k2

1
k2
Michaelis reasoned that If k2 is the smallest
rate constant, the overall velocity of the
reaction is
vo k2(ES)
Problem We cannot measure ES
Solution Substitute ES into equation
2
Evaluating and Using the
Michaelis-Menten Equation
3
S
vo
vo
S
Y axis
(Dependent Variable)
Rectangular Hyperbola
X axis
(Independent variable)
4
S vo
First order with S
vo
One-half Vmax
S Km
vo
Zero Order with S
S Km
5
Vmax k2ET
Vmax is first order with enzyme
Slope kcat
When the velocity Vmax k2 kcat
Vmax
E4
vo
E3
E2
Enzyme
E1
S
6
Vmax
Km S at one-half Vmax
7
Picture it this way
8
TWO DEFINITIONS OF KM
Rate Constant Definition
Affinity
No units
Substrate Definition
Km S that gives 1/2 Vmax
Km S that fills half the sites on the enzyme
Km has units of substrate concentration
9
Calculate the Km of an enzyme. When S is 2
micromolar, vo 3 micromoles per minute. At
saturation Vmax 10 micromoles per minute.
What does the answer tell you?
Setup
3(Km 2) 20
3Km 6 20
3Km 14
Km 4.7 ?M
When the S is 4.7 ?M, the enzyme is
half-saturated with S
When the S is 4.7 ?M, Efree ES
Km holds the same literal meaning as pKa and P50
10
An enzyme has a Km of 10 ?M. At what S will
the reaction be at Vmax? a) 20 ?M b) 5 ?M c)
cannot tell
Reason The effect of S on velocity is
hyperbolic, not linear
An enzyme has a Km of 10 ?M. When S equals 5
?M 15 ?moles of S are consumed per minute. At
what S will the reaction be at Vmax
a) 20 ?M b) 45 ?M c) cannot tell
What is the Vmax of the above reaction
a) 20 ?M b) 45 ?moles/min c) cannot tell
11
How close to Vmax will a reaction be when S
1) Km 2) 10 Km 3) 100 Km
Relative Max velocity
1)
One-half Vmax 1/2
2) 90.9 Vmax 10/11
3) 99 Vmax 100/101
12
Based on the Kinetic analysis we can conclude
There are two phases of an enzyme-catalyzed
reaction
1. Binding the substrate as determined by Km

• Modifying the substrate and releasing the
  product as determined by K2

The two phases will become more apparent
when we study inhibitors