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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
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Very short answer question.
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The graphs below show three types of natural selection. The shaded areas marked with arrows show the individuals in the population who are not selected. The dotted vertical lines show the statistical means.
 |
 |
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| character Graph A |
character Graph B |
character Graph C |
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State Hardy Weinberg's principle.
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.
