Bacterial growth

Question

Marco works in a pharmaceutical laboratory and has to study the action of an antibiotic on the behavior of a certain type of bacterium that duplicates every 15 minutes.

a. If there is 1 bacterium at the beginning of the experiment, how many bacteria will be there after 7 hours (assuming that no factors take over to alter the growth rate)?

b. After 7 hours the antibiotic is used on the obtained culture of bacteria; it is observed that from this instant the number of bacteria in the culture halves every 20 minutes. How many bacteria will be there after 2 hours since the antibiotic was used? Leave all results expressed as powers of 2.

I have read that for bacterial growth the number of microorganisms in an exponentially growing population is always given by $2^{n}$, where $n$ is the number of generations. I do not understand this part.

The given solution:

For part a, generally, the number $N_t$ of cells at time $t$ and the initial number $N_0$ of cells are related by $N_t=N_0\times 2^n. \tag 1$ We have $n=\frac{7\ \text{hours}}{15\ \text{minutes}}=\frac{420\ \text{minutes}}{15\ \text{minutes}}=28,$ so at the end we have $2^{28}$ bacteria.

For part b, we have $n'=\frac{20\ \text{hours}}{20\ \text{minutes}}=\frac{120\ \text{minutes}}{20\ \text{minutes}}=6,$ so at the end we have $$2^{28-6}=\frac{2^{28}}{2^6}=2^{22}$$ bacteria.

I do not understand the last step above. Is it more correct to write $2^{28}/2^6=2^{22}$ bacteria? Are the above formulas and steps correct? I await other solutions or a different formulation of the problem.

Answer 1

a certain type of bacterium that duplicates every 15 minutes.

I have read that for bacterial growth the number of microorganisms in an exponentially growing population is always given by $2^{n},$ where $n$ is the number of generations. I do not understand this part.

generally, the number $N_t$ of cells at time $t$ and the initial number $N_0$ of cells are related by $N_t=N_0\times 2^n. \tag 1$

That boldfaced part above isn't correct: the number of organisms equals the product of $N_0$ and $2^n,$ rather than just $2^n.$ Formula $(1)$ is just saying this: after $n$ rounds of doubling, the number of organisms becomes multiplied by two, $n$ times.

In the above, $n$ being the the number of generations refers to the number of rounds of the bacteria doubling in population size. On the other hand, if the population increases by $70\%$ after each round (that is, each specified period), then $N_t=N_0\times 1.7^n. \tag 2$

b. from this instant the number of bacteria in the culture halves every 20 minutes.

For part a, we have $2^{28}$ bacteria.

For part b, we have $n'=\frac{20\ \text{hours}}{20\ \text{minutes}}=\frac{120\ \text{minutes}}{20\ \text{minutes}}=6,$ so at the end we have $$2^{28-6}=\frac{2^{28}}{2^6}=2^{22}$$ bacteria.

I do not understand the last step above. Is it more correct to write $2^{28}/2^6=2^{22}$ bacteria?

Part b is no longer using formula $(1);$ here, every round/period, the bacteria are now halving instead of doubling, so the formula becomes $N_t=N_0\times 0.5^n, \tag 3$ where $N_0$ is now Part a's answer. So, this time, we have $$N_t=2^{28}\times0.5^6.$$

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Addendum

Formula $(1)$ refers specifically to population doubling. Observe that we can generalise it: the number of bacteria immediately after the $n$th period is $$N_n=(1+r)^nN_0,$$ where $r$ is the population-size increase per period. So,

• $r=1$ corresponds to the population increasing by $100\%,$ i.e., the population doubling,
• $r=0.59$ corresponds to the population increasing by $59\%,$
• $r=-0.5$ corresponds to the population decreasing by $50\%,$ i.e., the population halving.

Answer 2

Ans a) We can use the formula $N=N_0e^{kt}$

where $N$ is initial number of bacteria. they have given the time for bacteria to double ($t$) hence $$k=\frac{\ln 2}{t}$$ after plugging in the values you get,$$N=e^{\frac{\ln 2}{15}\times 7\times 60}$$ which will come out to be $2^{28}$.

Ans b) Now the bacteria is halfling very 20 mins. Hence the net growth constant will be $$k=-\frac{\ln 2}{20}$$

Now the initial number of bacteria is $2^{28}$ so, $$N=2^{28}e^{\frac{-\ln 2}{20}\times 2\times 60}=2^{28-6}=2^{22}.$$