Bacterial Growth Problem with Mutation

This post is about a bacterial growth problem, you know, those ones they made you do in your college algebra class? Only this time, we’re mixing it up a bit. This time, the bacteria MUTATES. Mwa ha ha ha…

image of bacteria

Anyway, here is the problem:

In a population of bacteria with doubling time 1 hour, one bacterium underwent some mutation that changed its doubling time to 1 minute. The population doubled its size 30 minutes after the mutation happened. Find the size of the population at the moment when the mutation happened and 1 hour after the mutation happened.

This problem was submitted to me by a user, and I was really excited about it because I actually haven’t seen a bacterial growth problem quite like this before. To get started, remember that for these kinds of problems, we want to use the exponential growth equation:

\begin{equation*}
P(t) = P_0e^{rt}, \quad r > 0
\end{equation*}

In this equation, \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, and \(r\) is the growth rate. To set up the problem correctly, it’s easiest to think of the moment of mutation as time zero. This way, to answer the first part of the problem, we only need to find \(P_0\). Also, it helps to view the bacteria in two separate sub-populations: the mutated bacterium, and the “normal” bacteria. Here is a diagram:

petri dish

Now we can see that the total population is merely the sum of the two sub-populations, i.e.

\begin{equation*}
P(t) = \text{population of mutated bacteria} + \text{population of non-mutated bacteria}
\end{equation*}

To do this, let’s model the non-mutated bacteria first. Now, the only thing we know about this bacteria is its doubling rate. Also, because they lost one population member to mutation, the initial population has decreased by 1. So we have

\begin{equation*}
2(P_0 – 1) = (P_0 – 1)e^{60r}
\end{equation*}

This is true because the non-mutated bacteria starts off with an initial population of \(P_0-1\), and it doubles in one hour, or 60 minutes. (I chose the time to be in minutes because our other time amounts in the problem are given in minutes). Simplifying this equation, we get

\begin{equation*}
2^{1/60} = e^r
\end{equation*}

Now, we could take the natural log of both sides and solve for \(r\), but I’m not going to . This is because if I did, I would have to deal with nasty decimals, and I don’t want to. In fact, as you will see later, we actually don’t need to solve for \(r\).

Now, for the mutated bacteria, we have that the initial population is \(1\), and the doubling time is 1 minute, so using the growth equation, we get

\begin{equation*}
2 = e^{r’}
\end{equation*}

where \(r’\) is the growth rate of the mutated bacteria. Again, we don’t need to solve for \(r’\).

Total Population

For the total population, we just add the equations for the two sub-populations:

\begin{equation*}
P(t) = e^{r’t} + (P_0 – 1)e^{rt}
\end{equation*}

Now plug in the values for \(e^{r’}\) and \(e^r\) that we found above to get

\begin{equation*}
P(t) = 2^t + (P_0 – 1)2^{t/60}
\end{equation*}

The problem told us that the total population doubled in 30 minutes, so

\begin{equation*}
P(30) = 2P_0 = 2^{30} + (P_0 – 1)2^{1/2}
\end{equation*}

Now we just solve for \(P_0\). I will leave the details to the reader, but it comes to be

\begin{equation*}
P_0 = \frac{2^{30} – \sqrt{2}}{2-\sqrt{2}} \approx 1.8 \times 10^9
\end{equation*}

Now we can answer the last part of the problem, which was to find the total population after one hour. To do this, just plug in \(60\) into the above equation. We get

\begin{equation*}
P(60) = 2^{60} + (1.8\text{E}9 – 1)*2 \approx 1.2\times 10^{18}
\end{equation*}

That’s a lot of bacteria. Just be grateful they are too small to see!


Rent Textbooks at Knetbooks.com
Bookmark the permalink.

Leave a Reply

Your email address will not be published. Required fields are marked *