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Fundamentals I
August 26, 2008
11:00 – 12:00
Dr. Baggott
Basic Concepts in Catalysis
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Goes into a spiel about testing big concepts…he relates this to a forest. You will not be tested on the
little trees in the forest, only on what’s important and will be on national boards.
Unless a tree (small detail) is very important, don’t know it. Important things will be pointed out.
Understand the forest and a couple of trees.
Healthcare is being more scientific and less-practice oriented.
For any chemical reaction, there is an equilibrium constant. The equilibrium constant for AB is
Keq= [B] products/[A] reactants;
[ ]= concentration
Units of concentration: micromoles/liter, millimoles/liter. In biological systems, concentration is not
usually over millimoles/liter.
The keq can be anything; for example
K1= 10 & k-1 = 1
so Keq = [10]/[1] = 10.
The equilibrium constant (Keq) is actually a function of rate constants…just accept this. It is a ratio
of rate constants.
Don’t need to go through the math….math is in the book.
CONCEPT: a catalyst put into this system of A B does NOT change the Keq…it can’t. The keq is a
function of the temperature, the nature of the reaction, and a bunch of other things. The catalyst
can’t change the keq.
If the catalyst can’t change the keq, what does it change? It changes the rate constants. If k1= 10
and k-1= 1 so keq = 10. Add a catalyst and the catalyst will ALWAYS increase both of these
proportionately to give you the same keq.
Example test question: k1 is increased to 1000 by a catalyst, what happens to k-1? The ratio always
has to be 10, so k-1 =100 so the keq will stay fixed at 10. The catalyst accelerates the forward and
backward reaction such that the keq DOES NOT change. Catalyst allows you to speed up the rate of
the reaction.
Said in a different way, rearranging Keq = [B]/[A] = k1/k-1 = 10 (if we throw in a catalyst the
equilibrium constant will stay at 10 regardless…even if k1 or k-1 goes way up or way down…the
catalyst accelerates the forward and backward reaction so that the equilibrium constant is not the
same
Catalysts
o Speed up rate of reaction; they do not change the keq.
o Catalysts are not used up; no chemical alternation (just like a sparkplug in a car; Are
sparkplugs used up every time they are used to catalyze the combustion of gas in your
cylinder? No. They are used over and over again.)
o Catalysts lower the energy of the transition state, from the reactant to the product. This
speeds up the reaction.
o Enzymes are proteins; they are biological catalysts; lower the free energy of the transition
state; bind substrates and products in the active site. Most enzymes utilize a cofactor or a
prosthetic group to perform their activity.
What is meant by lowering the energy of the transition state? Draws picture on the board.
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No one knows what the x-axis, the progress of the reaction, means. Accept this and move
on.
o x-axis – progress of reaction
o y-axis – free energy
o average energy (dashed line) of substrates (higher on graph) and products (lower on graph)
o delta G starts at substrate line and goes to transition state
o adding a catalyst lowers the free energy of the transition state
o In an uncatalyzed reaction, you must go through some sort of transition state.
o The transition state does not resemble the product, does not resemble the reactants, and
does not resemble the substrate.
o In biological systems, the reactant is the substrate. They are equivalent.
o All spontaneous reactions that occur from a pure substrate will go downhill (referring to the
shape of the graph- from high to low). They go from a higher energy to lower energy.
Between the high energy and low energy is the transition state.
o If a catalyst is added, the transition state is lowered; the reaction occurs quickly and more
rapidly.
TEST QUESTION: When an enzyme used acid/base catalysis, what does it do? Lowers free energy
of the transition state no matter what mechanism they use.
KEY CONCEPT: Most enzymes have a molecular weight of 10000 or more. No enzymes are known
with a molecular weight less than 10,000. The reason for this is because it can’t create an active
site. If you have too small of a polypeptide, you can’t create an active site.
Most enzymes are spherical. (Draws a circle with an oval shape inside the circle to represent the
active site.) The active site is a groove and takes up no more than 10-15% of the surface area. THE
ACTIVE SITE IS NOT IN THE CORE OF THE ENZYME, IT’S ON THE SURFACE.
The enzyme positions catalytic groups, usually side chains of amino acids or cofactors, at the active
site.
A substrate in solution cannot catalyze a reaction because it is not close enough the enzyme’s active
site; the substrate must be PHYSICALLY bound to the active site along with its coenzymes and
cofactors to work on conversion of reactants to products.
Lots of enzymes have cofactors; most of them are transition metals, sometimes they are alkali
metals like potassium.
Cofactors = metal ions, mostly transition metal ions.
Coenzymes
o Small, non-protein molecules that help the enzyme with its reaction.
o Coenzymes are BOUND at the active site to help the enzyme do its work; there is an
equilibrium though. Some coenzyme is bound to the active site, some is in free solution.
Some substrates are bound and some are in solution.
o Coenzymes that we will be discussing will be biotin, panotothenic acid, riboflavin, NAD,
pyridoxine, folate, thiamine.
o Sometimes, the coenzymes have prosthetic groups.
o Prosthetic group- means that there is a covalent bond between the coenzyme and the
enzyme. This prevents the coenzyme from diffusing out into the solution.
Apoenzyme: enzyme without its coenzyme or cofactor bound to it.
Holoenzyme: Apoenzyme with bound coenzymes or cofactors. When apoenzyme binds its
coenzyme or cofactor, it becomes a holoenzyme.
Vitamin: a precursor to an enzyme; there are a few exceptions, but this is a good general
hypothesis.
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Patient asks what’s a vitamin? A vitamin is a precursor and usually there is 1 to many
metabolic steps to convert a vitamin into the coenzyme.
o The coenzyme is not ingested by humans; we eat the precursor to the coenzyme (the
vitamin). The body does some metabolism to the vitamin and converts it to a coenzyme.
Reaction types. There are only six types in the body.
1. Oxidation/reduction: a substrate is oxidized and it’s reduced. A substrate that is reduced
becomes oxidized as a product. For everything oxidized, there must be something reduced.
They are all bisubstrate reactions.
2. Transferase: you transfer a group. One group from one metabolite is transferred to another
metabolite. Could be a 1 carbon group, a 2 carbon group, etc.
3. Hydrolases: Just adding or subtracting water from a substrate.
4. Lyases. Removing or creating a double bond. Lyse means remove or create double bonds.
5. Isomerases: optical isomerase (RS configuration), geometrical isomerases (cis trans)
6. Ligases: most important, you form covalent bonds. Almost always use energy from ATP.
Human body does not utilize or do anything with triple bonds.
Also, in this class, we will only be talking about humans. When we get to metabolism, we will be
talking about human or mammalian metabolism.
All enzymes can be named using their substrates plus an ending of oxidase, reductase, transferase,
hydrolase, dehydratase, lyase, isomeras (mutase) or ligase.
Examples:
a. X + ATP  ADP + X – P
b. Name: ATP: X phospho-transferase
c. X(ox) + Y(red) X(red) + Y(ox)
However, we do not use this naming system or the systematic name. Won’t be tested on systematic
naming system.
Three general ways for an enzyme to catalyze a reaction:
 All three of these ways lower the energy of the transition state.
 Known to occur in systems that do not involve enzymes.
 So we know that enzymes utilize mechanisms that are used in organic or inorganic
chemistry.
1. Acid-Base: In the active site of an enzyme, it may have side chains of amino acids
that can serve as an acid or a base.
a. Acid = proton donor
b. Base= proton acceptor
 Side chain of glutamatic acid, if it’s protinated, it is used as an acid.
 If the carboxyl group is ionized, it will become a base.
 Lysine, Arginine
 Histidine: the protinated form, a portion of which occurs at neutral pH of 7,
is a proton donor; the un-protinated is a proton acceptor.
 To a lesser extent, serine and tyrosine.
The big three would be glu, asp (he only said the abbreviations here), lysine and arginine and histadine.
If you go into research, you will see many enzymes that use histadine. Also glutamine and aspartic acid,
and aspartate – you will see many enzymes bind at their active site to facilitate acid base catalysis.
All acid base catalysis does is lower the free energy of the transition state.
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So we have this transition state in the un catalyzed reaction, if we put a proton on it, that lowers the
free energy of the transition state. That is what acid base catalysis does. Base would be to take away a
proton which would lower the free energy of the transition state.
2. Covalent catalysis – you’ve seen that with chromotrysin for a diagrammatic image of what covalent
catalysis is. You have an enzyme that covalently binds its substrate (indicated by 3 dotted lines) during
the catalytic sequence; it binds covalently with its substrate, coverts its substrate to product and then
you get enzyme plus product.
E + S  E lll S  E lll P  E + P
I told you enzymes are never modified – well they are briefly modified during covalent catalysis, but as it
turns out in the whole cycle, they are never modified. You start off with a free enzyme and end up with
a free enzyme. (These represent a small portion of the cycle; they are temporarily covalently made.)
Modified, but the full reaction sequence allows them to regenerate.
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Figure 9-37 (chymotrypsin), Figure 9-38
Chromotrypsin binds a peptide here, forms a covalently modified acyl-chromotrysin; the acyl group is
then released. In the case of chromotrypsin, the covalently modified intermediate here is an acyl
coenzyme.
3. Metal ion – usually you have a transition metal at the active site. It kelates the substrate and the
products to lower the activation energy of the transition state. It is there to kelate or form weak bonds
with the substrate and the products at the active site of the enzyme and facilitates catalysis by lowering
free energy.
Good example of that would be carboxyl peptidase A which uses zinc.
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Figure 9-27 (proposed tetrahedral transition state in the peptide-bond hydrolysis by
carboxpeptidase A…Zinc).
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Bottom line: table 8-5
Bottom line – add up rate enhancement or how much rate constants can be increased by these various
forms of catalysis:
Acid base + covalent+ Metal ion (these are not mutually exclusive; some enzymes use all three of these
things) You would get a 106 increase in the rate, or about a million times. That doesn’t even come close
to most of these things; not that they are not using all of these mechanisms, but there must be a 4th
method.
(Referring to table) 4th mechanism is unique to enzymes; the way we can illustrate the 4th mechanism is
by an organic reaction. We will do this because its fun 
Here is a reaction, an organic chemical reaction (refers to 11.12) to form a cyclic ester. This is a benzene
ring, this is an acid, and this is a hydroxyl group. When the acid and the hydroxyl group condense to
release water, they form an ester. If you look at a space filling molecule – you would see that the
carboxyl group is dangling out in space. It has one, two –methylene groups on it – it is dangling out in
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space. And say this is the hydroxyl group – every once in a while, every once in a while, bang – it hits the
carboxyl group and splits off a molecule of water to create the cyclic ester.
The rate constant here is about 10-6. Very slow because it is just wandering around.
Let’s then put some methyl groups on this molecule and methyl groups on one of the methylene linkage
to the carboxylic acid. And, let’s induce (put strain on this molecule.) What this does is – the carboxyl
group can no longer just rotate around and fling around in space – it is put right next to the hydroxyl
group. Right next to it. Right there. Right there where it should be. The rate constant goes from 6x10-6
to 1.5x10-5. The increase is 2x1011. And that is similar to what enzymes do. They force the substrate
into a confirmation that resembles a transition state or resembles something that has to react and
increases the rate 1012, 1014, 1017 is the most fastest.
Q: What is the number?
A: (Just so we have this on record incase he asks us on the test) Number is not important. Number is
not important. You will never be asked a number. That is not even a tree, that is a leaf. The concept is
important that the enzyme forces the substrate into a confirmation that resembles the transition state.
This won’t be asked on the test; it is just something that organic chemists devised to explain rate
enhancement that an enzyme can do.
Now let’s get to an enzyme; this is not enzyme. Lets see what an enzyme can do. We will use our
“stickase” enzyme. (Draws on board.)
Stickase enzyme breaks sticks. The enzyme is much bigger than the substrate; the transition
state is a bent sit almost broken. If the enzyme has an active site which is complementary to the
substrate that does not force the substrate into a confirmation that resembles a transition state,
the enzyme can not catalyze the reaction. The reaction is to break the stick. So if I have my
hands out, here is the stick – the stick perfectly fits into the active site of the enzyme, I am not
catalyzing this reaction. The stick is just there. What is the other choice? The other choice is
that the active site of the enzyme does not resemble the substrate, it resembles the transition
state. In this case, the transition state is an almost broken stick. So when I force the stick to
bind into this (almost broken) binding site, I force it into a really strained position that
represents the transition state. So if the active site of the enzyme is not this, but this (which he
does by pushing the center of the stick) you broke the stick. That is what we wanted to catalyze.
If he pushes it in like this, he breaks the stick. If he puts it in like this, the stick doesn’t break.
Enzymes don’t break sticks – so what do they do to force the substrate into a transition state. The
substrate in solution is surrounded by a bunch of water. When it binds to the active site of the enzyme,
all the waters are stripped. This destabilizes the enzyme. The substrate is no longer that stable, it wants
to have all the waters of hydration. This is one method – removing solvation shell. (Figure 16.5)
The substrate in solution has all kinds of motion. It has bond motion, bond rotation motion, and
translation motion to 3 different planes – X, Y, and Z. When it is bound to the enzyme, these bond
rotation motions are decreased, or even eliminated. That also produces stress on the substrate. So this
is a highly disordered substrate rotating, translating. This is a highly ordered substrate at the active site
of the enzyme.
Points to image – this is more stressed or more distorted that this.
(Explaining – inside the active site is more stress and distortion.
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Figure 16.6
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o Electrostatic destabilization in ES complex
The substrate may have charged groups on it like ionized carboxylic acids, or protonated groups like
lysine or histadine (which are positively charged at physiological pH.) When that substrate binds to the
enzyme, that enzyme may force like charges to be next to each other. That puts a huge strain on the
substrate. Does it a produce a strain on the enzyme? No – the enzyme is much much bigger than the
substrate. It’s like an ant trying to move a car; the substrate can’t stress the enzyme; the enzyme can
only stress the substrate.
So those are the types of things the enzymes does to force the substrate into a confirmation that
resembles the transition state. (Everything is better in color…you learn better in color, don’t you? No –
not right)
We are going to talk about the concept of an initial velocity, Vi. We are going to use a one substrate to
one product reaction because if you don’t understand the concept of Vi, you are not going to
understand all the stuff I am going to talk to you about during enzyme kinetics.
Vi – sometimes initial rate (will always have initial in front of it). Symbolized by V of i, where i is initial. If
we look at the formation of product,
Conditions for Vi: always that the substrate is in much greater concentration than the enzyme.
You can not do a Vi with the enzyme concentration = to the substrate; substrate has to be
higher.
S-- P
[S] >>> [E}
And this is reasonable because this is the way it is found in our bodies. We are going to let the substrate
be much great [] than the enzyme. We are going to have a test tube of substrate and a mechanism in
which we can measure the concentration of product verses time. And we are going to add the enzyme,
mix it up and follow the formation of product verse time. (Draws graph.)
We added the enzyme at time 0. Enzyme added.
(draw in graph)
What is Vi? Is Vi: ∆P/∆t?
No! The Vi is early portion of the reaction where no more than 5-10% of the substrate has been
converted to product. So it is the early portion of the reaction which is the Vi. And what is Vi? It is just
(in calculus) – the differential appearance of P over the differential change in time.
Vi = ∆P/ ∆t
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Everyone understand. We have to have conditions, has to measured here. Substrate > than enzyme.
First early portion of the reaction.
Okay, for all enzymes, then, if we plot Vi verse [E] (of course here, w/out enzyme the reaction is zero.)
(draw in graph)
If we have enzyme concentration increasing in this direction (upward), we will get a straight line as we
increase enzyme.
As we are increasing the enzyme and holding the substrate constant, isn’t the enzyme concentration
going to be greater than the substrate concentration? No – the enzyme concentration can vary from
1x10-9 – 1x10-8; the substrate can be 1x10-4, which makes it still higher. Even at this concentration of
enzyme, the substrate is still higher. The rate of the reaction will be proportionate to the enzyme
concentration using Vi measurement.
One characteristic of enzymes – the more you add, the faster the reaction goes. This is just like adding
more and more catalyst – the faster the reaction goes.
When we did this, we help substrate constant. Let’s do another experiment where we vary the
substrate concentration, but hold the enzyme concentration constant. So E is constant and we are going
to measure Vi verse variable substrate concentrations. All of our substrate concentrations are going to
much much greater than our enzyme concentrations. We have to keep this substrate concentration
much greater.
This type of experiment gives us 2 and only 2 possibilities.
2! Not 3, not 4 – just 2!
(draw in hyperbole and sigmoid)
Either my experiments will form a right handed hyperbole or a sigmoid curve. Sigmoid can be steep or
shallow and still be a sigmoid curve.
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The first is hyperbolic kinetics and is associated with Michaelis–Menten kinetics; the sigmoid type curve
is associated with an allosteric enzyme. A very large enzyme which will we discuss briefly but not go into
because the mathematics are horrible. Those are the 2 possibilities. Those are the only 2 possibilities.
Plot of the Vi verse substrate concentration can have points on hyperbole or sigmoid curve. Hyperboles
means Michaelis–Menten kinetics, a sigmoid curve means an allosteric enzyme.
Extra stuff:
 The initial velocity of an enzyme catalyzed reaction is directly proportional to the enzyme
concentration.
o A  B d(B)/dt = vi  [E]
o The initial velocity of an enzyme catalyzed reaction plotted verses the substrate
concentration is either a hyperbole of sigmoid
o [S] >>>>>>> [E]
 y-axis – product
 x-axis – time
 linear shape, then parabola (9-12 portion) that straightens out
 5-10% of the substrate that has been converted to product
 the early portion is the initial velocity
o 2 conditions of initial velocity
 measure early portion of reaction
 substrate has to be much greater than enzyme
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[E] = constant
o y-axis – initial velocity
o x-axis – [S]
o only 2 possibilities here
 right handed hyperbola (Michealas-Mikenten)
 sigmoid curve (associated with an allosteric enzyme always)
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