Question

In a population where the A allele is at a frequency of 0.6 and the a allele is at a frequency of 0.4, what frequencies of AA, Aa, and aa genotypes, respectively, are expected under Hardy Weinberg Equilibrium?

A) Frequency of AA genotype = 0.36, Aa genotype = 0.48, aa genotype = 0.16
B) Frequency of AA genotype = 0.64, Aa genotype = 0.32, aa genotype = 0.04
C) Frequency of AA genotype = 0.48, Aa genotype = 0.36, aa genotype = 0.16
D) Frequency of AA genotype = 0.64, Aa genotype = 0.16, aa genotype = 0.20

Answer 1

Under Hardy-Weinberg equilibrium with allele frequencies of p=0.6 and q=0.4, the expected genotype frequencies are 0.36 for AA, 0.48 for Aa, and 0.16 for aa.

Step-by-step explanation:

In a population under Hardy-Weinberg equilibrium, the allele frequencies are represented by p for the dominant allele (A) and q for the recessive allele (a). The expected genotype frequencies can be calculated using the equations p² for the homozygous dominant genotype (AA), 2pq for the heterozygous genotype (Aa), and q² for the homozygous recessive genotype (aa), where p + q = 1. In the given scenario with allele frequencies of p = 0.6 and q = 0.4, the expected genotype frequencies would be p² = (0.6)² = 0.36 for AA, 2pq = 2(0.6)(0.4) = 0.48 for Aa, and q² = (0.4)² = 0.16 for aa.