10/22/98

My assignment was to confirm Crichton's allegations, or disprove them.The mathematics of uncontrolled growth are frightening. A single cell of the bacterium E. coli would, under ideal circumstances, divide every twenty minutes. That is not particularlry disturbing until you think about it, but the fact is that bacteria multiply geometrically: one becomes two, two becomes four, four becomes eight, and so on. In this way, it can be that in a single day, one cell of E. coli could produce a super-colony equal in size and weight to the entire planet earth.

-M. Crichton (1969), The Andromeda Strain. (Dell, New York, p. 247)

First of all, we can tackle this problem two ways: the "pattern finding" way as I call it, or using some more sophisticated techniques. I will employ both ways to illustrate.

We are given that the average mass of an E. coli bacterium is 10^{-12} gm, where gm is of
course grams. We are also given that the mass of the earth is 5.9763*10^{24} kg, where kg
is kilograms. To make this easier, I will covert this kg amount into grams. There are 1,000 gms in
one kg, so we get:

5.9763*10^{24}*1,000 = 5.9763*10^{27} gms, which is the mass of the earth.

We are also given that a single cell of bacterium divides every twenty minutes. So, if it did this for a day, it would have divided 72 times, because there are 24*60 = 1,440 minutes in a day, and 1,440/20 = 72.

From all of this, we can obtain a function that describes what is going on. This is the function we want to look at:

T(t) = number of bacterium(ia) * total weight of bacterium(ia)

We are given that T(t) = 1*10^{-12}. So, in words, this means that at time 0, when 0 divisions
have taken place, we have 1 bacterium, which gives us a total mass of 10^{-12} gms.

We can continue this pattern to get:

- T(1) = 2*10
^{-12} - T(2) = 4*10
^{-12} - T(3) = 8*10
^{-12} - T(4) = 16*10
^{-12}

etc.

In general, T(t) = 2^{t}*10^{-12}

So, we know that it can possibly divide 72 times in one day, so we can compute:

T(72) = 2^{72}*10^{-12}, and using a calculator, we get 4,722,366,483 gms.

This amount is significantly less than the mass of the earth, which from above is 5.9763*10^{27} gms.

So Crichton was incorrect when he said that after one day, one cell of E. coli could produce a weight equal to the weight of the earth upon dividing.

The above steps are the "pattern finding" way of solving this problem. Once a pattern was determined for what is going on, the result was easily obtained. Below I will continue to solve the problem another way.

The above work is part of a larger picture called the Law of Natural Growth.

It says, that:

dT/dt = kT

and T(0) = W

and lastly, that

T(t) = W*e^{kt}

where e is the constant 2.71828...

Wow. A lot of stuff! In plain talk, the growth rate of the bacteria is proportional to the population of the bacteria, and the initial time and weight will assist us in finding the function that describes the total weight after t divisions.

There are a few things we need to take care of first. We need to figure out what k is.

So, we go back to what we were given in the problem, that:

T(0) = 1

From the Law of Natural Growth, we get:

T(t) = 1*e^{kt
}

So, T(1) = e^{kt} = 2

Doing some algebra, T(1) = e^{k} = 2

ln(e^{k}) = ln(2)

k = ln(2), which is about .693147...

So, our general equation that gives the number of bacteria at time t is:

T(t) = e^{ln(2)*t}

Ahh, but we are forgetting something. We still need to put the bacteria mass in this equation. Doing this, we get:

T(t) = e^{ln(2)*t}*10^{-12}

Ok, now we know that the earth's mass is 5.9763*10^{27} gms. So, setting T(t) equal to the earth's
mass should give us the number of divisions it took the
bacteria to get to a point where their mass was the same as the earth's. If the number of divisions is greater
than 72, it will tell us a number of things, namely that Crichton is wrong, and that the
mass of the bacteria after 1 days worth of divisions is significantly less than the mass of the earth. Here we go:

e^{ln(2)*t}*10^{-12} = 5.9763*10^{27}

e^{ln(2)*t} = 5.9763*10^{27}/10^{-12}

e^{ln(2)*t} = 5.9763*10^{39}

Taking the natural log of both sides we get:

ln(2)*t = 91.5886202741

Dividing by ln(2), we get:

t = 132.13 divisions

So, for the bacteria to reach the mass of the earth, they would have to divide a little over 132 times! If they divide every 20 minutes, this makes the number of minutes it would take them to do it 2,640. And if there are 1,440 minutes in a day, then 2,640/1,440 would be the number of days it would take the bacteria to do it, which is 1.8333...

So for Crichton to be correct, he should have said it would take the bacteria 2 days, or a bit less.

I hope you enjoyed this write-up!

If you enjoyed *any* of my content, please consider supporting it in a variety of ways:

- Check out a random article at http://statisticool.com/random.htm
- Buy what you need on Amazon from my affiliate link
- Share my Shutterstock photo gallery
- Sign up to be a Shutterstock contributor
- Search Statisticool.com: