In a population where the frequency of the recessive allele (q) is 0.3, what percentage of individuals are expected to be heterozygous according to Hardy-Weinberg equilibrium?

• A. 21%
• B. 9%
• C. 49%
• D. 42%

Answer: (D)

Verified step by step guidance

1. Recall that according to the Hardy-Weinberg equilibrium, the allele frequencies are represented as \(p\) for the dominant allele and \(q\) for the recessive allele, where \(p + q = 1\).
2. Given the frequency of the recessive allele \(q = 0.3\), calculate the frequency of the dominant allele \(p\) using the equation \(p = 1 - q\).
3. The frequency of heterozygous individuals (carrying one dominant and one recessive allele) is given by the term \$2pq$ in the Hardy-Weinberg equation.
4. Substitute the values of \(p\) and \(q\) into the heterozygous frequency formula: \$2 \times p \times q$.
5. Multiply the result by 100 to convert the frequency into a percentage, which represents the expected percentage of heterozygous individuals in the population.