In a population of 100 individuals, we have 25 people that are homozygous AA, 50 people that are heterozygous Aa, and 25 people that are homozygous aa. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of the aa genotype in this population? Write your answer as a decimal.

Question

BlackTom AI · Accepted Answer

To determine the frequency of the aa genotype under Hardy-Weinberg equilibrium, we can use the genotype counts provided. 
First, note the total population is 100 individuals, with 25 AA, 50 Aa, and 25 aa. The frequency of aa is the number of aa individuals divided by the total population: 25/100 = 0.25. 
Alternatively, we can derive allele frequencies to confirm consistency. The number of A alleles is (2×25) from AA plus (1×50) from Aa = 50 + 50 = 100 A alleles. The number of a alleles is (0×25) from AA plus (1×50) from Aa plus (2×25) from aa = 0 + 50 + 50 = 100 a alleles. Since there are 200 total alleles (2 per individual), p = freq(A) = 100/200 = 0.5 and q = freq(a) = 100/200 = 0.5. Under Hardy-Weinberg, q^2 = (0.5)^2 = 0.25, which matches the observed aa frequency of 25/100 = 0.25. 
Thus, the frequency of the aa genotype in this population is 0.25.

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