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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) (0.70) x (0.30)
B) (0.70) - (0.30)
C)2 x (0.70) x (0.30)
D)2 x (0.70 - 0.30)

Respuesta :

scme1702
The Hardy-Weinberg equation is as follows:

p² + 2pq + q² = 1

Where p is the frequency of the homozygous dominant genotype, q is the frequency of the homozygous recessive genotype and 2pq is the frequency of the heterozygous genotype. Using the 2pq term here,

Frequency of heterozygous = 2 * 0.7 * 0.3

Therefore, option C is correct.

The expression is the correct way to calculate the frequency of individuals that are heterozygous is 2 x (0.70) x (0.30).

Explain how?

According to HWE.

p² + 2pq + q² = 1

Where p is the frequency of the dominant genotype, q is the frequency of the recessive genotype and 2pq is the frequency of the heterozygous genotype. Using the 2pq term here,

Frequency of heterozygous = 2 * 0.7 * 0.3

Hence, option C is correct,According to HWE.

To learn more about Hardy-Weinberg equation click here:

https://brainly.com/question/12143126

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