Transcript Prof. Kamakaka`s Lecture 6 Notes

Km Vmax Kcat
Km is [S] at 1/2 Vmax
It is a constant for a given enzyme at a particular temp and pressure
Km is unique to each Enzyme and Substrate. It describes properties of enzymesubstrate interactions. Dependent on temp, pH etc. Independent of enzyme conc.
It is an ESTIMATE of equilibrium constant for substrate binding to enzyme
Small Km= tight binding, large Km=weak binding
It is a measure of substrate concentration required for effective catalysis
Vmax is THEORETICAL MAXIMAL VELOCITY
Vmax is constant for a given enzyme. It is directly dependent on enzyme conc. It is
attained when all of the enzyme binds the substrate. (Since these are equilibrium
reactions enzymes tend towards Vmax at high substrate conc but Vmax is never
achieved. So it is difficult to measure)
To reach Vmax, ALL enzyme molecules have to be bound by substrate
Kcat is a measure of catalytic activity- direct measure of production of product
under saturating conditions.
Kcat is turnover number- number of substrate molecules converted to product per
enzyme molecule per unit time
Catalytic efficiency = kcat/km
Allows comparison of effectiveness of an enzyme for different substrates
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Enzyme Km examples
Hexokinase prefers glucose as a substrate over ATP
Km values of enzymes range from 10-1M to 10-7M for their substrates. It also
varies depending on substrate, pH, temp, ionic strength etc.
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Kcat
Catalase is very efficient-it generates 40 million molecules of product per second.
Fumarase is not efficient-it generates only 800 molecules/per second
kcat = Vmax / [E]T
Turnover number
Number of reaction processes each active site catalyzes per unit time
Measure of how quickly an enzyme can catalyze a specific reaction
For M-M systems kcat = k2
Kcat is turnover number for the enzyme
number of substrate molecules converted into product per unit time by that enzyme 3
Kcat/Km
The catalytic constant, kcat is
kcat = Vmax/[ET]
for simple reactions, kcat = k2
When [S] << Km, very little ES is formed, such that [E] is
almost equal to [ET].
In this case, vo = k2[ET][S]/Km + [S]
Rate
constant of rxn E + S ---> E + P
Specificity
Gauge
constant
becomes = (k2/Km)[ET][S] = (kcat/Km)[ET][S]
of catalytic efficiency
Catalytic
perfection ~ 108 -109 M-1 s-1 (close to diffusion)
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Kcat/Km
Rate
constant of rxn E + S ---> E + P
Specificity
Gauge
constant
of catalytic efficiency
Catalytic
perfection ~ 108 -109 M-1 s-1 (close to diffusion)
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Enzyme cofactors
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Coenzymes
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How do ENZYMES carry out catalysis?
•Entorpy reduction- holds substrates in proper position
Bringing two reactants in close proximity (reduce entropy & increase effective reactant concentrat)
•Substrate is desatbilized when bound to enzyme favoring reaction-(change of solvent, chargecharge interactions strain on chemical bonds).
•Desolvation of substrate- H bonds with water are replaced by H bonds with active site
Enzymes form a covalent bond with substrate which stabilizes ES complex (Transition state is
stabilized)
Enzyme also interacts non-covalently via MANY weak interactions
Bond formation also provides selectivity and specificity
(H bonds- substrates that lack appropriate groups cannot form H bonds and will be poor substrates)
(Multiple weak interactions between enzyme and substrate)
Free energy released by forming bonds is used to activate substrate (decrease energy barrier/lower
activation energy of reaction)
Induced fit-binding contributes to conformation change in enzyme
Whats the Bill?
5.7 kJ/mol is needed to achieve a 10x increase in rate of a reaction
Typical weak interactions are 4-30 kJ/mol
Typical binding event yields 60-100 kJ/mol
MORE THAN ENOUGH ENERGY!!!
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A Hypothetical reaction
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Breaking a stick
Imagine you have to break a stick. You hold
the two ends of the stick together and apply
force. The stick bends and finally breaks.
You are the catalyst. The force you are
applying helps overcome the barrier.
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A stickase with a pocket complementary in structure to the stick (the substrate) stabilizes the substrate. Bending is
impeded by the attraction between stick and stickase.
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An enzyme with a pocket complementary to the reaction transition state helps to destabilize the stick, contributing to catalysis of
the reaction. The binding energy of the interactions between stickase and stick compensates for the energy required to bend the
stick.
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Role of binding energy in catalysis. The system must acquire energy equivalent to the amount by which DG‡ is lowered.
Much of this energy comes from binding energy (DGB) contributed by formation of weak noncovalent interactions between
substrate and enzyme in the transition state.
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Lock/Key or Induced Fit
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Lock/Key- Complementary shape
The enzyme dihydrofolate reductase with its substrate NADP+
NADP+ binds to a pocket that is complementary to it in shape and ionic properties, an illustration of "lock and key" hypothesis
of enzyme action. In reality, the complementarity between protein and ligand (in this case substrate) is rarely perfect,
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Induced Fit
Hexokinase has a U-shaped structure (PDB ID 2YHX). The ends pinch toward each other in a conformational change
induced by binding of D-glucose (red).
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Substrate specificity
The specific attachment of a prochiral center (C)
to an enzyme binding site permits enzyme to
differentiate between prochiral grps
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Enzyme-substrate
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Catalysis
•
•
•
•
•
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Catalysis
•Acid-Base Catalysis- donate or accept protons/electrons from and to substrate
•Covalent Catalysis-transient covalent link between substrate and enzyme side
chain
•Metal-Ion Catalysis-Metal in active site donate or accept protons with substrate
•Proximity & Orientation Effects (reduction in entropy-two mol brought together and
oriented in specific manner)
•Transition State Preferential Binding
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The active sites of enzymes
contain amino acid R groups.
Active site is lined with
hydrophobic residues
R Groups
Polar amino acid residues in
active site are ionizable and
participate in the reaction.
Anion/cation of some amino
acids are involved in catalysis
Lysozyme: Cleaves glycosidic bonds in carbohydrates
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Covalent Catalysis
All or part of a substrate is transiently covalently bound to the enzyme to form a reactive
intermediate
Group X can be transferred from A-X to B in two steps via the covalent ES complex -EX
A-X+ E <-----> X-E + A
X-E + B <-----> B-X+ E
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Two mechanisms for acid catalysis
Specific acid catalysis:
- A proton is transferred to the substrate in a rapid preequilibrium;
Subsequently, the protonated substrate reacts further to form the
product
Specific acid–base catalysis means specifically, –OH or H+ accelerate the reaction.
The reaction rate is dependent on pH only.
The rate is only dependent on the pH, not on [HA]
General acid catalysis:
- Proton transfer occurs in a slow, rate determining step;
Subsequently, the protonated substrate rapidly reacts to give the
product.
the reaction rate is dependent on all acids/bases present, dependent on the buffer
concentration, at constant pH.
Enzymes often use general acid or base catalysis: General acid/base catalysis by enzymes
• They work at neutral pH, so low [H+] and [OH-]
• High localized concentration of general acid/base
• Correct orientation of the acidic/basic group around the substrate
• Optimum catalysis at pH around pKa
General acid-base catalysis involves a molecule besides water that acts as a proton
donor or acceptor during the enzymatic reaction. It facilitates a reaction by
stabilizing charges in the transition state through the use of an acid or base, which
donates protons or accepts them, respectively.
Nucleophilic and electrophilic groups are activated as a result of the proton addition
or removal and causes the reaction to proceed.
:B
B
+HA
B
(1)
H A
B H
(2)
A
BH + A
(3)
X
H
:X-
Side chains of various amino acids act as general acids or general basis
Amino acid residues such as His often have a pKa that is close to neutral pH and are
therefore able to act as a general acid or base catalysts
B+
H
Without a catalyst the intermediate converts
back to the reactants and does not proceed
forward (high barrier).
Donation of a proton by water or an acid helps
the process move forward.
The active sites of enzymes contain amino
acid R groups, that participate in the
catalytic process as proton donors or proton
acceptors.
Catalysts
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Proton donor/acceptor (Nucleophile/electrophile)
Asp and Glu are negatively charged at pH7.0 and their side chains are acidic.
These side chains ACCEPT protons which neutralize the charge.
Lys, Arg, His are positively charged at pH 7.0 and their side chains are basic.
These side chains DONATE protons to neutralize their charge.
Asp/Glu
Lys/Arg
COO- + H+ <-----------> COOH
NH3+ <----------> NH2 + H+
Nucleophiles
R-OH <---> R-O: + H+ (hydroxyl)
R-SH <---> R-S: + H+ (sulphydryl)
R-NH3 <---> R-NH2: + H+ (amino)
Electrophiles
H+
M+
Proton
Metal ion
R
+C
O
Carbonyl
R’
Nucleophiles-groups rich in and capable of donating electron (attracted to nucleus)
Electrophile- group deficient in electron (attracted to electron)
Reactions are promoted by proton donors (general acids) or proton acceptors (general bases). The active sites of29
some
enzymes contain side groups, that can participate in the catalytic process as proton donors or proton acceptors.
Acid base catalysis
RNaseA cleavage of RNA
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Metal Ion catalysis
Roles of metals in catalysis:
• As “super acid”: comparable to H+ but stronger
• As template: metal ions are able to coordinate to more than 2
ligands and can thereby bring molecules together
• As redox catalyst: many metal ions can accept or donate
electrons by changing their redox state
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Enzyme Inhibition
S
P
S
EE
E+S
E
ES
P
E
EP
E
E+P
What happens if a reactant does not leave the active site?
–Enzyme is blocked (inhibited) from further interactions
–Why inhibit?
To control [S] or [P]
–Increase [S] unreacted
–Decrease [P] formed
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Enzyme Inhibition
Many molecules inhibit enzymes
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Enzyme Inhibition
Many molecules inhibit enzymes
Reversible
Competes with substrate
Does not compete with substrate
Irreversible (covalently bound to enzyme)
Reversible
Competitive
Uncompetitive
Mixed
Noncompetitive
Irreversible
Suicide
inactivators
To figure out what kind of inhibitor we have,
conduct two sets of rate experiments:
([E] constant in each case)
• [S] constant, test effect of increasing [I] on Vo
• [I] constant, vary [S]
• Plot the results as 1/Vo vs. 1/[S]
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Competitive Inhibitor
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Competitive Inhibitor
Most
common
Inhibitor
competes with natural substrate for binding to active site
Inhibitor
similar in structure to natural substrate and binds active site of enzyme
(reducing effective enzyme conc)
Binds
May
If
more strongly
or may not react
reacts, does so very slowly
Gives
info about active site through comparison of structures
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Drug targets
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Gleevec
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Gleevec: How it works
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HIV protease structure
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Protease Inhibitors
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Protease + Inhibitor
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





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I
and S compete for E
Increasing
[I]
Increases
Reduces
Need
[EI]
[E] available for substrate binding
to keep [I] high to ensure inhibition
Dosage
S
As
overcome inhibitor effects; saturate
[I] increases, KM increases
More
S required to reach ½ Vmax
Rate does not increase rapidly with [S] due to inhibition but the rate reaches the
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same maximal rate, just at a higher substrate conc
Reversible Inhibition (competitive)
1/v
Vmax
-Inh
-Inh
+inh
vo
1/2 Vmax
+Inh
a
1/Vmax
Km
Km
(app)
[Substrate]
1/[S]
-1/Km -1/Km (app)
Inhibitor competes with substrates for binding to active site
Inhibitor is similar in structure to substrate
binds more strongly
reacts more slowly
Increasing [I] increases [EI] and reduces [E] that is available for substrate binding
Need to constantly keep [I] high for effective inhibition (cannot be metabolized
away in body)
Slope
is larger
Intercept
KM
does not change (Vmax is the same)
is larger
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Uncompetitive Inhibitor
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Uncompetitive Inhibitor
Binds only to ES complex but not free
enzyme
Binds at location other than active site
Does not look like substrate. Binding of
inhibitor distorts active site thus
preventing substrate binding and catalysis
Cannot be competed away by increasing
conc of substrate (Vmax is affected by
[I])
Increasing [I] lowers Vmax and lowers
Km.
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Increasing
[I]
Lowers
Vmax (y-intercept increases)
Lowers
KM (x-intercept decreases)
Ratio
of KM/Vmax is the same (slope)
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Mixed
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Mixed
Inhibitor binds E or ES
Binds site distinct from active site
Increasing [I]
Lowers
Raises
Vmax (y-intercept increases)
KM (x-intercept increases)
Ratio
of KM/Vmax is not the same
(slope changes)
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Non-competitive inhibition
Inhibitor binds ES or E
It is a special case of mixed inhibition where Vmax is lowered when [I] increases but Km does not
change
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Reversible Inhibition (non-competitive)
+Inh
1/v
Vmax
-Inh
Vmax
(app)_
-Inh
1/Vmax
(app)
+inh
vo
1/2 Vmax
1/2 Vmax
(app)
1/Vmax
-1/Km
Km
Km
(app)
1/[S]
[Substrate]
Vmax is decreased proportional to inhibitor conc
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Reversible Inhibition (non-competitive)
A inhibitor binds the enzyme but not in its active site. It affects the Kcat because
substrate can still bind the active site.
Rate of catalysis is affected
+Inh
1/v
Vmax
-Inh
Vmax
(app)_
-Inh
1/Vmax
(app)
+inh
vo
1/2 Vmax
1/2 Vmax
(app)
1/Vmax
-1/Km
Km
Km
(app)
1/[S]
[Substrate]
Vmax is decreased proportional to inhibitor conc
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Uncompetitive and mixed inhibition only seen for enzymes with two or more substrates
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Example
When a slice of apple is cut, it turns brown- enzyme o-diphenol oxidase oxidizes phenols in the apple
Lets determine max rate at which enzyme functions (Vmax), and Km
1
When it acts alone (we will use catechol as substrate). Enz converts this to o-quinone which is
dark and can be measured via absorbance at 540 nm
2
when it acts in presence of competitive inhibitor para hydroxy benzoic acid which bind active site
but is not acted upon
3
when it acts in the presence of a non-competitive inhibitor- phenylthiourea which binds copper in
the enzyme which is necessary for enzyme activity
Make a supernatant of the apple-enzyme. Measure color produced (product)
Set up 4 tubes with different conc of cathecol and a fixed amount of enzyme (apple pulp).
Measure change in absorbance at 1 min intervals for several minutes and record average change in
absorbance.
Absorbance is directly proportional to product, we can measure rate of reaction (velocity)
TubeA
[S]
1/[S]
Vi (DOD)
1/Vi
mM
TubeB
mM
TubeC
mM
TubeD
mM
1/Vmax=10
Vmax=0.1
-1/Km=-0.8
Km=1.25mM
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Example
When a slice of apple is cut, it turns brown- enzyme o-diphenol oxidase oxidizes phenols in the apple
Lets determine max rate at which enzyme functions (Vmax), and Km
1
When it acts alone (we will use catechol as substrate. Enz converts this to o-quinone which is dark
and can be measured via absorbance at 540 nm
2
when it acts in presence of competitive inhibitor para hydroxy benzoic acid which bind active site
but is not acted upon
3
when it acts in the presence of a non-competitive inhibitor- phenylthiourea which binds copper in
the enzyme which is necessary for enzyme activity
Make a supernatant of the apple-enzyme. Measure color produced (product)
Set up 4 tubes with different conc of cathecol and a fixed amount of enzyme (apple pulp).
Measure change in absorbance at 1 min intervals for several minutes and record average change in
absorbance.
Absorbance is directly proportional to product, we can measure rate of reaction (velocity)
TubeA
TubeB
TubeC
TubeD
[S]
4.8 mM
1.2mM
0.6mM
0.3mM
1/[S]
0.21
0.83
1.67
3.33
Vi (DOD)
0.081
0.048
0.035
0.02
1/Vi
12.3
20.8
31.7
50.0
1/Vmax=10
Vmax=0.1
-1/Km=-0.8
Km=1.25mM
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Example
Each tube also has a fixed amount of PHBA (competitive inhibitor)
TubeA
[S]
mM
TubeB
mM
TubeC
mM
TubeD
mM
1/[S]
Vi (DOD)
1/Vmax=10
Vmax=0.1
-1/Km=-0.4
Km=2.5 mM
1/Vi
Each tube has a fixed amount of phenylthiourea (non competitive inhibitor)
TubeA
[S]
1/[S]
Vi (DOD)
mM
TubeB
mM
TubeC
mM
TubeD
mM
1/Vmax=20
Vmax=0.05
-1/Km=-0.8
Km=1.25 mM
1/Vi
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Example
Each tube also has a fixed amount of PHBA (competitive inhibitor)
TubeA
TubeB
TubeC
TubeD
[S]
4.8 mM
1.2mM
0.6mM
0.3mM
1/[S]
0.21
0.83
1.67
3.33
Vi (DOD)
0.060
0.032
0.019
0.011
1/Vi
16.7
31.3
52.6
90.9
1/Vmax=10
Vmax=0.1
-1/Km=-0.4
Km=2.5 mM
Each tube has a fixed amount of phenylthiourea (non competitive inhibitor)
TubeA
TubeB
TubeC
TubeD
[S]
4.8 mM
1.2mM
0.6mM
0.3mM
1/[S]
0.21
0.83
1.67
3.33
Vi (DOD)
0.040
0.024
0.016
0.01
1/Vi
25
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1/Vmax=20
Vmax=0.05
-1/Km=-0.8
Km=1.25 mM
Irreversible
Inhibitor
Binds
covalently, or
Destroys
functional group
necessary for enzymatic
activity, or
Very
stable noncovalent
binding
Suicide
Inactivators
Starts
steps of chemical
reaction
Does
not make product
Combines
irreversibly
with enzyme
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Regulation of enzymes
Catalytic activity is increased or decreased by
1)
2)
3)
Enzyme synthesis or degradation
Covalent modification
Non-covalent binding and conformational change (allosteric)
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Allosteric regulation
Usually located early in multi-enzyme reaction pathway
Kinetics differ for allosteric enzymes- sigmoidal curve and K1/2 instead of Km

Usually large; multiple subunits
Comparable


to Hb
Site for allosteric modulator (R = regulatory) generally different from active site (C
= catalytic)
Can be positive or negative
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Aspartate transcarbamoylase (nucleotide synthesis)
Catalytic subunit
Regulatory subunit
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Feed back Inhibition
Sigmoidal kinetic curves -Positive Homotropic enzyme
End product inhibition
Heterotropic modulator
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Covalent regulation
Modifying groups are attached to an enzyme • by a covalent bond…
Glycogen phosphorylase
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Special Regulation by Degradation
Digestive enzymes: Trypsinogen and Chymotrypsinogen
Hormonal regulation: Insulin is synthesized as pro-insulin
Fibrous proteins: Collagen is synthesized as pro-collagen
Blood clotting: Fibrinogen and pro-thrombin
Known as Zymogens (for proteases) or Proproteins
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