Question

The generation time of E. coli is 20 minutes. If there are \( 10^6 \) E. coli present in an exponentially growing synchronous culture, then the average time (in minutes) required to obtain a final population of \( 4 \times 10^6 \) E. coli is ................

Hint:

For exponential growth, the population doubles every generation time. Use the formula \( N(t) = N_0 \times 2^{t/T} \) to calculate the required time for a given population size.

Answer 1

Correct Answer: 40

Step 1: Understanding the exponential growth formula.
The exponential growth of bacteria is given by the formula: \[ N(t) = N_0 \times 2^{t/T}, \] where \( N(t) \) is the population at time \( t \), \( N_0 \) is the initial population, and \( T \) is the generation time.

Step 2: Substituting the known values.
Given that the initial population is \( 10^6 \), the final population is \( 4 \times 10^6 \), and the generation time is 20 minutes, we need to solve for \( t \) when: \[ 4 \times 10^6 = 10^6 \times 2^{t/20}. \] Simplifying: \[ 4 = 2^{t/20}. \]

Step 3: Solving for \( t \).
Taking the logarithm of both sides: \[ \log_2(4) = \frac{t}{20}. \] Since \( \log_2(4) = 2 \), we have: \[ 2 = \frac{t}{20}. \] Solving for \( t \): \[ t = 40 \, \text{minutes}. \]

Step 4: Conclusion.
The correct answer is \( \boxed{40} \, \text{minutes} \).