Final answer:
The population's genotype frequencies (A1A1 = 16, A1A2 = 67, A2A2 = 17) show a significant deviation from the Hardy-Weinberg equilibrium as evidenced by a chi-square value of 11.58, which is higher than the critical value. Thus, the population is not in equilibrium, suggesting that evolutionary forces or non-random mating may be at play.
Step-by-step explanation:
The student is asking about a population genotypes' deviation from the Hardy-Weinberg equilibrium, which is a fundamental principle in Biology that describes the expected frequencies of genotypes in a population that is not evolving. Given that p = q = 0.5 for a locus, we would expect the genotypic frequencies to follow the Hardy-Weinberg equation: p² + 2pq + q² = 1. For p = 0.5, this would predict that the genotypic frequencies for A1A1, A1A2, and A2A2 would be 0.25, 0.50, and 0.25, respectively. However, the observed genotype frequencies in this population (A1A1 = 16, A1A2 = 67, A2A2 = 17) deviate from these expectations.
A chi-square test has been applied to determine if the observed frequencies significantly depart from the expected frequencies under Hardy-Weinberg conditions. The calculated chi-square value of 11.58 is greater than the critical value of 3.84, with a p-value less than 0.05. This indicates that the null hypothesis, that the population is in Hardy-Weinberg equilibrium, should be rejected. Consequently, the population appears to be evolving or the data reflection condition(s) that are not met for Hardy-Weinberg equilibrium.