Review Session 7 Key

Review Session 7 Key - Review session problems for Nov. 13...

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1 Review session problems for Nov. 13 2011 KEY Topics: Genetic drift, gene flow, measurement of heritability 1. In the population of vipers that was the subject of a Hardy-Weinberg example problem/clicker question in class, there were 140 non-poisonous snakes (genotype tt), 680 mildly poisonous snakes (genotype Tt) and 1180 very poisonous snakes (genotype TT). Chi squared analysis led to the conclusion that this population is not in Hardy Weinberg equilibrium. Which of the following can explain why not? Circle all that apply: a. The population is too small b. Non-poisonous snakes migrate into the population in every generation c. T mutates to t at a high frequency d. Non-poisonous snakes are less fit than poisonous ones e. >There is inbreeding within the population a-d all predict that allele frequencies would change over time within this population, but we have no information about how allele frequencies may be changing in this population over time – all we know is the allele and genotype frequencies at a single timepoint. Thus, the only violation of H-W we can assess is inbreeding, which does appear to be occurring in this population, as follows: q = f(aa) + 1/2f(Aa) = 140/2000 + (1/2)(680/2000) = 0.24 so p = 0.76 If mating is random, expected f(aa) = q 2 = (0.24)2 – 0.0625 x 2000 snakes = 115 non poisonous; expected f(Aa) = 2pq = 2(0.24)(0.76) = 0.6624 x 2000 snakes = 730; expected f(AA) = p 2 = (0.76)2 = 0.5776 x 2000 snakes = 1155. Note that actual numbers of homozygotes of both types exceeds the numbers expected in random mating conditions. 2. An isolated population of 23 black birds is joined one day by another bird of the same species that is heterozygous for a recessive allele conferring white feathers (all birds in the population of 23 are homozygous for the dominant black allele). a. Assuming random mating within the population, what will be the frequency of the recessive allele in the next generation? p C = mp D + (1 – m)p R Here, m = 1/24 = 0.04, p D = 0.5, and p R = 1 So p C = (0.04)(0.5) + (0.96)(1) = 0.98 and q c = 0.02 You could also just add up the alleles: 46 black alleles from recipient population, plus one black and one white allele from immigrant makes 47 black alleles and one white allele (48 alleles total)… 1/48 = 0.02 b. What is the probability that eventually, all the birds in this population will have white feathers,
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Review Session 7 Key - Review session problems for Nov. 13...

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