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