Question

$9$ percent of all cicadas exhibit the homozygous recessive condition known as "flippant wings," the gene frequency for that gene in the general population is

• A. Cannot be determined
• B. $91$ percent
• C. $0.9$
• D. $0.3$
• E. $0.03$

Answer 1

Answer: (D)

$0.3$

Hardy-Weinberg equation is $p^2$ + 2pq + $q^2$ = 1. Here p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. Thus, $p^2$ represents the frequency of the homozygous dominant genotype AA, $q^2$ is the frequency of the homozygous recessive genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. In given question, frequency of homozygous recessive genotype ($q^2$)= 9% or 0.09. Frequency of recessive allele (q)= 0.3. Correct answer is D.