Problem Set 5 Evolution1. Below are some data for oyster viability based on their genotypes at the catalase locus.AAABBBTotalZygotes140353229722Adults14346200560Assuming that only viability selection is occurring:A.Calculate the relative fitnesses of each genotype, setting wAA= 1. Which genotype has the highest fitness?B.Calculate the average excess of each allele. C.What do these average excess values tell you about the course of natural selection.Answers:A. First calculate the absolute fitnesses. Because only viability selection is occurring, these fitnesses are equal to the probability of surviving to adulthood for each genotype. So AbsolutewAA= 14/140 = .10 wAB= 346/353 = .98 wBB= 200/229 = .87To make these relative fitnesses, divide them all by wAA:RelativewAA= .10/.10 = 1 wAB= .98/.10 = 9.8 wBB= .87/.10 = 8.7 B. For this, you need the allele frequencies before selection and W, the average population fitness. p=140/722 + ½353/722 = .1939 + .2445 = .4384 = .44q=229/722 + ½353/722 = .3172 + .2445 = .5617 = .56W = .442(1) + 2(.44)(.56)(9.8) + .56
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