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A particular population of chickens has a frequency of the dominant allele as 0.70. What 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 individuals that are heterozygous in a population, we can use the Hardy-Weinberg principle. This principle states that the frequencies of alleles and genotypes in a population will remain constant from generation to generation in the absence of other evolutionary influences. According to this principle:

- Let \( p \) be the frequency of the dominant allele (in this case, \( p = 0.70 \)).
- Let \( q \) be the frequency of the recessive allele.

Since the total frequency of alleles in the population must add up to 1, we have:
[tex]\[ q = 1 - p \][/tex]
[tex]\[ q = 1 - 0.70 \][/tex]
[tex]\[ q = 0.30 \][/tex]

The frequency of heterozygous individuals, who have one dominant allele and one recessive allele, can be found using the formula:
[tex]\[ 2pq \][/tex]
where \( p \) is the frequency of the dominant allele and \( q \) is the frequency of the recessive allele.

Substituting the values of \( p \) and \( q \) into the formula:
[tex]\[ 2 \times 0.70 \times 0.30 \][/tex]

This will give us the frequency of heterozygous individuals in the population. By multiplying:
[tex]\[ 2 \times 0.70 \times 0.30 = 0.42 \][/tex]

Hence, the correct way to calculate the frequency of individuals that are heterozygous is given by option:
C. [tex]\( 2 \times (0.70) \times (0.30) \)[/tex]