ITK-329 Kinetika & Katalisis
Chapter 6
Homogeneous Reaction Catalysis:
ENZYMATIC REACTION FUNDAMENTAL
Development of Kinetic Expression in
Enzymatic Reactions
Dicky Dermawan
www.dickydermawan.net78.net
dickydermawan@gmail.com
104
General Properties of Enzymes
# accelerate specific reaction
# very selective
# work on mild condition
k1
E+S
(ES)*
k2
k3
(ES)* + W
P+E
W = Solvent, usually water,
Considered constant in concentration
Example: Urease
O
NH2CNH2 + Urease
O
NH2CNH2-Urease
O
NH2CNH2-Urease
O
NH2CNH2 + Urease
O
NH2CNH2-Urease + HOH
2NH3 + CO2 + U
105
Development of Kinetic Expression
k1
E+S
(ES)*
k2
(ES)* + W
k3
PSSH :
Et
rES  k1ES  k 2 ES  k3 ESW  0
[E]t  [E]  [ES]
Enzyme balance:
ES
O
NH2CNH2
W = Pelarut (konstan)
P+E

[E]  [E]t  [ES]
k1 ([E]t  [ES])[S]  [ES](k 2  k 3[ W])
k1[S][E]t k1[S][ES]  (k 2  k 3[ W])[ES]
k1[S][E]t
 [ES] 
k1[S]  k 2  k 3[ W]
 rs  k1ES  k 2 ES

k1  k 3  [S][E]t  [ W]
 rs 
k1[S]  k 2  k 3[ W]
106
Michaelis – Menten Equation
 rp  rs
 rp  k 3  [E]t  [ W] 
Michaelis – Menten Equation:
where
 rs   rP 
[S]
[S] 
k 2 k3[ W ]
k1
Vmax  [S]
[S]  K m
Vmax  k 3  [E]t  [W]
Property at low substrate concentration:
[S] <<<< Km
 rs 
Property at high substrate concentration:
k 2  k 3  [ W]
Km 
k1
Vmax
[S]
Km
First order!!!
[S] >>>> Km
 rs 
Vmax
km
Zero order!!!
107
Plot of Michaelis – Menten Equation
-rs = rp
Vmax
2
Estimating Vmax and KM
Burke Lineweaver Plot
 1  [S]  K m


  rs  Vm ax  [S]
 1 
1
K
1

 
 m .

  rs  Vm ax Vm ax [S]
Alternatively, Eadie plot can be constructed
Km
[S]
1
 rs

1
Vm ax
tan 
Km
Vmax
1
[S ]
108
Example : 7-7
Determine the Michaelis – Menten Parameters for the reaction:
k
1

k 3 , H 2O
urea  urease
[urea  urease] * 
 2 NH3  CO 2  urease

k
2
The rate of reaction is given as a function if urea concentration
in the following table
Curea [kmol/m3]
-rurea [kmol/m3/s]
0.2
0.02
0.02
0.005
0.002
1.08
0.55
0.38
0.2
0.09
Use:
a. Burke – lineweaver plot
b. Eadie plot
Compare the results and determine which method
fits better
109
P7-11B
Beef liver catalase has been used to accelerate the
decomposition of hydrogen peroxide to yield water and oxygen.
The concentration of hydrogen peroxide is given as a function
of time for a reaction mixture with pH of 6.76 and maintain at
30oC,
(a)
(b)
t (min)
0
10
20
50
100
C H2O2 (mol/liter)
0.02
0.01775
0.0158
0.0106
0.005
Determine the Michaelis-Menten parameters
Vmax and Km
If the total enzyme concentration is tripled,
what will the substrate concentration be after
20 min ?
110
Integration of Rate Equation
Ex. 7-8
Calculate the time needed to convert 80% of the ure xammonia
and carbon dioxide in a 0.5 L batch reactor. The initial
concentration of urea is 0.1 mol/L, and the urease concentration
is 0.001 g/L. The reaction is to be carried out isothermally. At
this temperature, if total enzyme concentration used is 5 g/L:
 rs 
Ex. 7-8
1.33 [S] mol
0.0266 [S] L.s
If the reaction is carried out in 15 minutes using enzyme
111
concentration of 0.0002 g/L, what will be the conversion?
P7-10C Integration of Rate Equation
112
Enzymatic Reaction Inhibition
In addition to pH, another factor that greatly influences the rates of
enzyme-catalyzed reactions is the presence of an inhibitor.
The most dramatic consequences of enzyme inhibition are found in living
organisms, where the inhibition of any particular enzyme involved in a
primary metabolic sequence will render the entire sequence inoperative,
resulting in either serious damage or death of the organism. For example,
the inhibition of a single enzyme, cytochrome oxidase, by cyanide will
cause the aerobic oxidation process to stop; death occurs in a very few
minutes.
There are also beneficial inhibitors such as the ones used in the treatment
of leukemia and other neoplastic diseases.
113
Enzymatic Reaction Inhibition
The three most common types of reversible inhibition
occurring
in
enzymatic
reactions
are
competitive,
uncompetitive, and noncompetitive.
The enzyme molecule is analogous to the heterogeneous
catalytic surface in that it contains active sites.
When competitive inhibition occurs, the substrate and inhibitor
are usually similar molecules that compete for the same site
on the enzyme.
Uncompetitive inhibition occurs when the inhibitor deactivates
the enzyme-substrate complex, usually by attaching itself to
both the substrate and enzyme molecules of the complex.
Noncompetitive inhibition occurs with enzymes containing at
least two different types of sites. The inhibitor attaches only to
one type of site and the substrate only to the other.
114
Enzymatic Reaction Inhibition:
Competitive Inhibition
Competitive inhibition is of particular importance in
pharmacokinetics (drug therapy). If a patient were
administered two or more drugs simultaneously which react
within the body with a common enzyme, cofactor, or active
species, this could lead to competitive inhibition of the
formation of the respective metabolites and produce serious
consequences.
In this type of inhibition another substance, I, competes with
the substrate for the enzyme molecules to form an inhibitorenzyme complex, (E I).
k
1
Mechanism:
S+E
ES
k2
k3
I+E
EI
k4
ES
k5
P+E
115
Competitive Inhibition
k1
Mechanism:
S+E
I = Inhibitor = Penghambat laju reaksi
ES
k2
k3
I+E
EI
k4
ES
k5
P+E
d[ES]
PSSH :
 0  k1  [S]  [E]  k 2  [SE]  k 5  [SE]
dt

d[EI ]
PSSH :
 0  k 3  [I]  [E]  k 4  [EI ]
dt

k1
 [S]  [E]
k 2  k5
k
[EI]  3  [I]  [E]
k4
[ES] 
Enzyme balance:
[E]t  [E]  [EI]  [ES]  [E]t  [E] 
k3
k [E]  [S]
[I]  [E]  1
k4
k 2  k5
[ E ]t
[E] 
1
k3
k1
[I]  116
 [S]
k4
k 2  k5
Competitive Inhibition
(cont’)
k1
Mechanism:
S+E
SE
k2
k3
I+E
EI
k4
 rp  k 5  [ES]
 rp  k 5 
; substituting expression for [ES]:
ES
KM 
k 5  [S]  [E]t
k 2  k5 k 2  k5 k3
  [I]  [S]

k4
k1
k1
k 2  k5
k1
P+E
k1
 [S]  [E] ; substituting expression for [E]:
k 2  k5




k 5  k1  [S]  [E]t
[ E ]t
k 5  k1

 [S]  

k 2  k5
1  k 3 [I]  k1  [S]  (k  k )  (k  k )  k 3  [I]  k  [S]
1
5
2
5
2

 k 4
k4
k 2  k5
 rp 
k5
 rp 
k 5  [E]t  [S]
k
K M  K M  3  [I]  [S]
k4
k 
:  1 
 k1 
117
Competitive Inhibition
(cont’)
k1
Mechanism:
k
Ki  4
k3
k 5  [E]t  [S]
 rp 
k
K M  K M  3  [I]  [S]
k4
 rp 
k 5  [E]t  [S]
[ I]
KM  KM 
 [S]
KI
S+E
k2
k3
I+E
Vmax  [S]
 [ I] 
  [S]
K M 1 
K
I

Vmax  k5  [E]t
EI
k4
SE
 rp 
SE
k5
 rp  Vmax 
P+E
[S]
K M '  [S]
 [ I] 

K M '  K M 1 
 KI 
…… Jika penambahan suatu zat I ke dalam kultur
menyebabkan kenaikan harga KM tanpa perubahan harga
Vmax, maka I menginhibisi secara kompetitif.
118
Enzymatic Reaction Inhibition:
Uncompetitive Inhibition
Here, the inhibitor does not compete with the substrate
for the enzyme; instead, it ties up the enzymesubstrate complex by forming an inhibitor-enzymesubstrate complex, thereby restricting the breakdown of
the (E S) complex to produce the desired product.
Mechanism: S + E
k1
ES
k2
k3
I + ES
EIS
k4
ES
k5
P+E
119
Uncompetitive Inhibition
k1
Mechanism: S + E
ES
k2
k3
I + ES
ES
k4
k5
EIS
P+E
d[ES]
P SSH :
 0  k1  [S]  [E]  k 2  [ES]  k 3  [ES]  [I]  k 4  [EIS]  k 5  [ES]
dt
k1  [S]  [E]  k 3  [I]  [ES]
k1  [S]  [E]  k 4  [EIS]
[
ES
]

[ES] 
k 2  k 3  [ I]  k 5
k 2  k 3  [ I]  k 5
P SSH :
d[EIS]
 0  k 3  [I]  [ES]  k 4  [EIS]
dt
k
[EIS]  3  [I]  [ES]
k4
[ES] 
k1
 [S]  [E]
k 2  k5
[ES] 
k 3  [ I]
k1  [S]  [E]
 [ES] 
k 2  k 3  [ I]  k 5
k 2  k 3  [ I]  k 5


k 3  [ I]
k1  [S]  [E]
1 
  [ES] 
k 2  k 3  [ I]  k 5
 k 2  k 3  [ I]  k 5 


k 2  k5
k1  [S]  [E]

  [ES] 
k 2  k 3  [ I]  k 5
 k 2  k 3  [ I]  k 5 
120
Uncompetitive Inhibition (cont’)
Mechanism: S + E
Enzyme balance:
[E] t  [E] 
ES
k2
[E ] t  [E ]  [ES]  [EIS]
[E] t  [E] 
k1
k3
I + ES
k
k1
 [S]  [E ]  3  [I]  [ES]
k 2  k5
k4
EIS
k4
ES
k5
P+E
 k1

k
k1
 [S]  [E ]  3  [I]  
 [S]  [E ] 
k 2  k5
k4
 k 2  k5


 k1

k
k
3
1
[E ] t  [E ]  1 
 [S]   [I]  
 [S]  
k 2  k5
k4
 k 2  k5


[ E] 
[ E ]t
 k1 
k
k1
  [S]
1
[S]  3 [I]  
k 2  k5
k4
k

k
5
 2
k1
[ES] 
 [S]  [E]
k 2  k5
[ES] 
k1
 [S]
k 2  k5
[ E ]t
 k1 
k
k1
 121
1
[S]  3 [I]  
[S]
k 2  k5
k4
 k 2  k5 
Uncompetitive Inhibition (cont’)
k1
[ES] 
 [S]
k 2  k5
[ E ]t
 k1 
k3
k1
  [S]
1
[S]  [I]  
k 2  k5
k4
 k 2  k5 
k1  [S]  [E]t
[ES] 
k 2  k 5   k1 [S]  k 3 [I]  k1  [S]
k4
[S]  [E]t
[ES] 
 k 2  k5 
k

  [S]  3 [I]  [S]
k4
 k1 
[ES] 
[ES] 
Mechanism: S + E
k1
ES
k2
k3
I + ES
k 
:  1 
 k1 
ES
Ki 
k4
k3
k4
k5
KM 
EIS
P+E
k 2  k5
k1
[S]  [E]t
[ I]
K M  [S] 
 [S]
Ki
[E ]t  [S]
 [ I] 

K M  [S]  1 
K
i

122
Uncompetitive Inhibition (cont’)
Mechanism: S + E
k1
ES
k2
Ki 
Finally:
k4
k3
 rp  k 5  [E]
 rp  k 5 
 rp
I + ES
ES
[E]t  [S]
 [ I] 

K M  [S]  1 
 Ki 
Vmax  [S]

 [ I] 

K M  [S]  1 
 Ki 
k3
Vmax  k5  [E]t
Vmax
[S]

KM
 [ I] 
 [S]
1 

 K i  1  [I] 
 K 
i

V "[S]
 max
K M " [S]
 rp 
 rp
Inhibisi uncompetitive mengakibatkan penurunan
harga Vmax dan KM secara bersama-sama
EIS
k4
k5
P+E
KM 
Vm ax" 
Km"
k 2  k5
k1
Vm ax
 [ I] 
1 

 Ki 
Km
 [ I] 
1 

K
i

123
Non Competitive Inhibition
In noncompetitive inhibition, the substrate and inhibitor molecules react with
different types of sites on the enzyme molecule, and consequently, the
deactivating complex, IES, can be formed by two reversible reaction paths:
1. After a substrate molecule
attaches to the enzyme
molecule at the substrate
site, the inhibitor molecule
attaches to the enzyme at
the inhibitor site.
2. After an inhibitor molecule
attaches to the enzyme
molecule at the inhibitor
site, the substrate
molecule attaches to the
enzyme at the substrate
site.
Mechanism:
k1
S+E
ES
k2
k3
I+E
EI
k4
k5
EI + S
EIS
k6
k7
ES + I
EIS
k8
ES
k9
P+E
124
Non Competitive Inhibition (cont’)
Use “rate limiting” concept assuming reaction (5) as rate limiting step:
 rp  k9 [ES]
Other reactions are in equilibrium with equilibrium constant of KS, KI, KS’ and
KI’ for reaction (1) to (4), respectively.
Mechanism:
k1
S+E
ES
k2
k3
I+E
EI
k4
k5
EI + S
EIS
k6
k7
ES + I
EIS
k8
ES
k9
k1 [S] [E]  k 2 [ES]
 [ES]  KS [S] [E]
k3 [I] [E]  k 4 [EI]
 [EI]  KI [I] [E]
k5 [EI] [S]  k 6 [EIS]  [EIS]  KS ' KI [I] [E] [S]
k 7 [ES] [I]  k8 [EIS]  [EIS]  KI ' KS  [I] [E]  [S]
P+E
Further, assume KS= KS’ and KI = KI’ to obtain:
 rp 
Vmax  [S]
 [ I] 
(S  K m )  1  
 Ki 
KM 
1
1
; Ki 
KS
KI
125
Non Competitive Inhibition (cont’)
Further, assume KS= KS’ and KI = KI’ to obtain:
Enzyme balance:
[E] t  [E]  [ES]  [EI]  [EIS]
[E] t  [E]  KS  [S]  [E]  K i  [I]  [E]  KS  K i  [I]  [E]  [S]
[ E ]t
[E] 
1  KS  [S]  K i  [I]  KS  K i  [S]  [I]
Mechanism:
From
 rp  k9  [ES]
use
KM 
k1
S+E
ES
k2
k3
I+E
EI
k4
k5
EI + S
1
1
; Ki 
KS
KI
EIS
k6
k7
ES + I
EIS
k8
ES
k9
P+E
to obtain:
 rp 
Vmax  [S]
 [ I] 
(S  K m )  1  
Ki 
 126
Enzymatic Reaction Inhibition:
Comparison & Summary
a. Competitive Inhibition
S+E
I+E
k1
k2
k3
k4
k5
ES
c. Non Competitive Inhibition
k1
ES
S+E
ES
k2
k3
EI
P+E
I+E
EI
k4
k5
EI + S
EIS
k6
k7
b. Uncompetitive Inhibition
ES + I
k8
k1
S+E
ES
k2
EIS
ES
k9
P+E
k3
I + ES
ES
k4
k5
EIS
127
P+E
Enzymatic Reaction Inhibition:
Comparison & Summary
No Inhibition
 Competitive Inhibition
 Uncompetitive Inhibition
 Noncompetitive Inhibition
Vmax [S]
[S]  K m
 rp 
 rp 
VMax [S]
 [ I] 

[S]  K m  1 
 Ki 
 rp 
KM
 rp 
Vmax  [S]

[ I] 

 [S]  1 
Ki 

Vmax  [S]
 [ I] 

(S  K m )  1 
K
i

128
Enzymatic Reaction Inhibition:
Comparison in Plot
Burke – Lineweaver
Eadie
(a)
? (b)
? (c)
?
129
P7-14B
130
Special Case:
Inhibisi oleh Substrat
Mechanism:
k1
S+E
SE
k2
k3
S + ES
SE
k4
k5
SES
P+E
d[ES]
k  [S]  [E]  k 4  [SES]
 k1  [S]  [E]  k 2  [SE ]  k 3  [S]  [ES]  k 4  [SES]  [ES]  1
dt
k 2  k 5  k 3  [S]
d[SES]
 k 3  [S]  [ES]  k 4  [SES]
dt

[SES] 
k3
 [S]  [ES]
k4
131
Inhibisi oleh Substrat (cont’)
k

k1  [S]  [E]  k 4   3  [S]  [ES]
k4

[ES] 
k 2  k 5  k 3  [S]
[ES] 
k 3  [S]  [ES]
k  [S]  [E]
 1
k 2  k 5  k 3  [S] k 2  k 5  k 3[S]

Enzyme balance:
[ES] 
k1
[S]  [E]
k 2  k5
[E ]t  [E]  [ES]  [SES]
 [E] 
 k1

k
k1
[S]  [E]  3  [S]  
[S]  [E] 
k 2  k5
k4
 k 2  k5

[ E ]t
[E] 
1
 k1

k
k1
[S]  3  [S]  
[S] 
k 2  k5
k4
 k 2  k5 
132
Inhibisi oleh Substrat (cont’)
 rp  k 5  [ES]




[ E ]t
k1


[S]  
 k5 

k 2  k5


1  k1 [S]  k 3  [S]   k1  [S]  

 k k
 k 2  k5
k
4

 2 5

k 5  k1  [S]  [E]t

k
(k 2  k 5 )  k1  [S]  3  [S]  k1  [S]
k4
k 5  [E]t  [S]
 rP 
k3
k 2  k5
 [S]  [S] 2
k4
k1
 rs 
Vmax  [S]
 [S] 
K m  [S]  1  
 Ki 
133
Curve Fittings: Substrate-inhibited
Enzymatic Reaction
Perkirakan harga Vmax, KM, dan Ki berdasarkan plot hasil
percobaan dari reaksi enzimatik yang diinhibisi
substrat di bawah ini.
134