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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.)

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  • $\begingroup$ 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? $\endgroup$
    – Remi.b
    Jul 17, 2017 at 19:04
  • $\begingroup$ @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%. $\endgroup$
    – user66081
    Jul 17, 2017 at 19:05
  • $\begingroup$ But where did you get these numbers from? For example, I can't see the number 13.93 in the linked wikipedia articles. $\endgroup$
    – Remi.b
    Jul 17, 2017 at 19:06

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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.

How about HAR1?

Without looking for more accurate statement form the scientific literature, the same wikipedia article you cite says

HAR1A is active in the developing human brain between the 7th and 18th gestational weeks. It is found in the dorsal telencephalon in fetuses. In adult humans, it is found throughout the cerebellum and forebrain; it is also found in the testes.[1] There is evidence that HAR1 is repressed by REST in individuals with Huntington's disease, perhaps contributing to the neurodegeneration associated with the disease

It already gives you a vague idea of potential selection pressures on those sequences in the human genome.

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  • $\begingroup$ The rate is higher in sequences that undergo positive selection is a non-explanation to me. What is the mechanism? Is the mutation rate under "neutral selection" actually dampened (namely 13-fold?)? $\endgroup$
    – user66081
    Jul 17, 2017 at 19:12
  • $\begingroup$ 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? $\endgroup$
    – Remi.b
    Jul 17, 2017 at 19:17
  • $\begingroup$ Ok, so the "mechanism" is statistical: mutations do not easily result in substitutions because, being diluted in the population, they are weeded out. So in absence of selection pressure, the substitution rate is lower than the mutation rate. Yes? Maybe I'm not using the terminology correctly. $\endgroup$
    – user66081
    Jul 17, 2017 at 19:24
  • $\begingroup$ 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. $\endgroup$
    – Remi.b
    Jul 17, 2017 at 19:34
  • $\begingroup$ Ok, thank you. I upvote for potentially useful pointers, but cannot accept the current answer due to lack of clarity. $\endgroup$
    – user66081
    Jul 17, 2017 at 20:02

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