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%