Q104. An enzyme catalyzes the conversion of 30 nM of a substrate to product at reaction velocity of 0.9 nM s-1. [Et] = 20 nM and Km = 10 μM. Kcat /Km (in s-1). The value of n is _________ (in integer).
Kcat/Km equals 1,504,500 s-1 for this enzyme, so n is 1504500.
Problem Solution
The Michaelis-Menten equation gives initial velocity v = Vmax × [S] / (Km + [S]), where [S] = 30 nM = 3 × 10-8 M, v = 0.9 nM s-1 = 9 × 10-10 M s-1, Km = 10 μM = 10-5 M, and [Et] = 20 nM = 2 × 10-8 M.
Since [S] << Km (3 × 10-8 M vs. 10-5 M), the reaction follows first-order kinetics, but solve exactly: Vmax = v × (Km + [S]) / [S] ≈ 3.009 × 10-7 M s-1.
kcat = Vmax / [Et] = 15.045 s-1, so kcat/Km = 1.5045 × 106 s-1; round to integer n = 1504500.
Step-by-Step Calculation
Convert units to M: [S] = 3 × 10-8 M, Km = 10-5 M.
Vmax = (9 × 10-10) × (10-5 + 3 × 10-8) / (3 × 10-8) = 3.009 × 10-7 M s-1.
kcat = 3.009 × 10-7 / 2 × 10-8 = 15.045 s-1; kcat/Km = 15.045 / 10-5 = 1.5045 × 106 s-1.
No Options Explained
This fill-in-the-blank question lacks multiple-choice options, so no alternatives to evaluate. The direct calculation uses standard Michaelis-Menten parameters without assumptions beyond given data.
Michaelis-Menten Basics
Vmax derives from v = Vmax [S]/(Km + [S]); here [S] << Km approximates v ≈ (Vmax/[Et]/Km)[Et][S], but exact solves precisely.
Units align: nM to M ensures Kcat/Km in s-1 M-1, but problem specifies s-1 output.
Detailed Derivation
Compute Vmax = 0.9 × (10 + 0.03)/0.03 = 300.9 nM s-1 (in nM terms).
kcat = 300.9 / 20 = 15.045 s-1.
Kcat/Km = 15.045 / 10,000 = 1.5045 × 10-3 nM-1 s-1 = 1.5045 × 106 s-1 (since 1 μM = 1000 nM).
Value n = 1504500 fits integer fill-in.


