BAHAN KULIAH BIOKIMIA POWER POINT BAGIAN 1 /BIOCHEMISTRY POWER POINT LECTURES PART 1 | Karya Tulis Ilmiah

Enzyme Kinetics and Catalysis II
3/24/2003

Kinetics of Enzymes
Enzymes follow zero order kinetics when substrate
concentrations are high. Zero order means there is no increase
in the rate of the reaction when more substrate is added.
Given the following breakdown of sucrose to glucose and
fructose
Sucrose + H20

Glucose + Fructose
H

H

H

H OH

O

HO

H

H

H

O

HO

OH

H

HO

H
H

OH

OH

OH
OH

H
HO

H

k1

k2

E  S  ES   E  P
k -1

E = Enzyme S = Substrate P = Product
ES = Enzyme-Substrate complex
k1 rate constant for the forward reaction
k-1 = rate constant for the breakdown of the ES to
substrate
k2 = rate constant for the formation of the
products

When the substrate concentration becomes large
enough to force the equilibrium to form completely
all ES the second step in the reaction becomes rate
limiting because no more ES can be made and the
enzyme-substrate complex is at its maximum value.

d  P
v
k 2  ES
dt

[ES] is the difference between the

rates of ES formation minus the
rates of its disappearance. 1

d  ES
k1  E  S  k  1  ES  k 2  ES
dt

Assumption of equilibrium
k-1>>k2 the formation of product is so much
slower than the formation of the ES complex.
That we can assume:

k  1  E  S
Ks 

 ES
k1
Ks is the dissociation constant for the ES complex.

Assumption of steady state

Transient phase where in the course of a reaction the
concentration of ES does not change

2

d  ES
0
dt

 E  T  E    ES

3

Combining 1 + 2 + 3

k1   E  T -  ES  S  k -1  k 2  ES
rearranging

 ES  k -1  k 2  k1  S  k1  E T  S
Divide by k1 and solve for [ES]


E  T  S
 ES 
K M   S

Where

k -1  k 2
KM 
k1

k 2  E  T  S
 d  P 
vo 
 k 2  ES 
K M   S
 dt  t 0
vo is the initial velocity when the reaction is just starting out.

And

Vmax k 2  E  T

Vmax  S
vo 
K M   S

is the maximum velocity

The Michaelis - Menten
equation

The Km is the substrate concentration where vo equals
one-half Vmax

The KM widely varies among different enzymes

The KM
can be expressed as:

KM

k 1 k2
k2
  K s 
k1 k1
k1

As Ks decreases, the affinity for the substrate
increases. The KM can be a measure for substrate
affinity if k2k-1 or the ratio

Then

k cat
k1
KM

k1k 2

k 1  k2

is maximum

Or when every substrate that hits
the enzyme causes a reaction to
take place. This is catalytic
perfection.

Diffusion-controlled limit- diffusion rate of a substrate
is in the range of 108 to 109 M-1s-1. An enzyme lowers
the transition state so there is no activation energy
and the catalyzed rate is as fast as molecules collide.

Reaction Mechanisms
A: Sequential Reactions
• All substrates must combine with enzyme
before reaction can occur

Bisubstrate reactions

Random Bisubstrate Reactions

Ping-Pong Reactions
• Group transfer reactions
• One or more products released before all
substrates added

Kinetic data cannot unambiguously
establish a reaction mechanism.

Although a phenomenological description can be
obtained the nature of the reaction intermediates
remain indeterminate and other independent
measurements are needed.

Inhibition kinetics
There are three types of inhibition kinetics competitive,
mixed and uncompetitive.
•Competitive- Where the inhibitor competes with the

substrate.

Competitive Inhibition


E  I
KI 
 EI
Vmax  S
vo 
K M   S


I 
 1  
 KI 

HIV protease inhibitors

Competitive Inhibition: Lineweaver-Burke Plot

Uncompetitive Inhibition

Uncompetitive Inhibition: Lineweaver-Burke Plot

Mixed inhibition

Mixed inhibition is when the inhibitor binds to the
enzyme at a location distinct from the substrate
binding site. The binding of the inhibitor will either
alter the KM or Vmax or both.


E  I
KI 
 EI


ES I
KI 
 ESI

Vmax  S
vo 
K M    S



I 
  1  

K
I 


The effect of pH on kinetic parameters