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Biochemistry · Population genetics: Hardy-Weinberg calculation
Sample itemAn autosomal recessive disorder affects 1 in 10,000 live births in a population in Hardy-Weinberg equilibrium. What is the approximate carrier frequency?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
C. About 1 in 50
Disease frequency q squared is 1/10,000, so q is 1/100, which is 0.01. The frequency of the normal allele p is 0.99. Carriers are 2pq, which is 2 times 0.99 times 0.01, about 0.0198, or roughly 2%, which is 1 in 50.
Option by option
A. About 1 in 100
1 in 100 is the allele frequency q, not the carrier frequency.
B. About 1 in 10,000
1 in 10,000 is the disease frequency q squared.
C. About 1 in 50 (correct)
2pq is about 0.02, which is 1 in 50.
D. About 1 in 200
1 in 200 would result from halving q instead of doubling it to get 2pq.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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