Question

In a population that is in hardy-weinberg equilibrium for two alleles, d and d, 72% of the population shows the dominant trait. what is the frequency of the recessive allele, d?

Answer 1

72% of the population shows D trait.

To find: The frequency of allele for the d trait.

Method: Use of Hardy-Wineberg equation at equilibrium to solve it.

Solution:

Let p denote the frequency of D trait allele and q denote the frequency of the d trait allele.

p + q = 1

(p + q)² = 1

p² + 2pq + q² = 1

or q² = 1 - (p² + 2pq)

Since D is a dominant trait, it'll be expressed in both the homozygous and heterozygous individuals that make up 72% of the population.

∴ (p² + 2pq) = = 0.72

∴ q² = 1 - 0.72 = 0.28

q = √0.28 = 0.53

Hence the frequency of the recessive d trait is given by q = 0.53.

Step-by-step explanation:

Answer 2

Given: 72% of the population shows D trait.

To find: The frequency of allele for the d trait.

Method: Use of Hardy-Wineberg equation at equilibrium to solve it.

Solution:

Let p denote the frequency of D trait allele and q denote the frequency of the d trait allele.

p + q = 1

(p + q)² = 1

p² + 2pq + q² = 1

or q² = 1 - (p² + 2pq)

Since D is a dominant trait, it'll be expressed in both the homozygous and heterozygous individuals that make up 72% of the population.

∴ (p² + 2pq) =
`(72)/(100)` = 0.72

∴ q² = 1 - 0.72 = 0.28

q = √0.28 = 0.53

Hence the frequency of the recessive d trait is given by q = 0.53.