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

This can also be solved using the continuous exponential growth equation instead of counting doublings directly.

The specific growth rate $\mu$ is related to the doubling time $t_d$ by $$\mu = \frac{\ln 2}{t_d} = \frac{0.6931}{1} = 0.6931\ hr^{-1}$$

Growth only occurs during the active phase, which lasts $8 - 1 = 7$ hours (the first hour being the lag phase with zero growth). The exponential growth equation is $$N = N_0\,e^{\mu t}$$

Substituting $N_0 = 2$, $\mu = 0.6931\ hr^{-1}$, and $t = 7$ hours: $$N = 2\,e^{0.6931 \times 7} = 2\,e^{4.8518}$$

Since $0.6931 \times 7 = 4.852 \approx \ln(128)$, this exponential term evaluates to 128, giving $$N = 2 \times 128 = 256$$

The continuous-growth-rate method and the simple doubling method give identical answers, as expected, since they describe the same physical process. \[\boxed{N = 256\ cells}\]