In a population that meets the assumptions of Hardy-Weinberg equilibrium, 81% of the individuals are homozygous for a recessive allele. What percentage of the individuals would be expected to be heterozygous for this gene in the next generation? A. 18 B. 81 C. 1D. 9

A. 18

Explanation:

Given that the Hardy-Weinberg equilibrium:

p² + 2pq + q² = 1

and p + q = 1

We can determine the percentage of the individuals would be expected to be heterozygous for this gene in the next generation.

Let the homozygous recessive individuals, let say "aa" = q² = 81%

Let the homozygous dominant individuals = AA = p²

Let the heterozygous individuals = Aa = 2pq

The frequency of 'a' allele which means q² = 0.81,

by definition; if

q² = 0.81

q = 0.9

Since q equals to the frequency of the 'a' allele, then the frequency is 90%

The frequency of the 'A' allele from the homozyous dominant individual can be determined since q = 0.9

p+q = 1

p+0.9 = 1

p = 1 - 0.9

p = 0.1

Frequency of A allele by definition is said to be 10%

Now, to determine the percentage of the individuals that is expected to be heterozygous for the next generation, we have;

2pq = (2 * 0.9 * 0.1)

= 0.18

= 18%