Question

In a batch culture experiment, 2 cells of a microorganism were added to a growth medium. During the experiment, a lag phase of 1 hour was first experienced by the added cells, in which no cell multiplication occurred. Afterwards, cells started multiplying exponentially with a doubling time of 1 hour.

After total 8 hours of experiment, the number of cells present in the growth medium was ______ (answer in integer).

Hint:

Subtract the 1-hour lag phase from the total 8 hours to find how long exponential growth actually lasts, then apply N = N0 x 2^(number of doublings).

Answer 1

Correct Answer: 256

Step 1: Identify the two distinct phases of growth.
The total experiment runs for 8 hours. The first hour is a lag phase where no cell division occurs, so the cell count stays at the starting value of 2 cells.

Step 2: Find the duration of active exponential growth.
Growth only happens after the lag phase ends, so the exponential growth phase lasts \[8 - 1 = 7\ hours\]

Step 3: Find the number of doublings in that time.
Since the doubling time is 1 hour, the population doubles once every hour of active growth, giving \[n = \frac{7\ hours}{1\ hour/doubling} = 7\ doublings\]

Step 4: Apply the doubling formula.
Starting from \(N_0 = 2\) cells, after 7 doublings: \[N = N_0 \times 2^n = 2 \times 2^7 = 2 \times 128 = 256\]

This matches the expected value of 256 cells exactly.