Question

60% of alleles of a gene in a population are recessive. The proportion of heterozygous individuals would be

• A. $0.16$
• B. $0.24$
• C. $0.36$
• D. $0.48$
• E. $0.6$

Answer 1

The correct option is D $0.48$

According to the Hardy-Weinberg law, the allele and genotype frequencies in a population remain constant under absence of factors responsible for evolution. It states that 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 allele is (q) = (60% or 0.6). Since, p + q = 1; thus the frequency of dominant allele = p = 1 - q = 1 - 0.6 = 0.40. Proportion of heterozygous individuals = 2pq = 2 x 0.6 x 0.4 = 0.48.

Therefore, the correct answer is option D.