Question

In a batch experiment with cells following Monod growth kinetics, the initial substrate concentration is doubled from $S_0$ to $2S_0$ and it is observed that the initial doubling time in the log phase hardly changes. This tells us that

• A. $S_0 >> K_s$
• B. $S_0 = K_s$ (approximately)
• C. $S_0 << K_s$
• D. The growth in substrate is inhibited.

Hint:

Growth rate is constant/maximum when substrate is in excess ($S >> K_s$).

Answer 1

The Correct Option is A

Step 1: Concept
Monod kinetics is defined by $\mu = \mu_{max} \frac{S}{K_s + S}$, where $\mu$ is the specific growth rate.

Step 2: Meaning
Doubling time ($t_d$) is inversely proportional to $\mu$ ($t_d = \ln 2 / \mu$).

Step 3: Analysis
If doubling the substrate ($S$) does not change the doubling time, it means $\mu$ is already at its maximum and independent of $S$. This occurs when $S$ is much larger than $K_s$, such that $\frac{S}{K_s + S} \approx 1$ and $\mu \approx \mu_{max}$.

Step 4: Conclusion
The observation implies that the system is saturated with substrate, or $S_0 >> K_s$. Final Answer: (A)