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Wright BIO 2110 - BIO2110f13_exam3answers

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Name _________________________ BIO 2110 – Exam 3 6 December 2013 10 questions – 10 points/question see last page for potentially useless formulae1. Leg length variation in opiliones results from allelic variation in a single gene. Dominance relationships for alleles at this gene are shown below. In one population of opiliones, 360 have long legs, 280 have short legs and 360 have no legs. From the frequencies of these phenotypes, determine all allelic and genotypic frequequencies for this population. [Assume this population is in equilibrium.] AA Aa Aα - long legs aa aα - short legs αα - no legs phenotype number long legs 360 short legs 280 no legs 360 p2 + 2pq + 2pr + q2 + 2qr + r2 = 1 r2 = 0.36 r = 0.6 q2 + 2qr = 0.28 q2 + 1.2q – 0.28 = 0 q = 0.2 p = 1 – (0.2 + 0.6) = 0.2 p2 = 0.04 2pq = 0.08 2pr = 0.24 q2 = 0.04 2qr = 0.24 r2 = 0.362. Color variation in helioflowers results from allelic variation in a single gene. Dominance relationships for the alleles of this gene are shown below. Helioflowers are self-fertile, which leads to significant levels of inbreeding. The distribution of phenotypes in one population of helioflowers is shown below. From these phenotypic frequencies, determine the inbreeding coefficient for this population. genotype phenotype CRCR - red CRCY - orange CYCY - yellow phenotype number red 432 orange 336 yellow 232 p = [2 x 432) + 336]/2000 = 0.6 q = 0.4 2pq = 0.48 Hobs = 336/1000 = 0.336 FIS = 1 – (0.336/0.48) = 0.33. Draw graphs that indicate how phenotypic distributions would be impacted by: i. directional selection, ii. disruptive selection, and iii. stabilizing selection Assume all initial distributions to be normal.4. The presence or absence of eyes in cave fish is determined by allelic variation at a single gene. Fish with EE and Ee genotypes have eyes. Fish with ee genotypes lack eyes. Given the initial allelic frequencies and relative fitnesses shown below, determine the frequencies of cavefish with and without eyes after two generations of selection. genotype phenotype EE Ee - eyes present ee - eyes absent initial frequency of E allele, p0 = 0.9 initial frequency of e allele, q0 = 0.1 WEE = 0.2 WEe = 0.2 Wee = 1.05. The average nucleotide divergence, d, between S. kargeaux and S. pradille is 0.0016 substitutions per nucleotide. Effective population sizes, Ne, for these species are 40,000 and 36,000, respectively. Assuming a mutation rate, µ, of 10-8 substitutions per nucleotide per generation, what is the divergence time of these species? d = 0.0016 substitutions per nucleotide µ = 10-8 substitutions per nucleotide per generation Ne S.k. = 40,000 Ne S.p. = 36,000 d = 2tµ t = d/2µ t = 0.0016/(2 x 10-8) = 80,000 generations6. Jellyfish in Ongeim’l Tketau (Lake Jellyfish) have low toxin levels. Jellyfish in the adjacent lagoon have high toxin levels. This results from allelic variation in a single gene. Below are given dominance relationships for alleles in this gene, initial allele frequencies for lake and lagoon populations of jellyfish and migration rates between populations. Ongeim’l Tketau (Jellyfish Lake) i. Determine how allele frequencies in the lake population will change over two generations of migration. (Assuming no other departures from Hardy-Weinberg equilibrium.) ii. What would happen to allelic frequencies in the lake population if this rate of migration (and no other departures from H-W eq.) continued over a large number of generations (e.g. > 1,000)? ii. Explain how allelic frequencies in the lake population might remain constant despite migration from the lagoon population. (i.e. what other departure from H-W equilibrium could counteract the impact of migration?) genotype phenotype TT highly toxic Tt highly toxic tt non-toxic p = frequency of T allele p1 n+1 = p1(1 – m21) + p2(m21) q = frequency of t allele gen p1 q1 Lake Population (population 1) p0 = 0.01 q0 = 0.99 Lagoon Population (population 2) p0 = 0.95 q0 = 0.05 migration lagoon  lake, m2  1 = 0.10 lake  lagoon, m1  2 = 0.007. Shown below is a comparison of 10 sequences obtained from a single population. Based on these comparisons, what levels of linkage disequilibrium are present for each haplotype in this population? [Segregating sites indicated by shading. Sequences grouped by haplotype.] AGATA AGAGT GAGAC CCTCA CTAGC AGATA AGAGT GAGAC CCTCA CTAGC AGATA AGAGT GAGAC CCTCA CTAGC AGATA AGAGT GAGAC CCTTA CTAGC AGATA AGAGT GAGAC CCTTA CTAGC AGATA AGAGT GAGAC CCTTA CTAGC AGATA AAAGT GAGAC CCTCA CTAGC AGATA AAAGT GAGAC CCTTA CTAGC AGATA AAAGT GAGAC CCTTA CTAGC AGATA AAAGT GAGAC CCTTA CTAGC HA1B1 = 0.3 HA1B2 = 0.3 HA2B1 = 0.1 HA2B2 = 0.3 PA1 = 0.3 + 0.3 = 0.6 PA2 = 0.1 + 0.3 = 0.4 PB1 = 0.3 + 0.1 = 0.4 PB2 = 0.3 + 0.3 = 0.6 EA1B1 = (0.6)(0.4) = 0.24 EA1B2 = (0.6)(0.6) = 0.36 EA2B1 = (0.4)(0.4) = 0.16 EA2B2 = (0.6)(0.4) = 0.24 DA1B1 = 0.3 – 0.24 = 0.06 DA1B2 = 0.3 – 0.36 = -0.06 DA2B1 = 0.1 – 0.16 = -0.06 DA2B2 = 0.3 – 0.24 = 0.068. Prepupation caterpillar weight is a quantitative trait. Hungry caterpillars reach a weight of 230 mg prior to pupation. Sated caterpillars reach a weight of 10 mg prior to pupation. In table 17 below are shown weights and genotypes of backcross progeny derived from crosses between hungry and sated inbred lines of caterpillars. To identify QTL associated with prepupation weight in caterpillars, backcross progeny need to be divided into two groups at each marker. Weight distributions for these groups then are compared to each other. How would you divide the backcross progeny to make these comparisons at marker M3? How would you divide the backcross progeny to make these comparisons at marker M4? (i.e. for markers M3 and M4, identify two groups and identify which backcross progeny belong in each group.) M3 HH: 001, 002, 004, 005, 010, 011, 014, 016, 017, 0,18, 020 AVG = 180.7 HS: 003, 006, 007, 008, 009, 012, 013, 015, 019 AVG = 169,6 M4 HH: 002, 005, 009, 010, 014, 015, 017, 018, 019, 020 AVG = 176.1 HS: 001,003, 004, 006, 007, 008, 011, 012, 013, 016 AVG = 175.39. In two populations of ostriches, egg diameter is 560 mm. F1 progeny derived from crosses between these populations also lay eggs with a diameter of 560 mm. However, nine classes of F2 progeny are observed with egg diameters ranging from 520 mm


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