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A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following: The number of homozygous recessive fruit flies. - Biology (Theory)

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प्रश्न

A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following:

The number of homozygous recessive fruit flies.

संख्यात्मक
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उत्तर

In Hardy Weinberg, p2 + 2pq + q2 = 1, where 'p2' is the frequency of the homozygous dominant genotype (AA), '2pq' is the frequency of the heterozygous genotype (Aa), and 'q2' is the frequency of the homozygous recessive genotype (aa).

Given:

q = 0.4

We know p + q = 1

p = 1 − q

p = 1 − 0.4

p = 0.6

number of homozygous recessive fruit flies is

As, q2(aa) = (0.4)2

= 0.16

= 0.16 × 200

= 32

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A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following:

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The number of homozygous dominant fruit flies.


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