Intuiting the Strength of Negative Selection

One of the most nonintuitive things to me in population genetics is that large populations are better and faster at eliminating negative alleles. I still do not have a grasp on why this is, though I can certainly observe it in simulations.

Can any one explain why that is?


I don’t see why they should be faster either. Are they? Are you saying that some fraction of a population (a 3rd say) with a deleterious allele will get lost more quickly, on average, in a larger population than the same fraction is lost in a smaller one?

1 Like

This is my understanding, but I may be misreading quotes like this:

our results indicate that the increased efficacy of natural selection in a growing population results in a faster elimination of newly introduced mutations that are strongly deleterious.

Perhaps @Joe_Felsenstein can clarify for us what is really going on.

It says there the population is growing. I don’t think that’s the same as two stable populations with one being bigger than the other.

1 Like

Yup, the growing part is important. I don’t have time to read through this paper, though these are knowledgeable folks. Effects:

  1. As a population gets larger there are more copies of a gene in the population available to mutate, hence more mutations, though not more per individual.
  2. Population growth means that mildly deleterious mutants have more than 1 descendant copy, on average, so don’t get lost as quickly.
  3. There is less noise from genetic drift in a large population, but
  4. the individual mutants are each at lower initial frequency, and that nearly compensates for point 3

And, as Rumraket noted, it is important to keep separate in our
minds the effects of large population size and the effects of population growth.


Independent of growth, mutations with selection coefficients less than k/Ne (the k as some constant) are not easily eliminated. That means negative selection is more effective as Ne increases.

Why is that?

Both negative and positive selection are more effective in populations of larger size. Basically that is because there is more time for selection to shift gene frequencies directionally, before genetic drift eliminates that variability. That is, there is more time to have an overall average shift in gene frequencies toward those of higher fitness, before random fixation or loss occurs.


So my best intuition on this is that there is less stochastic elimination (or preservation) of variants in the opposite direction of selection. Is that correct?

If so, I can see a path to derive the the formula…but it is still against my intuition that it would be a sharp drop in selection efficiency at that point.

For that matter, would you please point me to a concise derivation?

We’re just talking good ol’ neutral theory here, right?

1 Like

Concise derivation? No, but the derivation(s) are sketched in my freely downloadable text Theoretical Evolutionary Genetics (just use the name as a search term). It is in Chapter VII.
A simpler way to explore this is to use Motoo Kimura’s 1962 formula for fixation probability with selection and genetic drift. When there is 1 copy of the allele initially, out of 2N copies of the gene in all, and the selection coefficient favoring the allele is s, the fixation probability is (1 - exp(-2s)) / (1 - exp(-4Ns)), where s can be either positive or negative. As s approaches zero from either side, the fixation probabilities approach 1/(2N), the value for a neutral mutant. Compute some values. Plot the fixation probability against s, say on a logarithmic scale on the s axis. Report the values to us.


Yes. I know the finding, but the intuition eludes me.

Thank you.

Well that is a simple proof, just making use of limits. No need to plot values, as it’s pretty easy to derive. Except I have to know how Kimura derived that formula now!

Thanks for you pointers. I’ll look into it and see if I can make sense of it.

More general formula is U(p) = (1 - exp(-4Nsp)) / (1 - exp(-4Ns)) where p is initial gene frequency.

It’s derivation is in Chapter VII.


Specifically, section VII.8 of my book.

1 Like

As far as intuition goes, I think it can be decomposed into two reasons. One has been mentioned by @Joe_Felsenstein. If the deleterious allele is to escape extinction, it has to achieve high frequency in the population, and in a larger population that requires a longer trajectory than in a small population; with more steps, the global effect of the fluctuations at each step is smaller. That is, you’re looking at the average effect over a larger number of samples. The other reason is that, for a given allele frequency, the sampling fluctuation (i.e. the amount of genetic drift) in each generation is smaller: σ(f) = sqrt(f(1-f)/n). Since the allele is only likely to escape extinction if s << σ, larger n means smaller probability of escaping.


Okay. That makes perfect sense to me. That’s the general direction I was thinking, but more clear. It’s the law of averages.

So why is there a steep cutoff in efficacy as a function of the selection coefficient? (My guess is this has to do with extreme value distributions).

The cutoff is not extremely steep. An example is when there is a favorable mutation with selection coefficient s in a population of effective size N. For smallish s the fixation probability of the mutant is approximately 1/(2N) + s. You can get a sense from this that selection becomes effective as s exceeds 1/(2N) but it is not an extremely sharp cutoff.


How steep is the cutoff?

1 Like

I’m busy this morning. Stop asking questions, folks, and instead compute some numbers with the fixation probability formula and display them here. I’m done.


Was doing just this. Thanks for the help. I got what I needed.

Can you post a table of fixation probabilities here for the others who were concerned about this?

1 Like