Question

A population is in Hardy- Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele A is 0.7, the genotype frequency of Aa is:

• A. 0.42
• B. 0.7
• C. 0.21
• D. 0.36

Answer 1

The correct option is A 0.42

For a gene with two alleles, A (dominant) and a (recessive), 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+q=1
${(p+q)}^2$=1
$p^2$ + 2pq + $q^2$ = 1 where, $p^2$ = frequency of AA (homozygous dominant) individuals, 2pq = frequency of Aa (heterozygous) individuals and $q^2$ = frequency of aa (homozygous recessive) individuals.
​​​​​​​The equation can be used to calculate allele frequencies if the number of homozygous recessive individuals in the population is known.
Here, p = 0.7 [p+q=1]
0.7+q=1,
hence q = 0.3
Therefore 2pq (frequency of heterozygote)= 2 × 0.7 × 0.3 = 0.42