Questions: Hardy-Weinberg Equilibrium

Question 1

A population has allele frequencies p = 0.6 (dominant A) and q = 0.4 (recessive a). What is the expected frequency of heterozygotes (Aa) under Hardy-Weinberg equilibrium?

• A. 0.36
• B. 0.16
• C. 0.48
• D. 0.24

Answer: (C)

The frequency of heterozygotes is 2pq = 2(0.6)(0.4) = 0.48. A common error is computing pq = 0.24 and forgetting to multiply by 2, which accounts for both Aa and aA configurations.

Question 2

If a population is in Hardy-Weinberg equilibrium, it means that no evolution is occurring and few can occur in the future.

• T. True
• F. False

Answer: False

HWE describes a snapshot in which allele frequencies are currently stable because the five conditions are met. It makes no claim about the future — as soon as any condition is violated (e.g., migration brings new alleles), frequencies will shift and evolution resumes. HWE is a null model, not a permanent state.

Question 3

A population shows observed genotype frequencies that deviate significantly from Hardy-Weinberg expectations. What does this tell you?

Model answer: One or more evolutionary forces are acting on the population — such as natural selection, genetic drift, non-random mating, gene flow, or mutation.

HWE predicts stable frequencies only when all five conditions hold. A significant deviation is evidence that at least one condition is violated. Identifying which condition helps diagnose the evolutionary mechanism at work — for example, excess heterozygotes suggest heterozygote advantage (balancing selection), while a deficit suggests inbreeding.