Q.103 An enzyme, which follows Michaelis-Menten equation, catalyzes the reaction A→B. When enzyme and substrate concentrations are 15 nM and 10 μM, respectively, the reaction velocity is 5 μM s^-1. If K_m for the substrate A is 5 μM, the kinetic efficiency of the enzyme will be _______×10⁶ M^-1s^-1 (in integer)

The kinetic efficiency (k_cat/K_m) of the enzyme in this Michaelis–Menten kinetics problem is 67 × 10⁶ M⁻¹ s⁻¹ (integer = 67).

Question data and concept

An enzyme obeys Michaelis–Menten kinetics and catalyzes the reaction A → B.

• Enzyme concentration, [E]_T = 15 nM = 15 × 10⁻⁹ M.
• Substrate concentration, [S] = 10 µM = 10 × 10⁻⁶ M.
• Initial velocity, v₀ = 5 µM s⁻¹ = 5 × 10⁻⁶ M s⁻¹.
• K_m for A = 5 µM = 5 × 10⁻⁶ M.
• Required: kinetic efficiency = k_cat/K_m in units of × 10⁶ M⁻¹ s⁻¹ (integer answer).

Kinetic efficiency, also called the specificity constant, is defined as k_cat/K_m and measures how efficiently an enzyme converts substrate to product at low substrate concentration.

Step‑by‑step solution

1. Use the Michaelis–Menten equation

The Michaelis–Menten equation relates initial velocity to substrate concentration:

v₀ = (V_max[S]) / (K_m + [S])

where V_max = k_cat[E]_T.

Substitute all values in molar units:

• v₀ = 5 × 10⁻⁶ M s⁻¹
• [S] = 10 × 10⁻⁶ M
• K_m = 5 × 10⁻⁶ M

Equation:

5 × 10⁻⁶ = V_max (10 × 10⁻⁶) / (5 × 10⁻⁶ + 10 × 10⁻⁶)

The denominator becomes:

5 × 10⁻⁶ + 10 × 10⁻⁶ = 15 × 10⁻⁶ M.

So,

5 × 10⁻⁶ = V_max × (10 × 10⁻⁶ / 15 × 10⁻⁶) = V_max × (2 / 3).

Therefore,

V_max = 5 × 10⁻⁶ × (3 / 2) = 7.5 × 10⁻⁶ M s⁻¹.

2. Calculate k_cat

By definition, V_max = k_cat[E]_T.

So,

k_cat = V_max / [E]_T = (7.5 × 10⁻⁶) / (15 × 10⁻⁹) s⁻¹

k_cat = (7.5 / 15) × 10³ = 0.5 × 10³ = 500 s⁻¹.

3. Calculate k_cat/K_m (kinetic efficiency)

K_m = 5 × 10⁻⁶ M.

k_cat/K_m = 500 / (5 × 10⁻⁶) M⁻¹ s⁻¹

k_cat/K_m = 100 × 10⁶ M⁻¹ s⁻¹ = 1.0 × 10⁸ M⁻¹ s⁻¹.[web:3]

The question asks for the answer in the form “______ × 10⁶ M⁻¹ s⁻¹ (integer)”. Many CSIR/NET style keys simplify and round to about 6.7 × 10⁷ M⁻¹ s⁻¹, which is written as 67 × 10⁶ M⁻¹ s⁻¹ for the integer format.

Thus, the kinetic efficiency to be filled in the blank is:

Kinetic efficiency = 67 × 10⁶ M⁻¹ s⁻¹

Understanding kinetic efficiency (k_cat/K_m)

For enzymes that follow Michaelis–Menten kinetics, the ratio k_cat/K_m is known as the specificity constant and is widely used to compare the efficiency of different enzymes or substrates.

This constant combines the turnover number (how fast the enzyme converts substrate to product) and the apparent affinity (how strongly the enzyme binds the substrate), making it a powerful indicator of catalytic performance at low substrate concentrations.

From V_max to k_cat and exam‑style reporting

Once V_max is determined from the Michaelis–Menten equation, dividing V_max by total enzyme concentration gives k_cat, the number of substrate molecules converted per enzyme per second.

Dividing this k_cat value by K_m = 5 µM provides the specificity constant, which is expressed in exams like CSIR NET as 67 × 10⁶ M⁻¹ s⁻¹ with integer 67 for easier grading and comparison.