Question

At a particular locus, the frequency of the 'A' allele is 0.6 and that of 'a' is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

• A. 0.24
• B. 0.16
• C. 0.48
• D. 0.36

Answer 1

Answer: (C)

0.48

• In a stable population, for a gene with two alleles, 'A' (dominant) and 'a' (recessive), if the frequency of 'A' is p and the frequency of 'a' is q, then the frequencies of the three possible genotypes ($AA$, $Aa$ and $aa$) can be expressed by the Hardy-Weinberg equation:
  

$p^2$ + $2pq$ + $q^2$ = $1$

• where $p^2$ = Frequency of $A$ (homozygous dominant) individuals
• $q^2$ = Frequency of $aa$ (homozygous recessive) individuals
• $2pq$ = Frequency of $Aa$ (heterozygous) individuals.
  
• So, $p$ = $0.6$ and $q$ = $0.4$ (given)
• Then, $2pq$ (frequency of heterozygote) = $2\times 0.6\times 0.4$ = $0.48$.

So, the correct answer is option C.