Having thought about this for a week, I think I’ve realized what the disagreement was about, and where I went wrong in expressing what I was thinking. I didn’t distinguish between long-term averages and short-term averages, which was a mistake. My statements were wrong if we only knew short-run frequency but I believe were correct if we knew long-run frequency.
Consider the prevalence of a gene in a finite population in each generation as a Markov chain. If you start off in state i, that is i individuals having the gene, the sum of the transition probabilities to lower numbers represents the chance that there are less individuals with that gene in the next generation, the sum of the transition probabilities to higher numbers represents the chance that there are more individuals in that gene in the next generation, and the remainder (the transition probability back to i) is the chance that nothing changes.
Selection pressure is related to the chance that the gene becomes less frequent compared to the chance that it becomes more frequent, and depends on the frequency. One can easily imagine situations in which a gene is pushed towards a frequency that isn’t 0 or 1, but instead, say, .3, and so has positive selection pressure below that and negative selection pressure above it. (Frequency = Prevalence / Population Size)
For a deleterious gene, it can easily be the case that for all positive states the chance of transitioning downwards is greater than the chance of transitioning upwards. Even in that case, the stationary distribution of states on that chain will have a positive mean (because there are no negative states). Consider a two-state system, with p(1|0)=.1 and p(0|1)=1. The stationary distribution is (10/11, 1⁄11), with mean 1⁄11.
As the the gene become less deleterious- the selection pressure becomes less negative- the stationary distribution will spread upwards. We expect the long-term mean will increase, and are less surprised to find the system in a state where a larger population is carrying that gene than we were before. In the same example, if we change p(0|1) to .9 then the stationary distribution is now (.9,.1), with mean .1.
Under such a view, it’s obvious that when you decrease the selection pressure on a rare, deleterious condition, the long-term average of individuals with such a condition will increase, but will not grow to dominate the population.
Having thought about this for a week, I think I’ve realized what the disagreement was about, and where I went wrong in expressing what I was thinking. I didn’t distinguish between long-term averages and short-term averages, which was a mistake. My statements were wrong if we only knew short-run frequency but I believe were correct if we knew long-run frequency.
Consider the prevalence of a gene in a finite population in each generation as a Markov chain. If you start off in state i, that is i individuals having the gene, the sum of the transition probabilities to lower numbers represents the chance that there are less individuals with that gene in the next generation, the sum of the transition probabilities to higher numbers represents the chance that there are more individuals in that gene in the next generation, and the remainder (the transition probability back to i) is the chance that nothing changes.
Selection pressure is related to the chance that the gene becomes less frequent compared to the chance that it becomes more frequent, and depends on the frequency. One can easily imagine situations in which a gene is pushed towards a frequency that isn’t 0 or 1, but instead, say, .3, and so has positive selection pressure below that and negative selection pressure above it. (Frequency = Prevalence / Population Size)
For a deleterious gene, it can easily be the case that for all positive states the chance of transitioning downwards is greater than the chance of transitioning upwards. Even in that case, the stationary distribution of states on that chain will have a positive mean (because there are no negative states). Consider a two-state system, with p(1|0)=.1 and p(0|1)=1. The stationary distribution is (10/11, 1⁄11), with mean 1⁄11.
As the the gene become less deleterious- the selection pressure becomes less negative- the stationary distribution will spread upwards. We expect the long-term mean will increase, and are less surprised to find the system in a state where a larger population is carrying that gene than we were before. In the same example, if we change p(0|1) to .9 then the stationary distribution is now (.9,.1), with mean .1.
Under such a view, it’s obvious that when you decrease the selection pressure on a rare, deleterious condition, the long-term average of individuals with such a condition will increase, but will not grow to dominate the population.