In a population of Mendel's garden peas, the frequency of the dominant A (yellow flower) allele is 80%. Let p represent the frequency of the A allele and q represent the frequency of the a allele. Assuming that the population is in Hardy-Weinberg equilibrium, what are the genotype frequencies?
A)64% AA, 32% Aa, 4% aa
B)50% AA, 25% Aa, 25% aa
C)16% AA, 40% Aa, 44% aa
D)75% AA, 15% Aa, 10% aa
E)80% AA, 10% Aa, 10% aa

Question

Quizplus · Accepted Answer

In Hardy-Weinberg equilibrium, the genotype frequencies can be calculated using the equations p^2 + 2pq + q^2 = 1, where p^2 represents the frequency of the homozygous dominant genotype (AA), 2pq represents the frequency of the heterozygous genotype (Aa), and q^2 represents the frequency of the homozygous recessive genotype (aa). Given p = 0.8 (or 80%), q = 1 - p = 0.2 (or 20%). Therefore, p^2 = 0.64 or 64% (AA), 2pq = 2(0.8)(0.2) = 0.32 or 32% (Aa), and q^2 = 0.04 or 4% (aa).

In a Population of Mendel's Garden Peas, the Frequency of the Dominant