Question

The catalytic efficiency of an enzyme following Michaelis-Menten kinetics is defined by

• A. \(k_{Cat}\)
• B. \(V_{max} / k_{Cat}\)
• C. \(k_{Cat} / K_M\)
• D. \(k_{Cat} / V_{max}\)

Hint:

At low substrate concentration the Michaelis-Menten equation collapses to a bimolecular rate law with rate constant kCat/KM; this ratio is the catalytic efficiency (specificity constant).

Answer 1

The Correct Option is C

Concept:
Catalytic efficiency measures how well an enzyme performs when substrate is scarce, combining how fast the enzyme turns over substrate with how tightly it binds substrate.

Step 1: Start from the Michaelis-Menten equation.
\[ v = \frac{k_{Cat}[E]_0[S]}{K_M + [S]} \]
Here \(k_{Cat}\) is the turnover number (how many substrate molecules one enzyme molecule converts per second at saturation), \([E]_0\) is total enzyme concentration, and \(K_M\) is the substrate concentration at which the reaction runs at half its maximum rate, a rough inverse measure of how tightly the enzyme binds substrate.

Step 2: Look at the low substrate limit.
When substrate is scarce, \([S] \ll K_M\), so \(K_M + [S] \approx K_M\), and the rate equation simplifies to
\[ v \approx \left(\frac{k_{Cat}}{K_M}\right)[E]_0[S] \]

Step 3: Identify the catalytic efficiency term.
In this simplified equation, \(v\) behaves like a simple second order reaction between free enzyme and free substrate, and the single combined constant multiplying \([E]_0[S]\) is \(k_{Cat}/K_M\). This ratio is called the specificity constant or catalytic efficiency, since it captures both how fast the enzyme reacts (numerator) and how easily it grabs the substrate at low concentration (inverse of the denominator). Enzymes that are almost perfectly efficient, limited only by how fast substrate molecules can diffuse to them, have \(k_{Cat}/K_M\) values around \(10^8\) to \(10^9 \; M^{-1}s^{-1}\).

Step 4: Rule out the other options.
Option (A), \(k_{Cat}\) alone, only tells you the turnover rate at saturation; it ignores how well the enzyme binds substrate at low concentration, so it cannot be the full efficiency measure.
Option (B), \(V_{max}/k_{Cat}\), is dimensionally just the total enzyme concentration \([E]_0\) (since \(V_{max} = k_{Cat}[E]_0\)), not a measure of efficiency at all.
Option (D), \(k_{Cat}/V_{max}\), reduces to \(1/[E]_0\), again just the inverse of enzyme concentration, with no connection to substrate binding or catalytic efficiency.

Step 5: Final Answer.
\[ \boxed{\text{Catalytic efficiency} = \frac{k_{Cat}}{K_M}} \]