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I was reading the textbook Probability Models for DNA Sequence Evolution by Durrett. In chapter 1, he discusses the Wright Fisher model and the coalescent theory which I am interested in.

He defines heterozygosity as the probability that two copies of an allele are the same. In theorem 1.3, he derives the expected value of heterozygousity, but what he seems to derive is the expected value of the probability that two allelles have a common ancestor. This does not seem to be the same as heterozygosity to me. Am I missing something here?

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The following borrows from Gillespie, J.H., 1998. Population Genetics: A Concise Guide. The Johns Hopkins University Press. It is a good concise guide, and reduces to some extent the mathematics.

Theorem 1.3 is used to show the decay of heterozygosity in a population due to genetic drift.

Consider alleles in a population. Two alleles drawn at random can differ by state, meaning they have two different DNA sequences. In other words, they are different alleles. Or the randomly drawn can have the same state, meaning they are the same allele. For alleles that are identical by state, they could differ by origin, meaning they are the same allele from two different chromosomes. The two alleles in a homozygous individual differ by origin (sampled from maternal and paternal homologs) but are identical by state. Or, two identical alleles could differ by descent, meaning they are the same because they trace ultimately back to a common ancestor.

To calculate heterozygosity, you can first account for homozygosity, the frequencies of alleles that are identical by state. First consider two alleles identical by origin. If you draw two alleles identical by origin when t=1, the probability that the two alleles are copies of same original allele (from the same chromosome) is

$$\frac{1}{2N}$$

because there was only a single copy of that allele on that one chromosome in the population prior to the round of mating, so it is 1 out of the 2N alleles at time point 0 ($t=0$). The probability that two alleles identical by state are not identical by origin then is

$$1-\frac{1}{2N}.$$

Therefore, total homozygosity ($G'$) in the population (identical by state) after one round of random mating is

$$G' = \frac{1}{2N} + (1-\frac{1}{2N})G,$$

where $G$ is the homozygosity prior to the round of mating. Of course, anything not homozygous is heterozygous. So, heterozyosity ($H$) is simply,

$$H=1-G.$$

By substituting the formula for $G$ into the above equation, the equation for $H$ becomes,

$$H' = 1-G' = (1-\frac{1}{2N})H$$

where $H$ is the heterozygosity prior to the round of mating. This formula represents the change in heterozygosity after one round of random mating. This can now be expanded for several rounds of random mating. For example, the change from $t=0$ to $t=1$ is

$$H_1 = (1-\frac{1}{2N})H_0.$$

The change from $t=1$ to $t=2$ is,

$$H_2 = (1-\frac{1}{2N})H_1$$

Therefore, the change in heterozygosity from $t=0$ to $t=2$ is

$$H_2 = (1-\frac{1}{2N})^2H_0.$$

This can be expanded to any time $t$ with the equation,

$$H_t = (1-\frac{1}{2N})^tH_0,$$

which you will recognize as Durrett's Theorem 1.3.

In short, in the Wright-Fisher model, two alleles that have a common ancestor are by definition identical by state and so cannot contribute to heterozygosity.

I hope this helps. I was a bit rushed and have not explained this often (mostly to myself in my mind) so if I can clarify something, please let me know.

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  • $\begingroup$ Thank you for the brilliant answer. I think you have one typo though. Just above the second equation, I think it should be the probability of two alleles do not have the same origin is 1- 1/2N, rather than the probability of two alleles with the same state having different origins. $\endgroup$ – Devil Sep 3 '14 at 23:56

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