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 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.

Hint: Use the population data and the Hardy–Weinberg equations, p + q = 1 and p2 + 2pq + q2 = 1, to calculate the genotype and allele frequencies.

Answer 1

To find the frequency of the dominant allele (C) for curly hair and heterozygous genotype (Cc) in a college class, use Hardy-Weinberg equations. Calculating q from the number of straight-haired individuals, then determining p, and finally 2pq for the heterozygous frequency, gives a dominant allele C frequency of 0.223 and heterozygous genotype Cc frequency of 0.346.

Step-by-step explanation:

In a college genetics class, calculating the frequency of the dominant allele (C) for curly hair and the heterozygous genotype (Cc) can be accomplished by using the Hardy-Weinberg principle.

As given, 52 out of 131 students have curly hair. Assuming curly hair (C) is dominant, the phenotype of individuals with at least one dominant allele (CC or Cc) will be curly hair. Although we cannot differentiate between homozygous dominant (CC) and heterozygous (Cc) individuals solely from observation, we can compute frequencies using Hardy-Weinberg equations p + q = 1 and p² + 2pq + q² = 1, with p representing the frequency of allele C and q representing the frequency of allele c.

Since only individuals with the homozygous recessive genotype (cc) will have straight hair, and we know that 79 students have straight hair (131 - 52 = 79), we can calculate the recessive allele frequency (q²) as 79/131. Therefore, q² = 0.603, giving us q = √0.603 = 0.777. Now using p + q = 1, we find p = 1 - q = 1 - 0.777 = 0.223.

The frequency of the heterozygous genotype (Cc) is given by 2pq, which equals 2 × 0.223 × 0.777, approximately 0.346. Thus, the frequency of the dominant allele C is 0.223, and the frequency of the heterozygous genotype Cc is approximately 0.346, rounded to the nearest thousandth.