Question

A population is in genetic equilibrium/Hardy-Weinberg equilibrium for a gene with 2 alleles (dominant allele is ‘A’ and recessive allele ‘a’). If the frequency of allele ‘A’ is 0.6, then the frequency of genotype ‘Aa’ is:

• A. 0.21
• B. 0.42
• C. 0.48
• D. 0.32

Hint:

Hardy-Weinberg Equilibrium Calculation

Answer 1

The Correct Option is C

According to Hardy-Weinberg equilibrium, the frequency of heterozygotes (Aa) is given by 2pq, where:
- p is the frequency of allele A (p = 0.6)
- q is the frequency of allele a (q = 1−p=0.4)
The frequency of genotype Aa is:
2pq = 2(0.6)(0.4) = 0.48