ہارڈی وینبرگ کے توازن کی مساوات ______ ہے؟
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A. None of these
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B. P + 2pq + q = 1
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C. P² + 2pq + q² = 1
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D. P + 2pq + q = 0
Explanation
- The Hardy-Weinberg equilibrium equation is P² + 2pq + q² = 1.
- It representing the genetic frequencies of a population.
- Here, P² represents homozygous dominant.
- q² homozygous recessive.
- 2pq heterozygous individuals.
جب آبادی ہارڈی وائنبرگ کے توازن میں ہے، تو درج ذیل میں سے کون سا سچ ہے؟
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A. Mating is random
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B. Population size to be limited
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C. Individuals migrate randomly
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D. None of these
Explanation
- In Hardy-Weinberg equilibrium, random mating occurs, meaning individuals do not mate based on selective factors.
- This equilibrium also assumes no mutation, migration, or natural selection, and a sufficiently large population size to maintain genetic stability.
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A. None of these
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B. 0.42
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C. 0.21
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D. 0.49
Explanation
- The frequency of A1 is 0.7.
- The frequency of A2 is 0.3 (1 - 0.7).
- The proportion of heterozygous flies (A1A2) is 2pq.
- It is 2(0.7)(0.3) = 0.42.
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A. 0.32
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B. 0.20
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C. 0.42
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D. None of these
Explanation
- The frequency of allele A is 0.8 (1 - 0.2).
- The frequency of heterozygous genotype (Aa) is 2pq.
- It is 2(0.8)(0.2) = 0.32.
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A. 18
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B. 90
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C. None of these
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D. 36
Explanation
- The frequency of allele a is 0.1.
- So the frequency of allele A is 0.9 (1 - 0.1).
- The heterozygous genotype (Aa) frequency is 2pq.
- It is 2(0.9)(0.1) = 0.18 or 18%.
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A. 16
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B. 8
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C. 32
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D. None of these
Explanation
- The frequency of allele a is 0.4.
- So the frequency of homozygous aa is calculated by squaring 0.4 (0.4² = 0.16).
- This gives 16% of the population being homozygous for allele a.
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