Final answer:
The population is not in Hardy-Weinberg equilibrium because the observed genotype frequencies in the next generation (HH = 14, Hh = 63, and hh = 74) do not match the expected frequencies based on allele frequencies of H = 0.2 and h = 0.8 in a population of 150 (HH = 6, Hh = 48, hh = 96). The deviation from expected frequencies indicates that the conditions for equilibrium have not been met, signaling evolution within the population.
Step-by-step explanation:
To determine if a population is in Hardy-Weinberg equilibrium, we must analyze whether the observed genotype frequencies align with the expected frequencies. The expected frequency of HH based on the given allele frequencies (H=0.2, h=0.8) would be p², or 0.2² = 0.04, while Hh would be 2pq, or 2(0.2)(0.8) = 0.32, and hh would be q², or 0.8² = 0.64. Multiplying these by the total population size (150) gives us the expected numbers: HH = 6, Hh = 48, hh = 96.
Comparing these to the observed numbers in the next generation: HH = 14, Hh = 63, and hh = 74, we can see that the observed frequencies (14/150, 63/150, 74/150) don't match the expected frequencies (6/150, 48/150, 96/150). Therefore, the population is not in Hardy-Weinberg equilibrium.
Hardy-Weinberg equilibrium reflects a state of genetic stability where allele frequencies remain constant from generation to generation under five conditions: no mutations, no gene flow, random mating, no genetic drift, and no selection. Since the question presents changes in genotype frequencies from one generation to the next, we can infer that one or more of these conditions have not been met, resulting in evolution within the population.