Download Biochemistry 304 2014 Student Edition Enzymes and Enzyme

Enzyme Catalysis and Enzyme
Kinetics
Student Edition
5/23/13 Version
Dr. Brad Chazotte
213 Maddox Hall
[email protected]
Web Site:
Pharm. 304
Biochemistry
http://www.campbell.edu/faculty/chazotte Fall 2014
Original material only ©2005-14 B. Chazotte
Goals- Catalysis
•Understand the nomenclature of enzymes.
•Understand the roles of cofactors and coenzymes and be familiar
with examples of each.
•Understand the meaning of activation energy and the concepts of
transition state theory & catalytic rate enhancement.
•Be generally familiar with the types of enzyme catalytic
mechanisms.
•Learn the biologically important electrophiles & nucleophiles.
Enzymes
Protein molecules that utilize the chemical and physical properties of their
component amino acids to facilitate biochemical reactions that otherwise would
be difficult to accomplish under physiological conditions.
•An enzyme provides a specific environment that allows the biochemical reaction to
occur more rapidly.
•The distinguishing feature of an enzyme-catalyzed reaction is that it occurs within
a pocket on the protein.
•This pocket is termed the active site and the molecule(s) that binds to the active
site and is converted is called the substrate.
•The molecule(s) produced by the reaction is called the product.
Lehninger 2005 Figure p.191
Enzymes: General Properties
Enzymatically catalyzed reactions:
• are typically 106 to 1012 times more rapid than the corresponding
uncatalyzed reactions.
• occur under relatively mild conditions under 100° C, 1 Atm, and
~pH 7.0.
•occur with a high degree of specificity with respect to both the
reactants and the products.
•can be controlled by non-substrate molecules, i.e. allosteric
control, covalent enzyme modification; or the synthesis or
degradation of the enzyme (amount of enzyme present).
Voet, Voet & Pratt 2002 p. 282
International Classification on
Enzymes (OTHLIL)
To be
systematic the
IUMBMB
developed a
numerical
coding and
naming
system. Here
are the 6 major
classes
Enzymes are commonly named by adding the suffix –ase to the substrate the
enzyme converts or a phase that describes the reaction catalyzed by the enzyme.
Lehninger 2000 Table 8.3
Enzyme Commission Number
An example:
The four-part enzyme classification number is a series of 4 numbers
separated by periods.
1.9.3.1
Cytochrome c oxidoreductase
trivial (common) names : cytochrome oxidase, Complex IV
1.
Class
Oxidoreductases
1.9.
Subclass
Acting on a heme group of donors
1.9.3
Sub-subclass With oxygen as an acceptor
See enzyme classification website @ http://expasy.ch/sprot/enzyme.html
Specificity of Enzymes
Geometric complementarity – the enzyme’s binding site
has a structure complementary to the substrate it needs to
bind.
Electronic complementarity – amino acids that form the
enzyme’s binding site are arranged to specifically interact
and attract the substrate molecule.
Stereospecificity – binding of chiral substrates and the
catalysis of their reactions is highly specific due in large part
to the inherent chirality of the L-amino acids that comprise
the enzyme.
EnzymeSubstrate
Complex
Illustration
Voet, Voet & Pratt 2013 Fig 11.1
Cofactors and Coenzymes
Some enzymes require some help.
Help can be in the form cofactors which are either one or more
inorganic ions or a complex organic or metalloorganic molecules
called a coenzyme.
Cofactors table
Holoenzyme: catalytically active complete enzyme together
with its bound coenzyme and/or metal ions.
Apoenzyme (aproprotein): protein part of such an enzyme.
Prosthetic group: a coenzyme or metal ion that is covalently or very tightly
bound to the protein.
Lehninger 2000 Table 8.2
Coenzymes as Transient Carriers
Table
Human Deficiency Disease
Perncious anemia
Pellagra
Megaloblastic anemia
Beriberi
What do enzymes do?
Enzymes as catalysts change the rate of a chemical
reaction but do not alter the equilibrium.
Transition State Theory,
Activation Energy (Ea),
& The Reaction Coordinate
Thermodynamics Again?
Transition state theory was developed to describe chemical
reactions by applying thermodynamic equilibrium concepts.
Transition State Diagram
Consider the bimolecular Rx (a) :
HA – HB + HC  HA + HB-- HC
Consider the generic bimolecular Rx (b) : A + B  X‡  P + Q
rate = k ~ e-ΔG‡/RT ; ΔG‡ = Ea
Voet, Voet & Pratt 2013 Fig 11.5
Transition State Diagram:
Effect of Catalyst
A catalyst functions by lowering the activation energy of a
reaction, the energy barrier for the reactants to become
products.
Voet, Voet & Pratt 2013 Fig 11.7
Catalytic Rate Enhancement*:
a ΔG‡ Calculation
The enhancement of catalyzed vs uncatalyzed is given by: k= eΔΔG‡/RT
ln k = ΔΔG‡/RT
RT ln k = ΔΔG‡
for a 100-fold change in rate at 25 ºC
(8.314514 J ºK-1 mol-1 x 298.15 ºK) x ln(100) = ΔΔG‡
(2478.97 J mol-1 ) x (4.605) = ΔΔG‡
11,416 J mol-1 =
10-fold:
11.42 kJ mol-1 = ΔΔG‡
5.71 kJ mol -1
compare H-bond ~20 kJ mol-1
1,000,000-fold:
34.25 kJ mol -1 covalent bond ~300-500 kJ mol-1
* Enhancement meaning the absolute value – hence the omitted “-” sign
Voet, Voet & Pratt 2013 Chap 11
Effect of
‡
ΔG
versus ΔG
Important Concepts!
A+B P+Q
ΔG‡ effects the likelihood of reactants going to products AND
products to reactants - speeds up both reactions!
ΔG effects the likelihood of reactant going to products OR
products going to reactants based on the free energy
difference between reactants and products.
thermodynamics
ΔG < 0 forward reaction favored
A+BP+Q
CATALYTIC MECHANISMS
Enzymes are effective as catalysts due to:
1)
their ability to rearrange covalent bonds using their
various amino acid side chains, metal ions and coenzymes.
2)
their ability to specifically bind the substrate molecule in an
enzyme-substrate complex and to use noncovalent interactions for
binding to significantly lower the free energy.
Types of Enzyme Catalytic
Mechanisms
1. Acid-Base Catalysis
2. Covalent Catalysis
3. Metal Ion Catalysis
4. Electrostatic Catalysis
5. Proximity and Orientation Effects
6. Preferential Binding of the Transition
State Complex
Acid-Base Catalysis
Mechanism: Keto-Enol Tautomerization
ketone
Uncatalyzed
Hydroxyl
“ol”
“ene”
acid
General
Acid
Catalyzed
General
Base
Catalyzed
base
δ ≡ partial charge
Voet, Voet & Pratt 2013 Fig 11.8
Types of Acid-Bases Catalysis
• Specific Acid-Base Catalysis: Non-enzymatic
reactions: Uses only the H+ on OH- ions present in
water. (No other molecules involved) Ions are
transferred between water and the intermediate faster
than the intermediate breaks down to reactants.
• General Acid Catalysis: A process in which partial
proton transfer from an acid lowers ΔG‡ and
accelerates the reaction.
• General Base Catalysis: A process in which partial
proton extraction by a base lowers ΔG‡ and
accelerates the reaction.
Amino Acids in General Acid-Base
Catalysis
Lehninger 2005 Figure 6.9
pH and Enzyme Activity
Lehninger 2005 Figure 6.17
RNase A Mechanism:
An Acid-Base Catalysis
Base catalysis
2’,3’Cyclic
ribonucleotide
intermediate
Acid catalysis
Voet, Voet & Pratt 2013 Fig 11.10
Covalent Catalysis:
Decarboxylation of Acetoacetate
uncatalyzed
carbonyl
elimination
decarboxylation
covalent bond
electron withdrawal
Transition state
Typical Stages:
1.
Nucleophilic Rx between catalyst & substrate to form covalent bond.
2.
Withdrawal of electrons from reaction center by the (now) electrophilic catalyst.
3.
Elimination of the catalyst (reverse of first step)
Voet, Voet & Pratt 2013Fig 11.11
Covalent Catalysis
• Covalent Catalysis (nucleophilic catalysis) definition:
accelerates the reaction via the transient formation of a
catalyst-substrate covalent bond. The typical way this
happens is a nucleophile group on the enzyme forms a
covalent bond with an electrophile group on the substrate.
• Important – The more stable the covalent bond formed
in the transition state, the less easily it can degrade.
Therefore good candidates for covalent catalysis like
imidazoles and thiols have high polarizability (mobile
electrons) have high nucleophilicity and also can form
good leaving groups.
Biologically Important
Electrophiles & Nucleophiles
“The nucleophilicity of a substance is closely related to its basicity”
Voet, Voet & Pratt 2013 Fig 11.12
Metal Ion Catalysis:
Carbonic Anhydrase
Participate in catalytic
process by:
•Binding to substrates to
properly orient them for
Rx
•Mediating redox Rx via
reversible metal ion
oxidation state changes.
CO2 + H2O HCO3- + H+
•Electrostatic stabilization
or neg. charge shielding
Active site opening
Nearly 1/3 of all enzymes utilize a metal ion for their catalytic function.
Voet, Voet & Pratt 2008 Fig 11.13
Electrostatic Catalysis
The charge distribution around enzyme active sites
appears to be arranged to stabilize the transition state of
the catalyzed reactions.
Proximity & Orientation Effects
Enzymes tend to be catalytically efficient.
•They bring substrate into contact with catalytic groups & multiple
substrates with each other. ~ 5-fold boost
•They bind their substrates in the proper orientation to promote the
reaction ~100-fold boost
•They freeze out relative translational and rotational motion of catalytic
groups and substrates – in transition state little relative motion of
catalytic groups ~107-fold boost
Voet, Voet & Pratt 2006 p. 328
Preferential Transition State Binding
S
ES
“An enzyme may bind the transition state with greater affinity than
the substrates or products” This increases the molecules in the
transition state, thus proportionally increasing the reaction rate
Lehninger 2013 Figure 11.15
Kinetics
The study of reaction rates
Goals -Kinetics
•Know the difference between 1st and 2nd order reactions and their half-lives.
•Know the nomenclature for enzyme kinetics.
•Understand the difference between rapid-equilibrium and steady-state approaches to
enzyme kinetics.
•Know the Michaelis-Menten equation and how to plot it.
•Understand the concepts of Km and Vmax.
•Understand the Lineweaver-Burke Plot, be able to plot it and extract kinetic
constants. Be aware that there are other types of kinetic plots
•Understand enzyme reaction reversibility and how that affects the kinetic equation.
•Understand basic types of multisubstrate enzymes (sequential and random, BiBi and
ping-pong) and how that effects their kinetic equations.
•Know the various types of enzyme inhibition, how their L-B plots look, and how to
get quantitative information from those plots.
•Understand how pH and temperature (Arrhenius eq.) affect enzyme activity
Setting Up A Kinetic Analysis
1. Write differential equations for each enzyme species.
2. Write the velocity equations.
3. Substitute in the differential equations the expression for
the enzyme species.
Internet Explorer Demonstration
Chemical Kinetics – Elementary Rxs
Reaction Stoichiometry
A→P
substance A; product P
A → I1 → I2 → P intermediates In
Rate Constant k
The proportionality constant at constant temperature describing the
rate of an elementary reaction. k is proportional to the frequency at which
the reacting molecules come together.
Reaction Velocity
The instantaneous rate of product appearance or reactant
disappearance.
Reaction Order
The number of molecules that must simultaneously come together
(collide) to generate a product, i.e. the molecularity
Reaction Order – First Order
First Order Reaction A → P
First Order Equation:
d[P]
v=
dt = -
d[A]
dt =
k[A]
The reaction at time t is proportional
to the concentration of A.
Ao is the initial reactant concentration
k is in s-1 units
For first order: The half-life is independent of [A0], i.e. it is a constant
Voet, Voet & Pratt 2013 Fig 12.1
Determining order & rate constant for
Irreversible 1st order Rx
Matthews et al.1999 Figure 11.1
Second Order Reactions (Bimolecular)
2A → P
Second Order Equation:
d[A]
v=
dt = k[A]2
1
[A]
k is in M-1 S-1 units
1
=
[Ao]
+ kt
Half-life : t½ = 1/k[Ao]
A+B→P
Second Order Equation:
d[A]
d[B]
v = - dt = - dt = k[A][B]
k is in M-1 S-1 units
Pseudo first order (for B if [A] >>[B] & vice versa)
Rate Equation
Describes reaction progress as a function of time.
Derivation: (where t = time)
d[A]
[A]
⌠ [A]
= d ln[A] = - k dt
Integrate from [Ao] to [A]
⌠t
⌡ [ Ao] d ln [A] = - k ⌡0 dt
ln [A] = ln[Ao] – kt or
[A] = [Ao] e-kt
Half Life Calculation
First Order
k t½ = ln [A0]/2
[Ao]
t½ = ln 2/k = 0.693 / k
Second Order
t½ = 1/(k [Ao])
Half Life: t½
Enzyme Kinetics
The study of enzyme reaction rates
Lehninger 205 Figure 8.6
Enzyme Kinetic Definitions:
E Enzyme;
[E] enzyme concentration:
ES enzyme substrate complex
P, Q, R,… “products”
[E]T total enzyme
[ES] complex concentration
A, B, C …. “substrates”
I, J, K… “inhibitors”
k rate constant k1 forward rate constant k-1 reverse rate constant
kp the catalytic rate constant.
v = reaction velocity
vo = initial reaction velocity, when [P] ~ 0
Two Approaches to Enzyme
Kinetics
Problem: One does not know the concentration of ES during a
reaction. Therefore, one has to make certain assumptions for
calculations
Rapid Equilibrium (Henri-Michaelis-Menten)
Early components of the reaction are at equilibrium. Permits one
to express [ES] in terms of [E], [S], & Ks, i.e. an equilibrium
expression. One still needs a velocity equation to insert expression
for [ES].
Steady-State (Briggs-Haldane) {most used approach}
Shortly after the reaction starts [ES] would reach a steady-state
and this would be close to the equilibrium level.
Lehninger 2005 Figure p.203
Henri-Michaelis-Menten I
E + S
k1
k-1
Unireactant Enzyme
ES k2
E
+ P
E, S, and ES equilibrate very rapidly compared to formation of E + P
Instantaneous velocity depends on [ES]  v = k2 [ES]
(1)
The total enzyme is written: [E]T = [E] + [ES]
(2)
Divide equation (1) by [E]T
v
[E]T
=
k2 [ES]
[E] + [ES]
(3)
Henri-Michaelis-Menten II
E + S
k1
k-1
ES
k2
E + P
With rapid equilibrium assumption [ES] can be expressed in terms of
[S], [E] and Ks (the dissociation constant of the ES complex)
Ks = [E] [S] = k-1
 [ES] = ([S]/ Ks) [E]
(4, 5)
[ES]
k1
Substitute for [ES] into eq. 3 using eq 5.
v
[E]T
=
k2 ([S]/Ks)[E]
[E] + ([S]/Ks)[E]
(6)
Henri-Michaelis-Menten III
E + S
k1
k2
E + P
k-1
Cross multiple by k2 and cancel out [E] (on right side)
v
=
([S]/Ks)
k2 [E]T
ES
1 + ([S]/Ks)
If v = kp [ES] then kp [E]T = Vmax (enzyme is saturated)
v
Vmax
v
Vmax
(6)
(7)
= ([S]/Ks)
1 + ([S]/Ks)
(8)
= [S]
Ks + [S]
(9)
A Simple Enzyme Reaction and the
Steady-State (Progress Curves)
Voet, Voet & Pratt 2013 Fig 12.2
Steady-State (Briggs-Haldane) I
E + S
k1
Unireactant Enzyme
ES k2
E
+ P
k-1
If the rate ES forms E + P is >> the rate ES goes back to E + S (K2 >
k-1) and [S] >> [E]T then E, S, and ES will not be in equilibrium.
Instantaneous velocity is
v
[E]T
=
v = k2 [ES]
(1)
k2 [ES]
[E] + [ES]
(from before)
(2)
If [ES] is constant, i.e. d[ES]/dt = 0 , then the rate that ES forms is
equal to the rate that ES decomposes.
Steady-State (Briggs-Haldane) II
E + S
k1
k-1
k2
ES
E + P
If [ES] is constant, i.e. d[ES]/dt = 0 , then the rate that ES forms is
equal to the rate that ES decomposes.
k1
ES formation: E + S → ES
kp
ES decomposition: ES → E + P
(3)
and
k-1
ES → E + S
(4)
rate of ES formation: = k1 [E][S]
(5)
rate of ES decomposition: = k-1 [ES] + k2 [ES]
(6)
=
(k-1 + k2) [ES]
(6a)
Steady-State (Briggs-Haldane) III
E + S
k1
ES
k-1
d[ES]/dt = 0 ,
kp
E + P
At steady-state.
rate of ES formation = rate of ES decomposition
k1 [E][S]
=
(7)
(k-1 + kp) [ES]
(8)
Solve eq. 8 for [ES]
k1[E][S]
[ES]
(9)
= (k-1 + kp)
define Michaelis constant
(for above reaction)
k-1 + kp
km =
k1
[S]

[ES] = km [E]
Steady-State (Briggs-Haldane) IV
E + S
k1
ES
k-1
v
= ([S]/Km)
Vmax
1 + ([S]/Km)
=
k2
E + P
v
= [S]
Vmax
Km + [S]
(10a,b)
Can rearrange ES expression
k-1 + k2
Km =
k1
[S] [E]

When [S] = Km
=
[ES]
Km
v=
Km + Km Vmax = ½Vmax
Vmax = k2 [E]T the highest velocity when all of the enzyme is [ES], i.e. saturated.
Steady-State (Briggs-Haldane) V
Differential Equations for Unireactant Enzyme
E + S k1k-1
ES k2
E + P
One can write 4 differential equations to describe the above reaction
d[E]
dt
=
(k-1 + k2) [ES] - k1 [E] [S]
d[ES]
dt
=
k1 [E] [S] - (k-1 + k2) [ES]
d[S]
dt
=
(k-1) [ES] - k1 [E] [S]
d[P]
dt
=
(k2) [ES]
Also know that:
[E]T =
[E] + [ES]
(11)
(12)
(13)
(14)
(15)
Segal Enzyme Kinetics 1975 p. 28
Steady-State Kinetic Plot
Initial Velocity vs Substrate [ ] in a
Simple Enzyme Reaction
o
Michaelis-Menten equation
Voet, Voet & Pratt 2013 Fig 12.3
Lehninger 2000 Figure 8.12
HTML DISPLAY
Michaelis-Menten Kinetics
12-1b_MichaelisMenten\MichaelisMenten.htm
Significance of Km
Operational Definition: The substrate concentration at which the
reaction velocity is half maximal, i.e. when Km = [S] then vo = ½Vmax
1.
It establishes an approximate value for the intracellular level of a substrate.
2.
Since it is constant for a given enzyme/substrate, its numerical value provides
a means of comparing enzymes from different organisms or from different
tissues of the same organism, or from the same tissue at different
developmental stages (Vmax is not a constant but depends on kp and [E]T.) Km
will vary with temperature and pH.
3.
Km can be altered by ligand binding – one mode of enzyme regulation.
4.
If Km is known the assay conditions can be altered so that [S] >> Km so that
Vmax can be determined which is a measure of [E]T.
5.
It indicates the relative “suitability” of alternate substrates for an enzyme. The
substrate with the lowest Km has the highest affinity for the enzyme. The
“best” substrate has the highest Vmax/Km ratio.
Segal Biochemical Calculations 1976 p. 218
Catalytic Efficiency & Turnover Number
Define the catalytic constant: kcat = Vmax/[E]T
This is also known as the turnover number – the number of reaction
processes (number of moles of substrate transformed per minute per mole of
catalytic site under optimal conditions - turnovers) that each active site
catalyzes per unit time (Note: some enzymes have more than one active site)
However, more complex enzymes have a more complicated expression for
kcat, i.e. more rate constants.
When [S] << Km very little ES is formed and then [E] ≈ [E]T
Vo ≈ (k2/Km)[E]T [S] ≈ (kcat/Km) [E] [S]
Under these conditions (kcat/Km) is a measure of the enzyme’s catalytic
efficiency since this apparent second order rate constant (depends on BOTH [E]
and [S]) the rate of the reaction depends on how often E and S encounter each
other in solution.
Most efficient enzymes have this ratio near the
diffusion-controlled limit: 108 – 109 M-1s-1
kcat/Km ≤ k1
Kinetic Data Analyses
o
Michaelis-Menten
Michaelis-Mentenequation
equation
Lineweaver-Burke Plot
y = (m) x + b
1 Km 1 + 1
vo Vmax [S]
Vmax
Voet, Voet & Pratt 2013 Fig 12.4
Eadie-Hofstee Plot
y
v
[S]
=
m * x
Km* v +
+
b
Vmax
Km
Matthews et al.1999 Figure 11.17
Reversibility I
Effect of Product on Forward Reaction
E +S
k1
ES
k-1
vf =
When [S] = 0
(final)
EP
k-2
When [P] = 0
(initial)
k2
vr =
Vmaxf [S]
Kms + [S]
Vmaxr [P]
Kmp + [P]
k3
k-3
E+P
The direction of the reaction
will depend upon the ratio of
[P]/[S] relative to Keq
Reversibility II
Effect of Product on Forward Reaction
E +S
k1
ES
k-1
k2
k3
EP
k-2
E+P
k-3
From rapid equilibrium assumptions set up net velocity eqs.
vnet = k2 [ES] - k-2 [EP]
vnet
[E]t
k2 [ES] - k-2 [EP]
[E] + [ES] + [EP]
Know [ES] = ([S]/Ks) [E]
vnet
[E]t
and
[EP] = ([P]/Kp) [E]; substitute in
k2 ([S]/Ks) [E] - k-2 ([P]/Kp) [E]
[E] + ([S]/Ks) [E] + ([P]/Kp) [E]
Factor [E]
out

Reversibility III
Effect of Product on Forward Reaction
E +S
k1
ES
k-1
vnet
vnet
k2
k3
EP
k-2
E+P
k-3
k2 [E]t([S]/Ks) - k-2 [E]t ([P]/Kp)
1+
([S]/Ks)] +
([P]/Kp)
Vmaxf ([S]/Kms) - Vmaxr ([P]/Kmp)
1+
([S]/Kms)] +
([P]/Kmp)
Final equation in terms of steady-state
Multiply
right side
by [E]t
Multisubstrate Enzymes
Voet, Voet & Pratt 2013 Figure 12.5
Sequential Reaction Mechanisms
“Reactions in which all substrates must combine with the
enzyme before a reaction can occur and products are released”
E
A
+
B
P
+
Q
Sequential Mechanisms: Types
Ordered
Bi-Bi
Random
Bi-Bi
Voet, Voet & Pratt 2013 p. 367
Bisubstrate Rx Steady-State Kinetic
Analysis: Random BiBi
Lehninger 2000 Figure 8.14a
Ping-Pong Non-Sequential Reaction
Mechanisms
“Group transfer reactions in which one or more products are
released before all susbtrates have been added.”
Non-Sequential Mechanism
MgATP
MgADP
glucose
Glucose-6-P
PingPong
Bi-Bi
V
Vmax
[A][B]
Kma[A] + Kmb[B] + [A][B]
Voet, Voet & Pratt 2013 p. 367
Bisubstrate Rx Steady-State Kinetic
Analysis: BiBi Ping Pong
Lehninger 2000 Figure 8.14b
King-Altman Diagrams
KING-ALTMAN DIAGRAM EXAMPLE PART A
k1[bc1ox]
Qr
Qrbc1ox
k-1
k4
k-4
k-2
k2
k-3[bc1ox]
QO
QObc1r
k3
Qr
QT
=
k2k3k4 +k-1k3k4 + k-3[bc1r]k-2k-4 + k-2k-1k4
D
Qrbc1ox
QT
=
k3k4k1[bc1ox] +k4k1[bc1ox]k-2 + k-2k-3[bc1r]k1[bc1ox] + k-4 k-3k-2
D
Qobc1r
QT
=
k4k1[bc1ox] +k1[bc1ox]k2k-3[bc1r] + k2k-4k-3[bc1r] + k-1 k-4 k-3[bc1r]
D
Qo
QT
=
KING-ALTMAN DIAGRAM EXAMPLE PART B
k1[bc1ox] +k2k3k-4 + k3k--4k-1 + k-2k-1k-4
D
Velocity Equation :
v
ET =
v= k1[Qr] [bc1ox] - k-1 [Qbc1ox]
k1k2k3k4 [bc1ox]
- k-1k-2k-3k-4 [bc1r]
k4(k2k3+k-1k3+k-1k-2) + k-4(k2k3+k3k-1+k-1k-2) + k1(k3k4+k4k-2+k2k4+k2k3)[bc1ox] + k-3(k1k2+k-1k-2)[bc1ox] [bc1r] + k-3(k1k-2+k-2k-4+ k-1k-4)[bc1r]
Enzyme Inhibition
Inhibitor - Any substance that reduces the velocity of an
enzyme-catalyzed reaction.
Reversible Types:
•Competitive
•Uncompetitive
•Mixed (Noncompetitive)
Inactivator - Any substance that irreversibly binds to an enzyme.
(appears similar to noncompetitive inhibition.)
Competitive Inhibition
Kmapp = Km [1 + ([I]/Ki)]
KI = [E][I] / [EI]
v = k2 [ES] vmax = k2[E]T
v
k2 [ES]
[E]T [E] + [ES] + [EI]
v
([S]/Km) [E]
k2[E]T [E] + ([S]/Km) [E] + ([I]/KI) [E]
v
([S]/Km)
Vmax 1 + ([S]/Km) + ([I]/KI)
Voet, Voet & Pratt 20 Figure 12.8
[S]
Km(1 + ([I]/KI)) +[S]
[S]
αKm + [S]
Competitive Inhibition
Matthews et al.1999 Figure 11.20
Voet, Voet & Pratt 2013 Figure 12.7
k1
k-1
Uncompetitive Inhibition: M-M enzyme in
L-B Plot
k
2
KI = [ES][I] / [ESI]
v = k2 [ES] vmax = k2[E]T
v
k2 [ES]
[E]T [E] + [ES] + [ESI]
Voet, Voet & Pratt 2013 Fig 12.9
v
([S]/Km) [E]
k2[E]T [E] + ([S]/Km) [E] + ([S][I]/KmKI) [E]
v
([S]/Km)
Vmax 1 + ([S]/Km) + ([S][I]/Km KI)
vmaxI = vmax / /(1 + ([I]/KI))
v
[S]
VmaxI
(Km/(1 + ([I]/KI)) +[S]
[S]
αKm + [S]
Mixed Inhibition:
M-M enzyme in L-B Plot
KI = [E][I]/[EI] and K’I = [ES][I]/[ESI]
v
k2 [ES]
[E]T [E] + [ES] +[EI] + [ESI]
v
[S]
Vmax Km (1+ ([I]/KI)) + [S] (1+ ([I]/K’I))
v
VmaxI
[S]
αKm + α’ [S]
Voet, Voet & Pratt 2013 Fig 12.10
Michaelis-Menten Equations:
Effects of Inhibitors Table
An example for competitive inhibitor
the X intercept is:
1/Km app =1/αKm = 1 /(1+[I]/Ki)Km
Lehninger 2005 Table 6.9
Voet, Voet & Pratt 2013 Table 12.2
Temperature Effects Enzyme
Activity
Kinetics, Thermodynamics, and
Transition State Theory I
A+B
K‡
X‡
k’
P+Q
dP/dt = k [A][B] = k’ [X‡] under rapid equilibrium assumption write:
K‡ = [X‡] / [A][B]
ΔG‡ = -RT ln K‡
equilibrium constant expression
relate equil. const. to free energy (thermodynamics)
dP/dt = k’ e-ΔG‡/RT [A] [B]
k’ = 
substituting into the first equation
define  vibra. freq. bond breaks as X‡ to products
and  prob. X goes to products rather than reactant. (0 -1)
ν = ε/h
h is Planck’s const. ε is the avg. energy of the vibration for
the decomposition on X‡
Kinetics, Thermodynamics, and
Transition State Theory II
A+B
ε = kBT
K‡
X‡
k’
P+Q
from Stat. Mech. The energy of a classical oscillator at T.
(the available thermal energy) where
kB = Boltzman constant 6.6261 x 10-34 J s
k’ = kBT/h
via substitution
k = (kBT/h) e-ΔG‡/RT = A e -ΔG‡/RT
Arrhenius Equation
log A
k = Ae-Ea/RT
Slope
ln k = ln A –Ea/RT or
-Ea/2.303R
log k = log A –Ea/2.303RT
Matthews et al.1999 Figure 11.3
pH Effects Enzyme Activity
Lehninger 2004 Figure 6.17
Enzyme Regulation
Control of Enzyme availability
Cells control the rate of enzyme synthesis and
degradation and thus can control the amount of enzyme
present.
Control of Enzyme activity
•
via binding of small molecule “allosteric
effectors” that alter catalytic activity.
•
via covalent modification of an enzyme, e.g.
phosphorylation and dephosphorylation
Voet, Voet & Pratt 2013 0.381
Allosteric Effect
2.0 mM
0.4 mM
Voet, Voet & Pratt 2013 Fig 12.11
Kinetics and Mechanism
The steady-state kinetic analysis of a reaction cannot
unambiguously establish the reaction’s mechanism
Bisubstrate Rx: Examples
Transferase
Redox
Voet, Voet & Pratt 2008 Fig 12.5
Cytochrome P450 & Drug Metabolism
• “Differences in reactions to drugs arise from genetic differences
among individuals as well as differences in their disease states,
other drugs they are taking, age, sex, and environmental factors”
• Cytochromes P450: a major function is to detoxify xenobiotics –
involved in metabolic clearance of a many of drugs in use today.
• Cytochromes P450: superfamily of enzymes – occur in nearly all
living organisms.
• Human genome has 57 isoforms -33% of isoforms are in liver
• Nomenclature: P450 isozymes are named by the letters CYP followed by a
number designating its family, an uppercase letter designating its subfamily,
and another number, e.g. CYP2D6
Voet, Voet & Pratt 2012 p.389
Cytochrome P450 & Drug Metabolism II
Typical reaction type:
RH + O2 + 2 H+ + 2e- → ROH + H2O
• Different Cyt P450 are SPECIFIC for particular types of a wide variety of
typically lipophilic compounds, RH.
•
PAH polycyclic aromatic hydrocarbons (carcinogenic – in smoke, broiled meat)
•
PCB polycyclic biphenyls (electrical insulators; plasticizers)
•
steroids (P450s also involved in their biosynthesis)
•
drugs (many types)
Convert Xenobiotics to a more water soluble form to facilitate excretion
• New hydroxyl group can be enzymatically conjugated to glucuronic acid, glycine,
sulfate & acetate – further increases water solubility
Good: Allow xenobiotics to be detoxified and/or cleared.
Bad: Sometimes converts an innocuous compound to a toxic agent , e.g. acetaminophen
metabolism at high dosages
Cytochrome P450 & Drug Metabolism III
Cyt P450s often mediate drug-drug interactions.
Text Example: Two drugs A & B.
Case 1: Drug A metabolized by or inhibits Cyt P450 isozyme.
Same isozyme metabolizes drug B.
Both drugs given together – Bioavailability of drug B increases. Why?
What if drug B has a low therapeutic index?
Case 2: Drug A induces increased expression of the particular isozyme of Cyt P450
Both drugs given together – Bioavailability of Drug B decreases. Why?
Case 3: Additionally Drug B is metabolized to a toxic product.
Both drugs given together. What happens to rate of Drug B metabolism? Why?
What is the resultant problem?
End of Lectures