Biology · Introductory biology · Worked example
Find the carrier frequency of a recessive allele
A recessive trait appears in 1 of every 2,500 people in a population. Assuming Hardy–Weinberg equilibrium, estimate the allele frequencies and the fraction of people who are carriers.
Read the trait frequency as q²
Only aa individuals show a recessive trait, so under Hardy–Weinberg equilibrium their frequency is q².
Take the square root
q is the square root of q², and p = 1 − q.
Find the carrier frequency
Carriers are heterozygotes, with frequency 2pq: about 3.9%, or roughly 1 person in 26.
Compare carriers with affected people
Carriers outnumber affected individuals 98 to 1. Most copies of a rare recessive allele sit unseen in heterozygotes, which is why selection against the trait removes the allele only slowly.
State what was assumed
The estimate assumes Hardy–Weinberg equilibrium: random mating, no selection, mutation or migration at this gene, and a large population. Counting phenotypes alone cannot test that assumption, because carriers and AA individuals look alike.
Result
q = 0.02 and p = 0.98, so about 3.9% of people, roughly 1 in 26, are carriers: 98 carriers for every affected person.
Your turn
A recessive phenotype appears in 9% of a population. Estimate q, p and the carrier frequency.
Show the answer and explanation
q = 0.3, p = 0.7 and 2pq = 0.42: 42% of the population are carriers.
q² = 0.09, so q = √0.09 = 0.3 and p = 0.7. Then 2pq = 2(0.7)(0.3) = 0.42.
Keep exploring
The Hardy–Weinberg tool opens with q² = 0.0004 and a population of 2,500. It predicts 2,401 AA, 98 Aa and 1 aa.
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