Answer key: Hardy-Weinberg Activity
For teachers. This page is hidden from search engines and not linked from student pages, but it is not password-protected, so please don’t share the link with students. Student activity: Hardy-Weinberg Activity.
Goal: Calculate allele and genotype frequencies with p + q = 1 and p² + 2pq + q² = 1, estimate how common carriers are, and use a chi-square test to decide whether a population is in equilibrium.
Completed results table
| Population | p | q | 2pq (carriers) | χ² (counts only) |
|---|---|---|---|---|
| q² = 0.09 | 0.7 | 0.3 | 0.42 (42%) | — |
| Cystic fibrosis (q² = 1 in 2,500) | 0.98 | 0.02 | 0.0392 (3.92%) | — |
| Counts 49 · 42 · 9 | 0.7 | 0.3 | 0.42 (42%) | 0 |
| Counts 30 · 60 · 10 | 0.6 | 0.4 | 0.48 (48%) | 6.25 |
Answers
In a population, 9% of individuals show a recessive trait. Show how to find q, p and the frequency of carriers.
Only aa individuals show the trait, so q² = 0.09 and q = √0.09 = 0.3. Then p = 1 − 0.3 = 0.7, and carriers are 2pq = 2 × 0.7 × 0.3 = 0.42, or 42% of the population.
Cystic fibrosis affects about 1 in 2,500 newborns of European ancestry. What fraction of people are carriers? Give it as “about 1 in …”.
q² = 1 ÷ 2,500 = 0.0004, so q = 0.02 and p = 0.98. Carriers: 2pq = 2 × 0.98 × 0.02 = 0.0392, about 3.92% or roughly 1 in 26 people.
Why are there so many more carriers of cystic fibrosis than people who have it?
Most copies of a rare recessive allele are hidden in heterozygotes. With q = 0.02, carriers (2pq = 0.0392) outnumber affected people (q² = 0.0004) by 98 to 1, so selection against the affected phenotype removes the allele only very slowly.
For the counts 30 AA, 60 Aa and 10 aa, calculate p and the counts you would expect if the population were in equilibrium.
p = (2 × 30 + 60) ÷ (2 × 100) = 120 ÷ 200 = 0.6, so q = 0.4. Expected counts: AA = 0.36 × 100 = 36, Aa = 0.48 × 100 = 48 and aa = 0.16 × 100 = 16.
Use χ² to decide whether the 30 · 60 · 10 population is in Hardy-Weinberg equilibrium. What might cause the excess of heterozygotes?
χ² = (30 − 36)²/36 + (60 − 48)²/48 + (10 − 16)²/16 = 1 + 3 + 2.25 = 6.25. With df = 1 the critical value is 3.841, and 6.25 is larger, so the population is not in equilibrium. Selection favoring heterozygotes (heterozygote advantage), non-random mating or migration could produce extra Aa individuals.
List the five conditions for Hardy-Weinberg equilibrium. Which one does heterozygote advantage break?
A very large population, random mating, no mutation, no migration (gene flow) and no natural selection. Heterozygote advantage is a form of natural selection, so it breaks the “no selection” condition.