*understanding*. It'll be in two parts; this first part is designed to answer the first question -

*what*the equation is. We'll derive the equation from first principles - it's pretty easy - because that's a great way to understand it.

Although this may not mean much at the moment,

**the Hardy-Weinberg equation describes the relation between gene allele frequencies and genotype frequencies within a population at genetic equilibrium.**

Now let's work our way towards understanding it.

Take two alleles of a gene, one of which is dominant (**A**) and the other is recessive (**a**). (The full theory actually doesn't require us to have one dominant and one recessive allele, or even to limit ourselves to two alleles. But it's much simpler to understand if we follow this explanatory convention.)

Now imagine a pool (yes, actually a pool) filled with the balls representing the two alleles - some '

**A**'s, the rest consisting of '**a**. What are the options for the possible phenotypes? In other words, if we dipped a cup into the pool and pulled out two alleles at random, what are the possibilities? Clearly, we could have:*'***AA**,**Aa**,**aA**, or**aa**.Ok, so far so good; nothing too interesting there. But now we ask ourselves a second question: if we happen to

*know*the frequency of the

**A**allele in the pool, can we predict what the odds of the getting any one of the particular 4 genotypes will be? In other words, if there are 100 balls (alleles) in the pool, and we know that 90 are '

**A**', can we predict the odds of getting (scooping up), say,

**AA**with our two balls?

Both Hardy and Weinberg independently saw that this was indeed quite possible - and useful. Following the conventional notation, let's call

**A**'s frequency '

*p*'

*.*Similarly, we'll call

**a**'s frequency '

*q*'. Using the above figures

*, p*would 0.9 (or 90/100), and

*q*would be 0.1 (or 10/100). Notice, by the way, that the

*frequencies*of both the alleles, no matter what they are, must obviously total 1. This is common sense - try it with different figures if you're not convinced. The algebraic representation of this would be:

(frequency of

**A**) + (frequency of

**a**) = 1,

Which is to say:

*p + q*= 1This equation will come in handy later. But back to our example. What are the odds of the first ball I scoop up out of the pool at random being

**A**? Clearly 0.9, or

*p.*The same goes for the second ball, when considered in isolation. Now comes the only slightly tricky bit of algebra: what are the odds of me getting both balls to be

**A**- which would correspond with the genotype

**AA**? The equation is:

0.9

*x*0.9 = 0.81or, in our general algebraic notation, for any value of *p*,

*p x p = p*

^{2}Similarly, the odds of randomly getting both balls being

**a**would be, specifically:0.1

*x*0.1 = 0.01, or in general form:*q x q = q*

^{2}The odds of getting genotype

**Aa**would, in identical fashion, be:

*p x q*, or

*pq*

And, to complete the story,

**aA**'s odds would be

*q x p*, or

*qp.*

But, come to think of it, the genotype

**Aa**is exactly equivalent to

**aA**, and

*pq*is obviously the same as

*qp*, so we can combine the two terms:

*pq + qp = 2pq*

We're nearly there now. Let's collect all the possible genotype frequencies. We have

*p*, 2

^{2}*pq*and q

^{2}as our options. Since together the genotypes that these symbols represent must make up 100% of the total genotypes, the frequencies must add up to 1. In other words:

*p*1^{2}+ 2pq + q^{2}=...which is, of course, the Hardy-Weinberg equation.

So let's summarise so far. This is the take-home message:

*The Hard-Weinberg equation describes the possible genotypes in terms of the allele frequency that goes into making them:*

- The homozygous dominant genotype (AA in this example) is represented as
*p*^{2}**A***allele*squared). - The heterozygous genotype (Aa in this example) is represented as
*2pq*. - The homozygous recessive genotype (aa in this example) is represented as
*q*.^{2}

This may seem a pointeless exercise, but writing the equation like this actually has several outstanding advantages.

That'll be the topic of the next post.

That'll be the topic of the next post.

That was a great intro! I have a lab practical with this stuff on it & I have been studying for days with no luck of understanding..suddenly it seems sooo easy!

ReplyDelete=) thanks doc

Glad to hear it!

ReplyDelete