Suppose the human trait for earlobe type is controlled by a simple dominant and recessive relationship at one locus. free earlobes, e, is the dominant allele, and attached earlobes, e, is the recessive allele. in a college genetics class, the professor takes a tally of students who have free earlobes and of students with attached earlobes. in this class of 117 students, 37 have free earlobes. calculate the frequency of the dominant allele, e, and the heterozygous genotype, ee. express the frequencies in decimal form rounded to the nearest thousandth. assume the class is in hardy–weinberg equilibrium for this trait.

The frequency of the E allele is 0.173

The frequency of the Ee genotype is 0.286

Explanation:

Free earlobes (E) is dominant over attached earlobes (e).

The class is in Hardy–Weinberg equilibrium for this trait, so the genotypic frequencies are:

Where p is the frequency of the E allele and q is the frequency of the e allele.

In a class of 117 students, 37 have free earlobes (EE + Ee genotypes) and thus 80 students have attached earlobes (ee).

The frequency of the ee genotype (q²) is then 80/117, so we can calculate q:

q² = 80/117

q= √0.684

q= 0.827

The frequency of the e allele is 0.827

p + q = 1, so p = 1-q = 0.173

The frequency of the E allele is 0.173

The frequency of the heterozygous genotype Ee is:

freq Ee = 2pq = 2 × 0.827 × 0.173 =

freq Ee =  0.286

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