Popgen practice problems Fall 2009 KEY

Popgen practice problems Fall 2009 KEY - POPULATION...

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POPULATION GENETICS - PRACTICE PROBLEMS KEY FALL 2009 These problems are designed to help you understand the concepts covered in lecture and section. However, merely completing this set of practice problems is not sufficient preparation for the exam – be sure to review the lecture slides as well! You should start getting in the habit of showing your work and stating the assumptions you make in each problem. Problem 1: For the scenarios given below, circle the more important evolutionary force. N e = 4 , s = 0 . 04: drift selection N e = 4 , m = 0 . 6: drift migration m = 1 , s = 0 . 04: migration selection Problem 2: You study evolution by natural selection in guppies found in Trinidad. In one of your streams, there are two populations of guppies separated by a waterfall. In the population below the waterfall, adult guppies are often eaten by a cichlid fish. There are no natural predators above the waterfall. You want to know if predation on adults has caused the guppies to mature earlier. Pretend that age to maturity is controlled by a single gene with two alleles: A and a .You collect genotype data for your two populations ( above and below the waterfall) for two generations: Above the waterfall genotype: AA Aa aa generation 0: 4 3 3 generation 1: 6 3 1 Below the waterfall genotype: AA Aa aa generation 0: 8 11 11 generation 1: 16 8 6 (A) Is the population above the waterfall in generation 1 in HWE? (use a χ 2 test to check) First, calculate the expected genotype numbers: p = freq( A ) = 6 + 0 . 5(3) 10 = 0 . 75 q = freq( a ) = 1 - p = 0 . 25 expected # AA individuals = p 2 (total # individuals) = 0 . 75 2 (10) = 5 . 625 expected # Aa individuals = 2 pq (total # individuals) = 2(0 . 75)(0 . 25)(10) = 3 . 75 expected # aa individuals = q 2 (total # individuals) = 0 . 25 2 (10) = 0 . 625 1
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2 FALL 2009 Then calculate the test statistic: χ 2 = X i (observed - expected) 2 expected = (6 - 5 . 625) 2 5 . 625 + (3 - 3 . 75) 2 3 . 75 + (1 - 0 . 625) 2 0 . 625 = 0 . 4 χ 2 < 3 . 841, the critical value for df = 1. Therefore, we CANNOT reject the null hypothesis that this population is in HWE at a significance level of p=0.05. (B) Can the observed change in allele frequencies in the population above the waterfall be explained by drift alone? Answer this question in two ways: by calculating binomial prob- abilities and by calculating the 95% confidence interval. We first need to calculate the allele frequencies in generation 0: p = freq( A ) = 4 + 0 . 5(3) 10 = 0 . 55 q = freq( a ) = 1 - p = 0 . 45 To answer this question by summing up the binomial probabilities of observing an allele frequency of equal to or more extreme than our data: we want to know the probability of observing 2(6) + 3 = 15 or more A alleles: n =# alleles sampled = 20 & p = allele frequency of parents = 0 . 55 Prob(at least 15 “ A ” in 20 alleles) = 20 X k =15 n k p k (1 - p ) n - k = 20 X k =15 20 k 0 . 55 k (0 . 45) 20 - k = 20 15 0 . 55 15 (0 . 45) 5 + 20 16 0 . 55 16 (0 . 45) 4 + ... + 20 20 0 . 55 20 (0 . 45) 0 = 0 . 0553 This probability is greater than 0.05, so we CANNOT reject our null hypothesis of drift.
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