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E + S ES P + E k 2 Michaelis reasoned that If k 2 is the smallest rate constant, the overall velocity of the reaction is v o = k 2 (ES) Problem: We cannot measure [ES] Solution: Substitute [ES] into equation Evaluating and Using the Michaelis-Menten Equation

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E + S ES P + E

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

Evaluating and Using

the

Michaelis-Menten Equation

Vmax

variables

Km

Dependent

variable

Constants

pO2

vo =

q =

P50 +

pO2

[S]

Vmax [S]

C + [S]

Km + [S]

[S]

vo

vo =

[S]

Y axis

(Dependent

Variable)

Rectangular Hyperbola

X axis

(Independent variable)

vo =

Vmax [S]

Vmax [S]

Vmax [S]

Vmax [S]

2[S]

[S]

Km + [S]

Km

S << Km

vo =

First order

with [S]

vo =

One-half

Vmax

S = Km

vo =

Zero Order

with [S]

S >> Km

Vmax = k2[ET]

Vmax is first order with enzyme

Slope = kcat

When the

velocity = Vmax

k2 = kcat

Vmax

E4

vo

E3

E2

Enzyme

E1

[S]

vo = Vmax

Vmax

vo =

Vmax

2

2

Vmax [S]

vo =

Km

S

S

S

[S] = Km

S

S

Vmax

Km = [S] at

one-half Vmax

HIGH [S]

LOW [S]

Picture it this way

k2 + k-1

Km=

k1

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

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?

vo =

10 moles/min x 2 M

3 moles/min =

Km + 2 M

Vmax [S]

Km + [S]

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

Vo

[S]

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

[S]

Vo

=

Vmax

Km + [S]

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

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