A population of mice has black or brown hair. The gene for black hair color (B) is dominant over brown (b). If 16% of the mice are homozygous black and 24% are heterozygous black, what would be the frequency of each allele?
(p + q = 1, p2 + 2pq + q2 = 1)

is it (A. p=0.65, q=0.35 (B. p=0.25, q=0.75 (C. p=0.4, q=0.6

Question

A population of mice has black or brown hair. The gene for black hair color (B) is dominant over brown (b). If 16% of the mice are homozygous black and 24% are heterozygous black, what would be the frequency of each allele?
(p + q = 1,  p2 + 2pq + q2 = 1)

is it (A. p=0.65, q=0.35 (B. p=0.25, q=0.75 (C. p=0.4, q=0.6

matt101 · Answer

In this question you're assuming Hardy-Weinberg equilibrium. In other words, if B and b are the two alleles, B + b = 1 and B^2 + 2Bb + b^2 = 1.

If 16% of mice are homozygous black, B^2 = 0.16, meaning B = 0.4 and b = 1 - 0.4 = 0.6. Answer is C. Notice how you don't even need to know that 24% of the mice are heterozygous.

anonymous · Answer

The answer is C