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 $"Hardy-Weinberg equation"$ which is:

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

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

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

Using $p+q = 1$ where:

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

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

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

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

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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