If a homozygous dominant genotype is 46% what is the dominant allele frequency?

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If a homozygous dominant genotype is 46% what is the dominant allele frequency?

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Answer 1 · 2017-03-25T22:46:49

Well, let's check.

Explanation:

Assuming Hardy-Weinberg conditions are met, we will use the $Hardy-Weinberg equation$ $"Hardy-Weinberg equation"$ which is:

$p^{2} + 2 p q + q^{2} = 1$ $p^2+2pq+q^2=1$

Where:
$p^{2} = frequency of homozygous dominant genotype$ $p^2 = "frequency of homozygous dominant genotype"$
$2 p q = frequency of heterozygous genotype$ $2pq = "frequency of heterozygous genotype"$
$q^{2} = frequency of homozygous recessive genotype$ $q^2 = "frequency of homozygous recessive genotype"$

If a homozygous dominant genotype is $46 %$ $46%$ , that means $p^{2} = 0.46$ $p^2 = 0.46$

Using $p + q = 1$ $p+q = 1$ where:

$p = frequency of dominant allele$ $p = "frequency of dominant allele"$
$q = frequency of recessive allele$ $q = "frequency of recessive allele"$

we will find the frequency of the dominant allele, meaning we will solve for $p$ $p$ .

If we have $p^{2}$ $p^2$ and we need to find $p$ $p$ , then using basic algebra:

$p^{2} = 0.46$ $p^2 = 0.46$
$\sqrt{p^{2}} = \sqrt{0.46}$ $sqrt(p^2) = sqrt0.46$
$p = 0.68$ $p = 0.68$

the dominant allele frequency turns out to be 0.68 or 68%

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If a homozygous dominant genotype is 46% what is the dominant allele frequency?

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