Question

Consider a population of sheep to be in Hardy-Weinberg equilibrium. The allele for black wool (\(p\)) has a frequency of \(0.81\) while the allele for white wool (\(q\)) has a frequency of \(0.19\). Then the percentage of heterozygous individuals in the population is

• A. \(4\%\)
• B. \(15\%\)
• C. \(31\%\)
• D. \(66\%\)

Hint:

Hardy-Weinberg genotype frequencies are \[ \boxed{ p^2:\;2pq:\;q^2. } \] Remember, \[ \boxed{ \text{Heterozygous frequency}=2pq. } \]

Answer 1

The Correct Option is C

Step 1: Use the Hardy-Weinberg equation. For a population in equilibrium, \[ \boxed{ p^2+2pq+q^2=1. } \] The frequency of heterozygous individuals is \[ \boxed{2pq.} \]

Step 2: Substitute the given values. Given, \[ p=0.81,\qquad q=0.19. \] Therefore, \[ 2pq = 2(0.81)(0.19) = 0.3078 \approx 0.31. \] Thus, \[ \boxed{31\%} \] of the population is heterozygous. Hence, \[ \boxed{(C)} \] is the correct answer.