Answer :
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:
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%.