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homework3_answers - Bios 334 Evolution 2005 Name Homework 1...

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Bios 334 Evolution 2005 Name_______________________________ Homework 1. Due 10:00 AM Monday Sept. 19. Full 10 pts for reasonable on-time effort. 1. A natural population of adult mice has the following genotype frequencies at a possible disease locus: freq( BB ) = 0.9000 freq( Bb ) = 0.1000 freq( bb ) = 0.0000 This provides two pieces of evidence consistent with the b allele being disadvantageous. Explain (Hint: calculate the allele frequencies and the Hardy Weinberg frequencies). p = D+H/2 = 0.9000 + 0.0500 = 0.9500 q = R+H/2 = 0.0000 + 0.0500 = 0.0500 The b allele is much rarer than the B allele, consistent with it being disadvantageous (though there are other possible explanations). Hardy-Weinberg frequencies: BB p 2 = 0.95 2 = 0.9025 Bb 2pq = 2(0.95)(0.05) = 0.0950 bb q 2 = = 0.05 2 = 0.0025 These do not match the observed frequencies; in particular bb 's are not observed when they should be present, suggesting that they have died. 2. You have discovered a gene affecting beak size in a Galapagos finch. The A allele tends to produce larger beaks than the a allele, and these leads to the following numbers of offspring for each genotype: R AA = 4.00 R Aa = 3.60 R aa = 3.60 (a) What are the relative fitnesses of the three genotypes? You could divide the three relative fitnesses by any constant, but it is most commonly done by dividing by the highest fitness, giving: W AA = 1.00 W Aa = 0.90 W aa = 0.90 Or, if you divide by the lowest fitness, you get: W AA = 1.11 W Aa = 1.00 W aa = 1.00 (b) What is the selection coefficient, s? If you followed the first choice above, s = W-1 = -0.10 for the Aa and aa
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homework3_answers - Bios 334 Evolution 2005 Name Homework 1...

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