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What is the probability that there are no differences if you compare the DNA of the ancient sample to a single DNA sequence from a modern individual? Assuming the individual was diploid and lived T generations ago.

How would you go about finding this probability? I would assume that drift and maybe selection would change the sequence and the probability of no change over time would be extremely low.

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Expected number of substitutions

Let's assume a haploid mutation rate of $\mu$ and a population of constant size $N$. For simplicity, I will assume an absence of selection because otherwise, we would need to talk about what kind of selection you want to talk about.

There are therefore $2 N \mu t$ mutations since this last ancestor. A fraction of $\frac{1}{2N}$ of them drifted to fixation. Therefore, there were $\frac{2 N \mu t}{2N} = \mu t$ mutations fixed.

If we consequence a sequence with a mutation rate of $\mu = 10^{-4}$ (about 10 kilo bases if is a human genome) and $t = 10^5$ generations. Then, there were 10 substitutions (on average).

Of course, the assumption of no selection is likely wrong and violation of this assumption could well drastically affect this estimate. However, you would need to perfectly describe what type of selection pressures (and what selection coefficient and dominance coefficient) you want to consider for more calculations.

Probability of k substitutions

As events of substitutions are independent, the probability that exactly $k$ substitutions happen given the expected number of substitutions ($\mu t$) is given by the Poisson distribution with rate $\mu t$. The probability $P(k | \mu, t)$ of $k$ substitutions had happen in $t$ generations, given the mutation rate $\mu$ is therefore

$$P(k | \mu, t) = \frac{\left(\mu t\right)^k e^{-\left(\mu t\right)} }{k!}$$

Per consequence, the probability of no mutation is obtained by replacing $k$ to zero

$$P(k=0 | \mu, t) = e^{-\left(\mu t\right)}$$

Again, if $\mu = 10^{-4}$ (about 10 kilo bases if is a human genome), then the probability that zero mutation had happen in $t = 10^5$ generations is $e^{-10^{-4} 10^5} ≈ 5\cdot 10^{-5}$

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  • $\begingroup$ Thanks, that makes sense. So would the probability there are no changes just 1-μt? I.e. 1-the probability of mutation, $\endgroup$ – Frank fras Dec 18 '17 at 4:29
  • $\begingroup$ @Frankfras I only gave you the expected number of substitutions. To compute the probability of $k$ substitutions, you need the Poisson distribution. See edit for more info. $\endgroup$ – Remi.b Dec 18 '17 at 15:25

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