Question

In a population that is in Hardy Weinberg equilibrium, the frequency of a recessive allele for a certain hereditary trait is 0.20. What percentage of the individual in the next generation would be expected to show the dominant trait?

• A. 16%
• B. 32%
• C. 64%
• D. 96%

Answer 1

Answer: (D)

96%

• As the population is in Hardy Weinberg equilibrium, the frequencies of alleles will remain the same for the next generation also.
• According to the question, the frequency of recessive allele is (q = 0.2) Frequency of dominant allele= p= 1-q= 1-0.2=0.8.
• Hardy Weinberg law states that the sum of all genotype frequencies is represented as the binomial expansion of the square of the sum of p and q.

This sum is equal to one - (p + q)2 = p2 + 2pq + q2 = 1.

• Hence, the frequency of homozygous dominant genotype in total population

"(p2) = 0.64 or 64%".

• And the frequency of heterozygous dominant genotype in total population

"(2pq) = 2 X 0.8 X 0.2 = 0.32 or 32%".

• And the frequency of dominant individuals in next population = 64+32= 96%.

The correct option is D.