What is an example hardy-weinberg equation practice problem?

A question could look like:

If $mathbf98$ out of $mathbf200$ 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: $p+q=1$

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

Genotypes: $p^2+2pq+p^2=1$

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

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 $98/200=0.49$. 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 $2pq$. So, we must find both $p$ and $q$.

Finding $mathbfq$:

$q^2=0.49$

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 $mathbfp$:

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:

$"frequency of heterozygous genotypes"=2pq$

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

$"frequency of heterozygous genotypes"=2(0.3)(0.7)=0.42$

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