Probability That Two Carriers Have an Affected Child (Hardy-Weinberg)

A common genetics question reads: an autosomal recessive disorder affects 1 in 650 births; what is the probability that two unrelated people who choose each other as partners have an affected child? The answer chains three steps: allele frequency, carrier frequency, then Punnett probability. Each step is a single line of arithmetic.

The three steps

1. q² is the disease frequency. Here q² = 1/650, so q = the square root of 1/650, about 1/25.5, or roughly 0.0392.
2. Carrier frequency is 2pq. With p = 1 - q, about 0.9608, 2pq = 2 x 0.9608 x 0.0392, about 0.0753, or roughly 1 in 13.3.
3. Combine. For a child to be affected, both partners must be carriers, and then both must pass the allele.

The full calculation

• Chance partner 1 is a carrier: about 0.0753.
• Chance partner 2 is a carrier: about 0.0753.
• Chance two carriers have an affected child: 1/4.
• Product: 0.0753 x 0.0753 x 0.25, about 0.00142, or roughly 1 in 705.

Because answer choices are usually spread widely, this lands closest to the choice near 1/700 in a list such as 1/3,650, 1/78,500, 1/120 and 1/700.

A shortcut check

For a rare recessive disease, p is close to 1, so carriers are about 2q. Then the probability is about (2q)(2q)(1/4) = q², which is the same as the population frequency of 1/650. That makes sense: if people mate at random, a random child is affected with probability q². The small difference from 1/650 comes from using p = 1 - q instead of p = 1. So the shortcut gives about 1/650, and the exact answer about 1/705. Both point to the same answer choice.

When the question changes

• Partners are not random (they are siblings or cousins). The carrier probability for the second partner is no longer 2pq; use the pedigree relationship instead.
• One partner is a known carrier. Skip his term: chance = 2pq x 1/4.
• One partner is unaffected with an affected sibling and unaffected parents. That person is a carrier with probability 2/3, not 2pq.
• Both are known carriers. The answer is 1/4.

Common traps

• Taking q = 1/650. That is q². Take the square root first.
• Using q instead of 2pq for carriers. The carrier class is heterozygotes.
• Forgetting the 1/4. Two carriers do not always have an affected child.
• Assuming Hardy-Weinberg when the question states otherwise. Random mating, no selection and no migration are required.

One-minute recap

• q² is disease frequency; q is its square root.
• Carrier frequency is 2pq, about 2q for rare alleles.
• Probability of an affected child from random partners is (2pq)(2pq)(1/4).