Lecture 14&15
Date: May 3&5
Chemistry 24b
Spring Quarter 2004
Instructor: Richard Roberts
Michaelis-Menten Kinetics
k
k
k2
1
3


→ E +P



→



→
EP ←
ES ←
E+S ←







k
k
k
−2
−1
−3
Turnover is controlled by k2.
Satisfied by initial rates or velocities and intial [ S] >> E0 , whence [ P] → 0 .
Rate = velocity = v ≅ k2 [ES ] =
where
and
k2 E0 [ S]
Vmax [S ]
=
KM + [S ]
K M + [S ]
k−1 + k2
⇐ Michaelis constant
k1
= k2 E0
k2 is sometimes called kcat
KM =
Vmax
hyperbolic kinetics
Vmax
v
1V
2 max
KM
[S]
Figure 7-1
Kinetic Plots
v =
where
V max [S]
K M + [S]
hyperbolic
v ≡ initial rates
[S] ≡ initial substrate concentrations >> E 0
Chemistry 24b — Lecture 14&15
2
Linearized Plots
(1) Lineweaver-Burk: 1/v vs 1/[S] — Double Reciprocal
1
1
1
KM
=
+
⋅
V max
v
V max [S]
1
v
Slope = KM /Vmax
1/Vmax
1/[S]
−1
KM
Figure 7-2
(2) Eadie-Hofster: v/[S] vs v — Single Reciprocal
v
V
−v
= max
[S]
KM
Vmax /KM
v
[S]
Slope = −KM−1
(for normal KM's)
0
Vmax
Figure 7-3
v
Chemistry 24b — Lecture 14&15
3
(3) Dixon: [S]/v vs [S]
[ S]
[S ]
KM
=
+
v
V max
V max
[S]
v
Slope = 1
Vmax
KM
Vmax
[S]
0
Figure 7-4
Inhibition of Enzymatic Reactions
(1)
Competitive Inhibition
A competitive inhibitor is a molecule that resembles the substrate and occupies the
catalytic site because of its similarity in structure, but is completely unreactive. By
occupying the active site, the inhibitor prevents normal substrates from binding and being
catalyzed. Operationally, competitive inhibitors bind reversibly to the active site.
Hence, inhibition can be reversed by (1) diluting the inhibitor, or (2) swamping the
system with excess substrate.
Reaction Mechanism
→
→
k1
k2





→
 ES ←
→
 E + P
E +S ←
k−1
k−2
+
I
EI
Define
KI =
[ E ][ I ]
[EI ]
Chemistry 24b — Lecture 14&15
4
Expectation
A competitive inhibitor increases KM , but does not affect V max (because of sufficiently
high [S], S will displace I).
Hyperbolic Plots
Vmax
[I] = 0
v
inhibitor
KM'
KM
[S]
Figure 7-5
v =
V max [S]
=
 [I] 
[S] + KM 1+
 K 
V max [S]
[S] + KM
′
I
 [I] 
where KM
′ = KM 1 +
 K I 
⇐ modified K M
Lineweaver-Burk Plots
inhibitor
1
v
[I] = 0
1/Vmax
−1
KM
−1
KM'
Figure 7-6
1/[S]
Chemistry 24b — Lecture 14&15
5
Eadie-Hofster Plot
Vmax /KM
v
[S]
Slope = −KM−1
inhibitor
Vmax
0
v
Figure 7-7
Dixon Plot
parallel
[S]
v
inhibitor
KM
Vmax
[I] = 0
Slope = 1
Vmax
[S]
0
Figure 7-8
Chemistry 24b — Lecture 14&15
6
Classical Example of Competitive Inhibition
Enzyme: succinic dehydrogenase (SDH), which catalyzes the oxidation of succinic acid
to fumaric acid.
COOH
CH2
COOH
SDH
CH
Here
CH 2
HC
COOH
substrate ≡ succinic acid
inhibitor ≡ malonic acid
COOH
succinic acid
COOH
CH2
fumaric acid
COOH
Here KI = 1 × 10 −5 M , which means that at [malmate] = 10 −5 M , the apparent
affinity of the enzyme for succinate decreases by a factor of 2!
(2)
Noncompetitive Inhibition
A noncompetitive inhibitor is one that binds reversibly to the enzyme, but not at the
active site itself, so that the substrate can still bind at the active site, but there’s no
catalyzed transformation.
This type of inhibition cannot be overcome by a large amount of substrate, thus
noncompetitive inhibition.
Reaction Mechanism
k1
k2





→
 ES ←
→
 E + P
E +S ←
k−1
k−2
+
+
I
I
→
→
→
→
.
EI
ES ⋅ I
Define
 [E ][ I ]
KI = 
 [ EI ]  eq
 [ES ][ I ]

KSI = 
 [ES ⋅ I ] eq
Chemistry 24b — Lecture 14&15
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Expectation
The rate or velocity decreases to the extent that E is complexed by inhibitor, irrespective
of EI or ES ⋅ I . Thus, a noncompetitive inhibitor decreases V max without affecting the
apparent KM . The situation is not overcome by swamping the system with substrate.
Hyperbolic Plots
Vmax
v
no inhibitor
Vmax
'
inhibitor
1V
max
2
1V ma
' x
2
[S]
KM
Figure 7-9
Typical Case
KI = KSI , i.e., the affinity of the inhibitor site for the inhibitor does not depend on
whether E is bound with S. In this instance
v =
where Vmax
′
=
V max [S]
=

 1 + [I]  ([S] + K )
M

KI 
1

 1 + [I] 

KI 
⋅ V max
Vmax
′ [S]
KM + [S]
Chemistry 24b — Lecture 14&15
8
Lineweaver-Burk Plots
1
v
inhibitor
1
Vmax
'
no
inhibitor
1
Vmax
1/[S]
−KM−1
Figure 7-10
Eadie-Hofster Plot
v
[S]
no inhibitor
[I]
lines parallel
v
0
Figure 7-11
Chemistry 24b — Lecture 14&15
9
Dixon Plot
(both slope and intercept increase)
[I]
1
Vmax
'
[S]
v
KM
Vmax
'
no inhibitor
Slope =
KM
Vmax
1
Vmax
[S]
Figure 7-12
More General Case
KI ≠ KSI allosteric interaction between catalytic and inhibitor sites.
v =
where
V ′max [S]
k2 E 0 [S]
1
⋅
=

 α 
 α 
[I] 
KM + [S]
K + [S]
1+
β 
β M

KSI 

[I] 
1+

 α 
KI 
=
β

 1 + [I] 
KSI 

and
Both V max and KM can be affected!
V ′max =
Vmax
β
Chemistry 24b — Lecture 14&15
Case of
10
KSI >> KI
This is an interesting situation where the allosteric interaction between catalytic and
inhibitor sites is so strong that binding is mutually exclusive.
v =
If
then
V max [S]
β  α 
K + [S]
β M
1
K SI >> [I] ,
v =
β → 1
V max [S]
=

[I] 
1+
K + [S]

KI  M
V max [S]
KM
′ + [S]
which is the result for competitive inhibition!
Irreversible Modification
Permanent or irreversible modification of the active site often yields a situation that
behaves like the case of simple noncompetitive inhibition!
For example, irreversible modification by the chemical alkylating agent iodoacetamide,
which reacts with exposed sulfhydryl groups such as a cysteine to form a covalently
modified
Enzyme — S — CH2 — C — NH 2
O
or irreversible modification by carbodiimide of carboxyl groups from glutamate and
aspartate.
Chemistry 24b — Lecture 14&15
11
R
O
O
NH
R'
Enzyme — C — O — C
Enzyme — C — N
N
C
R'
NH
O
R
Oftentimes, chemically modified enzyme is inactive catalytically.
So,
v ∝ [Enzyme ]active , i.e., enzyme not chemically modified and unreacted enzyme behave
normally with KM identical to the situation prior to the addition of “inhibitor.”
In fact
v =
f unmodified V max [S]
K M + [S]
like in simple noncompetitive inhibition.
Well-Studied Enzyme System That Behaves According to Michaelis-Menten
Kinetics
Ferriprotoporphyrin and (Mn)2 protein catalase; catalyze the decomposition of H2O2.
2 H2 O2 → 2 H2 O + O2
Table 7-1. Velocities and Energies for Protein Catalases
Catalyst
None
HBr
Fe2+/Fe3+
Hematin or Hb
Fe(OH)2 TETA+ b
Catalase
Velocitya,
−d [H 2 O2 ]
dt
10–8
10–4
10–3
10–1
103
107
,M
S
−1
Ea (kJ/mole)
71
50
42
—
29
8
Chemistry 24b — Lecture 14&15
a
b
for [H 2O 2 ] = 1 M ,
triethylenetetramine
12
[catalyst] active sites = 1 M
uncatalyzed
Ε
71 kJ
8 kJ
H2O2
−94.7 kJ
H2O + 1/2 O2
Figure 7-13
0
∆ G298
= − 103.10 kJ mol −1 for H2 O2 (aq) → H2 O (l) +
0
∆ H298 = − 94.64 kJ mol
1
O2 (g)
2
−1
Turnover number (S −1 ) ≡ maximum velocity divided by the concentration
of enzyme active sites
Chemistry 24b — Lecture 14&15
13
Table 7-2. Turnover Numbers for Various Enzymes and Substrates
Enzyme
Substrate
Catalase
Acetylcholinesterase
Lactate dehydrogenase (chicken)
Chymotrypsin
Myosin
Fumarase
H2O2
Acetylcholine
Pyruvate
Acetyl-L-tyrosine ethyl ester
ATP
L-Malate
Fumarate
CO2
HCO3−
Carbonic anhydrase (bovine)
*Typically, turnover number ~ 103 S−1 within a factor of 10.
Turnover No. (S−1)*
9 × 106
1.2 × 104
6 × 103
4.3 × 102
3
1.1 × 103
2.5 × 103
8 × 104
3 × 104
Chemistry 24b — Lecture 14&15
14
Two Intermediate Complexes
The reaction catalyzed by catalase is essentially irreversible. Therefore, it is not
necessary to worry about EP. Most enzymatic reactions are readily reversible, so an
enzyme-product complex can often be detected.
Good Example
COO
H
COO
C
−
fumarase
C
+ H2O
OOC
H—C—H
HO — C — H
C
−
−
H
COO
−
fumarate
L-malate
20%
80%
Equilibrium
at room
temperature
k
k2
k3
1
→ ES ←

→ E P←

→E + P
E + S ←







k
k
−1
k −3
−2
So in this case, you cannot ignore EP! More complete treatment leads to the following
results:
(1) Forward Reaction: [P]0 = 0
 d[P]
vF = 
= k3 [EP]
dt  0
k2 k3 E0 [S]
=
 k−1k −2 + k−1k3 + k 2k3

+[S]
 k1(k2 + k−2 + k3 )

(k2 + k−2 + k 3 ) 
which can be reduced to
with
vF =
V F [S]
F
+ [S]
KM
VF =
k2 k3 E0
k2 + k−2 + k3
and
F
KM
=
k−1 k−2 + k −1k3 + k2 k3
k1 (k 2 + k−2 + k3 )
Chemistry 24b — Lecture 14&15
15
(2) Reverse Reaction: [S]0 = 01
d [S]
vR = 
= k−1 [ES]
 dt  0
V R [P]
K MR + [P]
k−1 k−2 E0
VR =
k−1 + k2 + k−2
=
with
and
K MR =
k−1 k−2 + k −1k3 + k 2 k3
k −3 (k −1 + k 2 + k−2 )
(3) Net Velocity
R
F
[S] − V R KM
[P]
VF K M
v =
F R
R
F
[P]
K M K M + KM [S] + K M
Note that at equilibrium, v = 0 and hence
R
F
[S]eq = V R K M
[P]eq
V F KM
and
 [P]
V KR
= F FM
K = 
 [S]  eq
V R KM
There are too many kinetic constants here to be sorted out by steady state kinetics.
We must appeal to the methods of rapid kinetics.
1Alberty
and Peirce, J. Am. Chem. Soc. 79, 1526 (1957).