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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.
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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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 |
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| character Graph A |
character Graph B |
character Graph C |
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