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ENZYMES & BIOENERGETICS
LECTURE NOTES
4
Mechanisms of
Enzyme Action
Enzymes can hold two
molecules together
5
Mechanisms of
Enzyme Action
Enzymes can
move
electrons
around
6
Mechanisms of
Enzyme Action
Enzymes can
physically
bend
molecules
Enzymes reduce the activation energy
Reactions are broken down into smaller steps
DGact
DGact
Reactions are broken down into smaller steps
DGact
DGact
Make ATP
Reactions are broken down into smaller steps
Reaction rate is controlled by enzymes at each step
DGact
DGact
Make ATP
Enzyme Kinetics, Inhibition,
and Control
So good for you…
Many drugs are enzyme inhibitors.
Lipitor: HMG-CoA reductase, inhibits a liver enzyme that is important in
biosynthesis of cholesterol (>$100 billion total sales since 1996).
Viread & Emtriva: reverse transcriptase inhibitors, anti-retrovirus (HIV).
Saquinavir: protease inhibitor, anti-retrovirus (HIV).
15
Reaction Rates (reaction velocities): To
measure a reaction rate we monitor the
disappearance of reactants or appearance of
products.
e.g., 2NO2 + F2 → 2NO2F
initial velocity =>
[product] = 0,
no back reaction
o Zero Order. The rate of a zero-order reaction is independent
of the concentration of the reactant(s). Zero-order kinetics
are observed when an enzyme is saturated by reactants.
o First Order. The rate of a first-order reaction varies linearly on
the concentration of one reactant. First-order kinetics are
observed when a protein folds and RNA folds (assuming no
association or aggregation).
o Second Order. The rate of a second-order reaction varies
linearly with the square of concentrations of one reactant (or
with the product of the concentrations of two reactants).
Second order kinetics are observed for formation of double-
stranded DNA from two single-strands.
17
Use experimental data to determine the
reaction order.
If a plot of [A] vs t is a straight line, then the reaction is zero order.
If a plot of ln[A] vs t is a straight line, then the reaction is 1st order.
If a plot of 1/ [A] vs t is a straight line, then the reaction is 2nd order.
First Order Reaction
-6
-5
-4
-3
-2
-1
0
0.0E+00 2.0E+04 4.0E+04 6.0E+04
Time (in seconds)
ln
[A}
(in
mol
/
L
Second Order Reaction
0
50
100
150
200
250
0 500 1000
Time (in seconds)
1/[A]
(L
/mol)
Protein Folding: 1st order reaction
DNA annealing: 2nd order reaction
• Each elementary
step has reactant(s),
a transition state,
and product(s).
Products that are
consumed in
subsequent
elementary reaction
are called
intermediates.
• Kinetics is the study of reaction rates (time-dependent
phenomena)
• Rates of reactions are affected by
– Enzymes/catalysts
– Substrates
– Effectors
– Temperature
– Concentrations
Why study enzyme kinetics?
• Quantitative description of biocatalysis
• Understand catalytic mechanism
• Find effective inhibitors
• Understand regulation of activity
General Observations
• Enzymes are able to exert their influence at
very low concentrations ~ [enzyme] = nM
• The initial rate (velocity) is linear with
[enzyme].
• The initial velocity increases with [substrate]
at low [substrate].
• The initial velocity approaches a maximum at
high [substrate].
Initial velocity
The initial velocity increases with [S] at low [S].
The initial velocity approaches
a maximum at high [S].
The initial velocity increases with [S] at
low [S].
[velocity =d[P]/dt, P=product]
• Start with a mechanistic model
• Identify constraints and assumptions
• Do the algebra ...
• Solve for velocity (d[P]/dt)
Michaelis-Menten Kinetics
Simplest enzyme mechanism
- One reactant (S)
- One intermediate (ES)
- One product (P)
1. First step: The enzyme (E) and the
substrate (S) reversibly and quickly
form a non-covalent ES complex.
2. Second step: The ES complex
undergoes a chemical
transformation and dissociates to
give product (P) and enzyme (E).
3. v=k2[ES]
4. Many enzymatic reactions follow
Michaelis–Menten kinetics, even
though enzyme mechanisms are
always more complicated than the
Michaelis–Menten model.
5. For real enzymatic reactions use kcat
instead of k2
.
The Enzyme-Substrate Complex (ES)
• The enzyme binds non-covalently to the substrate
to form a non-covalent ES complex
– the ES complex is known as the Michaelis
complex.
– A Michaelis complex is stabilized by molecular
interactions (non-covalent interactions).
– Michaelis complexes form quickly and
dissociate quickly.
Michaelis-Menten Kinetics
E + S  ES  E + P
• The enzyme is either free ([E]) or bound ([ES]): [Eo] = [ES] + [E].
• At sufficiently high [S] all of the enzyme is tied up as ES (i.e., [Eo] ≈ [ES],
according to Le Chatelier's Principle)
• At high [S] the enzyme is working at full capacity (v=vmax).
• The full capacity velocity is determined only by kcat and [Eo].
• kcat = turnover #: number of moles of substrate produced per time per
enzyme active site.

kcat
velocity = v =
d[P]
dt
= kcat [ES]
vmax = kcat [E0 ]
kcat =
vmax
[E0 ]
kcat and the reaction velocity
Michaelis-Menten Kinetics
E + S  ES  E + P
• For any enzyme it is possible (pretty easy) to determine kcat.
• To understand and compare enzymes we need to know how
well the enzyme binds to S (i.e, what happens in the first
part of the reaction.) kcat does not tell us anything about
how well the enzyme binds to the substrate.
• so, … (turn the page and learn about KD and KM).

kcat
Michaelis-Menten Kinetics
Assumptions
1. k1,k-1>>k2 (i.e., the first step is fast and is
always at equilibrium).
2. d[ES]/dt ≈ 0 (i.e., the system is at steady
state.)
3. There is a single reaction/dissociation step
(i.e., k2=kcat).
4. STot = [S] + [ES] ≈ [S]
5. There is no back reaction of P to ES (i.e. [P] ≈
0). This assumption allows us to ignore k-2. We
measure initial velocities, when [P] ≈ 0.
d[ES]
dt
= rate of formation of ES -
rate of breakdown of ES
» 0 (at steady state)
Michaelis-Menten Kinetics
The time dependence
of everything (in a
Michaelis-Menten
reaction)
Now: we derive the Michaelis-Menten Equation
d[ES]/dt = k1[E][S] –k-1[ES] – k2[ES] (eq 12-14 VVP)
= 0 (steady state assumption, see previous graph)
solve for [ES] (do the algebra)
[ES] = [E][S] k1/(k-1 + k2)
Define KM (Michealis Constant)
KM = (k-1 + k2)/k1 => [ES] = [E][S]/KM
rearrange to give KM = [E][S]/[ES]
substitute [E]=[E]0 -[ES]
([E]0 -[ES])[S]
[ES]
= KM eqs 12-20 VVP
multiply both sides by [ES]
KM [ES]= ([E]0 -[ES])[S]
solve for [ES]
[ES]=
[E]0[S]
Km +[S]
eq 12-22 VVP
multiply both sides by k2 (this gives get the velocity of the reaction)
dP
dt
= v = k2[ES]=
k2[E]0[S]
KM +[S]
eq 12-23 VVP
and remember that k2[E]0 = vmax
v =
vmax[S]
KM +[S]
Michaelis Menten Equation eq 12-25 VVP
KM = [E][S]/[ES]
v =
vmax[S]
KM +[S]
Michaelis Menten Equation eq 12-25 VVP
When [S] = KM then,
v =
vmax[S]
[S]+[S]
=
vmax
2
This is saying that when KM =[S], the reaction runs at half maximum velocity.
KM is the substrate concentration required to reach half-
maximal velocity (vmax/2).
Significance of KM
• KM = [E][S]/[ES] and KM = (k-1 + k2)/k1.
• KM is the apparent dissociation constant of the ES complex. A
dissociation constant (KD) is the reciprocal of the equilibrium constant
(KD=KA
-1). KM is a measure of a substrate’s affinity for the enzyme (but
it is the reciprocal of the affinity).
• If k1,k-1>>k2, the KM=KD.
• KM is the substrate concentration required to reach half-maximal
velocity (vmax/2). A small KM means the sustrate binds tightly to the
enzyme and saturates (max’s out) the enzyme.
• The microscopic meaning of Km depends on the details of the
mechanism.
The significance of kcat
• vmax = kcat Etot
• kcat: For the simplest possible mechanism, where ES is the only
intermediate, and dissociation is fast, then kcat=k2, the first order rate
constant for the catalytic step.
• If dissociation is slow then the dissociation rate constant also contributes
to kcat.
• If one catalytic step is much slower than all the others (and than the
dissociation step), than the rate constant for that step is approximately
equal to to kcat.
• kcat is the “turnover number”: indicates the rate at which the enzyme
turns over, i.e., how many substrate molecules one catalytic site converts
to product per second.
• If there are multiple catalytic steps (see trypsin) then each of those rate
constants contributes to kcat.
• The microscopic meaning of kcat depends on the details of the
mechanism.
Significance of kcat/KM
• kcat/KM is the catalytic efficiency. It is used to rank enzymes. A big
kcat/KM means that an enzyme binds tightly to a substrate (small KM),
with a fast reaction of the ES complex.
•
• kcat/KM is an apparent second order rate constant
v=kcat/KM[E]0[S]
• kcat/KM can be used to estimate the reaction velocity from the total
enzyme concentration ([E]0). kcat/KM =109 => diffusion control.
• kcat/KM is the specificity constant. It is used to distinguish and describe
various substrates.
Data analysis
• It would be useful to have a linear plot of
the MM equation
• Lineweaver and Burk (1934) proposed the
following: take the reciprocal of both
sides and rearrange.
• Collect data at a fixed [E]0.
v =
vmax[S]
KM +[S]
Michaelis Menten Equation eq 12-25
take the reciprocal
1
v
=
KM +[S]
vmax[S]
=
KM
vmax[S]
+
1
vmax
Graph
1
v
versus
1
[S]
the y (1/v) intercept (1/[S] = 0) is 1/vmax
the x (1/[S]) intercept (1/v = 0) is -1/KM
the slope is KM/vmax
Lineweaver-Burk-Plot
the y (1/v) intercept (1/[S] = 0) is 1/vmax
the x (1/[S]) intercept (1/v = 0) is -1/KM
the slope is KM/vmax
Enzyme inhibition
Reversible Irreversible
• competitive
• noncompetitive
• mixed
• suicide inactivators
Sarin
Aspirin
Enzyme inhibition
Reversible Irreversible
• competitive
• noncompetitive
• mixed
• suicide inactivators
Sarin
Aspirin
Enzyme inhibition
Reversible Irreversible
• competitive
• noncompetitive
• mixed
• suicide inactivators
Sarin
Aspirin
Table 12-2
Enzyme Inhibition
Competitive Inhibition
Substrate and inhibitor compete for the same site
Competitive Inhibition
Substrate and inhibitor compete for the same site
If [S]>>[I]
Vmax is the same, Km h
Vmax
1/Km
Competitive Inhibition
Substrate and inhibitor compete for the same site
If [S]>>[I]
Vmax is the same, Km h
Vmax
What happens to the slope? Vmax? Km?
Page 378
Competitive Inhibition
[E][I]
[EI]
= KI inhibitor dissociation constant
[E]0 =[E] -[ES]-[EI] total enzyme concentration
a =1+
[I]
KI
1
v
=
aKM +[S]
vmax[S]
=
aKM
vmax[S]
+
1
vmax
Competitive Inhibition
Figure 12-6
Competitive Inhibition
Figure 12-7
Competitive Inhibition
Page 380
Product inhibition:
ADP, AMP can competitively inhibit enzymes that
hydrolyze ATP
Competitive Inhibition
Box 12-4c
Competitive Inhibition
Page 381
Uncompetitive Inhibition
[ES][I]
[ESI]
= KI inhibitor dissociation constant
[E]0 =[E] -[ES]-[ESI] total enzyme concentration
1
v
=
KM
vmax[S]
+
a
vmax
Uncompetitive Inhibition
Figure 12-8
Uncompetitive Inhibition
Page 382
Mixed (competitive and uncompetitive) Inhibition
Figure 12-9
Mixed (competitive and uncompetitive) Inhibition
Table 12-2
Noncompetitive
Inhibition
What happens to Vmax & Km?
a. Vmax same, Kmh
b. Vmax same, Kmi
c. Vmaxh, Km same
d. Vmaxi , Km same
X
http://en.wikipedia.org/wiki/Non-competitive_inhibition#mediaviewer/File:Non-competitive_inhibition.svg
No inhibitor
EI cannot be used;
Vmax = k2[E]with no inhibitor
http://alevelnotes.com/Enzyme-Inhibitors/148
Competitive and Noncompetitive Inhibition
Vmax/2
V
Add the curves for a competitive and non-
competitive inhibitor
http://alevelnotes.com/Enzyme-Inhibitors/148
Competitive and Noncompetitive Inhibition
Vmax/2
V
http://alevelnotes.com/Enzyme-Inhibitors/148
Competitive and Noncompetitive Inhibition
Vmax/2
V
1
Vmax
Substrate only
Add the curves for a competitive and non-
competitive inhibitor
1
Vmax
Noncompetitive (Km
constant, Vmax i)
Competitive (Vmax
constant, Km h)
Substrate only
Add the curves for a competitive and non-
competitive inhibitor
Lineweaver-Burk only applies to Michaelis-
Menten kinetics
Task:For an enzyme that follows Michaelis-Menten kinetics, by
what factor does the substrate concentration need to increase
to change the rate of the reaction from 20% to 80% Vmax?
]
[
]
[
max
S
K
S
V
V
m 

A. two-fold
B. 4-fold
C. 8-fold
D. 16-fold
E. 24-fold
Allosteric Enzymes
Modulator can activate or
inhibit the enzyme
How MM kinetic measurements are
made
*
*
Page 374
Real enzyme mechanisms
Figure 12-10
Shift at physiological concentrations of substrate
Allosteric
“T” form
Slight h[S] (in pink region) causes transition from “T” to “R”
Michaelis-Menten
“R” form
Figure 12-11
Figure 12-13
Page 390
Page 391
Figure 12-14b
Figure 12-15
Figure 12-16
Figure 12-17
Figure 12-18
Figure 12-20
Phosphorylation is the
most common reversible
covalent modification of
an enzyme
Kinase adds Pi to a
molecule
Phosphatase removes
Pi from a molecule
About 10,000 enzymes in a cell, one-third
of them are activated or deactivated
through phosphorylation.
Why Phosphorylation?
•Free energy, DG, is large
•Phosphorylation can change the reaction rate by 104
•Amplifies – single kinase phosphorylates hundreds of
molecules
• Pi has 2 negative
charges
• Phosphate group can
form 3 H-bonds
..ctd
• If Km increases, then the affinity between the
enzyme and substrate decreases (you will
need more substrate for the same rate of
reaction). Thus phosphorylation inhibits the
activity.
Enzyme Regulation by protein cleavage
Hydrolysis of
specific peptide
bonds
Inactive Precursor Active Enzyme
Activation site
Inactive Enzyme
Irreversible
Pepsin Regulation
Hydrolysis of
specific peptide
bonds
Pepsin: Active Enzyme
Activation site
Inactive Enzyme
Irreversible
Pepsinogen

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ENZYMES- bme.pptx

  • 2. 4 Mechanisms of Enzyme Action Enzymes can hold two molecules together
  • 3. 5 Mechanisms of Enzyme Action Enzymes can move electrons around
  • 4. 6 Mechanisms of Enzyme Action Enzymes can physically bend molecules
  • 5. Enzymes reduce the activation energy
  • 6. Reactions are broken down into smaller steps DGact DGact
  • 7. Reactions are broken down into smaller steps DGact DGact Make ATP
  • 8. Reactions are broken down into smaller steps Reaction rate is controlled by enzymes at each step DGact DGact Make ATP
  • 10. So good for you… Many drugs are enzyme inhibitors. Lipitor: HMG-CoA reductase, inhibits a liver enzyme that is important in biosynthesis of cholesterol (>$100 billion total sales since 1996). Viread & Emtriva: reverse transcriptase inhibitors, anti-retrovirus (HIV). Saquinavir: protease inhibitor, anti-retrovirus (HIV).
  • 11. 15 Reaction Rates (reaction velocities): To measure a reaction rate we monitor the disappearance of reactants or appearance of products. e.g., 2NO2 + F2 → 2NO2F initial velocity => [product] = 0, no back reaction
  • 12. o Zero Order. The rate of a zero-order reaction is independent of the concentration of the reactant(s). Zero-order kinetics are observed when an enzyme is saturated by reactants. o First Order. The rate of a first-order reaction varies linearly on the concentration of one reactant. First-order kinetics are observed when a protein folds and RNA folds (assuming no association or aggregation). o Second Order. The rate of a second-order reaction varies linearly with the square of concentrations of one reactant (or with the product of the concentrations of two reactants). Second order kinetics are observed for formation of double- stranded DNA from two single-strands.
  • 13. 17 Use experimental data to determine the reaction order. If a plot of [A] vs t is a straight line, then the reaction is zero order. If a plot of ln[A] vs t is a straight line, then the reaction is 1st order. If a plot of 1/ [A] vs t is a straight line, then the reaction is 2nd order. First Order Reaction -6 -5 -4 -3 -2 -1 0 0.0E+00 2.0E+04 4.0E+04 6.0E+04 Time (in seconds) ln [A} (in mol / L Second Order Reaction 0 50 100 150 200 250 0 500 1000 Time (in seconds) 1/[A] (L /mol)
  • 14. Protein Folding: 1st order reaction DNA annealing: 2nd order reaction
  • 15. • Each elementary step has reactant(s), a transition state, and product(s). Products that are consumed in subsequent elementary reaction are called intermediates.
  • 16. • Kinetics is the study of reaction rates (time-dependent phenomena) • Rates of reactions are affected by – Enzymes/catalysts – Substrates – Effectors – Temperature – Concentrations
  • 17. Why study enzyme kinetics? • Quantitative description of biocatalysis • Understand catalytic mechanism • Find effective inhibitors • Understand regulation of activity
  • 18. General Observations • Enzymes are able to exert their influence at very low concentrations ~ [enzyme] = nM • The initial rate (velocity) is linear with [enzyme]. • The initial velocity increases with [substrate] at low [substrate]. • The initial velocity approaches a maximum at high [substrate].
  • 19. Initial velocity The initial velocity increases with [S] at low [S].
  • 20. The initial velocity approaches a maximum at high [S]. The initial velocity increases with [S] at low [S]. [velocity =d[P]/dt, P=product]
  • 21. • Start with a mechanistic model • Identify constraints and assumptions • Do the algebra ... • Solve for velocity (d[P]/dt)
  • 22. Michaelis-Menten Kinetics Simplest enzyme mechanism - One reactant (S) - One intermediate (ES) - One product (P)
  • 23. 1. First step: The enzyme (E) and the substrate (S) reversibly and quickly form a non-covalent ES complex. 2. Second step: The ES complex undergoes a chemical transformation and dissociates to give product (P) and enzyme (E). 3. v=k2[ES] 4. Many enzymatic reactions follow Michaelis–Menten kinetics, even though enzyme mechanisms are always more complicated than the Michaelis–Menten model. 5. For real enzymatic reactions use kcat instead of k2 .
  • 24. The Enzyme-Substrate Complex (ES) • The enzyme binds non-covalently to the substrate to form a non-covalent ES complex – the ES complex is known as the Michaelis complex. – A Michaelis complex is stabilized by molecular interactions (non-covalent interactions). – Michaelis complexes form quickly and dissociate quickly. Michaelis-Menten Kinetics
  • 25. E + S  ES  E + P • The enzyme is either free ([E]) or bound ([ES]): [Eo] = [ES] + [E]. • At sufficiently high [S] all of the enzyme is tied up as ES (i.e., [Eo] ≈ [ES], according to Le Chatelier's Principle) • At high [S] the enzyme is working at full capacity (v=vmax). • The full capacity velocity is determined only by kcat and [Eo]. • kcat = turnover #: number of moles of substrate produced per time per enzyme active site.  kcat velocity = v = d[P] dt = kcat [ES] vmax = kcat [E0 ] kcat = vmax [E0 ] kcat and the reaction velocity Michaelis-Menten Kinetics
  • 26. E + S  ES  E + P • For any enzyme it is possible (pretty easy) to determine kcat. • To understand and compare enzymes we need to know how well the enzyme binds to S (i.e, what happens in the first part of the reaction.) kcat does not tell us anything about how well the enzyme binds to the substrate. • so, … (turn the page and learn about KD and KM).  kcat Michaelis-Menten Kinetics
  • 27. Assumptions 1. k1,k-1>>k2 (i.e., the first step is fast and is always at equilibrium). 2. d[ES]/dt ≈ 0 (i.e., the system is at steady state.) 3. There is a single reaction/dissociation step (i.e., k2=kcat). 4. STot = [S] + [ES] ≈ [S] 5. There is no back reaction of P to ES (i.e. [P] ≈ 0). This assumption allows us to ignore k-2. We measure initial velocities, when [P] ≈ 0. d[ES] dt = rate of formation of ES - rate of breakdown of ES » 0 (at steady state) Michaelis-Menten Kinetics
  • 28. The time dependence of everything (in a Michaelis-Menten reaction)
  • 29. Now: we derive the Michaelis-Menten Equation d[ES]/dt = k1[E][S] –k-1[ES] – k2[ES] (eq 12-14 VVP) = 0 (steady state assumption, see previous graph) solve for [ES] (do the algebra) [ES] = [E][S] k1/(k-1 + k2) Define KM (Michealis Constant) KM = (k-1 + k2)/k1 => [ES] = [E][S]/KM rearrange to give KM = [E][S]/[ES]
  • 30. substitute [E]=[E]0 -[ES] ([E]0 -[ES])[S] [ES] = KM eqs 12-20 VVP multiply both sides by [ES] KM [ES]= ([E]0 -[ES])[S] solve for [ES] [ES]= [E]0[S] Km +[S] eq 12-22 VVP multiply both sides by k2 (this gives get the velocity of the reaction) dP dt = v = k2[ES]= k2[E]0[S] KM +[S] eq 12-23 VVP and remember that k2[E]0 = vmax v = vmax[S] KM +[S] Michaelis Menten Equation eq 12-25 VVP KM = [E][S]/[ES]
  • 31. v = vmax[S] KM +[S] Michaelis Menten Equation eq 12-25 VVP When [S] = KM then, v = vmax[S] [S]+[S] = vmax 2 This is saying that when KM =[S], the reaction runs at half maximum velocity.
  • 32. KM is the substrate concentration required to reach half- maximal velocity (vmax/2).
  • 33. Significance of KM • KM = [E][S]/[ES] and KM = (k-1 + k2)/k1. • KM is the apparent dissociation constant of the ES complex. A dissociation constant (KD) is the reciprocal of the equilibrium constant (KD=KA -1). KM is a measure of a substrate’s affinity for the enzyme (but it is the reciprocal of the affinity). • If k1,k-1>>k2, the KM=KD. • KM is the substrate concentration required to reach half-maximal velocity (vmax/2). A small KM means the sustrate binds tightly to the enzyme and saturates (max’s out) the enzyme. • The microscopic meaning of Km depends on the details of the mechanism.
  • 34. The significance of kcat • vmax = kcat Etot • kcat: For the simplest possible mechanism, where ES is the only intermediate, and dissociation is fast, then kcat=k2, the first order rate constant for the catalytic step. • If dissociation is slow then the dissociation rate constant also contributes to kcat. • If one catalytic step is much slower than all the others (and than the dissociation step), than the rate constant for that step is approximately equal to to kcat. • kcat is the “turnover number”: indicates the rate at which the enzyme turns over, i.e., how many substrate molecules one catalytic site converts to product per second. • If there are multiple catalytic steps (see trypsin) then each of those rate constants contributes to kcat. • The microscopic meaning of kcat depends on the details of the mechanism.
  • 35. Significance of kcat/KM • kcat/KM is the catalytic efficiency. It is used to rank enzymes. A big kcat/KM means that an enzyme binds tightly to a substrate (small KM), with a fast reaction of the ES complex. • • kcat/KM is an apparent second order rate constant v=kcat/KM[E]0[S] • kcat/KM can be used to estimate the reaction velocity from the total enzyme concentration ([E]0). kcat/KM =109 => diffusion control. • kcat/KM is the specificity constant. It is used to distinguish and describe various substrates.
  • 36. Data analysis • It would be useful to have a linear plot of the MM equation • Lineweaver and Burk (1934) proposed the following: take the reciprocal of both sides and rearrange. • Collect data at a fixed [E]0.
  • 37. v = vmax[S] KM +[S] Michaelis Menten Equation eq 12-25 take the reciprocal 1 v = KM +[S] vmax[S] = KM vmax[S] + 1 vmax Graph 1 v versus 1 [S] the y (1/v) intercept (1/[S] = 0) is 1/vmax the x (1/[S]) intercept (1/v = 0) is -1/KM the slope is KM/vmax
  • 38. Lineweaver-Burk-Plot the y (1/v) intercept (1/[S] = 0) is 1/vmax the x (1/[S]) intercept (1/v = 0) is -1/KM the slope is KM/vmax
  • 39. Enzyme inhibition Reversible Irreversible • competitive • noncompetitive • mixed • suicide inactivators Sarin Aspirin
  • 40. Enzyme inhibition Reversible Irreversible • competitive • noncompetitive • mixed • suicide inactivators Sarin Aspirin
  • 41. Enzyme inhibition Reversible Irreversible • competitive • noncompetitive • mixed • suicide inactivators Sarin Aspirin
  • 43. Competitive Inhibition Substrate and inhibitor compete for the same site
  • 44. Competitive Inhibition Substrate and inhibitor compete for the same site If [S]>>[I] Vmax is the same, Km h Vmax 1/Km
  • 45. Competitive Inhibition Substrate and inhibitor compete for the same site If [S]>>[I] Vmax is the same, Km h Vmax What happens to the slope? Vmax? Km?
  • 47. [E][I] [EI] = KI inhibitor dissociation constant [E]0 =[E] -[ES]-[EI] total enzyme concentration a =1+ [I] KI 1 v = aKM +[S] vmax[S] = aKM vmax[S] + 1 vmax Competitive Inhibition
  • 50. Page 380 Product inhibition: ADP, AMP can competitively inhibit enzymes that hydrolyze ATP Competitive Inhibition
  • 53. [ES][I] [ESI] = KI inhibitor dissociation constant [E]0 =[E] -[ES]-[ESI] total enzyme concentration 1 v = KM vmax[S] + a vmax Uncompetitive Inhibition
  • 55. Page 382 Mixed (competitive and uncompetitive) Inhibition
  • 56. Figure 12-9 Mixed (competitive and uncompetitive) Inhibition
  • 58. Noncompetitive Inhibition What happens to Vmax & Km? a. Vmax same, Kmh b. Vmax same, Kmi c. Vmaxh, Km same d. Vmaxi , Km same X http://en.wikipedia.org/wiki/Non-competitive_inhibition#mediaviewer/File:Non-competitive_inhibition.svg No inhibitor EI cannot be used; Vmax = k2[E]with no inhibitor
  • 59. http://alevelnotes.com/Enzyme-Inhibitors/148 Competitive and Noncompetitive Inhibition Vmax/2 V Add the curves for a competitive and non- competitive inhibitor
  • 62. 1 Vmax Substrate only Add the curves for a competitive and non- competitive inhibitor
  • 63. 1 Vmax Noncompetitive (Km constant, Vmax i) Competitive (Vmax constant, Km h) Substrate only Add the curves for a competitive and non- competitive inhibitor Lineweaver-Burk only applies to Michaelis- Menten kinetics
  • 64. Task:For an enzyme that follows Michaelis-Menten kinetics, by what factor does the substrate concentration need to increase to change the rate of the reaction from 20% to 80% Vmax? ] [ ] [ max S K S V V m   A. two-fold B. 4-fold C. 8-fold D. 16-fold E. 24-fold
  • 65. Allosteric Enzymes Modulator can activate or inhibit the enzyme
  • 66. How MM kinetic measurements are made * *
  • 67. Page 374 Real enzyme mechanisms
  • 69. Shift at physiological concentrations of substrate Allosteric “T” form Slight h[S] (in pink region) causes transition from “T” to “R” Michaelis-Menten “R” form
  • 80. Phosphorylation is the most common reversible covalent modification of an enzyme Kinase adds Pi to a molecule Phosphatase removes Pi from a molecule About 10,000 enzymes in a cell, one-third of them are activated or deactivated through phosphorylation.
  • 81. Why Phosphorylation? •Free energy, DG, is large •Phosphorylation can change the reaction rate by 104 •Amplifies – single kinase phosphorylates hundreds of molecules • Pi has 2 negative charges • Phosphate group can form 3 H-bonds
  • 82. ..ctd • If Km increases, then the affinity between the enzyme and substrate decreases (you will need more substrate for the same rate of reaction). Thus phosphorylation inhibits the activity.
  • 83. Enzyme Regulation by protein cleavage Hydrolysis of specific peptide bonds Inactive Precursor Active Enzyme Activation site Inactive Enzyme Irreversible
  • 84. Pepsin Regulation Hydrolysis of specific peptide bonds Pepsin: Active Enzyme Activation site Inactive Enzyme Irreversible Pepsinogen