No history yet

Mutation Selection Balance

The Push and Pull of Evolution

In any population, two fundamental forces are locked in a constant tug-of-war. Mutation continuously introduces new alleles, while natural selection works to weed out the ones that are harmful. This dynamic creates a stable state known as mutation-selection balance, an equilibrium where the rate at which a deleterious allele is created is matched by the rate at which it's removed.

To understand this, let's define two key rates. The forward mutation rate, denoted by uu, is the rate at which a standard allele (let's call it AA) mutates into a different, often deleterious, allele (aa). The back mutation rate, vv, is the reverse process, where allele aa mutates back into AA. In reality, forward mutations that create non-functional alleles are far more common than back mutations that restore function. Therefore, we can usually assume that uu is much larger than vv.

This is known as mutation-selection balance.

If only mutation were at play, the frequency of allele aa, which we call qq, would change each generation. The gain in qq comes from AA alleles mutating to aa, which happens at a rate of uu times the frequency of AA (pp). The loss of qq comes from aa alleles mutating back to AA, at a rate of vv times the frequency of qq. The net change in qq from mutation alone is:

Δqmutation=upvq\Delta q_{\text{mutation}} = up - vq

When the population reaches equilibrium from mutation alone, Δqmutation\Delta q_{\text{mutation}} becomes zero. At this point, the frequency of aa, which we call its equilibrium frequency q^\hat{q}, is determined by the ratio of the mutation rates. But in the real world, selection doesn't sit idle.

Calculating the Equilibrium

Natural selection acts to remove deleterious alleles. The strength of this removal is measured by the selection coefficient, ss. Equilibrium is reached when the introduction of new alleles by mutation is perfectly balanced by their removal by selection. In other words, the change from mutation plus the change from selection equals zero: Δqmutation+Δqselection=0\Delta q_{\text{mutation}} + \Delta q_{\text{selection}} = 0.

The exact calculation depends on whether the deleterious allele is recessive or dominant. Let's start with a completely recessive deleterious allele, meaning it only impacts the fitness of homozygous individuals (aaaa).

For a harmful recessive allele, selection can only 'see' and act upon the small fraction of the population that is homozygous recessive (aaaa). Heterozygotes (AaAa) carry the allele without any negative effect, shielding it from selection.

At equilibrium, the rate of creation of new aa alleles (which is approximately uu, since most alleles are AA and thus p1p \approx 1) must equal the rate of their removal. Alleles are removed when they appear in homozygous recessive individuals, who have a fitness of w=1sw=1-s. The loss of alleles due to selection in this case is approximately sq2sq^2. Setting these two rates equal gives us:

usq^2u \approx s\hat{q}^2

Solving for the equilibrium frequency, q^\hat{q}, gives us a simple and powerful result for recessive alleles.

q^=us\hat{q} = \sqrt{\frac{u}{s}}

Now consider a deleterious dominant allele. In this case, selection acts against both the homozygous dominant (AAAA) and the heterozygous (AaAa) individuals. Because the allele is harmful even in a single copy, it's much more exposed to selection. The change in allele frequency due to selection is approximately sq-sq. At equilibrium:

usq^u \approx s\hat{q}

This leads to a different equilibrium frequency for dominant alleles.

q^=us\hat{q} = \frac{u}{s}

The Genetic Load

The presence of these deleterious alleles, even at low equilibrium frequencies, reduces the average fitness of the population compared to a hypothetical 'perfect' population with no harmful mutations. This reduction in fitness is called the genetic load or mutation load (LL).

The calculation is straightforward. For a recessive allele, the only genotype with reduced fitness is aaaa, which has frequency q^2\hat{q}^2. The reduction in fitness for these individuals is ss. The load is the total reduction in fitness across the population:

L=sq^2L = s\hat{q}^2

Remember that for a recessive allele, q^2=u/s\hat{q}^2 = u/s. Substituting this into the load equation gives a surprising result.

L=s(us)=uL = s \left( \frac{u}{s} \right) = u

This means that a mildly deleterious allele and a lethal one impose the same genetic load on the population, assuming they have the same mutation rate. A more harmful allele (larger ss) will have a lower equilibrium frequency (q^\hat{q}), but because its effect on fitness is so severe, the overall burden on the population remains the same.

Lesson image

Let's check your understanding of these concepts.

Quiz Questions 1/5

What is mutation-selection balance?

Quiz Questions 2/5

For a completely recessive deleterious allele, the equilibrium frequency (q^\hat{q}) is determined by the formula u/s\sqrt{u/s}, where uu is the mutation rate and ss is the selection coefficient. What does this imply?

This balance between mutation creating variation and selection filtering it is a cornerstone of population genetics, explaining why harmful genetic conditions persist and shaping the genetic makeup of all species.