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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 carrier 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
= 1 − 0.4
= 0.6
Number of carrier fruit flies is
As, 2pq (Aa)
= 2 × 0.6 × 0.4
= 0.48
= 0.48 × 200
= 96
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संबंधित प्रश्न
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In the Hardy-Weinberg equilibrium equation, the homozygous mutant is represented as ______.
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.
 |
 |
 |
| character Graph A |
character Graph B |
character Graph C |
- What names are given to the types of selection shown in graphs A, B and C?
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