Final answer:
The frequency of the AA genotype within a population following Hardy-Weinberg equilibrium, where p equals the frequency of allele A, is calculated as p². When p is 0.4, the frequency of the AA genotype is p², which is 0.16 or 16%.
Step-by-step explanation:
The concept in question relates to Hardy-Weinberg equilibrium, which is a principle of population genetics that describes the relationship between allele and genotype frequencies within a sexually reproducing population. According to this principle, allele frequencies (p and q) should always add up to 1 (p + q = 1). Consequently, genotype frequencies (p² for homozygous dominant, 2pq for heterozygous, and q² for homozygous recessive) should add up to 1 as well (p² + 2pq + q² = 1).
In the scenario provided, if p = 0.4 (frequency of allele A), we can calculate q as 1 - p, which equals 0.6 (frequency of allele a). Then, to find the frequency of the AA genotype, we square the frequency of the allele A (p²), resulting in 0.4² = 0.16. This implies that the frequency of the AA genotype is 16% in the population.