Question
In a population, the frequency of the dominant allele A is 0.6 and that of the recessive allele a is 0.4. What would be the frequency of the heterozygous genotype in a random mating population at equilibrium?
Answer: Option D
:
D
Hardy-Weinberg equation is used calculate the frequency of occurrence of alleles in a population.
It has been given that the frequency of dominent allele 'p' = 0.6 and recessive allele 'q' = 0.4.
Applying the Hardy-Weinberg Equation, p2+2pq+q2=1
The frequency of the heterozygous genotype (Aa) = 2pq
Substituting for p and q, we get
2pq = 2 x 0.4 x 0.6 = 0.48
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:
D
Hardy-Weinberg equation is used calculate the frequency of occurrence of alleles in a population.
It has been given that the frequency of dominent allele 'p' = 0.6 and recessive allele 'q' = 0.4.
Applying the Hardy-Weinberg Equation, p2+2pq+q2=1
The frequency of the heterozygous genotype (Aa) = 2pq
Substituting for p and q, we get
2pq = 2 x 0.4 x 0.6 = 0.48
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