The genotype frequencies in a population with no evolution occurring are AA = 0.25,Aa = 0.5,aa = 0.25.By removing all AA individuals before they reproduce and assuming the population will remain under Hardy-Weinberg conditions,what will be the frequency of aa individuals in the next generation?
A)0.25
B)0.33
C)0.5
D)0.67
E)0.75

Question

Quizplus · Accepted Answer

First, we need to calculate the frequency of the Aa heterozygotes in the population. Since Aa individuals have a frequency of 0.5, and AA individuals are being removed, we can simply divide 0.5 by the frequency of the remaining alleles (a):

Aa frequency = 0.5 / (2*0.25 + 0.5) = 0.5 / 1 = 0.5

Next, we can use the equation p + q = 1 to solve for the frequency of the aa homozygotes:

q = frequency of aa = 1 - p

Since we know the frequency of Aa is 0.5, we can use that to solve for the frequency of the A allele:

p = frequency of A = 1 - q

Substituting in the values we know:

0.5 = 1 - q

q = 0.5

So the frequency of aa individuals in the next generation would be 0.5 or 50%. Therefore, the correct answer is D (0.67 is incorrect since it would be the frequency of Aa after AA individuals are removed, not the frequency of aa).

The Genotype Frequencies in a Population with No Evolution Occurring