Answer:
0.549 is the frequency of the F allele.
0.495 is the frequency of the Ff genotype.
Explanation:
<em>FF </em>or <em>Ff</em> genotypes determine freckles, <em>ff</em> 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=
q= 0.451
<em>q </em>is the frequency of the recessive f allele.
Given <em>p</em> 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.