MCAT Biology Question 178: Answer and Explanation

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Question: 178

13. In a particular Hardy–Weinberg population, there are only two eye colors: brown and blue. Of the population, 36% have blue eyes, the recessive trait. What percentage of the population is heterozygous?

A. 24%
B. 48%
C. 60%
D. 64%

Correct Answer: B

Explanation:

Using the information given in the question stem, we can determine that the percentage of the population with blue eyes (genotype bb) = 36% = 0.36 = q²; therefore, q = 0.6. Because this is a Hardy–Weinberg population, we can assume that p + q = 1, so p = 1 – 0.6 = 0.4. The frequency of heterozygous brown eyes is therefore 2pq = 2 × 0.4 × 0.6 = 0.48 = 48%.

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