Final answer:
The frequency of individuals with the Aa genotype in a population where allele A has a frequency of 90% and allele a has a frequency of 10% is calculated using 2pq, resulting in a frequency of 0.18.
Step-by-step explanation:
To calculate the frequency of individuals with the genotype Aa, we use the Hardy-Weinberg principle, which describes the genetic equilibrium within a population. The principle is represented by the equation p² + 2pq + q² = 1, where p is the frequency of the dominant allele (A), and q is the frequency of the recessive allele (a). In the given scenario, p = 0.9 and q = 0.1. Therefore, the frequency of the Aa genotype, 2pq, can be calculated as 2 × 0.9 × 0.1 = 0.18. This corresponds to the genotype frequency of heterozygous individuals within the population.