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Say you have a genotype A that produces x offspring and another genotype B that produces y offspring, where x>y. These x offspring are of genotype A but with modest differences in fitness due to mutation and these y offspring are of genotype B but with modest differences in fitness due to mutation. How would you model how many offspring these x offspring can themselves produce and how many offspring these y offspring can themselves produce as a function of the quantity their original parents produced?

Obviously, you could say that each descendant of genotype A produces x and each descendant of genotype B produces y, but that would unrealistically benefit genotype A, since realistically, one of B’s offspring could have a chance to be fitter than A’s by developing a favorable mutation or one of A’s offspring could have a chance to be less fit than B’s by developing an unfavorable mutation.
You could say that each descendant of genotype A and B produce (x+y)/2, but this would not be fair to genotype A, since it is fitter, so its offspring would probably be fitter than B’s offspring.

You could use the breeder’s equation and say that each of A’s offspring produce xh+m(1-h), where h is the heritability of fitness and m is the population mean, and that each of B’s offspring produce yh+m(1-h). But this is also unfair to genotype B because all of its offspring are less fit than all of A’s offspring, since x>y. So, what’s a way of modeling how resultant mutations might occur, such that B has a chance of producing some offspring that are fitter than A’s offspring?

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  • $\begingroup$ A few questions: 1) you seem to be conflating fitness and number of offspring. How sure are you that this is a good idea, if only on a conceptual level? 2) "m" is population mean of what, fitness=number of offspring? Does this mean "m" changes as the makeup of the population changes, or is it just the mean of "x" and "y"? 3) if "h" is nil then A's offspring produce "x" offspring and if it's 1 they produce "m"; shouldn't it be the other way around? $\endgroup$
    – Oosaka
    May 6, 2017 at 1:21
  • $\begingroup$ 4) You seem to ignore the variation you postulated within your populations. It all depends on that; for example if the 2nd-gen fitnesses are normally distributed with a mean of the parent gen fitness, then "but what about B's offspring that are fitter than A's" doesn't necessarily matter, since B's less-fit offspring would be less fit than A's least fit, and A's fitter offspring would be fitter than B's fittest (i.e it will come down to the mean, which is x and y). And if the distributions between the two are different then all bets are off. Surely this needs to be in the model somewhere. $\endgroup$
    – Oosaka
    May 6, 2017 at 1:25
  • $\begingroup$ (for example you said each descendant of A producing x and each descendant of B producing y unfairly benefits A, but if the variation in fitness is that half the offspring produce "parent fitness + z" and half produce "parent fitness - z", then do the calculation you'll find it's equal to if each descendant of A produced x and each descendant of B produced y). $\endgroup$
    – Oosaka
    May 6, 2017 at 1:31
  • $\begingroup$ Thank you for the questions. 1) My understanding of fitness is that it has a number of definitions depending on context and I think my usage of the term is okay. Fitness covaries with the ability to produce offspring. 2) Yes, m is population mean of fitness (or number of offspring). m should change as the makeup of the population changes. I am not suggesting I should use heritability in this way though. Just gave an example of what I'm trying to get at. 3) You're right. I apologize. Will edit it. $\endgroup$
    – sterid
    May 6, 2017 at 1:36
  • $\begingroup$ 4) A's fittest offspring wouldn't necessarily be fitter than B's fittest. It would just be more likely. B's less fit offspring would not necessarily be less fit than A's less fit (especially since A would have more offspring). And yes, that would also be unfair to B because it wouldn't give B the possibility of having more grandoffspring or greatgrandoffspring like some kind of random process would. $\endgroup$
    – sterid
    May 6, 2017 at 1:40

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So, what’s a way of modeling how resultant mutations might occur, such that B has a chance of producing some offspring that are fitter than A’s offspring?

Here's one possible solution that fits your criteria. You could have something like this (with "d" being the distance between "x" and "y", i.e. d=x-y and "n" being the n-th offspring, so it varies between 1 and x for A and 1 and y for B) :

A: n -> x + r(n), where r(n) is a random number between x-d and x+d
B: n -> y + r(n), where r(n) is a random number between y-d and y+d

(I write r(n) not because r is a function n, but because a different number is drawn for each n. I also didn't specify r's distribution, that would also need to be decided when making an actual model)

This is your situation where some offspring of B have a chance of being fitter than offspring of A, and the least-fit offspring of B can even be fitter than the fittest offspring of A. I think it fits your criteria and thus answers your question. The sum of the grandoffspring is basically the basic x or y squared (that you would get if each offspring had the same fitness as their parent) + the sum of all the "r"s.

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  • $\begingroup$ OK thank you for the effort you made. Can I upvote your answer but not give it a check because I would like to leave the question open for other ways of modeling this (perhaps ways people have seen in the literature)? Can there be multiple "accepted" answers in a thread? $\endgroup$
    – sterid
    May 6, 2017 at 3:12

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