Final answer:
Using the Hardy-Weinberg equation, the frequency of the AA genotype (represented as p²) when p = 0.4 is calculated to be 0.16. Since the observed frequency for SS, which might correspond to AA, is 0.09, there is a discrepancy that indicates the population may not be in equilibrium. This suggests evolutionary forces might be at play, affecting the population's allele frequencies.
Step-by-step explanation:
To determine if this population is in Hardy-Weinberg equilibrium, we use the Hardy-Weinberg principle which states that the frequencies of alleles and genotypes in a population's gene pool will remain constant over generations unless acted upon by evolutionary forces. The equilibrium condition assumes no mutations, random mating, no gene flow (immigration or emigration), no genetic drift, and no selection. The mathematical representation of the principle is given by the equation p² + 2pq + q² = 1, where p is the frequency of the dominant allele, and q is the frequency of the recessive allele. To find the frequency of the genotype AA (p²), we need to square the frequency of the dominant allele p, which is provided as 0.4. Doing the calculation, p² = (0.4)² = 0.16.
Since the observed frequency of the AA genotype is not mentioned in the question, we cannot directly compare it to the calculated frequency of the AA genotype (p²) to determine equilibrium. However, assuming the frequencies provided for SS and ss correspond to AA and aa genotypes respectively, we see a discrepancy between the observed value (SS = 0.09) and the expected value under Hardy-Weinberg equilibrium (AA = 0.16). This suggests the population may not be in equilibrium. Differences in observed and expected frequencies can imply that evolutionary forces such as selection, mutation, or genetic drift may be acting on the population.