Why, you might ask, is this at all relevant? Here's why:
Cystic fibrosis (CF) is a devastating disease that is not uncommon amongst Europeans. In some European countries, about one in every 2500 children is born with the condition. It is an autosomal recessive disease, and the two alleles can be labelled C (dominant, normal variant), and c (recessive, potentially disease-causing). What are the odds of being a carrier for the CF gene in one of these countries?
You may be asking how it is possible to work this out, but it is actually rather easy. To start with, let's remind ourselves that in order to show signs of the illness in an autosomal recessive condition, both your alleles need to be the recessive, disease-causing ones. One dominant normal gene is enough to secure normality for yourself. However, if you happen to be a carrier (i.e. Cc), you won't be affected yourself, but you could very well pass on the defective CF gene to one of your children. And if your wife also passes one on, your child could be affected.
So, back to our equation of the moment:
Recall that p2 in the equation represents the genotype CC, q2 represents cc and 2pq represents Cc. Thus, the question is really asking us for find out the value of 2pq, the middle term. But hold, on we were told that the incidence of CF is one in 2500, or 0.0004. In other words,
q2 = 0.0004
Does this help us on our quest for the value of 2pq? Yes, of course: by simple maths (finding the square root of both sides), we can deduce one of the two variables:
q = 0.02
(Mathematician's corner: The actual answer is of course ±0.02, but a frequency of less than zero makes no logical sense, and so we'll stick with +0.02.)
Now we need to solve for the other variable, p. We'll need the help of that other important equation I mentioned in the last post:
p + q = 1
Plugging our known value of q in to the equation leaves us with:
p + 0.02 = 1
And so p = 0.98
Which of course, makes 2pq = 0.039 (roughly)
Done! Thanks to the Hardy-Weinberg equation, we now come to the somewhat startling conclusion that the genotype of Cc has a frequency of 0.039. In other words, a 3.9% of the population are carriers of the cystic fibrosis gene.
We could now, of course, easily work out the frequency of the CC (non-affected, non-carrier) genotype easily:
CC = p2 = (0.98)2 = 0.96 (= 96%)
So the Hardy-Weinberg equation, by cleverly describing genotype frequencies in terms of the their constituent allele frequencies, enables us to find the frequency for genotypes that we otherwise would have no clue about.
What to do next? I've found some very helpful websites that'll talk you through further examples.
First of all try this problem where they kindly walk you slowly through an example much like the above one.
If you've got the hang of it, sharpen your skills with this case example of albinism.
And if you've got no problems with either, you understand the Hardy-Weinberg equation. Congratulations!
There is one last topic we must briefly touch on before putting the whole matter to bed once and for all. That'll be the subject of the final post on this topic: the limits of the equation (or, "when the equation doesn't apply").