# Question 741a4

Sep 28, 2017

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

#### Explanation:

The Hardy-Weinberg equilibrium shows allele frequency in a population.

p represents the percentage of alleles that are of the dominant trait. q represents the percentage of alleles that are of the recessive trait.

Therefore, $p + q$ should equal 100%# or $1$.

Since ${1}^{2} = 1$, we can square $\left(p + q\right)$ and have it still equal $1$.

${\left(p + q\right)}^{2} = {p}^{2} + 2 p q + {q}^{2}$

${p}^{2}$ is equal to the percent of organisms homozygous for the dominant allele. $2 p q$ is equal to the percent of organisms that are heterozygous. ${q}^{2}$ is equal to the percent of organisms homozygous for the recessive trait.

When given any of these percentages- except for the percent of organisms that are heterozygous- one may solve for all the other values. This can tell you the genotypic composition of a population.