Hardy-Weinberg Question #2

Question:

Albinism is a recessive trait. If the frequency of albino (homozygous recessive) rats in a population is .16, what are the frequencies of homozygous dominant genotypes, and heterozygous genotypes?


Answer:

The most important thing to understand about the Hardy-Weinberg equation is that it writes the GENOTYPE frequencies of a population (group) in terms of the frequency of the alleles that make it up.

Some clarification:
(1) An allele is a version of a particular gene.
(2) Frequency refers to the "popularity", in this case, of the allele. For instance, if there are 100 alleles and 60 of them are "A", then A's frequency is 60/100 = 0.6

So recall that the Hardy-Weinberg equation is:

p² + 2pq + q² = 1

Also note that, by convention, 'q' is the recessive allele, and 'p' is the dominant one. Therefore the homozygous dominant is the p² term, the heterozygous genotypes are the 2pq term, and the homozygous recessive is the q² term.

The last bit of background information you need to know is that for a recessive allele to show it's effect, the genotype of the animal can't have a dominant allele on board as well. In other words, to be an albino, you must be homozygous recessive for the albinism gene. So,

q² = 0.16

Therefore q = 0.4

The other crucial equation is:

p + q = 1

Now, we already know q, so we can substitute it in:

p + 0.4 = 1

Therefore q = 0.6

Now, we know both p and q. Therefore we can work out the rest of the question:

Homozygous dominant = p² = (0.6)² = 0.36

Heterozygous = 2pq = 2(0.4)(0.6) = 0.48

Note that you can check your answers by substituting them into the original Hardy-Weinberg equation:

p² + 2pq + q² = 1
0.36 + 0.48 + 0.16 = 1