Answer:
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