Final answer:
The frequency of the aa phenotype in a Hardy-Weinberg equilibrium can be determined by subtracting the combined frequency of AA and Aa from 100%. Example calculation for AA genotype with a known p value also demonstrates how genotype frequencies can be predicted.
Step-by-step explanation:
According to the Hardy-Weinberg equilibrium, the frequency of phenotypes for genotypes AA and Aa in a population that is in equilibrium can be used to determine the frequency of the aa phenotype. The Hardy-Weinberg equation is p² + 2pq + q² = 1, wherein p is the frequency of the dominant allele A, and q is the frequency of the recessive allele a. If the combined frequency of AA and Aa is 75%, this implies that the frequency of the aa phenotype must make the total add up to 100%. Thus, it is simply calculated by subtracting the combined frequency of AA and Aa from 100, giving us the remainder, which is the frequency of the aa phenotype.
To work out an example with actual numbers, if p = 0.4 for the AA genotype as given, then the frequency of the AA genotype (p²) would be 0.16 (which is 0.4× 0.4). Subsequently, using the values of p and q, the frequency of the aa genotype (q²) can be found by subtracting p from 1 to get q, and then squaring q. The equilibrium implies that the allele frequencies in the population are stable, which means that the genotype frequencies also become predictable.