# What is an example hardy-weinberg equation practice problem?

Jan 22, 2016

A question could look like:

If $m a t h b f 98$ out of $m a t h b f 200$ individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?

An explanation, walked through:

The Hardy-Weinberg equilibrium is a mathematical relationship of the alleles and genotypes in a population that meets certain characteristics. The relationships are as follow:

Alleles: $p + q = 1$

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

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

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

From the question, we know that $98$ of $200$ individuals express the recessive phenotype. This means that these $98$ also have the homozygous recessive genotype, the frequency of which is equal to ${q}^{2}$.

To determine what the actual frequency is, simply divide $\frac{98}{200} = 0.49$. We now know that ${q}^{2} = 0.49$.

However, we wish to find the frequency of the population that is heterozygous, which is equal to $2 p q$. So, we must find both $p$ and $q$.

Finding $m a t h b f q$:

${q}^{2} = 0.49$

Take the square root of both sides.

$q = 0.7$

(This means that 70% of the alleles in the system are recessive alleles.)

Now that we've found the value of $q$, we can find the value of $p$ using the allele equation.

Finding $m a t h b f p$:

Through the equation $p + q = 1$, substitute in $q = 0.7$.

$p + 0.7 = 1$

Subtract $0.7$ from both sides to see that

$q = 0.3$

Finding the frequency of heterozygotes:

$\text{frequency of heterozygous genotypes} = 2 p q$

Substitute the known values for $p$ and $q$:

$\text{frequency of heterozygous genotypes} = 2 \left(0.3\right) \left(0.7\right) = 0.42$

Converting this into a percent, we see that 42% of the population is heterozygous.