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Enzyme
y
Kinetics
Reginald H. Garrett
Charles M. Grisham
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Chapter 13
݅ᑣ٫ Դৣ
Essential Questions
• Before this class, ask your self the following
questions:
–
–
–
–
–
What are enzymes?
How do enzymes work?
How many enzymes you know?
What is kinetics?
What is enzyme inhibitor?
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ibokjbAnbjm/oupv/fev/ux
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Outline
• Part I: Get to know enzymes
– Enzyme
y
Features
– Enzyme nomenclature
– Activity
– Unusual enzymes..
• Part II: Enzyme kinetics
• Part III: Inhibitor and bimolecular kinetics
3
Virtually All Reactions in Cells Are
Mediated by Enzymes
• Enzymes catalyze thermodynamically favorable
reactions causing them to proceed at
reactions,
extraordinarily rapid rates (see Figure 13.1)
• Enzymes
En mes provide
pro ide cells with
ith the abilit
ability to e
exert
ert
kinetic control over thermodynamic potentiality
• Living systems use enzymes to accelerate and
control the rates of vitally important biochemical
reactions
• Enzymes are the agents of metabolic function
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Figure
gu e 13.1
3 Reaction
eact o p
profile
o e sshowing
o
g tthe
e large
a ge free
ee
energy of activation for glucose oxidation. Enzymes lower
G‡, thereby accelerating rate.
13.1 What Characteristic Features Define
Enzymes?
•
•
•
•
•
Catalytic
y power
p
Specificity
R
Regulation
l ti
y
nomenclature
Enzyme
Coenzymes and cofactors
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Specificity
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Catalytic Power
• Catalytic power is defined as the ratio of the
enzyme-catalyzed
t l
d rate
t off a reaction
ti to
t the
th
uncatalyzed rate
• Enzymes can accelerate reactions as much
as 1016 over uncatalyzed rates
• Urease is a g
good example:
p
– Catalyzed rate: 3x104/sec
– Uncatalyzed rate: 3x10 -10/sec
– Ratio is 1x1014
7
• Enzymes selectively recognize proper
substrates
subs
a es o
over
e o
other
e molecules
o ecu es
• Enzymes produce products in very high
yields - often much greater than 95%
• Specificity is controlled by structure - the
unique fit of substrate with enzyme controls
the selectivity for substrate and the product
yield
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Regulation
90% yield in each step; 35% over 10 steps
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Figure 13.3 Yields in biological
reactions must be substantially
greater than 90%.
• Enzymes are the Agents of
Metabolic Function
• Regulation of enzyme
activity ensures that the
rate of metabolic reactions
is appropriate to cellular
requirements
Figure 13.2 The breakdown of glucose
by glycolysis provides a prime example
of a metabolic pathway.
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Enzyme nomenclature
• Common name:
– Suffix –ase to substrate
• Urease: urea hydrolyzing enzyme
• Phosphatase: hydrolyzing phosphoryl group
Enzyme
y
Nomenclature Provides a Systematic
y
Way of Naming Metabolic Reactions
– What is the activity of following enzymes?
• Catalase
• Trypsin
• Pepsin
• Systematic classification
• 6 major classes of reaction
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Coenzymes
y
and Cofactors
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Example:
• Classification of the enzyme catalyzing the
following reaction:
• ATP + D-glucose Æ ADP + D-glucose-6phosphate
• Phosphate group transferred
• Nonprotein components essential to enzyme
activity
• Cofactor = vitamin?
• Prosthetic group?
• Holoenzyme
y
vs. apoenzyme
p
y
– Transferase
T
f
((class
l
2)
• transferring P-containing group (subclass 7)
– With an alcohol group as acceptor (sub-subclass 1)
» Entry 2: glucokinase (E.C.2.7.1.2.)
» Entry 1: hexokinase (E
(E.C.2.7.1.1)
C 2 7 1 1)
• What is kinase?
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Why
y Enzymes
y
Are So Specific?
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• The “Lock and key” hypothesis
• The “Induced fit” hypothesis
• Induced fit favors formation of the transition
transitionstate
• Specificity and reactivity are often linked.
C t l i off Hexokinase
Catalysis
H
ki
– hexokinase example
• binding of glucose in the active site induces a
conformational change in the enzyme
• causes the two domains of hexokinase to close around
the substrate, creating the catalytic site
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Enzymatic Activity and pH
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• Enzyme-substrate recognition and catalysis
are greatly dependent on pH
• Enzymes have a variety of ionizable side
chains that determine its secondary
y and
tertiary structure and also affect events in the
active site
• Substrate may also have ionizable groups
• Enzymes are usually active only over a limited
range of pH
Enzymatic
E
ti Activity
A ti it is
i St
Strongly
l
Influenced by pH
Figure 13
13.11
11 The pH activity profiles of four different enzymes
enzymes.
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The Response of Enzymatic
Activity to Temperature
• Rates of enzyme-catalyzed reactions generally
increase with increasing
g temperature
p
• However, at temperatures above 50Ʊ to 60Ʊ C,
enzymes typically show a decline in activity
• Two effects here:
– Enzyme
E
rate
t typically
t i ll d
doubles
bl iin rate
t ffor ever
10ƱC as long as the enzyme is stable and
active
ti
– At higher temperatures, the protein becomes
unstable and denaturation occurs
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The effect of temperature on enzyme activity
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Unusual Enzymes
• Rib
Ribozymes - segments
t off RNA th
thatt di
display
l
enzyme activity in the absence of protein
– Examples:
E
l
RN
RNase P and
d peptidyl
id l
transferase
• Abzymes
y
- antibodies raised to bind the
transition state of a reaction of interest
– For a good review of abzymes, see Science,
Vol. 269, pages 1835-1842 (1995)
– Transition states are covered in more depth
in Chapter 14
RNA Molecules That Are Catalytic Have
Been Termed “Ribozymes”
Figure 13.25 RNA splicing in
Tetrahymena rRNA maturation
maturation.
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RNA Molecules That Are Catalytic Have
Been Termed “Ribozymes”
RNA Molecules That Are Catalytic
y Have
Been Termed “Ribozymes”
Figure 13
13.26
26 (a) The 50S subunit from H.
H marismortui
marismortui. (b) The
aminoacyl-tRNA (yellow) and the peptidyl-tRNA (orange) in the
peptidyl
p
p y transferase active site.
Figure 13.27 The peptidyl transferase reaction.
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Figure 13
13.28
28 (a)
Intramolecular
hydrolysis of a
hydroxy ester yields
a -lactone.
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Antibody
y Molecules Can Have Catalytic
y
Activity
13.8 Is It Possible to Design An Enzyme to
Catalyze Any Desired Reaction?
• A known enzyme can be “engineered”
engineered by in vitro
mutagenesis, replacing active site residues with
new ones that might catalyze a desired reaction
• Another approach attempts to design a totally
new protein with the desired structure and
activity
– This latter approach often begins with studies “in
in
silico” – i.e., computer modeling
– Protein folding and stability issues make this
approach more difficult
– And the cellular environment may provide
complications not apparent in the computer modeling
(b) The cyclic
phosphonate ester
analog of the cyclic
transition state.
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13 8 Is It Possible to Design An Enzyme to
13.8
Catalyze Any Desired Reaction?
Figure 13
13.29
29 cis
cis-1
1,2
2-Dichloroethylene
Dichloroethylene (DCE) is an
industrial solvent that poses hazards to human health.
Site-directed mutations have enabled the conversion of a
bacterial epoxide hydrolase to catalyze the chlorinated
epoxide
id h
hydrolase
d l
reaction.
ti
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End of Part 1
• Ask yourself…
– What are g
general features of an enzyme?
y
– Why and how an enzyme could be so
specific?
– Are all enzymes proteins?
– What
Wh t kind
ki d off enzyme can you b
buy and
d use iin
your daily life? Where are they come from?
– Can we design an enzyme by ourselves?
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13.2 Can the Rate of an Enzyme-Catalyzed
y
y
Reaction Be Defined in a Mathematical Way?
• What is Kinetics?
– the branch of science concerned with the rates of
reactions
ti
• What can we learn from enzyme kinetics?
– to
t determine
d t
i the
th maximum
i
reaction
ti velocity
l it and
d
binding affinities for substrates and inhibitors
• Why we have to learn about enzyme kinetics?
– To know insights of enzyme mechanisms and
metabolic pathways
– This information can be exploited to control and
manipulate the course of metabolic events
(Pharmaceutical purpose)
Several kinetics terms to understand
•
•
•
•
•
rate or velocity
rate constant
rate law
order of a reaction
molecularity of a reaction
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Before we learn enzyme kinetics
kinetics…
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• Let’s review some general ideas…
– Chemical kinetics
– Activation energy
– Transition state
– Reaction rate
Chemical Kinetics Provides a Foundation
for Exploring Enzyme Kinetics
• Enzyme kinetics is based on chemical
kinetics:
• First-order reaction = unimolecular
reaction Æ molecularity =1
– The simple elementary reaction of AP
A P is a
first-order reaction
– Examples
• Intramolecular rearrangement reaction
• Isotope decay
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The Time-Course
Time Course of a First
First-Order
Order Reaction
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Chemical Kinetics: First-order
First order reaction
• Consider a reaction of overall stoichiometry as
shown:
Ao P
d [ P]
v
dt
[[ A]
v
dt
d [ A]
dt
k[ A]
Figure 13.4 Plot of the course of a first-order reaction. The
h lf ti
half-time,
t1/2 is
i th
the titime ffor one-half
h lf off th
the starting
t ti amountt off
A to disappear.
• The rate is proportional to the concentration of A
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Chemical Kinetics: Second-order
Second order reaction
Summary of Chemical Kinetics
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• Bimolecular reaction = second-order reaction
• Molecularities = 2
• A+BÆP+Q
– the rate law is
v = k [A][B]
• 2AÆP+Q
v = k [A]2
• Molecularities > 2 is rare and never
greater 3!
g
• R
Remember:
b Ki
Kinetics
ti cannott prove a
reaction mechanism. They can only rule
out various alternative hypotheses!
• Question: What is the unit of rate constant (k)?
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Two ways to increase chemical reaction rate!
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Energy diagram for a chemical reaction (AP)
Transition state
• The transition state sits at the apex of the
energy profile
fil iin th
the energy di
diagram
• A typical enzyme-catalyzed reaction must pass
through a transition state
proportional
p
to the
• The reaction rate is p
concentration of reactant molecules with the
transition-state energy
gy
(a) raising the temperature from T1 to T2
(b) adding a catalyst.
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Activation energy
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• Free energy of activation, 'G‡
– the energy required to raise the average energy of
1 mol of reactant to the transition state energy (at a
given temperature).
• Decreasing G‡ increases the reaction rate
gy is related to the rate
• The activation energy
constant by Arrhenius equation:
k
13 3 What Equations Define the Kinetics of
13.3
Enzyme-Catalyzed Reactions?
• Enzyme is not involved in the chemical kinetics
of the reaction!
– True or False?
• A simple first-order reactions
– display a plot of the reaction rate vs. reactant
concentration as a straight line (Figure 13.6)
– If the same reaction with an enzyme involved, is
anything changed?
Ae ''G / RT
Note the difference between 'G and 'G‡
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Enzyme did change the chemical kinetics
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• Truth is more complicated!
• At low concentrations of the substrate
– The rate is proportional to S
S, as in a first-order
first order
reaction
• At higher
hi h concentrations
t ti
off substrate
b t t
– The enzyme reaction approaches zero-order
kinetics
Figure 13.6
A plot of v vs. [A] for the unimolecular chemical reaction,
A P yields a straight line having a slope equal to kk.
AP,
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Saturation Effect
We need a new equation for enzymecatalyzed chemical kinetics!
• Louis Michaelis and Maud Menten
invented a new equation which became
q
of enzyme
y
the fundamental equation
kinetics.
• What can we learn from M
M-M
M equation?
– We could know two major indexes of the
enzyme: Km and Vmax!
As [[S]] increases,, kinetic behavior changes
g from
st
1 order to zero-order kinetics
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Two More Assumptions of M
M-M
M equation
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Th Mi
The
Michaelis-Menten
h li M t Equation
E
ti
• A
Assumptions
ti
off M-M
M M enzyme reaction
ti
1. It assumes the formation of an enzymesubstrate complex
2. It assumes that the ES complex is in
rapid equilibrium with free enzyme
3. Breakdown of ES to form products is
assumed to be slower than
1) formation of ES and
2) breakdown of ES to re
re-form
form E and S
4. [ES] Remains Constant:
also known as steady state
assumption
ti by
b Briggs
Bi
&
Haldane in 1925
5 Rate measurement is
5.
finished right after
Substrate added
–
•
D[ES]/dt = 0
To ignore E + P Æ ES
These assumption would
simplify the following
calculating
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Understanding Km
• Km is
i th
the "kinetic
"ki ti activator
ti t constant”
t t” derived
d i d
from rate constants
• Learning Point!
– Three meanings of Km!
– How to get Km of an enzyme?
Km Step 1
• According to M-M assumption, enzyme
presented as…
reaction could be p
• The
e sy
synthesis
t es s rate
ate o
of ES
S
Vf = k1 [E] [S]
• Therefore, the dissociation of ES
Vd = k-11 [ES] + k2 [ES]
• Please pay attention!
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Km Step 2
• Briggs & Haldane assumed that Vf = Vd
under steady-state!
y
• Therefore….
Vf = k1 [E] [S] = k-1 [ES] + k2 [ES] =V
Vd (Eq1)
(E 1)
• Please
ease remember
e e be
– [E]: concentration of enzyme without substrate
binding!
g
– [ES]: concentration of enzyme binding with substrate!
– We don’t know [[E]] and [ES],
[ ], actually.
y We onlyy know
[ET]: concentration of total enzyme!
Km Step 3
• Because [ET] = [E] + [ES]
Therefore [E] = [ET] - [ES]
• Try to reduce unknown value in Eq1
k1 [E] [S] = k-1 [ES] + k2 [ES]
k1([ET] - [ES]) [S] = (k-1 + k2) [ES]
• Move all constants to one side, then….
([ET] - [ES]) [S]
k-1 + k2
k1
[ES]
Km (Eq2)
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Km
k-11 + k2
k1
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The First Meaning of Km
Sum of ES dissociation rate constants
ES synthesis rate constant
• An enzyme with bigger Km:
– Dissociation rate constants are larger or
synthesis rate constant is smaller
– Literally,
Li
ll substrate
b
and
d enzyme are diffi
difficult
l to
bind!
• What if an enzyme with small Km?
• Therefore,
Therefore Km is related to the enzymeenzyme
substrate binding capacity!
Km Step 4
• In all kinds of kinetics studies, the most
p
issue is reaction rate,, “V” !
important
V = k2 [ES]
(Eq3)
• We
W don’t
d ’t kknow [ES] exactly,
tl but
b t we
know…
Km
( T] - [ES])
([E
S ) [S]
S
[[ES]]
• Tryy to move [[ES]] to one side…
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The Second Meaning of Km
• After [ES] moved to one side
side, Eq2
becomes..
[ET] y [S]
[ES]
(Eq4)
Km + [S]
• Now combine Eq3 and Eq4, you will get
V
k2 [ET] y [S]
(Eq5)
Km + [S]
k2 [ET] y [S]
V
(Eq5)
Km + [S]
• As you see in Eq5, V increases when [S]
increased.
increased
• If [S] becomes very high, what happened
t V?
to
– Enzyme
y
will be saturated!
– Km + [S] Æ [S]
– V Æ Vmax
• Km is related to the reaction rate!
– An enzyme with smaller Km may has higher reaction
rate!
Km Step 5
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Km Step 6
• Now
N
E
Eq5
5b
becomes
k2 [ET] y [S]
Vmax
[S]
k2 [ET]
(Eq6)
• We could also combine Eq5 and Eq6,
p
make it simpler!
V
Vmaxy [[S]]
Km + [S]
The Third Meaning of Km
• From Eq7, we could redefine the meaning
of Km as..
Km
[S]y(Vmax – V)
V
• That
at means
ea s when
e
Vmax
V
[S]y((
[S]
Vmax
– 1) (Eq8)
V
- 1 = 1,, Km = [S]
V
– Also means max = 2 Æ V = ½ Vmax
V
(Eq7)
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• When reaction rate is ½ Vmax, the
corresponding [S] equals to Km!
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Take a look at the Km values for some
enzymes and their substrates
O h iindexes
Other
d
ffor an enzyme
• kcat, the turnover number
– A measure of catalytic activity
– Defines the activity
y of one molecule of enzyme
y
– If the M-M model fits, k2 = kcat = Vmax/Et
– Values of kcat range from less than 1/sec to
many millions per sec
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The Turnover Number Defines the Activity
of One Enzyme Molecule
O h iindexes
Other
d
ffor an enzyme
• The Ratio kcat/Km
– Defines the catalytic efficiency of an
enzyme
– An estimate of "how perfect" the enzyme is
– kcat/Km is an apparent
pp
second-order rate
constant
– It measures how the enzyme performs when S
i llow
is
– The upper limit for kcat/Km is the diffusion limit the rate at which E and S diffuse together
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The Ratio kcat/Km Defines the Catalytic
y
Efficiency of an Enzyme
How to measure the Km?
• The nature of Michaelis-Menten
Michaelis Menten equation:
– Combination of 0-order and 1st-order kinetics
– Describes
D
ib a rectangular
t
l h
hyperbolic
b li
dependence of v on S
• According to Eq8…
½ Vmax
But this is not good for
calculating
calculating…
[S] here is Km
Catalytic perfection
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Because of Vmax
• Vmax is a theoretical maximal velocity and is
a constant
• Vmax is NEVER achieved in reality
• For
F example:
l
To reach a rate of 99% Vmax
V
Vmax
0.99
Linear Plots Can Be Derived from the
Michaelis-Menten Equation
V
• That means you have to prepare
substrate [S] = 99 Km
• It is not possible practically.
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Km + [S]
(Eq7)
• Rearrange to obtain the Lineweaver-Burk
equation:
ti
1
v
[S]
Km + [[S]]
Vmaxy [S]
§ Km
¨
© Vmax
·§ 1 ·
1
¸¨
¸
[
S
]
©
¹ Vmax
¹
• A plot
l t off 1/
1/v versus 1/[S] should
h ld yield
i ld a straight
t i ht
line
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The Lineweaver-Burk double-reciprocal plot
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Y = aX + b
H
Hanes-Woolf
W lf Linear
Li
Plot
Pl t
• Begin with Lineweaver
Lineweaver-Burk
Burk and divide both
sides by [S] to obtain:
[S ]
v
§ 1
¨
© Vmax
·
Km
[
S
]
¸
Vmax
¹
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Hanes Woolf Plot is Better - Why?
Hanes-Woolf
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• Hanes-Woolf plot
p
• [S]/v versus [S]
• ‫ٱ‬smaller and more
consistent
i t t errors across the
th
plot
More about kinetics
• pH may effect on Km or Vmax or both in
y
kinetics
enzyme
• What if the plot is not linear?
– Regulatory enzymes (allosteric enzymes)
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End of Part 2
• Ask yourself…
– How many
y indexes you
y have learned to
describe the kinetic properties of an enzyme?
– What is M
M-M
M equation? Do you know what’s
what s
different from enzyme kinetics and chemical
kinetics?
– Do you understand the three meanings of Km
and know how to conduct from Eq1 to Eq8?
– Do you understand how to plot LineweaverB k plot,
Burk
l t and
d estimate
ti t Km and
d Vmax from
f
it?
13.4 What Can Be Learned from the
Inhibition of Enzyme Activity?
• Enzymes may be inhibited reversibly or
irreversibly
• Reversible inhibitors
– may bind at the active site or at some other site
• Competitive inhibition
• Noncompetitive inhibition
• Uncompetitive inhibition
• Irreversible inhibitors
– Covalent
C
l t modification
difi ti
• Kinetically similar to ___cometitive inhibition!
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Irreversible inhibitor
Example:
• Penicillin is an irreversible
inhibitor of glycopeptide
transpeptidase
Reversible Inhibitors May Bind at the
Active S
Site or at S
Some O
Other S
Site
– Suicide inhibitor
– catalyzes
t l
an essential
ti l step
t
in bacterial cell all synthesis
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Competitive inhibition
• Inhibitor, I, also binding to enzyme and
p
for the same binding
g site with
compete
substrate
E + S
E + I
K1
K-1
K3
K-3
K2
ES
Kinetic of competitive inhibition
Step 1
• Because [[ET] = [E]
[ ] + [ES]
[ ] and
([ET] - [ES]) [S]
E + P
k1
[ES]
EI
• So…
k-1 + k2
[ES] =
Km (Eq2)
[E] [S]
Km
(Eq9)
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• Because
E + I
K3
K-3
EI
• Vf = Vd under steady-state!
• Therefore….
Vf = k3 [E] [I] = k-3 [EI]
[EI] =
k3
k-3
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Lin’s B
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Kinetic of competitive
p
inhibition
Step 2
[E] [I] = 1/kI [E] [I]
(Eq10)
Kinetic of competitive inhibition
Step 3
• Knowing [ET] = [E] + [ES] + [EI]
[ET] = [E] +
[E] [S]
[E] [I]
+
Km
KI
(Eq12)
• Put [E] to one side
side….
(E 11)
(Eq11)
[E] =
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KI Km [ET]
(KI Km + KI [S] + Km [I])
(Eq13)
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Lin’s B
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Kinetic of competitive inhibition
Step 4
• Rate is still the most important!
V = k2 [ES]
• According to Eq9
[ES] =
[E] [S]
Km
• it could be changed to
to…
k2 [[E]] [S]
[ ]
V=
Km
(Eq9)
• Combine Eq13 and Eq14 to redefine rate!
V =
k2 KI Km [ET] [S]
Km(KI Km + KI [S] + Km [I])
(Eq15)
• Because Vmax = k2 [ET] …
V
(Eq14)
Vmaxy [S]
[S] + Km ( 1 + [I]/KI)
(Eq16)
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Effect of Competitive Inhibitors on
Lineweaver-Burk plot
V
Kinetic of competitive inhibition
Step 5
Vmaxy [S]
[S] + Km ( 1 + [I]/KI)
M
[I]
[I],
• More
– Less rate!
– Larger Km!
– Why?
Example of Competitive Inhibition
• Succinate Dehydrogenase (SDH)
• Vmax unchanged!
– Why?
y
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•
•
•
•
S and I bind to different sites on the enzyme
I could bind to E or ES
2 KI
2 types:
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Noncompetitive inhibition
ES + I
KI
KI’
• Inhibitor binds to E or ES equally
– KI = KI’
• Km not change!
– Why?
Wh ?
• Vmax smaller!
– Pure noncompetition
– Mixed noncompetition
E + I
Pure Noncompetitive Inhibition
– Why?
EI
IES
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Mixed Noncompetitive Inhibition
• Binding of I influences binding of S
• 2 situations:
Uncompetitive Inhibition
• I combines only with ES and affect k2
(kcat)!
• Km smaller!
– Why?
Wh ?
• Vmax smaller!
– Why?
• Slope is not changed!
– Why?
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Lin’s B
Biochemistryy Lecture
Bimolecular Reactions Catalyzed
by Enzymes
• Most enzymes catalyze reactions involving
two (or more) substrates
• 2 Reaction mechanisms
– Sequential (single-displacement)
reactions
– Ping-pong (double-displacement)
reactions
ti
Single displacement Reactions
Single-displacement
• Two distinct classes:
– Random: either substrate may
y bind first,,
followed by the other substrate
– Ordered: a leading substrate binds first,
first
followed by the other substrate
• Similar
Si il tto noncompetitive
titi iinhibition!
hibiti !
– Why?
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• All possible binary enzyme-substrate and enzymeproduct complexes are formed rapidly and
reversibly
• Conversion of AEB to PEQ is the Rate-Limiting Step
• If A has no influence on B binding Æ purely random!
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Random Single
Random,
Single-Displacement
Displacement Reactions
Example of Random, Singledisplacement enzyme
• Creatine Kinase (important in muscle)
+ ATP
Creatine
Kinase
+ ADP
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Lin’s B
Biochemistryy Lecture
Lineweaver Burk Plot for single
Lineweaver-Burk
singledisplacement bisubstrate enzyme
Purely random
Ordered Single
Ordered,
Single-Displacement
Displacement Reaction
• The leading substrate (obligatory or
compulsory substrate), A, must bind first
followed by B.
• Reaction
R
ti b
between
t
A and
d B occurs iin th
the
ternary complex and is usually followed by
an ordered release of the products, P and Q.
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An Alternative way of Portraying the
Ordered, Single-Displacement Reaction
• A and P are competitive for enzyme!
• A and B are not competitve for enzyme.
Example for Ordered, SingleDisplacement Reaction
• Alcohol dehydrogenase (ADH)
– Use NAD+ ((nicotinamide adenine dinucleotide))
as coenzyme or compulsory substrate
NAD+ + CH3CH2OH
• No A
A, no B binding!
(A)
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(B)
NADH + H+ + CH3CHO
(Q)
(P)
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Double Displacement Reaction
• Aka. Ping-Pong reactions
– Formation of a covalently
y modified enzyme
y
intermediate.
– The product of the enzyme’s
enzyme s reaction with A
(called P in the above scheme) is released
prior to reaction of the enzyme with the
second substrate, B.
An Alternative Presentation of the DoubleDisplacement (Ping-Pong) Reaction
• A and Q compete for E
E
• P and B compete for E’
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Lineweaver-Burk
Lineweaver
Burk Plot for the Double
Displacement Reaction
Similar to
uncompetitive
i hibiti
inhibition,
why?
h ?
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Example for the Double Displacement
Reaction
• Glutamate:aspartate
Aminotransferase
– Amino acid
metabolism
– Enzyme bond
coenzyme
– Pyridoxal
y
p
phosphate
p
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Lin’s B
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End of Part 3
• Ask yourself…
– What is the difference of reversible and
irreversible inhibition?
– What is competitive,
competitive noncompetitive
noncompetitive,
uncompetitve inhibition? What are they look
like in Lineweaver-Burk
Lineweaver Burk plot?
End of this chapter
• You should know…
–
–
–
–
–
–
–
–
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What characteristic features define enzymes?
How can enzymes be so specific?
Are all enzymes proteins?
Is it possible to design an enzyme to catalyze any desired
reaction?
Can the rate of an enzyme-catalyzed reaction be defined in a
mathematical way?
What equations define the kinetics of enzyme-catalyzed
reactions?
What can be learned from the inhibition of enzyme activity?
What is the kinetic behavior of enzymes catalyzing bimolecular
reactions?
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