Question

Suppose the human trait for hair type is controlled by a simple dominant and recessive relationship at one locus. Curly hair, C, is the dominant allele, and straight hair, c, is the recessive allele. In a college genetics class, the professor takes a tally of students who have curly hair and of students with straight hair. In this class of 131 students, 52 have curly hair. Calculate the frequency of the dominant allele, C, and the heterozygous genotype, Cc. Express the frequencies in decimal form rounded to the nearest thousandth. Assume the class is in Hardy–Weinberg equilibrium for this trait.

Answer 1

Answer:

The frequency of the dominant allele, C $= 0.2234$

and the heterozygous genotype, Cc$= 0.1736$

Explanation:

Given -  

Curly hair, C, is the dominant allele, and straight hair, c, is the recessive allele

Thus, C is dominant over c

The genotype frequency of curly hair is $52$ in $131$ students

so, remaining $131-52 = 79$ are straight hair students.  

Since straight hair is a recessive allele , the genotypic frequency of "cc" is  

$\frac{79}{131}\\0.6030$

Frequency of "cc" is represented by $q^{2}$

Thus frequency of allele "q"  

$= \sqrt{0.6030} \\= 0.7765$

As per Hardy Weinberg's I equation -  

$p+q=1\\p + 0.7765=1\\p = 0.2234$

As per Hardy Weinberg's II equation -  

$p^{2} +2pq + q^{2} =1\\0.2234^2+2pq+0.6030=1\\2pq= 0.1736$

Hence,  

The frequency of the dominant allele, C $= 0.2234$

and the heterozygous genotype, Cc$= 0.1736$