24   BLANK PAGE Turn over 25   9 A biodiversity hotspot is a region that is both highly diverse and threatened with destruction. (a) Biodiversity can be measured by the number of (1) A different genes in a population. B different species in a habitat. C homozygotes in a habitat. D individuals in a population. (b) Biodiversity hotspots have at least 1500 endemic plant species. These hotspots have lost at least 70% of their natural vegetation. Explain how protection of these hotspots can affect global biodiversity. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26   (c) A population of white-tailed deer Odocoileus virginianus was introduced from North America into Finland in 1934. There were four females and one male. The population has since increased rapidly. A study in 2012 compared the genetic diversity of the population in Finland with the population in North America. Ten genes were compared from 72 deer in each population. The allele richness (number of alleles in the population) of each gene was measured. A χ 2 test was carried out to compare the two populations. (i) The null hypothesis for this investigation is (1) A the allele richness in the population from Finland is higher than the USA. B the allele richness in the population from the USA is higher than Finland. C the population in the USA is more genetically diverse than the population in Finland. D there is no difference in allele richness between the two populations. Turn over 27   (ii) The values from the population in the USA can be taken as the expected values and a χ 2 test can be carried out. Gene Allele richness (Finland population) X (observed) Allele richness (USA population) Y (expected) X − Y (X − Y)2 (X − Y)2 / Y 1 6 14 8 64 4.57 2 4 5 1 1 0.2 3 7 13 6 36 2.77 4 7 15 8 64 4.27 5 6 6 0 0 0 6 8 12 4 16 1.33 7 2 3 1 1 0.33 8 4 4 0 0 0 9 4 9 10 3 4 Complete the table to calculate the χ 2 value, using the formula ( ) 2 2 O E E χ − = ∑ (3) Answer ................................................................ 28   (iii) The table shows the critical values of chi-squared at different levels of probability. Degrees of freedom Probability p=0.10 p=0.05 1 2.706 3.841 2 4.605 5.991 3 6.251 7.815 4 7.779 9.488 5 9.236 11.070 6 10.645 12.592 7 12.017 14.067 8 13.362 15.507 9 14.684 16.919 10 15.987 18.307 Deduce the effect of a small founder population on the allele richness in the population of white-tailed deer in Finland. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 29   (d) The table shows the heterozygosity index for each population of white‑tailed deer. Heterozygosity index for Finland population Heterozygosity index for North American population 0.692 0.742 Climate change may affect the habitat of both populations of deer. Explain which population is more likely to adapt to changing conditions. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 9 = 14 marks)

30   10 Some plants are adapted to grow in shady conditions. A study was carried out to compare the effect of growing in different light intensities on rates of photosynthesis of plants. The plants are rated for levels of shade tolerance. Plant seedlings of nine species were grown in either 25% or 5% of full sunlight. All other abiotic factors were controlled. These seedlings were grown for six weeks and then exposed to full sunlight for 15 minutes. The rate of photosynthesis was measured during the exposure to full light. The table shows the results of this investigation for four species of plant. Species Shade tolerance Rate of photosynthesis / a.u. Seedlings grown in 25% of full sunlight Seedlings grown in 5% of full sunlight A Intolerant 410 415 B Intermediate 300 275 C Tolerant 180 210 D Very tolerant 150 215 (a) (i) Determine which species had the greatest percentage change in rate of photosynthesis when grown in lower light intensities. (2) Answer ................................................................ Turn over 31   (ii) The rate of growth for all nine species of plant was also measured. The graph shows the effects of the rate of photosynthesis on the rate of growth of these plants. 500 400 300 200 100 0 Rate of photosynthesis / a.u. Rate of growth / a.u. 200 150 100 50 0 25% light 5% light Explain the effect of light intensity during the first six weeks of growth on the growth rate of these species of plant. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 32   *(b) Differences in the rate of photosynthesis may be due to the light‑dependent reactions in chloroplasts. Devise an investigation to compare the rate of the light‑dependent reaction in shade tolerant and shade intolerant plants grown at different light intensities. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 10 = 12 marks) TOTAL FOR PAPER = 100 MARKS

4   2 The human brain is an organ that has a range of functions. (a) A student produced a table linking three descriptions of brain function to a labelled diagram of the brain. Label Description of function of labelled region in brain 1 Site where more synapses are formed during the critical window for vision 2 Required for the fine motor skills to draw a straight line 3 Area that sends impulses to the sinoatrial node (SAN) in response to a lowering of blood pH Which diagram of the brain correctly matches the table? (1) 1 3 2 2 1 3 Diagram A Diagram B 3 1 2 3 2 1 Diagram C Diagram D Turn over 5   (b) There are various types of scan that can be used to study the brain. (i) Describe why a region of the brain might appear lighter in an image obtained by a functional magnetic resonance image (fMRI) scan. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) One scan combines positron emission tomography (PET) and computed tomography (CT). This scan can be used to show whether a person has cancer. Describe why a combined PET and CT scan may be better for diagnosing cancer than a PET or CT scan on its own. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks)

6   3 Some lipoproteins may reduce fertility in mammals. (a) An investigation was carried out using mice of the same breed. Female mice were divided into three groups, P, Q and R. Each group was given a different treatment. The table shows the treatment given to the three groups of female mice. Group Treatment P Genetically modified (GM) only Q Genetically modified and supplied with drug K R Supplied with drug K only The GM mice have an increased concentration of high‑density lipoproteins (HDLs) in their blood plasma. Drug K was used to lower the blood plasma HDL levels. The mice in each group were allowed to breed with non‑genetically modified male mice and the number of offspring born was recorded. The graph shows the results. P Q R Group 7 6 5 4 3 2 1 0 Mean number of offspring born per month per female Turn over 7   (i) The mice were all of one breed to improve data (1) A accuracy B precision C quantity D validity (ii) State a null hypothesis for this investigation. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Comment on the effect of blood plasma HDL levels on the fertility of these female mice. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 8   (b) Which of the following are always the same for both mammalian sperm and egg cells of the same species? 1. The total mass of nuclear DNA 2. They both release enzymes by exocytosis (1) A Neither of them B 1 only C 2 only D 1 and 2 (c) The diagram shows a mammalian sperm cell. 1 2 4 3 Which of the labelled parts contains enzymes that phosphorylate glucose during respiration? (1) A 1 B 2 C 3 D 4 (Total for Question 3 = 8 marks) Turn over 9   4 A person arrived at hospital having eaten some poisonous berries. (a) An electrocardiogram (ECG) was recorded for this person. The diagram shows part of the ECG obtained. Z 0.2 seconds (i) Tachycardia for an adult is defined as a heart rate above 100 beats per minute. Calculate the percentage increase above 100 beats per minute for the heart rate of this person. (2) Answer ..............................................................% (ii) The length of time labelled Z on the ECG was shorter than for a person with a normal heart rate. Which of the following is shown by label Z on the diagram? (1) A the time that the atria are contracting B the time for the atrioventricular (AV) valves to open C the time that the atrioventricular node (AVN) delays depolarisation D the time that the ventricles are contracting 10   (b) One of the poisons in the berries can bind to acetylcholine receptors on the surface membrane of cells in the SAN. This prevents acetylcholine binding to the receptors if it is in low concentration. This person was treated with a drug that stopped acetylcholinesterase from being released into the synaptic gap. Explain how this drug lowered the heart rate of this person. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) The poison also caused the pupils of this person to dilate. Describe the interaction of the muscles in the eye that led to this dilation of the pupils. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 9 marks) Turn over 11   5 A spirometer can be used to study aspects of the human respiratory system. (a) The diagram shows a trace produced using a spirometer. 1 3 4 2 Which of the following parts of the spirometer trace could be used to determine the maximum volume of air that can be exhaled? (1) A 1 only B 2 and 3 combined C 3 only D 3 and 4 combined 12   (b) The availability of oxygen in the air is lower at high altitudes because of the lower air pressure. Populations of humans that have lived at high altitude for many generations have adapted to these conditions. One adaptation these people have is a larger thorax than people living at lower altitudes. A student made the following hypothesis: People with this adaptation will have a lower breathing rate as their tidal volume will be greater than those without the adaptation. (i) Devise a valid investigation, using a spirometer, to test this hypothesis. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 13   (ii) Another adaptation to living at high altitude is to have more red blood cells circulating in their blood. Explain how a bone marrow stem cell can give rise to red blood cells. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 9 marks)

14   6 The nervous system is one of the features found only in animals. (a) The graph shows the change in potential difference across the cell surface membrane of a neurone when it is stimulated at point W. 0 2 4 6 8 12 10 W X Time / milliseconds 60 40 20 0 –20 –40 –60 –80 –100 Potential difference / millivolts (i) The threshold intensity is the minimum electrical stimulus that will cause a neurone to form an action potential. Which potential difference could be the threshold potential for this neurone? (1) A –80 millivolts B –70 millivolts C –60 millivolts D 40 millivolts Turn over 15   (ii) Explain the changes in the sodium and potassium ion concentrations in the cytoplasm of the neurone from point W to point X on the graph. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Explain why the neurone becomes hyperpolarised after point X on the graph. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 16   (b) The junction between the end of a motor neurone and a muscle myofibril is a modified synapse. The diagram shows this neuromuscular junction. muscle cell nucleus synaptic cleft acetylcholine vesicles motor neurone axon myofibril motor neurone endplate mitochondria Describe how the neurotransmitter crosses this synapse. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 10 marks) Turn over 17   7 Athletes compete in a range of environmental conditions. (a) An investigation studied the effects of environmental conditions on athletes. A group of nine athletes ran for 50 minutes in two sets of environmental conditions: • warm temperature and high humidity • cool temperature and low humidity. All other variables were kept constant including the level of exercise. The core temperature of each athlete was recorded during the exercise and for 70 minutes following the exercise. The results are shown in the graph. 120 100 80 60 40 20 0 Time / minutes 40.0 39.5 39.0 38.5 38.0 37.5 37.0 36.5 Mean core temperature / °C Key warm temperature and high humidity cool temperature and low humidity (i) Explain why there was a change in core temperature during the first 20 minutes of exercise. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 18   (ii) Using the results of this investigation, a student concluded that during exercise: • the human thermoregulatory system only works in cool environmental temperatures when the humidity is low. Comment on this conclusion. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Compare and contrast the changes in core body temperature, in the two environments, after the exercise was completed. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 19   (b) In another investigation, samples of blood were taken from athletes during exercise and after exercise. The concentration and activity of neutrophils, a type of phagocyte, in the blood were measured. The table shows the results. Measurements During exercise Immediately after exercise Mean concentration of neutrophils in blood / 106 cells cm–3 3.57 3.42 Mean activity of neutrophils / a.u. 28.4 22.0 (i) The mean body mass of the athletes was 70 kg and the mean volume of blood per athlete was 5 dm3. Calculate how many fewer neutrophils there are in the blood per kilogram of body mass immediately after exercise. (3) .............................................................. kg–1 (ii) State how these results support the statement that exercise increases the risk of infection. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 7 = 13 marks)

20   8 Cells can be classified as either prokaryotic or eukaryotic. (a) Which row identifies the type of DNA found in prokaryotic cells? (1) A B C D Plasmid DNA is Chromosomal DNA is double‑stranded double‑stranded double‑stranded single‑stranded single‑stranded double‑stranded single‑stranded single‑stranded (b) Eukaryotic cells contain a range of membrane‑bound organelles including the Golgi apparatus and the rough endoplasmic reticulum (rER). (i) Name two other organelles that are bound by a single membrane. (1) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 21   (ii) The table shows the percentage of cell membranes that make up Golgi apparatus and rER in the cells from two organs. Organ Percentage of cell membranes (%) Golgi apparatus rER Liver 7 35 Pancreas 10 60 Explain the differences in the percentage of membranes that comprise the two organelles in these cells. (5) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 22   (c) In one study, measurements were taken of a single Golgi apparatus. The measurements were: • total number of cisternae = 6 • height of each cisterna = 15 nm • distance between adjacent cisterna = 25 nm Calculate the total height of the stack of cisternae in this Golgi apparatus. Give your answer in micrometres. (2) .............................................................. μm (d) Beetroot cells contain the pigment betalain. When beetroot cells are placed in alcohol, the concentration of betalain in the cells changes. Explain why alcohol affects the concentration of betalain in these cells. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 12 marks) Turn over 23   9 The inheritance of coat colour in mice has been investigated. Some scientists crossed mice that had yellow coats. The offspring had either yellow coats or non‑yellow coats. (a) (i) Explain how the scientists knew which allele for coat colour was recessive. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) The crosses led to a total of 1599 offspring being produced. Predict the number of yellow and non‑yellow offspring produced. (2) Offspring with a yellow coat Offspring with a non‑yellow coat 24   (iii) The table shows the actual number of offspring with each coat colour. Offspring with a yellow coat Offspring with a non‑yellow coat 1064 535 Explain why there are differences between the predicted and actual numbers of these mice. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 25   *(b) There is a gene in humans similar to the mouse coat colour gene. This human gene comprises 74 775 base pairs. The mRNA produced from this gene codes for a protein made up of 132 amino acids. Explain the differences in the gene coding for this protein and the mRNA produced from it. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 9 = 13 marks)

26   10 All mammals have an internal skeleton that includes bone and cartilage. Bones contain calcium ions. (a) Explain one way in which cartilage at the ends of the bones in the knee joint reduces wear and tear. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) In humans, calcium ions are important for structure as part of bone material. These ions are also involved in physiological processes in soft tissue. Soft tissue is non‑bony material. The table shows some data relating to humans. mean mass of an adult 80 kg mean percentage of body mass that is bone 4% mean mass of calcium ions per adult 1000 g mean percentage of calcium ions present in soft tissue 1% (i) Calculate the calcium ion concentration in the soft tissue of a human. (3) .............................................................. µg g–1 Turn over 27   *(ii) Calcium ions are also found in the tissue of plants. The calcium ion concentration in one species of plant has been measured. It was found to be 170 µg g–1 of plant tissue. A student made the conclusion that: Calcium ions are more important in the tissue of plants than in animals. Evaluate this conclusion. (6) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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.................................................................................................................................................................................................................................................................................... 28   (c) Milk from humans contains both calcium ions and lactose. Lactose is a disaccharide. Describe how lactose is formed from two monosaccharides. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 10 = 13 marks) TOTAL FOR PAPER = 100 MARKS

6   2 The diagram shows an RNA virus that causes influenza. (Source: © Kateryna Kon/Shutterstock) (a) Influenza cannot be treated with antibiotics. Explain why antibiotics cannot be used to treat influenza. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 7   (b) The surface of this virus capsid is covered with ‘spike’ proteins. The diameter of the capsid of this virus is 240 nm. 240 nm b = 20.0 nm h = 22.4 nm Diagram of the surface of the virus showing each spike protein as a triangle The diagram shows the area occupied by one spike protein Calculate the maximum number of spike proteins that can be packed on the surface of one virus particle. The formula for calculating the surface area of a sphere is 4πr2 The formula for calculating the area of a triangle is h b × 2 (3) Answer .............................................................. 8   (c) Changes in the RNA of influenza produce new strains of the virus with an altered spike protein. Devise a procedure to determine the similarity of the strains of influenza in saliva samples collected from different people. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 2 = 9 marks) Turn over 9   3 Eukaryotic cells contain membrane‑bound organelles. (a) The photograph, obtained using an electron microscope, shows an organelle. X (Source: © Science History Images/Alamy Stock Photo) (i) Label a part of this organelle where carbon fixation occurs. (1) (ii) Give the name of the component labelled X. (1) .................................................................................................................................................................................................................................................................................... (iii) Describe how the structure of a membrane in the part labelled X is related to its function. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 10   (iv) Describe how GP is produced by carbon fixation in this organelle. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The graph shows the changes in biomass, gross primary productivity and the energy used in respiration during succession from grassland to mature forest. 80 60 40 20 0 9 8 7 6 5 4 3 2 1 0 Time / years Energy / a.u. respiration gross primary productivity biomass (i) Add shading to the graph to show the area that represents net primary productivity. (1) Turn over 11   (ii) Describe what is meant by the term succession. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (iii) Deduce the effect of succession on the level of carbon dioxide in the atmosphere. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 3 = 13 marks)

12   4 The photograph shows a plant called lantana (Lantana camara). (Source: © Peter Vrabel/Alamy Stock Photo) The leaves of lantana contain chemicals known to have antimicrobial properties. The antimicrobial activity of lantana leaf extracts prepared using three different solvents, A, B and C, was compared. Fresh lantana leaves were dried and powdered. The dried leaf material was mixed with the solvent and then the extract was purified and dried. The mass of extract obtained from 5 g of powdered leaf, using each solvent, was measured. Solvent used Mean mass of extract ± SD / µg A 501.3 ± 3.5 B 721.3 ± 1.5 C 245.6 ± 4.0 (a) The mass of dried and powdered lantana leaves is 10.5% of the mass of fresh leaves. Calculate the mass of fresh leaves needed to produce 1 mg of extract using solvent A. Give your answer to three significant figures. (2) .............................................................. g Turn over 13   (b) The antimicrobial properties of the extracts produced using these solvents are shown in the table. Solvent used to prepare extract Dry mass of extract / µg Bacteria tested Klebsiella pneumoniae (Gram negative) Micrococcus luteus (Gram positive) Mean diameter of the zone of inhibition / mm A 5.0 8.3 7.1 10.0 10.5 7.0 B 5.0 14.5 12.2 10.0 18.1 18.0 (i) Deduce the effect of using different solvents on the effectiveness of the extracts against these two bacteria. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 14   (ii) Devise a method that could be used to collect the data in the table. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 4 = 8 marks) Turn over 15   5 Habituation is a learning response observed in many types of animal. (a) Explain the importance of the habituation response in an animal. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) The ‘light-off jump’ assay can be used to study habituation in fruit flies. When the light is turned off the flies jump into the air and attempt to fly away. The strength of this response can be measured by recording the noise made by the flies. In one experiment, fruit flies with reduced expression of a protein called synapsin were compared with a control group. Synapsin is a protein that inhibits the binding of presynaptic vesicles to the cell membrane. The results of this experiment are shown in the graph. 100 90 80 70 60 50 40 30 20 10 0 Number of times light was turned off 80 70 60 50 40 30 20 10 0 Mean percentage of maximum jump response (%) reduced synapsin normal synapsin 16   (i) State two variables associated with the light-off stimulus that need to be controlled in this experiment. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (ii) Determine the effect of reduced synapsin on the habituation of fruit flies in this experiment. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 17   (iii) Explain how reduced expression of synapsin could produce these results. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (c) Scientists are using fruit fly habituation to investigate the role of genes associated with human autism spectrum disorders (ASD). Before they can do this, the scientists first identify genes linked to human ASD. Describe how genes linked to human ASD could be identified. (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 5 = 12 marks)

18   6 The red flour beetle (Tribolium castaneum) is a food pest found in flour. The photograph shows a red flour beetle. A B (Source: © Nigel Cattlin/Alamy Stock Photo) magnification ×15 (a) Calculate the length of the red flour beetle between A and B. (1) .............................................................. mm (b) The main food source for adult red flour beetles is starch. Red flour beetles lay eggs that contain all the nutrients required for the development of the embryo. The main food source in the eggs is glycogen. Describe how starch can be used to make glycogen. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 19   (c) The graph shows the changes in mass of glucose and glycogen in the eggs of the red flour beetle during embryo development. 80 70 60 50 40 30 20 10 0 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 Time after egg laid / hours Mass per egg / μg glucose glycogen (i) Calculate the percentage of glycogen used from 4 to 8 hours after the eggs were laid. (2) .............................................................% 20   (ii) A student suggested that between 20 and 24 hours the cells in the developing embryo switched from anaerobic to aerobic respiration. Describe how a respirometer could be used to test this suggestion. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 6 = 9 marks) Turn over 21   *7 The human genome codes for approximately 20 000 different proteins. The pie chart shows the proportion of proteins carrying out different functions. transcription factors and other nucleic acid binding molecules 20% receptors and control of receptor activity 11% carrier and transporter 17% communication and signalling inside cells 10% enzymes 41% other less than 1% defence 1% The 20 amino acids used to make proteins can be classified according to the properties of their side chains (R groups). Table 1 shows the number of amino acids with these properties. Property of the side chain group Number of amino acids with the property Non-polar 9 Polar, uncharged 6 Negatively charged 3 Positively charged 2 Table 1 22   Table 2 shows three amino acids, used to synthesise proteins, that have unique properties. Amino acid Comment on structure Cysteine The side chain contains a thiol group (-S-H) that is chemically reactive. Glycine The side chain is a hydrogen atom which is much smaller than any other side chain. This allows tight coiling of polypeptide chains. Proline The side chain forms a peptide bond with the nitrogen in the amino group. This makes a polypeptide chain more rigid. Table 2 Discuss the importance of the amino acid side chain to the structure, function and location of proteins. (9) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 23   .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 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(Total for Question 7 = 9 marks)

24   8 The scientific article you have studied is adapted from Scientific American. Use the information from the scientific article and your own knowledge to answer the following questions. (a) Explain why natural selection took place in Darwin’s Galapagos finches (paragraph 3). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (b) Explain how inflammation and the immune response can cause damage to tissues (paragraph 4). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 25   (c) Describe how tumour shrinkage could be observed (paragraph 6). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (d) “Why is aging, smoking or radiation exposure associated with cancer?” (paragraph 8). Explain why “These things cause mutations”. (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (e) Explain why a cancer cell needs to ‘use the resources of the tissue immediately around it’ (paragraph 12). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 26   (f) Explain how ‘a cancer‑causing mutation’ could increase the rate of division in stem cells (paragraph 14). (3) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (g) Describe how a cell that is ‘a key player in the body’s immune system’ differs from a stem cell (paragraph 14). (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... Turn over 27   (h) Describe how the effects of ‘competitive release’ could be demonstrated (paragraphs 18 and 19). (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (i) Describe how evasion mechanisms can enable pathogens to become drug‑resistant (paragraph 20). (2) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 28   (j) Discuss the ethical issues relating to the use of mice in experiments such as those described in the article (paragraph 25). (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... (Total for Question 8 = 30 marks) TOTAL FOR PAPER = 100 MARKS

page 02 SECTION 1 — 20 marks Attempt ALL questions 1. SDS‑PAGE separates proteins by: A shape B size C charge D isoelectric point. 2. Which row in the table describes functions of compartments involved in the synthesis and transport of proteins? Compartment Smooth endoplasmic reticulum Rough endoplasmic reticulum Golgi apparatus A post‑translational modification lipid synthesis protein transport B lipid synthesis protein transport post‑translational modification C lipid synthesis post‑translational modification protein transport D post‑translational modification protein transport lipid synthesis

page 03 3. The figure gives information about enzymes involved in glycogen metabolism in humans. protein kinase A phosphorylase kinase (inactive) phosphorylase kinase (active) protein phosphatase 2 glycogen phosphorylase (inactive) glycogen phosphorylase (active) phosphorylase phosphatase glycogen glucose Which row in the table describes events when the enzyme protein kinase A is activated? Phosphate group attached to glycogen phosphorylase Glycogen converted to glucose A yes yes B yes no C no no D no yes [Turn over

page 04 4. Which of the following are null hypotheses? 1. Blackbirds show no food preference based on colour. 2. As people age, they do not respond to certain painkillers. 3. As concentration of inhibitor increases, enzyme activity decreases. 4. Increasing light intensity will not affect the rate of germination of cress seeds. A 1 and 4 only B 2 and 3 only C 2 and 4 only D 1, 2 and 3 only 5. Statins are drugs used to treat people with cardiovascular disease. One large study investigating the effectiveness of a statin drug randomly assigned patients to two groups: one group received the statin; the other group was given a placebo. One measure of the effectiveness of a treatment is the relative risk, which can be calculated using the formula: percentage of death in treatment group Relative risk percentage of death in control group  After a period of approximately six years, 91% of the patients given the statin and 87% of patients given the placebo survived. The relative risk for those given the statin is: A 0.69 B 0.96 C 1.05 D 1.44

page 05 6. The diagram shows a protein molecule and the cell membrane. phospholipid bilayer protein Which row in the table identifies features of this protein? Protein type Nature of R groups on protein surface Integral Peripheral Hydrophobic Hydrophilic A ✓ ✓ B ✓ ✓ ✓ C ✓ ✓ ✓ D ✓ ✓ 7. The response to an increase in insulin concentration in the blood involves the following events. 1. Recruitment of GLUT 4 to the cell membrane. 2. A phosphorylation cascade inside the cell. 3. Insulin receptor changes conformation. 4. Phosphorylation of the insulin receptor. 5. Binding of insulin to its receptor. The correct sequence of these events is: A 5, 4, 3, 1, 2 B 2, 4, 3, 5, 1 C 5, 3, 4, 2, 1 D 5, 3, 1, 2, 4 [Turn over

page 06 8. The diagram shows events in the response to light in the rod cells of animals. outer membrane of rod cell inside of rod cell photoactivated rhodopsin activation of X by rhodopsin channel protein Z broken down by Y activation of Y by X X Y X Z Z When a photon of light hits a molecule of rhodopsin it activates molecule X, which disassociates from rhodopsin and activates another molecule, Y. Activated Y breaks down molecule Z, which in turn affects the function of a channel protein. Which row in the table identifies molecules X, Y and Z? Molecule X Molecule Y Molecule Z A cGMP transducin PDE B transducin PDE cGMP C PDE transducin cGMP D transducin cGMP PDE

page 07 9. A respirometer was used to measure the rate of respiration in maggots as shown in the diagram. The soda lime was present to absorb the carbon dioxide produced by the maggots during respiration. The manometer was used to measure the decrease in gas volume over a fixed time period. live maggots cotton wool soda lime manometer containing coloured fluid It was found that the decrease in gas volume, and hence the oxygen consumed, in the respirometer was 0.24 cm3 per gram of maggots. The experiment was repeated with identical conditions except that the soda lime was replaced with an equal volume of plastic beads, allowing the measurement of the net change in both carbon dioxide and oxygen. The net change in volume in this experiment was found to be a decrease of 0.07 cm3 per gram of maggots. The respiratory quotient is calculated using the equation: volume of carbon dioxide produced Respiratory quotient volume of oxygen consumed  Calculate the respiratory quotient for the maggots in this experiment. A 0.29 B 0.71 C 1.41 D 3.43 10. The diagram represents a cell in which stage of mitosis? A Metaphase B Prophase C Telophase D Anaphase [Turn over

page 08 11. The cotton harlequin bug has highly variable coloration; individuals may be all‑orange or have blue‑green iridescent patches of variable size. The bright orange colour induces avoidance behaviour in birds but not in arthropod predators such as the praying mantis. The praying mantis lacks a red photoreceptor and has difficulty distinguishing the all-orange bugs from a green background. The graph shows results from an experiment that was carried out to compare the ability of praying mantises to detect all-orange and iridescent bugs of different sizes from different distances. detection distance (cm) Key 70 40 20 10 5 11 12 13 14 15 16 17 bug size (mm) iridescent bugs all-orange bugs trend line for iridescent bugs trend line for all-orange bugs The following conclusions were drawn: 1. Both colour and size of bugs affect detection distance. 2. Iridescent bugs are mostly larger than all-orange bugs. 3. The negative correlation between bug size and detection distance shows that smaller bugs were detected from further away. Which of the conclusions are consistent with the data shown? A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3

page 09 12. Which row in the table shows a model organism correctly classified? Model organism Taxonomic group A Caenorhabditis elegans Bacteria B Escherichia coli Nematoda C Arabidopsis thaliana Chordata D Drosophila melanogaster Arthropoda 13. Which of the following descriptions of animal behaviour is anthropomorphic? A Woodlice consistently prefer damp environments. B Rats may be very cautious about eating new foods presented to them. C Lekking behaviour in a male black grouse encourages females to mate. D A greedy herring gull was recently observed swallowing a whole blackbird. [Turn over

page 10 14. Male long‑billed hermit hummingbirds (Phaethornis longirostris) defend a territory (favoured perch). At the same time, they have to forage successfully for nectar‑producing flowers over great distances. The ability to recall rewarding locations efficiently (spatial memory) allows more time to be spent defending the territory. An experiment was carried out to compare the importance of this spatial memory to the importance of three physical phenotypic traits: body size, bill‑tip length and load‑lifting ability. The importance of each trait was represented by an ‘effect size’. The larger the effect size the greater the effect on its ability to defend a territory. The figure shows the effect size for all four traits. spatial memory body size bill-tip length load- lifting trait effect size −0.5 0.0 0.5 1.0 1.5 Key adults adults and juveniles Which of the following conclusions can be drawn from the data? A There is no difference between adults and juveniles. B The ability to recall the position of nectar‑producing flowers increases the ability to defend a territory. C No conclusion can be drawn as none of the differences in the data are significant. D Adults always have a smaller territory.

page 11 15. Four students carried out an experiment to measure the glucose concentration of a carton of fruit juice. Each student measured the values four times. The table shows the readings obtained. Student Concentrations of glucose measured (mg/100 ml) Mean (mg/100 ml) 1 10 16 8 18 13 2 15 13 12 15 14 3 17 10 12 17 14 4 10 9 10 10 10 The actual concentration of glucose in the fruit juice was 12 mg/100 ml. Which statement best describes the data set? A Accurate, but not precise B Precise, but not accurate C Accurate and precise D Neither accurate or precise 16. The figure shows some of the stages in the life cycle of the malarial parasite, Plasmodium. Which of the four stages described is a gametocyte? Stage A enters human bloodstream when bitten by infected mosquito Reproduces asexually in liver cells to form Stage B which then penetrates red blood cells Stage C formed inside red blood cells and released into bloodstream when red blood cells burst Taken up by another mosquito and forms Stage D. Sexual reproduction occurs leading to formation of Stage A which moves to the salivary glands of the mosquito [Turn over

page 12 17. Graphs 1 and 2 show how the rate of glucose uptake by the liver is affected by the concentrations of glucose and insulin in the blood. concentration of insulin in the blood (units/cm3) Graph 1 concentration of glucose in the blood (mg/100 cm3) 15 10 5 0 0 20 40 60 80 100 Graph 2 concentration of insulin in the blood (units/cm3) rate of glucose uptake by liver (μg/g of liver/hour) 0 2 4 6 0 5 10 15 20 25 The mass of glucose taken up in 30 minutes by a 1.4 kg liver when the concentration of glucose in the blood is 60 mg/100 cm3 is: A 2 μg B 4 μg C 2800 μg D 5600 μg.

page 13 18. The table shows information concerning some characteristics of three species of MacArthur’s warblers, which are closely related species of insectivorous songbirds living in spruce forests in north‑eastern USA. Species of MacArthur’s warblers Setophaga fusca Setophaga coronata Setophaga castanea Foraging zone treetops trunk area of trees middle interior portions of trees Nesting sites high in trees near tips of branches horizontal branches in trees up to 15 m lower portions of spruce and fir trees Food sources insect larvae, spiders and spruce budworm insects, insect larvae and spruce budworm insects, spiders and spruce budworm Nesting period May–July May–July May–July The strongest evidence of resource partitioning between these three species comes from the analysis of their: A foraging zone and nesting period B nesting sites and food sources C nesting period and food sources D foraging zone and nesting sites. [Turn over

page 14 19. During an investigation into foraging behaviour, a student studied prey selection by shore crabs feeding on mussels. The student observed the crabs feeding and measured the length of each mussel predated on by the crabs over several days. The results were compared with another study where the energy expended by a crab to open a mussel of a specific size was compared to the energy content of the mussel to allow the energy gained per unit of foraging time to be calculated. The results are shown in the graph. average number of mussels eaten per day length of mussel (mm) energy gain (Joules per unit of foraging time) Key mussels eaten energy gain 7 6 5 4 3 2 1 0 0 5 10 15 20 25 30 35 0 1 2 3 4 5 6 7 What conclusion can be drawn from the results? A Shore crabs don’t select prey on the basis of size. B As prey size increases, foraging becomes less efficient. C As prey size increases, foraging becomes more efficient. D Shore crabs tend to select prey sizes that make foraging more efficient. 20. Which of the following characteristics is most likely to be found in an r‑selected species? A Produce a large number of small offspring B Most offspring reach adulthood C Have a longer generation time D High level of parental care [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET] page 15 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

page 16 [BLANK PAGE] DO NOT WRITE ON THIS PAGE Acknowledgement of copyright Question 14 Ondrej Prosicky/shutterstock.com

page 20 Published: May 2023 Change since last published: Question 10 (b) • ½ box tolerance for plotting. Removed from Additional Guidance column

  4 P71908A 12. Which electron donor is used in photosynthesis – hydrogen sulphide, iron or water – ultimately depends on the energy of the photons that are absorbed by the chlorophyll. In the case of purple bacteria, their chlorophyll can only absorb low-energy infrared rays. This provides enough energy to extract electrons from hydrogen sulphide and iron, but not from water. To extract electrons from water requires extra energy, which must be acquired from higher-energy photons. To do this requires a change in the structure of chlorophyll, so it can absorb red-light photons instead of infrared light. 13. According to Robert Blankenship of Arizona State University and Hyman Hartman of the Institute for Advanced Studies in Biology at Berkeley, California, tiny changes in the structure of bacterial chlorophylls can lead to large shifts in their absorption properties. Two small changes to the structure of bacteriochlorophyll a (which absorbs at 870 nm) are all that it takes to generate chlorophyll d, which absorbs at 716 nanometres. In 1996, an article in Nature by Hideaki Miyashita and colleagues of the Marine Biotechnology Institute in Kamaishi, Japan, reported that chlorophyll d is the main photosynthetic pigment in a bacterium called Acaryochloris marina, which splits water to generate oxygen. Thus, an intermediate between bacteriochlorophyll and plant chlorophyll is not only plausible: it actually exists. From chlorophyll d another trifling change is all that is required to produce chlorophyll a, the principal pigment in plants, algae and cyanobacteria, which absorbs light at 680 nanometres. 14. Technically, then, the evolutionary steps required to get from bacteriochlorophyll to plant chlorophyll are simply achieved. The question remains, why? A chlorophyll that absorbs light at 680 nanometres is less good at absorbing light at 870 nanometres. It is therefore less efficient at splitting hydrogen sulphide, and so bacteria carrying it are at a competitive disadvantage compared with the bacteria that kept their original chlorophyll. Even worse, switching chlorophylls to split water poses the problem of what to do with the toxic oxygen waste, as well as any leaking free-radical intermediates – the same as those produced by radiation. Without foresight, how did life manage to cope with its dangerous new invention? 15. Chlorophyll extracts electrons from water one at a time. To generate oxygen from water, it must absorb four photons and lose four electrons in succession, each time drawing an electron from one of two water molecules. The overall water-splitting reaction is: 2H2O → O2 + 4H+ + 4e- Only in the final stage is oxygen released. The rate at which chlorophyll extracts electrons depends on how quickly the photons are absorbed. As the successive steps cannot take place instantly, a series of potentially reactive free-radical intermediates must be produced, if only transiently. 16. If photosynthesis is to work at all, the reactive intermediates from water must be sealed inside a structure that immobilizes them, preventing them from escaping before oxygen is released. Needless to say, they are sealed in such a cage, this is how photosynthesis works. The cage is made of proteins and is called the oxygen-evolving complex (or sometimes the water-splitting enzyme). Water is bound tightly inside the protein cage while the electrons are extracted one at a time. But this is no ordinary cage. Its structure conceals a secret that is much older than the hills, which transports us back to the time before oxygenic photosynthesis evolved, to a time more than 2.7 billion years ago, before there was any oxygen in the atmosphere. This structure is the key to life on Earth, for without it the Earth would have remained as sterile as Mars.   5 Turn over P71908A 17. The structure of the oxygen-evolving complex is very similar to that of an antioxidant enzyme called catalase. In fact, the oxygen-evolving complex looks as if it evolved from two catalase enzymes lashed together. If so, then catalase must have evolved before the oxygen-evolving complex. If so, the chronology must be as follows. Catalase evolved on the early Earth, in an atmosphere devoid of oxygen. One day, two catalase molecules became bound together to form a cage that enabled the safe splitting of water: the oxygen-evolving complex. This cage allowed the evolution of oxygenic photosynthesis. As a result, the atmosphere filled with oxygen. Life was put under serious oxidative stress. Luckily it could cope: it already had at least one antioxidant enzyme that could protect it – catalase. How convenient! But wait a moment. If catalase came before photosynthesis, then even if there was no atmospheric oxygen, there must have been oxidative stress. Is this plausible? To answer this question, we must take a look at how catalase works. 18. There are several different types of catalase. Most animal cells have a form that has four haem molecules embedded in its core. In contrast, some microbes have a different sort of catalase, which contains manganese instead of haem at its core. Despite their different structures, both enzymes are equally fast, and are correctly called catalase, in the sense that they work in the same way – they both catalyse the reaction of two molecules of hydrogen peroxide with each other to form oxygen and water: 2H2O2 → 2H2O + O2 This simple reaction mechanism reveals a great deal about conditions on the Earth 3.5 billion years ago. It is the exact equivalent of the natural reaction between two molecules of hydrogen peroxide, but is speeded up 100 million times by the enzyme. The need for two molecules of hydrogen peroxide means that catalase is extremely effective at removing hydrogen peroxide when concentrations are high, when it is easy to bring two molecules together. It works less well at low concentrations of hydrogen peroxide, when it is harder to find two molecules close together. Catalase is thus swift to remove high concentrations of hydrogen peroxide, but is poor at mopping up trace amounts or at maintaining a stable low-level equilibrium. 19. Today, most aerobic organisms have a second group of enzymes, known as the peroxidases, which can dispose of trace amounts of hydrogen peroxide. These enzymes work better at low levels of hydrogen peroxide because they act in a fundamentally different way. Rather than bringing two molecules of hydrogen peroxide together, they use antioxidants such as vitamin C to convert a single molecule of hydrogen peroxide into two molecules of water, without generating any oxygen. Most aerobic cells have both sets of enzymes, and break down hydrogen peroxide using both mechanisms. Catalase is used for bulk removal, peroxidase for subtle adjustments. 20. Catalase would presumably have been present in the photosynthetic bacteria that generated energy by splitting hydrogen sulphide or iron salts in the era before oxygenic photosynthesis. ln fact, hydrogen peroxide has some parallels with these early photosynthetic fuels. To remove electrons from hydrogen peroxide requires a similar input of energy to that required to remove electrons from hydrogen sulphide, and so could have been achieved using the same bacteriochlorophyll. Hydrogen peroxide would therefore have been a good source of hydrogen for photosynthesis. And, while far less plentiful than hydrogen sulphide and iron salts, it was nonetheless formed most readily in the surface waters, closest to the full power of the Sun. If this scenario is true, then catalase could have doubled as a photosynthetic enzyme. Because splitting hydrogen peroxide generates oxygen, this recruitment of catalase to photosynthesis also bridges the evolutionary gap between anoxygenic and oxygenic photosynthesis.

  6 P71908A 21. If catalase was acting as a photosynthetic enzyme, then it would be natural for a number of catalase molecules to cluster around the photosynthetic apparatus. In these circumstances, it would be simple for two catalase molecules to became associated as a complex: the prototype oxygen-evolving complex. At first it would have continued to use hydrogen peroxide as an electron donor, but given the right energy input, this complex could split water. We know that three small changes in the structure of bacteriochlorophyll can transform its properties, enabling it to absorb high-energy light at a wavelength of 680 nm. We now have a prototype oxygen-evoIving complex (the nutcracker that can physically split water) and a chlorophyll that can provide enough energy for it to do so (or the hand that presses the nutcracker). Thus, with no foresight and no disadvantageous steps, we have taken a path leading from anoxygenic photosynthesis to oxygenic photosynthesis. Adapted from: Nick Lane. ‘Green Planet’. In Oxygen. Oxford, Oxford University Press, 2002. pp131–146.   7 P71908A BLANK PAGE   8 P71908A BLANK PAGE

4   2 The human circulatory system is composed of the heart and blood vessels. (a) The graph shows pressure changes in the aorta, left atrium and left ventricle during the cardiac cycle. 20 15 10 5 0 –5 Blood pressure / kPa 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Time / s Left ventricle Left atrium Aorta Q R S P (i) At which part of the graph is the left atrioventricular valve starting to close? (1) A P B Q C R D S (ii) At which part of the graph is a semilunar valve opening? (1) A P B Q C R D S Turn over 5   (iii) Describe what happens during diastole. 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(b) The photograph shows a cross‑section of part of a human aorta, as viewed under a light microscope. The diameter of the lumen can be reduced to maintain blood pressure in the aorta. X Y Magnification ×18 © JOSE CALVO/SCIENCE PHOTO LIBRARY (i) Calculate the diameter of the lumen along the line XY. Give your answer to two significant figures. (2) Answer .............................................................. mm 6   (ii) State the feature of the wall of the aorta that causes a reduction in the diameter of the lumen. (1) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 2 = 7 marks) Turn over 7   3 Starch and amylase are molecules found in some cells. (a) Amylase is a biological catalyst. State what is meant by the term biological catalyst. 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(b) The hydrolysis of starch by amylase was investigated. The diagram shows the hydrolysis of starch into maltose. substrate product enzyme (amylase) 8   The concentration of maltose in the solution was measured at regular intervals. The graph shows the results of this investigation. 300 250 200 150 100 50 0 Concentration of maltose / mg dm–3 0 10 20 30 5 15 25 Time / minutes (i) Which of the following shows the type of molecules in this reaction? (1) A B C D (ii) Which of the following is the type of bond that is broken when starch is hydrolysed by amylase? (1) A glycosidic B hydrogen C ionic D peptide Substrate Product Polypeptide Disaccharide Polysaccharide Disaccharide Polypeptide Monosaccharide Polysaccharide Monosaccharide Turn over 9   (iii) Which of the following shows the correct unit for the rate of this reaction? (1) A mg dm3 min1 B mg dm–3 min1 C mg dm3 min–1 D mg dm–3 min–1 (c) Explain how the structure of starch makes it a good storage molecule. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 10   (d) The diagram shows part of another molecule. OH H O H H H H H2COH OH OH H O O O H H OH H H O H H O H OH H H2COH OH H O H2COH The enzyme amylase cannot hydrolyse this molecule. Deduce why amylase cannot hydrolyse this molecule. 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(Total for Question 3 = 9 marks) Turn over 11   BLANK PAGE

12   4 The phenotype of an organism is affected by its genotype. (a) The shape of the eye in fruit flies is controlled by a single gene. Fruit flies that are homozygous for the allele R have round eyes. Fruit flies that are homozygous for the allele B have bar eyes. The diagram shows the eyes on the heads of the fruit flies. round-eyed fruit fly bar-eyed fruit fly wide bar-eyed fruit fly eye (i) State what is meant by the term allele. 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Turn over 13   (ii) When a round‑eyed fruit fly is crossed with a bar‑eyed fruit fly, offspring are produced that have wide bar eyes. Explain why offspring are produced with the wide bar eye shape. 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(b) Tay‑Sachs disease is a genetic condition. Individuals who are homozygous recessive for this condition usually survive for up to four years. (i) Determine the probability of two individuals who are heterozygous for the Tay‑Sachs gene having a child with this condition. Use a genetic diagram, and the alleles T and t. (2) Probability .............................................................. 14   (ii) An embryo, created by IVF, can be screened for Tay‑Sachs disease before being placed in the mother’s uterus. Name this type of genetic screening. (1) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (iii) Explain why the mother may prefer this type of prenatal screening to amniocentesis. 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(Total for Question 4 = 10 marks) Turn over 15   BLANK PAGE

16   5 Cardiovascular disease (CVD) can occur in the blood vessels of organs such as the brain and the heart. The diagram shows a human heart and the location of a blood clot in one of the blood vessels in this heart. A B C D location of blood clot in vessel E (a) (i) Which labelled blood vessel carries oxygenated blood under lower pressure than the aorta? (1) A B C D (ii) Which labelled blood vessel carries blood from the right ventricle? (1) A B C D Turn over 17   (b) The endothelium of vessel E was damaged and collagen fibres in the vessel wall were exposed. This resulted in the formation of a blood clot. The blood clot blocked vessel E. This caused a heart attack. (i) Sketch an outline on the diagram to show the area of heart muscle that would receive a reduced blood supply. (1) (ii) Describe how this blood clot could have been formed. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 18   (c) The risk of a heart attack in humans can be reduced by treatment with platelet inhibitors. The effectiveness of a platelet inhibitor (PI) drug was investigated. One group of patients took a new platelet inhibitor (PI) drug whereas the control group were given no treatment. The graph shows the percentage of these patients who had a heart attack. 15 10 5 0 Percentage of patients who had a heart attack (%) 0 100 200 300 400 PI group control group Days Explain the results of this investigation. 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(Total for Question 5 = 10 marks) Turn over 19   6 Smoking cigarettes can result in a lung disease called emphysema. The diagram shows stages in the development of emphysema. Healthy lungs Healthy alveolus Inflammatory response triggered Inflammatory chemicals dissolve alveolar septum Harmful particles trapped in alveolus Emphysema Large air cavity forms in alveolus (a) Explain why emphysema affects the rate of gas exchange between the alveolus and the blood. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 20   *(b) Forced expiration volume (FEV) is the maximum volume of air a person can breathe out in one second. Emphysema reduces FEV. The effects of ageing and long‑term cigarette smoking on FEV and the development of emphysema have been investigated. The graph shows the results of these investigations. start of symptoms of emphysema severe disability due to emphysema 4 3 2 1 0 FEV / dm3 25 45 65 85 non-smoker smoker Age / years Scientists also investigated the changes in the heart and pulmonary blood vessels of patients with emphysema. The diagrams show the heart and pulmonary blood vessels in a patient with emphysema and a patient without emphysema. with emphysema without emphysema Turn over 21   Evaluate the effects of smoking on the respiratory system, heart and pulmonary circulation. (6) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 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(Total for Question 6 = 10 marks)

22   7 One component of the cell surface membrane is the phospholipid bilayer. The diagram shows a phospholipid bilayer. Phospholipid A Phospholipid B (i) A cell surface membrane also contains cholesterol molecules. Name two other types of organic molecule that would be found in a cell surface membrane. (2) 1 ................................................................................................................................................................................................................................................................................. 2 ................................................................................................................................................................................................................................................................................. (ii) The diagram shows two types of phospholipid, A and B. It is possible for phospholipid molecules to have different shapes. Explain what causes these two phospholipids to have different shapes. 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Turn over 23   (b) A phospholipid is a modified triglyceride. Compare and contrast the structures of a phospholipid and a triglyceride. 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(c) The diagram shows a simplified low‑density lipoprotein (LDL). triglyceride cholesterol phospholipid This spherical lipoprotein has a diameter of 27.50 nm. Which of the following shows the circumference (C) of this lipoprotein? Use the formula C = 2πr (1) A 86.39 nm B 172.79 nm C 593.96 nm D 2375.83 nm 24   (d) The concentration of LDL in the blood is a risk factor for cardiovascular disease (CVD). The effect of LDL concentration in the blood on the number of deaths from CVD, in males aged from 25 to 40, was investigated. The graph shows the results of this investigation. 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 Number of deaths per 1000 males per year < 100 100–129 130–159 160–189  190 Concentration of LDL in the blood / mg 100 cm–3 (i) Describe the relationship shown in the graph. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... Turn over 25   (ii) The UK population in 2021 was 68 269 157. A student concluded that 81 923 people with an LDL concentration in the blood of 130–159 mg per 100 cm3 died from CVD in 2021. Comment on the validity of this conclusion. Use the information in the graph to support your answer. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 7 = 14 marks)

26   8 The effect of the concentration of sucrose solution on the mass of potato cylinders was investigated. The cylinders were cut from the same potato and had the same volume. The initial mass of each cylinder was recorded. The cylinders were left in the same volume of each solution for the same length of time. The final mass of each cylinder was recorded. The diagram shows the five test tubes, A, B, C, D and E, that were set up in this investigation. The investigation was then repeated twice. potato cylinder sucrose solution A B C D E 27   Turn over (a) The table shows the results of this investigation. Test tube Concentration of sucrose solution / mol dm–3 Mean initial mass of potato cylinder / g Mean final mass of potato cylinder / g A 0.0 2.79 3.82 B 0.4 2.75 2.67 C 0.6 2.68 2.31 D 0.8 2.71 2.21 E 1.0 2.79 2.01 (i) For test tube E the range of the initial mass was 2.79 ± 0.04 g and the range of the final mass was 2.01 ± 0.16 g. Calculate the largest percentage change in mass for potato cylinders in 1.0 mol dm–3. (2) Answer ..............................................................% (ii) Explain the change in mass for potato cylinders in test tube E. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 28   (b) Describe how this investigation could be modified to determine the concentration of sucrose solution that does not change the mass of the potato cylinders. (3) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 29   (c) Beetroot cell vacuoles contain a red pigment called betalain. A student investigated the effect of ethanol on the permeability of beetroot membranes. Test tubes were set up as shown in the diagram, each with a different concentration of ethanol. 20 cm3 ethanol beetroot disc The test tubes were placed in a thermostatically controlled water bath at 25 °C for 30 minutes. The beetroot discs were removed from the ethanol. (i) Calculate the volume of a beetroot disc with a diameter of 1 cm and a height of 5 mm to one decimal place. Include the appropriate unit. Use the formula V = πr2h (2) Answer .............................................................. Turn over 30   (ii) The relative colour of the ethanol after 30 minutes was measured using a colorimeter. The results of the investigation are shown in the graph. Relative colour of the ethanol solution Concentration of ethanol Explain the results of this investigation. (2) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... 31   (iii) Devise a valid investigation to compare the effects of ethanol and propanol on the permeability of beetroot membranes. (4) ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... (Total for Question 8 = 15 marks) TOTAL FOR PAPER = 80 MARKS 32   BLANK PAGE

26 (A400U20-1) 26 © WJEC CBAC Ltd. 8. Graphs 8.1–8.4 show the concentrations of four hormones found in the plasma of a mature ewe (female sheep). The hormone concentrations are plotted over a time period of 16 days. The days are numbered either side of the peak concentration of LH secretion. 0 – 8 – 6 – 4 – 2 0 +2 +4 +6 +8 4 2 10 8 6 Time / days Oestrogen concentration / au Graph 8.1 0 – 8 – 6 – 4 – 2 0 +2 +4 +6 +8 4 2 10 8 6 Time / days Progesterone concentration / au Graph 8.2 0 – 8 – 6 – 4 – 2 0 +2 +4 +6 +8 4 2 10 8 6 Time / days Follicle stimulating hormone concentration / au Graph 8.3 0 – 8 – 6 – 4 – 2 0 +2 +4 +6 +8 4 2 10 8 6 Time / days Luteinising hormone concentration / au Graph 8.4 (A400U20-1) Turn over. 27 27 Examiner only © WJEC CBAC Ltd. If sheep are given progesterone for eight days and then the treatment is stopped, they will ovulate eight to twelve days later. In this way, the oestrus cycle of a flock of sheep can be synchronised. Use Graphs 8.1, 8.2, 8.3 and 8.4 to explain the role of oestrogen in the control of the timing of ovulation in the sheep. Explain why the sheep will ovulate after the doses of progesterone are stopped. Suggest why is it useful for a farmer to be able to determine when ovulation will occur and suggest why some people are against injecting reproductive hormones into sheep that are used for food. [9 QER] 28 (A400U20-1) 28 © WJEC CBAC Ltd. Examiner only (A400U20-1) Turn over. 29 29 Examiner only © WJEC CBAC Ltd. 30 (A400U20-1) 30 Examiner only © WJEC CBAC Ltd. END OF PAPER 9 (A400U20-1) Turn over. 31 31 © WJEC CBAC Ltd. Examiner only Question number Additional page, if required. Write the question number(s) in the left-hand margin. 32 (A400U20-1) 32 © WJEC CBAC Ltd. Question number Additional page, if required. Write the question number(s) in the left-hand margin. Examiner only

(B400U20-1) 16 © WJEC CBAC Ltd. 3. Tomato plants, Solanum lycopersicum, are grown commercially for their edible fruit. They produce bright yellow flowers from which the fruit develop. Sucrose is a disaccharide transported in phloem tissue and used by tomato plants as a source of energy. Image 3.1A shows a light micrograph of Image 3.1B shows an electron phloem tissue. micrograph of a sieve plate. (a) The mechanism of transport in phloem is described as the mass flow hypothesis. Increasing hydrostatic pressure in the sieve tubes at the source in the leaves drives the transport of sucrose to sinks elsewhere in the plant. (i) Name two regions of a tomato plant that are considered ‘sinks’ for sucrose. [1] ................................................................. ................................................................. sieve plate sieve tube 16 Examiner only (B400U20-1) Turn over. B 40 0 U201 17 17 Examiner only © WJEC CBAC Ltd. 17 (ii) The rate of flow of solutes in the phloem sieve tubes of tomato plants may be up to ten times faster than the flow of solutes through other cells. Using Images 3.1A and 3.1B and your knowledge of sieve tube structure, describe and explain one structural feature that allows a relatively fast flow rate to be achieved in the sieve tubes. [2] (iii) Sucrose is loaded into the phloem in the leaves where it is formed. Explain why loading sucrose at the source results in increased hydrostatic pressure in the sieve tubes at that point. [2] (iv) The rate of uptake and translocation of sucrose in phloem was found to be inhibited by: • low temperature • phosphate deficiency • cyanide. Explain why each of these observations appear to contradict the mass flow hypothesis, which describes a passive process. [3] 18 (B400U20-1) © WJEC CBAC Ltd. (b) An experiment to investigate the rate of transport of organic molecules in the phloem of tomato plants used aphids to sample phloem contents. Aphids are insects that insert a tubular mouthpart called a stylet into the phloem, through which it feeds on phloem solution. Plants may respond to phloem damage by blocking some pores in the sieve plates and restricting flow. Aphids can suppress this response. Image 3.2 shows a diagram of an aphid feeding on phloem. Image 3.2 The experiment was set up as shown in Image 3.3 Image 3.3 • Aphid colonies were placed at 25 cm intervals from a start point (0) near the leaves on a branch and allowed to feed. • The aphids were detached from their stylets which were left in place penetrating the phloem. • Leaves were covered by a sealed glass chamber and provided with 14CO2 containing the radioisotope 14C. Organic compounds became radioactive as 14C was incorporated during and after photosynthesis. • Phloem solution, collected from the stylet at regular intervals, was tested for radioactivity. The time taken for radioactivity to reach each colony position was recorded. epidermis mesophyll stylet companion cell sieve tube 0 1 2 3 position of aphid colonies glass chamber containing radioactive carbon dioxide (14CO2) 18 (B400U20-1) Turn over. 19 © WJEC CBAC Ltd. 19 Table 3.4 shows the time taken for the radioactivity to travel between positions on the branch. Table 3.4 (i) Calculate the mean rate of flow of the solution in the phloem between position 0 and position 3. Give your answer to 3 significant figures. [3] Rate = ............................................... cm hr–1 (ii) Suggest one advantage and one disadvantage of using several aphids at each position. [2] (iii) For ethical reasons, the use of micro-injection needles inserted into the phloem was suggested as an alternative to using aphids for sampling the phloem. Suggest one reason why the use of micro-injection needles may provide less precise results. [1] Direction of movement of radioactivity between positions. (25 cm intervals) Time taken for radioactivity to travel between positions / hours 0 to 1 1.2 1 to 2 1.4 2 to 3 1.5 14 Examiner only 20 (B400U20-1) Examiner only © WJEC CBAC Ltd. 4. (a) Rabbits and foxes are mammals that consume different proportions of macronutrients. Image 4.1 shows the digestive systems of a rabbit (a herbivore) and a fox (a carnivore). Image 4.1 (i) Most products of digestion are absorbed in the ileum which is the longest section of the small intestine. With reference to its diet, explain why a fox has a proportionally shorter ileum than a rabbit. [2] (ii) The caecum and large intestine are proportionally larger in a rabbit than in a fox and contain micro-organisms similar to those found in the rumen of cattle. State one function carried out by these micro-organisms. [1] Rabbit small intestine large intestine caecum Fox 20 (B400U20-1) Turn over. 21 Examiner only © WJEC CBAC Ltd. 21 (b) Image 4.2 is a photomicrograph of part of the ileum. An epithelial cell is labelled. Image 4.2 (i) The specimen in Image 4.2 has been viewed using a light microscope. Suggest the magnification of the objective lens that was used to provide the visual detail seen in this photograph. [1] Magnification = ×............................................................................................................. (ii) Describe the process of glucose transport from the lumen of the small intestine into and then out of epithelial cells as it is absorbed along the pathway from X to Y, indicated by the arrow in Image 4.2. [3] lumen epithelial cell X Y 22 (B400U20-1) Examiner only © WJEC CBAC Ltd. (iii) The lumen of the small intestine contains water from food and digestive secretions. Describe how the transport of glucose into the epithelial cells affects the absorption of water from the lumen. [2] 9 22 (B400U20-1) Turn over. 23 © WJEC CBAC Ltd. 23 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE

24 (B400U20-1) Examiner only 5. Terrestrial organisms obtain oxygen from air. They have internal gas exchange surfaces that provide a large surface area for sufficient diffusion of oxygen and carbon dioxide. (a) (i) State one reason, other than protection, why it is an advantage for gas exchange surfaces to be internal in all terrestrial organisms. [1] (ii) Explain the purpose of ventilation in the lungs of mammals. [1] (b) Graph 5.1 shows changes in volume and pressure in an adult human lung while breathing. Graph 5.1 © WJEC CBAC Ltd. – 0.8 – 0.6 – 0.4 – 0.2 0.2 0 inspiration expiration pleural pressure alveolar pressure lung volume Pressure / kPa 0 0.25 0.5 Volume change / dm3 24 (B400U20-1) Turn over. 25 Examiner only (i) Describe how muscles of the thorax, together with the pleural membranes, cause the change in alveolar pressure seen in Graph 5.1 during inspiration. [4] © WJEC CBAC Ltd. 25 Question continued overleaf 26 (B400U20-1) Examiner only Graph 5.2 illustrates the changes in lung volume for an adult human while breathing at rest. The arrow on Graph 5.2 represents the volume change as this individual breathes in or out. Graph 5.2 © WJEC CBAC Ltd. 0 0 10 20 30 1 3 2 Time / seconds Lung volume / dm3 4 5 (ii) Using the data in Graph 5.2 calculate the total volume of air inspired in one minute when this individual is breathing normally at rest. [2] Total volume inspired in one minute = ........................................................... dm3 26 (B400U20-1) Turn over. 27 Examiner only © WJEC CBAC Ltd. The volume of air inspired and expired is equal, however, the composition of gases varies. Table 5.3 shows the percentage of three gases in air from three points of the breathing cycle. Table 5.3 (iii) The percentage of carbon dioxide in exhaled air is lower than in alveolar air. Explain why. [1] (iv) Exhaled air has a slightly higher percentage of nitrogen than inhaled air despite this gas being inert in mammals. Use the information in Table 5.3 to suggest a reason for the difference. [1] Gas Percentage of gas mixture / % Inhaled (atmospheric) air Exhaled air Alveolar air Oxygen 20.96 16.2 14.0 Carbon dioxide 0.04 4.2 6.4 Nitrogen 79.0 79.6 79.6 27 28 (B400U20-1) Examiner only © WJEC CBAC Ltd. (c) Exposure to cigarette smoke has been shown to have a damaging effect on the lungs and reduce respiratory efficiency. Images 5.4 A and B show two electron micrographs of lung tissue at the same magnification. Image 5.4 A tissue from a lung that has had long-term exposure to cigarette smoke. Image 5.4 B lung tissue that has not been exposed to cigarette smoke. With reference to Images 5.4 A and B, describe one difference in the appearance of the two lung samples and explain how this would decrease gas exchange efficiency. [2] Image 5.4 A Exposed to smoke Image 5.4 B Not exposed to smoke 12 28 (B400U20-1) Turn over. 29 © WJEC CBAC Ltd. 29 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE 30 (B400U20-1) © WJEC CBAC Ltd. 30 6. Image 6 is a photograph of an adult endoparasitic worm, Hirudinella ventricosa, attached to the folds inside the stomach of its primary host, which is a large fish. Image 6 Hirudinella’s thick body wall has a cuticle covered in mucus. The anterior end has two suckers. It feeds on the host’s blood through mouthparts that penetrate the stomach lining. The parasite is hermaphrodite (has male and female reproductive organs). One adult can release many eggs into the host’s digestive tract from which they are egested in faeces. Eggs hatch into larvae which swim in sea water then penetrate the secondary host, a marine snail. Several larval stages reproduce asexually inside the snail. Using the above information together with your knowledge of parasitic nutrition, state what is meant by the term ‘endoparasite’ and distinguish between the terms ‘primary host’ and ‘secondary host’. Suggest how the features of this parasitic life cycle increase the chance of Hirudinella infecting its primary host. Describe the problems encountered by an adult Hirudinella in the stomach of its host and explain how it is adapted to survive in such conditions. [9 QER] 30 mm 30 mm Examiner only (B400U20-1) Turn over. 31 © WJEC CBAC Ltd. 31 Examiner only 32 (B400U20-1) © WJEC CBAC Ltd. 32 Examiner only (B400U20-1) Turn over. 33 © WJEC CBAC Ltd. 33 Examiner only END OF PAPER 9 Examiner only Question number Additional page, if required. Write the question number(s) in the left-hand margin. (B400U20-1) 34 © WJEC CBAC Ltd. 34 (B400U20-1) 35 © WJEC CBAC Ltd. 35 Question number Additional page, if required. Write the question number(s) in the left-hand margin. Examiner only (B400U20-1) 36 © WJEC CBAC Ltd. 36 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE

page 02 SECTION 1 — 20 marks Attempt ALL questions 1. The proteome is A all the genes expressed as proteins B the entire set of proteins expressed by the genome C the number of proteins expressed due to alternative RNA splicing D the number of proteins expressed due to alternative RNA splicing and post‑translational modification. 2. Transmembrane proteins carry a signal sequence, which halts translation and directs the ribosome synthesising the protein to dock with the A plasma membrane B endoplasmic reticulum C transport vesicle D golgi apparatus. 3. Cell components can be separated by a technique called differential centrifugation. This technique uses a series of centrifugation steps at specific centrifugation forces (g) for a given time. After each step, the supernatant is removed and the cell components in the pellet identified. The supernatant is re‑centrifuged at a higher centrifugation force for a longer time. The table gives information about a differential centrifugation experiment. Step Centrifuge conditions Cell component(s) in pellet Force (g) Time (minutes) 1 600 10 nucleus, cytoskeleton 2 15 000 15 mitochondria 3 100 000 60 plasma membrane, endoplasmic reticulum fragments Plasma membrane would be present in the supernatant of which centrifugation step(s)? A 1 only B 2 only C 1 and 2 only D 3 only

page 03 4. Which row in the table is correct for positive modulators? Binding site Effect of binding on enzyme affinity for substrate A allosteric increases B active increases C allosteric decreases D active decreases 5. Which row in the table describes events in signalling by the hormone insulin? Location of insulin binding site Mechanism of signal transduction A intracellular G‑protein activation B intracellular phosphorylation C extracellular G‑protein activation D extracellular phosphorylation [Turn over

page 04 6. A newly identified drug designed to treat cancer by inhibiting cell cycle progression was tested on cancer cells in vitro. The distribution of the cancer cells across the different phases of the cell cycle was then investigated by measuring the DNA content of the cells. Results for untreated cells are shown in Figure 1 and for treated cells in Figure 2. DNA content is displayed with arbitrary units where 2n units represents the DNA content of a non‑dividing diploid cell. Figure 1 Figure 2 untreated cells treated cells (with drug) relative amount of DNA relative amount of DNA number of cells number of cells 2n 4n 2n 4n At which phase(s) of the cell cycle does this drug act? A G1 only B S only C G1 and G2 D G2 or M 7. The three Rs of replacement, reduction and refinement are used in animal studies to avoid, reduce or minimise the harm to animals. Which of the following is an example of replacement? A Use of most appropriate anaesthetic regimes for pain relief during procedures. B Smaller mammals such as mice or rats instead of larger mammals. C Substitution of in vivo studies with tissue culture to test the effect of a new drug. D Less invasive techniques instead of surgical techniques. 8. Active cyclin‑CDK complexes regulate the cell cycle by A phosphorylation of specific proteins B proteolytic cleavage of specific proteins C dephosphorylation of specific proteins D acting as transcription factors.

page 05 9. Leucism is the lack of melanin (dark) pigments in all or part of the skin, hair, or feathers. It is found in many animal species, and in birds it results in the partial or complete absence of melanin in one or more feathers. The figures show data from a study of leucism in a population of red‑necked nightjars conducted in a national park in Spain from 2016–2020. Nightjars are considered adults when they are two years old and over. Figure 1 Year Number leucistic Total number Frequency (%) Males Females Males Females Males Females 2016 2 2 64 107 3.13 1.87 2017 2 4 74 98 2.70 4.08 2018 2 0 75 108 2.67 0.00 2019 3 3 78 72 3.85 4.19 2020 0 5 43 88 0.00 5.68 Figure 2 numbers above bars are the number of leucistic birds in each age category and sample sizes in brackets bird age (years) frequency of leucism (%) 40 35 30 25 20 15 10 5 0 0 1 2 3 4 5 6 7 8 0 (508)1 (101) 4 (35) 3 (17) 3 (8) 2 (17) 1 (10) 0 (4) 0 (5) Which of the following conclusions about leucism in nightjars cannot be drawn from the data shown? A Leucism occurs at similar frequencies overall in males and females. B The frequency of leucism is low in older birds because leucistic individuals have a greater risk of predation. C The frequency of leucism shows considerable annual variation. D The frequency of leucism increases markedly from the juvenile to the adult stage. [Turn over

page 06 10. The alveoli of vertebrate lungs have thin walls, constructed of a single layer of epithelial cells, which allow rapid diffusion of gases. The rate of diffusion of oxygen across exchange surfaces can be calculated using the following formula: 1 2 C C Rate P A T ( )     where: P is the permeability constant A is the surface area C1 is the higher concentration C2 is the lower concentration T is the thickness of the exchange surface. A section of alveolar epithelium has a surface area of 2.2 μm2 and is 1.0 μm thick. The permeability constant of the epithelium for oxygen is 0.012 s−1. The concentration of oxygen on one side of the epithelium is 2.3 × 10−16 mol μm−3 and the concentration on the other side is 9.0 × 10−17 mol μm−3. What is the rate of diffusion of oxygen across this alveolar epithelium? A 1.8 × 10−1 mol μm−2 s−1 B 2.3 × 10−16 mol μm−2 s−1 C 1.8 × 10−18 mol μm−2 s−1 D 3.7 × 10−18 mol μm−2 s−1 11. Red‑green colour deficiency is a sex‑linked condition caused by a recessive allele. The diagram shows the inheritance of this condition in a family. Key affected male affected female unaffected male unaffected female R S T U Using the information given in the diagram, which of the females must be carriers of red‑green colour deficiency? A S only B R and S only C R, S and T only D R, S, T and U

page 07 12. Steps 1 to 6 describe some events that occur during meiosis. Some terms are represented by the letters J, K, L and M. 1. Chromosomes replicate to form two identical chromatids. 2. Homologous chromosomes pair up and points of contact form between J chromatids of a homologous pair. 3.  K takes place to form new combinations of alleles of linked genes. 4. Chromosomes of each homologous pair attach to spindle fibres and separate to opposite poles. 5.  L occurs and two daughter cells form. 6. Two daughter cells undergo further division separating M chromatids of each chromosome. Which row in the table identifies J, K, L and M? J K L M A non‑sister cytokinesis crossing over sister B sister crossing over cytokinesis non‑sister C non‑sister crossing over cytokinesis sister D sister cytokinesis crossing over non‑sister [Turn over

page 08 13. Which of the following formulae would allow absolute fitness to be calculated? A number of surviving offspring per individual of a particular genotype number of surviving offspring per individual of the most successful genotype B number of surviving offspring per individual of the most successful genotype number of surviving offspring per individual of a particular genotype C frequency of a particular genotype before selection frequency of a particular genotype after selection D frequency of a particular genotype after selection frequency of a particular genotype before selection 14. In pea plants, the allele for yellow peas is dominant to the allele for green peas. A sample of 100 peas contained 84 yellow and 16 green peas. What is the frequency of the yellow allele in this population? A 0.36 B 0.40 C 0.60 D 0.84 15. An experiment investigated the effect of an insecticide on local wild bee populations. Scientists sampled each species of bee in proportion to the contribution its population size made to the total number of bees of all species present in the area. What name is given to this type of sampling strategy? A Random sampling B Stratified sampling C Systematic sampling D Point count sampling

page 09 16. Eukaryotic chromosomes end in repetitive non-coding sequences of DNA called telomeres. Telomeres protect the coding DNA but become shorter with each round of DNA replication. The shortening of telomeres explains why many cells have a limited number of cell divisions and is associated with the ageing processes in organisms. Rapidly dividing cells may use an enzyme called telomerase to increase the lengths of telomeres. Parasitic infections often increase the rate of cell division. A student made the following predictions: 1. Parasitic infections can lead to faster ageing. 2. Telomerase activity will be low in cancer cells. 3. There is a positive correlation between telomere length and age. 4. Telomerase activity will be high in yeast cultures where there is a high reproductive rate. Which of these predictions are consistent with the information given? A 1 and 2 B 1 and 4 C 2 and 3 D 3 and 4 [Turn over

page 10 17. A student was planning an investigation into the effect of temperature on the rate of respiration in yeast; this was measured using a carbon dioxide probe. As part of their pilot studies, they used the probe to take three readings from one sample. This would be most useful for checking A precision B accuracy C reliability D validity. 18. The diagram shows a boxplot for a set of data. R S T U V Which statement about the dataset represented by this boxplot is true? A T is the mean value B R minus V is the interquartile range C 50% of the data are above U D 75% of the data are below S 19. Which row in the table describes natural killer cells? Type of response Mechanism of action Specific Non‑specific Induce apoptosis in parasite‑infected cells Use enzymes in lysosomes to digest parasites A ✓ ✗ ✓ ✗ B ✓ ✗ ✗ ✓ C ✗ ✓ ✓ ✗ D ✗ ✓ ✗ ✓

page 11 20. Trypanosoma gambiense is a parasite that causes chronic infections in humans and can evade the immune system. The graph shows the number of trypanosomes over time in a human infected with Trypanosoma gambiense. weeks from time of infection trypanosomes/mm3 blood 1500 1000 500 0 21 25 30 From the data shown, the most probable mechanism used by this parasite to evade the immune system is A antigenic variation B suppression of the host immune system C mimicry of host antigens D latency. [END OF SECTION 1. NOW ATTEMPT THE QUESTIONS IN SECTION 2 OF YOUR QUESTION AND ANSWER BOOKLET.] page 12 [BLANK PAGE] DO NOT WRITE ON THIS PAGE

4 (1400U30-1) Examiner only 2. Most terrestrial plants use chlorophylls a and b to construct pigment-protein complexes which harvest light. Graph 2.1A shows an absorption spectrum for a terrestrial plant and the Graph 2.1B shows corresponding action spectrum. (a) Describe the relationship between the absorption spectrum and the action spectrum and state a suitable conclusion which explains the relationship. [2] © WJEC CBAC Ltd. 400 500 600 700 400 500 430 640 670 600 700 400 500 600 700 Wavelength / nm Wavelength / nm Wavelength / nm Absorption spectrum Action spectrum chlorophyll a chlorophyll b Rate of photosynthesis / au Absorption / au 04 Graph 2.1A Graph 2.1B (1400U30-1) Turn over. 140 0 U3 01 0 5 5 Examiner only (b) Different taxa contain different photosynthetic pigments. Diatoms are aquatic photosynthetic organisms. Their chloroplasts contain chlorophyll a, but instead of chlorophyll b they contain chlorophyll c. Graph 2.2 shows the absorption spectrum of chlorophyll c. (i) Describe the main difference between the absorption spectrum for chlorophyll b and the absorption spectrum for chlorophyll c. [1] (ii) With reference to the peaks labelled on the action spectrum in Graph 2.1B, predict how the action spectrum for diatoms would differ from that of terrestrial plants. [1] © WJEC CBAC Ltd. Wavelength / nm Absorption / au 400 500 600 700 Graph 2.2 05 6 (1400U30-1) Examiner only Image 2.3 shows the depths that different wavelengths of light are able to penetrate water. Image 2.3 (c) Use information from Graph 2.2 and Image 2.3 to explain why diatoms living at a depth of 200 m have chlorophyll c instead of chlorophyll b. [1] © WJEC CBAC Ltd. Wavelength of light / nm Depth of water / m 06 (1400U30-1) Turn over. 140 0 U3 01 07 7 Examiner only Image 2.4 shows an electronmicrograph of a single diatom. (d) Using information from Image 2.4, classify diatoms into their Domain. Give a reason for your choice. [2] Domain ……………………………………………………………………....…………………………………………………………………………………………………………… Reason …………………………………………………………………………………………………………………………………………………………………………………….. © WJEC CBAC Ltd. 1 µm Image 2.4 chloroplast nucleus 07 8 (1400U30-1) Examiner only (e) Diatoms are responsible for about 40% of marine productivity and because the oceans cover about 70% of the Earth’s surface they make a great contribution to global productivity. Blue whales feed on tiny crustaceans called krill, by filtering seawater through sheets in their mouths called baleen. Krill feed on diatoms. Net primary productivity (NPP) for diatoms has been estimated to be 50 g m –3 day –1 and secondary productivity for krill has been estimated to be 5 g m –3 day –1. Image 2.5 shows a simplified food chain, the numbers shown are in g m –3 day –1. R represents respiration and E represents excretion. (i) Calculate the rate at which krill use diatom biomass for respiration (R). [1] Rate = ……….……….....………. g m –3 day –1 (ii) It is estimated that a single blue whale needs to consume 8 000 kg of biomass per day. The biomass of krill was estimated as 25 g m –3. Calculate the volume of water a whale needs to filter per day to take in 8 000 kg of biomass. Give your answer in standard form. [3] Volume of water = ……….……….....………. m 3 day –1 © WJEC CBAC Ltd. Image 2.5 08 diatoms blue whale krill R E 50 5 11 1 (1400U30-1) Turn over. 140 0 U3 01 0 9 9 Examiner only 3. ATP is described as the universal energy currency of cells. (a) (i) Describe why ATP is described as a universal currency. [2] Image 3.1 summarises the production and use of ATP in muscle cells. (ii) Complete the diagram in Image 3.1. [3] © WJEC CBAC Ltd. Image 3.1 ........................................... + oxygen ............................................................. relaxed muscle ATP .......................... + .......................... carbon dioxide + ............................................. energy energy 09 10 (1400U30-1) Examiner only (b) Image 3.2 shows diagrammatic representations of membranes found in two organelles where ATP synthesis takes place. (i) Name the organelles in which the membranes (shown in Image 3.2) would be found. [1] A ……………………………….………………………………. B ……………………………….………………………………. (ii) In Image 3.2, X, Y and Z represent membranes and compartments found in organelles A and B. Complete the Table 3.3 to name the membranes and compartments represented in Image 3.2 for each organelle. [3] Table 3.3 © WJEC CBAC Ltd. Image 3.2 Name of membrane / compartment Letter Part represented in image Organelle A Organelle B X membrane ......................................................... ......................................................... Y compartment enclosed by membrane ......................................................... ......................................................... Z compartment surrounding membrane ......................................................... ......................................................... 10 Membrane found in organelle B Membrane found in organelle A (1400U30-1) Turn over. 140 0 U3 01 11 11 Examiner only (iii) Describe how the components in the membrane of organelle B are involved in the synthesis of ATP by chemiosmosis. [5] © WJEC CBAC Ltd. 11 12 (1400U30-1) Examiner only The experiment shown in Image 3.4 is considered to be evidence supporting the chemiosmotic hypothesis. It was carried out on membranes isolated from organelles of type A in Image 3.2 and made into vesicles. The isolated membranes were placed in buffer solutions, the lower the pH the higher the concentration of protons. (c) Explain why the experiment shown in Image 3.4 supports the chemiosmotic theory. [4] © WJEC CBAC Ltd. Image 3.4 ATP ADP + Pi H+ pH 4 pH 8 ADP + Pi pH 4 pH 4 membrane isolated from organelle A ATP detected rapid change of external pH to pH 8 placed in buffer pH 4 internal pH equalised with external pH 18 12 (1400U30-1) Turn over. 140 0 U3 01 13 13 Examiner only 4. The size of any population at a given time is determined by the equation: Number of individuals = (birth rate + immigration) – (death rate + emigration) In field studies which monitor population size over a period of time the number of individuals often stays constant. (a) Using the terms in brackets from the above equation, write another equation which shows the relationship between the terms when the population size remains constant. [1] ……………………………………………..…………...................……………. = ……………....…………………………....……..…………...........……………. Scientists monitored the population of frogs in a woodland surrounding a pond. The capture-mark-recapture method was used to determine the number of adult frogs, as follows: • 19 frogs were caught • marked by clipping off one toe • they were then released back into the pond • a week later the scientists collected as many frogs as they could over three consecutive days • the results are shown in Table 4.1 • captured frogs from the three consecutive days were not released until after the third collection. Table 4.1 Result of collections following release of marked frogs © WJEC CBAC Ltd. Date Total no. of frogs captured No. of marked frogs Day 1 48 5 Day 2 45 5 Day 3 50 7 Total 143 17 13 (b) (i) From the figures given in the method and Table 4.1 estimate the total number of frogs in the woodland, using the following formula: [2] Where, • N = number in population • M = number initially captured and marked • n = total number subsequently captured • m = number of marked individuals recaptured. Estimated number of frogs in the woodland = .......................................... N = Mn m 14 (1400U30-1) Examiner only (ii) Explain why the chosen method of marking the frogs might have affected the estimate of the frog population. [1] (c) Between capture and release the adult frogs were kept, ten to a tank, partially submerged in water collected from the pond. The frogs in one of the tanks developed red patches on their legs. The scientists suspected they were suffering from ‘red-leg disease’, caused by the bacterium, Aeromonas hydrophila, a Gram-negative bacillus. The scientists took a swab from the leg of one of the frogs, performed a Gram stain and examined the sample under the microscope. Describe the shape and colour of the bacteria they would have seen if the frog had been suffering from red-leg disease. [2] Shape ………………………………………….............................................………… Colour ………………………………………….............................................………… (d) In order to study the survival rates of the larval stage of the frogs (tadpoles), two smaller ponds of equal volume were created from the existing pond using polyethylene sheets. Pond 1 was stocked with 5000 tadpoles per m3 and pond 2 was stocked with 1000 tadpoles per m3. The scientists took 20 samples of water from each pond every ten days and counted the number of tadpoles in each sample. They used the mean counts to calculate the number of tadpoles per m3. Their results are shown in Graph 4.2, the straight lines drawn in black are tangents to the curves. © WJEC CBAC Ltd. 14 (1400U30-1) Turn over. 140 0 U3 01 15 15 Examiner only © WJEC CBAC Ltd. 0 1000 2000 3000 4000 5000 6000 0 10 20 30 40 50 Pond 2 Pond 1 Graph 4.2 Days after adding tadpoles Mean number of tadpoles per m3 (i) Calculate the rate of decline in number of tadpoles for pond 1 at day 10. Give your answer to two significant figures. [3] Rate of decline …………………..……………..……….. tadpoles m–3 day–1 (ii) Using the information from Graph 4.2 it was concluded that a density-dependent factor was causing the change in the tadpole populations. State the evidence for this conclusion and suggest what the factor might have been. [3] 15 16 (1400U30-1) Examiner only (e) The scientists also carried out a laboratory experiment to investigate the effect of selection on rate of development as measured by body length. They used 3 different tanks, which contained 10 dm3 of water. • Tank 1 – low density (1 tadpole per dm3) • Tank 2 – high density (5 tadpoles per dm3) • Tank 3 – high density with selection against small individuals (two of the smallest tadpoles were removed each week). All tanks were kept under the same environmental conditions. The length of the tadpoles in each tank was measured every 10 days and a mean calculated. The results of the experiment are shown in Graph 4.3 (i) With reference to Graph 4.3, state two conclusions that can be drawn about the effect of density and selection on the rate of development of tadpoles in the three tanks. [2] © WJEC CBAC Ltd. 6 7 8 9 10 11 12 13 14 15 0 10 low density Key: high density high density with selection 20 30 40 50 Graph 4.3 Time / days Mean body length / mm 16 (1400U30-1) Turn over. 17 Examiner only (ii) Another group of scientists said that it was not valid to use the results from tank 2 and tank 3 to make a conclusion about the effect of selection. Suggest why it might not be valid to compare these two tanks and describe how the method could be changed to enable a more valid conclusion to be made. [2] © WJEC CBAC Ltd. 16 17 X Y 18 (1400U30-1) Examiner only 5. The mammalian kidney has a role in two physiological processes, excretion and homeostasis. Image 5.1 shows a single kidney nephron. The strategy that the kidney uses for excretion is ultrafiltration followed by selective reabsorption. (a) (i) Use labelled lines on Image 5.1 to show the sites of: [2] I. ultrafiltration II. selective reabsorption © WJEC CBAC Ltd. Image 5.1 afferent arteriole 18 1 0 0.5 1.0 1.5 2.0 2.5 3.0 2 3 4 5 6 (1400U30-1) Turn over. 19 Examiner only (ii) Structures labelled X and Y in Image 5.1 are involved in homeostasis. Name structures X and Y and the homeostatic process in which they are involved. [2] X …………………………………………………………………………………………………… Y …………………………………………………………………………………………………… Homeostatic process ……………………………………………………………………………………………………...................................... The use of micropipettes has allowed samples of fluid to be withdrawn from specific points along kidney tubules of experimental animals. Samples of filtrate were taken from five positions along the proximal convoluted tubule and the concentrations of urea and chloride ions were measured. The results are shown in Graph 5.2. (b) (i) Use Graph 5.2 to explain the change in concentrations for urea and chloride ions along the tubule. [2] © WJEC CBAC Ltd. Graph 5.2 Distance along proximal convoluted tubule / au From Bowman’s capsule urea chloride Concentration / au 19 1 0 0.5 1 1.5 2.0 2 3 4 5 6 20 (1400U30-1) Examiner only © WJEC CBAC Ltd. The experiment was repeated but concentrations of glucose were measured with and without oligomycin. Oligomycin is a chemical compound that specifically inhibits respiration. The results are shown in Graph 5.3. (ii) Use Graph 5.3, to explain the change in concentration along the tubule for glucose with and without the respiratory inhibitor oligomycin. [4] Graph 5.3 20 Concentration / au From Bowman’s capsule Glucose with oligomycin Glucose Distance along proximal convoluted tubule / au (1400U30-1) Turn over. 21 Examiner only A third experiment was carried out to study how structure X, shown in Image 5.1, is involved in the formation of urine. Image 5.4 shows regions of tubule on either side of structure X. Samples were withdrawn from two locations, 1 and 2. The concentration of sodium ions (Na+) was measured in each location. (c) (i) Explain why these locations were chosen for sampling. [1] (ii) The Na+ concentration in the sample taken from location 1 was higher than the sample taken from location 2. Use your knowledge of the function of both limbs of the structure in Image 5.4 to explain this result. [4] © WJEC CBAC Ltd. Image 5.4 direction of flow direction of flow location 2 location 1 15 21 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE (1400U30-1) 22 © WJEC CBAC Ltd. 22 (1400U30-1) Turn over. 23 Examiner only © WJEC CBAC Ltd. 6. Image 6.1 shows a natural carbon cycle unaffected by human activity. The arrows represent processes which transfer carbon from one form to another. (a) (i) Process W involves the fixation of carbon in green plants. Name the two reactants involved and the enzyme that catalyses this process. [2] Reactants ……………………………………………………………………………………………… ……………………………………………………………………………………………… Enzyme ……………………………………………………………………………………………… (ii) I. On Image 6.1, label one of the arrows with an X to show a process which involves micro-organisms. II. Explain the role of micro-organisms in this process. [2] (iii) Add an arrow labelled C to Image 6.1 to represent the transfer of carbon as a result of human activity. [1] Image 6.1 W carbon dioxide (atmosphere) organic compounds (animals) organic compounds (coal,gas,oil) inorganic compounds (shells) limestone (rocks) carbonates (oceans) organic compounds (plants) organic compounds (dead plants and animals) 23 24 (1400U30-1) © WJEC CBAC Ltd. (b) Nine global systems have been identified as being key regulators of the Earth’s stability. Values have been proposed that represent boundaries or thresholds. Table 6.2 shows two of the nine systems together with their threshold values and current values and Image 6.3 displays the threshold values and current values as a circular graph. Planetary System Parameters Threshold values Current value Climate change Atmospheric carbon dioxide concentration (ppm by volume) 350 387 Nitrogen How much nitrogen is removed from the atmosphere for human use (tonnes × 106 / year) 35 121 biodiversity land use freshwater ocean acidification Z Y aerosol ozone Threshold chemical pollution Image 6.3 Table 6.2 – Planetary Boundaries 24 (1400U30-1) Turn over. 25 Examiner only © WJEC CBAC Ltd. (i) Use the information in Table 6.2 to name the two missing planetary systems labelled Y and Z in Image 6.3. [1] Y ....................................................................... Z ...................................................................... (ii) Use Image 6.3 to state what the two planetary systems in Table 6.2 have in common with each other and with the Land-use system. [1] (c) Explain what is meant by a safe operating space for humanity, describe where that is shown in Image 6.3, and describe the consequences of exceeding planetary boundaries. [3] 10 25 26 (1400U30-1) 7. Image 7.1 shows the pathway from a tooth to an area of the brain which generates the sensation of pain. It also shows the site of action of two local anaesthetics used in dentistry. Image 7.2 shows a reflex arc. © WJEC CBAC Ltd. pain centre brain tooth ganglion ganglion spinal cord spinal cord Site of action of local anaesthetic B which prevents the entry of calcium ions into the presynaptic membrane Site of action of local anaesthetic A which prevents the passage of sodium ions into neurons Image 7.1 Image 7.2 26 (1400U30-1) Turn over. 27 Examiner only Compare and contrast the pathway shown in Image 7.1 with the reflex arc in Image 7.2. Using your knowledge of the generation of action potentials, suggest how anaesthetic A will prevent pain. Using your knowledge of synaptic transmission, suggest how anaesthetic B could also prevent pain. [9 QER] © WJEC CBAC Ltd. 27 28 (1400U30-1) Examiner only © WJEC CBAC Ltd. 28 (1400U30-1) Turn over. 29 Examiner only © WJEC CBAC Ltd. END OF PAPER 29 9 Question number Additional page, if required. Write the question number(s) in the left-hand margin. Examiner only (1400U30-1) 30 © WJEC CBAC Ltd. 30 (1400U30-1) 31 Examiner only Question number Additional page, if required. Write the question number(s) in the left-hand margin. © WJEC CBAC Ltd. 31 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE (1400U30-1) 32 © WJEC CBAC Ltd. 32