Error catastrophe

The question arises: how many mutations can be made during each replication before the population of viruses begins to lose self-identity?

Basic mathematical model

Consider a virus which has a genetic identity modeled by a string of ones and zeros (e.g. 11010001011101....). Suppose that the string has fixed length L and that during replication the virus copies each digit one by one, making a mistake with probability q independently of all other digits.

Due to the mutations resulting from erroneous replication, there exist up to 2^L distinct strains derived from the parent virus. Let x_i denote the concentration of strain i; let a_i denote the rate at which strain i reproduces; and let Q_ij denote the probability of a virus of strain i mutating to strain j.

Then the rate of change of concentration x_j is given by

${\displaystyle {\dot {x}}_{j}=\sum _{i}a_{i}Q_{ij}x_{i}}$

At this point, we make a mathematical idealisation: we pick the fittest strain (the one with the greatest reproduction rate a_j) and assume that it is unique (i.e. that the chosen a_j satisfies a_j > a_i for all i); and we then group the remaining strains into a single group. Let the concentrations of the two groups be x , y with reproduction rates a>b, respectively; let Q be the probability of a virus in the first group (x) mutating to a member of the second group (y) and let R be the probability of a member of the second group returning to the first (via an unlikely and very specific mutation). The equations governing the development of the populations are:

${\displaystyle {\begin{cases}{\dot {x}}=&a(1-Q)x+bRy\\{\dot {y}}=&aQx+b(1-R)y\\\end{cases}}}$

We are particularly interested in the case where L is very large, so we may safely neglect R and instead consider:

${\displaystyle {\begin{cases}{\dot {x}}=&a(1-Q)x\\{\dot {y}}=&aQx+by\\\end{cases}}}$

Then setting z = x/y we have

${\displaystyle {\begin{matrix}{\frac {\partial z}{\partial t}}&=&{\frac {{\dot {x}}y-x{\dot {y}}}{y^{2}}}\\&&\\&=&{\frac {a(1-Q)xy-x(aQx+by)}{y^{2}}}\\&&\\&=&a(1-Q)z-(aQz^{2}+bz)\\&&\\&=&z(a(1-Q)-aQz-b)\\\end{matrix}}}$.

Assuming z achieves a steady concentration over time, z settles down to satisfy

${\displaystyle z(\infty )={\frac {a(1-Q)-b}{aQ}}}$

(which is deduced by setting the derivative of z with respect to time to zero).

So the important question is under what parameter values does the original population persist (continue to exist)? The population persists if and only if the steady state value of z is strictly positive. i.e. if and only if:

${\displaystyle z(\infty )>0\iff a(1-Q)-b>0\iff (1-Q)>b/a.}$

This result is more popularly expressed in terms of the ratio of a:b and the error rate q of individual digits: set b/a = (1-s), then the condition becomes

${\displaystyle z(\infty )>0\iff (1-Q)=(1-q)^{L}>1-s}$

Taking a logarithm on both sides and approximating for small q and s one gets

${\displaystyle L\ln {(1-q)}\approx -Lq>\ln {(1-s)}\approx -s}$

reducing the condition to:

${\displaystyle Lq<s}$

Information-theory based presentation

See also