At a particular locus, frequency of allele A is 0.6 and that of allele a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

Question

At a particular locus, frequency of allele A is 0.6 and that of allele a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium? (A) 0.36 (B) 0.16 (C) 0.24 (D) 0.48

Tardigrade Admin · Accepted Answer

In a stable population, for a gene with two alleles, 'A' (dominant) and 'a' (recessive), if the frequency of 'A' is

p

and the frequency of 'a' is q, then the frequencies of the three possible genotypes (AA, Aa and aa) can be expressed by the Hardy-Weinberg equation:

p^{2} + 2 pq + q^{2} = 1

where

p^{2}

= Frequency of

AA

(homozygous dominant) individuals

q^{2}

= Frequency of

aa

(homozygous recessive) individuals

2 pq

= Frequency of Aa (heterozygous) individuals
so,

p = 0.6

and

q = 0.4

(given)

∴

2 pq

(frequency of heterozygote)

= 2 \times 0.6 \times 0.4

= 0.48

Q. At a particular locus, frequency of allele $A$ is $0.6$ and that of allele a is $0.4$ . What would be the frequency of heterozygotes in a random mating population at equilibrium?

$0.36$

$0.16$

$0.24$

$0.48$

Q. At a particular locus, frequency of allele A is 0.6 and that of allele a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

0.36

0.16

0.24

0.48

Q. At a particular locus, frequency of allele $A$ is $0.6$ and that of allele a is $0.4$ . What would be the frequency of heterozygotes in a random mating population at equilibrium?

$0.36$

$0.16$

$0.24$

$0.48$