Question

At ( t = 0 ), the bacterial cell number is 10,000 cells/mL. At ( t = 480 ) minutes, the cell number increased to 320,000 cells/mL. The mean generation time during this exponential growth period, rounded off to the nearest integer is ____ minutes.

Hint:

Mean generation time provides insights into the growth rate of bacteria and is critical in microbiological studies involving the dynamics of bacterial population growth under various conditions.

Answer 1

To determine the mean generation time, we use the formula for exponential growth: \[ N_t = N_0 \cdot 2^{t/G} \]

where:

• \( N_t \) is the final population size,
• \( N_0 \) is the initial population size,
• \( t \) is the time period,
• \( G \) is the generation time.

Step 1: Given values.

• \( N_0 = 10,000 \)
• \( N_t = 320,000 \)
• \( t = 480 \) minutes

Step 2: Rearranging the formula to solve for \( G \). \[ 2^{t/G} = \frac{N_t}{N_0} \Rightarrow \frac{t}{G} = \log_2\left(\frac{N_t}{N_0}\right) \Rightarrow G = \frac{t}{\log_2\left(\frac{N_t}{N_0}\right)} \] Step 3: Plugging in the numbers. \[ G = \frac{480}{\log_2\left(\frac{320,000}{10,000}\right)} = \frac{480}{\log_2(32)} \]

Since \( \log_2(32) = 5 \) (because \( 2^5 = 32 \)):

\[ G = \frac{480}{5} = 96 \text{ minutes} \] Conclusion:

Explanation: The mean generation time is calculated as the time it takes for the bacterial population to double. In this case, the population doubles **5 times** to increase from **10,000 to 320,000** in **480 minutes**, resulting in a mean generation time of approximately **96 minutes**.