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Unformatted text preview: c) Which population is at HardyWeinberg 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 chisquare calculation. ! 2 = .241 3. State your conclusion based on the data. population is in HWE (31 = 2 dof; 0.90 > p > 0.50)...
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 Spring '09
 Freeling
 Genetics, Evolution, Population Genetics

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