Can a microbe grow faster later by growing slower now?

This question popped up a few years ago during a conversation with my PhD advisor. I think it makes a good riddle or exercise problem, so I’m posting it here, along with a solution I came up with. My solution doesn’t feel intuitive, even though the problem seems borderline trivial, so I’m curious if anyone has a less “technical” way of thinking about this.

Imagine a population of microbes that begins inducing a protein P needed for a newly encountered environment. The microbes grow more slowly at lower expression levels of P, and faster at higher P levels. Additionally, the microbes could “choose” (via regulatory circuits) to grow slower than is allowed by their current level of P. This, in turn, could allow P to induce faster, as the turnover of P due to dilution by growth will be decreased. Therefore, the question is, should microbes ever “choose” to grow slower than is possible in order to allow a beneficial gene/protein to induce faster?

Here is a calculation to show that the answer is no, the microbes cannot grow faster overall by temporarily “choosing” to grow slower. At the end I’ll also mention a few fun papers with similar types of analyses.

Let P be the level of the protein and assume that it is produced at a constant rate \alpha and turned over by dilution due to growth at rate r. This is equivalent to stating that \frac{dP}{dt} = \alpha - r P and therefore that P(t) = \frac{\alpha}{r} \left ( 1 - e^{-r t} \right ).

Consider a simple special case where cells grow exponentially with constant rate r until the protein level reaches P_{max}, and then with constant growth rate r_{max} thereafter. Also let r < r_{max}.

Now imagine 2 growth strategies, illustrated in the picture below:

  1. A “standard” strategy where growth rate is r_1 < r_{max} while P < P_{max} and r_{max} at the moment T_1 that P \geq P_{max}.
  2. A “slow growth” strategy where r_2 < r_1 while P < P_{max} and r_{max} at the moment T_2 that P \geq P_{max}.

Photo Oct 07, 12 42 56 PM

The question now becomes, which strategy leads to more growth until time T_1? The amount of growth can be expressed as a fold-increase in biomass F since time zero:

  • For strategy 1, F_1 = e^{r_1 T_1}.
  • For strategy 2, F_2 = e^{r_2 T_2} e^{r_{max} \left (T_1 - T_2 \right)}.

We want to know if it can ever be true that F_2 > F_1, which is equivalent to r_2 T_2+ r_{max} ( T_1 - T_2 ) > r_1 T_1. Now note that the actual value of r_{max} doesn’t matter because we are simply trying to compare outcomes for r_2 < r_1 < r_{max}. Therefore we can assume for convenience that r_{max} = 1 without loss of generality. This lets us rearrange the inequality to

T_2 \left ( r_2 - 1 \right ) > T_1 \left ( r_1 - 1 \right ).

Using the equation for the increase in protein level P(t) above, we can solve for T as

T  = -\frac{1}{r}\log \left ( 1 - C r \right ),

where C = P_{max}/\alpha. Thus the inequality becomes

\left ( \frac{1}{r_2} - 1 \right ) \log \left ( 1 - C r_2 \right ) > \left ( \frac{1}{r_1} - 1 \right ) \log \left ( 1 - C r_1 \right ).

To figure out if this inequality holds, let’s look at the function g(r) = \left ( \frac{1}{r} - 1 \right ) \log \left ( 1 - C r \right ). Note that the domain is constrained: by definition, r < r_{max}. Additionally, C < r_{max} because P_{max} must be less than \alpha / r — otherwise the cell will never reach the definition of “maximal” induction). Given our stipulation above that r_{max}=1, we can see all the relevant behavior of g(r) by plotting it for various values of C < 1.


It is clear that g(r) is monotonically increasing with r, and therefore g(r_2) > g(r_1) can never be true for r_2 < r_1. Therefore, the cell cannot ever gain more biomass under the second “slow growth” strategy. QED.

A problem with this solution is that we focused on a special case where the growth rate has a step function shape, so technically we haven’t said anything about the general case where r(P) could have more complicated shapes. However, I think as long as r increases monotonically with P, the basic intuition of this proof can probably be generalized, maybe by treating the time intervals above as infinitesimal, or using variational calculus (although I’m not quite sophisticated enough to make these types of arguments rigorously).

Another problem, at least aesthetically, is that this required doing a bunch of algebra and making a plot in the end. I wonder if there is a way to solve this “in my head” by looking at it the right way.

As for the case where r(P) is non-monotonic with P, this could be where the “slow down and delay” strategy is useful — for example, if there is a particular range of values of P that are detrimental to fitness, and it is best to induce P past that range as quickly as possible. I won’t attempt analyzing this now though. UPDATE: I thought about this a little in terms of a simple piece-wise growth rate function same as above, but where there is a range of P values where growth rate is at a local minimum. I think even here, the delay strategy still will not be beneficial. This makes it an even more interesting challenge to come up with a proof for arbitrary r(P).

Finally, I should note that this is only a really basic (but I think novel) example of an “optimal gene induction” question is part of a much broader (although still kind of niche) field in biology. I find this type of problem fun to think about because it creates an analogy between the problems faced by cells and problems faced by engineers, e.g. optimizing takeoff trajectories of airplanes. Some nice papers containing this type of thinking are this theoretical analysis of gene regulatory functions given potential costs of gene expression, and this combined theoretical and experimental investigation of how stem cells differentiate in intestinal tissue.

This entry was posted in Literature review, Quantitative principles. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s