If a population with two alleles is at Hardy-Weinberg equilibrium and the frequency of one allele is 0.7, what is the heterozygote frequency?
A) 0.7
B) 0.42
C) 0.3
D) 0.21
E) Insufficient information is given to answer the question.

Question

Quizplus · Accepted Answer

At Hardy-Weinberg equilibrium, the frequency of the homozygous dominant genotype (AA) is p^2, where p is the frequency of one allele. The frequency of the homozygous recessive genotype (aa) is q^2, where q is the frequency of the other allele. The frequency of the heterozygous genotype (Aa) is 2pq.

Since the frequency of one allele is 0.7, let's call that allele A and the other allele a. Therefore, p = 0.7 and q = 0.3.

The frequency of the homozygous dominant genotype (AA) is p^2 = 0.49, and the frequency of the homozygous recessive genotype (aa) is q^2 = 0.09.

The frequency of the heterozygous genotype (Aa) is 2pq = 2(0.7)(0.3) = 0.42.

Therefore, the heterozygote frequency is 0.42 or choice B.

If a Population with Two Alleles Is at Hardy-Weinberg Equilibrium