Problem 4. A) Is the population at H-W equilibrium for the A gene? How about the B gene? (Show your work) Determine allelic frequencies: f(A) = p = ( 202 +202 + 101 + 372 + 372 + 186 + 166 + 166 + 83 ) / 2960 = .625 f (a) = q = 1 – p = 1 - .625 = .375 Now find the genotypic frequencies and the number of expected individuals out of 1480 to have this genotype and compare it to the observed number: f (AA) = p 2 = .625 2 = .39 X 1480 = 577 expected and 740 observed f (Aa) = 2pq = 2 X .625 X .375 = .47 X 1480 = 696 expected and 370 observed f (aa) = q 2 = .375 2 = .14 X 1480 = 207 expected and 370 observed Expected and observed are not similar so the A gene is not at H-W equilibrium. Do the same thing for the B gene and you will find that the number of expected and the number of observed are nearly identical. Thus, the population is at H-W equilibrium for the B gene. B) After one generation of random mating, what fraction of the next generation will be AA (independent of the B gene)? Since we are at H-W equilibrium allele frequencies do not change.
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