In a certain population, the allele causing sickle cell anemia has an allele frequency of 0.2. If the population is in genetic equilibrium for this allele, what fraction of the population would be carriers for the allele (heterozygous individuals)?
A) 0.24
B) 0.32
C) 0.42
D) 0.48
E) 0.80

Question

Quizplus · Accepted Answer

The frequency of carriers (heterozygous individuals) can be calculated using the Hardy-Weinberg principle, which is 2pq, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. Given q = 0.2, p = 1 - q = 0.8. Therefore, 2pq = 2(0.8)(0.2) = 0.32.

In a Certain Population, the Allele Causing Sickle Cell Anemia