Question

Read the passage and answer the following question.

The number of people in certain regions of Africa that suffer from sickle cell anemia is $16$ percent. This genetic disorder is caused by a mutation in which homozygous recessive results in sickle cell anemia, but in the heterozygote condition causes the sickle cell trait. The gene frequency for that allele and the percentage of people that are heterozygotes is

• A. $0.6,24$
• B. $0.4,24$
• C. $0.6,48$
• D. $0.4,48$
• E. $0.6,72$

Answer 1

The correct option is D $0.4,48$

According to the Hardy Weinberg law, the sum of all genotype frequencies can be represented as the binomial expansion of the square of the sum of p and q. This sum is equal to one. (p + q)2 = $p^2$ + 2pq + $q^2$ = 1. Here, "p" is the frequency of dominant allele, $p^2$ is frequency of homozygous dominants, q is frequency of recessive allele and $q^2$ is frequency of homozygous recessive individuals. The "2pq" in equation shows frequency of heterozygotes in the population. In the question, the genotype frequency of recessive population is ($q^2$) = (16% or 0.16). Hence, frequency of recessive allele= q= √0.16= 0.4. Since p+q=1; thus the frequency of dominant allele= p= 1-q= 1-0.4= 0.60. Frequency of homozygote 2pq= 2 x 0.6 x 0.4= 0.48= 48%. Thus, the correct answer is option D.