Question

A population of 100 individuals has a frequency of allele 'A' of 0.3 and a frequency of allele 'a' of 0.7. The frequency of the heterozygous genotype (Aa) is 0.49. Is this population in Hardy-Weinberg equilibrium? Justify.

(a) Yes; because the frequencies match expectations.
(b) No; because the frequencies do not match expectations.
(c) Not enough information to determine.
(d) Hardy-Weinberg equilibrium does not apply.

Answer 1

The population is not in Hardy-Weinberg equilibrium as the observed heterozygous genotype frequency (0.49) does not match the expected frequency calculated using Hardy-Weinberg equation (0.42).

Step-by-step explanation:

The population is not in Hardy-Weinberg equilibrium. The frequencies provided for allele 'A' (0.3) and allele 'a' (0.7) indicate that in a population in Hardy-Weinberg equilibrium, the frequency of the heterozygous genotype (Aa) should be 2pq, which is 2(0.3)(0.7) = 0.42, not 0.49 as given. Therefore, the discrepancy between the expected frequency (0.42) and the observed frequency (0.49) tells us that the population does not meet Hardy-Weinberg expectations. For a population to be in Hardy-Weinberg equilibrium, it must meet five conditions: no mutations, random mating, no gene flow, large population size (no genetic drift), and no selection. When any of these conditions are not met, it may lead to changes in allele frequencies, indicating that evolution is occurring. In this case, the incorrect frequency of the heterozygotes suggests that one or more of the Hardy-Weinberg conditions are not being met.