In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.2. What is the percentage of the population that is heterozygous for this allele?
A) 0.2
B) 2.0
C) 4.0
D) 16.0
E) 32.0

Question

Quizplus · Accepted Answer

In a Hardy-Weinberg population, the frequency of two alleles A and a can be represented as p and q, respectively. Given that the frequency of allele a (q) is 0.2, the frequency of allele A (p) would be 1 - 0.2 = 0.8. The percentage of the population that is heterozygous (Aa) can be calculated using the formula 2pq. Therefore, 2 * 0.8 * 0.2 = 0.32 or 32%.

In a Hardy-Weinberg Population with Two Alleles, a and A