Question

For an exponentially growing microbial culture, the specific growth rate $(\mu)$ is related to its doubling time $(t_d)$ by which one of the following relations?

• A. $t_d = \mu$
• B. $t_d = \mu^2$
• C. $t_d = \dfrac{1}{\mu}\ln 2$
• D. $t_d = \dfrac{1}{2}\ln \mu$

Hint:

For exponential growth, doubling time is always inversely proportional to the specific growth rate $\mu$.

Answer 1

The Correct Option is B

For exponential microbial growth, the biomass concentration follows \[ X(t) = X_0 e^{\mu t}. \] The doubling time $t_d$ is defined as the time required for biomass to become twice its initial value: \[ X(t_d) = 2X_0. \] Substitute into the growth equation: \[ 2X_0 = X_0 e^{\mu t_d}. \] This simplifies to: \[ 2 = e^{\mu t_d}. \] Taking natural logarithm: \[ \ln 2 = \mu t_d \quad \Rightarrow \quad t_d = \frac{\ln 2}{\mu}. \] Thus the correct relation is option (C).
Final Answer: $t_d = \frac{\ln 2}{\mu}$