Question

A bacterial population in the log-phase grows from \( 4 \times 10^6 \) cells to \( 8.64 \times 10^6 \) cells in 20 minutes. The doubling time of the bacterium is ............ minutes (round off to 1 decimal place)

Hint:

The doubling time is found using the formula \( t_d = \frac{t}{\log_2(N_t/N_0)} \), where \( t \) is the time taken and \( N_t \), \( N_0 \) are the final and initial populations, respectively.

Answer 1

Correct Answer: 17.9

Step 1: Use the doubling time formula.
The doubling time \( t_d \) can be calculated using the formula for exponential growth: \[ t_d = \frac{t}{\log_2 \left( \frac{N_t}{N_0} \right)} \] where \( t \) is the time interval, \( N_t \) is the final population, and \( N_0 \) is the initial population.
Step 2: Substituting the values.
Given \( N_0 = 4 \times 10^6 \), \( N_t = 8.64 \times 10^6 \), and \( t = 20 \) minutes, we calculate: \[ \log_2 \left( \frac{8.64 \times 10^6}{4 \times 10^6} \right) = \log_2 (2.16) \approx 1.09 \] Now, calculate the doubling time: \[ t_d = \frac{20}{1.09} \approx 18.3 \, \text{minutes} \]
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{10.0} \) minutes.