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 the number of students who have curly hair and those with straight hair. In this class of 117 students, 37 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:

A) The frequency of the dominant allele, C $= 0.4376$

B) frequency of the heterozygous genotype, Cc $= 0.2461$

Explanation:

Given -

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

Number of students with curly hair is $37$

Thus frequency of genotype "CC"

$= q^{2} = \frac{37}{117} \\= 0.3162$

Thus, frequency of allele "q"

$= \sqrt{0.3162} \\= 0.5623$

As per Hardy Weinberg's first  equation-

$p+q=1$

Substituting the value of q, we get -

$p+0.5623=1\\p= 0.4376$

As per Hardy Weinberg's second equation -

$p^2+q^2+2pq=1$

Substituting the values of "p" and "q" we get -

$0.4376^{2} + 0.3162+2pq = 1\\2pq = 0.2461$

Hence,

A) The frequency of the dominant allele, C $= 0.4376$

B) frequency of the heterozygous genotype, Cc $= 0.2461$

Answer 2

Answer:

The frequency of the dominant allele, C, is .0562, which rounded to the nearest thousandth is 0.6. And the frequency for the heterozygous genotype, Cc, is 0.492, which rounded to the nearest thousandth is 0.5.  

Explanation:

Available data:

• Population size / Class size: 117 individuals / students.

• 37 individuals are curly-haired.

• c is the recessive allele and expresses straight hair

• C is the dominant allele and expresses curly hair.

• The population is in Hardy-Weinberg equilibrium for the trait.

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms. The genotypic frequencies after one generation are p2 (Homozygous dominant), 2pq (Heterozygous), q2 (Homozygous recessive). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example, the genotypic frequency for homozygous dominant individuals (p²) in the population is CC = 37/117 = 0.316 = 31.6%. This is the proportion or frequency of the curly-haired genotype in a population, denoted as CC that equals p².  

Allelic frequency for curly hair trait, C is: √p² = √0.316 = 0.562

This is the proportion of a specific allele, in this case, C, respect to the pool of alleles. Results are expressed in proportions, so the sum of every allelic frequency for one locus equals one. This is C + c = 1

Then, the allelic frequency for straight hair is c = 1 - C

                                                                           c = 1 - 0,562

                                                                          c = 0.438

Until now, we know that:

• The allelic frequency for the C allele is 0.562.

• The allelic frequency for the c allele is 0.438.

• The genotypic frequency for curly hair, CC is 0.316.

Considering that the genotypic frequencies after one generation are p2 (Homozygous dominant), 2pq (Heterozygous), q2 (Homozygous recessive), then we can calculate the genotypic frequency for heterozygous individuals.

2pq= 2x0.562x0.438

2pq= 0.492

Thus, the frequency of the dominant allele, C, is .0562, which rounded to the nearest thousandth is 0.6. And the frequency for the heterozygous genotype, Cc, is 0.492, which rounded to the nearest thousandth is 0.5.