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I'm a microbiologist, but I'm teaching some ecology in my intro-level course, so when we got to population growth, I thought I'd use the example of a microbial population. But, I hit a strange problem that I thought maybe an ecologist could help me understand:

Imagine a population of bacteria that can divide every 30 minutes. In one hour, every cell produces four cells. Since no one really dies, the intrinsic growth rate (r) is 4. The exponential growth equation, dN/dt = rN works fine to show the growth of the population: starting with one cell, in one hour it's 4, then in two hours rN = 4*4 = 16, in three hours rN = 16*4 = 64 and so on. At 16 hours, we get to about 4 billion bacteria, which is exactly what the microbiologist expects.

But, we then tell the students that to make it easy to predict future numbers, we can do some math and get N(t) = N(0) e^rt. If N(0) = 1, t = 16 hours and r = 4 bacteria/hour/cell, this gives e^64 which is about 6x10^27. Yikes! Kinda far from 4 billion.

So...why doesn't this work? I feel like either I'm not understanding r correctly (it would have to be down around 0.6 for this to work out, I think) or maybe there's a limit on the equation and it just doesn't work for the absurdly high r of bacteria?

Thanks for the help...

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Considering your assumption:

I'm just looking at the exponential part, where the simple exponential equation works. If we assume there's sufficient nutrients for bacteria to grow unchecked for a number of hours (more-or-less true in a real culture)

In your original model you are using discrete states and fixed time steps. So, if 30 min is one time step then after n-steps you have $2^n$ or $4^{(n/2)}$ cells. You are basically counting double steps, which is fine. For a continuous approximation you should estimate the rate like this.

You have the first-order growth equation:

$$\frac{dN}{dt}=rN$$

When you integrate this equation (definite integral) you'll get:

$$ln\left(\frac{N_f}{N_i}\right)=r\tau$$

Where $N_f$ is final number of cells, $N_i$ is initial number of cells and $\tau$ is the time interval.


When the cell number doubles then $\dfrac{N_f}{N_i}=2$ and $\tau=30\ \text{min}$ (doubling time).


Then your rate constant would be: $\dfrac{ln(2)}{30}\approx0.023\text{ min}^{-1}$

After 16 hours (16×30min) your number of cells would be: $e^{0.023\times16\times60}\approx 4.27\times10^9$ (approx 4 billion).

So what is basically wrong? You haven't rescaled your rate. Since the exponent is $e$ and you know that $a^x=e^{x.ln(a)}$, all you need to do is scale the rate constant with $ln(a)$ (a=2). Moroever, in a continuous model, you don't have fixed time steps. So you scale the constant with the doubling time as well.

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  • $\begingroup$ @Jon If this answer addressed your problem, please consider accepting it by clicking on the check mark/tick to the left of the answer, turning it green. This marks the question as resolved to your satisfaction, and awards reputation both to you and the person who answered. $\endgroup$ – WYSIWYG May 5 '16 at 14:21
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Real bacteria population will likely reach some carrying capacity that will prevent them to grow exponentially. As a consequence, the exponential model will be a good fit for early growth only but after a while, one will need to use some other model (typically a logistic model).

Logistic model

Here, I quickly present one standard model of logistic growth for countinuous (and not discrete) time.

$$\frac{dN}{dt} = r N(t) \left(1-\frac{N(t)}{K}\right)$$

, where $K$ is the carrying capacity. From the above model, you can see that when $N(t)$ is low compared to $K$, then the differential equation is well approximated by the exponential model you described $\frac{dN}{dt} \approx r N(t)$. Setting $\frac{dN}{dt}=0$ shows that an equilibrium is reached when $N(t)=0$ (unstable equilibrium) or $N(t)=K$ (stable equilibrium).

The equivalent model in discrete generation can yield to all kind of behaviour when $r$ reaches high enough values including cyclic variation and chaos.

More advanced models

There are other more realistic models. For example, real bacteria populations could as well consume their resource at a rate that is higher than the rate at which resources are produced and population size would then decay to eventually reach extinction.

You will here the description of a model of population growth that includes predation populations.

Build your own model of population growth

These models are not so hard to build up t fit your needs. The first chapters of A Biologist's Guide to Mathematical Modeling in Ecology and Evolution (by Otto and Day) will help you to build and analyze your own models of population biology if you are interested.

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  • $\begingroup$ This is helpful in terms of understanding the bigger picture, and I appreciate the reference to what looks like a really helpful book. But for this example, I'm just looking at the exponential part, where the simple exponential equation works. If we assume there's sufficient nutrients for bacteria to grow unchecked for a number of hours (more-or-less true in a real culture), then I think the dN/dt equation and the e^rt equation should give the SAME approximation of population size at a given time. Why do they give such different results? $\endgroup$ – Jon May 4 '16 at 20:34
  • $\begingroup$ Either your $r$ is too big (but I think it is not too a bad approximation) or the assumptions that there is no competition for access to resource is violated. $\endgroup$ – Remi.b May 4 '16 at 20:38
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The quantity $r$ must be a frequence, measured in hours $^{-1}$. So it cannot be measured in bacteria/hour/cell. If the number of bacteria is multiplied by 4 each hour, this means that $\exp r= 4, $ or $r= \ln 4 $ , or 1, 39 per hours.

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