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

Mar 25, 2017

Well, let's check.

#### Explanation:

Assuming Hardy-Weinberg conditions are met, we will use the $\text{Hardy-Weinberg equation}$ which is:

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

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

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

Using $p + q = 1$ where:

$p = \text{frequency of dominant allele}$
$q = \text{frequency of recessive allele}$

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:

${p}^{2} = 0.46$
$\sqrt{{p}^{2}} = \sqrt{0.46}$
$p = 0.68$

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