(SEPT 2022-11)
72. An enzyme has a K_m of 5 x 10^-5 M and a V_max of 100 µmoles.lit^-1. min^-1(K_m is the Michaelis constant and V_max is the maximal velocity). What is the velocity in the presence of 1 x 10^-4 M substrate and 2 x 10^-4 M competitive inhibitor, given that the K_i for the inhibitor is 2 x 10^-4 M?
(1) 0.005 µmoles. Lit^-1. Min^-1
(2) 50 µmoles. Lit^-1. Min^-1,
(3) 5 µmoles. Lit^-1. Min^-1,
(4) 500 µmoles. Lit^-1. Min^-1

The correct answer is (2) 50 µmoles. Lit⁻¹. Min⁻¹.

Introduction

Enzyme kinetics is fundamental for understanding how enzymes catalyze reactions and how inhibitors affect these rates. In competitive inhibition, the inhibitor competes with substrate for binding at the enzyme’s active site, modifying the apparent affinity but not the maximal velocity (VmaxVmax). This article walks through a practical example calculation to determine enzyme velocity in the presence of competitive inhibitor using the Michaelis-Menten framework.

Problem Data

• Km=5×10−5 MKm=5×10−5M

• Vmax=100 μmol⋅L−1⋅min−1Vmax=100μmol⋅L−1⋅min−1

• Substrate concentration [S]=1×10−4 M[S]=1×10−4M

• Inhibitor concentration [I]=2×10−4 M[I]=2×10−4M

• Inhibitor constant Ki=2×10−4 MKi=2×10−4M

Formula for Velocity with Competitive Inhibition

v=Vmax[S]Km(1+[I]Ki)+[S]v=Km(1+Ki[I])+[S]Vmax[S]

Calculation Details

1. Calculate the apparent KmKm:

Km,app=Km(1+[I]Ki)=5×10−5×(1+2×10−42×10−4)=5×10−5×2=1×10−4 MKm,app=Km(1+Ki[I])=5×10−5×(1+2×10−42×10−4)=5×10−5×2=1×10−4M

2. Plug values into velocity formula:

v=100×1×10−41×10−4+1×10−4=100×10−42×10−4=1002=50 μmol⋅L−1⋅min−1v=1×10−4+1×10−4100×1×10−4=2×10−4100×10−4=2100=50μmol⋅L−1⋅min−1

Interpretation

• The presence of inhibitor increases the apparent KmKm to 1×10−4 M1×10−4M, decreasing substrate affinity.

• Since substrate concentration is equal to the new KmKm, initial velocity is half the maximum.

• This aligns with the expected competitive inhibition kinetics.

Summary Table

Conclusion

Competitive inhibition changes enzyme kinetics by increasing the apparent KmKm without affecting VmaxVmax. The enzyme velocity reduced to 50 µmol/L/min from 100 µmol/L/min when the inhibitor was present, demonstrating the principle of substrate-inhibitor competition in enzyme catalysis.