Enzyme kinetics. Read Lehninger Ch. 6, pp.200-213.
Objectives: • Explain the initial velocity assuption and steady-state assuption as
they pertain to the !ichaelis-!enten e"uation.
• #istinguish $et%een & and &d.
• Exaine 'inetic data and estiate & and (ax values.
• Estimate Vmax and Km from both a Michaelis-Menten and Lineweaver-
Burk plot. • Distinguish between Vmax and kcat.
• Describe catalytic perfection in terms of kcat /Km(catalytic efficiency).
• Understand how steady-state kinetic studies can be used to study enzyme mechanism.
• Describe how Vmax and Km are affected by the presence of competitive,
uncompetitive, and mixed inhibitors. • Examine a Michaelis-Menten and Lineweaver-Burk plot and indicate the type of inhibition.
• Draw both Michaelis-Menten and Lineweaver-Burk plots for
competitive, uncompetitive, and mixed inhibitors. •
 
The relationship between [S] and the reaction rate is described by theMichaelis-Menten Equation:
v = Vmax [S] / (Km + [S])
v = rate Vmax = maximum rate
[S] = substrate concentration Km = Michaelis constant
 
 
k1
k-1
k2
k-2
Consider the conversion of substrate, S, to the product, P, by the enzyme, E:
To describe this enzyme catalyzed conversion, we can make two assumptions to simplify the above relationships:
2.Steady-state assumption
state phase
With [S] >> [E], overall there is a constant ratio of [ES] and [E]
Steady-State kinetics
 
k1
k-1
k2
k-2
Consider the conversion of substrate, S, to the product, P, by the enzyme, E:
To describe this enzyme catalyzed conversion, we can make two assumptions to simplify the above relationships:
2.Steady-state assumption
k1
k-1
k2
k-2X
The Michaelis Constant, Km
Km = rate(s) of breakdown of ES/ rate of formation of ES
Km = (k-1 +k2) / k1
Km is roughly equivalent to Kd only when k-1>> k2 
Remember Kd= k-1 / k1 
([S] << Km)
([S] >> Km)
 
We can experimentally determine Vmax & Km
 
MM equation: v = Vmax [S] / (Km + [S])
 
turnover number, kcat: number of substrate molecules
converted to product per enzyme molecule per unit time. (often expressed in units of sec-1)
In our simple case, kcat= Vmax / [Etotal]
E + S ES E + P
k1
k-1
k2
k-2X
 
v = kcat[Etotal] [S] / (Km + [S])
When [S] << Km, then
v = (kcat / Km) [Etotal] [S]
 
E + S ES E + P
k1
k-1
k2
k-2
Effects of pH on enzymatic activity
 
We can experimentally determine Vmax & Km
 
k1
k-1
k2
k-2X
The Michaelis Constant, Km
Km = rate(s) of breakdown of ES/ rate of formation of ES
Km = (k-1 +k2) / k1
Km is roughly equivalent to Kd only when k-1>> k2 
Remember Kd= k-1 / k1 
- can overcome inhibition by increasing [S]
- theapparent Km (the Km observed in the
presence of the inhibitor) is increased by the factor,
- therefore, v = Vmax [S] / (Km + [S])
where = 1 + [I]/K I  and K
I = [E][I]/[EI]
Vmax unchanged
 
- binds at a site distinct from the substrate binding site
- binds only to the ES complex
- theapparent Vmax & Km are both decreased
by ′ at high [S].
where ′ = 1 + [I]/K I  and K
I = [ES][I]/[ESI]
Effect of uncompetitive inhibitor on LWB plot
Lowers apparent Vmax& Km
- binds at a site distinct from the substrate binding site
- binds either to E or the ES complex
- theapparent Vmax & Km are both affected.
- v = Vmax [S] / (Km + ′[S])
where = 1 + [I]/K I  and ′ = 1 + [I]/K
I
Effect of mixed inhibition on LWB plot
affects Vmax & Km
Pure noncompetitive inhibition: = ′
Irreversible inhibition
An irreversible inhibitor will combine with or destroy a functional group on an enzyme that
is essential for activity
peptidase (GP).
 
covalent catalysis/general acid-base catalysis
Michaelis-Menten Equation
catalytic efficiency (kcat /Km)
competitive vs. uncompetitive vs. mixed inhibition
irreversible or covalent inhibition What can kinetics can tell us about enzyme mechanism?
(pH dependence, burst kinetics, steady state kinetics, inhibition, etc…)