Fixation Of Alleles And The Red Queen's Race

Suppose we have a species with two alleles, A and B, of some gene; and that it is succeptible to some form of parasite (virus, bacterium, or what have you) which also has a gene with two alleles, which we shall also call A and B; such that the "A" form of the parasite is more effective at infecting the "A" form of the host species; and the "B" virus more effective at infecting "B" hosts.I have sometimes seen it suggested that such a situation would lead to an equilibrium, in both host and parasite, between the two sorts of allele. This is very far from the truth.It is, of course, true that such a situation has a point of equilibrium, but this is unstable. However close to equilibrium we start, the allele frequency of any of the alleles (host or parasite, A or B) when plotted against time, describes a curve which tends from having a shallow and sinusoidal shape (if we start reasonably close to equilibrium) to a "square curve" the amplitude of which tends to the maximum possible value --- that is, as time goes on, the peaks of the graph of allele frequencies tend to 100%, the troughs to 0%.The graph below was produced by taking natural selection to be completely deterministic and the population to be continuous rather than discrete: this makes the effect most clear visually. In any real, finite population, then, one allele will eventually be fixed in the gene pool of the host species, and one will be fixed in the gene pool of the parasite. The winner of this leg of the Red Queen's Race will, of course, be the species that achieves fixation second.Simulations show that this applies also in cases where the host (as is usual in nature) has two copies of each gene, and one of the alleles is taken to be recessive.As we have some geneticists on this forum, perhaps they could tell me if this is a new observation or old hat. In any case, I found it interesting and I hope someone else does.Edited by AdminJar, : fix image width

Thread moved here from the Proposed New Topics forum.

Bump.I'd really like to hear from the geneticists on the forum. Are my results true, and are they original?

It would be easier to understand your question if you told us what assumptions went into the modeling of this host-parasite interaction as well as why do you think those would be thr right assumptions, and what method did you use to obtain the plot. This plot is particular interesting because it represents a pendulum-like behaviour (characteristic of a stable equilibrium, as oposed to exponential growth which is characteristic of unstable equilibrium) but the amplitude of the oscilation is increasing. I can`t tell what`s the meaning of that withou first having a better look into how you got that plot to begin with

What you are seeing is very similar to the cyclic population curves in predator-prey relationships, where one lags slightly behind the other. And it makes sense that it should be so in the host-parasite relationship: as the parasite 'A' genotype increases in frequency, the more infections of 'A hosts occur, and presumbaly more deaths, resulting in a drop in the frequency of the host 'A' allele. With fewer hosts to infect, the 'A' parasite genotype will decrease in frequency as well, which will allow more 'A' hosts to survive later, increasing the population, and starting the rises in both curves again.Of course, this is assuming the parasite has a detrimental effect on the host. Edited by Allopatrik, : No reason given.Edited by Allopatrik, : No reason given.

---

Natural Selection is not Evolution-- R.A. Fisher

The graph you present makes it look like the 'A' allele reaches fixation (100%). Maybe its just my tired eyes. While fixation is possible, it is very unlikely that the frequency would drop after that, since fixation of an allele means all other alleles at that locus have been eliminated from the population. Unless you specify that the 'B' allele is a recurring mutation of some kind (and at a very high rate as well), it would be highly unlikely to see cyclic fixations occurring. Edited by Allopatrik, : No reason given.

---

Natural Selection is not Evolution-- R.A. Fisher

I assumed that the parasite was bad for the host. If it was good for the host, that would be symbiosis, not parasitism.That was my only assumption.I messed around with the figures relating to how much effective parasitism hurts the host, and how much it benefitted the parasite, and whatever I did, I found the same sort of curve --- whatever figures you take, and however close you start to equilibrium, it goes away from the sinosidal curve wavering about the middle and towards the "square curve" with the top and bottom at 100% and 0%.I did this two ways. As I said in the OP, this graph is produced by assuming that natural selection is deterministic, i.e. acting on an infinite population. Before I did that, I modelled the same thing with a simulated small finite population, so that my model incorporated genetic drift and so that natural selection wouldn't always win. I didn't believe the results I got, so I did it again with an infinite population, and ended up with the same results.Perhaps I should post the actual algorithms I used, hold on a bit.---Thanks for asking questions. I was really pissed off that this sank unnoticed.--- Will the moderators forgive me if I use the word "bollocks"?What the graph shows is a dynamic which drives both allele frequencies further and further away from equilibrium.As I pointed out, there is an equilibrium, but it is unstable. The graph doesn't show an approach to an equilibrium unless you read the "time" axis backwards.Edited by Dr Adequate, : No reason given.Edited by Dr Adequate, : Correcting my grammar.Edited by Dr Adequate, : And my speeling. Speling. Spelling.

Yes, it looks a bit similar to predator-prey curves. But it doesn't actually work that way.Your common-sense view of what ought to happen is common sense, it's just not what actually happens when you do the math. I, too, was expecting a convergence to a stable equilibrium, but what you get is the graph I've shown, which gets further and further away from an unstable equilibrium, until we reach fixation of the alleles. Well, that's just what I mean by "parasite". If the parasite was beneficial, that would be symbiosis; and I'd have called it a "symbiote", and done different math. I used the word "parasite", rather than saying "virus" or "bacterium" especially because I wanted a word which implied that the organism was hurting its host.

Yes, it is.The graph shows a closer and closer approximation of both alleles to fixation, and does not suggest whether "A" or "B" will be fixed, or whether the host or the parasite will win. I don't think that you've followed my posts.When the "B" allele of the host is losing, that favors the "B" version of the parasite. Which favors the "A" version of the host. Which favors the "A" version of the parasite. Which favors the "B" allele of the host. Which means that the "B" allele of the host is winning. Oh, but we started off with the premise that it was losing.In the Red Queen's Race, losing is a winning strategy, and winning is a losing strategy.The point of my original post is that the result of this is that you do not get an equilibrium between the "A" and "B" alleles. One of them does in fact win.Edited by Dr Adequate, : No reason given.Edited by Dr Adequate, : No reason given.Edited by Dr Adequate, : No reason given.

There is no doubt about that. What I`m trying to get to is to find out whether this instability is a real effect, or an artifact from the procedure used to solve the problem (numeric instability is much more commom then some people realise)

quote:

---

Your common-sense view of what ought to happen is common sense, it's just not what actually happens when you do the math. I, too, was expecting a convergence to a stable equilibrium, but what you get is the graph I've shown, which gets further and further away from an unstable equilibrium, until we reach fixation of the alleles.

---

I never said one could expect a stable equilibrium. Where did you get that idea? The cycles seen in predator-prey relationships can last for long periods, but that depends on a lot of conditions. Predatory inefficiency, alternative food sources for the predator, etc., can enhance stability. Changes to them can disrupt that stability, resulting in one becoming extinct.It seems to me you have some implicit assumptions in your model (most notably an effect on the fitness/fertility of the host by the parasite, which isn't necessarily true for a parasite). That's ok, but it would be nice to know what those are. You did ask for comments, after all.

quote:

---

Well, that's just what I mean by "parasite". If the parasite was beneficial, that would be symbiosis; and I'd have called it a "symbiote", and done different math. I used the word "parasite", rather than saying "virus" or "bacterium" especially because I wanted a word which implied that the organism was hurting its host.

---

Parasitism is a form of symbiosis. And it usually is seen as being detrimental to its host, but would that detrimental effect necessarily apply to its fertility? If not, then your graph would look a lot different. It would also look different depending on the degree of effect on fitness the parasite has on the host. So, like I said above (and somebody else asked in another reply), it would be nice to know what your assumptions are.A Edited by Allopatrik, : No reason given.

---

Natural Selection is not Evolution-- R.A. Fisher

You were comparing the Red Queen's Race with predetor-prey oscillation, where there's an oscilation around an equilibrium. I'm sorry if I misunderstood you. I did, and I'm very sorry if my responses seem caustic or rebarbative. Or vice versa.But I am modelling the case where parasitism hurts the host. I agree that my model does not spply to symbiotic relationships. I am modelling the case where the "A" parasite hurts the "A" host, and the same for the "B" parasite and the "B" host. In the context of the theory of evolution, "detrimental" means a negative effect on fertility. That is, at least, what I'm modelling: that reproductive success for the parasite tends to reproduce the probability of reproductive success for the host.As I said, I'm not trying to model symbiosis, I'm trying to model parasitism. The graph turns out the same in the qualitative sense (i.e. it goes from a sinusoidal curve to a square curve) so long as you assume that parisitism hurts the host. That's my only assumption.As I said, I didn't believe this myself. I fiddled around with the figures for hours trying to find a set of figures that would make the equilibrium stable, because my intuition (i.e. my inborn stupidity) told me that this should be the case. I wrote two different programs because I thought that if I got it right, the result would be equilibrium.It isn't. It's fixation.

I guess I'd just better post the program.The way I annotate my programs is that the code comes first and the explanation comes after it.For example: ---begin
o:=0;
n:=1;
v[o]:=0.55;
h[o]:=0.50;{v and h stand for virus and hostThe choice of v and h as 0.5 and 0.55 is because the most conservative case is that we start close to equilibrium.}ch:=strtofloatdef(Edit5.text,0);
cv:=strtofloatdef(Edit6.text,0);{The previous two lines allow me to input the degree of the reproductive difficulty that an "A" type virus inflicts on an "A" type host; and the difficulties that a "B" type host imposes on an "A" type virus. Now I look at it, I find that the program assumes that "A" and "B" are mirror images: however, this is a conservative assumption.}for lp := 1 to 1000 do
begin
v[n] := (v[o]*h[o] + v[o]*cv*(1-h[o])) /
(v[o]*h[o] + v[o]*cv*(1-h[o]) + (1-v[o])*cv*h[o] + (1-v[o])*(1-h[o]));
h[n] := (h[o]*ch*v[o] + h[o]*(1-v[o])) /
(h[o]*ch*v[o] + h[o]*(1-v[o]) + (1-h[o])*v[o] + (1-h[o])*ch*(1-v[o]));{n stands for "new", o stands for "old".}image1.Canvas.Pixels):=clBlack;
image1.Canvas.Pixels):=clRed;
o:=1 - o;
n:=1 - n;
end;{We draw the graph and do the housekeeping.}end;Edited by Dr Adequate, : No reason given.Edited by Dr Adequate, : No reason given.

Would it be possible to walk us through the derivation of this algorithm?v[n] := (v[o]*h[o] + v[o]*cv*(1-h[o])) /
(v[o]*h[o] + v[o]*cv*(1-h[o]) + (1-v[o])*cv*h[o] + (1-v[o])*(1-h[o]))

---

Natural Selection is not Evolution-- R.A. Fisher

You modeling efforts are very impressive and I have enjoyed this thread a lot. Comments on your OP have been good, too. Just wondering: Should it matter if your parasite-host or predator-prey model accounts for K-selection or r-selection? (A parasite-host relationship would seem to involve r-selection, while a predator-prey relationship more likely involves K-selection.) Or possibly your model assumes both kinds of selection without needing to differentiate them?”HM