In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. what is the frequency of the dominant allele?

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

a

explanation:

the answer would be 2, if crossing over doesn’t occur 2 out of 4 of the daughter cells will have that allele.

hope this : )