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 proportion of heterozygotes, so we would expect to see a higher proportion of two and/or four
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This note was uploaded on 01/16/2012 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.

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

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