Question

In a batch culture, biomass concentration increases from \(0.5~g/L\) to \(4~g/L\) in 2 hours. The specific growth rate (\(\mu\)) is

• A. \(0.693~h^{-1}\)
• B. \(1.039~h^{-1}\)
• C. \(1.386~h^{-1}\)
• D. \(2.079~h^{-1}\)

Hint:

For exponential microbial growth, remember: \[ \boxed{ \mu=\frac{\ln X_2-\ln X_1}{t} } \] Also, \[ \boxed{ \ln\left(\frac{X_2}{X_1}\right) = \ln X_2-\ln X_1 } \] This shortcut makes calculations much easier in competitive examinations.

Answer 1

The Correct Option is B

Concept: During the exponential phase of microbial growth, biomass increases exponentially. The specific growth rate is calculated using \[ \boxed{ \mu=\frac{\ln X_2-\ln X_1}{t} } \] where
• \(X_1\) = Initial biomass concentration
• \(X_2\) = Final biomass concentration
• \(t\) = Time interval

Step 1: Write the given values.
\[ X_1=0.5~g/L \] \[ X_2=4~g/L \] \[ t=2~h \]

Step 2: Apply the exponential growth equation.
\[ \mu=\frac{\ln(4)-\ln(0.5)}{2} \] Using logarithmic properties, \[ \mu=\frac{\ln\left(\frac{4}{0.5}\right)}{2} \] \[ =\frac{\ln(8)}{2} \] Since \[ \ln(8)=2.079 \] therefore, \[ \mu=\frac{2.079}{2} \] \[ =1.0395~h^{-1} \] \[ \boxed{\mu\approx1.039~h^{-1}} \]

Step 3: Identify the correct option.
The calculated specific growth rate is \[ \boxed{1.039~h^{-1}} \] Hence, \[ \boxed{Option (B) is correct \]