Question

A bacterial strain is grown in nutrient medium at 37$^\circ$C under aerobic conditions. The medium is inoculated with \( 10^2 \) cells from a seed culture. If the number of cells in the culture is \( 10^5 \) after 10 hours of growth, the doubling time of the strain (rounded off to the nearest integer) is _________ h.

Hint:

In exponential growth, use the formula \( N_t = N_0 \times 2^{\frac{t}{T_d}} \) to find the doubling time \( T_d \). Take the logarithm to solve for \( T_d \).

Answer 1

Correct Answer: 1

The population growth of bacteria can be modeled using the exponential growth equation: \[ N_t = N_0 \times 2^{\frac{t}{T_d}}, \] where:
- \( N_t \) is the number of cells at time \( t \),
- \( N_0 \) is the initial number of cells,
- \( T_d \) is the doubling time, and
- \( t \) is the time of growth.
Step 1: Identify the known values:
- Initial number of cells: \( N_0 = 10^2 = 100 \),
- Number of cells after 10 hours: \( N_t = 10^5 = 100000 \),
- Time of growth: \( t = 10 \, \text{hours} \).
Step 2: Substitute values into the growth equation:
\[ 100000 = 100 \times 2^{\frac{10}{T_d}}. \] Step 3: Solve for \( T_d \):
First, divide both sides by 100: \[ 1000 = 2^{\frac{10}{T_d}}. \] Take the logarithm (base 2) of both sides: \[ \log_2 1000 = \frac{10}{T_d}. \] Since \( \log_2 1000 \approx 9.97 \), we get: \[ 9.97 = \frac{10}{T_d}. \] Solve for \( T_d \): \[ T_d = \frac{10}{9.97} \approx 1.003 \, \text{hours}. \] Step 4: Round the answer to the nearest integer:
Thus, the doubling time is approximately: \[ \boxed{1} \, \text{hour}. \]