E + S ES P + Ek2

vo = k2(ES)

Michaelis reasoned that If k2 is the smallest rate constant, the overall velocity of the reaction is

Problem: We cannot measure [ES]

Solution: Substitute [ES] into equation

Evaluating and Using the

Michaelis-Menten Equation

Vmax [S] Km + [S]

vo = [S][S]

variablesvo

Dependentvariable

Vmax Km

Constants [S] C + [S]

vo = pO2

P50 + pO2

Rectangular Hyperbola(DependentVariable)

(Independent variable)

Y axis

X axis

Vmax [S] Km + [S]

vo =

Vmax [S] Km

vo =

Vmax [S] 2[S]

vo =

Vmax [S] [S]

vo =

S << Km

S = Km

S >> Km

First orderwith [S]

First orderwith [S]

One-halfVmax

One-halfVmax

Zero Orderwith [S]

Zero Orderwith [S]

Vmax = k2[ET]

Vmax is first order with enzyme

vo

[S]

E1

E2

E3

E4

Vmax

Enzyme

Slope = kcat

When the velocity = Vmax

k2 = kcat

When the velocity = Vmax

k2 = kcat

Vmax [S] Km

vo =

Vmax

2vo =

S S SS S

vo = Vmax

Vmax

Vmax

2Km = [S] at one-half Vmax

[S] = Km

Picture it this way

vo = Vmax

vo =Vmax2

vo =VmaxKm

(S)

LOW [S] HIGH [S]

TWO DEFINITIONS OF KM

Rate Constant Definition

Km = k2 + k-1

k1

Affinity

No unitsSubstrate 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?

Vmax [S]

Km + [S]

vo =

3 moles/min =10 moles/min x 2 M

Km + 2 M 3(Km + 2) = 20

3Km + 6 = 203Km = 14Km = 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]

Setup:

Km holds the same literal meaning as pKa and P50Km holds the same literal meaning as pKa and P50

An enzyme has a Km of 10 M. At what [S] will the reactionbe 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 M15 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

How close to Vmax will a reaction be when [S] =

1) Km 2) 10 Km 3) 100 Km

Vo

Vmax=

[S]

Km + [S]

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

2. Modifying the substrate and releasing

the product as determined by K2

The two phases will become more apparent when we study inhibitors