# HAR1 speedy mutation

From wiki:

Human accelerated region 1 (HAR1) is a segment of the human genome found on the long arm of chromosome 20. It is a Human accelerated region. [..] These 49 regions represent parts of the human genome that differ significantly from highly conserved regions of our closest [living relatives, the chimpanzees] in terms of evolution.

What could explain those differences?

A back of the envelope computation shows that in the course of 8mio years, at a rate of 64 mutations per generation, one can expect 1% of mutated basepairs. Indeed, the HAR1 of length 106 bp has 1.08 substitutions in the chimp. But humans have 13.93. (Data: see the table in this article.)

• Can you please show how you computed the 1%? When you state the HAR1 of length 106 bp has 1.08 substitutions in the chimp, you mean substitution compared to what species? But humans have 13.93 substitutions compared to chimps I suspect? Or maybe you made an attempt at reconstructing the hypothetical sequence of the ancestor and compared to it? Or maybe you referred to some paper who sequenced a fossilized genome? – Remi.b Jul 17 '17 at 19:04
• @Remi.b: yes, I was wondering also, but I think both numbers refer to the most recent common ancestor. At 64 mutations per generation over 400'000 generations, that's roughly 25mio mutations among 3bn bp, or roughly 1%. – user66081 Jul 17 '17 at 19:05
• But where did you get these numbers from? For example, I can't see the number 13.93 in the linked wikipedia articles. – Remi.b Jul 17 '17 at 19:06

Substitution rate at neutral sequences

The rate of substitutions at neutral sequences is given by the mutation rate. It si a very simple and classic result. It is because the probability of each new neutral mutation to fix is $\frac{1}{2N}$ (which you can calulate from a Moran branching process as well as from a simple Wright-Fisher model of genetic drift) and there are $2N\mu$, new mutations at each generation (where $N$ is the population size and $\mu$ the haploid mutation rate), resulting in a substitution rate of $2N\mu\frac{1}{2N} = \mu$.

Substitution rate at non-neutral sequences

A beneficial mutation has a probability higher than $\frac{1}{2N}$ to fix, while a deleterious mutation has a probability lower than $\frac{1}{2N}$ to fix. Therefore, the more beneficial mutations there are, the higher is the substitution rate. There are a number of approximations to the fixation probability of a non-neutral mutation. For example, using diffusion equations, one can approximate the probability of fixation of a deleterious mutation with selection coefficient $s$ by $\frac{1-e^{-\frac{4s}{N}} }{1-e^{-4Ns}}$ from some Kimura paper.

This rate of substitution therefore depends upon the selection scenario. It is lower than $\mu$ in conserved sequences and it is high than $\mu$ in sequences that undergo positive selection. So, the reason for such fast evolving sequences is just that they are under positive selection.

• I am not sure I understand what is unclear to you but let's try to make our way through it. A beneficial mutation has a probability higher than $\frac{1}{2N}$ to fix, while a deleterious mutation has a probability lower than $\frac{1}{2N}$ to fix. Therefore, the more beneficial mutations there are, the higher is the substitution rate. Does that clear things up? – Remi.b Jul 17 '17 at 19:17
• The exact number of substitutions is definitely is stochastic process. In absence of selection, the substitution rate is $\mu$. If mutations are often beneficial, then the substitution rate is higher than $\mu$. If mutations are deleterious, then the substitution rate is lower than $\mu$. You said So in absence of selection pressure, the substitution rate is lower than the mutation rate.. Well it depends upon whether selection is purifying or positive. – Remi.b Jul 17 '17 at 19:34