Consider a population of 20 aa, 32 Aa, and 48 AA individuals. Selection acts against the lethal recessive homozygote (aa) before adulthood. After selection, survivors mate randomly. What are the genotypic frequencies of the next generation?
A) 25% aa, 50% Aa, and 25% AA
B) all AA
C) 4% aa, 16% Aa, and 64% AA
D) 4% aa, 64% Aa, and 16% AA

Question

Quizplus · Accepted Answer

Selection is acting against aa individuals, so all of the aa individuals will be eliminated. 20 aa individuals are eliminated, leaving 32 Aa individuals and 48 AA individuals. The surviving Aa individuals can produce both A and a gametes with a frequency of 0.5 for each, while the AA individuals can only produce A gametes. Therefore, the A frequency in the next generation will be (32 x 0.5 + 48) / (32 + 48) = 0.67, and the a frequency will be 0.33. Using the Hardy-Weinberg equation, we can calculate the genotypic frequencies in the next generation:

AA = p^2 = (0.67)^2 = 0.4489, or 44.89% 
Aa = 2pq = 2 x 0.67 x 0.33 = 0.4422, or 44.22% 
aa = q^2 = (0.33)^2 = 0.1089, or 10.89%

Therefore, the genotypic frequencies of the next generation are 4% aa, 16% Aa, and 64% AA.

Consider a Population of 20 Aa, 32 Aa, and 48