Enzymes II
Andy Howard
Introductory Biochemistry
21 October 2010
Biochem: Enzymes II
10/21/2010
What we’ll discuss
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Enzymes
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Enzymes (concluded)
 Bisubstrate reactions
Enzyme kinetics
 Measurements and
Michaelis-Menten
calculational tools
kinetics: overview
Kinetic Constants  Inhibition
Kinetic
 Why study it?
Mechanisms
 The concept
Induced Fit
 Types of inhibitors
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[B]
Kinetics, continued
t
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In most situations more product will be
produced per unit time if A0 is large than if it is
small, and in fact the rate will be linear with the
concentration at any given time:
d[B]/dt = v = k[A]
where v is the velocity of the reaction and k is a
constant known as the forward rate constant.
Here, since [A] has dimensions of concentration
and d[B]/dt has dimensions of concentration /
time, the dimensions of k will be those of
inverse time, e.g. sec-1.
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More complex cases
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More complicated than this if >1 reactant
involved or if a catalyst whose concentration
influences the production of species B is
present.
If >1 reactant required for making B, then
usually the reaction will be linear in the
concentration of the scarcest reactant and
nearly independent of the concentration of the
more plentiful reactants.
In fact, many enzymes operate by converting a
second-order reaction into a pair of first-order
reactions!
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Bimolecular reaction
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If in the reaction
A+DB
the initial concentrations of [A] and [D] are
comparable, then the reaction rate will be
linear in both [A] and [D]:
d[B]/dt = v = k[A][D] = k[A]1[D]1
i.e. the reaction is first-order in both A and
D, and it’s second-order overall
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Forward and backward
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Rate of reverse reaction may not
be the same as the rate at which
the forward reaction occurs.
If the forward reaction rate of
reaction 1 is designated as k1,
the backward rate typically
designated as k-1.
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Multi-step reactions
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In complex reactions, we may need to keep
track of rates in the forward and reverse
directions of multiple reactions.
Thus in the conversion A  B  C
we can write rate constants
k1, k-1, k2, and k-2
as the rate constants associated with
converting A to B, converting B to A,
converting B to C, and converting C to B.
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[ES]
Michaelis-Menten
kinetics
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t
A very common situation is one in which for
some portion of the time in which a reaction is
being monitored, the concentration of the
enzyme-substrate complex is nearly constant.
Thus in the general reaction
E + S  ES  E + P
where E is the enzyme, S is the substrate, ES is
the enzyme-substrate complex (or "enzymeintermediate complex"), and P is the product
We find that [ES] is nearly constant for a
considerable stretch of time.
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Michaelis-Menten rates
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Rate at which new ES molecules are being
produced in the first forward reaction is
equal to the rate at which ES molecules are
being converted to (E and P) and (E and S).
Formation of ES is first-order in both S and
available [E]
Therefore: rate of formation of ES from left =
vf = k1([E]tot - [ES])[S]
because the enzyme that is already
substrate-bound is unavailable!
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Equating the rates
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We started with the statement that the
rate of formation of ES and the rate of
destruction of it are equal
Rate of disappearance of ES on right
and left is
vd = k-1[ES] + k2[ES] = (k-1+ k2)[ES]
This rate of disappearance should be
equal to the rate of appearance
Under these conditions vf = vd.
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Derivation, continued
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Thus since vf = vd by assumption,
k1([E]tot - [ES])[S] = (k-1+ k2)[ES]
Km  (k-1+ k2)/k1 =
([E]tot - [ES])[S] / [ES]
[ES] = [E]tot [S] / (Km + [S])
But the rate-limiting reaction is the
formation of product: v0 = k2[ES]
Thus v0 = k2[E]tot [S] / (Km + [S])
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Maximum velocity
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What conditions would produce the
maximum velocity?
Answer: very high substrate
concentration ([S] >> [E]tot),
for which all the enzyme would be
bound up with substrate. Thus
under those conditions we get
Vmax = v0 = k2[ES] = k2[E]tot
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Using Vmax in
M-M kinetics
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Thus since
Vmax = k2[E]tot,
v0 = Vmax [S] / (Km+[S])
That’s the famous Michaelis-Menten
equation
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Graphical interpretation
0.01
Michaelis-Menten kinetics
0.009
Initial velocity v0, Ms-1
0.008
0.007
Vmax = 0.01 Ms -1
0.006
Km = 0.03M
0.005
[E]tot = 10 -7M
0.004
kcat = 10 5 s-1
0.003
0.002
0.001
0
0
0.1
0.2
0.3
0.4
0.5
0.6
Substrate conc, M
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0.7
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Physical
meaning of Km
As we can see from the plot, the
velocity is half-maximal when [S] = Km
Trivially derivable: if [S] = Km, then
v0 = Vmax[S] / ([S]+[S]) = Vmax /2
We can turn that around and say that
the Km is defined as the concentration
resulting in half-maximal velocity
Km is a property associated with
binding of S to E, not a property of
turnover
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Michaelis
Constant:
British
Christian
hiphop band
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kcat
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We’ve already discussed what Vmax is;
but it will be larger for high [E]tot than
otherwise.
A quantity we often want is the maximum
velocity independent of how much
enzyme we originally dumped in
That would be kcat = Vmax / [E]tot
Oh wait: that’s just the rate of our ratelimiting step, i.e. kcat = k2
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Physical meaning of kcat
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Describes turnover of substrate to
product:
Number of product molecules produced
per sec per molecule of enzyme
More complex reactions may not have
kcat = k2, but we can often approximate
them that way anyway
Some enzymes very efficient:
kcat > 106 s-1
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Specificity constant, kcat/Km
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kcat/Km measures affinity of enzyme for a
specific substrate: we call it the specificity
constant or the molecular activity for the
enzyme for that particular substrate
Useful in comparing primary substrate to
other substrates (e.g. ethanol vs.
propanol in alcohol dehydrogenase)
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Dimensions
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Km must have dimensions of concentration
(remember it corresponds to the
concentration of substrate that produces
half-maximal velocity)
Vmax must have dimensions of concentration
over time (d[A]/dt)
kcat must have dimensions of inverse time
kcat / Km must have dimensions of inverse
time divided by concentration, i.e.
inverse time * inverse concentration
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Typical units for
kinetic parameters
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Remember the distinction between
dimensions and units!
Km typically measured in mM or µM
Vmax typically measured in mMs-1 or µMs-1
kcat typically measured in s-1
kcat / Km typically measured in s-1M-1
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Kinetic Mechanisms
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If a reaction involves >1 reactant or >1 product,
there may be variations in kinetics that occur
as a result of the order in which substrates are
bound or products are released.
Examine eqns. 13.48, 13.49, 13.50, and the
unnumbered eqn. on p. 430 in G&G, which
depict bisubstrate reactions of various sorts. As
you can see, the possibilities enumerated
include sequential, random, and ping-pong
mechanisms.
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Historical thought
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Biochemists, 1935 - 1970 examined effect
on reaction rates of changing [reactants] and
[enzymes], and deducing the mechanistic
realities from kinetic data.
In recent years other tools have become
available for deriving the same information,
including static and dynamic structural
studies that provide us with slide-shows or
even movies of reaction sequences.
But diagrams like these still help!
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Sequential, ordered
reactions

W.W.Cleland
Substrates, products must bind in
specific order for reaction to complete
A
B
P
Q
_____________________________
E
EA (EAB) (EPQ) EQ E
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Sequential, random reactions
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Substrates can come in in either order, and
products can be released in either order
A
B
P Q
EA
EQ
__
E
(EAB)(EPQ)
E
EB
EP
B
A
Q P
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Ping-pong mechanism
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First substrate enters, is altered, is
released, with change in enzyme
Then second substrate reacts with
altered enzyme, is altered, is released
Enzyme restored to original state
A
P
B
Q
E EA FA F
FB FQ E
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Induced fit
Daniel
Koshland

QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Conformations of enzymes
don't change enormously
when they bind substrates,
but they do change to some
extent. An instance where
the changes are fairly
substantial is the binding of
substrates to kinases.
Cartoon from
textbookofbacteriology.net
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Kinase reactions
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unwanted reaction
ATP + H-O-H ⇒ ADP + Pi
will compete with the desired reaction
ATP + R-O-H ⇒ ADP + R-O-P
Kinases minimize the likelihood of this
unproductive activity by changing
conformation upon binding substrate so that
hydrolysis of ATP cannot occur until the
binding happens.
Illustrates the importance of the order in which
things happen in enzyme function
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iClicker quiz, question 1
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The Michaelis constant Km has
dimensions of
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(a) concentration per unit time
(b) inverse concentration per unit time
(c) concentration
(d) inverse concentration
(e) none of the above
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iClicker quiz question 2
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kcat is a measure of
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(a) substrate binding
(b) turnover
(c) inhibition potential
(d) none of the above
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Hexokinase conformational
changes
G&G Fig. 13.28
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Measurements and
calculations
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The standard Michaelis-Menten
formulation is v0=f([S]), but it’s not linear
in [S]. We seek linearizations of the
equation so that we can find Km and kcat,
and so that we can understand how
various changes affect the reaction.
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Lineweaver-Burk
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Dean
Burk
Simple linearization of Michaelis-Menten:
v0 = Vmax[S]/(Km+[S]). Take reciprocals:
1/v0 = (Km +[S])/(Vmax[S])
= Km /(Vmax[S]) + [S]/(Vmax[S])
1/v0 = (Km/Vmax)*1/[S] + 1/Vmax
Thus a plot of 1/[S] as the independent
Hans
variable vs. 1/v0 as the dependent
Lineweaver
variable will be linear with Y-intercept =
1/Vmax and slope Km/Vmax
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How to use this
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Y-intercept is useful directly:
computeVmax = 1/(Y-intercept)
We can get Km/Vmax from slope and then
use our knowledge of Vmax to get Km; or
X intercept = -1/ Km
… that gets it for us directly!
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Demonstration that the
X-intercept is at -1/Km
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X-intercept means Y = 0
In Lineweaver-Burk plot,
0 = (Km/Vmax)*1/[S] + 1/Vmax
For nonzero 1/Vmax we divide through:
0 = Km /[S] + 1, -1 = Km/[S], [S] = -Km.
But the axis is for 1/[S], so the intercept is
at 1/[S] = -1/ Km.
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Graphical form of L-B
1/v0, s L mol-1
1/Vmax,
s L mol-1
Slope=Km/Vmax
1/[S], M-1
-1/Km, L mol-1
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Are those values to the left of
1/[S] = 0 physical?
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No. It doesn’t make sense to talk about
negative substrate concentrations or
infinite substrate concentrations.
But if we can curve-fit, we can still use
these extrapolations to derive the kinetic
parameters.
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Advantages and
disadvantages of L-B plots
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Easy conceptual reading of Km and Vmax
(but remember to take the reciprocals!)
Suboptimal error analysis
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[S] and v0 values have errors
Error propagation can lead to significant
uncertainty in Km (and Vmax)
Other linearizations available
(see homework)
Better ways of getting Km and Vmax available
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Don’t fall into the trap!
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When you’re calculating Km and Vmax
from Lineweaver-Burk plots,
remember that you need the
reciprocal of the values at the
intercepts
If the X-intercept is -5000 M-1, then
Km = -1/(X-intercept) =(-)(-1/5000 M-1)
= 2*10-4M
Remember that the X intercept is
negative, but Km is positive!
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Sanity checks
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Sanity check #1:
typically 10-7M < Km < 10-2M (table 13.3)
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Typically kcat ~ 0.5 to 107 s-1 (table 13.4),
so for typical [E]tot =10-7M,
Vmax = [E]totkcat = 10-6 Ms-1 to 1 Ms-1
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If you get Vmax or Km values outside of
these ranges, you’ve probably done
something wrong
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iClicker quiz: question 3
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The hexokinase reaction just described
probably operates according to a
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(a) sequential, random mechanism
(b) sequential, ordered mechanism
(c) ping-pong mechanism
(d) none of the above.
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iClicker quiz #4
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If we alter the kinetics of a reaction by
increasing Km but leaving Vmax alone, how
will the L-B plot change?
Answer
X-intercept
a
Moves toward origin Unchanged
b
Moves away from
origin
Unchanged
c
Unchanged
Moves away from
origin
d
Unchanged
Moves toward origin
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Y-intercept
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iClicker question 5
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Enzyme E has a tenfold stronger affinity
for substrate A than for substrate B.
Which of the following is true?
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(a) Km(A) = 10 * Km(B)
(b) Km(A) = 0.1 * Km(B)
(c) Vmax(A) = 10 * Vmax(B)
(d) Vmax(A) = 0.1 * Vmax(B)
(e) None of the above.
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Another physical significance
of Km
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Years of experience have led biochemists to a
general conclusion:
For its preferred substrate, the Km value of an
enzyme is usually within a factor of 50 of the
steady-state concentration of that substrate.
So if we find that Km = 0.2 mM for the primary
substrate of an enzyme, then we expect that the
steady-state concentration of that substrate is
between 4 µM and 10 mM.
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Example:
hexokinase
isozymes
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QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Mutant human type I
hexokinase
PDB 1DGK, 2.8Å
110 kDa monomer
Hexokinase catalyzes
hexose + ATP  hexose-6-P + ADP
Most isozymes of hexokinase prefer glucose;
some also work okay mannose and fructose
Muscle hexokinases have Km ~ 0.1mM so they
work efficiently in blood, where [glucose] ~ 4 mM
Liver glucokinase has Km = 10 mM, which is
around the liver [glucose] and can respond to
fluctuations in liver [glucose]
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Using kinetics to
determine mechanisms
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In a reaction involving substrates A and B, we
hold [B] constant and vary [A].
Then we move to a different [B] and again vary
[A].
Continue through several values of [B]
That gives us a family of Lineweaver-Burk plots
of 1/v0 vs 1/[A]
How those curves appear on a single plot tells
us which kind of mechanism we have.
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L-B plots
for ordered
sequential
reactions
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http://www-biol.paisley.ac.uk/
kinetics/Chapter_4/chapter4_3.html
Plot 1/v0 vs. 1/[A] for various [B] values;
flatter slopes correspond to larger [B]
Lines intersect @ a point
in between X intercept and Y intercept
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L-B plots
for pingpong
reactions
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Again we plot 1/v vs 1/[A] for various [B]
Parallel lines (same kcat/Km);
lower lines correspond to larger [B]
http://www-biol.paisley.ac.uk/kinetics/
Chapter_4/chapter4_3_2.html
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Using exchange reactions
to discern mechanisms
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Example: sucrose phosphorylase and
maltose phosphorylase both cleave
disaccharides and add Pi to one product:
Sucrose + Pi  glucose-1-P + fructose
Maltose + Pi  glucose-1-P + glucose
Try 32P tracers with G-1-P:
G-1-P + 32Pi  Pi + G-1-32Pi
… so what happens with these two
enzymes?
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Sucrose & maltose
phosphorylase
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Sucrose phosphorylase does
catalyze the exchange;
not maltose phosphorylase
Sucrose
This suggests that SucPase uses
phosphorylase
double-displacement reaction;
Bifidobacterium
MalPase uses a single-displacement
113 kDa dimer
Sucrose + E  E-glucose + fructose PDB 1R7A, 1.77Å
E-glucose + Pi  E + glucose-1-P
EC 2.4.1.7
Maltose + E + Pi  Maltose:E:Pi
Maltose:E:Pi  glucose-1P + glucose
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Why study inhibition?
• Let’s look at how enzymes get inhibited.
• At least two reasons to do this:
• We can use inhibition as a probe for
understanding the kinetics and properties of
enzymes in their uninhibited state;
• Many—perhaps most—drugs are inhibitors of
specific enzymes.
• We'll see these two reasons for
understanding inhibition as we work our
way through this topic.
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The concept of inhibition

An enzyme is a biological
catalyst, i.e. a substance that
alters the rate of a reaction
without itself becoming
permanently altered by its
participation in the reaction.

The ability of an enzyme
(particularly a proteinaceous
enzyme) to catalyze a reaction
can be altered by binding small
molecules to it
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Inhibitors and accelerators

Usually these alterations involve a
reduction in the enzyme's ability to
accelerate the reaction; less commonly,
they give rise to an increase in the
enzyme's ability to accelerate a reaction.
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Why more inhibitors
than accelerators?
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Natural selection: if there were small molecules
that can facilitate the enzyme's propensity to
speed up a reaction, nature probably would
have found a way to incorporate those
facilitators into the enzyme over the billions of
years that the enzyme has been available.
Most enzymes are already fairly close to
optimal in their properties; we can readily mess
them up with effectors, but it's more of a
challenge to find ways to make enzymes better
at their jobs.
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Distinctions we can make
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Inhibitors can be reversible or irreversible
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If they’re reversible, where do they bind?
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At the enzyme’s active site
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At a site distant from the active site.
If they’re reversible, to what do they bind?
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To the unliganded enzyme E

To the enzyme-intermediate complex or the
enzyme-substrate complex (ES)
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To both (E or ES)
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Types of inhibitors
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Irreversible
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Inhibitor binds without possibility of release
Usually covalent
Each inhibition event effectively removes a
molecule of enzyme from availability
Reversible
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Usually noncovalent (ionic or van der Waals)
Several kinds
Classifications somewhat superseded by
detailed structure-based knowledge of
mechanisms, but not entirely
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