Sitting out a lag phase

In a constant environment, a population of bacteria is growing exponentially after some lag phase, both in number as in biomass, i.e. the change in biomass is proportional to the biomass itself. Almost all chemical reactions in the synthesis of biomass are enzyme mediated. Yet enzymes in isolation do not increase autocatalytically. This contradictions first calls for a closer analysis. Suppose that we have two enzymes each of which increases its substance by the addition of something derived from the working of the other. Then we shall have

Think e.g. of the interplay of nucleic acids and proteins, each of which playing a decisive role in the synthesis of the other. In order to obtain the solution of (13.1), we first write it as

If is the matrix with eigenvectors of in its columns, and the diagonal matrix with eigenvalues, so , the solution of (13.2) can be written like

The eigenvalues of are easily found to be , and

with . Substitution finally results in

(13.3)

with and . This leads to

			(13.4)

Ultimately, the second term in (13.4) vanishes and the ratio becomes constant at value and the population grows exponentially. So the cyclic autosynthetic reactions as modelled in (13.1) is consistent with exponential growth of biomass, indeed. For the purpose of, e.g., relating the chemical composition of cells to that of the medium, one needs cells growing at steady state (here: growing exponentially). After inoculation, the culture usually shows a lag phase (i.e. a deviation from exponential growth, which becomes apparent by plotting the logarithm of the number of cells against time). So we have to wait a while before taking the sample material. Now, let us study the length of the lag phase, , after a (momentary) transition from an environment in which the population was growing exponentially at rate , into and environment in which the population eventually will grow exponentially at rate . If both and are proportional to $\mu$, there will be no lag phase at all, i.e. . It seems realistic to assume that only one, say , is increasing with $\mu$, while the other remains constant, so . From a mathematical point of view, we have to wait infinitely long for exponential growth, since the second term in (13.4) vanishes only asymptotically. Being practical, let us accept exponentiality, if the relative error of the number of cells is less than a small fraction , i.e. , where represents the first term in (13.4), i.e.

. This leads to

The same can be done for , of course, but this will lead again to relation (13.5). When we choose a certain value for , we can solve the length of the lag phase $t$, given and . The solution is

The range of values for and is restricted to , because for negative values, the only steady state is that of being extinct, while values larger than the maximum growth rate are biologically impossible. A natural scaling for the length of the lag phase is therefore the dimensionless variable , expressed in terms of the dimensionless arguments . We arrive at

A plot of (13.7) is given in Fig.13.1. If and differ only a little bit, i.e. when , the apparent lag phase is zero, which is the result of our acceptance of a relative error of . This is a bit artificial, which becomes obvious when we could slowly increase or decrease $\mu$. We would have to wait a time zero for each incremental change, in other words, we would not have to wait at all for any change. The relative errors would built up this way, far beyond our setting of , of course. Apart for this artifact, some rather counter-intuitive results are obvious from Fig.13.1. For, the length of the lag phase in an up shift, is not equal to that of a down shift, i.e. . Further: Starting from a small initial growth rate, we have to wait longer to reach steady states for a bit higher new growth rate, than for a much higher one. Finally: We have to wait really long in case of a significant down shift. This illustrates that it is extremely difficult to standardize the cells to conditions of a small growth rate. This is important, because this is the usual condition outdoors, which we need to reach when we want, e.g, to exclude growth for the study of maintenance. These results are possibly easier to remember, when we express the lag phase in the number of division intervals in the new situation, where . In steady state we have for the number of bacteria at time , which equals , for time . If we divide the lag time in (13.6) by this division interval, we arrive at

The role of and in (13.8) is now symmetric, and the largest number of division intervals in the lag phase, i.e. when or equals 0, is . See Fig.13.1. The model (13.1) can easily be extended to more than two types of enzyme, in which case the population growth rate, $\mu$, equals the geometric mean of the rate of increase of the different enzyme types, . It remains rather easy, because of the linearity. Such a system can be conceived as a first approximation to the more usual hyperbolic rate functions, where the substrate concentrations are small with respect to the saturation constants (see example 10). Although the presented modelling of population growth in terms of chemical reactions is considered standard in microbiology (, the classical results given in Hinshelwood (1952) are followed here), a warning seems appropriate for skipping levels of organisation (here: the cell), and a too loose reference to variables that can be measured directly or indirectly. Further reading for a classical and still relevant work on enzyme performance in growing cells: [#!Hins52!#].

Figure: Contours for the length of the lag phase following a transition from one exponential growth rate to another, accepting a relative error of 0.05. The growth rates are expressed as fractions of the maximal growth rate. The contours of values 5, 2.5, 1, 0.5 and 0 are shown for the time lag times the maximum growth rate times .