Question

Suppose the human trait for freckles is controlled by a simple dominant and recessive relationship at one locus. Freckles, F, is the dominant allele, and a lack of freckles, f, is the recessive allele. In a college genetics class, the professor takes a tally of students who have freckles and of students with a lack of freckles. In this class of 123 students, 98 have freckles. Calculate the frequency of the dominant allele, F, and the heterozygous genotype Ff. 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:

0.549 is the frequency of the F allele.

0.495 is the frequency of the Ff genotype.

Explanation:

FF or Ff genotypes determine freckles, ff determines lack of freckels.

In this class of 123 students, 98 have freckles (and 123-98= 25 do not).

If the class is in Hardy-Weinberg equilibrium for this trait, then the genotypic frequency of the ff genotype is:

q²= 25/123

q²=0.203

q= $\sqrt{0.2$

q= 0.451

q is the frequency of the recessive f allele.

Given p the frequency of the dominant F allele, we know that:

p+q=1, therefore p=1-q

p=0.549 is the frequency of the F allele.

The frequency of the Ff genotype is 2pq. Therefore:

2pq=2×0.549×0.451

2pq=0.495 is the frequency of the Ff genotype.