The presence of recessive trait in a large population is found to be 16%. The frequency of dominant trait in that population is:-
Let's start with the basic Hardy-Weinberg equations first.
p+q=1 and p2+2pq+q2=1
With "p" being the dominant allele and "q" being the recessive allele
We 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% hh
1 - 0.16 = 0.84 (84%) sq = 0.7056 // (84%) dominant.
So, the correct option is 'Option A'.