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

Bacteria dividing with no deaths follow simple doubling growth. The flask starts with $100$ cells and we want to know roughly how many doublings, or generations, are needed to reach $10^5$ cells.

1. Work out the fold increase needed: the population must grow from $100=10^2$ cells to $10^5$ cells, so it must multiply by $\dfrac{10^5}{10^2}=10^3=1000$ times over.
2. Recall what doubling does: after $g$ generations of simple binary fission, the starting count is multiplied by $2^{g}$. So the equation to solve is $2^{g}=1000$.
3. Use a handy doubling fact: doubling ten times multiplies a number by $2^{10}=1024$, which is close enough to $1000$ for an approximate answer. So $g\approx 10$ fits well.
4. Sanity check the other choices: $g=1000$ or $g=100000$ generations would multiply the starting count by astronomically large factors, nowhere near just $1000$-fold. $g=1$ only doubles the count once, giving $200$ cells, far short of $10^5$.

So counting doublings rather than working with logarithms directly, ten generations of division take the flask from $100$ bacteria to around $10^5$ bacteria.

Let's summarize:

• Doubling growth follows $N=N_0\times2^{g}$.
• A thousand-fold increase needs about $10$ doublings, since $2^{10}\approx1000$.

The population reaches $10^5$ cells after about $10$ generations.

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