AP Biology: Hardy-Weinberg Applications Practice Questions
Practice Hardy-Weinberg applications with original AP Biology multiple-choice questions. Each item connects allele-frequency math, model assumptions, and biological interpretation with clear explanations.
Free original practice questions for study purposes. Open Exam Prep is an independent study resource and is not affiliated with, endorsed by, or sponsored by the makers of AP Exams.
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1. In a population at Hardy-Weinberg equilibrium, the frequency of the recessive phenotype is 0.04. What is the frequency of the recessive allele?
Explanation. The recessive phenotype frequency is q^2. If q^2 = 0.04, then q = 0.20.
2. If p = 0.65 for allele A at a two-allele locus, what is q for allele a?
Explanation. For a two-allele locus, p + q = 1. Therefore q = 1 - 0.65 = 0.35.
3. A population has p = 0.6 and q = 0.4. Under Hardy-Weinberg equilibrium, what genotype frequency is expected for heterozygotes?
Explanation. The expected heterozygote frequency is 2pq, so 2(0.6)(0.4) = 0.48.
4. In a sample of 200 diploid organisms, genotype counts are AA = 98, Aa = 84, and aa = 18. What is the frequency of allele A?
Explanation. There are 400 allele copies. A copies = 2(98) + 84 = 280, so p = 280/400 = 0.70.
5. A population's observed genotype frequencies are close to p^2, 2pq, and q^2 predictions for several generations. Which interpretation is best supported?
Explanation. Close agreement with expected frequencies suggests the modeled locus may be near Hardy-Weinberg equilibrium, but it does not prove all assumptions for the whole genome or future generations.
6. A student observes fewer heterozygotes than expected from 2pq. Which process could explain this pattern?
Explanation. Inbreeding or other nonrandom mating can reduce heterozygosity relative to Hardy-Weinberg expectations.
7. Which statement correctly describes Hardy-Weinberg equilibrium?
Explanation. Hardy-Weinberg equilibrium provides expected genotype frequencies when no evolutionary forces affect a two-allele locus under the model assumptions.
8. The recessive phenotype frequency for a trait is 16% in a population at equilibrium. What is the expected frequency of heterozygous carriers?
Explanation. If q^2 = 0.16, then q = 0.4 and p = 0.6. The heterozygote frequency is 2pq = 2(0.6)(0.4) = 0.48.
9. A large population is randomly mating, but individuals with genotype aa have lower survival before reproduction. Which Hardy-Weinberg assumption is violated?
Explanation. Different survival among genotypes means selection is acting, violating the no-selection assumption.
10. Why is a very small sample risky when estimating Hardy-Weinberg genotype frequencies?
Explanation. Small samples are more affected by random sampling error, which can distort estimated allele and genotype frequencies.
11. A population has p = 0.2 and q = 0.8. Under Hardy-Weinberg equilibrium, which genotype frequency is expected to be highest?
Explanation. With q = 0.8, aa frequency is q^2 = 0.64. AA is 0.04 and Aa is 0.32, so aa is highest.
12. Researchers compare observed genotype counts with Hardy-Weinberg expected counts and find a large, repeatable difference. What should they do next?
Explanation. A repeated mismatch suggests at least one model assumption or sampling condition may not hold, so researchers should evaluate biological and methodological explanations.
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FAQ
What Hardy-Weinberg formulas should I know for AP Biology?
Know p + q = 1 for allele frequencies and p^2 + 2pq + q^2 = 1 for expected genotype frequencies at a two-allele locus.
How do I start a Hardy-Weinberg calculation?
Identify whether the problem gives allele frequencies, genotype counts, or a recessive phenotype frequency. Then solve for p and q before calculating expected genotype frequencies.
What does it mean if observed and expected genotype frequencies differ?
A difference can indicate selection, drift, gene flow, mutation, nonrandom mating, or sampling bias. The model points to a question; it does not identify the cause by itself.