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?
A) 0.24
B) 0.32
C) 0.42
D) 0.48
E) 0.80

Question

Quizplus · Accepted Answer

In a population in Hardy-Weinberg equilibrium, the frequency of carriers (heterozygotes) is calculated using 2pq, where p is the frequency of the normal allele and q is the frequency of the sickle cell allele. If q = 0.2, then p = 1 - q = 0.8. Thus, 2pq = 2 * 0.8 * 0.2 = 0.32.

In a Certain Population, the Allele Causing Sickle Cell Anemia