Within-Host Mutation and Selection of Viral Variants

Why Mutation and Selection Must Be Considered Together

A viral population can contain multiple variants with different sensitivities to a drug or immune response. A variant that is initially rare may increase in frequency if its production and clearance give it a relative advantage. Mutation can also continually supply that variant, even when it is disadvantaged. These are distinct processes: mutation creates variant genomes, while selection changes their relative representation through differences in net growth.

The distinction matters when interpreting viral trajectories. A resistant variant can become a larger fraction of the population while total viral abundance is falling. Conversely, a variant may have a relative advantage but remain rare for some time if it starts at low abundance. The model below separates these effects using variant-specific production and clearance rates.

A Two-Variant Model

Let $W(t)$ and $M(t)$ denote the abundances of a wild-type viral variant and a mutant variant. Each variant produces progeny at a per-capita rate $b_i$ and is cleared at a per-capita rate $c_i$, where $i$ is $W$ or $M$. These rates have units of inverse time, such as day$^{-1}$. A fraction $\mu$ of progeny produced by the wild type is assumed to carry the mutation that produces the mutant. Here $0 \leq \mu \leq 1$ is a probability per wild-type production event, not a rate per unit time. Assume no reverse mutation.

Definition: The net per-capita growth rate of variant $i$ is $r_i=b_i-c_i$. Under the assumptions above, the two-variant model is $$ \frac{dW}{dt}=(b_W-c_W-\mu b_W)W,\qquad \frac{dM}{dt}=(b_M-c_M)M+\mu b_W W. $$ The term $\mu b_W W$ is the mutation influx into the mutant population. The wild-type population loses that fraction of its progeny, while the mutant population receives it.

This model treats the rates and the mutation probability as constant over the period being considered. It also assumes that variant abundances are large enough for deterministic differential equations to be a useful approximation, and that production and clearance are proportional to abundance. It does not explicitly represent target-cell depletion, changing immune pressure, or stochastic extinction of a rare lineage. If those processes are important, the model may not describe the observed dynamics adequately.

The net growth-rate difference $s=r_M-r_W$ is the mutant’s selection advantage in this model. A positive $s$ means the mutant has a higher net per-capita growth rate than the wild type under the specified conditions. It does not by itself mean the mutant abundance is increasing: the mutant’s own net growth rate $r_M$ may still be negative.

Deriving the Change in Mutant Frequency

Define total viral abundance as $N=W+M$, and mutant frequency as $f=M/N$, provided $N>0$. The frequency describes the mutant’s share of the two-variant population. It is not the same quantity as mutant abundance $M$.

The frequency equation follows by differentiating $f=M/N$. Since mutation transfers progeny from the wild-type category to the mutant category, it does not add or remove total progeny in this model:

\begin{aligned} \frac{dN}{dt} &=\frac{dW}{dt}+\frac{dM}{dt}\\ &=(r_W-\mu b_W)W+r_M M+\mu b_W W\\ &=r_W W+r_M M. \end{aligned}

Using $f=M/N$, $W/N=1-f$, and $df/dt=(1/N)(dM/dt-f\,dN/dt)$, substitute the model equations:

\begin{aligned} \frac{df}{dt} &=\frac{1}{N}\left(r_M M+\mu b_W W -f(r_W W+r_M M)\right)\\ &=\mu b_W(1-f)+f(1-f)(r_M-r_W)\\ &=(1-f)(\mu b_W+s f). \end{aligned}

This expression makes the two contributions explicit. Mutation supplies mutants at a frequency-dependent rate $\mu b_W(1-f)$, while selection contributes $s f(1-f)$. The factor $1-f$ appears because, in this two-variant model, selection between the variants ceases to change frequency when either variant is absent. The mutation influx also vanishes when $f=1$, because there is then no wild-type population to supply new mutants.

Key result: In the absence of mutation, a mutant with $s>0$ increases in frequency whenever $0<f<1$, while a mutant with $s<0$ decreases. With one-way mutation from wild type to mutant, a disadvantaged mutant can persist at a nonzero frequency because mutation continually supplies it.

Worked Examples

Worked Example: Selection Without Mutation

Suppose two variants have the following hypothetical rates. The wild type has $b_W=4.0$ day$^{-1}$ and $c_W=3.6$ day$^{-1}$; the mutant has $b_M=3.5$ day$^{-1}$ and $c_M=2.8$ day$^{-1}$. First consider the selection-only comparison, with mutation ignored over the interval. Does a mutant starting at 1.00% increase in frequency over four days?

The net rates are $r_W=4.0-3.6=0.4$ day$^{-1}$ and $r_M=3.5-2.8=0.7$ day$^{-1}$. Thus $s=0.7-0.4=0.3$ day$^{-1}$. With mutation absent, the two abundances grow exponentially, so their ratio obeys

\frac{M(t)}{W(t)} =\frac{M(0)}{W(0)}e^{s t}.

An initial frequency of $0.0100$ corresponds to odds $M(0)/W(0)=0.0100/0.9900=0.01010$. After four days, the odds are $0.01010e^{0.3\times4}=0.01010(3.320)=0.03354$. Converting back to frequency gives $f(4)=0.03354/(1+0.03354)=0.03245$, or about 3.25%.

The mutant rises from 1.00% to about 3.25% because its net growth rate is higher by 0.3 day$^{-1}$. This calculation isolates selection; it is not the full mutation model. It also assumes the rates remain constant over all four days.

Worked Example: Mutation-Selection Balance

Suppose a mutant is disadvantaged by $s=-0.20$ day$^{-1}$, while wild-type production is $b_W=5.0$ day$^{-1}$ and the mutation probability is $\mu=0.001$. Find the nonzero equilibrium mutant frequency predicted by the model, and check whether a mutant frequency of 1.00% is moving toward or away from that equilibrium.

At an interior equilibrium, $df/dt=0$ and $f<1$. Therefore the frequency equation requires $\mu b_W+s f=0$. Here the mutation supply term is $\mu b_W=0.001(5.0)=0.0050$ day$^{-1}$, so

f^*=-\frac{\mu b_W}{s} =-\frac{0.0050}{-0.20} =0.0250.

The predicted equilibrium frequency is 2.50%. At $f=0.0100$, the frequency derivative is

\frac{df}{dt} =(1-0.0100)\bigl(0.0050-0.20(0.0100)\bigr) =0.9900(0.0030) =0.00297\ \text{day}^{-1}.

The derivative is positive, so the frequency is increasing at 1.00%. For this interior balance to exist below 1, the condition is $0<\mu b_W<-s$. The equilibrium is stable: the derivative of the frequency equation with respect to $f$, evaluated at $f^*$, is $s(1-f^*)<0$. A small displacement from $f^*$ therefore produces a change directed back toward the equilibrium in this constant-rate model.

Worked Example: A Resistant Variant Under Treatment

During a hypothetical treatment, suppose the wild type has $b_W=5.0$ day$^{-1}$, $c_W=5.4$ day$^{-1}$, and the mutant has $b_M=4.5$ day$^{-1}$, $c_M=4.6$ day$^{-1}$. Let $\mu=0.001$ and let the mutant initially make up $f=0.00200$ of the viral population. Determine whether its frequency and its absolute abundance are initially increasing.

The net rates are $r_W=-0.400$ day$^{-1}$ and $r_M=-0.100$ day$^{-1}$, giving $s=0.300$ day$^{-1}$. The mutant has a relative advantage, although both variants have negative net growth rates. From the frequency equation,

\frac{df}{dt} =(1-0.00200)\left(0.001(5.0)+0.300(0.00200)\right) =0.998(0.00560) =0.00559\ \text{day}^{-1}.

The mutant frequency is increasing at that instant. Its absolute abundance obeys $dM/dt=r_M M+\mu b_W W$. Because $M=fN$ and $W=(1-f)N$,

\frac{1}{N}\frac{dM}{dt} =(-0.100)(0.00200)+(0.0050)(0.998) =0.00479\ \text{day}^{-1}.

Thus mutant abundance is also initially increasing in this example, partly because the mutation influx exceeds the mutant’s own loss. Meanwhile, total abundance has relative growth rate $r_W(1-f)+r_M f=-0.399$ day$^{-1}$, so the total viral population is declining. The result illustrates why frequency, variant abundance, and total viral burden must be reported separately.

Assumptions, Diagnostics, and Alternatives

The model is useful when the main question is how mutation and differences in variant-specific net growth affect a two-variant population over a period with approximately stable conditions. It is not a general account of viral evolution. Before interpreting fitted or simulated trajectories, check whether its assumptions are plausible:

Approximately constant rates: treatment exposure, target-cell availability, and immune pressure can change $b_i$ and $c_i$. A sustained trend in residual differences or a changing selection advantage may indicate that constant rates are inadequate.
Proportional production and clearance: if viral growth saturates, target cells become depleted, or clearance depends on abundance, exponential per-capita dynamics may fail. Compare predicted and observed trajectories across the abundance range rather than relying on an early-time fit alone.
Reliable variant measurements: a rising estimated frequency should be interpreted in light of sampling variation, detection limits, and the possibility that the two-variant description omits relevant variants. Frequencies near zero are especially sensitive to measurement and sampling noise.
Deterministic approximation: when mutant numbers are very small, the mutation influx and lineage survival are stochastic. A deterministic model can predict a smooth increase even though an individual rare lineage may be lost by chance.

A useful alternative is to compare variant-specific exponential growth rates over an interval in which mutation is negligible. This directly estimates a relative selection advantage from the change in the log abundance ratio, as in the first worked example. The full mutation model is more appropriate when continuous mutational input could materially affect the mutant frequency. More detailed within-host models can make production depend on target-cell availability or immune effectors; those are preferable when the corresponding feedbacks are central to the question, but they require additional data and parameters.

The model’s limitations are substantive. Its mutation probability is treated as fixed, only one mutation direction is represented, and it does not distinguish the generation of a mutant genome from its successful establishment as a lineage. It also does not specify whether an observed variant has clinical consequences. A positive selection advantage under a given treatment or immune environment is a mechanistic comparison of growth rates, not proof of clinical resistance or immune escape.

Common Pitfall

A frequent error is to say, “The mutant frequency is increasing, so the mutant is replicating faster and total virus must be increasing.” The first inference is not justified: frequency can rise because the mutant declines more slowly, because mutation supplies it, or both. The second is also incorrect: total burden depends on the abundance-weighted growth of both variants. In the treatment example, the total population declines even as the mutant frequency rises.

Key takeaway: Compare $r_M$ with $r_W$ to assess relative selection, use the mutation term $\mu b_W W$ to account for mutational supply, and examine $dM/dt$ and $dN/dt$ separately to determine whether mutant abundance and total viral burden are increasing.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If $r_W=0.2$ day$^{-1}$, $r_M=0.5$ day$^{-1}$, and mutation is absent, what is the mutant’s selection advantage, and what happens to its frequency when both variants are present?
With $s=-0.10$ day$^{-1}$, $b_W=4.0$ day$^{-1}$, and $\mu=0.002$, calculate the interior mutation-selection equilibrium frequency and state whether it satisfies the condition for an equilibrium below 1.
Can mutant frequency increase while mutant abundance decreases? Explain using the distinction between relative frequency and absolute abundance.
Why does the model subtract $\mu b_W W$ from wild-type production and add the same amount to mutant abundance?

Solutions. (1) $s=r_M-r_W=0.5-0.2=0.3$ day$^{-1}$. With $0<f<1$, $df/dt=s f(1-f)>0$, so mutant frequency increases. (2) $\mu b_W=0.002(4.0)=0.0080$ day$^{-1}$, so $f^*=-0.0080/(-0.10)=0.0800$, or 8.00%. The condition is $0<\mu b_W<-s$; here $0.0080<0.10$, so the equilibrium lies below 1. (3) Yes. Frequency can rise if the wild type declines faster, even while $M$ itself is declining; $df/dt>0$ does not imply $dM/dt>0$. (4) A fraction $\mu$ of wild-type progeny is assumed to be mutant progeny. The terms represent transfer between variant categories, so mutation changes their composition but does not create or remove total progeny in this model.