Question

Consider that \(120\) cells of bacteria are inoculated in a nutrient rich media. If the doubling time of the bacteria is \(20\) minutes and assuming no cell death, the number of bacterial cells present after \(2\) hours will be _ _ _. (in integer)

Hint:

For exponential bacterial growth: \[ N=N_0\times2^n \] where \(n\) is the number of doublings.

Answer 1

Correct Answer: 7680

Step 1: List the data.
Start with $N_0 = 120$ cells, doubling time $20$ minutes, total time $2$ hours which is $120$ minutes.

Step 2: Count the doublings.
\[ n = \frac{120}{20} = 6 \] generations.

Step 3: Grow the population.
After $n$ doublings, $N = N_0 \times 2^n = 120 \times 2^6 = 120 \times 64$.

Step 4: Multiply.
\[ 120 \times 64 = 7680 \]

Step 5: Answer.
\[ \boxed{7680} \]