Enzymes catalyze biochemical reactions and lower their activation energy. They do this by holding molecules together, moving electrons between them, and physically manipulating their structures. Enzymes control reaction rates by forming enzyme-substrate complexes at each step. Kinetic studies examine how factors like temperature, concentrations, and effectors influence reaction rates. The Michaelis-Menten model describes how enzyme velocity increases with substrate concentration until plateauing at maximum velocity. Key parameters are KM, the substrate concentration at half-maximal velocity, and kcat, the enzyme's turnover number. Enzyme inhibitors are used as drugs and studied to understand regulation.

1. ENZYMES & BIOENERGETICS
LECTURE NOTES

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

9. Enzyme Kinetics, Inhibition,
and Control

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

42. Table 12-2
Enzyme Inhibition

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?

46. Page 378
Competitive Inhibition

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

48. Figure 12-6
Competitive Inhibition

49. Figure 12-7
Competitive Inhibition

50. Page 380
Product inhibition:
ADP, AMP can competitively inhibit enzymes that
hydrolyze ATP
Competitive Inhibition

51. Box 12-4c
Competitive Inhibition

52. Page 381
Uncompetitive 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

54. Figure 12-8
Uncompetitive Inhibition

55. Page 382
Mixed (competitive and uncompetitive) Inhibition

56. Figure 12-9
Mixed (competitive and uncompetitive) Inhibition

57. Table 12-2

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

60. http://alevelnotes.com/Enzyme-Inhibitors/148
Competitive and Noncompetitive Inhibition
Vmax/2
V

61. http://alevelnotes.com/Enzyme-Inhibitors/148
Competitive and Noncompetitive Inhibition
Vmax/2
V

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

68. Figure 12-10

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

70. Figure 12-11

71. Figure 12-13

72. Page 390

73. Page 391

74. Figure 12-14b

75. Figure 12-15

76. Figure 12-16

77. Figure 12-17

78. Figure 12-18

79. Figure 12-20

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