Question

A microbial culture (following Monod model for growth) has a maximum specific growth rate of 0.1 h\(^{-1}\), Monod constant of 1 mg L\(^{-1}\) and endogenous decay rate of 0.1 day\(^{-1}\). At limiting substrate concentration of 23 g L\(^{-1}\), the net specific growth rate will be ________ day\(^{-1}\). (rounded off to one decimal place)

Hint:

Since \(S \gg K_s\), \(\mu \approx \mu_{max}\); convert \(\mu_{max}\) to day\(^{-1}\) and subtract the endogenous decay rate \(k_d\).

Answer 1

Correct Answer: 2.3

Step 1: Write the Monod equation with endogenous decay.
The Monod model gives the gross specific growth rate as
\[ \mu = \frac{\mu_{max}\,S}{K_s + S} \]
The net specific growth rate accounts for cell loss due to endogenous decay (maintenance/death), so
\[ \mu_{net} = \frac{\mu_{max}\,S}{K_s + S} - k_d \]

Step 2: Convert all rate constants to the same time basis.
The answer is wanted in day\(^{-1}\), but \(\mu_{max}\) is given in h\(^{-1}\). Convert:
\[ \mu_{max} = 0.1\ \text{h}^{-1} \times 24\ \frac{\text{h}}{\text{day}} = 2.4\ \text{day}^{-1} \]
The decay rate is already in day\(^{-1}\): \(k_d = 0.1\) day\(^{-1}\).

Step 3: Convert the substrate concentration and Monod constant to the same units.
\(K_s = 1\) mg L\(^{-1}\), and \(S = 23\) g L\(^{-1}\) = \(23000\) mg L\(^{-1}\). Since both are now in mg L\(^{-1}\), they can be substituted directly.

Step 4: Compute the gross specific growth rate.
\[ \mu = \frac{2.4 \times 23000}{1 + 23000} = \frac{2.4 \times 23000}{23001} \]
\[ \mu = 2.4 \times 0.999957 \approx 2.39990\ \text{day}^{-1} \]
Because \(S \gg K_s\) (23000 mg/L vs 1 mg/L), the Monod term is essentially saturated, so \(\mu \approx \mu_{max}\).

Step 5: Subtract the endogenous decay rate.
\[ \mu_{net} = 2.39990 - 0.1 = 2.29990\ \text{day}^{-1} \]

Final Answer:
Rounded off to one decimal place,
\[ \boxed{\mu_{net} \approx 2.3\ \text{day}^{-1}} \]