Transcript Epjj Lecture 4

Lecture 5:Enzymes
Ahmad Razali Ishak
Department of Environmental Health
Faculty of Health Sciences
UiTM Puncak Alam
Introduction to Enzymes
Enzymes are biological catalysts.
Like all catalysts, enzymes lower the
energy needed to get a
reaction started. Enzymes are much
generally better at accelerating the rates
of reactions than
non-biological catalysts.
Diagram showing that less energy is
required to get an enzyme catalyzed
reaction started as compared to a
non-catalyzed reaction
Enzyme Catalysis (Cont’d)
 Consider the reaction
H2 O 2
H2 O +
O2
These six classes are:
1. Oxidoreductases - enzymes catalyzing oxidation
reduction reactions.
2. Transferases - enzymes catalyzing transfer of
functional groups.
3. Hydrolases - enzymes catalyzing hydrolysis reactions.
4. Lyases - enzymes catalyzing group elimination
reactions to form double bonds.
5. Isomerases - enzymes catalyzing isomerizations (bond
rearrangements).
6. Ligases - enzymes catalyzing bond formation reactions
couples with ATP hydrolysis
The Active Site of the Enzyme.
 Each enzyme has a unique active site.
 Active site = catalytic site.
 The enzyme binds its substrate(s) at
the active site and the enzyme
catalyzes chemical changes in the
substrate(s).
 Enzymes contain a large number of
amino acids, but most AA side chains
are used for forming the enzyme's
shape. Only a few AA side chains are
at the active site. These special AA
side chains:
1. Bind the substrate(s) and
2. Catalyze the reaction
3-D model of the enzyme ribonuclease
with the key amino acid side chains at
the active site shown in red. The active
site is a deep groove at the center of this
structure.
· Enzyme has large structure with
hundreds of AA side chains but only
a few are involved
in catalysis.
· Each enzyme has a unique active site.
· Key AA side chains are involved in
binding
and catalysis in the active site.
 Enzyme + Substrate = EnzymeSubstrate Complex
 E + S = ES Complex
Binding Models
 Two models have been developed to
describe formation of the enzymesubstrate complex
 Lock-and-key model: substrate binds
to that portion of the enzyme with a
complementary shape
 Induced fit model: binding of the
substrate induces a change in the
conformation of the enzyme that results
in a complementary fit
2 Modes of E-S Complex
Formation
Formation of Product
Enzyme Kinetics
Usefulness of enzyme kinetics:
· Common clinical assays to detect
enzymes
· Understanding metabolic pathways
· Measuring binding of substrates and
inhibitors to the active site of an
enzyme
· Understanding the mechanism of
catalysis of an enzyme
A Simple Mechanism for the
Enzyme Catalyzed Reaction.
 For catalysis to begin, the substrate must
bind to the enzyme, which results in the
formation of the enzyme-substrate complex
(ie E-S complex).
 The E-S complex forms rapidly in the first
part of the enzyme catalysis process and
the concentration of the E-S stays constant
at a steady-state level. For this reason, this
type of kinetics is called steady-state
kinetics.
E + S ↔ ES → E + P
These data show that
at low [S], the initial
velocity is more or less
proportional to the [S].
At high [S], the initial
velocity no longer
increases as more
substrate is added.
Thus, at high [S] the
enzyme is saturated
with substrate and
no increase in the
enzyme catalyzed rate is
observed.
The Michaelis-Menten Equation.
 The plot of Vo versus [S] can be represented
by an equation, which is known as the
Michaelis-Menten equation in honor of the
scientist who first described it. This equation,
sometimes called the M-M equation, is an
important one for you to know and understand.
Vo = Vmax [S] /Km
+ [S]
The constants in this
equation, Km and Vmax,
are defined:
Vmax = Maximum velocity
catalyzed by a fixed [E]
Km = the [S] which gives
1/2 Vmax
 The Km is sometimes called the
Michaelis Constant. The Km is an
intrinsic property of an enzyme
related to the binding constant for
forming the ES complex, which is an
equilibrium and can be defined by the
rate constants for its formation and
breakdown using the simple enzyme
To calculate the Km and Vmax, the Michaelis-Menten equation
is converted into a linear form
by taking the reciprocal of both sides of the equation.
This is called the Lineweaver-Burk
equation in honor of the first scientists to describe it.

This equation then takes on the form of the equation of a line. The y
values are 1/Vo, the x values are 1/[S]. The b value in the line
equation is the slope and equal to Km/Vmax, while the c value is the yintercept and equal to 1/Vmax.
Enzyme Inhibitors
 Inhibitors of enzymes: Two types are
considered - Competitive and NonCompetitive.
 Competitive Inhibitor has a
chemical similarity to the substrate
and competes with the substrate for
binding to the active site of the
enzyme. A good example to describe
competitive inhibition is the
mitochondrial enzyme, succinate
dehydrogenase:
(A)The reaction catalyzed by succinate
dehydrogenase is the oxidation of succinate
to fumarate. (B) Malonate and oxaloacetate
are competitive inhibitors of succinate
dehydrogenase.
 Both these competitive inhibitors, malonate and
oxaloacetate, look like succinate in their chemical
character. Both inhibitors are dicarboxylic acids like the
substrate succinate so they
have groups which can bind in the same places in the
active site of succinate dehydrogenase as the substrate.
 However, neither inhibitor has the capacity to undergo
the reaction and so the enzyme is inhibited.
 Since these inhibitors simply bind to the enzyme, when
the succinate concentration is high, they will be pushed
out of the site by the substrate and the enzyme will
catalyze the reaction as if no inhibitor were present.
An enzyme mechanism model of the action of a competitive
inhibitor (Ic) based on the standard model of a Michaelis-Menten
enzyme where E + S leads to the E-S complex, which leads to
product P:
Vo versus [S] plot comparing the kinetics of the
reaction in the absence of inhibitor and in the
presence of the competitive inhibitor (Ic).
At high [S], the initial velocity in the presence of
Ic will be about the same as it is in the absence
of the inhibitor. The concentration of S which will
be required to overcome the effect of the
competitive inhibitor will depend on the
[Ic] (ie. concentration of the competitive inhibitor)
and the Ki (ie. the binding constant of the
inhibitor to enzyme).
Here the uninhibited reaction gives the
standard double reciprocal plot from which
Km and Vmax can be calculated.
The reaction in the presence of the
competitive inhibitor yields apparent
constants for the enzyme which are called
the Km' and Vmax'. For the true
competitive inhibitor,
the Vmax' (apparent Vmax for
inhibited enzyme) will be the same as
the real Vmax, while the Km' (apparent Km
for the inhibited enzyme) will be greater than
the real Km. Thus, the -1/Km'
will be smaller than -1/Km.
Non-competitive Inhibition.
 Non-Competitive Inhibitor does
not compete with substrate and the
[S] has no influence on the degree of
inhibition of the enzyme's catalytic
rate. For example, enzymes with a
thiol ( -SH ) not at the active site can
be inhibited
Example of a heavy metal inhibiting an
enzyme by binding to a thiol group not at the
active site and inactivating the enzyme.
In this case
where the non-competitive inhibitor (Inc) react
the enzyme at a site other than the active
site, both the free enzyme (E) and the
enzyme-substrate complex (E-S) react with Inc
Clearly, in this case the reaction of the
non-competitive inhibitor is irreversible.
A Vo versus [S] plot for the
Non-competitive Inhibitor looks
very different than that for a
competitive inhibitor since increas
the [S] has no impact
Non competitive inhibitors decrease
Vmax but have no effect on Km.
Evaluating Enzyme Inhibitors to determine type and
their Ki.
To determine what type an inhibitor is:
1. Find Km and Vmax for uninhibited from 1/Vo vs 1/[S]
plot.
2. On same graph find Km' and Vmax' for inhibited reaction.
A. If Vmax = Vmax' then inhibitor is competitive type.
(Vmax and Vmax' should not be more than 10% different)
B. If Vmax does not equal Vmax', then if Km = Km',
inhibitor is non competitive type.
Assignment
1.PLOT Vo versus [S] to show that the
enzyme obeys Michaelis-Menten
Equation.
2.PLOT 1/Vo versus 1/[S]
to determine Km and Vmax.
1) Draw "vo vs. [S]" and "1/vo vs 1/[S]" plots
2) Determine the Km and Vmax from the double reciprocal plot
3) Decide what type of inhibitors "x" and "z" are
4) Calculate the Ki for "x" and "z"
Competitive Inhibitor : Km' = Km (1 + [I]/Ki)
Noncompetitive Inhibitor: Vmax' = Vmax / (1 + [I]/Ki)