Lecture 14
– Intro to enzymes
– This Friday: Seminar speaker Howard
Salis (148 Baker) 3PM. Extra credit
seminar
Thermodynamics
•
•
•
Determines if the reaction is spontaneous (does it occur).
Does not tell us how fast a reaction will proceed.
Catalysts (enzymes) can lower the activation barrier to get
from products to reactants.
Thermodynamics of Enzyme Function
• Catalysts lower the energy barrier
between reactants and products.
Free energy diagrams of simple chemical catalysis
Figure 2-21 Energetics of catalysis (cont.)
In the presence of a catalyst (red curve), which in this
case is acting by changing the pathway of the
reaction and introducing additional smaller activationenergy barriers, intermediate I1, formed by crossing
transition-state barrier TSc1, leads to transition-state
barrier TSc2.
Its free energy (ΔGc), although the highest point in the
reaction, is considerably lower than the free energy
(ΔGu) of the uncatalyzed transition state, TSu.
After formation of a second intermediate, I2, a third
transition state, TSc3, leads to product.
Because TSc2 is the highest transition state in the
catalyzed reaction, the rate at which the reactants
pass over this barrier determines the overall rate and
thus it is said to be the rate-determining transition
state of the catalyzed reaction.
The rate-determining step of this reaction is thus the
conversion of I1 to I2.
Figure 2-21
Energetics of
catalysis
Figure 2-21 Energetics of catalysis (cont.)
The transition state is the highest point in free energy
on the reaction pathway from substrate to product.
It is the top of the activation-energy barrier (see TSu in
Fig. 2-21).
Chemically, it is a species that exists for about the time
required for a single atomic vibration to occur (about 10-15
s).
In the transition state, the making or breaking of chemical
bonds in the reaction is not yet complete: the atoms are
"in flight".
The stereochemistry and charge configuration of the
transition state is thus likely to be quite different from that
of either the substrate or the product, although it may
resemble one more than it does the other.
Figure 2-21
Energetics of
catalysis
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.
U2-1 Enzyme Kinetics: General Principles
•
•
•
•
•
•
•
Enzymes function as biological catalysts to increase the rate
(speed) of chemical catalysis.
Reaction rates reflect key properties of enzymes and the
reactions they catalyze
Kinetics is the study of how fast chemical reactions occur.
The free energy change can tell us in which direction the reaction will
spontaneously occur.
The free energy change does not tell us how rapidly.
Some spontaneous reactions occur quickly, e.g. sec., others occur
almost imperceptibly over many years.
The rate of a chemical reaction or process, or the reaction rate, is the
change in the concentration of reacting species (or of the products of
their reaction) as a function of time (Fig. U2-1.1).
Figure U2-1.1 Reaction rates measure how fast processes
occur
(a) In this example, two reacting
species A and B (red and blue)
combine to form a product C
(purple).
(b) The concentration of product
molecules increases as the reaction
proceeds, and within two seconds
of the reacting species being mixed
together, the concentration of C
becomes 10 mM.
(c) The rate of the reaction during these
two seconds is therefore 5 mM s-1,
which is the slope of the line when
we plot the concentration of the
product against time.
Figure 6.6 A plot of initial reaction velocity versus the
concentration of enzyme [E].
Note that velocity
increases in a
linear fashion with
an increase in
enzyme
concentration.
Add more catalyst,
get faster reaction
rate.
-Not surprising!
Figure U2-1.2 Rate constants are measured
from reaction rates at different reactant
concentrations
• Follow the progress of the
reaction between molecules A
and B, as shown in Figure U21.1:
• Rate at which product C is
produced decreases as the
reaction proceeds (Fig. U21.2a).
• The rate decreases because
the concentration of reactants
decreases.
• The possibility also exists that
the reverse reaction (C → A +
B) becomes significant as the
concentration of product C
increases.
Figure U2-1.2 Rate constants are measured
from reaction rates at different reactant
concentrations
• To avoid these complications, we
can measure the initial rate of the
reaction (i.e, 5-10% of total reaction
time):
The rate before the concentration
of reactants decreases significantly
and before the accumulation of
product is able to interfere with the
reaction (Fig. U2-1.2b).
Rate constants are measured from reaction
rates at different reactant concentrations
•
•
The initial reaction rate for this example depends upon
the concentrations of the reactants (denoted as [A] and
[B]) according to the equation:
Rate ∝ [A] [B]
The reaction rate doubles if the concentration of A is
doubled, as the reactants collide twice as often.
•
A more useful measure is called the rate constant, k,
which tells us how the reaction rate varies with the
concentrations of the reactants:
Rate = k [A] [B]
•
Note: Kinetic rates use small letter k, not to be confused
with equilibrium constant, Keq
Figure U2-1.2 Rate constants are measured from
reaction rates at different reactant concentrations
• The units of k will depend on
the number of reacting
molecules.
• In this example (bimolecular
reaction) the units would be
M-1 s-1.
• For a unimolecular reaction,
such as the conversion of
molecule A to molecule B, the
units of k would be s-1.
The rate constant is measured from the slope of the line
when we plot the initial reaction rate at different
concentrations of one of the reactants (Fig. U2-1.2c).
Reaction orders
•
Rate equations show the frequency with which reacting molecules
come together dependent upon their concentration.
•
Rate = k[A]a[B]b…[Z]z
•
The order of a reaction is defined as the sum of exponents in the rate
equation.
First order
(unimolecular)
k is the rate constant
R
k1
P
Velocity (v) = -d[A] = d[P]
dt
dt
Rate of formation of P = k1[R]
Reaction orders
•
For a reaction that is reversible
First order
(unimolecular)
R
k1
k2
P
Rate of formation of P = k1[R] - k2[P]
Reaction orders
Second order
(bimolecular)
k1
A+B
P
k2
Rate of formation of P = k1[A][B] - k2[P]
2nd order forward and 1st order reverse
If A = B:
2A
k1
k2
P
Rate of formation of P = k1[A]2 - k2[P]
1st and 2nd order reactions are common
Reaction orders
Third order
(termolecular)
A+B+C
k1
k2
P
Rate of formation of P = k1[A][B][C] - k2[P]
3rd order forward and 1st order reverse,
nearly impossible and 4th order reactions
are not known
Reaction orders
The steady state assumption: assumes that the reverse reactions are
slow
A+B
k1
k2
[AB] + C
k3
P
k4
If k3 is slow, the rate of P formation = k3[C][AB].
However, [AB] is  k1[A][B]
Therefore, the net rate is k1[A][B][C] = 3rd order.
If k1 is slow, the rate will appear as k1[A][B] = 2nd
order.
Figure U2-2.1 The binding of substrate to an enzyme is
dependent on the substrate concentration
At low substrate concentrations,
only a small fraction of enzyme
molecules (blue) will have
substrate (red) bound at any
moment (top).
If the substrate concentration is high,
practically all the enzyme molecules
will contain bound substrate (bottom).
At these high substrate concentrations,
if a substrate molecule dissociates from
the enzyme, another substrate
molecule will almost immediately
collide with the enzyme to take its
place.
Enzyme kinetics
Catalytic activity maybe so fast that the reaction is not
rate limiting but rather the binding of the substrate to
the enzyme. Therefore, by studying the order of
binding, you have some idea of the reaction
mechanism.
Enzyme kinetics
Early observation:
•
Unusual effect of substrate concentration on the rate
of reaction (catalytic rate)
•
v0 - Initial velocity (rate of reaction) increases upon
adding more substrate (we are talking about molar
ratios of Enzyme E and substrate S);
•
But only up to a point!
•
Net behavior observed is a hyperbola.
•
Has asymptotic upper limit in the number of substrate
molecules processes per unit time per mole of
enzyme.
Figure 6.3 Experimental procedure to study the kinetics
of an enzyme-catalyzed reaction.
An identical
amount of
enzyme is
added to a set
of tubes
containing
increasing
amounts of a
substrate.
The reaction
rate or initial
velocity is
measured for
each reaction
mixture by
determining
the rate of
product
formation.
The rate (or velocity, v) of an
enzyme-catalyzed reaction
increases as the concentration of
substrate is increased.
The rate approaches a maximum
value (Vmax) as the substrate
concentration becomes so high that
all of the enzyme molecules are
occupied (saturated) with substrate.
The concentration of substrate at
which the rate is 1/2 Vmax is
denoted as Km (the Michaelis
constant).
(moles formed per second)
Figure U2-2.2 Graph of rate against total substrate
concentration for a typical enzyme-catalyzed reaction
(moles, molarity)
This curve can be fitted by the Michaelis-Menten equation:
v = Vmax [S]/(Km + [S]).
(Direct fit to model requires non-linear optimization.)
Vmax depends only upon the enzyme concentration and the rate constant, kcat.
Review of Enzyme Function
•
Enzymes generally function in the following
manner:
1. Recognize, bind specific chemical
compounds (“substrate(s)”) in solution.
2. Convert bound substrate to product via
lowering Gibbs free energy of intermediate
transition state between initial substrate, final
product.
3. Release weakly-bound product, prepare to
repeat new catalytic reaction cycle.
Michaelis-Menten Steady-State Enzyme Kinetics
Saturation kinetics with respect
to substrate concentration.
kcat = “turnover number” [s-1]
Km = “Michaelis constant” [M]
kcat/Km = “catalytic efficiency” [s-1 M-1]
Vmax = maximum kcat value possible when enzyme is saturated with
.
substrate
• Km is simply defined as the substrate conc. at which rate (“velocity”) v =
1/2 Vmax. Note: Km ≠ Kd for substrate binding!
Michaelis-Menten Equation
In 1913, based on experimental observations of
enzyme kinetics, Michaelis and Menten proposed
model:
E+S
k1
k2
ES
k3
E+P
k4
Where E is enzyme molecule
S is the substrate molecule
ES is the enzyme-substrate complex
P is the resulting product molecule
• Assumption (mostly valid in initial stages of forward
reaction and in hydrolysis reactions):
• k4 = 0 (no back reaction of product!)
Michaelis-Menten Enzyme Kinetics
E+S
k1
k-1
[ES]
k2
E+P
k-2
• ES is not the same as an activated complex
– Activated complex is energy maximum going from R to P
– Enzyme-Substrate (ES) is amount of substrate bound to
enzyme.
• Rate is of primary importance in [ES]. Maximum rate occurs
when all of the enzyme is in [ES] form (∞ [S])
– Assume steady state, therefore [S] = [S] initial (not - ES) and P
formation is irreversible.
Michaelis-Menten Enzyme Kinetics
E+S
k1
ES
k-1
For ES
k2
k-2
Free enzyme
Vform = k1[E][S]
but E = Etotal - ES
so, Vform = k1[Etotal - ES][S]
Vdeg = k2[ES] + k-1[ES]
E+P
E+S
k1
k2
ES
E+P
k-2
k-1
At the steady state, rate of formation = rate of degradation
so, vform = vdeg
vform = k1[Etotal - ES][S]
vdeg = k2[ES] + k-1[ES]
k1[Etotal - ES][S] = k2[ES] + k-1[ES]
[Etotal - ES][S] = k2+ k-1
k1
[ES]
k1
E+S
ES
k2
E+P
k-2
k-1
This defines KM
[Etotal - ES][S] = k2+ k-1 = K
M
k1
[ES]
[Etotal][S] - [ES][S] = KM
[ES]
[ES]
[Etotal][S] = KM + S
[ES]
E+S
k1
ES
k-2
k-1
[Etotal][S] = KM + S
[ES]
[Etotal][S] = [ES]
KM + S
Solve in
terms of
products
k2
vform = k2[ES]
= k2[Etotal][S]
KM + S
E+P
E+S
k1
ES
E+P
k-2
k-1
v = k2[ES]
k2
substitute
[Etotal][S] = [ES]
KM + S
v = k2[Etotal][S]
Rearrange the
terms
KM + S
[Etotal] = v(KM + S)
k2 [S]
vmax= k2[Etotal] = k2 v(KM + S)
k2 [S]
vmax= v(KM + S)
[S]
E+S
k1
ES
k-2
k-1
vmax= v(KM + S)
[S]
k2
Can be
rearranged in
terms of v
v0 = Vmax[S]
KM + [S]
E+P
Michaelis-Menten Equation
Basic Michaelis-Menten Equation:
v0 = Vmax[S]
KM + [S]
Where v0 = initial velocity concentration at substrate concentration
[S].
v0 = -D[Substrate]/Dt = D[Product]/Dt
Vmax = maximum velocity, rate of reaction
{moles per second }
[S] is the initial substrate concentration
KM = Michaelis constant
k k
KM =
2
k1
3
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
0.02
KM
Vmax [S]
- 20000
1
+
Vmax
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.