Question

An enzyme-catalyzed conversion of a substrate at 298 K proceeds by a Michaelis-Menten mechanism. The Lineweaver-Burk plot for the analysis of the experimental data has an intercept along the ( y )-axis of ( 0.357 , {mmol}^{-1} , {dm}^{3} , {s} ) and a slope of ( 2.10 , {s} ). The CORRECT Michaelis constant for the reaction is------- (rounded off to 2 decimal places):

• A. ( 5.88 , {mmol} , {dm}^{-3} )
• B. ( 5.88 , {mmol} , {dm}^{-3} , {s}^{-1} )
• C. ( 2.80 , {mmol} , {dm}^{-3} )
• D. ( 2.80 , {mmol} , {dm}^{-3} , {s}^{-1} )

Hint:

The Lineweaver-Burk plot provides the slope as ( frac{K_m}{V_{{max}}} ) and the intercept as ( frac{1}{V_{{max}}} ). Use these relationships to calculate ( K_m ) and ( V_{{max}} ).

Answer 1

The Correct Option is A

Step 1: Lineweaver-Burk equation.

The Lineweaver-Burk equation is given by:

\[ \frac{1}{v} = \frac{K_m}{V_{\text{max}}} \cdot \frac{1}{[S]} + \frac{1}{V_{\text{max}}} \]

where:

• \( K_m \): Michaelis constant.
• \( V_{\text{max}} \): Maximum reaction rate.
• \( [S] \): Substrate concentration.

From the equation, the slope of the Lineweaver-Burk plot is:

\[ \text{slope} = \frac{K_m}{V_{\text{max}}} \]

and the intercept on the \( y \)-axis is:

\[ \text{intercept} = \frac{1}{V_{\text{max}}}. \] Step 2: Calculating \( V_{\text{max}} \).

The intercept is given as:

\[ \text{intercept} = \frac{1}{V_{\text{max}}} = 0.357 \, \text{mmol}^{-1} \text{dm}^{3} \text{s}. \]

Thus:

\[ V_{\text{max}} = \frac{1}{0.357} \approx 2.80 \, \text{mmol} \, \text{dm}^{-3} \, \text{s}^{-1}. \] Step 3: Calculating \( K_m \).

The slope is given as:

\[ \text{slope} = \frac{K_m}{V_{\text{max}}} = 2.10 \, \text{s}. \]

Substituting \( V_{\text{max}} = 2.80 \, \text{mmol} \, \text{dm}^{-3} \, \text{s}^{-1} \):

\[ K_m = \text{slope} \times V_{\text{max}} = 2.10 \times 2.80 = 5.88 \, \text{mmol} \, \text{dm}^{-3}. \] Step 4: Conclusion.

The Michaelis constant for the reaction is:

\[ K_m = 5.88 \, \text{mmol} \, \text{dm}^{-3}. \]