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Wright BIO 2110 - BIO2110_exam3_fall2015

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Name ___________________________ BIO 2110 – Genetics Exam 3 11 December 2015 10 questions/10 points per question see last page for potentially useless formulae calculators permitted SHOW YOUR WORK1. Below is shown a comparison of four sequences from a single population. Segregating sites indicated with arrowheads. Based on this comparison, estimate the effective size, Ne, of this population. Show your work. Assume a mutation rate of 10-8 mutations per nucleotide per generation. ▼ ▼ ▼ ▼ ▼ 1: AGATC AGCCT AAGAT CATAC GATAC GGACT TTTCT TCCTG ACGAT CAGTT 2: AGAAC AGCCT AATAT CATAC GATAC GGACT TTTCT TTCTC ACGAT CAGTT 3: AGAAC AGCCT AATAT CATAC GATAC GGACT TTTCT TCCTC ACGAT CAGTT 4: AGAAC AGCCT AATAT CATCC GATAC GGACT TTTCT TTCTC ACGAT CAGTT2. In a population of saber-tooth tigers, allelic variation in a single gene determines fang length. Dominance relationships and phenotypic distribution of fang lengths are shown below. Based on these data, determine the frequencies of all alleles and genotypes in this population. Assume the population is in equilibrium phenotypic genotype fang length frequency GG Gg 9 cm 0.19 Gγ gg 13 cm 0.32 gγ γγ 17 cm 0.493. In geoducks, syphon length is determined by allelic variation in a single gene. In one population of geoducks, there are two alleles, B and b. The initial frequency of the b allele in this population, q0, is 0.8. The relative fitnesses of the different genotypes are shown below. i. determine the mean fitness of this population ii. determine the frequencies of both alleles after one generation of selection WBB = 1.0 WBb = 0.9 Wbb = 0.34. Below are shown the results of a QTL on study club size in mantis shrimp. Based on the results shown, what is the minimum number of genes on chromosome 7 impact mantis shrimp club size? What criterion was used to make this determination. 0 4 8 12 161234567position, Mb, on chromosome 7Lod5. Below are shown ten sequences from a population in which variation is observed in the A and B loci (indicated by arrowheads). These sequences are grouped by gamete type, i.e. G1 = A1B1, G2 = A1B2, G1 = A2B1, G1 = A2B2. Based on these data, determine linkage disequilibrium values for the A and B loci in this population. A1 = A B1 = T A2 = G B2 = C A locus B locus ▼ ▼ GATCA GTTCA TTCAA AATAT GTATA CATAC CCATC GTACT GACTA GTACT GATCA GTTCA TTCAA AATAT GTATA CATAC CCATC GTACT GACTA GTACT GATCA GTTCA TTCAA AATAT GTATA CACAC CCATC GTACT GACTA GTACT GATCA GTTCA TTCAA AATAT GTATA CACAC CCATC GTACT GACTA GTACT GATCA GTTCA TTCAA AATAT GTATA CACAC CCATC GTACT GACTA GTACT GATCA GTTCA TTCAA AATAT GTATA CACAC CCATC GTACT GACTA GTACT GATCA GTTCG TTCAA AATAT GTATA CATAC CCATC GTACT GACTA GTACT GATCA GTTCG TTCAA AATAT GTATA CACAC CCATC GTACT GACTA GTACT GATCA GTTCG TTCAA AATAT GTATA CACAC CCATC GTACT GACTA GTACT GATCA GTTCG TTCAA AATAT GTATA CACAC CCATC GTACT GACTA GTACT6. Two populations of apatosaurs differ in average mass. In one population, adults are 34 tons. In the other population, adults are 36 tons. Seven phenotypic classes, ranging from 32 to 38 ton, are obtained in F2 progeny derived from crosses between these populations. i. how many genes are involved in the regulation of body mass in apatosaurs? ii. if 3 F2 apatosaurs are obtained that weigh 32 tons, then how many apatosaurs would you expect in the other size classes? iii. what genetic structure in the P0 populations would allow for the results shown below (i.e. what were the genotypes of the P0s)? P0 34 tons x 36 tons F1 35 tons x 35 tons F2 32 tons 33 tons 34 tons 35 tons 36 tons 37 tons 38 tons7. Allelic variation in a single gene regulates alcohol tolerance in the hypotrichous ciliate, Euplotes crassus. Dominance relationships for this gene are shown below. Also shown below are the numbers of individuals in a population that exhibit different tolerance levels. Based on these data, how does the structure of this population depart from expectations for a population in Hardy-Weinberg equilibrium? Provide a quantitative measure to describe this departure from equilibrium. % alcohol number in genotype tolerance (LD50) phenotype AA 17 480 Aa 13 240 aa 9 2808. In a population of 3600 diploid organisms, what is the initial frequency of each new mutation that occurs? What is the neutral probability of fixation for these new mutations? What is the neutral probability of extinction for these new mutations?9. Yellow-spotted lizards can either have seven spots or nine spotted. Variation in spot number is regulated by allelic variation in a single gene. Dominant homozygotes and heterozygotes have seven spots. Recessive homozygotes have nine spots. Two populations of lizards vary in the frequencies of seven-spotted and nine-spotted individuals. Determined genotypic frequencies for these populations after one generation of migration. Initial population structure and migration rates are shown below. Assume migration is the only departure from Hardy-Weinberg equilibrium. Phenotype Population 1 Population 2 seven-spotted 510 840 nine-spotted 490 160 migration coefficients, m12 = 0.0 m21 = 0.110. The divergence, d, between two populations of red-winged locust is 0.00047. Assuming a mutation rate, µ, of 1 x 10-8, what is the divergence time (in generations) between these populations?e = mc2 WAAp2 + WAa2pq + Waaq2 = W p1 n+1 = (1 - m21)(p1 n) + (m21)(p2 n) F = 1 – (!!"!"#) p + q + r = 1 p2 + q2 + r2 + 2pq + 2pr + 2pr = 1 pn+1 = (WAAp2 + WAapq)/W θT = π θW = k/a a = 1 + ½ + 1/3 … 1/(n – 1) PA1 = GA1B1 + GA1B2 e = hν EA1B1 = (PA1)(PB1) DA1B1 = EA1B1 – GA1B1 p + q = 1 p2 + 2pq + q2 = 1 n = !"#!(!"#$!!"#$!!"#$$)!!.!"# θ = 4Neµ !!" !"!!!" d = 2µt Lod = Log10(odds) odds QTL = !"#$!!"#!"#$!!"!!"# n = (c – 1)/2 d = 2kt k = µ 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6


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Wright BIO 2110 - BIO2110_exam3_fall2015

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