Hardy Weinberg Practice Problems With Answer Key Explained
Hardy-Weinberg practice problems with an answer key work like this: you calculate allele frequencies (p and q) from the given genotype data, then use two equations — p + q = 1 for alleles and p² + 2pq + q² = 1 for genotypes — to find whatever the question asks, and you verify every result by checking that the frequencies sum to 1. The Hardy-Weinberg principle states that in a large, randomly mating population with no mutation, migration, or selection, allele and genotype frequencies stay constant across generations.
Core Hardy Weinberg Equations
The principle rests on two variables. The dominant allele frequency is p, the recessive allele frequency is q, and because every allele is one or the other, p + q = 1. Each term in the second equation maps to a genotype: p² is the frequency of the homozygous dominant genotype (AA), 2pq is the heterozygous genotype (Aa), and q² is the homozygous recessive genotype (aa). Together they satisfy p² + 2pq + q² = 1.
Calculations run in one of two directions:
| Given | Find p and q by | Then get genotypes from |
|---|---|---|
| Homozygous recessive frequency | q = √q², p = 1 − q | p² and 2pq |
| Heterozygote frequency | 2pq value, solving with p + q = 1 | q² = 1 − p² − 2pq |
A quick note on the math: the square root and squaring steps assume the population is in equilibrium, which is why verifying with both equations matters. Nature Education Scitable’s population genetics overview covers the assumptions in more depth.
A one-line worked example
If 16% of a population shows the recessive phenotype, q² = 0.16, so q = 0.4 and p = 0.6. Then p² = 0.36, 2pq = 2(0.6)(0.4) = 0.48, and the three genotypes sum to 0.36 + 0.48 + 0.16 = 1.
Practice Problems With Worked Answers
Problem 1 — Finding allele frequencies
In a population of 1,000 individuals, 360 are AA, 480 are Aa, and 160 are aa. What are p and q?
Answer: Count alleles. A alleles = 2(360) + 480 = 1,200 out of 2,000 total, so p = 1,200/2,000 = 0.6. Then q = 1 − 0.6 = 0.4. Check: a alleles = 2(160) + 480 = 800, and 800/2,000 = 0.4.
Problem 2 — From phenotype to genotype
A recessive disorder affects 1 in 10,000 people. Assuming Hardy-Weinberg equilibrium, what fraction are carriers?
Answer: q² = 0.0001, so q = 0.01 and p = 0.99. Carriers are 2pq = 2(0.99)(0.01) = 0.0198, or about 1 in 50 people. This is a classic demo that rare recessive alleles persist mostly in unaffected carriers. Khan Academy has video walkthroughs of this exact setup.
Problem 3 — Working backward from heterozygotes
If 42% of a population is heterozygous, what is the frequency of the recessive allele?
Answer: 2pq = 0.42, and q = 1 − p, so 2(1 − q)q = 0.42 → 2q − 2q² = 0.42 → q² − q + 0.21 = 0. Solving the quadratic gives q = 0.7 or q = 0.3. Both solutions are mathematically valid; the problem needs phenotype data to pick one. This is a genuine trap in exam questions — flag it rather than guessing.
Problem 4 — Checking for equilibrium
A population has genotype frequencies AA = 0.50, Aa = 0.20, aa = 0.30. Is it in Hardy-Weinberg equilibrium?
Answer: Allele frequencies: p = 0.50 + ½(0.20) = 0.60, q = 0.30 + ½(0.20) = 0.40. Expected genotypes under equilibrium: p² = 0.36, 2pq = 0.48, q² = 0.16. The observed values (0.50, 0.20, 0.30) differ sharply from expected, so the population is not in equilibrium — something like selection, non-random mating, or migration is at work.
Problem 5 — Applying it to a real trait
About 1 in 2,500 newborns has phenylketonuria (PKU), an autosomal recessive metabolic disorder. Estimate the carrier frequency.
Answer: q² = 1/2,500 = 0.0004, so q = 0.02 and p = 0.98. Carriers = 2pq = 2(0.98)(0.02) = 0.0392, roughly 1 in 25. That carrier estimate is a standard textbook figure, but clinical interpretation of any individual’s genetic risk belongs with a genetic counselor or physician.
Common Mistakes And Traps
The single most common error is plugging a percentage into an equation as if it were a proportion. If 9% of a population is recessive, write q² = 0.09, not 9. Squaring 9 gives 81, and the whole problem cascades wrong from there.
The second trap is confusing allele frequencies with genotype frequencies. The number 0.4 can be p, a genotype proportion, or something else entirely depending on context — always label whether a value describes alleles (p, q) or individuals (p², 2pq, q²).
A third frequent mistake is assuming equilibrium when nothing supports it. Hardy-Weinberg is a null model. If a question gives observed genotype counts that already violate p² + 2pq + q² = 1, you should report the deviation rather than force the numbers to fit. Related errors include forgetting to halve the heterozygote count when tallying alleles from genotypes, and taking the square root of p² instead of q² when only the recessive phenotype is given.
Finally, watch for multiply-valid quadratics like Problem 3, and for dominance questions where the dominant phenotype lumps AA and Aa together — you cannot split them from phenotype data alone without extra information.
Answer Key Verification Tips
Self-checking is fast if you use both equations as guardrails.
- p + q = 1. If your allele frequencies don’t sum to exactly 1, recheck the arithmetic before going further.
- p² + 2pq + q² = 1. Every genotype-answer key should sum to 1 (allowing for rounding). If it gives 0.94 or 1.07, you’ve made an error.
- Check the heterozygote maximum. 2pq can never exceed 0.5, since it peaks at p = q = 0.5. An answer with 2pq = 0.8 is impossible.
- Sanity-check magnitudes. If a rare recessive phenotype is 1 in 10,000, carriers should land near 1 in 50 — if your key says 1 in 5, something is off.
- Match phenotype to genotype. Recessive phenotype frequency must equal q² exactly; dominant phenotype frequency equals p² + 2pq, not p² alone.
Run these five checks on every problem and errors become obvious within seconds. For more structured walkthroughs, Khan Academy’s genetics unit and Nature Education Scitable both offer step-by-step derivations you can compare against your own work.
Key Takeaways
- The two governing equations are p + q = 1 (alleles) and p² + 2pq + q² = 1 (genotypes), where p² = AA, 2pq = Aa, and q² = aa.
- When only the recessive phenotype is known, set q² = that frequency, take q = √q², then p = 1 − q.
- Always convert percentages to proportions (9% → 0.09) before calculating.
- Answer keys should always sum to 1 and produce a heterozygote frequency no greater than 0.5.
- Hardy-Weinberg assumes no mutation, migration, selection, or non-random mating — if observed genotypes don’t match expected, report the deviation instead of forcing equilibrium.
