Lecture 15
– Enzymology
– Today, Howard Salis (148 Baker) 3PM.
Extra credit seminar-assignments due
Monday.
Enzyme-catalyzed reactions must involve formation of an
enzyme-substrate complex, followed by one or more
chemical steps
• Vmax and Km are two key measurable properties of
enzymes.
Vmax:
• kcat: the rate constant for the catalytic step carried out by
the enzyme.
• Vmax: the rate at which a given amount of enzyme
catalyzes a reaction at saturating concentrations of
substrate.
kcat [Etotal] = Vmax
• kcat is interpreted as a measure of the rate of chemical
conversion of substrate to product.
Enzyme-catalyzed reactions must involve
formation of an enzyme-substrate complex,
followed by one or more chemical steps
• Vmax and Km are two key measurable properties of
enzymes - Km
Km:
• Km is the ratio of the rate constants for the individual steps:
Km = (k-1 + kcat)/k1.
• If kcat is small compared to k-1, then Km = Ks, the
dissociation constant for the enzyme-substrate complex.
• Thus for many enzymes, Km can be interpreted as a
measure of the affinity of the enzyme for its substrate.
• Km also = [S] at which v = 1/2 Vmax
Some Rules on Interpreting Michaelis-Menten
Enzyme Kinetic Parameters (from A. Fersht
text)
• kcat is a 1st order rate constant that refers to the
properties and reactions of the enzyme-substrate,
enzyme-intermediate (‡) and enzyme product
complexes.
• Km is an apparent dissociation constant that may
be treated as the overall dissociation constant of
all enzyme bound species.
• kcat/Km is an apparent second order rate constant
that refers to properties of the free enzyme and
the free substrate.
Michaelis-Menten Enzyme Kinetics
E+S
k1
[ES]
E+P
k-2
k-1
Basic Michaelis-Menten
Equation:
If we take the inverse of MM eq:
v0 = Vmax[S]
1/v0 =
KM + [S]
1/v0 =
k2
KM
Vmax
1
1
+
Vmax
[S]
KM + [S]
Vmax[S]
Lineweaver-Burk Inversion
Use the equation for a straight line: y = ax + b
slope
1/V0 =
KM
Vmax
1
1
+
Vmax
[S]
Plot 1/v (= y) versus 1/[S] (= x)
slope = KM/Vmax
1/V
x-intercept = -1/KM
y-intercept
y-intercept = 1/Vmax
1/[S]
Figure U2-2.3 Lineweaver-Burk or
reciprocal kinetic plot of 1/v against 1/[S]
1. Original raw kinetic data
2. Transformed (inverse) data
Note: this L-B method weights data incorrectly, puts emphasis on
data near minimum rate (Vmax) : (e.g., 1/ 0.05 = 20, 1/0.1 = 10)
You should perform non-linear fit of actual raw data to equation with
computer software (KaleidaGraph, Sigmaplot, Orig, GraphPad
Prism, kinetic fitting packages, etc.)
Eadie-Hofstee Plot
Another inversion method: multiply MM eq by V0Vmax
v0 = Vmax[S]
KM + [S]
V0
+ Vmax
V0 = - KM
[S]
Plot v (= y) versus v/[S] (= x)
Vmax
v
slope = -KM
Vmax/KM
v/[S]
Hanes-Wilkinson Plot
Another inversion method: multiply MM eq by [S]
slope
v0 = Vmax[S]
KM + [S]
y-intercept
1
S/V0 =
[S] +
Vmax
KM
Vmax
Plot 1/v (= y) versus 1/[S] (= x)
[S]/V
-KM
Slope =1/Vmax
KM/Vmax
[S]
Enzyme Kinetics: General Principles
•
•
•
Remember, Vmax = maximal rate of an enzyme (Etotal = ES)
KM = (k-1 + k2)/k1
KM is the [S] at which V = 1/2 (Vmax)
Basic Michaelis-Menten
Equation:
If KM = [S]:
v0 = Vmax[S]
v0 = Vmax[S]
KM + [S]
v0 = Vmax[S]
2 [S]
[S] + [S]
= Vmax
2
Enzyme Kinetics: General Principles
k1
E+S
ES
k2
E+P
k-2
k-1
• If k-1 >> k2
for KM
k2+ k-1
k-1
=
=
k1
k1
= KS
Dissociation constant of the
enzyme and substrate
• If k2 >> k-1
E+S
k1
k-1
ES
k2
E+P
k-2
Enzyme reacts every time it interacts
with substrate. (see p. 480)
Example of Michaelis-Menten Enzyme Kinetics
Given this data, what is Vmax? What is KM?
[S] moles/L
2.00E-01
2.00E-02
2.00E-03
2.00E-04
1.50E-04
1.30E-05
1/[S]
5.00
50.00
500.00
5000.00
6666.67
76923.08
v (µmol/min)
60
60
60
48
45
12
1/v
0.01666667
0.01666667
0.01666667
0.02083333
0.02222222
0.08333333
First, graph [S] vs. v to make sure it obeys MM kinetics
v (µmol/min)
70
60
50
Vmax is 60 by inspection
40
30
20
10
0
1.00E-05
1.00E-04
1.00E-03
1.00E-02
[S]
1.00E-01
1.00E+00
Example of Michaelis-Menten Enzyme Kinetics
Given this data, what is Vmax? What is KM?
[S] moles/L
2.00E-01
2.00E-02
2.00E-03
2.00E-04
1.50E-04
1.30E-05
1/[S]
5.00
50.00
500.00
5000.00
6666.67
76923.08
v (µmol/min)
60
60
60
48
45
12
1/v
0.01666667
0.01666667
0.01666667
0.02083333
0.02222222
0.08333333
Since Vmax = 60 we can solve for KM, plug this into MM eq.
v0 = Vmax[S]
KM + [S]
KM = [S] Vmax
-1
v0
If v = 48, [S]= 2 X 10-4, KM = 5.0 X 10-5
If v = 12, [S]= 1.3 X 10-5, KM = 5.2 X 10-5
These should agree with one another!
Example of Michaelis-Menten Enzyme Kinetics
Given this data, what is Vmax? What is KM?
[S] moles/L
2.00E-01
2.00E-02
2.00E-03
2.00E-04
1.50E-04
1.30E-05
1/[S]
5.00
50.00
500.00
5000.00
6666.67
76923.08
v (µmol/min)
60
60
60
48
45
12
1/v
0.01666667
0.01666667
0.01666667
0.02083333
0.02222222
0.08333333
We can also check by Lineweaver-Burke plot
0.09
0.08
1/v
0.07
0.06
0.05
1/Vmax
0.04
1/V0 =
0.03
-1/KM
KM
Vmax [S]
-20000
1
+
Vmax
0.02
Scale is important
0.01
0
-40000
1
0
20000
40000
60000
1/[S]
80000
100000
Graph of rate against total substrate
concentration for a typical enzyme-catalyzed
reaction
• In this example, saturating
concentrations of substrate B
are converted to product more
slowly than saturating
concentrations of substrate A
• (Vmax is lower for substrate B
than for substrate A), but
substrate B binds more tightly to
the enzyme (reflected in its
lower Km).
• The value of kcat/Km is higher for
substrate B than A, indicating
that the enzyme is more specific
for substrate B.
Enzyme-catalyzed reactions can have multiple
steps with several intermediates
• Most enzyme-catalyzed reactions are more complex than we have presented
so far.
• Frequently multiple substrates are involved, and even when they are not, the
reaction often proceeds with the formation of one or more semi-stable species.
• These species, which do not dissociate from the enzyme and are on the
pathway from substrate to final product, are called intermediates.
• In such reactions, there will be one step that is slower than all the others.
• In theory this can be any step in the reaction, including substrate binding
(diffusion-controlled reactions) or product release, but often the slowest step is
a chemical step involving the formation of one of the intermediates.
• When the overall reaction rate is measured, what will be observed is the rate
of this slowest step. Hence, it is referred to as the rate-determining step of
the reaction (Figure U2-3.2).
Figure U2-3.2 The rate-determining step
•
•
The rate-determining step of a
reaction is the step with the
largest energy barrier.
In this example, the height of the
barrier between intermediates I1
and I2 is greater than between S
and I1 or I2 and P, and the rate
determining step is therefore I1
to I2.
Michaelis-Menten Enzyme Kinetics
What happens if we consider the back reaction of the products?
E+S
k1
ES
k2
E+P
k-2
k-1
VProduct = k2[ES] - k-2[P][E]
If we assume the steady state d[ES]/dt =0, so Vproduct = Vreactant
k1[S][E] + k-2 [P][E] = k-1[ES] + k2[ES]
Solving for ES gives us:
[ES] = (k1+ k-2[P])[E]
and we know:
k-1+ k2
[ES] = [Etotal -E]
2 equations,
2 unknowns
Michaelis-Menten Enzyme Kinetics
What happens if we consider the back reaction of the products?
E+S
k1
ES
k-1
Solving
k2
k-2
[E] = (k-1+ k-2)[Etotal ]
k1[S] + k-2[P] + k-1+ k2
[ES] = (k1 [S] + k-2[P])[Etotal ]
k1[S] + k-2[P] + k-1+ k2
E+P
Michaelis-Menten Enzyme Kinetics
Substitute back into the initial equation and simplify:
VProduct = k2[ES] - k-2[P][E]
[ES] = (k1 [S] + k-2[P])[Etotal ]
k1[S] + k-2[P] + k-1+ k2
[E] = (k-1+ k2)[Etotal ]
k1[S] + k-2[P] + k-1+ k2
VProduct = k2((k1 [S] + k-2 [P])[Etotal ]) - k-2[P]((k-1+ k2)[Etotal ])
k1[S] + k-2[P] + k-1+ k2
VProduct = ((k2k1[S] - k2k-2[P])- (k-1k-2[P] + k2k-2[P]))[Etotal ]
k1[S] + k-2[P] + k-1+ k2
Michaelis-Menten Enzyme Kinetics
Substitute back into the initial equation and simplify:
VProduct = (k2k1[S] - k-2k-1[P])[Etotal ]
k1[S] + k-2[P] + k-1+ k2
E+S
k1
ES
k2
k-1
If we define
for KM =
k2+ k-1
k1
E+P
k-2
for KP =
k2+ k-1
k-2
and for Vmax
forward
Vmax = k2[Etotal]
reverse
Vmax = k-1[Etotal]
Michaelis-Menten Enzyme Kinetics
Substitute and rearrange:
VProduct = (k2k1[S] - k-2k-1[P])[Etotal ]
k1[S] + k-2[P] + k-1+ k2
VProduct = k2k1[S][Etotal ] - k-2k-1[P ][Etotal ]
k1[S] + k-2[P] + k-1+ k2
for KM =
forward
k2+ k-1
k1
Vmax = k2[Etotal]
for KP =
reverse
k2+ k-1
k-2
Vmax = k-1[Etotal]
Michaelis-Menten Enzyme Kinetics
Substitute and rearrange:
VProduct = k2k1[S][Etotal ] - k-2k-1[P ][Etotal ]
k1[S] + k-2[P] + k-1+ k2
forward
reverse
VProduct = Vmaxk1[S]- k-2 Vmax[P ]
k1[S] + k-2[P] + k-1+ k2
forward
Vmax = k2[Etotal]
reverse
Vmax = k-1[Etotal]
Michaelis-Menten Enzyme Kinetics
Substitute and rearrange:
forward
reverse
VProduct = Vmaxk1[S]- k-2 Vmax[P ]
k1[S] + k-2[P] + k-1+ k2
Solve in terms of k1 and k-2:
for KM =
k2+ k-1
k1
k2+ k-1
k1 =
KM
forward
for KP =
k2+ k-1
k-2
k-2 = k2+ k-1
KP
reverse
VProduct = Vmax((k2+k-1)/KM)[S]- ((k2+k-1)/KP)Vmax[P ]
((k2+k-1)/KM)[ S] + ((k2+k-1)/KP)[P] + k-1+ k2
Michaelis-Menten Enzyme Kinetics
Substitute and rearrange:
reverse
forward
V = Vmax((k2+k-1)/KM)[S]- ((k2+k-1)/KP)Vmax[P ]
((k2+k-1)/KM)[ S] + ((k2+k-1)/KP)[P] + k-1+ k2
Divide by k2+k-1
V
forward
reverse
= ((Vmax/KM)[S]- (Vmax /KP) [P ] )
[S] /KM + [P] /KP + 1
Multiply by KmKp
forward
V
=
reverse
KPVmax[S]- KMVmax [P ]
KMKP + KP [S] + KM [P]
Michaelis-Menten Enzyme Kinetics
Substitute and rearrange:
forward
Vproduct =
reverse
KPVmax[S]- KMVmax [P ]
KMKP + KP [S] + KM [P]
But, V(dP/dt) = Vforward - Vreverse, so we break it down into the V forward
and V reverse reactions.
Vforward =
forward
Vmax[S]
KM + [S] + [P] (KM/ KP)
reverse
Vreverse =
-Vmax[P]
KP + [P] + [S] (KP/ KM)
Note at P=0, we get the Michaelis-Menton Equation
This derivation was for an enzyme that has two
possible reactions
• E+S
• E+P
• [P] inhibits the forward reaction by combining with E.
• Note that as [S] increases, then the maximum rate is Vmax
[P](KM/KP) is insignificant at high [S].
• Because the Product and Substrate are competing for the
same site, this is the model for competitive inhibition.
Enzyme reactions can be slowed by the presence
of inhibitors
• Other inhibitors bind noncovalently and reversibly to their
target enzymes.
• These are usually divided into three broad classes,
competitive, noncompetitive, and uncompetitive,
depending on their manner of binding.
• Kinetic analysis can distinguish among these inhibitors if
the reaction rate is measured against substrate
concentration at different inhibitor concentrations.
Figure U2-4.1 Competitive, noncompetitive
and uncompetitive inhibition
Figure U2-4.1a Competitive
inhibition
(a) A competitive inhibitor (I) binds to
the same site as does the substrate
(S; top).
The inhibitor changes the apparent Km
for the reaction, but not the Vmax
because enough substrate can keep
any inhibitor from binding.
A Lineweaver-Burk plot (bottom) for
the reaction at various concentrations
of the inhibitor reflects this behavior.
E
ES
EI
Competitive Inhibition
S+E
ES
P
v0 =
+I
Vmax[S]
KM + [S] +[I] (KM/KI)
EI
KI = [E][I]
[EI]
v (µmol/min)
70
Vmax
60
50
w/o I
w/ I
40
30
20
10
0
1.00E-05
1.00E-04
1.00E-03
1.00E-02
[S]
1.00E-01
1.00E+00
Competitive Inhibition
ES
S+E
v0 =
P
+I
Vmax[S]
KM + [S] +[I] (KM/KI)
1/V0 = KM 1
Vmax [S]
EI
1/V0 = KM 1
Vmax [S]
(1+ [I] ) + V1
KI
() + V1
max
I
w/o I
1/v0
-1/KM
-1
KM(1+ [I] )
KI
1/Vmax
-1
KM
1/[S]
-1/KM’ or apparent KM
max
Competitive inhibition
•  = 1 +[I]/KI
• As [I] increases, v decreases (1/v increases)
• As [I] increases, KM decreases (1/KM increases)
• Vmax is the same and the inhibition can be overcome by
high [S]
Figure U2-4.1b Noncompetitive inhibition
(b) A noncompetitive inhibitor (I) (green)
does not bind to the substrate binding
site and can bind to both the free enzyme
or the ES complex.
Usually a noncompetitive inhibitor
resembles one substrate (S2) in a twosubstrate reaction, as shown here, where
both substrates are present on the
enzyme at the same time.
In the simplest cases, noncompetitive
inhibitors don't change the Km for the first
substrate (S), because they don't affect
its binding.
But providing the concentration of S2 is
not high enough to out-compete all the
inhibitor, the inhibitor does reduce the
Vmax for the reaction.
E
ES
EI
Noncompetitive Inhibition (Mixed Inhibition)
ES
S+E
KI
+I
KI’
P
v0 =
+I
KM + ’[S]
 = 1 +[I]/KI
’ = 1 +[I]/KI’
EIS
EI + S
KI’ = [ES][I]
[ESI]
KI = [E][I]
[EI]
v (µmol/min)
70
Vmax[S]
Vmax
60
50
w/o I
40
w/ I
30
20
10
0
1.00E-05
1.00E-04
1.00E-03
1.00E-02
[S]
1.00E-01
1.00E+00
Noncompetitive Inhibition (Mixed Inhibition)
ES
S+E
KI
+I
EI + S
KI’
P
v0 =
Vmax[S]
+I
KM + ’[S]
EIS
1/V0 = KM 1
Vmax [S]
() + V’
max
I
w/o I
1/v0
-1/KM
1/Vmax
1/[S]
Noncompetitive inhibition
•  = 1 +[I]/KI, ’ = 1 +[I]/KI’
• As [I] increases, v decreases (1/v increases)
• As [I] increases, KM decreases (1/KM increases), but in
the simplest case does not change (very slight).
• Cannot be overcome by increasing [S]
• May intersect above, blow or even on the line (x or y axis)
• x-axis - no affect on KM (KM = KM’)
• y-axis - no affect on Vmax
Figure U2-4.1c
Uncompetitive inhibition
(c) An uncompetitive inhibitor (blue)
binds only to the enzyme-substrate
(ES) complex and slows down the
reaction probably by inducing a
conformational change in the
enzyme.
Both the apparent Km and Vmax are
affected proportionally by such an
inhibitor, leading to parallel
Lineweaver-Burk plots for different
inhibitor concentrations.
Uncompetitive Inhibition
ES
P
S+E
KI’
+I
KI’ = [ES][I]
[ESI]
EIS
v0 =
Vmax[S]
KM + ’[S]
1/V0 = KM 1
Vmax [S]
’ = 1 +[I]/KI’
+
I
w/o I
1/v0
1/Vmax
1/[S]
’
Vmax
Uncompetitive inhibition
• ’ = 1 +[I]/KI’
• Reacts only with ES complex
• As [I] increases, v decreases (1/v increases)
• As [I] increases, apparent KM increases (apparent 1/KM
decreases), but there is no effect on binding of E to S.
• Cannot be overcome by increasing [S]
• Relatively rare in single substrate reactions but can be
more common in complex cases.
Enzyme reactions can be slowed by the presence
of inhibitors
• A key parameter that can be obtained from such an
analysis is the affinity of the inhibitor for the enzyme, the
inhibition constant Ki.
• By convention, Ki is given as the dissociation constant for
the enzyme-inhibitor equilibrium:
Ki = [E][I]
[EI]
Ki’ = [ES][I]
[ESI]
• The lower the value of Ki the tighter the inhibitor binds. In
pharmacology, the value of Ki is often used as a measure of the
effectiveness of a drug.
• A compound with a very low Ki, say 10-9 M (nanomolar) or less, can
be given at very low doses and will still be able to bind its target.
Enzyme Inhibitor Classification
1. Competitive inhibitors:
•
Most common class of reversible inhibitor consists of
compounds that resemble the substrate.
•
Such molecules can fit into the substrate binding site,
thereby blocking access from substrate molecules.
These inhibitors compete with the substrate for the
active site.
•
Example: Many HIV protease inhibitors that have proven
to be effective in treatment of AIDS are competitive
inhibitors that were designed to resemble the peptide
substrate of the HIV protease.
Enzyme Inhibitor Classification
•
Not all inhibitors compete with the substrate for the active site of the
enzyme; other inhibitors bind to a separate site.
2. Noncompetitive inhibitors bind to both the free enzyme
and the enzyme-substrate complex.
•
If an enzyme has two substrates that must bind simultaneously for
the reaction to occur, an inhibitor might compete with one substrate
but not the other.
Enzyme Inhibitor Classification
•
Not all inhibitors compete with the substrate for the active site of the
enzyme; other inhibitors bind to a separate site.
3. Uncompetitive inhibitors bind only to the enzymesubstrate complex.
•
In effect, the inhibitor reduces the amount of ES that can
go on to form product.
•
Lineweaver-Burk plots characteristically show a series of
parallel lines (Fig. U2-4.1c).
•
Uncompetitive inhibitors often stabilize an alternative
conformation of the protein, and in this case they are
called allosteric inhibitors.
•
There are also allosteric activators: molecules that
activate enzymes by stabilizing a conformation of the
enzyme that is more active than the conformation that
exists in their absence.
Figure 3-10 Ligand-induced conformational change activates
aspartate transcarbamoylase
Binding of the allosteric activator ATP to its
intersubunit binding sites on the regulatory
subunits (that between R1, outlined in purple,
and R6 is arrowed) of the T state of ATCase
(top) causes a massive conformational
change of the enzyme to the R state
(bottom).
In this state the structure of the enzyme is
opened up, making the active sites on the
catalytic subunits (C) accessible to substrate.
Al and Zn in the lower diagram indicate the
allosteric regions and the zinc-binding region,
respectively; cp and asp indicate the binding
sites for the substrates carbamoyl phosphate
and aspartate, respectively.
The red and yellow regions are the
intersubunit interfaces that are disrupted by
this allosteric transition.
Allostery
• Does not follow Michaelis-Menton kinetics!
• EIS goes to products at the same rate as ES but with
lower affinity for [S]
• Doesn’t affect Vmax.
• Does affect KM
• Can operate in both directions and involves a second
binding site.
• Positive-activator, cooperative
• Negative-inhibitor, antagonistic
• Enzyme controlled by binding at second site
homotrophic (modifier is related to the substrate)
heterotrophic (related to substrate).
Enzyme-catalyzed reactions can have multiple
steps with several intermediates
• Multi-step reactions can have very complicated kinetics,
e.g., “double-displacement” or “ping-pong” enzymes.
• These are enzymes that use two or more substrates but
catalyze reactions that are strictly ordered in the sequence
in which substrates bind and products are released (Figure
U2-3.3a).
• Lineweaver-Burk plots of the velocity against one substrate
concentration at a series of fixed concentrations of other
substrate give a family of parallel lines (Fig. U2-3.3b).
Insert: double reciprocal plot of
observed initial
velocities versus CO2 concentration for
CA at different concentrations of
TAPS buffer: 5 mM, 10 mM, 20 mM, 50
mM.
Biochemistry 1999, 38, 13119-13128
Enzyme-catalyzed reactions can have multiple
steps with several intermediates
Example of Ping-Pong Enzyme: Aspartate aminotransferase: Catalyzes the conversion
of aspartate to glutamate with production of oxaloacetate and consumption of alphaketoglutarate.
The reaction sequence starts with the binding of aspartate to the enzyme followed by its
conversion to oxaloacetate, in the process of which the aspartate leaves behind its amino
group bound to a cofactor in the active site.
After oxaloacetate departs (the so-called ping step), alpha-ketoglutarate reacts with the
amino group and is converted to glutamate (the so-called pong step), bringing the enzyme
back to its original state.
The two substrates, aspartate and alpha-ketoglutarate, never encounter each other on the
enzyme. The kinetics are characteristically simple for this kind of reaction:
Lineweaver-Burk plots of the velocity against aspartate concentration at a series of fixed
concentrations of alpha-ketoglutarate, give a family of parallel lines (Fig. U2-3.3b).
Figure U2-3.3 Ping-pong or doubledisplacement kinetic behavior
•
•
(a) In ping-pong reactions, two substrates bind sequentially to an
enzyme. In this example, a chemical group (green) is transferred from
substrate A (red) to substrate B (blue).
(b) A Lineweaver Burk plot of 1/v against 1/[substrate A] at various
fixed concentrations of substrate B shows a set of parallel lines which
are diagnostic for the ping-pong reaction mechanism.
Enzyme reactions can be slowed by the presence
of inhibitors
• The rate of an enzyme-catalyzed reaction can be affected by
molecules that do not themselves participate in the chemical
reaction.
• Activators increase the reaction rate and inhibitors decrease the rate.
• Many drugs, including aspirin, penicillin, statins and Viagra are
enzyme inhibitors:
• they achieve their pharmacological effects by reducing the rate of a
key enzyme-catalyzed reaction.
• In some cases the effect is achieved by forming a dead-end covalent
complex between the inhibitor and enzyme.
• Penicillin and aspirin work this way: they form stable chemical bonds
with residues in the active sites of the enzymes they inhibit.
• Such “suicide inhibitors” permanently inactivate their target
enzyme molecules, and cells can only overcome their effects by
synthesizing fresh enzyme.