In a Hardy-Weinberg population with two alleles A1 and A2 that are in equilibrium, the frequency of allele A2 is 0.2. What is the frequency of individuals that are heterozygous for this allele?
A) 0.02
B) 0.04
C) 0.16
D) 0.32

Question

Quizplus · Accepted Answer

The frequency of allele A2 is given as 0.2, so the frequency of allele A1 is 1 - 0.2 = 0.8. The frequency of heterozygous individuals (A1A2) can be calculated using the formula 2pq, where p is the frequency of one allele (A1) and q is the frequency of the other allele (A2). Therefore, 2 * 0.8 * 0.2 = 0.32.

In a Hardy-Weinberg Population with Two Alleles A1 and A2