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Hardy-Weinberg Principle · Q28

Q.In a population of 1,000 individuals in Hardy-Weinberg equilibrium, a recessive allele aa causes a metabolic disorder when present in the homozygous condition (aaaa). A population screening finds that 9% of individuals are affected. Calculate

(a) the frequency of the recessive allele qq,
(b) the frequency of the dominant allele pp,
(c) the percentage of the population that are heterozygous (AaAa) carriers, and
(d) the actual number of carrier individuals in this population.
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Step 1. The affected (homozygous recessive, aa) frequency is given as 9% = 0.09 = q².

Step 2. Solve for q: q = √0.09 = 0.3.

Step 3. Since p + q = 1: p = 1 - 0.3 = 0.7.

Step 4. Heterozygote (carrier, Aa) frequency = 2pq = 2 × 0.7 × 0.3 = 0.42, i.e. 42%.

Step 5. Check: p² + 2pq + q² = 0.49 + 0.42 + 0.09 = 1.00. ✓ …

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