Question

There are 100 cats in a population that follows hardy-weinberg conditions. 4 cats display a recessive white coat color. how many of the cats are heterozygous?

Answer 1

We know that Hardy-Weinberg conditions include the following equations:

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

where $p+q=1$

And where p = dominant, and q = recessive; this means that $p^{2}$ is equal to the homozygous dominant, $2pq$ is the heterozygous, and $q^{2}$ is the homozygous recessive .

So we have 100 total cats, with 4 having the recessive white coat color. That means we have a ratio of $\frac{4}{100}$ or 0.04. Let that equal our $q^{2}$ value.

So when we solve for q, we get:

$q^{2}=0.04$

$q=\sqrt{0.04} =0.2$

Now that we have our q value, we can use the other equation to find p:

$p+q=1$

$p+0.2=1$

$p=0.8$

So then we can solve for our heterozygous population:

$2pq=2(0.8)(0.2)=0.32$

This is the ratio of the population. So we then multiply this number by 100 to get the number of cats that are heterozygous:

$0.32*100=32cats$

So now we know that there are 32 heterozygous cats in the population.