The correct option is A 84%
The Hardy-Weinberg equation is an algebraic equation which is an expression of the Hardy-Weinberg law. The frequencies of alleles and the genotypic ratios in a population can be calculated with the help of this equation.
According to Hardy Weinberg equation, if there are two alleles for a particular gene where A represents the dominant allele and a represents the recessive allele, then the frequency of A is p and a is q.
Combined frequencies of all alleles for a gene in a population is equal 1.
Therefore, p+q=1.
The frequency of homozygous dominant individuals (AA) in the population would be equal to pxp or p2
The frequency of homozygous recessive individuals (aa) in the population would be equal to qxq or q2
Since there are two ways of forming the heterozygote Aa, (allele A from the father and a from mother and vice versa) the frequency of heterozygous dominant individuals (Aa) in the population would be 2pq.
The sum of all three genotypic frequencies is p2+2pq+q2=1 which is a binomial expansion of (p+q)2
p = frequency of the dominant allele (A)
q = frequency of the recessive allele (a)
p2 = frequency of homozygous dominant individuals (AA)
q2 = frequency of homozygous recessive individuals (aa)
2pq = frequency of heterozygous dominant individuals (Aa)
As per the question given, let us denote p as the frequency of the dominant allele (C) and q as the frequency of the recessive allele (c).
The frequency of CC individuals is given as 36%, which means that p2 = 0.36 and p = 0.6.
Since p + q = 1, q = 1–0.6 = 0.4.
The question asks for the percentage of the population with curly hair; this includes both homozygous (CC) and heterozygous individuals (Cc).
The genotype frequencies can be found using the equation p2+2pq+q2
CC =p2 = (0.6)2 = 0.36 = 36% homozygous curly hair (CC)
Cc = 2pq = 2(0.6)(0.4) = 0.48 = 48% heterozygous curly hair (Cc)
cc = q2 = 0.42 = 0.16 = 16% straight hair (cc)
Therefore, the percentage of the population with curly hair is 36% +
48% = 84%.