Final answer:
The given genotypic frequencies do not satisfy the Hardy-Weinberg principle, and thus the population is not in equilibrium.
Step-by-step explanation:
The given genotypic frequencies of AA = 0.39, Aa = 0.56, and aa = 0.05 do not satisfy the Hardy-Weinberg principle and indicate that the population is not in Hardy-Weinberg equilibrium. The Hardy-Weinberg equilibrium describes a state in which allele frequencies in a population remain constant from generation to generation.
To determine if a population is in Hardy-Weinberg equilibrium, we can use the equation p^2 + 2pq + q^2 = 1, where p represents the frequency of the dominant allele (AA) and q represents the frequency of the recessive allele (aa). In this case, we have p^2 = 0.39 and q^2 = 0.05. By calculating the square root of both values, we find that p ≈ 0.625 and q ≈ 0.223. However, the observed frequency of the heterozygote genotype (Aa) is 0.56, which does not match the expected value of 2pq ≈ 0.278. This discrepancy indicates that the population is not in Hardy-Weinberg equilibrium.