Consider island C in 1980 with 546 adults. If the frequency of the black allele (f(y)) is 0.12, assume Hardy-Weinberg conditions and estimate the frequency of heterozygotes in this island population?

Question

Options

(A) 0.0<f(Yy)<0.2

(B) 0.2< f(Yy) <0.4

(C) 0.4< f(Yy) <0.6

(D) 0.6< f(Yy) <0.8

(E) 0.8< f(Yy) <1.0

I think the answer is B, but I'm not sure.

Answer 1

You are correct! Here's why:

The Hardy-Weinberg equation is p² + 2qp + q² = 1, where p is the frequency of one allele, and q is the frequency of another allele. Each animal will have two alleles

The first term, p², is the frequency of individuals who are homozygous for p (i.e. both of their alleles are p). The frequency of p is 0.12, so the frequency of pp is (0.12)².

The last term, q², is the frequency of individuals who are homozygous for q. If the frequency of p is 0.12, then the frequency of q must be 1 - 0.12 = 0.88. So, the frequency of q is 0.88, and the frequency of qq is (0.88)².

The middle term, 2qp, is the frequency of individuals who are heterozygous for p and q (i.e. one of their alleles is p and the other is q). There are two ways to be heterozygous: pq and qp, so the frequency of heterozygotes is pq + qp. Since pq = qp, we just write 2qp.

So, we can plug in our numbers:

2qp = 2 *0.12*0.88 = 0.2112, which is option B