Question

The presence of recessive trait in a large population is found to be 16%. The frequency of dominant trait in that population is:-

A
0.84
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B
0.42
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C
0.56
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D
0.96
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Solution

The correct option is A 0.84Let's start with the basic Hardy-Weinberg equations first.p+q=1 and p2+2pq+q2=1With "p" being the dominant allele and "q" being the recessive alleleWe know that 16% (or 0.16) show the recessive trait. This means that the fraction of the population with the recessive trait, q2, is 0.16.If 16% of the population shows the recessive trait, then that figure (16%) accounts for all the homozygous (hh) recessive individuals in that population.This is one of most classical cases of applying a quadratic equation to the life sciences. One must also understand that this is a classical “Mendelian” genetics problem.Heterozygyous individuals (Hh) [inferably] do not express the homozygous recessive condition for this particular allele, but follow a traditional dominant expression. Classical genetics relies on these formally declared values.The remaining 84% of the population does not show for the recessive trait.From H-W equation, p^2 + 2pq + q^2 = 1.0 (equilibrium)0.0256 sqrt = 0.16 // 16% hh1 - 0.16 = 0.84 (84%) sq = 0.7056 // (84%) dominant.So, the correct option is 'Option A'.

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