Question

A flask containing nutrient-rich media is seeded with 100 isogenic bacteria. Assuming that no bacteria die in the flask, after approximately how many generations will the population reach a size of \(10^5\)?

• A. \(10\)
• B. \(1000\)
• C. \(100000\)
• D. \(1\)

Hint:

Bacteria growing by doubling every generation means \(N=N_0\times 2^{g}\); find \(g\) such that this thousand-fold increase happens, noting \(2^{10}\approx 1000\).

Answer 1

The Correct Option is A

Step 1: Set up the growth equation.
Bacteria in a nutrient-rich flask, with no deaths, grow by simple binary fission: every cell splits into two cells each generation. If \(N_0\) is the starting number of cells and \(g\) is the number of generations that pass, the population after \(g\) generations is
\[ N = N_0 \times 2^{g} \]

Step 2: Plug in the given numbers.
Here the flask starts with \(N_0 = 100\) bacteria, and we want the population to reach
\[ N = 10^{5} = 100000 \]
So we need
\[ 100 \times 2^{g} = 10^{5} \]
Dividing both sides by \(100\),
\[ 2^{g} = \frac{10^{5}}{10^{2}} = 10^{3} = 1000 \]

Step 3: Solve for the number of generations.
Take \(\log_2\) of both sides:
\[ g = \log_2(1000) \]
Since \(2^{10} = 1024\), which is very close to \(1000\),
\[ g \approx 10 \]
So the flask needs about \(10\) generations of doubling for the population to grow from \(100\) cells up to around \(10^5\) cells.

Step 4: Check the other options and conclude.
Option (B), \(1000\) generations, and option (C), \(100000\) generations, are far too large, since even after only \(20\) doublings a population already multiplies by about a million times, so that many generations would produce a population size far beyond \(10^5\). Option (D), \(1\) generation, only doubles the count once, from \(100\) to \(200\), nowhere near \(10^5\). Only about \(10\) generations gives the roughly thousand-fold increase in cell number that is needed.
\[ \boxed{10} \]