Chris E. answered • 12/20/21

Teaching Assistant - Molecular/Developmental Introductory Genetics Lab

You are correct! Here's why:

The Hardy-Weinberg equation is p^{2} + 2qp + q^{2} = 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, p^{2}, 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, q^{2}, 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