Expert: Dana Krempels, Ph.D. - 3/19/2008

Question
I'm not sure if you can help me; my maths is awful and I haven't studied for about 20 years; I am currently undertaking a BSc in the UK and am doing pretty well....until I hit Hardy-Weinberg !! It doesn't matter what I look at I just don't understand it. Is there any simple way of understanding it and answering the questions. I tried understanding your notes and got lost with the questions on the squirrels. I just don't understand the mathematical formula. HELP!! I get lost when you ask about genotype frequencies and when you subtract for example 0.6 from 1. The more I read it the more confused I get.

Answer
Dear David,

I'm not sure I can help, but I'll try to describe this as simply as possible, and then you can ask me any questions you might have.

The Hardy-Weinberg equation is this:

p2 + 2pq + q2

(this is a text program, and so won't do exponents, but you know that the p2 and the q2 stand for p squared and q squared, respectively).

The formula is used as a way to predict the expected frequencies of the three possible genotypes found in a population with two alleles for a particular gene locus.  (Remember:  you're considering ONLY ONE GENE LOCUS when you are doing HW equilibrium calculations.)

The three quantities together must add up to 1.0, or 100% of all the individuals in the population.  For example, if you have a population of 100 individuals and the gene you're looking at has two alleles, A and a, then a certain proportion of the individuals will be AA, a certain proportion will be Aa, and the rest will be aa.  

Just for grins, let's arbitrarily say that you are able to look into the genes of your population, and you found that of the 100 individuals there were 25 who had the genotype AA, 50 who had the genotype Aa, and 25 who had genotype aa.  That adds up to 100.

The *frequency* of the AA genotype is expressed as p2.  That means if you have 25/100 individuals who are AA, then the frequency of that genotype in the population is 25%, or 0.25.

Similarly, 2pq (the frequency of genotype Aa) is 50% of the population, or 0.5 and q2 (the frequency of genotype aa) is 25%, or 0.25.

Notice again that p2 + 2pq + q2 add up to 1.0, or 100% of the individuals in the population.

The HW equilibrium equation is used to determine whether the gene locus you're looking at is undergoing evolution in the population you're studying.  At a given starting frequency of the two alleles, the HW equation can be used to predict the *expected* relative frequencies of all three possible genotypes in a two-allele system.

In the example I gave you, it just so happens that these are the proportions expected in the population if half of the alleles in the population (50%, or 0,5) are A and half (50% or 0.5) are a.

Notice that if you subtract the frequency of A (50% or 0.5) from the total possible (100%, or 1.0), then you get 0.5, which is the frequency of a, the recessive allele.  If you're talking about all the alleles in the population (A or a), then their two frequencies *must* add up to 1.0 (or 100% of the alleles in the population).

Let's start with that.  Please go over this first, and let me know if there's any confusion, and we can clear that up before we move on.

Since I'm "on vacation" until the end of March, please write to me at dana@miami.edu if you need an answer immediately.

Hope this helps.

Dana