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= [tex]\sqrt{0.2[/tex]
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.