Is the hardy weinberg equilibrium derived by using a model similar to the Wright Fisher model, just without assuming genetic drift and finite pop size? Both seem to use the same assumptions except Wright Fisher takes into account finite pop size and genetic drift. What is the model used to derive Hardy-Weinberg equilibrium called?


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When you talk about the 'Wright-Fisher model' I suppose you are referring to the 'Wright-Fisher model of genetic drift (WF)' involving the use of a binomial distribution.

You are correct that the two models make very similar assumptions. However those assumptions are generally speaking found in a large set of different models just because they often simplify the math. I think it is quite wise to consider these two models as very much independent.

Shared assumptions

Both the WF and the Hardy-Weinberg model (HW) assume

  • non-overlapping generation
  • panmictic population
  • absence of selection, migration and mutation
  • random mating
  • no sex-ratio bias

Of course, violation of some of these assumptions are relatively easy to model but such complication is often not taught in intro classes to population genetics.

Wright-Fisher model derivation

Let $2N$ be the total number of haplotypes, where $N$ is the population size. I here just assumed a diploid population but derivation for a haploid population is just as easy. Let $k$ be the number of occurrences of allele A, so that the frequency of the allele A is $p=\frac{k}{2N}$. Assuming sexual reproduction only, the number of occurrences of the allele A in the next generation is

$$k' = {2N \choose k} p^k (1-p)^{2N-k}$$

Quick side note: Effective population size

The variance of this distribution is $var(k') = 2N p (1-p)$. The variance of the allele frequency in the next generation is therefore $var(p') = var(\frac{k'}{2N}) = \frac{1}{4N^2} var(k') = \frac{2N p (1-p)}{4N^2} = \frac{p(1-p)}{2N}$.

Replacing $N$ by $N_e$ and solving for $N_e$ yields

$$N_e = \frac{p(1-p)}{2 var(p')},$$

which is the definition of the effective population size $N_e$.

Wright-Fisher model assumptions

The WF does not per say assume a finite population size, it is just that when the population size is infinite, the model loses all of its appeal!

Hardy-Weinberg model derivation

See here for a more complete discussion

The derivation of Hardy-Weinberg genotype frequencies is simple enough to not rely on any pre-existing model. Let's consider the simple version of the HW model, the one with a bi-allelic locus. The two alleles A and B exist at frequency $p$ and $1-p$ respectively. Imagine you have to draw an allele from this population. The probability to draw a given allele is equal to its frequency. Now draw a second allele. Assuming the population is large enough or assuming that selfing is allowed, the probability of drawing a given allele is still the same at the second draw than at the first draw.

As such the probability of drawing the allele A twice is simply $p p = p^2$. Similarly the probability of drawing the allele B twice is simply $(1-p)(1-p)=(1-p)^2$. The probability of first drawing the A allele and then the B allele is $p(1-p)$ and the probability of first drawing the B allele and then the A allele is $(1-p)p$. If we don't make the discrimination between AB and BA genotypes and just call them all AB, then the frequency of the AB genotypes is $2p(1-p)$. Of course $p^2 + 2p(1-p) + (1-p)^2 = 1$.

Hardy-Weinberg model assumptions

It is common to say that HW assume infinite population size. This is correct if you make from HW the interpretation that the genotype frequencies are exactly at $p^2$, $2p(1-p)$ and $(1-p)^2$ and that no evolutionary change will occur. But if you interpret HW as indicating the expected genotype frequencies from which deviation exist, then you do no make the assumption of infinite population size. Under this second interpretation, one can test deviation from HW expectations with a binomial test (or most often, when the population size is large enough with a $\chi^2$ goodness-of-fit test).

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