Question

The presence of freckles is due to a dominant allele. four percent of the individuals in a particular population lack freckles. use the hardy-weinberg formula to calculate the percentage of individuals in this population who are homozygous dominant for freckles

Answer 1

Let us assign the recessive allele for freckle to be f and the dominant F. According to the statement, we deduce that ff = 0.04.

The frequency of the recessive f allele is, therefore, is 0.2 (square root of 0.04 ).

Hardy-Weinburg p + q = 1

P + 0.2 = 1; p = 1- 0.2 = 0.8

Therefore using the Hardy-Weinburg equation of a population in equilibrium

P2+2pq+q2=1

Heterozygous individuals are;

2pq = 2*0.8*0.2 = 0.32

This is 32% of the population

Answer 2

Answer:

The percentage of individuals in this population who are homozygous dominant for freckles is 64%.

Explanation:

For freckles trait, F is the dominant allele and f the recessive.

4% of the individuals in the population lack freckles, which means that the genotypic frequency for the homozygous recessive allele for freckles equals 0.04. This represents the q² in the H-W formula. The square root of 0.04 is 0.2, the genic frequency for the recessive allele, which  represents q. According to H-W, p + q = 1, so p = 1 - q. Then 1 - 0.2 = 0.8, the dominant allelic frequency for the freckles trait, which represents p. Knowing the frequency of each allele, the frequency of the individuals´ homozygous dominant in the population can be calculated. FF = p2 = 0.8 x 0.8 = 0.64. Then the percentage of individuals in this population who are homozygous dominant for freckles is 64%.