Final answer:
The frequency of the "aa" genotype is 0.36. Using the Hardy-Weinberg principle, the frequency of allele 'a' is 0.6, and the allele 'A' is 0.4, allowing us to calculate the genotype frequencies of AA (0.16), Aa (0.48), and aa (0.36) in a population.
Step-by-step explanation:
The frequency of the "aa" genotype in a population where the homozygous recessive genotype (aa) is 36% means that 36 out of 100 individuals are aa.
To express this as a fraction, the frequency is 0.36. According to the Hardy-Weinberg principle,
for a gene with two alleles A and a, the frequency of the allele 'a' (q) can be calculated by taking the square root of the frequency of the homozygous recessive genotype, which in this case is √0.36 = 0.6. The frequency of the allele 'A' (p) would then be 1 - q, which equals 0.4.
The genotype frequencies for a population in Hardy-Weinberg equilibrium (p² + 2pq + q² = 1) are as follows: the frequency of AA (p²) is 0.16, the frequency of Aa (2pq) is 0.48, and the frequency of aa (q²) remains at 0.36.