Rules for deriving rate laws for simple systems
1. Write reactions involved in forming P from S
2. Write the conservation equation expressing the
distribution of total enzyme concentration [E]total
among the various species
3. Write the velocity dependence equation, summing
all the catalytic rates constants multiplied by the
concentration of the respective product-forming
species.
4. Divide the velocity dependence equation by the
conservation equation.
5. Express the concentration of each enzymic species
in terms of free enzyme concentration & substitute
6. Substitute & algebra
1. Write reactions involved in forming P from S
For a simple 1 step reaction, no inhibitor
k1
k3
E+S
ES

E
+
P
k
2
2. Write the conservation equation expressing the distribution of
total enzyme concentration [E]total among the various species
[E]T = [E] + [ES]
3. Write the velocity dependence equation,
summing all the catalytic rates constants
multiplied by the concentration of the
respective product-forming species.
v = k3[ES]
4. Divide the velocity dependence equation by
the enzyme conservation equation.
v = k3[ES]
[E]T [E] + [ES]
5. Express the concentration of each enzymic
species in terms of free enzyme concentration &
substitute
[ES] = k1[S]  [E] - k2[ES] - k3[ES]
Collecting terms:
[ES] = k1[S]  [E] - (k2+ k3)[ES]
Make steady state assumption:
@ steady state d[ES] = 0,  rate of formation = rate of breakdown
dt
k1 [S]  [E] = (k2 + k3)[ES]
k1[S][E]
Solve for [ES]; [ES] = k + k
2
3
Define KM= k2 + k3 then
k1
[S][E] = [ES]
Km
6. Substitute & Algebra
[S][E] = [ES]
Km
v = k3[ES]
[E]T= [E] + [ES]
Divide by [E]
v = k3 [S][E]
Km
[E]T= [E]+ [S][E]
Km
v = k3 [S]
Km
[E]T= 1+ [S]
Km
Substitute & Algebra (cont’d)
v = k3 [S]
Km
[E]T= 1+ [S]
Km
Multiply by Km
v = k3 [S]
[E]T= KM+ [S]
Define Vmax= k3[ET]
v = Vmax [S]
KM+ [S]
Expressing With Measurable
Quantities
The total concentration of enzyme [E]T can be
determined, but it is very difficult to measure [E] and
[ES].
Steady state rate equation: v = k1[E][S] = k-1[ES] + k2
[ES]
By defining the Michaelis Constant (KM) = (k-1 + k2)/k1
then
rearranging and substituting, [ES] can be expressed
as:
[ES] = [E]T [S]/(KM + [S])
Importance of Initial Velocity
In reality, vo is velocity measured before 10% of
S is consumed.
Before [ES] has built up- can assume reverse
reaction is negligible.
Working with vo minimizes complications
 Reverse reactions
 Inhibition of reaction by product
Implications of M-M Equation
Vo = Vmax [S] / (KM + [S])
1. From the M-M equation, when [S] = KM, vo = Vmax/2.
 This means that low values of KM imply the enzyme
achieves maximal catalytic efficiency at low [S].
2. The catalytic constant, kcat = Vmax / [E]T
 kcat is also called the turnover number of the enzyme, i.e.
the number of reaction processes (turn-overs) that each
active site catalyzes per unit time.
Turnover numbers vary over many orders of magnitude for
different enzymes in accord with need.
Implications of M-M Equation
(cont.)
3. When [S] << KM, very little ES is formed.
Under these conditions: vo ~ (kcat/Km)[E][S]
 The kcat/Km term is a measure of the enzyme’s
catalytic efficiency.
 The upper limit to kcat/Km is k1, I.e. the
decomposition of ES to E + P can occur no more
frequently than ES is formed.
 The most efficient enzymes have kcat/Km values
near the diffusion-controlled limit of 108-109 M-1 s-1.
They will catalyze a reaction almost every time a
substrate is bound in the active site- catalytic
perfection!!!
Practical Summary- Analysis of
Kinetics
 When [S] << KM, the reaction increases
linearly with [S]; I.e. vo = (Vmax / KM ) [S]
 When [S] = KM, vo = Vmax /2 (half maximal
velocity); this is a definition of KM: the
concentration of substrate which gives ½ of
Vmax.
 When [S] >> Km, vo = Vmax
Practical Summary- Vmax and Km
 KM gives an idea of the range of [S] at which a
reaction will occur. The larger the KM, the
WEAKER the binding affinity of enzyme for
substrate.
 Vmax gives an idea of how fast the reaction can
occur under ideal circumstances.
 Kcat / KM gives a practical idea of the catalytic
efficiency, i.e. how often a molecule of substrate
that is bound reacts to give product.
Inhibition of Enzymes
• Enzyme Inhibitor: a molecule that reduces the
effectiveness of an enzyme.
• Mechanisms for enzyme inhibition
 Competitive inhibition- substrate analogs bind in
the active site reducing availability of free enzyme
 Uncompetitive inhibition- inhibitor binds to the
substrate-enzyme complex and presumably
distorts the active site making the enzyme less
active
 Mixed or non-competitive inhibition- combination
of competitive and uncompetitive inhibition
 Inactivator- irriversible reaction with enzyme
Enzyme Inhibitors in Medicine
• Many current pharmaceuticals are enzyme
inhibitors (e.g. HIV protease inhibitors for
treatment of AIDS- see pages 336-337).
• An example: Ethanol is used as a competitive
inhibitor to treat methanol poisoning.
 Methanol is metabolized by the enzyme alcohol
dehydrogenase producing highly toxic
formaldehyde.
 Ethanol competes for the same enzyme.
 Administration of ethanol occupies the enzyme
thereby delaying methanol metabolism long
enough for clearance through the kidneys.