Hardy‑Weinberg Equilibrium Practice Problems with Answers
The Hardy‑Weinberg equilibrium (HWE) is a cornerstone concept in population genetics that describes how allele and genotype frequencies remain constant from generation to generation in the absence of evolutionary forces. Mastering HWE through practice problems not only reinforces the underlying mathematics but also builds intuition for when real populations deviate from the ideal model. Below you will find a clear explanation of the principle, the essential equations, a series of graded practice problems with detailed solutions, and tips to avoid common pitfalls.
Understanding Hardy‑Weinberg Equilibrium
Before diving into the problems, it helps to recall the assumptions that define a population in Hardy‑Weinberg equilibrium:
- No mutation – alleles do not change from one form to another.
- No migration – there is no gene flow into or out of the population.
- Infinite population size – genetic drift is negligible.
- Random mating – individuals pair by chance, not according to genotype.
- No natural selection – all genotypes have equal fitness.
When these conditions hold, the relationship between allele frequencies (p and q) and genotype frequencies (AA, Aa, aa) is given by the Hardy‑Weinberg equation:
[ p^{2} + 2pq + q^{2} = 1 ]
where
- p = frequency of the dominant allele (A)
- q = frequency of the recessive allele (a)
- p² = expected frequency of homozygous dominant genotype (AA) - 2pq = expected frequency of heterozygous genotype (Aa)
- q² = expected frequency of homozygous recessive genotype (aa)
Because p + q = 1, knowing either allele frequency lets you calculate the other and all three genotype frequencies.
Key Equations to Remember
| Symbol | Meaning | Equation |
|---|---|---|
| p | Frequency of dominant allele | (p = 1 - q) |
| q | Frequency of recessive allele | (q = 1 - p) |
| p² | Frequency of AA genotype | (p^{2}) |
| 2pq | Frequency of Aa genotype | (2pq) |
| q² | Frequency of aa genotype | (q^{2}) |
| Observed recessive phenotype frequency | Often equals q² (if recessive trait is fully penetrant) | (q = \sqrt{\text{observed recessive phenotype}}) |
These formulas are the workhorses for every Hardy‑Weinberg practice problem.
Practice Problems
Problem 1 – Basic Allele Frequency Calculation
In a population of 1,000 individuals, 160 show the recessive phenotype for a trait controlled by a single gene with two alleles (A dominant, a recessive). Assuming the population is in Hardy‑Weinberg equilibrium, calculate:
- The frequency of the recessive allele (q).
- The frequency of the dominant allele (p).
- The expected number of individuals with each genotype (AA, Aa, aa).
Solution 1. The recessive phenotype corresponds to genotype aa, whose frequency is q².
[
q^{2} = \frac{160}{1000} = 0.16 \quad\Rightarrow\quad q = \sqrt{0.16} = 0.40
]
q = 0.40 (40 % of alleles are recessive).
-
Since p + q = 1:
[ p = 1 - q = 1 - 0.40 = 0.60 ]
p = 0.60 (60 % of alleles are dominant). -
Expected genotype frequencies:
- AA: (p^{2} = (0.60)^{2} = 0.36) → 0.36 × 1000 = 360 individuals
- Aa: (2pq = 2(0.60)(0.40) = 0.48) → 0.48 × 1000 = 480 individuals
- aa: (q^{2} = 0.16) → 0.16 × 1000 = 160 individuals (matches the observed recessive count).
Problem 2 – Heterozygote Frequency from Phenotype Data
A survey of a large random‑mating population finds that 9 % of individuals exhibit a recessive genetic disorder. Assuming Hardy‑Weinberg equilibrium, what percentage of the population are carriers (heterozygotes) of the disorder allele?
Solution
The recessive disorder phenotype frequency = q² = 0.09.
[ q = \sqrt{0.09} = 0.30]
[ p = 1 - q = 1 - 0.30 = 0.70 ] Carrier (heterozygote) frequency = 2pq:
[ 2pq = 2(0.70)(0.30) = 0.42 = 42% ]
Answer: 42 % of the population are expected to be carriers.
Problem 3 – Testing for Equilibrium (Chi‑Square)
In a sample of 500 plants, the genotypes for a flower‑color gene are observed as follows:
- AA (red): 200
- Aa (pink): 250
- aa (white): 50
Test whether this population is in Hardy‑Weinberg equilibrium using a chi‑square goodness‑of‑fit test (α = 0.05).
Solution Step 1: Calculate allele frequencies from observed data.
Total alleles = 2 × 500 = 1000.
-
Number of A alleles = 2(AA) + 1(Aa) = 2(200) + 250 = 400 + 250 = 650 - Frequency of A (p) = 650 / 1000 = 0.65
-
Number of a alleles = 2(aa) + 1(Aa) = 2(50) + 250 = 100 + 250 = 350
-
Frequency of a (q) = 350 / 1000 = 0.35
(Notice p + q = 1, as expected.) Step 2: Compute expected genotype counts under HWE.
- Expected AA = p² × N = (0.65)² × 500 = 0.4225 × 500 = 21
