If a chi-square test for Hardy-Weinberg deviation yields p < 0.05, what does this imply about the population?

• A. The population deviates from Hardy-Weinberg equilibrium
• B. The population is in Hardy-Weinberg equilibrium
• C. There is no information about equilibrium
• D. Mutation is occurring at high rate

Answer: (C)

In Hardy-Weinberg, genotype frequencies are predicted from allele frequencies if certain conditions hold. A chi-square test compares what you actually observe to what HW predicts. If the p-value is below 0.05, the observed frequencies differ enough that this difference is unlikely to be due to random sampling; you reject the idea that the population is in Hardy-Weinberg equilibrium. So the population deviates from HW expectations. The result doesn’t pinpoint a specific cause—nonrandom mating, selection, migration, drift, mutation, or sampling error could all contribute. It also doesn’t support a claim that mutation is occurring at a high rate; that would be one possible cause among others and isn’t directly concluded from the p-value alone.