Kinetics: Reaction Order
• Reaction Order: the number of reactant molecules that
need to come together to generate a product.
• A unimolecular S  P reaction is first order.
• A bimolecular 2S  P reaction is second order.
• A bimolecular S1 + S2  P is second order, first order
in S1 and first order in S2.
• The reaction velocity for bimolecular reactions are not
linearly dependent on S.
2S  P
S1 + S2  P
v = k [S]2
v = k [S1] [S2]
First Order Rate Equations
Rate equation: describes the progress of a reaction
over time.
For a unimolecular reaction: v = d[P]/dt = -d[S]/dt= k[S]
rearranging gives: d[S]/[S] = d ln[S] = -k dt
Integrating from the starting concentration [S]o at time
to to the concentration [S] at a future time t gives:
 d ln[S] = -k  dt
ln[S] = ln[S]o –kt
This allows the calculation of [S] at any time t from the
starting concentration and the reaction rate constant.
First Order Rate Equations (cont.)
ln[S] = ln[S]o –kt
This equation for a first order reaction is linear
A plot of ln[S] vs. t is a straight line:
[Fig. 12-1]
 The intercept is the starting concentration [S]o
 Slope is the negative of the rate constant (k)
Half-life of the reaction (t½) is the time required for half
of S to be used up. Since the rate is intrinsic to the
reaction in a first order reaction, the slope of the line in
the plot never changes and the half-life is the same
regardless of the starting concentration.
t½ = ln2/k = 0.693/k
Enzyme Kinetics
E + S  ES  P + E
E- enzyme, S- substrate, ES- enzyme-substrate complex, P-product
k1 and k-1 are the forward and reverse rate constants for formation
of ES
k2 is the rate constant for the decomposition of ES to form P, and
we assume that the reverse reaction is negligible.
When [S] is high enough, all of the enzyme is in the ES form and the
second step is rate limiting. Above this [S], the progress of the
reaction is insensitive to the amount of S present, i.e. “the reaction
is zero order in substrate”.
Enzyme Reaction Velocity
For the complex enzymatic reaction: v = d[P]/dt = k2 [ES]
The overall rate of the production of ES is given by rate of
production from E and S minus the rate of
decomposition of ES either via the reverse reaction or
formation of product:
d[ES]/dt = k1 [E][S] - k-1[ES] - k2[ES]
Simplifying Assumptions
The velocity equation is too complex to be directly
evaluated, so assumptions must be made.
1. Equilibrium. If k-1 >> k2, the first step of the reaction
reaches equilibrium. Unfortunately, this situation
rarely holds.
2. Steady State. If [S] >> [E], [ES] remains constant
because the large excess of S molecules rapidly fill all
enzyme active sites: d[ES]/dt = 0. This holds true until
all of S is used up.
[Fig. 12-2]
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])
The initial velocity (vo) = k2 [ES] = k2 [E ]T [S]/(KM + [S])
The key point is that [E]T and [S] are measurable.
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
Maximal Velocity
When the enzyme is saturated with substrate, the
reaction is progressing at its maximal velocity, Vmax.
At saturation [E]T = [ES], and the equation for
reaction velocity simplies to Vmax = k2 [E]T
Combing with the expression for vo leads to the
Michaelis-Menten Equation of enzyme kinetics:
[Fig. 12-3]
Vo = Vmax [S] / (KM + [S])
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
Using Fig. 12-3, the plot of vo vs. [S]
 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.