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Posed with a question that has me stuck:

Is increased variance of a phenotypic trait in a population associated with slower evolution of that trait?

I have to either 'disagree or agree' and argue my point. And I disagree. For this question, I think that if the variance of phenotypic trait has a genetic basis (additive genetic variance =/= 0) then increased additive genetic variance should speed up it's evolution. The greater the additive genetic variance, the greater the response to selection in the next generation. So increased additive genetic variance should increase the rate of evolution of that trait.

However, at the same time, if there is variation in a phenotypic trait but it does not have a genetic basis (narrow sense heritability is zero) and the basis for variation is due to an environmental, dominance, or interactive effect, then increased variation would have no effect in the rate of evolution. Or so I think. I'm not too sure how the rate of evolution would be affected in the case of a non-genetic basis of variation.

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The question can be understood in different ways

Causality

I think the question is a bit unclear in using the term associated in terms of what is being causal. Is the interest to understand how do processes that affect speed of evolution also affect phenotypic variance or is the question `how does a change in phenotypic variance affect the speed of evolution.

What is meant by speed of evolution

Speed of evolution has no strict definition. Or rather it has two standard definitions and it can be measured in Haldanes or in Darwins (which are units).

Quite often by "speed of evolution" authors mean "response to selection over medium to long term" as for example in the wiki article for rate of evolution.

Below I am answering to two different questions.

How does processes that affect "speed of evolution" at neutral sequences also affect phenotypic variance

At neutral sequences, the rate at which mutations fixes is equal to the mutation rate (see this answer for explanation). And all else being equal (typically without changing the mutational effect), high mutation rate leads to high phenotypic variance.

How does a sudden change in the phenotypic variance affect the response to selection in regard to this trait?

I think you pretty much said it all. I will just rephrase what you just said with a few simple equations.

Breeder's equation and heritability

The response $R$ to selection $S$ is given by

$$R=S h_N^2$$

, where $h_N^2$ is the heritability in the narrow sense. This above equation is known as the breeder's equation and is a fundamental result from quantitative genetics. The heritability in the narrow sense is defined as

$$h_N^2 = \frac{V_A}{V_P}$$

, where $V_A$ is the additive genetic variance and $V_P$ is the phenotypic variance.

Partitioning phenotypic variance

So it feels like increasing $V_P$ would decrease $h_N^2$ and therefore the response to selection $R$, but that would be very naive as $V_P$ is not independent of $V_A$. It is quite standard to decompose the phenotypic variance into genetic ($V_G$) and environment variances ($V_E$) $\left(V_P = V_G + V_E\right)$. Of course in reality, we should take into account variance due to other factors such as epistasis or developmental noise for example and all of their covariances. The genetic variance can further be decomposed into additive genetic variance $V_A$ and dominance genetic variance $V_D$ $\left(V_G = V_A + V_D\right)$.

Interdependence between $V_P$ and $V_A$

$V_P$ is a function of $V_A$. So as currently stated, it is impossible to answer the question by a simple yes or no. The shortest correct answer would probably be "The response to selection will decrease if the increased variance is not (not mainly) related to increase in additive genetic variance".

To be perfectly accurate, let $\Delta_P$ be the increase phenotypic variance and $\Delta_A$ be the increased genetic additive variance such that $\Delta_P > \Delta_A$. All else being equal the increase in $V_P$ will cause faster response to selection if and only if $\frac{Va+\Delta_A}{V_P+\Delta_P} > \frac{Va}{V_P}$. Isolating the term of interest $\Delta_P$, we got

$$\Delta_P < \frac{\Delta_A V_P}{V_A} = \frac{\Delta_A}{h_N^2} $$

In other words, all else being equal, an increase of phenotypic variance of $\Delta_P$ above the original variance $V_P$ increases the response to selection if and only if this increase is lower than the increase in additive genetic variance $\Delta_A$ multiplied by the inverse of the original heritability (heritability before a change in phenotypic variance occured)."

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  • $\begingroup$ I think the question is supposed to be " How does a sudden change in the phenotypic variance affect the response to selection in regard to this trait? " $\endgroup$ – google_doggle Oct 8 '16 at 22:46
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    $\begingroup$ Perfect, this is on this part of the post that I spent most time! Do you feel like it answers your question? $\endgroup$ – Remi.b Oct 8 '16 at 23:24
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Depends on how you phrase the question, ie, on how you write the breeder's equation.

The breeder's equation can be written in two ways:

$R=h^2 \cdot S$

or

$R=V_A \cdot \beta$

where $R$ is response to selection, $h^2$ is narrow sense heritability, $S$ is selection differential, $V_A$ is additive genetic variance, and $\beta$ is selection gradient.$^*$

Total phenotypic variance is pretty much a denominator which gets shunted between the right-hand quantities of the two versions, since $h^2=\frac{V_A}{V_P}$ and $\beta = \frac{S}{V_P}$. Rewriting the two versions, they become

$R=\frac{V_A}{V_P} \cdot S$

and

$R= V_A \cdot \frac{S}{V_P}$

So to answer your question: for a fixed selection gradient $\beta$, only the additive genetic variance $V_A$ matters. But for a fixed selection differential $S$, it is the ratio $\frac{V_A}{V_P}$ (narrow sense heritability) which matters.


$^*$The selection differential is the covariance between relative fitness and the trait value: $S=cov(w,z)$. The selection gradient is the regression coefficient between relative fitness and trait value: $\beta=\frac{cov(w,z)}{cov(z,z)}$. Note that the denominator $cov(z,z)$ is just the covariance of trait value with itself, ie, the variance of trait value: hence, $\beta = \frac{S}{V_P}$.

The selection differential is often described as the difference between the overall first generation's mean phenotype, and the selected parents' mean phenotype; this is true for viability selection, where parents either simply do or don't reproduce, with no other differences in fecundity.

For multivariate evolution of correlated traits, the selection gradient is actually the partial regression coefficient of relative fitness and trait value: partial because it isolates the 'direct effect' due to selection on the trait (as opposed to the residual 'indirect effect' due to selection on other, correlated traits). The multivariate breeder's equation is $\vec{R}=\mathbf{G}\cdot\vec{\beta}$, where: $\vec{R}$ is a vector of responses to selection (each element being the response of one trait); $\vec{\beta}$ is a vector of selection gradients on those traits; and $\mathbf{G}$ is the additive genetic variance-covariance matrix (each element being the additive genetic covariance of one trait with another; the elements on the diagonal give the additive genetic covariance of the traits with themselves, ie, their additive genetic variances).

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Variance of a phenotype/genotype with respect to the rate of mutagenesis is an indicator of selection pressure. High variance would denote lower selection and vice versa. Note that this has to be normalized to the mutation rate; if you have high mutation rate then you would get high variance even in the presence of a moderate selection pressure.

With protein coding genes, the rate of selection is measured by comparing the number of non-synonymous mutation with the number of synonymous mutations (a gross indicator of general mutation rate). Se Ka/Ks ratio.

For more information you can have a look at this review by Gingerich (2009).

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