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: Write the given information.
Initial number of bacterial cells:
\[ N_0=120 \] Doubling time:
\[ 20\ \text{minutes} \] Total time:
\[ 2\ \text{hours}=120\ \text{minutes} \]

Step 2: Determine the number of generations.
Number of doublings is given by
\[ n=\frac{\text{Total time}}{\text{Doubling time}} \] \[ n=\frac{120}{20} \] \[ n=6 \]

Step 3: Recall the bacterial growth formula.
After \(n\) doublings:
\[ N=N_0\times2^n \]

Step 4: Substitute the values.
\[ N=120\times2^6 \] \[ =120\times64 \]

Step 5: Perform the multiplication.
\[ 120\times64=7680 \]

Step 6: Interpret the result.
Thus, after \(2\) hours, the bacterial population increases exponentially to \(7680\) cells.

Step 7: Final conclusion.
Therefore, the number of bacterial cells present after \(2\) hours is
\[ \boxed{7680} \]