24_PopulationGen

24_PopulationGen - 24 POPULATION GENETICS SOLUTIONS TO TEXT...

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24 POPULATION GENETICS SOLUTIONS TO TEXT PROBLEMS 24.1 In the European land snail Cepaea nemoralis, multiple alleles at a single locus determine shell color. The allele for brown ( C B ) is dominant to the allele for pink ( C P ) and to the allele for yellow ( C Y ). The dominance hierarchy among these alleles is C B C P C Y . In one population sample of Cepaea, the following color phenotypes were recorded: Assuming that this population is in Hardy-Weinberg equilibrium (large, randomly mating, and free from evolutionary processes), calculate the frequencies of the C B , C Y , and C P alleles. Answer: Equate the frequency of each color with the frequency expected in Hardy-Weinberg equilibrium, letting p = f ( C B ), q = f ( C P ), and r = f ( C Y ). Brown: f ( C B C B ) + f ( C B C P ) + f ( C B C Y ) = p 2 + 2pq + 2pr = 236/500 = 0.472 Pink: f ( C P C P ) + f ( C P C Y ) = q 2 + 2qr = 231/500 = 0.462 Yellow: f ( C Y C Y ) = r 2 = 33/500 = 0.066 Now solve for p, q, and r, knowing that p + q + r = 1. 2 0 066 so 0 066 0 26 r . , r . . = = = There are two approaches to solve for q. First, since q 2 + 2 qr = 0.462, one can substitute in r = 0.26, giving q 2 + 2 q (0.26) = 0.462. Recognize this as a quadratic equation and set it equal to 0, and solve for q. That is, solve the equation q 2 + 0.52 q - 0.462 = 0. Solving the quadratic equation for q, one has: ( 29 ( 29 ( 29 ( 29 2 4 1 0 462 0 467 2 1 0 52 0 52 . . . . q - = = 2 2 6 A second approach to solve for q is to realize that q 2 + 2 qr = 0.462 and r 2 = 0.066. Adding left and right sides of the equations together, one has q 2 + 2 qr + r 2 = 0.066 + 0.462 ( q + r ) 2 = 0.528 q + r = 0.726
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q = 0.726 - r = 0.726 - 0.26 = 0.467 Since p + q + r = 1, p = 1 - ( q + r ) = 1 - (0.26 + 0.467) = 0.273. 24.2 Three alleles are found at a locus coding for malate dehydrogenase (MDH) in the spotted chorus frog. Chorus frogs were collected from a breeding pond, and each frog’s genotype at the MDH locus was determined with electrophoresis. The following numbers of genotypes were found: a. Calculate the frequencies of the M 1 , M 2 , and M 3 alleles in this population. b. Using a chi-square test, determine whether the MDH genotypes in this population are in Hardy- Weinberg proportions. Answer: a. The tally for M 1 alleles is as follows: The total number of individuals is 254; thus the total number of alleles is 254 × 2, or 508. The frequency of M 1 alleles is 104/508 = 0.205. The frequency of the other alleles is obtained similarly so that f ( M 1 ) = 0.20 = p, f ( M 2 ) = 0.30 = q, and f ( M 3 ) = 0.50 = r. b. For three alleles with frequencies p, q, and r, a population in Hardy-Weinberg equilibrium will have p 2 + 2 pq + 2 pr + q 2 + 2 qr + r 2 = 1. To test the hypothesis that the population is in Hardy-Weinberg equilibrium, calculate the numbers of individuals expected in each class using this relationship and the values for p, q, and r obtained in (a). Calculate the value of χ 2 as shown in the following table. Since the six phenotypic classes are completely specified by three allele frequencies, the number of
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This note was uploaded on 04/26/2010 for the course HUM 3255 taught by Professor Richards during the Spring '10 term at Florida A&M.

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24_PopulationGen - 24 POPULATION GENETICS SOLUTIONS TO TEXT...

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