Answer:
The frequency of the dominant allele is 0.7.
Explanation:
If the population is in Hardy-Weinberg equilibrium, the genotypic frequencies are:
p² = freq (AA)
2pq = freq (Aa)
q² = freq (aa)
where p is the frequency of the dominant A allele and q is the frequency of the recessive q allele.
The dominant phenotype of the trait occurs 91% of the time, so
p² + 2pq = 0.91
And therefore the recessive phenotype appears in 9% of the cases:
q² = 1 - 0.91 = 0.09
The frequency of the recessive q allele is thus:
q = √0.09
q=0.3
And because we only have two alleles, the frequencies of both must add to 1.
p + q = 1
p = 1 - q
p = 1 - 0.3
p = 0.7
The frequency of the dominant allele is 0.7.
In the case when the population should be in the above equilibrium so here the genotypic frequencies should be
p² = freq (AA)
2pq = freq (Aa)
q² = freq (aa)
Here
p represents the frequency of the dominant A allele
and q represent frequency of the recessive q allele.
Since The dominant phenotype of the trait arise 91% of the time,
so
p² + 2pq = 0.91
Now
q² = 1 - 0.91 = 0.09
The frequency of the recessive q allele should be
q = √0.09
q=0.3
And since we only have two alleles, the frequencies of both must add to 1.
p + q = 1
p = 1 - q
p = 1 - 0.3
p = 0.7
hence, The frequency of the dominant allele is 0.7.
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