PS 2 answers, probs 1+2+3 - c) Which population is at...

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LS4 Problem Set 2, 2010 Population Genetics Corresponding Lectures: November 5, 8 and 10 Corresponding Reading: Hartwell et al., 757-773 Corresponding Quiz: November 15-19 (Along with positional cloning) If you want to try additional problems, there are many at the end of chapter 21 that pertain to this section of the course. Problem 1. a) What are the allele frequencies of M and N in this population? Freq M = [(550*2)+150]/(2*1000) = 0.625 Freq N = [(300*2)+150]/(2*1000) = 0.375 b) If this population is at Hardy-Weinberg equilibrium, how many individuals with a genotype of MN would you expect to see? 2pq = 2*0.625*0.375 = 0.469 0.469*1000 = 469 people Problem 2. a) What are the allele frequencies of R and r in population 1? p = Freq R = [(2*10)+74]/(2*100) = 0.47 q = Freq r = [(2*16)+74]/(2*100) = 0.53 b) What are the allele frequencies of R and r in population 2? p = Freq R = [(2*81)+18]/200 = 0.9 q = Freq r = [(2*1)+18)]/200 = 0.1
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Unformatted text preview: c) Which population is at Hardy-Weinberg equilibrium? Population 2 is at H.W. eq Population 1 expected genotype frequency p^2 = 0.47^2 = 0.2209 0.2209 *100 = 22 RR individuals 2pq = 2*0.47*0.53 = 0.4982 0.4982 * 100 = 50 Rr individuals q^2 = 0.53^2 = 0.2809 0.2809 * 100 = 28 rr individuals Doesnt match given observed values, not in H.W. eq Population 2 expected genotype frequency p^2 = 0.9^2 = 0.81 0.81*100 = 81 RR individuals 2pq = 2*0.9*0.1 = 0.18 0.18*100 = 18 Rr individuals q^2 = 0.1^2 = .01 .01*100 = 1 rr individual This matches the observed values, so population 2 is at H.W. eq Problem 3. 1. List the expected values for each genotypic class under your null hypothesis. A1A1 = 306 A1A2 = 170 A2A2 = 24 2. Perform the chi-square calculation. ! 2 = .241 3. State your conclusion based on the data. population is in HWE (3-1 = 2 dof; 0.90 > p > 0.50)...
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