Select the correct answer.

A particular population of chickens has a frequency for the dominant allele as 0.70 and a frequency for the recessive allele as 0.30. Which expression is the correct way to calculate the frequency of individuals that are heterozygous?

A. [tex]\((0.70) \times (0.30)\)[/tex]
B. [tex]\((0.70) - (0.30)\)[/tex]
C. [tex]\(2 \times (0.70) \times (0.30)\)[/tex]
D. [tex]\(2 \times (0.70 - 0.30)\)[/tex]



Answer :

To determine the frequency of heterozygous individuals in a population, we can use the Hardy-Weinberg equilibrium principle. The principle states that the frequency of genotypes in a population can be predicted from the allele frequencies. Specifically, for a two-allele system (alleles [tex]\( p \)[/tex] and [tex]\( q \)[/tex]):

- [tex]\( p^2 \)[/tex] represents the frequency of the homozygous dominant genotype.
- [tex]\( q^2 \)[/tex] represents the frequency of the homozygous recessive genotype.
- [tex]\( 2pq \)[/tex] represents the frequency of the heterozygous genotype.

Given:
- [tex]\( p = 0.70 \)[/tex] (frequency of the dominant allele)
- [tex]\( q = 0.30 \)[/tex] (frequency of the recessive allele)

The expression to determine the frequency of heterozygous individuals ([tex]\( 2pq \)[/tex]) is calculated as follows:

[tex]\[ 2pq = 2 \times p \times q \][/tex]

In this case:
[tex]\[ 2pq = 2 \times 0.70 \times 0.30 \][/tex]

Hence, the correct way to calculate the frequency of individuals that are heterozygous is:

C. [tex]\( 2 \times (0.70) \times (0.30) \)[/tex]

Thus, the correct answer is C.