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

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24-01-2020
Biology

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

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namordu namordu · Answer 1 · 2020-01-25T19:48:49+01:00

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

andromache andromache · Answer 2 · 2022-03-16T05:26:47+01:00

The frequency of the dominant allele is 0.7.

Hardy-Weinberg equilibrium:

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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