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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]

Sagot :

To determine the frequency of heterozygous individuals in a population, we need to use the principles of population genetics. In this scenario, the frequencies of the dominant and recessive alleles are given.

The heterozygous genotype appears when an organism has one dominant allele and one recessive allele. According to the Hardy-Weinberg principle, the frequency of heterozygous individuals (denoted as 2pq) can be calculated by:

[tex]\[2 \times \text{frequency of dominant allele} \times \text{frequency of recessive allele}\][/tex]

Given:
- The frequency of the dominant allele (p) = 0.70
- The frequency of the recessive allele (q) = 0.30

Thus, the correct way to calculate the frequency of heterozygous individuals is:

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

So, the correct answer is:
C. [tex]\[ 2 \times(0.70) \times(0.30) \][/tex]
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