Intro to Pharmacology and Toxicology
Topics
In
a chemical reaction, the energy states of the reactants (R) and products (P)
determine the ratio of cencentrations at equilibrium (i.e. the dissociation
or equilibrium constant). The energy of activation (Ea)
of a system determines how quickly the system approaches equilibrium.
A catalyst lowers the energy of activation for a system (without being consumed by the reaction), speeding the approach to equilibrium. Enzymes are proteins that act as catalysts for chemical reactions carried out in biological systems. Since the enzyme is not permanently changed in the process, a simple enzyme E catalyzed reaction with one substrate S can be represented as:
k1
kp
[E] + [S] 
[ES] 
[E] + [P]
k-1
where [ES] is the concentration of the enzyme-substrate complex. The velocity of the reactin can be measured as the rate of appearance of P or dissapearance of S:
v = d[P] / dt = - d[S] / dt
There are several assumptions built into the basic approach to enzyme kinetics:
For most of the time period measured in the [S]-[P] graph above:
d[ES] / dt = 0
This condition defines the steady-state, which in most enzymes occur within the first seconds to minutes of reaction. The steady-state is not equilibrium, which typically takes minutes to achieve in most assays. Equilibrium is the time when no net change in concentration of any component of the system can be measured.
The equation that describes the velocity of a simple enzymatic reaction is derived below.
k1
kp
[E] + [S]
[ES]
[E] + [P]
k-1
d[ES] /dt = formation - breakdown
d[ES] /dt = k1[E][S] - [ES](k-1 + kp)
Since a steady-state d[ES] /dt = 0,
Þ k1[E][S] - [ES](k-1 + kp) = 0
k1[E][S] = [ES](k-1 + kp)
Since [E] = [E]Total - [ES]
Þ k1([E]Total - [ES])[S] = [ES](k-1 + kp)
k1[S][E]Total - k1[S][ES] = [ES](k-1 + kp)
k1[S][E]Total = [ES](k-1 + kp) + k1[S][ES]
k1[S][E]Total = [ES]{(k-1 + kp) + k1[S]}
[ES] =
k1[S][E]Total
(k-1
+ kp)
+ k1[S]
[ES] =
[S][E]Total
(k-1+kp)/k1
+ [S]
Since v = kp[E]Total
Þ
v =
kp [S][E]Total
(k-1+kp)/k1
+ [S]
Now we can obtain the Michaelis-Menten equation by defining some constants. First recall that [E]<<<[S], so all the enzyme may be bound to substrate, achieving the maximal rate Vmax. Another constant is Km, the ratio of rate constants controlling the breakdown of [ES] divided by the rate constant contrilling its formation.
k1
kp
[E] + [S]
[ES]
[E] + [P]
k-1
Vmax = kp[E]Total ; Km = (k-1 + kp)/k1
Þ
v = Vmax
[S] = (concentration/time)(concentration)
= concentration
Km
+ [S]
concentration + concentration
time
The
Michaelis-Menten equation describes the relationship between the enzymatic reaction
and the substrate concentration. The independednt variable is [S] and the dependent
variable is v. The other two terms are constants defined by the reaction scheme.
When the substrate concentration equals Km ([S] = Km), velocity is half the maximum velocity:
v = Vmax
Km = Vmax
Km +
Km 2
When the substrate concentration is much lower than Km ([S]<<<Km), lots of enzyme molecules are available and the reaction is first order, with v linearly related to [S]:
v = Vmax
[S] » Vmax
[S] = (Vmax
/ Km)
[S]
Km +
[S] Km
At very high substrate concentration ([S]>>>Km), almost all molecules of enzyme are occupied and the reaction appears to be zero order, totally dependent on Vmax:
v = Vmax
[S] » Vmax
[S] = Vmax
Km +
[S] [S]
In other words, the velocity is largely independent of small changes in substrate concentration. Mathematically this is illustrated by the rectangular hyperbola shape of the graph. A rectangular hyperbola approaches a theoretical maximum or asymptote, the value of which here is Vmax.
Continue to "Enzyme
Systems" or take a quiz: [Q1] [Q2] [Q3] [Q4].
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1- What determines the equilibrium
concentrations in a chemical reaction?
The energy states of the reactants and products.
2- What is the energy of activation
of a reaction?
An amount of free energy needed to move from reactants to products by reaching
an energy barrier, it determines how quickly the system approaches equilibrium.
3- What is a catalyst?
A substance that lowers the energy of activation for a system, without being
consumed by the reaction, speeding the approach to equilibrium.
4- What are enzymes?
Proteins that act as catalysts for chemical reactions carried out in biological
systems.
5- Define velocity for a simple
enzyme catalyzed reaction.
A simple enzyme E catalyzed reaction with one substrate S can
be represented as:
k1
kp
[E] + [S]
[ES]
[E] + [P]
k-1
where [ES] is the concentration of the enzyme-substrate complex. The velocity of the reactin can be measured as the rate of appearance of P or dissapearance of S:
v = d[P] / dt = - d[S] / dt.
6- List 3 asumptions built into
the basic approach to enzyme kinetics.
[E] << [S], therefore [S] will not change significantly during measurement
and [S] ~ [S]total
kp << k1 or k-1,
therefore [E], [S] and [ES] equilibrate before significant amounts of product
are formed
[EP] is not formed
7- What are steady-state and equilibrium
for a simple enzyme catalized reaction?
For most of the time course of an enzymatic reaction d[ES]/dt = 0, a condition
that defines the steady-state that occurs in most reactuions within the first
seconds to minutes. Steady-state is not the same as equilibrium, which is the
time when no net change in concentration of any component of the system can
be measured, and tipically takes minutes to reach.
8- Derive the equation for velocity
of a simple enzymatic reaction given that .
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