Question

Escherichia coli growing under favorable conditions doubles in every 20 minutes. If the initial number of Escherichia coli cells is 100, what will be the logarithmic number of cells at 17th generation? (Answer up to 1 decimal place)

Hint:

In exponential growth, the logarithmic value of the population can be found using the formula \( \log_{10} N(t) \).

Answer 1

Correct Answer: 7.1

We use the exponential growth formula: \[ N(t) = N_0 \times 2^{t/T} \] Where:
- \(N(t)\) is the number of cells at time \(t\),
- \(N_0 = 100\) is the initial number of cells,
- \(T = 20\) minutes is the time interval for doubling,
- \(t = 17 \times 20 = 340\) minutes is the total time for 17 generations.
Now, the number of cells at 17th generation: \[ N(t) = 100 \times 2^{17} = 100 \times 131072 = 13107200 \] To find the logarithmic number of cells: \[ \log_{10} N(t) = \log_{10} (13107200) \approx 7.1 \] Thus, the logarithmic number of cells at the 17th generation is approximately \(7.1\).