Case study in population biology. Sequences and series, find general solution.

Question

Think about a population of individuals which all have two set of chromosomes (as human do). There is one gene that codes for a helping behaviour. This gene has two alleles (an allele is a variant of a gene. Example: Gene: Eye color, Allele1: blue eyes, Allele2: Braun eyes): Allele H (for Helping) and allele NH (for non-helping). I am looking for a function that describes the number of alleles of a given type in an population of infinite size at time t.

For example: N(NH, H, t) is the number of alleles NH in a population that was fixed for H(meaning that only Hexisted) at time t

At time t=-1 the whole population is fixed for allele H (N(NH,H,-1) = 0) by definition. At time t=0, a mutation had occured and one allele in the population is NH N(NH, H, 0) = 1 by definition. At time t=1, the number of copies of allele NH depends on the fecundity of individuals carrying the mutant allele NH.

In order to find the fecundity of the individuals carrying the different alleles we need to understand what these alleles cause. Hcauses their a helping behaviour toward their parents. an individual carrying one or two allele H helps its two parents and get a cost cin fecundity. The parents of one individual that carries one or two allele H have a benefit bcoming frmo the help they received from one offspring. Therefore the fecundity of someone depends on the number of parents (which is always 2 because nobody ever die in the population (by definition)), the number of offsprings (which vary), the benefit of the behaviour b, the cost of the behaviour c and their own genotype (allele combination) and the genotype of their offspring which itself. We'll consider that half of the offsprings of a mutant carry the mutant allele. The baseline fecundity (from which we substract c or add b) is g.

summary:

• b= benefit
• c= cost
• g= baseline
• Nobody dies!
• NH,NH individuals has no cost c (and might have benefits)
• H,H and H,NH individuals has a cost c (and might have benefits)
• the mutant allele is trasnmitted to half the offsprings

We can as well calculate the number of copies of an allele NH1 in a NH2population even if carrying the NH1 does not make any difference of behaviour than carrying the NH2allele

So there are 4 formulas to be found: $$N(NH1, NH2, t)$$ $$N(NH, H, t)$$ $$N(H, NH, t)$$ $$N(H1, H2, t)$$

I think that:

$$N(NH1, NH2, t) = 1 + \sum_{i=1}^{t} \sum_{j=1}^{i} \left(\frac{g}{2}\right)^j$$ Does it seem correct to you? Step by step: $$N(NH1, NH2, t=0) = 1$$ $$N(NH1, NH2, t=1) = 1 + g/2$$ $1$ is the first mutant individual. $g/2$ is its number of offsprings divided by 2 because only half the offsprings carry this NH1 alleles. $$N(NH1, NH2, t=2) = 1 + g/2 + g/2 + \left(\frac{g}{2}\right)^2 = 1 + g + \left(\frac{g}{2}\right)^2$$ $1$ is the first mutant, $g/2$ the alleles in the offspring created during the previous time step, $g/2$ is the copies produced by the first mutant at this time step, and (\frac{g}{2})^2 is the copies of alleles in the offspring of the offsprings of the first mutant!

The other formulas are harder to find for me. I guess it should looks roughly like the one below but I can't figure out. $$N(H,NH,t) = 1 + \sum_{i=1}^{t} \sum_{j=1}^{i} \left(\frac{g-c+b\cdot((i-j)\cdot(g-c)/2)}{2}\right)^j$$

Step by step:

$$N(H, NH, t=0) = 1$$ $$N(H, NH, t=1) = 1 + \frac{g-c}{2}$$ $$N(H, NH, t=2) = 1 + \frac{g-c}{2} + \frac{g-c + b\cdot\frac{g-c}{2}}{2} + \left(\frac{g-c}{2}\right)^2$$

What are the 4 formulas? (one is already found if I haven't made a mistake)

Thanks a lot!

UPDATE: A solution for N(H,NH,t) might be found by solving this question !