SURVEY OF BIOCHEMISTRY
Enzyme Kinetics and Inhibition
1
Rates of Chemical Reactions
Enzyme kinetics is the study of rates of reactions catalyzed
by enyzmes.
v=
A
k
P
d[A] d[P]
v

dt
dt
The rxn rate (velocity, v) can be described in several ways:
[1] disappearance of reactant, A
[2] appearance of product, P

These eqn’s relate velocity to concentration of reactants and products.
2
Rate Laws
Enzyme kinetics is the study of rates of reactions catalyzed
by enyzmes.
v=
A
k
P
d[A] d[P]
v

dt
dt
A rate law is an equation describing the velocity of a
chemical reaction.

Differential Rate Laws
Integrated Rate Laws
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Differential Rate Laws
• Differential rate laws correspond to
order of the reaction.
Order of Reaction
0
Rate Law
v=k
1
v = k [A]
2
v = k [A]2 or k[B]2 or
v = k [A] x [B]
4
Integrated Rate Laws:
First Order
• Integrated rate laws express the rxn
velocity in terms of time.
A
Rate of
Disappearance
of A
Rearranging…
products
- d[A]
dt
= k [A]
d[A]
[A]
first order rxn
= - k dt
5
Integrated Rate Laws:
First Order
Rearranging…
Integrate on
both sides
of eqn:


[A]t
[A]0
d[A]
[A]
1
[A]
= - k dt
dA
=

- k dt
(ln [A]t + constant) - (ln [A]0 + constant) = - kt

ln [A]t - ln [A]0 = - kt

6
Integrated Rate Laws:
First Order
ln [A]t - ln [A]0 = - kt
ln [A] = ln [A]0 - kt
7
Integrated Rate Law:
Other Versions of First Order
ln [A]t - ln [A]0 = - kt
Rearranging:
Take exponent
of both sides:
ln
[A]t
[A]0
[A]t
[A]0
= - kt
first order rxn
= e -kt
[A]t = [A]0 e -kt
first order rxn
8
Integrated Rate Law:
Second Order
• How does the integrated rate law change if
the order of the reaction is second order?
2A
Rate of
Disappearance
of A
products
- d[A]
Rearranging…
dt
= k [A]2
second order
rxn
d[A]
[A]2
= - k dt
Show
result on
board
9
Michaelis-Menten Equation
Many enzymes obey Michaelis-Menten kinetics behavior:
k1
E+S
k-1
ES
k2
E+P
Rate limiting step
d[P]
v
 k 2[ES]
dt
Problem:
[ES] is difficult to measure!
What can we do?
10
Michaelis-Menten Equation
k1
E+S
k-1
ES
k2
E+P
d[P]
 k 2[ES]
Recall v 
dt
I.
II.
Assume equilibrium is maintained in 1st step
Assume “steady state”

d[ES]
 k1 [E] [S]
dt
Formation
of ES
- k-1 [ES] - k2 [ES] = 0
Depletion of ES
See notes on board…
11
Michaelis-Menten Kinetics
V max [S]
v
Km  [S]

12
Lineweaver-Burk Plot
13
Enzyme Inhibition
• What is an inhibitor?
• Modes of Inhibition
– Competitive
binds to same site in E as S
– Uncompetitive
bind to different
– Noncompetitive
site
in
E
than
S
– Mixed
Note: Text does not distinguish “non” and “mixed”
14
Competitive Inhibition
• Competitive inhibitors bind to the
same site on E as S
15
Competitive Inhibition
16
Competitive Inhibition
17
Uncompetitive Inhibition
Uncompetitive inhibitors
bind directly to the ES
complex but not to the
free enzyme
18
Uncompetitive Inhibition
19
Mixed Inhibition
Mixed inhibitors can
bind to E or ES complex
S cannot bind if I is
already bound!
20
Mixed Inhibition
21
Noncompetitive Inhibition
Noncompetitive
inhibitors can bind to E
or ES complex
S can bind even if
I is already bound!
+I
See board for plot
22
Updates and Reminders
• Exam #2 in two weeks (June 26)
–
–
–
–
–
Chapter
Chapter
Chapter
Chapter
Chapter
7: Protein Function
11: Enzyme Catalysis
12: Kinetics & Inhibition
8: Carbohydrates
14: Introduction to Metabolism
• Suggested HW problems online this weekend
• Resources: What You Should Know
more coming soon
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