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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:
Frequency of the allele (A).
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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
= 1 − 0.4
= 0.6
Frequency of allele, p(A) = 1 − 0.4
= 0.6
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संबंधित प्रश्न
According to the Hardy-Weinberg principle, the allele frequency of a population remains constant. How do you interpret the change of frequency of alleles in a population?
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Very short answer question.
State the Hardy – Weinberg equilibrium.
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What is the role of natural selection in affecting Hardy-Weinberg equilibrium?
What does it indicate when the measured frequency of an allele differs from the expected Hardy-Weinberg value?
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