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.
Best Answer
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 $N_t=N_0\times 1.7^n. \tag 2$
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 $$N_t=2^{28}\times0.5^6.$$
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,