Enzymes/ Kinetics
Study of enzyme catalyzed reactions
Rate of reactions
Enzyme specificity
Mechanism of catalysis
There are several good reasons for adding kinetic modeling. Mechanism and
dynamics are intimately related, and kinetic modeling reveal new and useful
information about biological control systems.
Observation during steady state is
not informative about the process
Catalyst
Catalyst=
Enzyme= biocatalyst
e.g.
Fermentation of sugar to ethanol by yeast enzymes
Conversion of fatty acids to polyketide antibiotics by filamentous fungi
Conversion of milk to cheese by microorganisms
Conversion of sugar to CO2 by bakers yeast to make leavened bread
BioCatalyst
MOST Enzymes are Proteins
Enzyme advantages over chemical catalysis:
Enzyme concentration is very small and enzyme is almost always limiting
Enzyme function
S <--------->P
1
2
3
DGo =standard free energy change for chemical reactions
DG’o =standard free energy change for biological reactions
298o k (RT); 1 atm, 1M substrate, pH7.0
Active site- pocket in enzyme that binds substrates- most complimentary in
structure to transition state
Rationale for enzyme kinetics
Conformation of proteins and positions of side chains are important for enzyme-substrate
interactions and catalysis.
Forces involved in protein folding and structure are also involved in catalysis- enzymesubstrate specificity
Trypsin hydrolyses proteins by cleaving peptide bond adjacent to Lys/Arg.
Aspartate residue in trypsin active site mediates ionic interaction with Lys /Arg and this
arranges protein residue at which hydrolysis occurs.
To use enzymes in biotechnology, pharmaceutics or drugs that inhibit enzymes in medicine,
you NEED TO KNOW KINETIC PARAMETERS OF THE ENZYME REACTION.
We may want enzymes that WORK FAST- convert more substrate in a fixed unit of time. To
do this optimization we have to perform and analyze the enzyme catalyzed reaction.
You can adjust pH, temperature and add co-factors to optimize enzyme activity.
You cannot adjust substrate selectivity.
Just like chemical reactions, enzyme catalyzed reactions have kinetics and rates
Reaction kinetics is Michaelis-Menten kinetics.
Enzyme-substrate cycle
Free energy
Activation energy
Ground state= starting point (energetically speaking)
-
-
Free energy
Energy Barriers
[S]
[P]
-
Sucrose to CO2 and H2O
C12H22O11 + 12O2 <-----> 12CO2 + 11 H2O
Reaction has large negative DG’o
Therefore Product prevails at equilibrium
In your pantry you can store sucrose in the presence of oxygen. It does not
spontaneously convert into CO2 and water!
ACTIVATION energy of this reaction is HUGE
=
/
DG
In a cell sucrose is rapidly converted to CO2 and H2O
ENZYMES!!!
Activation energy barriers are required for ordered life!!!
Without barriers molecules would spontaneously interconvert- there would be no
regulation
Reaction coordinate
Overall std free
energy change
(S goes to P)
Catalyzed Vs uncatalyzed reaction
Rate enhancements
Enzymatic rate enhancements are 103 to 1017 x
How do ENZYMES carry out catalysis?
•
•
•
~
~
~
Whats the Bill?
5.7 kJ/mol is needed to achieve a 10x increase in rate of a reaction
Typical weak interactions are 4-30 kJ/mol
Typical binding event yields 60-100 kJ/mol
MORE THAN ENOUGH ENERGY!!!
A to B to C
A------------> B------------>C
DG
conc
The rate constant for B to C is smaller. B to C is slow (rate determining)
A is rapidly converted to B but B accumulates because its conversion to C is slow
Reaction is in only one direction because DG of B is lower than A and C is lower than B
B to C is rate limiting
time
Rxn coordinate
A to B to C
A------------> B------------>C
DG
conc
The rate constant for A to B is smaller. A to B is slow (rate determining)
A is not rapidly converted to B because its conversion to B is slow. B never accumulates
because it is rapidly converted to C
Reaction is in only one direction because DG of B is lower than A and C is lower than B
A to B is rate limiting
time
Rxn coordinate
A to B to C
A------------> B------------>C
<-----------
DG
conc
The rate constant for A to B is small. The rate constant for B to A is large. A to B is slow but
B to A is very fast and B to C is kinda fast.
A is not rapidly converted to B. B never accumulates because it is very rapidly reconverted to
A and C accumulation is not very rapid because most of B is reconverted back to A rather
than to C
Reaction is in two directions because DG of B is higher than A and C is lower than B and A
time
Rxn coordinate
A to B to C
A------------> B------------>C
<-----------
DG
conc
The rate constant for A to B is small. The rate constant for B to A is large. A to B is slow but B
to A is kinda fast and B to C is very fast.
A is not rapidly converted to B. B never accumulates because it is very rapidly converted to C.
or reconverted to A. Accumulation of C is rapid because most of B is converted to C rather
than reconverted to A
Reaction is primarily in one directions because while DG of B is higher than A, C is lower
than B and A
A to B is rate limiting
time
Rxn coordinate
Enzyme catalyzed Steady State
Enz + Sub <-------> Enz-Sub <-------> Enz + Prod
[S]
Conc
[P]
[ES]
[E]
Time
Pre
steady
state
Steady state
During steady state
formation of ES equals its
breakdown
k1[E][S]=k-1[ES]+k2[ES]
Km= k-1+k2/k1
Rate of a Reaction
What is the rate of a chemical reaction
S---->P
-
A+B----.>P is a bimolecular reaction (there are two reactants) and this is a second order
reaction
-
Sometimes second order reactions appear as first order. E.g. If conc of B is very large
and does not change much then reaction rate is entirely dependent on conc of A only
Rate of catalysis
Vo=
For a fixed amount of enzyme,
The rate of catalysis
E+S
k1
k-1
ES
All measurements are done at very high substrate conc and very low product
conc so the reverse reaction is rare. We can simplify the above reaction
scheme as
Rate constants
A
k1
k2
-----> B ------> C
<----k-1
If k-1 is greater than k2 (B reforms A faster than it forms C) then there will be a
rate-determining pre-equilibrium and overall rate of formation of C will depend on
ALL THREE RATE CONSTANTS
((((d[C]/dt= [k1][k2]/[k-1][A]))))
If k2 is vastly greater than k-1 (B forms C faster than it reforms A) then we can
effectively ignore the back reaction (B to A) and the only question then is whether
k1 or k2 is rate limiting.
E+S
k1
ES
k2
E+P
k-1
k1 is rate constant for formation of ES
k-1 is rate constant for conversion of ES to E+S
k2 is rate constant for formation of product = kcat
1)
Binding of substrate to enzyme is reversible
2)
Product release from enzyme is instantaneous
[ES]
Rate of formation = k1[E][S]
Rate of breakdown = k-1[ES] + k2[ES]
Assume steady state
–As ES is produced, it reacts
–[ES] remains constant
–Rate formation = rate breakdown
So, k1[E][S] = k-1[ES] + k2[ES]
k1[E][S] = (k-1 + k2)[ES]
Rearrange
[E][S] / [ES] = (k-1 + k2) / k1
Where (k-1 + k2) / k1 = KM (Michaelis constant)
Define total enzyme concentration [E]T = [E] + [ES]
Substitute for [E]
([E]T – [ES])[S] / [ES] = KM
Solve for [ES]
[ES] = [E]T[S] / KM + [S]
[ES] = [E]T[S] / KM + [S]
Since
V0 = k2[ES]
V0 = k2[E]T[S] / KM + [S]
Define
maximum velocity Vmax
Occurs
Enzyme
at high [S]
is all in ES form
[ES]
= [E]T
Vmax = k2[E]T
Therefore
V0 = Vmax[S] / KM + [S]
Rate of reaction increase with [S]
Rate levels off as approach Vmax
–More S than active sites in E
–Adding S has no effect
At V0 = ½ Vmax
–> [S] = KM
The classic Michaelis-Menten rate equation
Michaelis-Menten
The first order rate constant kcat is k2
Vmax is product of kcat and enzyme conc
Km is mixture of rate constants that describe formation and dissociation of enzymesubstrate complexes
Km gives a sense of the affinity of enzyme for substrate
When [substrate] is >>> Km, the reaction kinetics are equal to max rate (Vmax).
When [substrate] is <<< Km, the reaction rate is kcat/Km
All these analyzes are done at saturation kinetics
Kcat and Km are PRIMARY INDICATORS of how well an enzyme will react with a
particular substrate.
High kcat= fast reactions. low kcat= slow reactions
High Km = low affinity of enzyme for substrate. Low Km = high affinity of enzyme for
substrate
Michaelis-Menton Equation
-
–V0 =
What is the [ES]?
What do we measure?
--------
A+B<--->AB
At low [S], Km >> [S]
At high [S], [S] >> Km
Product (mol/lit)
[S]4
[S]3
Vo(4)
[S]2
Vo(3)
Vo(2)
[S]1
Vo(1)
Time
Initial Velocity
Velocity measured at very beginning of reaction when very little product is made
Initial velocity is measured at saturation kinetics- at high [S], enzyme is saturated with
respect to substrate
Vmax
[Product]
Vo
Time
[Substrate]
Vmax and Km
V0 = Vmax
[S]
[S]+Km
Km= k-1 + k2
k1
When V0=Vmax, Km= [S]
Km is unique to each Enzyme and Substrate. It describes properties of enzyme-substrate
interactions
Independent of enzyme conc. Dependent on temp, pH etc.
Vmax is maximal velocity POSSIBLE. It is directly dependent on enzyme conc. It is
attained when all of the enzyme binds the substrate. (Since these are equilibrium
reactions enzymes tend towards Vmax at high substrate conc but Vmax is never achieved.
So it is difficult to measure).
When an enzyme is operating at Vmax, all enzyme is bound to substrate and adding more
substrate will not change rate of reaction (enzyme is saturated). (adding more enzyme will
change the reaction).
Measuring Km and Vmax
1/vo
Vmax
vo
1/Vmax
[Substrate]
-1/Km
1/[S]
You can use a curve fitting algorithm to determine Km and Vmax from a V vs [S] plot (need
a computer)
Reaction rates are initial rates determined when the substrate is in vast excess and isn’t
changing much.
Alternatively you can convert the curve to a straight line via a double reciprocal plot
(1/Vmax and 1/[S])
Reciprocal
It is not easy to extrapolate a hyperbola to its limiting value
(computers can do this)
The Michaelis-Menten equation can be recast into a linear form
To obtain parameters of interest
Reciprocal form of equation
1 = Km
V Vmax
1
S
+ 1
Vmax
Y= m
x +
b
The y-intercept gives the Vmax value and the slope gives Km/Vmax
Vmax is determined by the point where the line crosses the 1/Vi = 0 axis (so the [S] is infinite).
Km equals Vmax times the slope of line. This is easily determined from the intercept on the X axis.
1/Vo
It is difficult to accurately measure Vmax
[S]
V0 = Vmax
[S]+Km
Reciprocal
1
Km
V0 = Vmax
1
S
+ 1
Vma
Vmax = k2[Et]
k2 is also called Kcat
Et is conc of active sites in enzyme
Km values of enzymes range from 10-1M to 10-7M for their substrates. It varies depending on substrate,
pH, temp, ionic strength etc.
Kcat is turnover number for the enzyme-number of substrate molecules converted into product per unit
time by that enzyme
Km and Vmax
Km is [S] at 1/2 Vmax
It is a constant for a given enzyme at a particular temp and pressure
It is an estimate of equilibrium constant for substrate binding to enzyme
Small Km= tight binding, large Km=weak binding
It is a measure of substrate concentration required for effective catalysis
Vmax is THEORETICAL MAXIMAL VELOCITY
Vmax is constant for a given enzyme
To reach Vmax, ALL enzyme molecules have to be bound by substrate
Kcat is a measure of catalytic activity- direct measure of production of product
under saturating conditions.
Kcat is turnover number- number of substrate molecules converted to product
per enzyme molecule per unit time
Catalytic efficiency = kcat/km
Allows comparison of effectiveness of an enzyme for different substrates
Enzyme Km examples
Hexokinase prefers glucose as a substrate over ATP
Kcat
Catalase is very efficient-it generates 40 million molecules of product per second.
Fumarase is not efficient-it generates only 800 molecules/per second
kcat = Vmax / [E]T
Turnover
number
Number of reaction processes each active site catalyzes per unit time
Measure of how quickly an enzyme can catalyze a specific reaction
For M-M systems kcat = k2
Kcat/Km
Rate
constant of rxn E + S ---> E + P
Specificity
Gauge
constant
of catalytic efficiency
Catalytic
perfection ~ 108 -109 M-1 s-1 (close to diffusion)
Enzyme cofactors
Coenzymes
How do ENZYMES carry out catalysis?
•Entorpy reduction- holds substrates in proper position
Bringing two reactants in close proximity (reduce entropy & increase effective reactant concentrat)
•Substrate is desatbilized when bound to enzyme favoring reaction-(change of solvent, chargecharge interactions strain on chemical bonds).
•Desolvation of substrate- H bonds with water are replaced by H bonds with active site
Enzymes form a covalent bond with substrate which stabilizes ES complex (Transition state is
stabilized)
Enzyme also interacts non-covalently via MANY weak interactions
Bond formation also provides selectivity and specificity
(H bonds- substrates that lack appropriate groups cannot form H bonds and will be poor substrates)
(Multiple weak interactions between enzyme and substrate)
Free energy released by forming bonds is used to activate substrate (decrease energy barrier/lower
activation energy of reaction)
Induced fit-binding contributes to conformation change in enzyme
Whats the Bill?
5.7 kJ/mol is needed to achieve a 10x increase in rate of a reaction
Typical weak interactions are 4-30 kJ/mol
Typical binding event yields 60-100 kJ/mol
MORE THAN ENOUGH ENERGY!!!
A Hypothetical reaction
Breaking a stick
Imagine you have to break a stick. You hold
the two ends of the stick together and apply
force. The stick bends and finally breaks.
You are the catalyst. The force you are
applying helps overcome the barrier.
A stickase with a pocket complementary in structure to the stick (the substrate) stabilizes the substrate. Bending is
impeded by the attraction between stick and stickase.
An enzyme with a pocket complementary to the reaction transition state helps to destabilize the stick, contributing to catalysis of
the reaction. The binding energy of the interactions between stickase and stick compensates for the energy required to bend the
stick.
Role of binding energy in catalysis. The system must acquire an amount of energy equivalent to the amount by which DG‡
is lowered. Much of this energy comes from binding energy (DGB) contributed by formation of weak noncovalent interactions
between substrate and enzyme in the transition state.
Lock/Key or Induced Fit
Lock/Key- Complementary shape
The enzyme dihydrofolate reductase with its substrate NADP+
NADP+ binds to a pocket that is complementary to it in shape and ionic properties, an illustration of "lock and key" hypothesis
of enzyme action. In reality, the complementarity between protein and ligand (in this case substrate) is rarely perfect,
Induced Fit
Hexokinase has a U-shaped structure (PDB ID 2YHX). The ends pinch toward each other in a conformational change
induced by binding of D-glucose (red).
Substrate specificity
The specific attachment of a prochiral center (C)
to an enzyme binding site permits enzyme to
differentiate between prochiral grps
Enzyme-substrate
Catalysis
•Acid-Base Catalysis- donate or accept protons/electrons from and to substrate
•Covalent Catalysis-transient covalent link between substrate and enzyme side
chain
•Metal-Ion Catalysis-Metal in active site donate or accept protons with substrate
•Proximity & Orientation Effects (reduction in entropy-two mol brought together and
oriented in specific manner)
•Transition State Preferential Binding
The active sites of enzymes
contain amino acid R groups.
Active site is lined with
hydrophobic residues
R Groups
Polar amino acid residues in
active site are ionizable and
participate in the reaction.
Anion/cation of some amino
acids are involved in catalysis
Lysozyme: Cleaves glycosidic bonds in carbohydrates
Covalent Catalysis
All or part of a substrate is transiently covalently bound to the enzyme to form a reactive
intermediate
Group X can be transferred from A-X to B in two steps via the covalent ES complex -EX
A-X+ E <-----> X-E + A
X-E + B <-----> B-X+ E
Without a catalyst the intermediate converts
back to the reactants and does not proceed
forward (high barrier).
Donation of a proton by water or an acid helps
the process move forward.
The active sites of enzymes contain amino
acid R groups, that participate in the
catalytic process as proton donors or proton
acceptors.
Catalysts
Proton donor/acceptor (Nucleophile/electrophile)
Asp and Glu are negatively charged at pH7.0 and their side chains are acidic.
These side chains ACCEPT protons which neutralize the charge.
Lys, Arg, His are positively charged at pH 7.0 and their side chains are basic.
These side chains DONATE protons to neutralize their charge.
Asp/Glu
Lys/Arg
COO- + H+ <-----------> COOH
NH3+ <----------> NH2 + H+
Nucleophiles
R-OH <---> R-O: + H+ (hydroxyl)
R-SH <---> R-S: + H+ (sulphydryl)
R-NH3 <---> R-NH2: + H+ (amino)
Electrophiles
H+
M+
Proton
Metal ion
R
+C
O
Carbonyl
R’
Nucleophiles-groups rich in and capable of donating electron (attracted to nucleus)
Electrophile- group deficient in electron (attracted to electron)
Reactions are promoted by proton donors (general acids) or proton acceptors (general bases). The active sites of some
enzymes contain side groups, that can participate in the catalytic process as proton donors or proton acceptors.
Acid-base catalysis
Reaction acceleration achieved by catalytic transfer of proton
A general base (B:) acts as proton acceptor to remove proton from OH, NH, CH (XH)
This produces a stronger nucleophilic reactant (X:)
A general base(B:) removes a proton from water thereby generating the equivalent of
OH-in neutral solution.
:B
X
H
:X-
B+
H
•A general acid (BH+) can donate a proton. A covalent bond may break more easily if one of its atoms is
protonated.
Acid base catalysis
RNaseA cleavage of RNA
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Enzyme Inhibition
S
P
S
EE
E+S
E
ES
P
E
EP
E
E+P
Enzyme Inhibition
Many molecules inhibit enzymes
Competitive Inhibitor







Competitive Inhibitor
Most
common
Inhibitor
competes with natural substrate for binding to active site
Inhibitor
similar in structure to natural substrate and binds active site of enzyme (reducing
effective enzyme conc)
Binds
May
If
more strongly
or may not react
reacts, does so very slowly
Gives
info about active site through comparison of structures
Drug targets
Gleevec
Gleevec: How it works
HIV protease structure
Protease Inhibitors
Protease + Inhibitor











Reversible Inhibition (competitive)
1/v
Vmax
-Inh
+Inh
-Inh
+inh
vo
1/2 Vmax
1/Vmax
Km
Km
(app)
[Substrate]
-1/Km -1/Km (app)
Inhibitor competes with substrates for binding to active site
Inhibitor is similar in structure to substrate
binds more strongly
reacts more slowly
Increasing [I] increases [EI] and reduces [E] that is available for substrate binding
Need to constantly keep [I] high for effective inhibition (cannot be metabolized away in body)
Slope
is larger (multiplied by a)
Intercept
KM
does not change (Vmax is the same)
is larger (multiplied by a)
1/[S]
Uncompetitive Inhibitor
Binds only to ES complex but not free enzyme
Binds at location other than active site
Does not look like substrate. Binding of inhibitor
distorts active site thus preventing substrate
binding and catalysis
Cannot be competed away by increasing conc of
substrate (Vmax is affected by [I])
Increasing [I] lowers Vmax and lowers Km.
Increasing
[I]
Lowers
Vmax (y-intercept increases)
Lowers
KM (x-intercept decreases)
Ratio
of KM/Vmax is the same (slope)
Mixed
Inhibitor binds E or ES
Increasing [I]
Lowers
Raises
Ratio
Vmax (y-intercept increases)
KM (x-intercept increases)
of KM/Vmax is not the same
(slope changes)
Non-competitive inhibition
Inhibitor binds ES or E
It is a special case of mixed inhibition where Vmax is lowered when [I] increases but Km does not
change
Reversible Inhibition (non-competitive)
A inhibitor binds the enzyme but not in its active site. It affects the Kcat because
substrate can still bind the active site.
Rate of catalysis is affected
+Inh
1/v
Vmax
-Inh
Vmax
(app)_
-Inh
1/Vmax
(app)
+inh
vo
1/2 Vmax
1/2 Vmax
(app)
1/Vmax
-1/Km
Km
Km
(app)
[Substrate]
Vmax is decreased proportional to inhibitor conc
1/[S]
Example
When a slice of apple is cut, it turns brown- enzyme o-diphenol oxidase oxidizes phenols in the apple
Lets determine max rate at which enzyme functions (Vmax), and Km
1
When it acts alone (we will use catechol as substrate. Enz converts this to o-quinone which is dark
and can be measured via absorbance at 540 nm
2
when it acts in presence of competitive inhibitor para hydroxy benzoic acid which bind active site
but is not acted upon
3
when it acts in the presence of a non-competitive inhibitor- phenylthiourea which binds copper in
the enzyme which is necessary for enzyme activity
Make a supernatant of the apple-enzyme. Measure color produced (product)
Set up 4 tubes with different conc of cathecol and a fixed amount of enzyme (apple pulp).
Measure change in absorbance at 1 min intervals for several minutes and record average change in
absorbance.
Absorbance is directly proportional to product, we can measure rate of reaction (velocity)
TubeA
[S]
1/[S]
Vi (DOD)
1/Vi
mM
TubeB
mM
TubeC
mM
TubeD
mM
1/Vmax=10
Vmax=0.1
-1/Km=-0.8
Km=1.25mM
Example
Each tube also has a fixed amount of PHBA (competitive inhibitor)
TubeA
[S]
mM
TubeB
mM
TubeC
mM
TubeD
mM
1/[S]
Vi (DOD)
1/Vmax=10
Vmax=0.1
-1/Km=-0.4
Km=2.5 mM
1/Vi
Each tube has a fixed amount of phenylthiourea (non competitive inhibitor)
TubeA
[S]
1/[S]
Vi (DOD)
1/Vi
mM
TubeB
mM
TubeC
mM
TubeD
mM
1/Vmax=20
Vmax=0.05
-1/Km=-0.8
Km=1.25 mM
Irreversible
Inhibitor
Binds
covalently, or
Destroys
functional group
necessary for enzymatic
activity, or
Very
stable noncovalent
binding
Suicide
Inactivators
Starts
steps of chemical
reaction
Does
not make product
Combines
irreversibly
with enzyme
Regulation of enzymes
Catalytic activity is increased or decreased
1) Enzyme synthesis or degradation
2) Covalent modification
3) Non-covalent binding and conformational change (allosteric)
Usually located early in multi-enzyme reaction pathway
Kinetics differ for allosteric enzymes- sigmoidal curve and K1/2 instead of Km
Usually
large; multiple subunits
Compare
Site
Can
to Hb
for allosteric modulator (R = regulatory) generally different from active site (C = catalytic
be positive or negative
xxxxxxx