Question

What is an example hardy-weinberg equation practice problem?

Answer 1

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.