ANTHONY CARRUTHERS, BLOCK 1
Enzyme Kinetics
Fluorescence (%)
10
0 mM
1 mM
2 mM
3 mM
4 mM
5 mM
12.5 mM
8
6
4
2
0
0.001
0.01
0.1
1
Time (s)
GLUCOSE BINDING TO A GLUCOSE SENSOR
10
100
Foreword
Graduate students often characterize “kinetics” as
uninteresting. I have observed several explanations for
this:
• The subject matter may be difficult to the
mathematically-challenged biology student.
• Students may think that they will never undertake a
kinetic analysis and thus the study of kinetics is
unnecessary.
• The student may find the subject matter accessible
but nevetheless uninteresting.
My goals in developing this guide to kinetics are:
The chemistries that drive biological processes are
reversible, occur on a time-scale of 10-12 to 109 sec
and are most frequently catalyzed by enzymes. The
study of reversible biological reactions, their timedependence and the mechanisms of enzymemediated catalysis is called “enzyme kinetics”.
Enzyme-kinetics is central to every biological process
that ever has or will be studied and is the basis for a
great many assays that are routinely undertaken in
every research laboratory. Understanding enzyme
kinetics is, therefore, important if we are to understand
biological processes and the limitations of the assays
we undertake in order to study these processes.
• To show that “kinetic analysis” is simply one more
very powerful tool that biologists use to study
biological problems.
• To show that the use of the tools is straightforward
when the underlying principles and assumptions
are appreciated.
• To emphasize that the student does NOT have to
memorize equations and derivations - they are
included in this guide for your reference.
• To provide examples of analyses.
• To emphasize key points that student should
understand.
• To provide formative, self-evaluation questions with
keys in order that the student can evaluate their
understanding of the concepts.
i
1.2
Equilibria
Fractional equilibration
C HAPTER 1
0.8
0.4
0.0
0.1
3. How do enzymes accelerate
reactions?
100.0
1.2
Fractional Equilibration
2. When can a chemical reaction occur
spontaneously (by itself)?
10.0
Time (min)
This chapter considers reversible
chemical reactions. We ask:
1. What is a reversible chemical
reaction and what are its
characteristics?
1.0
0.8
0.4
0.0
0.1
1.0
10.0
100.0
Time (min)
Loss (◦) and uptake (●) of sugar by human red cells (lower
graph) and human red cell ghosts (upper graph) at 4ºC.
S ECTION 1
Reversible Reactions
L EARNING OBJECTIVES
1. Reversible reactions never come to a halt they achieve an equilibrium in which the
forward and reverse reactions are
quantitatively balanced.
2. For any reversible reaction, there is a fixed
relationship between the conentration of
products formed and substrates remaining at
equilibrium.
Law of Mass Action
A reversible reaction is one in which a
product can be formed from starting
material (substrate) and substrate can be
formed from the ending material
(product). This reaction never stops but
reaches an equilibrium in which the rate
of product formation from substrate is
identical to the rate of substrate
formation from product.
In a reversible reaction, there is a fixed
relationship between the concentrations
of reactants (substrate, S) and products
(P) at a given temperature in the
equilibrium mixture.
3. This relationship is unique to each reaction
and reaction conditions.
3
For example, the reaction between
hydrogen and iodine
For the reaction
H2 + I2
Glucose + ATP
k1
Glucose-6-phosphate + ADP
[HI]2
K eq =
[H 2 ][I 2 ]
k-1
K eq =
2HI; [Glucose − 6 − phosphate]e [ADP]e k1
=
[Glucose]e [ATP]e
k−1
where [ ]eq denote equilibrium
concentrations, k1 and k-1 are constants
describing the rates of forward and
reverse reactions respectively and Keq is
the equilibrium constant.
or, in general
pA + qB + rC
xD + yE + zF;
[D]x [E] y [F]z
K eq =
[ A] p [B]q [C]r
The equilibrium constant for any specific
reaction (at a given temperature) is rather
like a fingerprint - it is unique.
4
Displacement of the position of
equilibrium
If the concentration of any one of the
substances is altered in an equilibrium
mixture, the concentrations of the other
substances must change so as to keep
Keq constant.
Formative self-evaluation questions
In the reaction
S
k1
k -1
P
1. Define Keq in terms of k1, k-1
2. Define Keq in terms of [S]eq, [P]eq
[Glucose − 6 − phosphate]e [ADP]e
K eq =
[Glucose]e [ATP]e
3. What would happen if the reaction
were at equilibrium and more S were
added?
For example, if [ADP] is increased in an
equilibrium mixture by adding
exogenous ADP, [glucose-6-phosphate]
will fall by combination of glucose-6phosphate and ADP to increase
[Glucose] and [ATP]. The net effect, of
course, is that Keq is unchanged.
4. What would happen if the reaction
were at equilibrium and more P were
added?
5
S ECTION 2
The chemical reaction S ⇌ P
In which direction will a
reversible reaction
proceed?
can be described by the
following energy diagram:
X
Energy
We return to the chemical
intermediate X and the term
∆GA later.
Example 1
∆GA
S
∆G
P
Progress of Reaction
This reaction (Example 1) will proceed in the net
direction left to right. Why? Because the chemical
potential (energy) of S > P. (Note ∆G is independent of
the path of the reaction).
Now consider Examples 2 and 3 below
Example 3
Example 2
1. The chemical potential of the reactants
determines the net direction of the reaction.
Energy
L EARNING O BJECTIVES
X
∆GA
P
S
2. The Gibbs free energy change for a reaction,
∆G, is the chemical potential of the product(s)
minus the chemical potential of the reactant(s)
3. The reaction proceeds in the direction of high
chemical potential to low chemical potential
(∆G < 0)
∆GA
∆G
S
Energy
X
P
∆G = 0
Progress of Reaction
Progress of Reaction
Reaction 2 will proceed, net right to left (chemical
potential of P > S). Reaction 3 above will proceed
equally in both directions (no net reaction) because the
chemical potential of S = P.
6
Chemical potential and ∆G
Forms of Energy
The chemical potential of a molecule is a measure of
the ability of the molecule to perform work.
Free Energy (G) - performs work at constant
temperature and pressure.
∆G is the change in Gibbs free energy for the reaction.
Heat Energy (enthalpy, H) - performs work only
through a change in temperature.
Consider the reaction:S ⇌ P
The “chemical potential” or “partial molar free energy”
of S, for example is given by:
µS = µºS + RT ln[S]
Entropy (S) is the unavailable energy. Changes in the
free energy (∆G) and enthalpy (∆H) of a system (the
reactants) and changes in the entropies (∆S) of the
system plus the surroundings (universe) are related in
the following manner:
We will discuss the meaning of µºs very shortly, but
you can see that the chemical potential of S is directly
proportional to its concentration.
The free energy change (∆G) for the reaction S ⇌ P is
given by:
∆G = µP - µS
At equilibrium, the rate of P formation is matched
exactly by the rate of S formation. Thus the abilities of
P to form S (to do work) and vice versa are identical.
Hence, at equilibrium,
µP = µS
Thus,
∆G = 0
∆G = ∆H - T∆S
Thus:
∆G α T
When T∆S > ∆H, ∆G < 0
When T∆S < ∆H, ∆G > 0
What is the importance of the entropy change?
The irreversible increase in Entropy gives direction to
the reaction!
Imagine we have two glass flasks connected by a
valve. One flask has an internal volume of 10 mL and
the second has an internal volume of 100 mL. The 10
mL flask is filled with an ideal gas.
7
When the valve is opened between the two flasks, the
gas immediately expands to occupy both flasks. We
have all observed this type of behavior in one form or
another and know this to be a spontaneous reaction
(∴ ∆G < 0).
⇌
Since with our ideal gas there are no interactions
between gas molecules (∆H = 0). Thus:
Formative self-evaluation questions
In the reversible reaction
S⇌P
1.
Define ∆G and the relationship between µS and µP
when the reaction is at equilibrium.
2.
Define ∆G and the relationship between µS and µP
when the reaction proceeds in a net direction from
left to right.
3.
Define the ∆G and the relationship between µS
and µP when the reaction proceeds in a net
direction from right to left.
4.
At equilibrium, what is the relationship between
∆H and T∆S?
∆G = -T∆S
The fall in free energy is due to increased ∆S. The
molecules redistribute to maximize system entropy.
Summary
1. A reaction occurs spontaneously only if ∆G is
negative (µP < µS).
2. A system is at equilibrium (forward and reverse
reactions are balanced) and no net change occurs
when ∆G = 0 (µP = µS)
3. The forward reaction cannot occur spontaneously
when ∆G is positive. An input of free energy is
required to drive the reaction.
4. ∆G depends upon the free energy of the products
(final state) minus that of the reactants (initial state).
i.e. ∆G is independent of the reaction mechanism.
5. The irreversible increase in entropy provides a
directional driving force for the reaction.
8
S ECTION 3
∆G and equilibria
To proceed with the following discussion, we must first
understand 3 conditions.
The starting condition refers to the concentration of
reactants at the beginning of the reaction.
The equilibrium condition refers to the concentration
of reactants the reaction is at equilibrium.
The standard condition refers to the concentration of
reactants under standard conditions.
In the reaction:
A+B⇌C+D
[C] [D]
G = G o + RTln [A] [B]
∆Gº is the standard free energy change
R = gas constant (1.98 cal/mol/degree)1 T = absolute temperature
LEARNING
O BJECTIVES [A], [B], [C] and [D] are the molar activities of the
reactants under starting conditions.
1. Understanding starting, standard and
equilibrium conditions.
2. The relationships between ∆G, ∆Gº and Keq.
3. Understanding that standard free energy
changes are additive and how this is exploited
in nature.
1one
calorie (cal) is that amount of heat required to raise the
temperature of 1 gram of water from 14.5 ºC to 15.5 ºC. One joule (J)
is that energy needed to apply a 1 newton force over a distance of 1
meter. 1 kcal (1000 cal) = 4.184 kJ (4184 J)
9
∆Gº is the free energy change under standard
conditions
i.e. Defining Keq as
[C] [D]
K eq = [A] [B]
∆Gº = -RT ln Keq
[A] = [B] = [C] = [D] = 1 M
= -2.303 RT log10 Keq
P = 1 Atmosphere
pH = 7.0
Rearranging gives us
T = 298ºK = 25ºC
Keq = e-∆Gº/RT
This condition does not, however, describe [A], [B], [C]
and [D] (and therefore ∆G) under the conditions of the
reaction but is used only to describe ∆Gº.
At equilibrium, ∆G = 0 and the ratio [C] [D]
[A] [B]
remains constant
(i.e. the forward reaction is balanced by the reverse
reaction)
Thus the free energy equation becomes
[C] [D]
0 = G o + RTln [A] [B]
and
[C] [D]
G o =- RTln [A] [B]
=10-∆Gº/(2.303 RT)
Substututing R and T (25ºC) we obtain
Keq =10-∆Gº/1.36
when ∆Gº has units of kcal/mol.
Thus an equilibrium constant of 10 corresponds to a
∆Gº of -1.36 kcal/mol (see Table below)
∆Go
Keq per M
1,000,000
10,000
100
10
1
0.1
0.01
0.0001
0.0000001
kcal/mol
-8.2
-5.5
-2.7
-1.4
0.0
1.4
2.7
5.5
9.5
kJ/mol
-34.2
-22.8
-11.4
-5.7
0.0
5.7
11.4
22.8
39.9
Note: H-bond energies range from 3 to 7 kcal/mol. van der Waal’s bond energies
are approximately 1 kcal/mol.
10
Standard Free Energy Changes are Additive
Thus
Consider the following reactions:
∆G˚total = +13.8 kJ/mol +(-30.5 kJ/mol) = -16.7 kJ/mol.
The overall reaction is therefore exergonic.
A ⇌ B ∆Gº1
≡
We can also compute Keq for each reaction.
A ⇌ C ∆Gºtotal
[G6P]
K eq1 = [glucose] [P] = 3.9 x 10 -3 M -1
i
B ⇌ C ∆Gº2
The ∆G˚ of sequential reactions are additive, thus
∆Gºtotal = ∆Gº1 + ∆Gº2
This principle of bioenergetics explains how an
endergonic reaction (Keq < 1) can be improved (more
product formed) by coupling it to a highly exergonic
reaction (Keq >>1) through a common intermediate.
Consider the synthesis of glucose–6–phosphate – a
reaction that occurs in all cells:
Glucose + Pi
Glucose–6–phosphate + H2O
∆G˚ = 13.8 kJ/mol
ATP + H2O
ADP + Pi
∆G˚ = -30.5 kJ/mol
These reactions share common intermediates (Pi and
H20) and may be expressed as the sequential
reactions:
1)
Glucose + Pi
Glucose–6–phosphate + H2O
2)
ATP + H2O
ADP + Pi
sum:
ATP + glucose
(note H20 is not included)
K eq2 =
[ADP] [P]i
5
=
3.9
x
10
M
[ATP]
[G6P] [ADP] [P]i
K eqtotal [glucose] [P] [ATP]
i
=Keq1 * Keq2 = 7.8 x 102
Thus by coupling ATP hydrolysis to G–6–P synthesis,
the Keq for G–6–P formation is raised by a factor of 2 x
105.
This strategy of coupling one reaction with a low Keq
(or ∆Gº > 0) to a second with a high Keq (or ∆Gº < 0) is
used by all living cells in the synthesis of metabolic
intermediates.
ADP + glucose–6–phosphate
11
Summary
1.
At equilibrium, ∆G = 0
2.
Exergonic reactions are characterized by Keq > 1
and ∆Gº < 0
3.
4.
5.
Endergonic reactions are characterized by Keq < 1
and ∆Gº > 0
An endergonic reaction can be made more
favorable (i.e. Keq increases) by coupling it to a
second exergonic reaction via common chemical
intermediates. Here the ∆Gº of the reactions are
summative.
At 25ºC, and equlibrium constant of 10
corresponds to a ∆Gº of -1.36 kcal/mol
Formative self assessments
We will solve an exam-type question to illustrate
relationships between starting and equilibrium levels
of reactants and reaction spontaneity (direction of net
flux).
We use the isomerization of dihydroxyacetone
phosphate (DHAP) to glyceraldehyde 3-phosphate
(G3P) as our example.
Question:
Keq for the reaction DHAP ⇌ G3P is 0.1 (log10 Keq
= -1.). Defining: ∆G = ∆Gº+ {1.36 x log10[G3P]/
[DHAP]} kcal/mol. When the starting
concentrations of DHAP and G3P are 0.2 mM and 2
µM respectively (log10[G3P]/[DHAP] = -2).
F.
∆G = 1.36 kcal/mol and ∆Gº = -2.72 kcal/mol
G. Under the starting conditions stated above, the
reaction cannot occur spontaneously.
H. ∆Gº = 2.72 kcal/mol and ∆G = -1.36 kcal/mol
I.
Under the starting conditions stated above, the
reaction can occur spontaneously.
J.
The net reaction will proceed from right to left.
Solution
A.
At equilibrium, [G3P]/[DHAP] = 0.1 thus Keq = 0.1
12
B. Thus ∆Gº = - 1.36 x log10 0.1 = -1.36 x -1 =
+1.36 kcal/mol.
C. When the initial concentrations of DHAP and G3P
are 2 x 10-4 M and 2 x 10-6 M respectively we can
substitute these concentrations and ∆Gº into the
equation to obtain
D. ∆G = 1.36 kcal/mol + (1.36 x -2) = -1.36 kcal/
mol
E.
The answer is, therefore, D.
The negative value for ∆G indicates that the reaction
can occur spontaneously when the species are
present at the concentration stated above.
Note however that no calculations were necessary.
Since Keq > starting [G3P]/[DHAP], the forward
reaction must occur spontaneously (more G3P must
be formed) since all reactions must proceed to
equilibrium. If starting [G3P]/[DHAP] had been > Keq
the reaction would have proceeded spontaneously
from right to left.
Answer the following:
1. In biochemical reactions:
A. A reaction can occur spontaneously only when
the sum of the entropies of the system and its
surroundings < 0.
B. The most important criterion that determines
whether a reaction can occur spontaneously is
∆Go.
C. The change in energy of a system is independent
of the path of a reaction.
D. At chemical equilibrium where no net change in
[products] or [reactants] can occur, ∆G > 0.
A reaction can occur spontaneously only if the
standard free energy change of the reaction is < 0.
2. When a reversible reaction between substrate
and product has achieved equilibrium:
A. The standard free energy change of the reaction is
always zero.
B. Subsequent addition of product will drive the
reaction to the right.
C. The chemical potentials of substrate and product
are equal.
D. The forward and reverse reactions halt.
E. Subsequent addition of an enzyme that
accelerates the reaction will drive the reaction to
the right.
13
3. Keq for the reaction A ⇌ Β is 0.01 (log10 Keq =
-2). Defining: ∆G = ∆Go+ {1.36 x log10 [B]/[A]} kcal/
mol. When the starting concentrations of A and B
are 1 mM and 0.1 mM respectively (log10[B]/[A] = -1).
A. ∆G = 1.36 kcal/mol and ∆Go = -2.72 kcal/mol
B. Under the starting conditions stated above, the
reaction can occur spontaneously.
C. ∆Go = 2.72 kcal/mol and ∆G = -1.36 kcal/mol
D. The reaction will proceed from left to right.
E. The reaction will proceed from right to left.
14
S ECTION 4
Explanations of Catalytic Action
Enzymes are biological
catalysts
Enzymes lower the free energy of activation necessary
for a reaction to occur.
How can an enzyme accelerate a reaction without
shifting its equilibrium? To understand this we return
to the "Transition state theory" (Eyring, 1935).
Reactant molecules must overcome an energy barrier
and pass through an activated complex before
proceeding on to the product of the reaction.
The isomerization reaction
S⇌P
is best represented by
S⇌X⇌P
where X is the activated complex or transition state.
In terms of an energy diagram
Example 1
L EARNING O BJECTIVES
Energy
1. Enzymes accelerate a chemical reaction but in
doing so are neither chemically transformed at
the completion of the reaction nor do they
alter the equilibrium of the reaction.
X
∆GA
S
P
∆G
Progress of Reaction
2. Enzymes introduce althernative reaction
pathways with lower energy barriers to
catalysis.
Reactant molecules that achieve only a fraction of the
activation energy (∆GA) fall back to the ground state.
Those that achieve the transition state energy are
committed to form product.
15
The fraction of S that achieves the transition state X at
any temperature T is given by
GA
S F = e - RT
How can an enzyme introduce alternative reaction
pathways?
A number of mechanisms are observed:
•
If T = 0ºC (273ºK) and ∆GA = 10,000 cal/mol
SF = 9.9 x 10-9
•
If ∆GA is in some way reduced to 1,000 cal/mol
SF = 0.158
A 10-fold decrease in ∆GA results in a 16,000,000-fold
increase in the fraction of S that can achieve the
transition state! In principle, the reaction would be
accelerated by 16,000,000-fold.
Enzyme thus provide alternative routes of reaction
which have lower energy barriers.
An enzyme could act in the following manner:
X
X
1. Covalent Catalysis
A nucleophile (electron-rich group with a strong
tendency to donate electrons to an electron-deficient
nucleus) on the enzyme displaces a leaving group on
the substrate. The enzyme-substrate bond is then
hydrolyzed to form product and free enzyme.
2. Acid-base Catalysis
e.g. Lysozyme cleaves the glycosidic bond between
C1 of N-acetylmuramic acid and C4 of Nacteylglucosamine of bacterial cell wall
polysaccharides. Glu35 of lysozyme donates a proton
to the oxygen of the polysaccharide glycosidic bond
thereby hydrolyzing the bond.
Energy
Energy
3. Proximity
X
S
X
X
S
P
Progress of Reaction
P
Progress of Reaction
Either way, less energy is needed to form transition
state species but the ground states of substrate S and
product P are unchanged (∆G is unchanged). Thus the
reaction is accelerated and the equilibrium is
unchanged.
An enzyme may bind two reactants and in doing so
increase their proximity. Reaction rate is related to the
number of collisions of correct orientation. When an
enzyme binds its substrates it insures that their
orientation is precisely that required for reactivity.
4. Molecular Distortion
The enzyme active site undergoes a conformational
change upon binding substrate distorting the
substrate into a conformation resembling the transition
state species.
16
Summary
Formative self-evaluation questions
1.
1.
What is the Transition state theory?
2.
By how much could a 20-fold reduction in ∆GA
from 10 kcal/mol to 0.5 kcal/mol accelerate a
reaction at 0ºC?
3.
Name 4 mechanisms by which enzymes lower
energy barriers for catalysis.
4.
Do enzymes affect Keq?
Reactant molecules must overcome an energy
barrier and pass through an activated complex
before proceeding on to the product of the
reaction.
2.
Enzymes accelerate reactions by lowering this
energy barrier by introducing alternative reaction
pathways
3.
Enzymes do not affect the ground state of
reactants and products therefore do not affect the
equilibrium of a reaction.
4.
Enzymes therefore accelerate forward and reverse
reactions equally.
5.
Enzyme introduce alternative reaction pathways
through covalent catalysis, acid-base catalysis,
proximity effects or by molecular distortion or
combinations thereof.
17
S ECTION 5
Key to formative
evaluations
3. When the reaction proceeds right to left, ∆G > 0
and µS < µP
4. At equlibrium, ∆H = T ∆S
Section 3
1. C (S increases; ∆G not ∆Gº; ∆G=0 at equilibrium)
2. C (∆G=0, not ∆Gº; more P produces more S; the
reaction continues; enzyme leaves Keq unaltered)
Section 1
1. Keq = k1/k-1
2. Keq = [P]eq/[S]eq
3. [S] would fall and [P] would increase so that Keq
remained unchanged.
4. [S] would rise and [P] would fall so that Keq
remained unchanged.
Section 2
1. At equilibrium, ∆G =0 and µS = µP
2. When the reaction proceeds left to right, ∆G < 0
and µS > µP
3. E. Keq = 0.01; starting [B]/[A] = 0.1 > Keq thus the
reaction must proceed right to left.
Section 4
1. Transition state theory states that reactant
molecules must overcome an energy barrier and
pass through an activated complex before
proceeding to product formation.
2. At 0 ºC & ∆GA = 10kcal/mol, SF = e-10,000/(1.987*273) =
9.8 x 10-9. When ∆GA falls to 0.5 kcal/mol, SF =
0.398. Thus SF increases 40.6 x106-fold.
3. Covalent catalysis, acid-base catalysis, proximity
effects or by molecular distortion.
4. No! Keq is unchanged.
18
C HAPTER 2
relative fluorescence (Fr)
Analysis of
time
dependent
processes
39
38
37
kobs
36
F
35
34
33
32
0
0.2 0.4 0.6 0.8
1
time in seconds
DNPA + NaOH → DNP + acetate
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1.2
0.8
0.6
0.4
0.2
0
10
100
1000
time, msec
10
4
Constants, equations and
glossary
Bmax ≡ Bm; sometimes Bmax ≡ Bm = [Et] (I will
explain when this is true)
Your challenge will be to understand when different
names mean the same thing or different terms.
In general you will encounter 2 types of kinetic
constants in this and subsequent chapters.
• Constants shown in lowercase (e.g. k1, k-1, kon, koff
etc) are typically rate constants with units of per
unit time (e.g. s-1) or per concentration per unit time
(e.g. M-1.s-1).
Some constants are given multiple names in this and
subsequent chapters. This reflects: 1) my poor editing;
2) the reality that multiple names for the same
constant are also found in the literature:
KS ≡ KD; KM ≡ Km ≡ KM(app); sometimes KM ≡ KM(app)
can be equal to KS ≡ KD (I will explain when this is
true)
kf ≡ kon; kr ≡ koff
Vmax ≡ Vm ≡ Vm(app)
• Constants shown in uppercase (e.g. KM, KD, KS, Ki,
Vmax etc) can have more complex meanings, have
units of concentration (M) or are rates (mol/sec) and
are related in some predictable way to multiple rate
constants
e.g. KM = (k-1+kp)/k1
You will also encounter a great many equations. You
do not need to memorize these (although you may do
so involuntarily particularly if you use them frequently).
My goal is for this pamphlet to serve as a useful
resource for you.
kp ≡ kcat (I will explain when this is true)
xx
Glossary
catalysis - the process by which an enzyme or
catalyst accelerates a reaction.
catalyst - an agent that accelerates a chemical
reaction but which is unchanged in amount or
chemistry at the end of the reaction.
chemical equilibrium - a reaction in which forward
and reverse reactions continue to proceed but are
quantitatively balanced
thermodynamic equilibrium - Keq = [Product]e/
[Substrate]e i.e. the ratio of product formed : substrate
remaining at equilibrium
thermodynamics of reaction rates - enzymes
introduce alternative reaction pathways in which the
Gibbs free energy of activation is reduced
velocity or rates - amount of material (substrate,
product, # cells etc) consumed or produced or
number of events occurring per unit time
enzyme - a biological catalyst
equilibrium - a state in which opposing forces are
balanced.
products - molecules produced by the action of
enzymes on substrates
reaction order - the dependence of the reaction rate
on [substrates]n when n is the number of substrates
which must interact to form a single molecule of
product.
substrates - molecules that are acted upon by
enzymes
xxi
S ECTION 1
Context
SAMs have been studied extensively by
SFG since 1991 (8), but ultrafast probing of a
flash-heated SAM requires some elaboration. In
the SFG technique we used, a femtosecond infrared (IR) pulse at 3.3 mm with a bandwidth of
150 cm−1 is incident on the SAM, coherently
exciting all the alkane CH-stretch transitions in
the 2850 to 3000 cm−1 range, along with
electrons in the Au skin layer, producing an
oscillating polarization in both the Au and the
SAM layers. At the same time, a picosecondduration 800-nm pulse (“visible”) with a bandwidth of 7 cm−1 is incident on the sample. The
visible pulse interacts with this oscillating polarization through coherent Raman scattering to
create a coherent output pulse at the IR + visible
frequency. This combined IR-Raman interaction
is forbidden (in the dipole approximation) in
centrosymmetric media because the secondorder susceptibility c(2) vanishes in such media.
Because the methylene –CH2- groups of the
alkane SAM form a nearly centrosymmetric
solid, the SFG signal that we observed originated
predominantly from the Au surface and the
terminal methyl –CH3 groups. The well-known
SFG spectrum obtained in ppp polarization (4),
malized ensemble–average IR dipole
moment (⟨m⟩/mIR)2, which is temperature dependent. (C) With an
instantaneous temperature jump to
1100 K, the methyl head groups
become orientationally disordered in less than 2 ps.
Some examples
W.Stuhmer et al.
A
Fig. 3. (A) SFG spectra
of C8 (n = 7) and C18
(n = 17) SAMs without
heating pulses (blue) and
with flash-heating to
800°C (red). (B) VRF for
a C8 monolayer. (C) VRF
for a C18 monolayer.
==
RCK1
1 SOpA
RCK3
1
J6pA
RCK4
4 0
0
p
RCK5
A~~~~~~4Op
1
,,,J@20pA
Fig.
20ms
chann
B
poten
G/Gm
788
10 AUGUST 2007
VOL 317
SCIENCE
www.sciencemag.org
1.2 T
Vibrational Response
Functions (VRFs) of SelfAssembled Monolayers. B
VRF for a C8 monolayer. C
VRF for a C18 monolayer
From: Wang et al.,
SCIENCE VOL 317, pp
787-790, 2007
2. The tools used to analyze these reactions,
however, are invariant and fall under the
general umbrella of “kinetic analysis”.
+
Normalized current
CTX-Biotin
1.0
0.5
0.0
Washout
0
200
-40
From: Stühmer et al. EMBO Journal
vol.8, pp.3235 - 3244,
Fig. 5. Conductance-voltage
relations1989.
of RCK channels. (A) Families
of outward currents in response to depolarizing
voltage steps. From
3
top to bottom RCKI, RCK3, RCK4, RCK5. The traces are responses
to 50 ms voltage steps from -50 to 40 mV in 10 mV intervals.
- Thiol Sugar Ensemble currents recorded
+Thiol
Sugar
from
macro-patches. Sampling at 10 kHz,
filtering at 3 kHz low pass. (B) Plots of normalized conductance
(G/Gm) versus test potential for different RCK channels (RCK1: open
circles; RCK3: crosses; RCK4: diamonds; RCK5: filled circles). To
obtain the conductance values the current at a particular test potential
was divided by the driving potential assuming a reversal potential of
- 100 mV. The lines showed the results of a non-linear least-squares
TCEP
fit of a Boltzmann isotherm (see Materials and methods) to the
conductance values. The maximal conductance (Gm) obtained by the fit
was used to normalize the data. The half-activation voltages in this
400
600 are -24
800mV (RCK1),
1000
1200
plot
-37
mV (RCK3), -30 mV (RCK4) and
-40 (s)
mV (RCK5).
Time
Sugar
- Thiol
Sugar of metabolically-labeled
control
Shaker
to voltage+ Thiol
from -60 to 0 mV are shown in Figure 7.
B Chemical
steps
channels with CTX–Biotin.
Washout(A) CTX–Biotin (10 nM) After TCEP
The
of
size
step
currents at 0 mV varied between
elementary
Initial
Initial
inhibits Shaker-IR K+ currents in CHO-K1 cells treated
0.46
and
1.02
pA
(RCK4)
pA
(RCK5). The single channel
with either 50 µM thiol sugar 1 (h) or 0.5% ethanol as a
current - voltage
relations were measured in cell attached
vehicle (s). Only metabolically-labeled
(+thiol sugar)
patchesbywith
normal after
frog
Washout
channels were irreversibly blocked
CTX–Biotin
aRinger's solution on the extracellular
side. For
all channels,
simple washout; inhibition was
completely
reversed
by an the current-voltage relation
50 ms
50 ms in
is linear
application
of 1 mM TCEP. Reaction
profiles
were range -20 to 20 mV. However, since
the voltage
monitored with a 200 ms, 40 this
mV is
pulse
every
15s. range for conductance estimation, we
a rather
narrow
measured
the
average
amplitudes at 0 mV membrane
- Thiol Sugar
+ Thiol Sugar
CFrom: Hua
Z, Lvov A, Morinpotential.
TJ, Kobertz
WR.
While
theChemical
RCK1, RCK3 and RCK5 channels have
I/I
I/I
control of metabolically-engineered
voltage-gated
K(+)
rather similar
1.0single-channel current amplitudes, that of the
1.0
channels. Bioorg Med Chem RCK4
Lett 2011.
channel is considerably lower (Table I).
1 nA
1. Biological reactions occur over intervals
ranging from psec (10-12 s) to days (10 s) or
even longer
A
1 nA
L EARNING OBJECTIVES
needed to be chemically reversible. Although disulfide bond formation between CTX and thiol-containing sialic acids on the cell surface is an obvious chemoselective and cell friendly reaction, we
chose to label CTX with a bismaleimide that had an internal disulfide bond because maleimides are inherently more stable in water
than MTS reagents. Moreover, this subtle difference would allow
for delivery of a small molecule probe to the modified K+ channel
subunit after cleavage with reductant, which would be useful in
subsequent biochemical, biophysical or imaging experiments. To
simplify the synthesis of the bismaleimide, derivatization of CTX,
and ensure delivery of a molecular probe to a K+ channel subunit,
we set out to
5 synthesize a symmetrical bismaleimide (Scheme 1)
from cystamine dihydrochloride 2 that would allow for the facile
incorporation of a molecular probe in the final step of the synthesis. The amino groups of cystamine 2 were capped with 2 equiv of a
doubly amino-protected, activated ester of L-lysine 3. Selective
deprotection of the Fmoc protecting groups gave the symmetric
diamine 4. Addition of 2 equiv of the NHS-ester of 3-(maleimido)propionic acid and N-Boc deprotection afforded bismaleimide
5, which was subsequently biotinylated with 2 equiv of NHS–Biotin to give biotin bismaleimide 6. CTX was then derivatized by
labeling a cysteine mutant of CTX (R19C)32 with 100-fold molar excess of biotin bismaleimide 6 to yield CTX–Biotin, which was purified by reverse phase HPLC as we have previously described.17,19
With CTX–Biotin in hand, we decided to change both the channel (Shaker–IR)33 and the expression system (CHO-K1 cells) to
demonstrate that our approach was versatile and could be used
to label glycosylated ion-conducting subunits. Inactivation-removed Shaker is similar to Q1 in that it is an archetypical volt-
Conductance-voltage relations of
RCK channels. (A) Outward
currents in response to depolarizing
voltage steps. From top to bottom
RCKI, RCK3, RCK4, RCK5. The
traces are responses to 50 ms
voltage steps from -50 to 40 mV in
10 mV intervals.
-80
Z. Hua et al. / Bioorg. Med. Chem. Lett. xxx (2011) xxx–xxx
+
max
Initial
Washout
0.5
max
Initial
After
TCEP
Pharmacology
0.5 of RCK channels
22
A profile of the pharmacological sensitivity of the different
RCK channels to the K+ channel blockers 4-aminopyridine
chann
Hold
inact
low
depe
subs
at 2
Tabl
curr
RCK
sensi
inac
sensi
RCK
bloc
effec
Dis
Comp
RCK
An
funct
chan
ident
pa
a
prope
sing
shou
chan
func
memb
prel
struc
De
slowl
foun
inact
spina
chann
1988
whic
The analysis of these time courses has 2 elements:
1. The time courses may be predictable from first
principles (e.g. an assumed reaction mechanism)
and thus an analysis yields important information
about mechanism.
Time (h)
Fig. I . Time course of 3-0-methylglucose uptake in isolated muscle fibres. Ordinate: ratio
of i n t r a ~ l l u l a ractivity t o extracellula~activity of 3-0-methy1glucose per equixra1ent
volume of bulk external solution, .4bscissa: time in hours. External sugar concentration,
Time course
of 3-0-methylglucose
uptake in isolated
muscle
of Balanus
1 m M Uptake
was measured using rc)nventional
and scintillator
probe (cells
0 )methods.
Kumber of points per conventional determination, five or more. The water content of
nubilis. Ordinate:
ratio of intracellular activity to extracellular activity of 3-0isolated fibres (70%) is shown by the continuous line above the points. The time a t
diame*r, 1352
p m ;k m pAbscissa:
rature.
half-equilibration
is shown by thedmhed
methylglucose
per equivalent
volumeline.
ofMean
bulkfibre
external
solution.
21 OC.
time in hours. External sugar concentration, 1 mM. Uptake was measured
using conventional (filled squares) and scintillator probe (open circles)
The calculated rate of sugar uptake a t 30 min is 2 pmol .cm-*. s-I. Assuming
methods.
The water content of isolated fibres (70%) is shown by the
uptake is not saturated, the permeability of the barnacle muscle fibre, '2 (em. s-I),
continuous
line above the
points.
timeflux,
at half-equilibration
to 3-0-methy~g~ucose
is related
t o The
the sugar
J (mol. ernv2.s-I), byis shown by
the dashed line. Mean fiber diameter, 1352 µm ; 21 ºC.
where So is the external sugar concentration (in01 . emp3).This corresponds to a value
From:
Carruthers, cm.
A. J.s-IPhysiol.
VOL 336, pp 377-396, 1983
which is some 3 4 orders of magnitude larger than the
of P of 2 x
of artificial
lipid bilayers to sugars (Jung & Snell, 1968; Lidgard & Jones,
Ribosomal R N A Stability permeability
and Growth
Conditions
19'75).
At equilibrium, the 3-0-methylglucose space of the fibre is 70 yo,which is in close
contentsof barnacle muscle (71 & 1 7; ; R = 5 ) .
sgreement with estimates of the xra%er
Assuming this water is not bound, these results shour that 3-0-methylglucose is not
accumulated by barnacle muscle and that the transfer of the sugar across the
sarco1emma is mediated by a passive, facilitated process.
EJat ofphloretia on mqzr uptuke. Re18tively low concentrations of phloretin inhibit
C""
Length of chose, h
Downloaded from www.jbc.org at Univeristy of Massachusetts Medical Center/The La
tion
45 min of exposure t o sugar. 3feasurements of sugar uptake in subsequent experiments
were made using an incubation period of 30 min in order to obtain accurate
measurements of t2heinitial rate of sugar uptake.
FIG. 2. First-order decay analysis of embryonic total RNA
during
normal
and slow larval growth conditions. The perFirst-order
decay analysis
of Drosophila embryonic total RNA during
centage of embryonic RNA remaining a t various times duringa chase
normal
and
slow
larval
conditions.
The
percentage
of embryonic
with light yeast (normalgrowth
growth
conditions,
solid
circles) or dense
RNAalgae
remaining
at various
times open
during
a chase
normalThe
growth
(slow growth
conditions,
circles)
has under
been plotted.
values in(solid
solid circles
from
thegrowth
experiment
shown in (open
Fig. circles)
conditions
circles)areorderived
during
slow
conditions
1 as well
as regression
four other independent
experiments
(not shown).
is plotted.
The
lines indicate
that the
stability ofThe
embryonic
values in open circles are derived from the experiment shown in Fig.
RNA4 increases
from
48
h
115
h
if
the
larval
growth
rate
as well as one other independent experiment (not shown).
The is reduced.
regression lines indicate that the stability
of embryonic RNA increases from48h ( r 2 = 0.915) to 115 h ( r 2 = 0.892) if the larval
From:
Winkles
al., J. Biol. Chem.,Vol 269, pp 7716-7720, 1985
growth
rate iset
reduced.
not depleted during the chase. The dense RNA samplemay
be slightly enriched in rRNA, as would be expected if nonrRNA species decay faster than rRNA. Therefore, the halflife of total RNA isa good estimate of rRNA half-life.
To investigate whether the stability of embryonic rRNA
varied under differentlarval growth conditions, we then measured the half-life of this RNA using the L + D protocol as
described under "Materials and Methods." A representative
experiment using a dense chase is shown in Fig. 4. Adult
females were fed a yeast paste containing [3H]uridinefor 46
b.Some of the embryoslaid during the last 12 h of this
radioactive labeling period were transferred into a medium
containing 50% 13C,15N-labeled Chlorella cells, 50%
13C,2H,'5N-labeledChlorella cells, and [14C]uridine. First-in-
2. The data can be analyzed using theory but the
method of analysis may involve:
a. linearization of the data followed by linear
regression to obtain the constants related to
the underlying mechanism
b. nonlinear fitting to obtain the constants related
to the underlying mechanism
c. quality analysis to determine whether the
analysis is appropriate.
Goals
1. This chapter will review the classification of
reaction orders and the limitations of this
classification.
2. The tools available to analyze reactions and the
quality or appropriateness of the underlying
analysis.
These time courses may be typical of the
measurements you may undertake in your own
research or observe when reading the work of other
researchers.
23
S ECTION 2
Reaction order
Rate of a reaction
The rate or velocity, v, of a reaction or process
describes how fast it occurs.
The velocity is expressed as a change in
concentration (C) per unit time (t),
dC
v = dt
but may also express the change in a population of
cells with time, the increase or decrease in the
pressure of gas with time or the change in absorption
of light by a colored solution with time.
The order of a reaction
describes how the velocity of the reaction depends
upon the concentration of reactants.
In the (irreversible) isomerization reaction
L EARNING O BJECTIVES
1. Reaction orders
1. Zero-order
2. First order
3. Second order
1. Class 1
2. Class 2 (pseudo first-order)
A
k1
B
the theory of chemical kinetics tells us that
d [B]
m
=
= k 1 [A]
k
1 [A]
dt
Since m = 1, this reaction is first order with respect to
A and since A is the only independent concentration
variable in the rate equation, the reaction is overall
first-order.
24
The units for this first order reaction are derived from
moles of product formed per second per mole of
reactant or,
M per sec = k 1 M
M per sec
= k 1 = per sec
M
The Order of a Reaction must be determined
experimentally
Understanding the stoichiometry of a reaction is not
sufficient to predict the rate law of the reaction. This is
illustrated in the table below.
⇌
In a reaction of the type
E+S
⇌
k1
E$S
⇌
⇌
The rate of the reaction is proportional to [E].[S].
d [ES]
1
1
=
k
1 [E] [S] = k 1 [E] [S]
dt
Because m = 1 for both species the reaction is firstorder with respect to E or S but is second order overall
as one single step is involved in the reaction of two
species.
The units of this second order reaction are derived
from moles of product formed per second per mole2 of
reactants or
molarity per sec = k1 (molarity)2
M per sec
= k 1 = per M per sec
M2
⇌
⇌
⇌
⇌
If the concentration of a reactant remains unchanged
by the reaction, it is frequently omitted in the rate-law
expression. For example, with the first reaction, a
more complete rate law is:
v = k[sucrose][H+][H2O]
(the reaction is 3rd order overall). However, H+ is a
catalyst and its [ ] is constant during the run; [H2O]
(solvent) is little changed because of its vast excess
(55.5 M). Thus the terms [H+] and [H2O] are omitted in
the rate law. If the reaction were carried out at varying
[H+] or in an inert solvent, a first-order dependence of
the reaction on [H+] and on [H2O] is seen.
25
The first column in the table indicates only
stoichiometry, NOT reaction mechanism or order.
Reaction 3 Dinitrogen pentoxide decomposition
N2O5
NO2 + NO3
NO3
NO2 + O
2O
O2
Hence, the first step is first order. 2N2O5 was
indicated to balance the equation. It could just as
easily have been written as: N2O5 ⇌ 2NO2 + ½O2
Reaction 4 Nitrogen dioxide decomposition to nitric
oxide and oxygen.
This involves formation of an intermediate thought to
be the one shown in the reaction below
Examples of reaction orders in nature
To understand how we can distinguish reaction orders
experimentally, we will examine the following
reactions1
•
Zero-order kinetics
•
First order kinetics
•
True Second order kinetics
•
Second order kinetics characterized by pseudofirst order behavior.
1The
illustrated reactions are available in the file: “CoreKinetics.pzf”
This file is a “GraphPad Prism file that contains the data for each type
of plot shown and the types of analysis made.
If you wish to plot these data yourself, you should download the file
from the Core Curriculum website:
2NO2
ONOONO
The first step is therefore second order.
2NO + O2
http://inside.umassmed.edu/uploadedFiles/gsbs/courses/
2012-2013_Core_Course_Files/CoreKinetics.pzf%20-%20for
%20students%20only.zip
and download the “GraphPad Prism” version 6 software from Prism 6 Win http://cdn.graphpad.com/downloads/prism/6/
InstallPrism6.exe
Prism 6 Windows serial number: GPW6-200512-LEM5-16772
Prism 6 Mac http://cdn.graphpad.com/downloads/prism/6/
InstallPrism6.dmg
Prism 6 Mac serial number: GPM6-200513-LEM5-F3EF2
26
most of its course (ALDH is saturated by ethanol &
NAD+).
A zero-order reaction
zero-order kinetics
Zero-order reaction
Substrate
Product
8
80
v (d[P]/dt)
[Substrate] or [Product]
10
100
60
40
20
6
0
0
100
4
200
[S] µM
d [ethanol] d [acetaldehyde]
v ==
= k0
dt
dt
2
0
0
2
4
6
8
10
TIME
Note that [substrate] decreases linearly with time and
[product] increases linearly with time.
The negative sign is used with reactant - ethanol because its concentration decreases with time. The
concentration of its product, acetaldehyde, increases
with time.
C0
An example of such a reaction is ethanol conversion
to acetaldehyde by the liver enzyme, alcohol
dehydrogenase (ALDH). The oxidizing agent is
nicotinamide adenine dinucleotide (NAD+) and the
reaction can be written:
C
ALDH
+
CH3CH2OH + NAD ⇌ CH3CHO + NADH + H+
At saturating [alcohol] (about 2 beers) and with NAD+
buffered via metabolic reactions that restore it rapidly,
the rate of this reaction in the liver is zero-order over
CH3CHO
0
CH3CH2OH
0
t
27
Theory of Zero-order Reactions
dC
dt = k 0
The units of k0 are molarity per sec. This is a “zeroorder reaction because there is no concentration term
in the right hand of the equation.
Defining C0 as the concentration at zero time and C as
the concentration at any other time, the integrated rate
law is:
C = C0 + k0 t
y = y-intercept + slope * x
This is the equation for a linear relation between the
independent (time) and dependent (concentration)
variables.
We can therefore subject the raw data to linear
regression analysis to obtain C0 (y-intercept) and k0
(the slope).
10
[Substrate] or [Product]
A zero-order reaction corresponds to the differential
rate law
Zero-order reaction
Substrate
Product
8
6
4
2
0
0
2
4
6
8
10
TIME
y = x-intercept + slope * x
Best-­‐fit values" "
"
Slope" "
"
"
Y-­‐intercept when X=0.0"
X-­‐intercept when Y=0.0"
Goodness of Fit" "
R squared"
"
"
"
"
"
"
Substrate" "
-­‐1 ± 0 "
"
10 ± 0"
"
10.00""
"
Product"
1 ± 0 0 ± 0""
0" "
"
1.000""
1.000
"
"
"
"
Units
mols/sec
mols
sec
General rules for zero-order reactions
1. Plot of St or Pt vs time produces a straight line with
slope = -k (for St) or k (for Pt)
2. k has units of mols produced or consumed per unit
time
3. Zero-order, enzyme catalyzed kinetics are typically
observed at saturating [S]
28
A first-order reaction
Theory of First-order Reactions
A first-order reaction corresponds to the differential
rate-law:
1stOrder
dC
dt = k 1 C
5
[A] or [B]
4
3
The units of k1 are time-1 (e.g. s-1). There are no
concentration units in k1 so we do not need to know
absolute concentrations - only relative concentrations
are needed.
[Substrate]
[Product]
2
The reaction
A
1
0
0
k1
B
has the rate law
2
4
6
8
10
TIME
Note here that [substrate] decreases in a curvilinear
fashion with time and [product] increases in a
curvilinear manner with time. This observation
indicates that the reaction is NOT zero-order. How can
we analyze this further?
d [A] d [B]
v =- dt = dt = k 1 [A]
where k1 is the rate constant for this reaction.
The velocity may be expressed in terms of either the
rate of disappearance of reactant (-d[A]/dt) or the rate
of appearance of product (d[P]/dt).
29
First Order reactions - loss of substrate
First Order reactions - loss of substrate
Theory
Integrated rate law
- d [A] = k 1 [A] 0 t
[A] = [A] 0 e -k t
1
Defining [A]0 as [A] at time = 0 and integrating between
A at time zero and time t gives
ln [A] =- k 1 t + ln [A] 0
Half-life
Defining [A] at t1/2 as [A]0/2
y
= slope x + intercept
ln2 0.693
t 1/2 = k = k
1
1
and because τ = 1/k1,
t1/2 = 0.693 τ
30
First Order reactions - product formation
First Order reactions - product formation
Theory
Integrated rate law
Defining [B]∞ as [B] at equilibrium and assuming that
all A is converted to B, [A] at any time t can be
calculated from [B] at time t as
[B] = [B] {1 - e -k t}
1
[A]t = [B]∞ - [B]t
Thus we obtain
ln ([B] - [B] t) =- k 1 t + ln [B]
y
= slope x + intercept
Half-life
Defining [B] at t1/2 as [B]∞/2
ln2 0.693
t 1/2 = k = k
1
1
and because τ = 1/k1,
t1/2 = 0.693 τ
31
Returning to our example of a first order reaction,
1stOrder
5
[A] or [B]
4
3
[Substrate]
This is characteristic of “first-order decay” as observed
with radioactive decay.
[Product]
2
A second clue comes from the measurement of halftimes. As [Substrate] declines from 5 - 2.5 mM, from
2.5 - 1.25 mM and from 1.25 to 0.625 mM, the time
required for each 50% reduction is unchanged at ≈1.4
sec.
1
1st Order
5
0
0
2
4
6
8
10
TIME
The raw data suggest that [substrate] falls from 5 mM
to an equilibrium value of 0 mM.
If we plot the log [substrate] vs time (or show the yaxis data on a log scale), we obtain
1stOrder
3
2
1.4 sec
1
0
0
10
1.4 sec
1.4 sec
1
2
3
4
5
6
7
8
9
10
TIME
[Substrate]
1
[A]
Constant decay times and the linear relationship
between log {[A]t - [A]∞} vs time indicate a first order
process. Let us now check this by applying a firstorder analysis to the data.
0.1
0.01
0
[Substrate]
4
2
4
6
8
10
TIME
This produces a linear plot which is consistent with 1st
order kinetics!
32
Non-linear regression analysis
To do this we subject the data to nonlinear regression
(the plot is nonlinear) using an appropriate equation for
first-order reactions.
The integrated rate law for first-order substrate loss is
[A] = [A] 0 e -k t
1
Nonlinear regression finds the values of those
parameters of the equation (k1 and [A]0) that generate
a curve that comes closest to the data. The result is
the best possible estimate of the values of those
parameters.
To use nonlinear regression, therefore, you must
choose a model or enter one. GraphPad Prism offers
a model for first-order reactions called “One-Phase
Decay”
The equation is:
Y=(Y0 - Plateau)*exp(-k*X) + Plateau
In which the parameters are defined as:
1.
Y0 is the Y value when X (time) is zero or [A]0 in this case.
2.
Plateau is the Y value at infinite time (0 for our data set).
3.
k is the rate constant k1 (per unit time).
4.
Span is the difference between Y0 and Plateau and is [A]0 in this
case because Y0-plateau = [A]0
Every nonlinear regression method follows these
steps:
1. Start with initial estimated values for each parameter in the
equation.
2. Generate the curve defined by the initial values. Calculate the
sum-of-squares - the sum of the squares of the vertical distances
of the points from the curve.
3. Adjust the parameters to make the curve come closer to the data
points - to reduce the sum-of-squares. There are several
algorithms for adjusting the parameters - Prism uses the
Marquardt algorithm.
4. Adjust the parameters again so that the curve comes even closer
to the points. Repeat.
5. Stop the calculations when the adjustments make virtually no
difference in the sum-of-squares.
6. Report the best-fit results. The precise values you obtain will
depend in part on the initial values chosen in step 1 and the
stopping criteria of step 5. This means that repeat analyses of the
same data will not always give exactly the same results.
33
General rules for 1st order reactions
1stOrder
[Substrate] or [Product]
5
4
1. First-order enzyme catalyzed kinetics are typically
observed at subsaturating [S]
[Substrate]
[Product]
2. Plot of log (St-S∞) vs time produces a straight line
with slope = -k
3
3. The half-time (t1/2) and k are invariant of the
starting value of St chosen.
2
1
0
0
4. Plot of log (P∞-Pt) vs time produces a straight line
with slope = -k
2
4
6
8
10
TIME
Y=(Y0 - Plateau)*exp(-k*t) + Plateau
One phase decay"Perfect fit"
Best-­‐fit values" "
Y0""
"
"
"
Plateau" "
"
"
k" "
"
"
"
Half Life""
"
"
Tau = 1/k"
"
"
Goodness of Fit"
"
Degrees of Freedom"
R square""
"
"
"
[Substrate]"
[Product]" "
Units
"
"
"
"
"
5.000""
0"
"
0.5000"
1.386 "
2.000""
"
"
"
"
"
0 5.000 0.5000 1.386 2.000 mM
mM
per sec
sec
sec
"
"
48" "
1.000""
"
"
48
1.000
5. t1/2 = 0.693/k
6. k has units of time-1 (e.g. s-1). There are no
concentration units in k so we need not know
absolute concentrations - only relative
concentrations are needed.
7. k may be obtained by direct curve fitting
procedures using nonlinear regression
8. The full equation for loss of substrate is
[S]t = {[S]0 - [S]∞} e-(k.t) + [S]∞
9. The full equation for product formation is
[P]t = [P]∞ (1 - e-(k.t))
10. When a first order reaction is reversible (as most
are), e.g.
A
k1
k2
B
The equations are unchanged but now
k = k1 + k2
34
Second-order reactions
Fall into 2 categories in which the rate law depends
upon:
2. the product of the concentrations of two different
reagents.
Class 1 reactions (A+A ⇌ P)
[Substrate]
[Product]
3
2
1
The differential rate law is 0
0
2
4
6
8
10
TIME
v=k[A]2
Although one or more reactants may be involved, the
rate law for many reactions depends only on the
second power of a single component. e.g.
2 proflavin ⟶ [proflavin]2
v = k [proflavin] 2
A–A–G–C–U–U
2 A–A–G–C–U–U
4
[A] or [B]
1. the second power of a single reactant species, or
2nd Order class 1
5
U–U–C–G–A–A
v = k [A 2 GCU 2] 2
[substrate] decreases and [product] increases in a
curvilinear fashion with time. This indicates that the
reaction is NOT zero-order. How can we analyze this
further?
The curves drawn through the points were computed
by nonlinear regression assuming first order kinetics
(one-phase decay equation). Note the systematic
deviations from the fit. This strongly suggests that this
reaction does not follow first order kinetics.
We can investigate this further by using GraphPad
Prism to plot the residuals of the fit (how each point
deviates from the calculated fit) vs time.
A non-random scatter of residuals around the origin
(perfect fit) would confirm a poor fit and that we
should consider either an error in data sampling or
another model for the data.
35
Nonlin fit of 2ndOrderIrrev:Residuals
0.4
8
0.3
[Substrate]
6
[Product]
[A]0/[A]
0.2
Class 1, 2nd order Transform
of data
0.1
4
1st order data
2nd order data
0.0
2
-0.1
-0.2
0
1
2
3
4
5
TIME
6
7
8
9
10
This plot therefore shows that this is not a 1st order
reaction
Theory of Class 1 Second-order Reactions
Defining [A] at zero-time = [A]0, it can be shown that
the integrated rate law is
1
1
[A] [A] 0 = k t
adding 1/[A]0 to both sides gives
1
1
k
t
=
+
[A] 0
[A]
multiplying both sides by [A]0 gives
0
0
2
4
6
8
10
TIME
Here we re-plot the data from the previous page (open
circles, a second order reaction) as well as data from a
true first order reaction (closed circles) as suggested
by the 2nd-order linearized equation.
As you can see, transformation of the 2nd order data
produces a straight line with slope [A]0 k and yintercept = 1.
The slope [A]0 k indicates that the rate of loss of [A]
will increase linearly with [A]0
This is infact observed
[A] 0
[A] = [A] 0 k t + 1
Thus one expects a linear relation between the
reciprocal of the reactant concentration and time.
36
Although we do not show it here, this analysis breaks
down when the reaction is reversible. Thus in the
reaction
How starting [A] affects rate of 2nd order reaction
15
1
2
3
4
5
6
7
8
9
10
[A]0/[A]
10
Increasing
[A]0
5
0
0
2
4
6
8
10
TIME
[A]0 k per sec
1.5
1.0
A0k (slope) vs A0 second order
Best-fit values
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
1/slope
kr
P
the kinetics more closely resemble 1st-order kinetics
when when kr ≥ kf/10
In fact, this general analysis of 2nd-order Class 1
kinetics derives from classical irreversible chemical
kinetics which have only limited application in biology.
2nd-order reactions - Class 2 (A+B⇌P)
[A]0 k per sec
0.1320 ± 2.842e-009
-3.974e-009 ± 1.763e-008
3.010e-008
7.576
AKA pseudo-first order
A reaction that is 2nd order overall is 1st order with
respect to each of the two reactants.
0.5
0.0
0
kf
A+A
For example, in the reaction
2
4
6
8
E+S
10
[A]o
General rules for 2nd order reactions (Class 1)
1. Standard 1st order analysis does not work
2. Plotting [A]0/[A] vs time produces a straight line
with slope [A]0 k
3. Plotting slope vs [A]0 produces a straight line with
slope k and y-intercept 0.
4. The units of k are concentration-1.time-1.
k1
k2
E$S
If the enzyme E were maintained at a constant low [ ]
(e.g. [E] < [S]/100) and the substrate were varied, the
reaction differential rate law is:
v=
d [ES]
dt = [E] k 1 [S] - k 2 [ES]
37
Let us review this by examining ligand (L) binding to a
receptor (R).
R+L
kf
kr
R$L
This can also be plotted with the x-axis (time) shown
as a log scale - this allows us to observe the data at
short time intervals more closely
Pseudo 1st Order
(note the forward and reverse rate constants have now been called kf
and kr but this name change is purely arbitrary - they could have been
called kon and koff or k1 and k2)
At zero-time, various concentrations of L (µM) were
mixed with 1 nM R. The time course of LR formation
was monitored at each [L].
Pseudo 1st Order
[L]
0.0010
10
5.995
0.0008
3.594
[LR] µM
2.154
0.0006
1.292
.774
0.0004
.464
.278
0.0002
.167
.1
0.0000
0
5
TIME
10
0.0010
10
5.995
0.0008
3.594
2.154
[LR] µM
Upon rapid mixing of R and L, the receptor may
undergo a fluorescence change allowing measurement
of ligand binding. Alternatively, it may be possible to
measure ligand binding by use of radiolabeled ligand
and filter-bound receptor. Either way, the time course
of ligand binding may be examined to determine
whether it displays first or second order kinetics.
[L]
0.0006
1.292
.774
0.0004
.464
.278
0.0002
.167
.1
0.0000
0.1
1
10
TIME
The data were fitted by nonlinear regression using
GraphPad Prism and the one-phase association
equation. The fit is excellent in each case (the
residuals < [LR]/100)
You can also see that the reaction becomes faster at
higher [L] i.e. k increases and t1/2 falls with increasing
[L].
Each curve fit produces a value of k (typically called
kobs because it is an experimentally observed k) for
each starting [L].
We can analyze this further by plotting kobs vs [L]
38
Theory for pseudo first-order reactions
kobs vs L
25
For our reaction
kobs per sec
kobs per sec
20
15
Best-fit values
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
1/slope
R+L
1.999 ± 0.0001861
0.5012 ± 0.0007352
-0.2507
0.5002
kr
R$L
The rate of LR formation is given by:
d [LR]
v = dt = [R] k f [L] - k r [LR]
10
5
0
0
kf
2
4
6
[L] µM
We will show below that:
1. The slope is kf
2. The y-intercept is kr
3. The x-intercept is -kr/kf
8
10
If [R]0 is the amount of receptor at t = 0, it can be
shown that the integrated rate law is:
[LR] = [R] 0 constant (1 - e -t(k + k [L]))
r
f
1. The time dependent component of this expression
is e-t (kr + kf [L]) = e-t kobs.
2. Thus kobs = (kr+kf[L])
3. In a plot of kobs versus [L], kobs increases linearly
with [L] (slope = kf) and the y-intercept = kr.
4. The x-intercept (when kobs = 0) = -kr/kf. Why?
0=kr + kf[L]; thus -kf[L] = kr; thus -[L] = kr/kf
5. Analysis of the time course of L binding to R at
varying [L] permits computation of kf, kr and kf/kr =
Keq for the reaction.
6. This is ONLY true when [L] >> [R]. Here, first-order
kinetics are observed because [L] does not
change significantly. If [L] ≈ [R] the system will
behave like a class 1 second order reaction.
39
S ECTION 3
Recap of Key points
3. A “class 2” 2nd order reaction is described
accurately by first order equations.
4. kobs for a “class 2” 2nd order reaction is kf[S]0 +
kr and when [S]0 is 0, kobs = kr.
Summary for Reaction order and kinetics
1. What is the difference between a first-order
reaction and a second-order reaction that behaves
like a first order reaction?
1. A true first-order reaction is characterized by a
rate-constant, k, that is independent of
[substrate] or [product]. t1/2 is independent of
[substrate].
2. A second-order reaction that behaves like a
first order reaction (e.g. see this plot) is called a
pseudo-first-order reaction. Its rate constant,
kobs, increases linearly with [S] (i.e. kobs = kr
+kf[S]). t1/2 falls with increasing [substrate].
2. What is the difference between a class 1 second
order reaction and a class 2 second-order
(pseudo-first-order) reaction?
1. A “class 1” 2nd order reaction is not described
accurately by first order equations but when 1/
[S] is plotted vs time, the plot is linear.
2. kobs for a “class 1” 2nd order reaction is k[S]0
and when [S]0 is 0, kobs = 0
1. The reaction order describes how the velocity of
the reaction depends upon the concentration of
reactants.
1. In the reaction, E+S ⇌ ES, the reaction is first
order with respect to [E] at fixed [S], first order
with respect to [S] at fixed [E] but second-order
overall.
2. In the reaction, S+S ⇌ P, the reaction is secondorder with respect to [S].
2. Zero-order reactions occur at a constant rate even
as substrate levels fall.
1. A plot of [S] vs time for a zero-order reaction is
linear with slope = -k
2. k has units of mol consumed or produced per
unit time.
3. First order reactions are non linear with time
1. A plot of St vs time is described by
[S]t = {[S]0 - [S]∞} e-(k.t) + [S]∞
2. Plot of Pt vs time is described by
[P]t = [P]∞ (1 - e-(k.t))
40
3. t1/2 = 0.693/k
4. k has units of time-1 (e.g. s-1).
5. The half-time (t1/2) and k are invariant of the
starting value of [S].
4. There are two classes of second order reactions.
1. In reactions where two molecules of a
substrate combine to form a product (Class 1
reactions), the reaction is non linear with time
2. If [S] is varied and [ES] is plotted as a
function of time, each curve is described by
first-order kinetics but now:
1. kobs increases and t1/2 decreases with [S]
2. kobs = kr + kf [S]
3. kf has units of M-1.s-1 and kr units of s-1
4. Keq can be obtained as kf / kr
1. Plots of [S]0/[S] vs time are linear with slope
= [S]0 k. This slope has units of per sec
2. Plots of [S]0 k vs [S]0 are linear with yintercept = 0 and slope = true k.
3. These rules break down for reversible
reactions.
4. k has units of concentration-1 time-1 (e.g.
M-1.s-1).
2. In reactions where two different molecular
species combine to form a product (Class 2
reactions), the reaction is non linear with time
1. If one species (e.g. an enzyme or receptor)
is held at a fixed and very low [ ] relative to
its substrate or ligand, the reaction is
pseudo-first order with respect to
[substrate] or [ligand].
41
S ECTION 4
Formative self-evaluation
questions
10. In a 1st order reaction, why is a log plot of Pt vs
time never the best approach to confirm a 1st order
reaction? At fixed [enzyme], what concentration of
[S] produce zero-order kinetics?
11. At fixed [enzyme], what concentration of [S]
produce zero-order kinetics?
Test your understanding of time-dependent processes
by answering the following questions
12. At fixed [enzyme], what concentration of [S]
produce first-order kinetics?
1.
What are the units of zero-, first- and secondorder rate constants?
13. What is the difference between Class 1 and Class
2 second-order kinetics?
2.
Does the stoichiometry of a reaction always
predict reaction order and mechanism?
14. Why is it that a Class 2 second order reaction can
behave like a first-order reaction?
3.
What is the defining characteristic of a zero-order
reaction?
15. How may we compute the rate constant k for a
Class 1 second-order reaction?
4.
How do you compute the value of a zero-order
rate constant?
16. How does the rate constant k for a Class 1
second-order reaction vary with [S]?
5.
Do first-order reactions show a linear dependence
on time?
17. When does this type of analysis break down?
6.
What is the t1/2 of a first-order reaction?
18. How does kobs vary with [S] for a Class 2 secondorder reaction?
7.
How does t1/2 of a true first-order reaction vary
with [S]?
19. How can we use this relationship to compute Kd, kf
and kr for a Class 2 second-order reaction?
8.
How is t1/2 related to the rate constant k of a firstorder reaction?
20. What are the major differences between first-order
kinetics and Class 2 second-order kinetics?
9.
In the 1st order reaction S ⇌ P, why is a log plot of
St vs time not always the best approach to
confirm a 1st order reaction?
21. Why, then are Class 2 second-order kinetics called
pseudo-first order kinetics?
42
S ECTION 5
Key to formative
evaluations
4. Plotting [S] vs time yields a straight line with slope
= -k. Plotting [P] vs time yields a straight line with
slope = k.
5. No. Plots of [S] or [P] vs time are curvilinear.
6. t1/2 or the half-time of a reaction is the time
required for [S] to decrease by 50%.
7. t1/2 for a true first-order reaction is independent of
the starting [S].
Section 2 - Reaction order
8. t1/2 = 0.693/k
1. The units are:
9. A log plot of [S]t vs time will only produce a straight
line if all of S is converted to P. What will work in all
cases is to plot log([S]t - [S]∞) vs time where [S]∞ is
that concentration of S that remains when the
reaction achieves equilibrium.
1. zero-order = mol/s (amount per unit time)
2. first-order = per sec (per unit time)
3. second-order = per M per sec (per amount per
unit time)
2. No. The order of a reaction and mechanism must
be determined experimentally. Stoichiometry
simply shows a balanced reaction.
3. A zero-order reaction proceeds ([S] falls or [P]
increases) linearly with time.
10. A log plot of [P]t vs time can never produce a
straight line because [P] increases with time. You
have to invert the [P]t vs time data so that it now
resembles [S]t vs time data. This can be achieved
by measuring [P]∞ (that concentration of P
produced when the reactions attains equilibrium)
then calculating [P]∞-[P]t and plotting that result vs
time. Thus a plot of log([P]∞-[P]t) vs time will
43
produce a straight line. This analysis assumes that
[P]0 (that [P] present at zero-time) is 0. If [P]0 > 0
then [P]0 must be subtracted from [P]∞ and [P]t for
this analysis to work.
11. Zero-order kinetics are observed when [S] >>> Km.
12. First-order kinetics are observed when [S] <<< Km.
13. A Class 1 second-order reaction describes a
reaction in which 2 molecules of a single molecular
species combine to form a product e.g. S + S ⇌ P.
A Class 2 second-order reaction describes a
reaction in which 2 molecules of different
molecular identities combine to form a product e.g.
E + S ⇌ P.
18. kobs of a Class 2 second-order reaction increases
linearly with [S] when the concentration of the
second reactant is held constant and is << [S].
19. A plot of kobs vs [S] produces a linear plot with
slope = kf, y-intercept = kr and x-intercept = -Kd
20. In a true first-order reaction, k and t1/2 are
independent of [S]. In a Class 2 second-order
reaction, kobs increases with [S] and t1/2 falls with
[S].
21. A Class 2 reaction therefore resembles a true firstorder reaction at any specific [S] but unlike a truefirst order reaction k and t1/2 vary with [S]. Hence
the moniker of “pseudo.”
14. In a Class 2 second-order reaction e.g. E + S ⇌ P,
when E is held at a low and constant level, the
reaction will show a first-order dependence on S
and vice-versa.
15. By plotting [S]0/[S] vs time. This produces a
straight line with slope [S]0 k.
16. k is independent of [S]0.
17. When the reaction is reversible.
44
C HAPTER 3
Steady-state
kinetics of
enzymecatalyzed
This chapter considers receptor-ligand
Lorem ipsum
sit amet,
equilibria
and dolor
enzyme
catalyzed
suspendisse
rhoncus
reactions. Wenulla
seekpretium,
to:
tempor placerat fermentum, enim integer
1. rationalize simple Michaelis-menten
ad vestibulum volutpat. Nisl rhoncus
behavior
turpis est, vel elit, congue wisi enim
nunc
ultricies
dolor
magna
tincidunt.
2. provide
a set
of sit,
tools
for routine
Maecenas
est maecenas ligula
analysisaliquam
of enzyme-catalyzed
nostra.
reactions
S
S
E
C
I
C
I
S
S
E
E
I
C
kp
I
E+P
Ki
EI
+
S
Ks
ES + I
C
kp
E+I
+
S
E+P
ESI
Several observation tell us that an enzyme E and its
substrate S combine to form a complex.
Context
1. The physical properties of an enzyme can change
upon binding S.
2. The spectroscopic characteristics of E and S can
change upon ES formation.
3. High specificity for ES formation is observed.
4. The ES complex may be isolated in pure form.
5. At constant [E], increasing [S] results in increased
product formation to a point where product
formation no longer increases. This saturation is
presumed to reflect the fact that all E is now in the
form ES. This is illustrated below.
A number of observations point to the existence of ES
L EARNING OBJECTIVES
prior to the release of product (P) and regeneration of
free
These are:
1. E.Enzymes
are biological catalysts
1. ES complexes have been directly visualized by EM
2.and
Enzymes
combine reversibly with substrates
X-ray crystallography.
to form products
Jack Griffith developed
techniques that let scientists
3. Enzymes show saturability
see the finer details of DNA. In
1971 he and Arthur Kornberg
published this photo, the first
electron microscope image of
DNA bound to a known protein
- DNA polymerase.
4. The behavior of simple enzymes can be
rationalized in terms of the Michaelis Menten
equation
120
v (d[P]/dt), rate of reaction
A key step in enzyme function is the formation of the
enzyme (E)/substrate (S) complex, ES.
Vm
100
80
60 0.5 V
m
40
20
0
Km
0
10
20
30
40
50
[S] mM
xlvi
100
v (d[P]/dt)
75
50
25
0
0.01
0.1
1
10
100
1000
[S] [S]
mM
µM
Vm is a theoretical, maximum value for v.
Km is that concentration of [S] producing a v of Vm/2.
The reaction velocity curve is a section of a single,
rectangular hyperbola which in this instance takes the
generic form
Vm [S]
v = K + [S]
m
This equation is called the Michaelis-Menten equation.
Our challenge in this chapter is to understand why the
phenomenon of enzyme-mediated catalysis is well
approximated by this relationship.
xlvii
S ECTION 1
The steps in enzyme-mediated catalysis
Michaelis-Menten Kinetics
Let us examine an enzyme catalyzed reaction.
1. Breaking down an enzyme-catalyzed reaction
into its various parts
2. Understanding enzyme-substrate interactions
1. Analyzing enzyme-substrate interactions
2. Introducing the catalytic step
3. Analyzing enzyme-catalyzed reactions
3. Inhibition of enzyme-catayzed reactions
1. Competitive
ES
kr
k2
E+P
k -p
Here enzyme E reacts with substrate S to form the
complex ES. ES is then converted to EP which
dissociates to E and product (P). The rate constants
kf, kr, k2, k-2, kp and k-p describe the various steps
involved in the reaction:
kr, k2, k-2, and kp are first-order rate constants
kf and k-p are second-order rate constants
This may be represented as a reaction coordinate - an
abstract one-dimensional coordinate representing
progress along a reaction pathway.
(ES..EP)†
(E...P)†
EP
6
+
(E..S)†
2. Noncompetitive
0
4
5
7
3
E+S
3. Uncompetitive
4. What does Km represent?
kp
EP
k -2
Energy
L EARNING O BJECTIVES
E+S
kf
2
ES
1
-
E+P
5. What does Vm represent?
6. Kinetic perfection
Reaction coordinate
48
Deriving the equations that describe this reaction
scheme is a very significant undertaking. Moreover,
the equation describing the rate of reaction in terms of
substrate and product levels and rate constants is
quite complex.
However, we can make a number of simplifying
assumptions in order to more readily obtain a solution
to this scheme. How do we do this?
•
Assume that the reverse reaction (P→ S) is
negligible. While enzymes accelerate both forward
and reverse reactions to the same extent, we (as
the biochemists working with this enzyme) can
establish experimental conditions that preclude or
minimize the reverse reaction.
•
•
•
e.g. adding an additional enzyme which
converts P into another species Q which cannot
react with our enzyme. Or, we can measure the
rate of reaction at very early time points where
the reverse reaction is insignificant.
With these assumptions in mind, the reaction scheme
now becomes:
E+S
kf
kr
ES
kp
E+P
There are two parts to this reaction:
1) Formation of ES
2) ES breakdown to product P and free enzyme E
The formation of ES is a second order process and the
breakdown of ES to E + S or to E + P are first order
processes.
The units are:
1.
kr = kp = per sec.
2.
kf = per M per sec. You will see later why these
units are important.
Assume only a single central complex (ES) exists.
i.e. ES breaks down directly to E + P.
Make certain that [S] >> [E]. Thus the instantaneous
interaction of S and E to form ES does not
significantly affect free [S] (although [S] will slowly
fall due to its conversion to P).
•
Typically, ([S]-[ES])/[S] ≥ 99.9%
49
Enzyme/substrate interactions ≡ Receptor/ligand
interactions
Substituting for [ES] we obtain
kf
[ES]
k r [E] [S]
[E] t =
k
[E] + k f [E] [S]
r
The first step in enzyme function is the formation of
the enzyme (E)/substrate (S) complex, ES.
Consider the reaction:
E+S
Cancelling [E] we find
kf
kr
ES
kf
[ES]
k r [S]
[E] t =
k
1 + k f [S]
r
The rate of ES formation is given by:
d [ES]
dt = k f [E] [S]
The rate of ES breakdown is given by:
and rearranging we obtain the fraction of enzyme
complexed with S at any given [S], [E]t as
[ES]
[S]
=
[E] t
kr
k f + [S]
- d [ES]
dt = k r [ES]
Thus at equilibrium
d [ES] - d [ES]
dt = dt ` k f [E] [S] = k r [ES]
Hence
k
[ES] = k f [E] [S]
r
If we seek to understand how much substrate is
bound (i.e. [ES]) at any given [S] and [E], we can
express [ES] as a fraction of total enzyme [E]t as:
[ES]
[ES]
=
[E] t
[E] + [ES]
What is this constant kr/kf ?
kr/kf has units of
per sec
per M per sec = M
When [S] = kr/kf, this means that
[ES]
[S]
[S]
=
=
[E] t
kr
2 [S] = 0.5
k f + [S]
This means that one-half of [Et] = [ES] when [S] = kr/kf
50
What is the significance of kr/kf?
In the reaction
E+S
kf
kr
ES
kf/kr = Keq = the equilibrium constant or association
constant for E and S interaction
kr/kf = 1/Keq = KS or Kd = the dissociation constant of
the ES complex.
When Kd is 1 µM, 50% of Et = ES when [S] = 1 µM
= 1 x 10-6M.
When Kd is 1 nM, 50% of Et = ES when [S] = 1 nM
= 1 x 10-9M.
A low value for Kd means that the ES complex is more
stable (less dissociates to E + S) thus at any [S], there
is a higher probability that ES is formed. Less S is
required to occupy one-half of the available binding
sites. The enzyme shows high affinity for S.
A low value for Kd (high affinity for S) results from
energetically favorable interactions between E and S
(e.g. H-bonding and multiple van der Waals).
You may recall from “Thermo” that Keq = e-∆Gº/RT
Thus at 20ºC, an equilibrium constant of 10
corresponds to a ∆Gº of -1.36 kcal/mol (see Table
below)
∆Go
Keq per M
1,000,000
10,000
100
10
1
0.1
0.01
0.0001
0.0000001
Kd M
0.000001
0.0001
0.01
0.1
1
10
100
10000
10000000
kcal/mol
-8.2
-5.5
-2.7
-1.4
0.0
1.4
2.7
5.5
9.5
kJ/mol
-34.2
-22.8
-11.4
-5.7
0.0
5.7
11.4
22.8
39.9
Note: H-bond energies range from 3 to 7 kcal/mol. van
der Waal’s bond energies are approximately 1 kcal/
mol.
51
How do we measure Kd for substrate binding and
enzyme binding capacity?
In this experiment we continue our analysis of ligand
binding started on page 37. Here we measure the
amount of ligand bound to the receptor R at 10 sec
(where [L], [R] and [LR] are in equilibrium; see page
37).
equilibrium binding
[LR] µM
2.848e-004
3.995e-004
5.262e-004
6.496e-004
7.557e-004
8.377e-004
8.959e-004
9.349e-004
9.599e-004
9.756e-004
0.0010
0.0008
[LR] µM
[L] µM"
0.100"
0.167"
0.278"
0.464"
0.774"
1.292"
2.154"
3.594"
5.995"
10.000"
There are other ways of computing Kd and Bmax. These
are summarized on the next several pages and involve
linearization of nonlinear data.
Defining Bmax as
where n is the number of substrate binding sites per
enzyme, we can rearrange the equation for ligand
binding to give the Michaelis-Menten equation for
ligand binding as
B max [S]
[S] b = K + [S]
d
0.0006
0.0004
Where [S]b is the concentration of substrate bound to
the enzyme.
0.0002
0.0000
B max = [E] t n
0
1
2
3
4
5
6
7
8
9
10
[L] µM
This plot can be analyzed by nonlinear regression in
Prism by using the “One-site binding - specific
binding” equation
Thus if [E]t is 10 µM and n = 2,
•
Bmax = 20 µM
•
when [S] = Kd, [S]b = 10 µM
Y=Bmax*X/(Kd + X)
Where Bmax is the enzyme’s capacity to bind S
The result yields:
[LR] µM
One site -- Specific binding
Best-fit values
Bmax
0.0010
Kd
0.25
µM
µM
52
1 ) Lineweaver Burk method
2 ) Hanes Woolf method
Kd 1
1
1
=
+
[S] b B max [S] B max
Taking the reciprocal of the binding equation we
obtain
[S]
Kd
Kd 1
1
1
=
+
=
+
[S] b B max [S] B max [S] B max [S] B max
This is a linear equation. Plotting 1/[S]b on the y-axis
and 1/[S] on the x axis, we obtain:
•
slope = Kd/Bmax,
•
y-intercept = 1/Bmax
and, when [S]b = -Kd
Kd
1
1
1
1
=
+
=
+
[S] b B max K d B max B max B max = 0
[S]
[S]
K d [S]
Kd
1
[S]
=
+
=
+
B max
[S] b B max [S] B max B max
This is a linear equation. Thus plotting [S]/[S]b on the
y-axis and [S] on the x axis, we obtain:
•
slope = 1/Bmax,
•
y-intercept = Kd/Bmax
When [S] = -Kd
[S]
Kd
Kd
=
+
[S] b B max B max = 0
x-intercept = 1/-Kd
•
Lineweaver Burk
x-intercept = -Kd
1/[LR] per µM
4000
Best-fit values
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
1/slope
Hanes-Wolf Transform of equilibrium binding
251.0 ± 0.09620
999.2 ± 0.3800
-3.980
0.003983
[LR] µM
10000
3000
Best-fit values
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
999.9 ± 0.03448
250.5 ± 0.1362
-0.2505
8000
slope = Kd/Bmax
2000
y-cept = 1/Bmax
1000
[L]/[LR]
1/[LR] per µM
x-cept = -1/Kd
•
Multiplying by [S]
6000
slope = 1/Bmax
y-cept = Kd/Bmax
x-cept = -Kd
4000
2000
-5
0
5
1/[L] per µM
10
0
5
10
[L] µM
53
3 ) Scatchard method
Which methods are used by today’s scientists?
B max [S]
[S] b = K + [S]
d
[S] b K d + [S]
[S] b K d
B
=
max =
[S]
[S] + [S] b
Dividing by Kd gives
B max [S] b [S] b
K d = [S] + K d
Significance of Bmax
which rearranges to:
[S] b B max
1
[S]
=
b
Kd
Kd
[S]
This is a linear equation. Plotting [S]b/[S] vs [S]b yields:
slope = -1/Kd,
•
•
y-intercept = Bmax/Kd
When [S]b = Bmax
[S] b B max B max
[S] = K d - K d = 0
[LR] µM
0.004
Best-fit values
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
1/slope
-3.984 ± 0.002759
0.003986 ± 2.095e-006
0.001000
-0.2510
0.003
0.0005
0.0010
The number of ligand binding sites per enzyme
If Bmax is measured as 10 µM bound ligand or
substrate and you know that [E]t is 5 µM,
0.001
0.000
0.0000
If Bmax is measured as 10 µM bound ligand or
substrate and you know that [E]t is 10 µM,
n = Bmax/[E]t = 1
slope = -1/Kd
y-cept =Bmax/Kd
x-cept = Bmax
0.002
When you perform a binding assay (measuring the
concentration of bound ligand as a function of free
[ligand]), you typically measure the maximum binding
capacity (Bmax) of your system and Kd. Your challenge
is then to understand how Bmax is related to [Et].
B max = [E] t n
Scatchard Transform of equilibrium binding
0.005
The plots on the previous pages gave us maximum
[ligand bound] or Bmax under the prevailing
experimental conditions.
Remembering the expression
x-intercept = Bmax
[LR]/[L]
•
Curve-fitting to the Michaelis-menten equation by
non-linear regression analysis is the preferred method
used today. Why? Inverting data amplifies errors at
low [S] where experimental error (signal to noise) is
frequently greatest and today’s computers are more
than fast enough to undertake the analysis.
0.0015
[LR] µM
54
The number of ligand binding sites per enzyme
Competitive antagonists displace substrates from
the enzyme
n = Bmax/[E]t = 2
This can be depicted in the following ways:
If Bmax is measured as 10 µM bound ligand or
substrate and you know that [E]t is 20 µM,
I
S
The number of ligand binding sites per enzyme
n = Bmax/[E]t = 0.5
With some enzymes, the functional unit
comprises only 1 enzyme molecule which
binds 1 substrate or ligand. Here, [Et] = Bmax
and n = 1.
Other enzymes form a complex that
contains 2 or more functionally noninteracting enzyme molecules - each of
which binds 1 substrate or ligand.
Expressing [E]t in terms of [subunit]t, [Et] =
Bmax and n = 1
enzyme
Subunit
A
ES
S+E+I
Ki
EI
Where Kd and Ki are the dissociation constants for S
and I binding to the enzyme, E, respectively.
Ligand binding in the presence of a competitive antagonist
I
Subunit Subunit
A
B
1.0
[ligand]b/[Enzyme]t
Subunit B
Subunit A
0.8
Yet other enzymes are complexes comprising
2 or more functionally interacting, but
otherwise identical enzyme molecules. Here,
two enzyme molecules may be required to
form a single binding site, and n = 0.5.
Kd
[I]=0
[I]=5
[I]=25
[I]=50
[I]=100
0.6
The curve is shifted
to the right (Kd
increases) but the
maximum amount of
binding (Bmax) is
unchanged.
0.4
0.2
0.0
1
10
100
[S] µM
1000
10000
55
Binding in the presence of a competitive antagonist is
described by:
[S] b =
B max [S]
[I]
K d (1 + K ) + [S]
i
Where I is the inhibitor and Ki the dissociation
constant for I binding to E.
Introducing catalysis
Let us now consider the breakdown of ES to
regenerate free enzyme E and release the product, P.
The overall scheme is visualized as
E+S
[I]
K dapp
Kd - 1 = Ki
[I]
Ki = K
dapp
Kd - 1
Thus computation of Kd in the absence of I and Kdapp
at any given [I] permits computation of Ki.
kr
ES
kp
E+P
Defining the rate of product formation, v as
Defining Kdapp as the “apparent” Kd measured in the
presence of inhibitor
[I]
K dapp = K d (1 + K )
i
kf
v = kp [ES]
We must again express [ES] in terms of known
quantities
Rate of [ES] formation
+d[ES]/dt = kf [E][S]
Rate of breakdown of [ES]
-d[ES]/dt = kr[ES] + kp [ES]
or
-d[ES]/dt = (kr + kp) [ES]
In the "steady state" the concentrations of
intermediates (e.g. ES) are unchanged, whereas [S] +
[P] can change. If we limit measurements of v to early
stages, [ES] does not change (there is no reverse
reaction)
d[ES]/dt = 0
56
[S]
v
=
kr + kp
k p [E] t
k f + [S]
thus
kf [E] [S] = (kr + kp) [ES]
hence
[E] [S] k r
[E] [S]
[ES] = k + k = k + k
r
p
r
p
kf
The following steps are algebraic tricks:
[E]t = [E] + [ES]
dividing the velocity equation by [E]t we obtain
k p [ES]
v
=
[E] t [E] + [ES]
divide both sides by kp
and defining:
Vm = kp [E]t
and
Km =
kr + kp
kf
we obtain the Michaelis Menten equation
Vm [S]
v = K + [S]
m
[ES]
v
=
k p [E] t [E] + [ES]
substituting for [ES]
kf
k r + k p [E] [S]
v
k p [E] t =
k
[E] + k +f k [E] [S]
r
p
canceling [E]
kf
k r + k p [S]
v
k p [E] t =
k
1 + k +f k [S]
r
p
dividing both the numerator & denominator by kf/(kr+kp)
57
Properties of the Michaelis Menten Equation
At very high [S], v≈V
How do we measure Vm and Km?
zero-order
kinetics
m
i.e. v does not increase with
increasing [S]
100
In this experiment we have measured the rate of
product formation in an enzyme catalyzed reaction
and express this rate, v, versus [S].
80
Enzyme
Vm [S]
v. K
m
i.e. v increases linearly with [S]
60
rate of reaction M/min
at low [S]
v (d[P]/dt)
80
40
20
0
first-order
kinetics
0
100
60
40
M/min
100.0
25.00
Vm
Km
20
200
[S] µM
However, Vm is rarely measurable because in
many instances it is not possible to add sufficient
quantities of substrate to saturate the enzyme.
You may then ask, if Vm is not measurable, how
does one determine Vm and Km for a reaction?
The very same tricks that helped us with ligand
binding also help us here.
Why? Because the Michaelis Menten equation
and the saturable ligand binding equation take
the same form:
0
0
20
40
60
[S] M
Non-linear regression (using Prism) using the
Michaelis-Menten equation yields Vm and Km.
There are other methods of calculating these
parameters by linearization of the data and by using
linear regression.
You should note, however, that linearization greatly
amplifies errors of analysis!
Const $ x
y = Const 1+ x
2
58
Lineweaver Burk Analaysis
V m [S]
v = K + [S]
m
Hanes-Wolf Analysis
The Lineweaver Burk equation is
[S]
Km
1
=
+
v
V m [S] V m [S]
1 Km 1
1
=
+
v
V m [S] V m
Multiplying both sides by [S] gives
thus
1 Km 1
1
=
+
v
V m [S] V m
plotting 1/v vs 1/[S] yields a straight line with slope =
Km/Vm and y-intercept = 1/Vm
0.15
Lineweaver Burk Enzyme
1/v min/ M
0.2500 ± 1.824e-009
0.0100 ± 2.711e-010
-0.0400
4.000
[S] K m
1
=
+
[S]
v
Vm
Vm
plotting [S]/v vs [S] yields a straight line with slope = 1/
Vm and y-intercept = Km/Vm
Hanes Wolf Enzyme
1.0
0.10
0.8
slope = Km/Vm
0.05
x-cept = -1/Km
y-cept = 1/Vm
-0.1
Thus
0.0
0.1
0.2
1/[S] per M
0.3
0.4
When [S] = -Km
-1
Km
1
1
1
=
+
=
+
=0
v
Vm K m Vm Vm Vm
Thus the X-intercept = 1/-Km
[S]/v per min
1/v min/ M
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
1/slope
[S] K m [S] [S]
v = V m [S] + V m
0.6
slope = 1/Vm
y-cept = Km/Vm
x-cept = -Km
0.4
0.2
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
1/slope
-20
-0.2
20
[S]
[S]/v
0.0100 ± 2.320e-010
0.2500 ± 7.331e-009
-25.00
100.0
40
60
When [S] = -Km
[S] K m - K m
v = Vm + Vm = 0
Thus the x-intercept = -Km
59
Enzyme Inhibition
Noncompetitive Inhibition
Enzymes are inhibited by specific molecules
S
S
Reversible Inhibitors - May occur naturally or may be
synthesized as drugs.
E
Competitive Inhibition
Schematic
C
I
King-Altman Diagram
C
I
S
I
E+I
+
S
EI
Ki
S
S
E
E
I
Ki
EI
+
S
Ks
ES + I
C
kp
E+I
+
S
ESI
C
kp
I
Ks
enzyme
E+P
ES
kp
E+P
A competitive inhibitor, I, competes for binding with S.
I is not transformed into product. The inhibitor, I,
resembles the substrate, S. I reduces the rate of
reaction + S by reducing the proportion of enzyme in
the form ES
E+P
S and I are not mutually exclusive but ESI is
catalytically inactive. When S binds, the enzyme
undergoes a conformational change which aligns the
catalytic center, C, with the susceptible bonds of S; I
interferes with the conformational change, but has no
effect on S binding.
60
Noncompetitive inhibition is a common theme in
feedback inhibition.
Uncompetitive Inhibition
For example, the biosynthesis of isoleucine from
threonine in bacteria involves 4 steps mediated by 4
different enzymes.
E
+
S
C
+S
S
Threonine
threonine
A
B
D
Isoleucine
-
+I
E
KI
E
ES + I
C
I
kp
ESI
C
kp
deaminase
The first reaction is catalyzed by threonine deaminase
(TD). This enzyme is noncompetitively inhibited by
isoleucine.
Thus as product increases, the first step in the
reaction decreases. As P falls due to shut down of
TD, so TD is released from inhibition due to
dissociation of P from the TD.P complex and the
reaction cycle proceeds again.
E+P
E+P
When S binds, the enzyme undergoes a
conformational change which aligns the catalytic
center, C, with the susceptible bonds of S and which
also exposes an inhibitor binding site.
Thus I can only bind to the ES complex to form ESI
which is catalytically inactive.
20% of enzymes show uncompetitive substrate
inhibition which is now recognized as an important
regulatory mechanism - Bioessays 32: 422–429, 2010 Reed,
Lieb & Nijout
61
For all 3 types of inhibition, the Michaelis Menten
equation takes the form:
V m (app) [S]
v= K
m (app) + [S]
Where:
Competitive Inhibition
V m (app) = V m
[I]
K m (app) = K m T 1 + K Y
i
The law of mass action tells us that sufficient [S]
should overcome inhibition caused at any [I] but it will
take more S to produce the equivalent v observed in
the absence of I. Thus Km(app) increases with [I].
Noncompetitive Inhibition
Vm
V m (app) =
[I]
T1 +
Y
Ki
K m (app) = K m
The law of mass action tells us that binding of S is
unaffected at all [I] so Km(app) is unchanged by [I] but
since [I] reduces the amount of catalytic E, Vm(app) falls
with [I].
Uncompetitive Inhibition
Vm
[I]
T1 +
Y
Ki
Km
K m (app) =
[I]
T1 +
Y
Ki
V m (app) =
The law of mass action tells us that [I] reduces the
amount of catalytic ES, thus Vm(app) falls with [I]. Since
S increases the amount of E accessible to I, [ESI]
increases with [S] so it takes less [I] to saturate ES as
[S] increases. Thus Km(app) falls with [I].In the following
examples, we simulate an enzyme E with Km and
Vmax of 25 µM and 100 µmol/min respectively.
We then inhibit the enzyme using 3 separate inhibitors
Competitive - Ki = 10 µM
Noncompetitive - Ki = 10 µM
Uncompetitive - Ki = 10 µM
Control and inhibited rates of catalysis were measured
at 2.5 - 100 µM substrate in the absence (control) or
presence (inhibited) of 10 µM inhibitor.
62
Noncompetitive Inhibition
Competitive inhibition
100
100
inhibited
v µmol/min
v µmol/min
80
Control
80
60
40
Control
Michaelis-Menten Perfect fit
Best-fit values
Vmax
100.0
Km
25.00
20
0
0
Control Inhibited
Michaelis-Menten Perfect fit
Best-fit values
Vmax
100.0
50.00
Km
25.00
25.00
60
40
20
inhibited
Perfect fit
0
0
100.0
50.00
50
[S] µM
Control
Inhibited
50
[S] µM
100
100
Hanes Wolf Competitive
Hanes Wolf Noncompetitive
control
Inhibited
Best-fit values
Slope
0.0100 ± 2.100e-010 0.0100 ± 2.572e-010
Y-intercept when X=0.0 0.2500 ± 1.047e-008 0.5000 ± 1.282e-008
X-intercept when Y=0.0 -25.00
-50.00
[S]/v
µM.min/µmol
control
Inhibited
1.5
[S]/v
µM.min/µmol
3
1.0
control
Inhibited
2
1
0.5
-25
-50
control
Inhibited
Best-fit values
Slope
0.0100 ± 2.100e-010 0.0200 ± 4.200e-010
Y-intercept when X=0.0 0.2500 ± 1.047e-008 0.5000 ± 2.093e-008
X-intercept when Y=0.0 -25.00
-25.00
0
50
0
25
50
75
100
[S] µM
100
[S] µM
Lineweaver Burk Noncompetitive
Lineweaver Burk competitive
0.25
control
inhibited
Best-fit values
Slope
0.2500 ± 1.371e-009 0.5000 ± 3.877e-009
Y-intercept when X=0.0 0.0100 ± 2.134e-010 0.0100 ± 6.035e-010
X-intercept when Y=0.0 -0.0400
-0.0200
0.25
0.20
control
inhibited
0.15
1/v
min/µmol
1/v
min/µmol
0.20
0.10
0.05
0.0
control
inhibited
Best-fit values
Slope
0.2500 ± 1.371e-009 0.5000 ± 3.877e-009
Y-intercept when X=0.0 0.0100 ± 2.134e-010 0.0200 ± 6.035e-010
X-intercept when Y=0.0 -0.0400
-0.0400
control
inhibited
0.15
0.10
0.05
0.1
0.2
1/[S] per µM
0.3
0.4
0.0
0.2
1/[S] per µM
0.4
63
Uncompetitive Inhibition
Control Inhibited
Michaelis-Menten Perfect fit Perfect fit
Best-fit values
Vmax
100.0
50.00
Km
25.00
12.50
100
v µmol/min
80
Km(app)
Vm(app)
Ki
µM
µmol/min
µM
None
25
100
Competitive
50
100
10
Noncompetitive
25
50
10
Uncompetitive
12.5
50
10
Inhibition
Control
Inhibited
60
40
20
0
0
50
[S] µM
100
Hanes Wolf Uncompetitive
[S]/v
µM.min/µmol
3
-25
control
Inhibited
Best-fit values
Slope
0.0100 ± 2.100e-010 0.0200 ± 2.749e-010
Y-intercept when X=0.0 0.2500 ± 1.047e-008 0.2500 ± 1.370e-008
X-intercept when Y=0.0 -25.00
-12.50
control
Inhibited
2
0
25
50
75
100
[S] µM
1/v min/µM
Best-fit values
Slope
Y-intercept when X=0.0
X-intercept when Y=0.0
0.10
control
inhibited
0.2500 ± 1.371e-009
0.0100 ± 2.134e-010
-0.0400
0.2500 ± 3.358e-009
0.0200 ± 5.227e-010
-0.08000
Vm
V m (app) - 1
For competitive inhibition, Ki is calculated as
However, there is another widely used method to
determine Ki experimentally for Competitive and
noncompetitive inhibitions.
control
inhibited
Measuring v at fixed [S] but with increasing [I] then
plotting 1/v vs [I] yields a “Dixon plot”
0.05
0.0
[I]
[I]
Ki = K
m (app)
Km - 1
Lineweaver Burk Uncompetitive
-0.1
For both noncompetitive and uncomptitive inhibition,
Ki is calculated as
Ki =
1
0.15
Determining Ki
0.1
0.2
1/[S] per µM
0.3
0.4
64
Derivation of Dixon plot for competitive inhibition
V m [S]
v= K
m (app) + [S]
Dixon plots for competitive, noncompetitive
and uncompetitive inhibitions
0.5
thus
Noncompetitive
[I]
T
Y
+
1
K
m
[S]
Ki
1 K m(app)
1
=
+
=
+
v
Vm
V m [S]
V m [S]
V m [S]
0.4
1/v min/µM
Competitive
Uncompetitive
0.3
hence
Km
1 T Km
1
Y
=
+
[I]
v
K i V m [S] V m [S] + 1
0.2
Therefore plotting 1/v vs [I] yields a straight line with
0.1
-60
-40
-20
0
K
slope = K V m[S]
i m
20
40
60
[I] µM
1 Km
y - intercept = V T [S]
+ 1Y
m
when 1/v = 0
Km
Km
[I] K [S] = T [S]
+ 1Y
i
The x-intercept for competitive inhibition yields
-Ki(1+[S]/Km)
because S and I compete for binding to the enzyme.
The x-intercept for non competitive inhibition yields
and
[S] K m
[S]
[I] = - K i K T [S]
+ 1 Y = -Ki T 1 + K Y
m
m
-Ki.
The x-intercept for uncompetitive inhibition yields
-Ki(1+Km/[S])
65
Dixon Derivation for noncompetitive inhibition
Dixon Derivation for unncompetitive inhibition
V m (app) [S]
v = K + [S]
m
thus
thus
[I]
[I]
T1 +
Y T1 +
Y
K
m
[S]
Ki
Ki
Km
1
+ V
v = V m(app) [S] + V m(app) [S] =
V m [S]
m
hence
and
[I] 1 K m
1 T
Y
T
Y
=
+
1
v
K i V m [S] + 1
1
1 1 T Km
1 T Km
Y
Y
=
+
+
[I]
1
v
K i V m [S]
V m [S] + 1
Therefore plotting 1/v vs [I] yields a straight line with
1 1 Km
slope = K V T [S]
+ 1Y
i m
1 Km
y - intercept = V T [S]
+ 1Y
m
when 1/v = 0
- 1
Km
1 1 Km
[I] K V T [S]
+ 1 Y = V T [S]
+ 1Y
i m
m
and
V m(app) [S]
v= K
m(app) + [S]
1 Km
1 Km
[I] V T [S]
+ 1 Y = - K i V T [S]
+ 1Y
m
m
thus
[I] = - K i
K m(app)
[S]
Km
1
=
+
=
v V m(app) [S] V m(app) [S] V m [S] +
hence
T1 +
[I]
Y
Ki
Vm
1
1 T Km
1
Y
=
+
+
1
[I]
v V m [S]
K i Vm
Therefore plotting 1/v vs [I] yields a straight line with
1
slope = K V
i m
1 Km
y - intercept = V T [S]
+ 1Y
m
when 1/v = 0
- 1
Km
1
[I] K V = V T [S]
+ 1Y
i m
m
and
[I] - K m
T
Y
K i = [S] + 1
thus the x-intercept
Km
x - cept = [I] = - K i T [S]
+ 1Y
and
-
Ki =
[I]
T Km + 1 Y
[S]
66
Examples of Competitive Inhibitors
Naturally occurring (source and targets)
Examples of Noncompetitive Inhibitors
Naturally occurring (source and targets)
Digitalis (fox glove; Na,KATPase)
Caffeic Acid (tomato; lipoxygenase)
Tetradotoxin (puffer fish; Na-Channel)
Caffeine (tea, coffee; cAMP phosphodiesterase,
glucose transporters)
Cytochalasin B (fungi; glucose transporters)
Atropine (deadly nightshade; Acetylcholine
receptor)
Synthetic (their targets and use)
Ibuprofen (cyclo-oxygenase inhibitor - antiinflammatory)
Sulfanilamide (dihydropteroate synthetase;
antibacterials - inhibit folate synthesis)
Neostigmine (acetylcholinesterase; prolong
neuromuscular transmission - treat myasthenia
gravis)
Synthetic (their targets and use)
Haloperidol (brain & endothelial cell Nitric Oxide
Synthase inhibitor - anti-psychotic)
Trichostatin A (Histone DeAcetylase; anti-cancer)
mycophenolic acid (Inosine monophosphate
transferase; Dengue virus)
see http://www.emdbiosciences.com/docs/docs/
LIT/ISB_USD.pdf
Indinavir (HIV protease II inhibitor; prevent HIV
transmission and full-blown AIDS)
see http://www.emdbiosciences.com/docs/docs/
LIT/ISB_USD.pdf
67
What are Km and Vm?
Km is the concentration of [S] at which 1/2 of the
active sites are filled with substrate. The fraction of
filled sites [ES]/[E]t
Thus, when [S] = Km
[ES]
[S]
Km
=
=
[E] t
K m + [S] 2K m = 0.5
Hence one-half of the enzyme exists as ES and the
rate of the reaction v, is
Vm reveals the turnover number (sometimes called kcat)
of an enzyme if [Et] is known because
Vm = kcat [E]t
if [Et] = 1 µM and Vm = 600 mmols/L/sec,
kcat = 6 x 105 sec-1 - Each round of catalysis is 1/kcat =
1.7 µsec.
The expression Vm = kcat [E]t applies to all enzymes.
For the simple enzyme we are considering, the
solution for kcat is
kcat = kp
0.5 [E]t kp or v = 0.5 Vm
We also saw above that
Km =
kr + kp
kf
For more complex enzymes with more intermediate
complexes, the algebraic solution for kcat is more
complex.
For example, in the reaction
Thus if kr >> kp,
k
Km = kr = Kd
f
where Kd is the dissociation constant for the ES
complex.
Thus in some reactions where kp << kr, Km = Kd
while in other enzyme-mediated reactions where kp
≥ kr, Km > Kd.
E+S
ks
k-s
ES
k1
EP
k2
kp
E+P
k-p
k 1 k -p
k cat = k + k + k
-p
2
1
However, for the purposes of the discussion that
follows we shall assume that kcat = kp
68
“Kinetic Perfection” and the kcat/Km Criterion
When [S] << Km,
For uncharged molecules in solution, the encounter
rate constant, ke can be calculated as
V m [S]
v. K
m
ke =
where
We also know that
Vm = kcat [E]t
Since for our enzyme kcat = kp
this means that
kp
v = K [S] [E] t
m
v is therefore directly proportional to kp/Km and [S] at
fixed [E]t.
Are there limits upon kcat/Km?
kp
kp kf
=
Km kp + kr
Examination of this equation indicates that kf is
limiting. When kr << kp (i.e. catalysis is much faster
than ES dissociation into E and S),
kp kf
kp kf
.
kp + kr
kp = kf
Thus the greatest value that kp/Km can attain is kf. kf is
the second order rate constant that describes
association of E and S to form the ES complex. kf
includes terms that describe the collisional frequency
of E and S.
4rN (D A + D B) (rA + rB) -1
-1
M
.sec
1000
kT
D A = 6rr h
A
D and r refer to the diffusion constants and reaction
radii respectively of molecules A and B and N is
Avagadro’s number.
For an enzyme molecule of radius = 30 Å and a
substrate molecule of radius = 5 Å, the encounter rate
constant is 109 M-1.sec-1.
Thus diffusion limits the rate of encounter of E and S
and an upper limit of kp/Km is 109 M-1.sec-1. Even this
requires that the substrate can encounter the enzyme
surface in any orientation.
kcat/Km ratios of some enzymes, e.g.
acetylcholinesterase and carbonic anhydrase are
between 108 - 109 M-1 sec-1 indicating they have
achieved kinetic perfection.
For these enzymes, because kp/Km ≈ 108 M-1 sec-1,
this means that kp is so fast relative to kr that the ES
complex breaks down faster to form product and E
than it dissociates back to S and E.
Their activity is limited only by the rate at which they
encounter substrate in solution. Any further gain can
69
only be achieved by decreasing diffusion times. This
can be achieved by sequestering substrates and
products in the confined volume of a multi-enzyme
complex, e.g. mitochondria.
For an interesting discussion of kinetic perfection,
read
Biochemistry 1988, 27, 1158-1167
Triosephosphate Isomerase Catalysis Is Diffusion
Controlled.
Stephen C. Blacklow, Ronald T. Raines T. Wendell
A. Lim, Philip D. Zamore, and Jeremy R. Knowles
70
S ECTION 2
Recap of Key points
B max [S]
[S] b = K + [S]
d
4. What is the relationship between [E]t and Bmax?
B max = [E] t n
1. What assumptions did we make in order to derive
the Michaelis-menten equation for an enzyme
catalyzed reaction?
1. The reverse reaction (P→ S) is negligible.
2. Assume only a single central complex (ES)
exists. i.e. ES breaks down directly to E + P.
3. [S] >> [E]. The instantaneous interaction of S
and E to form ES does not significantly affect
free [S].
Ligand Binding
2. What is the relationship between Kd and Keq?
1. For the reaction
E+S
kf
kr
ES
2. Keq = kf/kr and has units of M-1
1. where n is the number of substrate binding
sites per enzyme
5. How can we compute Kd and Bmax from
measurements of [S]b at varying [S]?
1. Nonlinear regression analysis of plots of [S]b vs
[S] using the Michaelis-menten equation to
obtain best estimates of Kd and Bmax.
2. Linear transformation of the [S]b vs [S] data:
1. Lineweaver-Burk (1/[S]b vs 1/[S]) gives
1. Kd = -1/x-intercept
2. Bmax = 1/y-intercept
2. Hanes-Wolf ([S]/[S]b vs [S]) gives
1. Kd = -x-intercept
2. Bmax = 1/slope
3. Scatchard ([S]b/[S] vs [S]b) gives
3. Kd is kr/kf and has units of M
1. Kd = -1/slope
4. Kd = 1/Keq
2. Bmax = x-intercept
3. What is the relationship between free substrate
and [substrate] bound to an enzyme?
6. How does a competitive inhibitor affect [S]b vs [S]
plots?
71
1. The inhibitor shifts the plot to the right. Bmax is
unchanged but Kd increases.
7. What is Ki for a competitive inhibitor?
1. Ki is the Kd for inhibitor binding to E - We call it
Ki in order not to confuse it with Kd for S
binding to E.
8. How do we compute Ki for a competitive inhibitor?
1. We obtain the dissociation constant for S
binding to E in the absence of inhibitor (true Kd)
and in the presence of inhibitor (measured as
Kdapp).
2.Ki is obtained as:
[I]
Ki = K
dapp
Kd - 1
Catalysis
9. What is the steady-state assumption?
1. During the period in which we measure a
reaction, the concentration of intermediates
(e.g. ES) do not change although [S] and [P]
may change somewhat.
10. What is the Michaelis-menten equation for
catalysis?
Vm [S]
v = K + [S]
m
11. How is Vm related to [E]t?
1. Vm = [E]t kcat
12. What is Km?
1. Km is that [S] where v = Vm/2
13. How is Km related to the rate constants describing
the reaction?
1. The answer is different for each specific kinetic
model.
2. For our reaction, Km = (kp+kr)/kf
14. How can we compute Km and Vm from
measurements of v at varying [S]?
1. Nonlinear regression analysis of plots of v vs [S]
using the Michaelis-menten equation to obtain
best estimates of Km and Vm.
2. Linear transformation of the v vs [S] data:
1.Lineweaver-Burk (1/v vs 1/[S]) gives
1. Km= -1/x-intercept
2. Vm = 1/y-intercept
3. Hanes-Wolf ([S]/v vs [S]) gives
1. Km = -x-intercept
2. Vm = 1/slope
15. How do inhibitors affect substrate/velocity plots?
1. Competitive inhibitors
1. Leave Vm unchanged
2. Increase Kmapp
72
2. Noncompetitive inhibitors
1. Decrease Vmapp
2. Leave Km unchanged
3. Uncompetitive inhibitors
1. Decrease Vmapp
2. For noncompetitive inhibition, Ki is calculated
as
x - cept = - K i
3. For uncompetitive inhibition, Ki is calculated
from
2. Decrease Kmapp
16. How do we compute Ki for an inhibitor?
1. By measuring Km and Vm for a reaction in the
absence of an inhibitor and Kmapp and Vmapp in
the presence of the inhibitor.
1. For competitive inhibition, Ki is calculated as
[I]
Ki = K
m (app)
Km - 1
2. For noncompetitive and uncompetitive
inhibition, Ki is calculated as
Ki =
[S]
x - cept = - K i T 1 + K Y
m
[I]
Vm
V m (app) - 1
2. By measuring the rate of a reaction v at fixed
[S] in the presence of increasing [I] and making
a Dixon plot (1/v vs [I])
1. For competitive inhibition, Ki is calculated
from
Km
x - cept = - K i T [S]
+ 1Y
17. When is an enzyme kinetically perfect?
1. kcat/Km for a kinetically perfect enzyme ≥ 108
M-1.s-1
18. What does kinetic perfection really mean?
1. It means that the enzyme is so catalytically
active that when the ES complex forms,
substrate is immediately converted to product
or to an intermediate species thereby
preventing the ES complex from dissociating
back to E and S.
19. There are a great many equations in this chapter.
Do I need to know them all?
1. No. They are provided to show how our
understanding of this field develops from basic
concepts. You should, however, try to
understand the derivations of the many
solutions presented.
73
2. Most equations are provided as tools that will
allow you to analyze the data you may collect
in the research setting.
20. Which equations should I learn?
1. The equation for reversible substrate binding to
an enzyme:
E+S
kf
kr
ES
1. Keq = kf/kr and has units of M-1
2. Kd is kr/kf and has units of M
3. Kd = 1/Keq
2. The equations for Vm and Bmax
1. Vm = kcat[E]t
2. Bmax =[E]tn
3. The Michaelis-Menten equation
Vm [S]
v = K + [S]
m
74
S ECTION 3
9.
What assumptions did we make in deriving the
Michaelis-menten equation for enzyme catalysis?
Formative self-evaluation
questions
10. What is the Michaelis-menten equation for
enzyme-mediated catalysis?
11. What is Km?
Test your understanding of time-dependent processes
by answering the following questions:
1.
What is the Michaelis-Menten equation for ligand
binding?
2.
What is Kd?
3.
How does Kd obtained from Michaelis-Menten
analysis of equilibrium ligand binding differ from
Kd obtained from pseudo-first-order analysis of
the time course of ligand binding to an enzyme?
4.
What information does Bmax provide if [E]t is
known?
5.
What graphical methods are available to compute
Bmax and Kd?
6.
Which of these methods is used most commonly
today and why?
7.
How does a competitive inhibitor affect ligand
binding?
8.
What is Ki for inhibitor inhibition of ligand binding?
12. What graphical methods are available to compute
Vm and Km for enzyme-catalyzed reactions?
13. Which of these methods is used most commonly
today and why?
14. How is Vm related to [E]t?
15. How are Km and Kd related?
16. When is Km almost identical to Kd?
17. When is Km > Kd?
18. How and why does a competitive inhibitor affect
Km and Vm?
19. If you were to linearize your v vs [S] data to
illustrate the effects of a competitive inhibitor on
your enzyme, how would the inhibitor affect the
slope, y-and x intercepts of a Hanes-Wolf plot?
20. How and why does a noncompetitive inhibitor
affect Km and Vm?
21. If you were to linearize your v vs [S] data to
illustrate the effects of a noncompetitive inhibitor
on your enzyme, how would the inhibitor affect
75
the slope, y-and x intercepts of a Lineweaver-Burk
plot?
22. How and why does an uncompetitive inhibitor
affect Km and Vm?
23. If you were to linearize your v vs [S] data to
illustrate the effects of an uncompetitive inhibitor
on your enzyme, how would the inhibitor affect
the slope, y-and x intercepts of a Hanes-Wolf plot
vs a Lineweaver-Burk plot?
24. Describe 2 methods for calculating Ki for inhibition
of an enzyme by competitive, noncompetitive and
uncompetitive inhibitors.
25. When can you state that an enzyme has achieved
kinetic perfection?
26. What does kinetic perfection imply in terms of
rates of catalysis vs ES dissociation to E and S?
76
S ECTION 4
Key to formative
evaluations
Section 2 - Michaelis-Menten Kinetics
1. The equation is:
4. Bmax/[E]t yields n - the number of substrate binding
sites per enzyme.
5. The methods are:
1. [S]b vs [S] data can be fitted directly by
nonlinear regression analysis to yield Bmax and
Kd.
2. [S]b vs [S] data can be linearized in the
following ways to obtain Bmax and Kd from the
intercepts and slope obtained from linear
regression analysis of the plots:
1. Lineweaver-Burk (1/[S]b vs 1/[S]) gives
B max [S]
[S] b = K + [S]
d
2. Kd is that [S] at which [S]b is Bmax/2. It is also
known as the dissociation constant of the ES
complex.
3. Kd obtained from Michaelis-Menten and from
pseudo-first-order analsis of binding kinetics
should be identical. The advantage of the latter
analysis is that you also obtain kf and kr. The
advantage of the former analysis is that you do not
have to perform complex time-course studies.
1. Kd = -1/x-intercept
2. Bmax = 1/y-intercept
2. Hanes-Wolf ([S]/[S]b vs [S]) gives
1. Kd = -x-intercept
2. Bmax = 1/slope
3. Scatchard ([S]b/[S] vs [S]b) gives
1. Kd = -1/slope
2. Bmax = x-intercept
6. Curve-fitting to the Michaelis-Menten equation by
direct non-linear regression analysis is the
77
preferred method because inverting data amplifies
errors at low [S] where experimental error (signal to
noise) is frequently greatest. Today’s computers
are more than adequate to perform this analysis.
7. A competitive inhibitor increases Kd for substrate
binding but leaves Bmax unchanged.
8. Ki is Kd for inhibitor binding to the enzyme.
9. We assumed:
2. v vs [S] data can be linearized in the following
ways to obtain Vm and Km from the intercepts
and slope obtained from linear regression
analysis of the plots:
1. Lineweaver-Burk (1/v vs 1/[S)] gives
1. Km= -1/x-intercept
2. Vm = 1/y-intercept
2. Hanes-Wolf ([S]/v vs [S]) gives
1. The reverse reaction (P→ S) is negligible.
1. Km = -x-intercept
2. That only a single central complex (ES) exists.
i.e. ES breaks down directly to E + P.
2. Vm = 1/slope
3. [S] >> [E]. Thus the instantaneous interaction of
S and E to form ES does not significantly affect
free [S].
10. The equation is:
Vm [S]
v = K + [S]
m
13. Curve-fitting to the Michaelis-Menten equation by
direct non-linear regression analysis is the
preferred method because inverting data amplifies
errors at low [S] where experimental error (signal to
noise) is frequently greatest. Today’s computers
are more than adequate to perform this analysis.
14. Vm = [E]t kcat
11. Km is that [S] at which v is Vm/2.
15. Kd = kr/kf and Km = (kp+kr)/kf.
12. The methods are:
16. When kp < kr, Km ≈ Kd.
1. v vs [S] data can be fitted directly by nonlinear
regression analysis to yield Vm and Km.
17. When kp ≥ kr, Km ≥ Kd
78
18. A competitive inhibitor increases Kmapp but leaves
Vm unchanged because binding of the inhibitor and
S are mutually exclusive.
19. A Hanes-Wolf plot of v vs [S] data in the absence
or presence of a competitive inhibitor will show
that inhibitor makes the x-intercept (-Km) more
negative but leaves the slope (1/Vm) unchanged.
20. A noncompetitive inhibitor reduces Vm but leaves
Kmapp unchanged because inhibitor binding blocks
catalysis but does not affect substrate binding.
21. A Lineweaver-Burk plot of v vs [S] data in the
absence or presence of a noncompetitive inhibitor
will show that inhibitor increases the y-intercept (1/
Vm), leaves the x-intercept (1/-Km) unchanged and
increases the slope (Km/Vm).
22. An uncompetitive inhibitor reduces Vm and Kmapp
because I can bind only to the ES complex but the
ESI complex is inactive. Thus S reduces Kd for I
binding and I reduces Vm.
23. A Hanes-Wolf plot of v vs [S] data in the absence
or presence of an uncompetitive inhibitor will show
that inhibitor makes the x-intercept (-Km) more
positive, increases the slope (1/Vm) and leaves the
y-intercept (Km/Vm) unchanged.
A Lineweaver-Burk plot of v vs [S] data in the
absence or presence of an uncompetitive inhibitor
will show that inhibitor increases the y-intercept (1/
Vm), makes the x-intercept (1/-Km) more negative
and leaves the slope (Km/Vm) unchanged.
24. If we obtain Vm and Km for a reaction in the
presence or absence on an inhibitor, Ki can be
calculated in the following way:
1. For competitive inhibition, Ki is calculated as
[I]
Ki = K
m (app)
Km - 1
2. For noncompetitive and uncompetitive
inhibition, Ki is calculated as
Ki =
[I]
Vm
V m (app) - 1
25. If we measure the rate of a reaction, v, at fixed [S]
and vary the inhibitor concentration [I], we can then
make a Dixon plot (1/v vs [I]) and obtain Ki as:
1. For competitive inhibition, Ki is calculated from
[S]
x - cept = - K i T 1 + K Y
m
79
2. For noncompetitive inhibition, Ki is calculated
as
x - cept = - K i
3. For uncompetitive inhibition, Ki is calculated
from
Km
x - cept = - K i T [S]
+ 1Y
26. An enzyme is said to have achieved kinetic
perfection when the ratio kcat/Km ≥ 108 M-1.s-1
27. This means that kp >> kr, Km >> Kd and the ES
complex never has an opprtunity to dissociate to E
+ S. The enzyme is said to be diffusion-limited.
80