I am pretty bad in thinking quantitative genetics models. I am trying to get some basic understanding of modelling the evolution of a quantitative trait. I am therefore asking for help to analyze a very simple model. I welcome any explanation of some other classic model of the evolution of quantitative traits.


Consider a haploid population of constant size $N$. The fitness of the individuals is determined exclusively by a single quantitative trait $z$. $l$ loci codes for this trait. The genetic value at each locus adds up to give the quantitative trait $z$. The mutation rate at each locus is $\mu$. A mutation changes the genetic value at any given locus by $+ \Delta z$ with probability $\frac{1}{2}$ and by $- \Delta z$ with probability $\frac{1}{2}$. If you prefer to consider the effects of a mutation to be drawn from a normal distribution with mean $=0$ and variance $= \sigma^2$, please feel free to do so. The fitness $w(z)$ of an individual is given as a gaussian function of its quantitative trait $z$


where $V$ is the strength of selection. The optimal phenotype is $0$, that is $w(0)=1$. The larger is $V$, the more severe it is to be at a given distance to the optimal phenotype. If you want to assume that the fitness $w(z)$ is some other function of the trait $z$, such as the even more simple $w(z)=z$, please feel free to do so. For simplicity, we assume no environmental variance on the quantitative trait.


What is the equilibrium mean fitness $\bar w$ of the population?

What is the equilibrium genetic variance (=phenotypic variance) for fitness $w$ and for phenotype $z$?

  • $\begingroup$ What have you gotten, so far? This reads like a homework question. $\endgroup$
    – blep
    May 6, 2015 at 0:51
  • $\begingroup$ I would have love receiving this kind of homework. It is not a homework question. Because I was trained in Europe, the last homework I had to do was probably during the first year of my bachelor degree or even in High school. For the moment, I pretty much developed the basis of a model (presented in the post) to think the evolution of quantitative trait but this is it, I fail to bring it any further to answer my question(s). $\endgroup$
    – Remi.b
    May 6, 2015 at 2:06
  • $\begingroup$ @dd3 This is a little complicated than a usual homework. $\endgroup$
    May 30, 2015 at 14:22

1 Answer 1


Trait z is represented by k genes: z1....zk (I am using k instead of l because the former is visually differentiable from 1 ). For simplicity let's assume that there is only one mutable site in a gene. So a mutation can impart a change of $\pm \Delta z$. Starting from initial state of z at zero, the system will proceed to equilibrium where the rate of forward mutation would be same as that of backward mutation; since the rates are same the deterministic steady state of z should be zero.

A mutation event can also be considered a binomially distributed RV. After n events, the mean number of forward mutations would be 0.5×n.

 Mean fitness = mean forward mutation — mean backward mutations = 0 

You can also model the mutation events as a simple random walk. The mean of that is 0 and the variance is n. But what about large n; I am not sure. You should read about random walks or ask in CrossValidated. With $t\to \infty$, you would get a lot of variants but variants would be filtered out by selection (as the optimum is at 0). I think selection can be incorporated in the random walk with death rate being proportional to the distance from 0. But again I do not have much experience with such hybrid models.

I tried a simulation with n=10000000 (using Monte Carlo); different colours denote different runs (or in another words different loci). Y-axis denotes the value of z.

enter image description here

  • $\begingroup$ This is interesting, I didn't think about modelling this with a diffusion process. A diffusion equation contains 2 terms, "diffusion" (equivalent of genetic drift) and "drift" (equivalent of selection). An issue with this methodology though is that we can only track the mean trait and the variance in mean trait (the RV mean trait is normally distribution if the process is a random walk) of the population and not the variance in the trait within a population. +1 $\endgroup$
    – Remi.b
    May 30, 2015 at 15:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .