Estimate Carrier Frequency and Couple Risk With Hardy-Weinberg

Hardy-Weinberg calculations appear in every genetics course and are used in public health to estimate how common carriers of recessive conditions are. This guide takes a recessive condition with an incidence of 1 in 2,500 births, close to the figure often quoted for cystic fibrosis in populations of northern European ancestry, and works from incidence to carrier frequency and couple risk. It then tests a set of observed genotype counts against the equilibrium. This is an educational exercise; real risk assessment for a family is a matter for genetic counselling.

Step 1: from incidence to allele frequency

For an autosomal recessive condition, affected people have genotype aa, so the incidence is q². With q² = 1 ÷ 2,500 = 0.0004, the Hardy-Weinberg calculator gives q = 0.02 and p = 0.98. The assumption is that the population mates at random with respect to this gene, which is reasonable for a single large population but not for one that mixes groups with different allele frequencies.

Step 2: carrier frequency

Carriers are heterozygotes: 2pq = 2 × 0.98 × 0.02 = 0.0392, 3.92% of the population, or about 1 person in 26. There are 98 carriers for every affected person, and 98% of all copies of the allele are in carriers. This is why the condition persists even though affected individuals historically had few children: selection acts on aa individuals, who hold only a tiny fraction of the alleles.

Step 3: risk for a couple

If two unrelated people from this population have a child, the chance that both are carriers is 0.0392² = 0.00154, about 1 in 650. Each child of two carriers has a 1 in 4 chance of being affected, so the risk for a child of a random couple is 1 in 2,600, close to the incidence of 1 in 2,500 we started from, which is a good consistency check. If one partner is known to be a carrier, for example because a sibling is affected and testing confirmed it, the risk changes completely: the other partner is a carrier with probability 0.0392, and the child’s risk is 0.0392 × 1/4 = 0.98%, about 1 in 100. A common mistake is to apply population carrier frequencies to relatives of affected people, who have much higher prior probabilities.

Step 4: testing observed counts

Hardy-Weinberg is also a test. Suppose a survey genotypes 3,000 people at a marker and finds 1,405 AA, 1,101 Aa and 494 aa. The calculator estimates p = 0.6518 and q = 0.3482 and expects 1,274.7 AA, 1,361.7 Aa and 363.7 aa. There are far fewer heterozygotes and more homozygotes than expected: χ² = 109.9 with 1 degree of freedom, p below 0.0001. Equilibrium is clearly rejected. A heterozygote deficit like this usually points to a sample that combines two populations with different allele frequencies, known as the Wahlund effect, to inbreeding, or to a genotyping assay that fails to detect one allele in heterozygotes. It does not by itself indicate selection.

Common mistakes

• Taking the square root of the wrong number. q comes from the frequency of the recessive phenotype, not the dominant one.
• Forgetting the 2 in 2pq. Heterozygotes can arise from either parent’s allele.
• Using population figures for relatives. Family history changes the prior probability.
• Applying the model to X-linked genes. In males the trait frequency is q, not q².
• Two degrees of freedom. The equilibrium test with two alleles has one.

Hardy-Weinberg checklist

1. Identify the recessive phenotype and its frequency in the population.
2. Take the square root to find q, then p = 1 − q.
3. Calculate p², 2pq and q², and check that they add up to 1.
4. For couple risk, multiply carrier probabilities and the 1 in 4 chance per child.
5. For observed counts, estimate p from the data and run a 1-degree-of-freedom chi-square test.
6. Interpret deviations by checking sampling, mating patterns and genotyping before invoking selection.