Lecture 6 (EXTRA) – Enzyme
Kinetics (Thunnissen)
Enzyme kinetics
The progress of any reaction can be expressed as a velocity (v), either the rate of disappearance of
the substrate (S) or the rate of appearance of the product (P):
V = - (d[S] / dt) = (d[P] / dt)
Not surprisingly, the more catalyst present, the faster the reaction.
When the enzyme concentration is held constant, the reaction velocity varies with the substrate
concentration, but in a nonlinear fashion. The shape of this velocity versus substrate curve is an
important key to understanding how enzymes interact with their substrates. The hyperbolic, rather
than linear, shape of the curve suggests that an enzyme physically combines with its substrate to
form an enzyme-substrate (ES) complex.
So the reaction can be more accurately written as:
E + S —> ES —> E + P
At low substrate concentrations, the enzyme quickly inverts all the substrate to product, but as more
substrate is added, the enzyme becomes saturated with substrate, which means that there are more
substrate molecules than enzyme molecules so not all the substrate can be converted to product in a
given time —> saturation kinetics.
Michaelis-Menten equation
For a unimolecular reaction A —> B the progress of the reaction can be mathematically described by
a rate equation in which the reaction rate (v) is expressed in terms of a constant (the rate constant)
and the reactant concentration [A]:
v = -(d[A] / dt) = k[A]
k is the rate constant and has units of 1/s (reciprocal seconds). Such a reaction is said to be first-
order because its rate depends on the concentration of one substance. A bimoleulcar or second-
order reaction involves two reactant (A + B —> C) and its rate equation is:
v = -(d[A] / dt) = -(d[B] / dt) = k[A][B]
Here, k is a second order rate constant and has units of M -1 • s-1
The velocity of a second-order reaction is therefore proportional to the product of the two reactant
concentrations.
In the simplest case, an enzyme binds its substrate (in an enzyme-substrate complex) before
converting it to product, so the overall reaction actually consists of first-order and second-order
processes, each with a characteristic rate constant:
The rate equation for product formation is:
v = (d[P] / dt) = k2[ES]
, To simplify our analysis, we choose experimental conditions such that the substrate concentration is
much greater than the enzyme concentration ([S] » [E]). Under these conditions, after E and S have
been mixed together, the concentration ofES remains constant until nearly all the substrate
molecules have been converted to product. This is said to maintain a steady state (it has a constant
value):
d[ES] / dt = 0
According to the steady-state assumption, the rate of ES formation must balance the rate of ES
consumption:
k1[E][S] = k-1[ES] + k2[ES]
At any point in the reaction, [E] and [ES] are difficult to determine, but the total enzyme
concentration [E]T is usually known. This results in the following formula:
( ([E]T – [ES])*([S]) ) / [ES] = (k-1 + k2) / k1
This brings us to the Michaelis constant Km which equals (k-1 + k2) / k1
Connecting all the formulas and simplifying/solving, we can make up an equation with only known
quantities: [E]T and [S]. Although some S is consumed in forming the ES complex, we can ignore it
because [S]T >> [E]T
Typically, kinetic measurements are made soon after the enzyme and substrate are mixed. Therefore
the velocity at the start of the reaction (at time zero) is expressed as v0 (the initial velocity).
The maximum reaction velocity (vmax) can be expressed as vmax = k2 [E]T since all the enzyme is in its
ES form when [S] is very high.
Therefore:
v0 = (vmax [S]) / (kM + [S])
This relationship is called the Michealis-Menten equation, which is the rate equation for an enzyme-
catalysed reaction and is the mathemical description of the hyperbolic curve.
The catalytic constant describes how quickly an enzyme can act
The catalytic constant (kcat) shows how fast the ES complex proceeds to E + P. For any enzyme
catalysed reaction:
Kcat = vmax / [E]T and for simple reactions kcat = k2
Kcat is the rate constant for the reaction when the enzyme is saturated with substrate, k cat is also
known as the turnover number because it is the number of catalytic cycles that each active site
undergoes per time unit.
Catalytic efficiency
An enzyme's effectiveness as a catalyst depends on how avidly it binds its substrates and how rapidly
it converts them to products. Thus, a measure of catalytic efficiency must reflect both binding and
catalytic events. The quantity kcat/kM satisfies this requirement.
An enzyme is said to have reached catalytic perfection when the overall rate is diffusion-controlled:
it catalyses a reaction as rapidly as it encounters its substrate.
Kinetics (Thunnissen)
Enzyme kinetics
The progress of any reaction can be expressed as a velocity (v), either the rate of disappearance of
the substrate (S) or the rate of appearance of the product (P):
V = - (d[S] / dt) = (d[P] / dt)
Not surprisingly, the more catalyst present, the faster the reaction.
When the enzyme concentration is held constant, the reaction velocity varies with the substrate
concentration, but in a nonlinear fashion. The shape of this velocity versus substrate curve is an
important key to understanding how enzymes interact with their substrates. The hyperbolic, rather
than linear, shape of the curve suggests that an enzyme physically combines with its substrate to
form an enzyme-substrate (ES) complex.
So the reaction can be more accurately written as:
E + S —> ES —> E + P
At low substrate concentrations, the enzyme quickly inverts all the substrate to product, but as more
substrate is added, the enzyme becomes saturated with substrate, which means that there are more
substrate molecules than enzyme molecules so not all the substrate can be converted to product in a
given time —> saturation kinetics.
Michaelis-Menten equation
For a unimolecular reaction A —> B the progress of the reaction can be mathematically described by
a rate equation in which the reaction rate (v) is expressed in terms of a constant (the rate constant)
and the reactant concentration [A]:
v = -(d[A] / dt) = k[A]
k is the rate constant and has units of 1/s (reciprocal seconds). Such a reaction is said to be first-
order because its rate depends on the concentration of one substance. A bimoleulcar or second-
order reaction involves two reactant (A + B —> C) and its rate equation is:
v = -(d[A] / dt) = -(d[B] / dt) = k[A][B]
Here, k is a second order rate constant and has units of M -1 • s-1
The velocity of a second-order reaction is therefore proportional to the product of the two reactant
concentrations.
In the simplest case, an enzyme binds its substrate (in an enzyme-substrate complex) before
converting it to product, so the overall reaction actually consists of first-order and second-order
processes, each with a characteristic rate constant:
The rate equation for product formation is:
v = (d[P] / dt) = k2[ES]
, To simplify our analysis, we choose experimental conditions such that the substrate concentration is
much greater than the enzyme concentration ([S] » [E]). Under these conditions, after E and S have
been mixed together, the concentration ofES remains constant until nearly all the substrate
molecules have been converted to product. This is said to maintain a steady state (it has a constant
value):
d[ES] / dt = 0
According to the steady-state assumption, the rate of ES formation must balance the rate of ES
consumption:
k1[E][S] = k-1[ES] + k2[ES]
At any point in the reaction, [E] and [ES] are difficult to determine, but the total enzyme
concentration [E]T is usually known. This results in the following formula:
( ([E]T – [ES])*([S]) ) / [ES] = (k-1 + k2) / k1
This brings us to the Michaelis constant Km which equals (k-1 + k2) / k1
Connecting all the formulas and simplifying/solving, we can make up an equation with only known
quantities: [E]T and [S]. Although some S is consumed in forming the ES complex, we can ignore it
because [S]T >> [E]T
Typically, kinetic measurements are made soon after the enzyme and substrate are mixed. Therefore
the velocity at the start of the reaction (at time zero) is expressed as v0 (the initial velocity).
The maximum reaction velocity (vmax) can be expressed as vmax = k2 [E]T since all the enzyme is in its
ES form when [S] is very high.
Therefore:
v0 = (vmax [S]) / (kM + [S])
This relationship is called the Michealis-Menten equation, which is the rate equation for an enzyme-
catalysed reaction and is the mathemical description of the hyperbolic curve.
The catalytic constant describes how quickly an enzyme can act
The catalytic constant (kcat) shows how fast the ES complex proceeds to E + P. For any enzyme
catalysed reaction:
Kcat = vmax / [E]T and for simple reactions kcat = k2
Kcat is the rate constant for the reaction when the enzyme is saturated with substrate, k cat is also
known as the turnover number because it is the number of catalytic cycles that each active site
undergoes per time unit.
Catalytic efficiency
An enzyme's effectiveness as a catalyst depends on how avidly it binds its substrates and how rapidly
it converts them to products. Thus, a measure of catalytic efficiency must reflect both binding and
catalytic events. The quantity kcat/kM satisfies this requirement.
An enzyme is said to have reached catalytic perfection when the overall rate is diffusion-controlled:
it catalyses a reaction as rapidly as it encounters its substrate.
