Question

2. Locus Fhas two alleles: F, and F2. In a sample of 1,200 individuals, 400 are F,F 600 are FiF2, and 200 are F2F2. A. What are the allele frequencies? (3 pt) B. Given the allele frequencies you just calculated, what genotypic frequencies would be expected under Hardy-Weinberg equilibrium? (3 pt)

Answer 1

2.

A.

Frequency of F1 allele = p = Number of F1 alleles/Total number of alleles

Or p = (2 $\times$ Number of F1F1 homozygote) + (Number of F1F2 heterozygote) / (2 $\times$ total number of individuals)

Or p = (2 $\times$ 400) + (600) / (2 $\times$ 1200)

Or p = 1400/2400 = 0.583 or 0.58

Frquency of F2 (i.e q) allele can be calculated in a similar way as above or from the value of p as:

According to Hardy Weinberg , p + q = 1

Or q = 1 - p = 1 - 0.58 = 0.42

Thus frequency of F1 allele is 0.58 and that of F2 allele is 0.42.

B.

The genotypic frequencies can be calculated as:

f(F1F1) = p² = 0.58 $\times$ 0.58 = 0.3364 or 0.34

f(F2F2) = q² = 0.42 $\times$ 0. 42 = 0.1764 or 0.18

f(F1F2) = 2pq = 2 $\times$ 0.58 $\times$ 0.42 = 0.4872 = 0.49