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7 January, 21:11

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)

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Answers (1)
  1. 7 January, 21:33
    0
    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.
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