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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
What does the following equation represent? Explain:
p2 + 2pq + q2 = 1.
What is the Founder's effect?
Multiple choice question.
In Hardy - Weinberg equation, the frequency of homozygous recessive individual is represented by:
A population will not exist in Hardly Weiberg equilibrium if ____________.
(p + q)2 = p2 + 2pq + q2 = 1 represents an equation used in ______.
For the MN-blood group system, the frequencies of M and N alleles are 0.7 and 0.3, respectively. The expected frequency of MN-blood group bearing organisms is likely to be ______.
State Hardy Weinberg's principle.
Give a mathematical expression for 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:
Frequency of the allele (A).
Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
