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Review Session 6 Key

Review Session 6 Key - Review session problems for Nov 6...

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1 Review session problems for Nov. 6 2011 KEY Topics: H-W equilibrium, non random mating, selection against a recessive trait, mutation, mutation/selection equilibrium 1. Imagine that the number of leaves in clover plants is controlled by alleles of a single gene with semi-dominant alleles F and f*. ff plants have four leaves, Ff plants have three, and FF have two. In a meadow with 735 clovers, 137 had four leaves, 357 had three leaves, and 241 had two leaves. Recall that H-W equilibrium occurs when mating is random so that f(aa) = q 2 , f(Aa) = 2pq and f(AA) = p 2 . a. Is the population in Hardy-Weinberg equilibrium? Hypothesis: population is in H-W equilibrium f(F) = f(FF) + ½f(Ff) = 241/735 + ½(357/735) = 0.571 = p q = 1 p = 0.429 q 2 = 0.184 X 735 = 135 2pq = 0.490 X 735 = 360 p 2 = 0.326 X 735 = 240 Chi squared = (O E) 2 /E = (137 135) 2 /135 + (357 360) 2 /360 + (241 240) 2 /240 = 0.0592 degrees of freedom = 1 so p >> 0.05 and we cannot reject the hypothesis and must conclude that the population is in H-W equilibrium. b. If there were inbreeding in this population, how would you expect the relative proportions of FF, Ff and ff to be different? Inbreeding increases the proportion of homozygotes and decreases the

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Review Session 6 Key - Review session problems for Nov 6...

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