Questions: Population Genetics and Hardy-Weinberg Equilibrium

Question 1

A population is in Hardy-Weinberg equilibrium for a gene with alleles A (frequency p = 0.6) and a (frequency q = 0.4). What is the expected frequency of heterozygotes (Aa)?

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

Answer: (B)

The frequency of heterozygotes is 2pq = 2(0.6)(0.4) = 0.48. A common error is computing pq = 0.24 and forgetting the factor of 2, which accounts for the two ways to be heterozygous (receiving A from mother and a from father, or vice versa).

Question 2

The equation p + q = 1 describes the frequencies of genotypes in a Hardy-Weinberg population.

• T. True
• F. False

Answer: False

p + q = 1 describes *allele* frequencies — p is the frequency of one allele, q of the other, and together they must sum to 1. *Genotype* frequencies are described by p² + 2pq + q² = 1, which is the expanded form of (p + q)². Confusing these two equations is one of the most common errors in population genetics problems.

Question 3

A population is found to have significantly more homozygous individuals than Hardy-Weinberg predicts. Name one evolutionary mechanism that could cause this deviation and explain why it would produce an excess of homozygotes.

Model answer: Non-random mating (inbreeding) causes individuals to preferentially mate with relatives, increasing the probability that both copies of a gene descend from the same ancestor and are therefore identical. This increases homozygosity beyond H-W expectations without changing allele frequencies. Alternatively, strong disruptive selection against heterozygotes would reduce 2pq below predicted levels.

This tests whether students understand H-W as a null model: deviations reveal which assumptions are violated. An excess of homozygotes specifically points to mechanisms that reduce heterozygosity — inbreeding, assortative mating, or selection against heterozygotes — rather than mechanisms like drift or mutation that shift allele frequencies.