This document provides an overview of enzyme kinetics concepts including: 1) Rate constants like k1 and k-1 describe the rates of individual reaction steps. The overall rate v depends on reactant concentrations according to rate laws. 2) Michaelis-Menten kinetics describe enzyme-catalyzed reactions using parameters like Km, Vmax, and kcat/Km. Km represents substrate binding affinity, Vmax is the maximum reaction rate, and kcat/Km is the catalytic efficiency. 3) Reversible inhibitors are classified as competitive, non-competitive, or uncompetitive depending on whether they bind the enzyme (E), enzyme-substrate complex (ES), or both.

1. Lecture 10 Enzyme Kinetics

2. Rate constants and reaction order
Rate constant (k) measures how rapidly a reaction occurs
k1
k-1
A B + C
Rate (v, velocity) = (rate constant) (concentration of reactants)
v= k1 [A]
1st order reaction (rate dependent on concentration of 1 reactant)
v= k-1[B][C]
2nd order reaction (rate dependent on concentration of 2 reactants)
Zero order reaction (rate is independent of reactant concentration)

3. Sample questions
The rate of a second order reaction depends on the concentration of
_________.
• (a) one substrate
• (b) two substrates
• (c) three substrates
• (d) none of the above

4. E + S E S E + P
k1
k-1
k2
k-2
E + S ES E + P

5. Initial Velocities
[S] = 1 mM
Hold [E] constant
[E]<<<<<[S]
d[P]/dT = Vo1 mM
[P]
time

6. Initial Velocities
d[P]/dT = Vo [S] = 10 mM 10 mM
[S] = 5 mM
[S] = 1 mM
d[P]/dT = Vo5 mM
d[P]/dT = Vo1 mM
[P]
time

7. Plot Vo vs. [S]
Vo 5 mM
Vo 1 mM
Vo 10 mM

8. Initial Velocity Assumption
1) Measurements made to measure initial velocity (vo). At
vo very little product formed. Therefore, the rate at
which E + P react to form ES is negligible and k-2 is 0.
Therefore
E + S E S
k1
k-1
k2
E + S ES E + P
k-2
E + P

9. Steady State Assumption
Steady state Assumption = [ES] is constant. The rate of ES
formation equals the rate of ES breakdown
E + S E S E + P
k1
k-1
k2
E + S ES E + P

10. Data from a single experiment
performed with at a single [S].
(single point on Vo vs. [S] plot)

11. Rate of ES formation
E + S E S
k1
E + S ES
Rate = k1 [E] [S]

12. Rate of ES breakdown
E S E + P
k2
ES E + P
E S E + S
k-1
ES E + S
Rate = (k2 [ES]) + (k-1[ES])
Rate = [ES](k2 + k-1)

13. If the rate of ES formation equals the rate of ES
breakdown
1) k1[E][S] = [ES](k-1+ k2)
2) (k-1+ k2) / k1 = [E][S] / [ES]
3) (k-1+ k2) / k1 = Km (Michaelis constant)

14. Not required to know
Michaelis-Menton Derivation
k1 k2
k-1
E + S ES E + P
1. The overall rate of product formation: v = k2 [ES]
2. Rate of formation of [ES]: vf = k1[E][S]
3. Rate of decomposition of [ES]:
vd = k-1[ES] + k2 [ES]
4. Rate of ES formation = Rate of ES decomposition
(steady state)
5. So: k1[E][S] = k-1[ES] + k2 [ES]

15. Michaelis-Menton Derivation
6. In solving for [ES], use the enzyme balance to
eliminate [E]. ET = [E] + [ES]
7. k1 (ET - [ES])[S] = k-1[ES] + k2 [ES]
k1 ET[S] - k1[ES][S] = k-1[ES] + k2 [ES]
8. Rearrange and combine [ES] terms:
k1 ET[S] = (k-1 + k2 + k1 [S])[ES]
k1 ET[S]
9. Solve for [ES] = -----------------------
(k-1 + k2 + k1 [S])
Not required to know

16. Michaelis-Menton Derivation
ET[S]
10. Divide through by k1: [ES] = -----------------------
(k-1 + k2)/k1 + [S]
11. Defined Michaelis constant: KM = (k-1 + k2) / k1
12. Substitute KM into the equation in step 10.
13. Then substitute [ES] into v = k2 [ES] from step1
and replace Vmax with k2 ET to give:
Vmax[S]
vo = -----------
KM + [S]
Not required to know

17. Vmax = velocity where all of the
enzyme is bound to substrate
(enzyme is saturated with S)
Km = [S] at ½ Vmax
(units moles/L=M)
(1/2 of enzyme bound to S)

18. Understanding Vmax
The theoretical maximal velocity
• Vmax is a constant
• Vmax is the theoretical maximal rate of the
reaction - but it is NEVER achieved in reality
• To reach Vmax would require that ALL enzyme
molecules are tightly bound with substrate
• Vmax is asymptotically approached as
substrate is increased

19. What does Km mean?
1. Km = [S] at ½ Vmax
2. Km is a constant; Km is a combination of rate
constants describing the formation and breakdown
of the ES complex
3. Km is usually a little higher than the physiological [S]

20. What does Km mean?
4. Km represents the amount of substrate required to
bind ½ of the available enzyme (binding constant of
the enzyme for substrate)
5. Km can be used to evaluate the specificity of an
enzyme for a substrate
6. Small Km means tight binding; high Km means weak
binding
Glucose Km = 8 X 10-6
Allose Km = 8 X 10-3
Mannose Km = 5 X 10-6
Hexose Kinase
Glucose + ATP <-> Glucose-6-P + ADP

23. Sample questions
• How does the Michaelis-Menten equation
explain why the rate of an enzyme-catalyzed
reaction reaches a maximum value at high
substrate?
• At high So, Km <<<< So (numerically), so the term Km +
So in the M-M equation becomes equal to So. Vo =
(Vmax So)/So, and So cancels. Therefore at high So
then, Vo = Vmax.

24. The turnover number
A measure of catalytic activity
• kcat, the turnover number, is the number of
substrate molecules converted to product
per enzyme molecule per unit of time,
when E is saturated with substrate.
• If the Michaelis-Menten model fits, k2 = kcat
= Vmax/Et

25. What does kcat mean?
1. kcat is the 1st order rate constant describing
ES  E+P
2. Also known as the turnover number because it
describes the number of reactions that a molecule of
enzyme can catalyze per second under optimal
condition.
3. Most enzyme have kcat values between 102 and 103 s-1
4. For simple reactions k2 = kcat , for multistep reactions
kcat = rate limiting step
k1
k-1
kcat
E + S ES E + P

27. The catalytic efficiency
What does kcat/Km mean?
• It measures how the enzyme
performs when S is low
• kcat/Km describes an enzymes
preference for different substrates =
specificity constant
• The upper limit for kcat/Km is the
diffusion limit - substrate diffuse into
the active site, or product diffuse out
• Catalytic perfection when kcat/Km =
diffusion rate

28. kcat/KM
kcat/KM is taken to be a measure of the efficiency
of an enzyme.
Rewriting kcat/KM in terms of the kinetic constants
gives:
kcat k1k2
---- = -----------
KM k-1 + k2
So, where k2 is small, the denominator becomes
k-1 and kcat/KM is small.

30. Short summary
• Km  substrate specificity; substrate binding
• kcat,  the turnover number
• kcat/Km  the catalytic efficiency

31. Sample questions
Which of the following kinetic parameters best describes how
well suited a specific compound functions as a substrate
for a particular enzyme?
• (a) Km
• (b) Vmax
• (c) kcat
• (d) kcat/Km

32. Sample questions
The rate-determining step of Michaelis Menten kinetics is
• A.the complex formation step
• B.the complex dissociation step to produce product
• C.the product formation step
• D.Both (a)and(c)

33. Limitations of Michaelis-Menten
model
1. Some enzyme catalyzed reactions show more complex behavior
E + S<->ES<->EZ<->EP<-> E + P
Michaelis-Menten can look only at rate limiting step
2. Often more than one substrate
E+S1<->ES1+S2<->ES1S2<->EP1P2<-> EP2+P1<-> E+P2 Must
optimize one substrate then calculate kinetic parameters for the
other
3. Assumes k-2 = 0
4. Assume steady state conditions

34. The dual nature of the
Michaelis-Menten equation
Combination of 0-order and 1st-order kinetics
• When S is low, the equation for rate is 1st
order in S
• When S is high, the equation for rate is 0-
order in S
• The Michaelis-Menten equation describes a
hyperbolic dependence of v on S

35. How do you get values for Vmax, Km and kcat?
• Can determine Km and Vmax experimentally
• Km can be determined without an absolutely pure
enzyme
• Kcat values can be determined if Vmax is known and
the absolute concentration of enzyme is known (Vmax
= kcat[Etotal]

36. Vo [S]
B
B
B
B
B B
B
0.25
0.2
0.15
0.1
0.05
0
0 1 2 3 4 5 6 7 8 9 10
V max
[S] Vo
0.5 0.075
0.75 0.09
2 0.152
4 0.196
6 0.21
8 0.214
10 0.23
Km
Km ~ 1.3 mM
Vmax ~ 0.25

37. Lineweaver-Burke Plots
(double reciprocal plots)
•Plot 1/[S] vs 1/Vo
•L-B equation for straight
line
•X-intercept = -1/Km
•Y-intercept = 1/Vmax
•Easier to extrapolate
values w/ straight line vs
hyperbolic curve

38. Sample questions
• For an enzyme (5 μM) , the following
initial velocities have been reported
depending on the substrate
concentration:
• (a) Draw a Michaelis-Menten plot for
this enzyme.
• (b) Draw a Lineweaver-Burke plot for
this enzyme.
• (c) Determine Km and Vmax for this
enzyme
• (d) Indicate in both graphs (a & b)
where Vmax and Km can be
recognized.
• (e) Calculate the turnover number and
the catalytic efficiency for this enzyme.
[Substrate],
mM
v0, mM/s
0.02 10.83
0.04 18.57
0.07 26.76
0.1 32.50
0.15 39.00
0.2 43.33
0.3 48.75
0.5 54.17
0.7 56.88
kcat = Vmax / [E]total catalytic efficiency: kcat/Km

39. Answer
• (a)
• (b)
• (c) Km and Vmax can be determined from the intercepts in the Lineweaver-
Burke plot:
1/Vmax = 0.015 s/mM Vmax = 66 mM/s
-1/Km = -10 mM Km = 0.1 mM
• (e) kcat = Vmax / [E]total = 65 mM/s ÷ 5 μM
= 65 mM/s ÷ 0.005 mM = 13000/s
catalytic efficiency: kcat/Km = 13000/s ÷ 0.1 mM
= 13000/s ÷ 0.0001 M
= 1.3×108 M/s

40. Enzyme Inhibition
• Inhibitor – substance that binds to an
enzyme and interferes with its activity
• Can prevent formation of ES complex or
prevent ES breakdown to E + P.
Reversible versus Irreversible
• Reversible inhibitors interact with an
enzyme via noncovalent associations
• Irreversible inhibitors interact with an
enzyme via covalent associations

42. Reversible Inhibitors
E + S <-> ES -> E + P
E + I <-> EI
Ki = [E][I]/[EI]
• Competitive
• Uncompetitive
• Non-competitive

43. Classes of Reversible Inhibition
Two real, one hypothetical
• Competitive inhibition - inhibitor (I) competes with the
substrate for the active site of the enzyme
• Non-competitive inhibition - inhibitor (I) binds to an enzyme
somewhere other than the active site. It can binds either ES
or E. A non-competitive inhibitor reacts with the enzyme-substrate
complex, and slows the rate of reaction to form the
enzyme-product complex.
• Uncompetitive inhibition - inhibitor (I) binds only to ES, not to
E. This is a hypothetical case that has never been
documented for a real enzyme, but which makes a useful
contrast to competitive inhibition

47. • Competitive inhibitor: Vmax stays the same, but Km increases
• Non-competitive inhibitor decreases the turnover number of the
enzyme rather than preventing substrate binding- Vmax decreases
but Km stays the same. This cannot be overcome with an increase
in substrate concentration.

48. Enzyme Inhibition
Noncovalent binding:
Competitive (I binds only to E)
Uncompetitive (I binds only to ES)
Noncompetitive (I binds to E or ES)
Covalent binding – irreversible
Group Specific
Substrate Analogs
(bound to the active site and prevent further reactions)

49. Competitive Inhibitor (CI)
•CI binds free enzyme
•Competes with substrate for enzyme binding.
•Raises Km without effecting Vmax
•Can relieve inhibition with more S

50. Competitive Inhibitors look like substrate
O
HO C NH2
S NH2
O
H2N
O
PABA
Sulfanilamide

51. Non-competitive Inhibitor (NI)
•NI can bind free E or ES complex
•Lowers Vmax, but Km remains the same
•NI’s don’t bind to S binding site therefore don’t effect Km
•Alters conformation of enzyme to effect catalysis but not
substrate binding

52. Uncompetitive Inhibitor (UI)
•UI binds ES complex
•Prevents ES from proceeding to E + P or back to E + S.
•Lowers Km & Vmax, but ratio of Km/Vmax remains the
same
•Occurs with multisubstrate enzymes

53. Sample questions
Which of the following binds to an enzyme at its active site?
• A) irreversible inhibitor
• B) reversible competitive inhibitor
• C) reversible noncompetitive inhibitor
• D) more than one correct response
• E) no correct response
An uncompetitive inhibitor binds to _____.
• (a) E
• (b) ES
• (c) P
• (d) a and b
• (e) a and c

54. Sample questions
A reversible inhibitor that can bind to either E alone or the ES complex
is referred to as a _____.
• (a) competitive inhibitor.
• (b) non-competitive inhibitor.
• (c) uncompetitive inhibitor.
• (d) suicide inhibitor.
• (e) irreversible inhibitor.

55. Sample questions
A competitive inhibitor of an enzyme is usually
• A.a highly reactive compound
• B.a metal ion such as Hg2+ or Pb2+
• C.structurally similar to the substrate.
• D.water insoluble
The enzyme inhibition can occur by
• A.reversible inhibitors
• B.irreversible inhibitors
• C.Both (a) and (b)
• D.None of these

56. Sample questions
In a Lineweaver-Burk Plot, competitive
inhibitor shows which of the following
effect?
• A.It moves the entire curve to right
• B.It moves the entire curve to left
• C.It changes the x-intercept
• D.It has no effect on the slope

57. Sample questions
Non-competitive inhibitor of an enzyme catalyzed
reaction
• A.decreases Vmax
• B.binds to ES
• C.both (a) and (b)
• D.can actually increase reaction velocity in rare
cases

58. Sample questions
A classical uncompetitive inhibitor is a compound that binds
• A.reversibly to the enzyme substrate complex yielding an
inactive ESI complex
• B.irreversibly to the enzyme substrate complex yielding an
inactive ESI complex
• C.reversibly to the enzyme substrate complex yielding an
active ESI complex
• D.irreversibly to the enzyme substrate complex yielding an
active ESI complex

59. Kinetics of Multisubstrate
Reactions
E + A + B <-> E + P + Q
• Sequential Reactions
a) ordered
b) random
• Ping-pong Reactions

60. Sequential Reactions
A B P Q
E EA (EAB) (EPQ) EQ E
A B
B A Q P
E
EA
EB
(EAB)(EPQ)
P Q
EQ
EP
E
Ordered
Random

61. Ordered Sequential

62. Ordered Random

63. Ping-Pong Reactions
A P B Q
E (EA)(FP) (F) (FB)(EQ) E
•In Ping-Pong reactions first product released
before second substrate binds
•When E binds A, E changes to F
•When F binds B, F changes back to E

65. Lineweaver-Burke Plot of
Multisubstrate Reactions
Increasing
[B]
Increasing
[B]
Sequential Ping-Pong
Vmax doesn’t change
Km changes
Both Vmax & Km change
1/Vo
1/[S]
1/Vo
1/[S]