Question

For a culture, \( \mu_{max}=0.8~h^{-1} \) and \( K_s=0.2~g/L \). At substrate concentration \( S=0.2~g/L \), \( \mu \) is

• A. \(0.2~h^{-1}\)
• B. \(0.4~h^{-1}\)
• C. \(0.6~h^{-1}\)
• D. \(0.8~h^{-1}\)

Hint:

Remember this important Monod relationship: \[ \boxed{ S=K_s \Longrightarrow \mu=\frac{\mu_{max}}{2} } \] Whenever the substrate concentration equals the half-saturation constant, the microorganism grows at half of its maximum growth rate.

Answer 1

The Correct Option is B

Concept: The relationship between microbial growth rate and substrate concentration is described by the Monod equation. It is analogous to the Michaelis-Menten equation used in enzyme kinetics. The Monod equation is \[ \boxed{ \mu=\mu_{max}\left(\frac{S}{K_s+S}\right) } \] where
• \( \mu \) = Specific growth rate
• \( \mu_{max} \) = Maximum specific growth rate
• \( S \) = Substrate concentration
• \( K_s \) = Half-saturation constant When \(S=K_s\), the organism grows at exactly half of its maximum growth rate.

Step 1: Write the Monod equation.
\[ \mu=\mu_{max}\left(\frac{S}{K_s+S}\right) \] Substitute the given values: \[ \mu=0.8\left(\frac{0.2}{0.2+0.2}\right) \]

Step 2: Simplify the expression.
\[ \mu=0.8\left(\frac{0.2}{0.4}\right) \] \[ =\;0.8\times0.5 \] \[ =\boxed{0.4~h^{-1}} \]

Step 3: Identify the correct option.
The calculated growth rate is \[ \boxed{\mu=0.4~h^{-1}} \] Hence, \[ \boxed{Option (B) is correct \]