Welcome, we will start our discussion on Enzymes.
We considered what proteins are. We have studied
protein structures the constituents of proteins
that are the amino acids and their properties.What
we are going to start today is our study on
enzymes which will cover about three lectures
and we will be studying about the Enzyme mechanisms
of a few enzymes which will be basically Ribonucleus,
Lysozyme and Serine proteins. As we go along
you will see what they mean.
Now you heard about enzymes and you know that
you need them in the body for digestive purposes,
you need them actually for any process that
goes on in the body. So whether it is your
respiratory process or it is your digestive
process or what ever other process is going
on all of them are enzymatic processes. They
are specific enzymes that are involved which
bring about specific reactions and you will
see how specificity is extremely important
in the way enzymes act.
So basically all Enzymes are protein catalysis.
The protein catalysts are catalytic in nature
but they are proteins. And what we have here
is you know about catalysts they alter the
rate of a chemical reaction without undergoing
a permanent change in structure. So this is
exactly what the protein is going to do also.
We will see how it is exactly is going to
change. It is going to change the rate of
the chemical reaction but the protein itself
is not going to undergo any change even though
it is going to bring about drastic change
in the reaction.
Now this is the typical picture that all of
you have seen, this typical diagram gives
you the free energy of A + B where these are
the reactants and your P + Q are the protons
in these case. So what do you have? You have
a ?G of reaction which is the G of (P) + G
of (Q) - G of (A) and G of (B). It is Gfinal
- Ginitial. And what is this? This is your
transition state your activation states that
this activation barrier has to be overcome
to get you from the reactants to the products.
Now if you look at this, this is a specific
nomenclature that we have put here which is
some times called as the daggers sign. So
we have a ?Gá that tells us this is the energy
of activation for the reactants to go to this
transition state.
Now, if we have a catalyst, how does this
reaction profile change? it changes in such
a manner that further reaction A + B is going
to P + Q. now we have a catalyzed portion
of the reaction or catalyzed reaction in which
our activation energy has decreased.
What does it do in decreasing? The reaction
is catalyzed which means that the rate has
been enhanced. But the ?G itself has nothing
to do with the rate. You have to remember
the ?G as a thermodynamic quantity but the
activation that we are considering is a kinetic
quantity. This distinction is extremely important
when we are considering enzyme catalysis.
Now what do we have in an enzyme? It is basically
a protein, a protein that has a specific structure
to it and what it looks like is if we look
at a specific protein say we have the protein
that is shaped like this, we have particular
cleft in our protein. Now if this is our enzyme
it is going to react with a particular substrate.
What is the substrate is going to do? The
substrate is going to fit into this active
site. Then what do have in the active site?
in the active site we have which is say this
region here is marked with particular residues
or specific residues that are going to act
on the substrate to give you the product but
the enzyme will remain as it is, remember
the enzyme is the catalyst.
So we have the Active site that is lined with
residues and sometimes contains a co-factor.
The Active site residues have several important
properties. So the residues that are in the
active site of enzymes have specific charges
so their charges could be partial, you could
have a dipoles, you could have a helix dipoles
because you have to remember that the specificity
of an enzyme is the reactant that is coming
to act on enzymes so that is going to have
specific groups. The specific groups will
interact with what ever residues lie in the
active site. So we have charge, we have pKa.
If you remember when we studied amino acid
properties I mention the pKa and Hydrophobicity
are very important properties because the
pKa is going to determine whether you have
a protonated or deprotonated or a zwitterionic
form amino acid. And if you remember I mentioned
that the Hestidine amino acid residue has
a pKa of 6 and it lines a lot of active sites
of enzymes. The sole region being the pKa
of Hestidine where we are talking about this
pKa value this pKa of Hestidine close to the
physiological pH of 7.4 so it can act not
only as a base it can also acts as a acid.
Another important property that we learnt
about amino acid was the Hydrophobicity. So
it is also going to be important in the active
site residue lining. Then the Flexibility,
because we should have residues move over
to make way for the substrate to come in and
definitely the reactivity.
So now we will go back to see what other properties
do we have. So look at the enzyme active sites
that we have here, these are three typical
enzymes Chymotrypsin, Trypsin and Elastase
where these are what we have in the active
sites of these proteins. We will see what
these proteins or what these enzymes do as
we proceed in the course but for now what
we have here is we have two Glycines under
Serine and here we have two Glycines and then
Aspartic acid. As we know the Aspartic acid
is an acidic amino acid residue, it has a
negative charge associated to it. So it is
likely that the substrate that it is going
to bind through is going to be the opposite
charged one which is a positive one here.
So what do we see here? We see here a Lysine.
So if this is our enzyme active site such
that it will accommodate a residue that is
counter in charge to the active site residue
and it will fix in the active site. So basically
we should have a fit of the specific substrate
that we are talking about in the enzyme active
site. Again this is another protein called
Elastase or another enzyme rather that we
can call as Elastase. It has two different
groups in its active site and it again acts
on a different type of amino acid. So each
of these active sites have their own properties
in deciding which substrate it is going to
act upon it.
Now if we look at the enzyme classification
in which we have Simple Enzymes and we have
Complex Enzymes. What do we mean by them?
For Simple Enzyme, we have an Enzyme that
is composed of a whole protein and example
of a simple enzyme would be Ribonucleus. Ribonucleus
is composed of a whole protein so it is a
simple enzyme.
If you have a Complex Enzyme, it is not the
protein alone that can act on the substrate
but it will be composed of a protein plus
a relatively small Organic molecule. A Complex
Enzyme is a holoenzyme which it is comprised
of an apoenzyme again which is the protein
part and a prosthetic group or a coenzyme.
So for the Complex Enzymes we have a holoenzyme
that is basically the whole enzyme, the holoenzyme
that is comprised of an apoenzyme and a prosthetic
group or a coenzyme.
What is the prosthetic group? A prosthetic
group is a small organic molecule bound to
the apoenzyme which is a protein by covalent
bonds. So if you have a holoenzyme that contains
a prosthetic group then it is attached to
the protein that is the apoenzyme by covalent
bonds.
In the contrast, when the binding between
the apoenzyme and non-protein component is
non-covalent in nature then it is called a
coenzyme. So that is the difference. So in
this simple very ordinary classification we
would have two types in which one is the Simple
Enzymes that comprises just of whole proteins
for example Ribonucleus.
The other one is Complex Enzymes that have
a protein part and along with a protein part
they have a small organic molecule moiety
either attached to it covalently or just associated
to it. If it is attached covalently then it
is called a prosthetic group or if it is just
a non-covalent association then it is called
a coenzyme.
There are other classifications of enzymes
which you will have to know about. They are
on the type of reaction that they work upon.
You have to recognize that these enzymes are
biological catalysts, they are proteins themselves
and they act on specific substrates.
Now we have classification and we have six
different groups of Enzymes based on what
a substrate or what reaction they are going
to catalyze. Now these are different groups
of Enzymes that we have here. If you look
at the names at the end they all of them are
suffix by as ëesí. So the first one an Oxidoreductases
from the name itself suggest that it would
be a redox type of reaction that it would
catalyze. So Oxidoreductases would act on
the chemical groupings either to add or remove
hydrogen atoms.
So from the name itself you can actually figure
out the function of the specific enzyme. So
what is the Oxidoreductases doing is it involved
in a redox reaction so it acts on groups to
add or remove hydrogen atoms. The next one
is Transferases. What is that going to do?
It is going to transfer, since you are looking
at specific reactions it is going to transfer
a functional group between the donor and the
acceptor so that would be a Transferases.
We have a special name for groups that transfer
phosphate they are called Kinases. Later on
when we will be doing a bioenergetics you
can see how Kinases are extremely important
in the transfer of phosphate from ATP to other
molecules because this ATP is our currency
of energy.
This something that probably you studied in
your 12th standard where you know that ATP
is our energy currency, it provides all the
energy. so there is a specific name for the
transfer of the high energy phosphate and
the phosphate transfers are done by Kinases.
So Transferases would transfer a functional
group between a donor and acceptor molecule
and the Kinases are the special groups of
Transferases that transfer phosphate groups.
Then we have Hydrolases. What does the name
mean? The name mean is hydrolysis. So we have
Hydrolases that would add water across a bond.
in the event of doing that it hydrolysis the
bond.
Then we have Lyases. What do Lyases do? They
add water, ammonia or carbon dioxide across
double bonds, or they remove these elements
to produce double bonds. So the Lyases are
the ones that are going to add the small molecular
moieties across double bonds or remove them
to form double bonds.
Next one is Isomerases. The name itself is
sounding like isomer. So what is that going
to do? It is going to bring about isomerization.
An isomerization reaction is going to be catalyzed
by an Isomerase. So it will carry out or catalyze
the isomerization say from L to D form. So
basically what it is doing? It is shifting
the chemical groups. If you have an asymmetric
carbon atom then you will know you can have
a L form or a D form in which only the difference
is in the position of the groups.
So this Isomerases can change the position
of the groups about the asymmetric carbon
atom giving rise to an either L isomer or
a D isomer so we have the name Isomerases.
The last one is Ligases. The ligation means
linking together. So a Ligase would catalyze
reactions in which two chemical groups are
joined or ligated with the use of energy from
ATP. So a Ligase would catalyze the reaction
where they would be chemical groups that are
joined together.
So these are the six different classes of
Enzymes that are used. We will see later on
what is called the universal enzyme classification.
Each enzyme has its own identity that sort
of looks like an IP address, like a computer
has an IP address your enzyme has an enzyme
classification in which looks exactly like
an IP address like 3.102.11.3 that is an enzyme
classification that would belong. So what
the first number actually is which group it
belongs to here.
So if it were one dot something it would be
Oxidoreductases. So that amounts to be enzyme
classification. So what are the six types
we have? We have a Oxidoreductases, a Transferases,
a Hydrolases, a Lyases, a Isomerases and a
Ligases. And what you need to know is which
reactions are catalyzed by which enzymes and
all you actually need to follow is what the
name means. If you understand what an Ligases
means then you know will that the chemical
groups are joined by this enzyme. If you know
what an Oxidoreductases means then you will
know that it is involved in a redox reaction.
Now we come to this thing that I was talking
about the ?G and the ?Gá. What do we have
here? We have not an A + B or P + Q anymore
we now have an enzyme and a substrate. The
enzyme and the substrate actually form an
enzyme substrate complex.
We will go into the details of exactly how
this is formed or what is actually going on
here but we have an activated enzyme substrate
complex here that is eventually will lead
us to the product. So we have a ?G associated
with that. What is this ?G? This is Thermodynamic
quantity and it is the free energy change
for the system. But the ?Gá that we have
here is the free energy of activation in a
reaction, it is related to the rate, or the
kinetics of the system.
You know that the Thermodynamics does not
provide any information about the rate of
the system. It is just going to give you the
?G thermodynamic quantity that is going to
tell you whether the reaction is spontaneous
or not. So if we just consider two reactions
here say we have a ?G for an uncatalyzed reaction
that corresponds to a 107 KJ and then we have
?Gá. If we put ëáí means it is an activation.
When I say this is there for the catalyzed
reaction is this going to be larger or smaller
than that, smaller because now my reaction
is catalyzed the energy for the activation
has decreased. So if I want to find out how
the rates have changed which reaction or which
equation do I used? Which expression do I
used? I used Arrhenius expression which is
k = A e -?Gá/RT.
So what can I do? I can find a k value. Now
this k is always for rate, K is for equilibrium.
It should never the other way round and you
should never make that mistake. The K is always
for equilibrium and small k is always for
the rate. So we have a k for the uncatalyzed
reaction, we have a k for catalyzed reaction.
How do I determine it? All I have to do is
plug in the values. Here A and if we consider
the pre exponential factor to be the same
so I have to be careful about units here which
are kilojoules 8.314 and by default we usually
use 298.
What about my catalyzed? A e- 46000/8.314x298
remember to use the right R also 298. So I
can find out how the rate has increased for
the catalyzed reaction as compared to the
uncatalyzed reaction. It turns out that this
is approximately 5 x 1010. So I am saying
that when I have reduced this energy of activation
I have increased the rate 1010 folds basically.
So I have a uncatalyzed reaction a catalyzed
reaction and then I have associated to it.
The change in the rate is due to reaction
being catalyzed whether the catalyst happens
to be an enzyme or any other catalyst it is
the same thing.
So what do we have here? We have our enzyme
plus substrate and we have our enzyme plus
product. So what is our catalyst? Our catalyst
is the enzyme. If you notice that what do
we have in the beginning, in the very first
I showed you that we had A + B going to P
+ Q, here I have E + S going to E + P. So
E is the catalyst that is transforming the
substrate to the product but itself it is
remaining unchanged. So it is not like a normal
reaction where we would have A + B going to
P + Q here it is E + S going to E + P where
E remains as it is in this reaction.
What do Enzymes do? Enzymes accelerate reaction
by lowering the free energy of activation
and they do this by binding the transition
state of the reaction better than the substrate.
Enzymes exhibit saturation kinetics.Now how
do I determine the Catalytic activity of an
Enzyme? There must be a measure that tells
us whether this Enzyme is good or bad. So
the measure of that activity of an enzyme
is called its turnover number.
How would you define a catalytic efficiency?
An efficient enzyme would be one that could
transform more of the substrate product that
would be an efficiency of an enzyme. So the
turnover number is defined as the maximum
number of moles of substrate that can be converted
to product per mole of catalytic site per
second. What do I mean by a catalytic site?
Remember I showed you some diagrams where
we have an enzyme active site, in that enzyme
active site there are substrates that bind
to the enzyme active catalytic site or the
active site and they convert the substrate
into product. As they do that the enzyme retains
it structure and activity so that it can convert
another substrate to product.
But there is a turnover number that is going
to tell us the catalytic efficiency and that
efficiency is measured by how many number
of moles we can transfer of the substrate
to the product, the conversion per mole of
catalytic site per second because you have
to remember it is a rate that we are considering.
So you are looking at the maximum number of
moles of substrate that can be converted to
product per mole of catalytic site per second.
Now, what happens to a protein when we heat
it? It denatures and enzyme is a protein.
So what do you have for your normal reactions
when you increase the temperature? Certainly
the rate increases.
For enzymes, you would expect that there would
be an optimum temperature at which the enzyme
is going to act because beyond that it is
going to get destroyed. If it gets destroyed
then the enzyme active site is destroyed as
a result of that what is going to happen is
the substrate cannot bind and obviously the
product cannot formed.
so we have an optimum temperature that if
we do a percent activity versus temperature
curve like this then we are going to have
an optimum temperature for the highest activity
of the enzyme. And you realize that at low
temperatures the enzyme is not going to function,
at high temperatures also the enzyme is not
going to function because it has to have and
optimum activity especially of course the
activity is also going to considering the
reactions that take place in our body then
it is going to be an optimum of 37∫C.
Then how about pH? If we add too much acid
to the protein then it is going to depend
upon the amino acid composition of the protein.
We have to remember that enzyme is a protein
so the amino acid composition of the active
site is extremely important in determining
which pH is going to act that. So if you look
at here there are certain graphs here corresponding
to the relative activity verses the pH. Here
the first one is Pepsin. if you remember you
studied in school about Enzymes, if you had
biology where Pepsin acts in our stomach where
the acid is at low pH. The pH of the acid
in your stomach is 2 so Pepsin acts at pH
= 2. It means that there are certain groups
in Pepsin that are active at a pH of 2.
If we go to any other enzyme usually there
are enzymes acted very low pH or very high
pH the normal activity that you would see
for example here you would see for Trypsin
where you would found an optimum around 7
to 7.5 That would be the pH optimum for a
usual enzyme. Pepsin acts in the stomach where
there is a high level of acidity.
So the effect of pH on its optimum activity
is at a pH of 2. In this case for example
Trypsin the optimum pH activity is around
a pH of 7. That is the normal case that we
would observe.Now, we will get into the Enzyme
Kinetics. What is happening here is we have
an Enzyme plus a Substrate that is going to
form the Enzyme substrate complex. We have
a pre equilibrium step in this Enzyme substrate
complex. This step is our pre equilibrium
step. This is a very simplistic way of writing
an Enzyme Kinetic reaction but nevertheless
we will consider the kinetics of this reaction
for the definition of certain terminologies.
Now the first reaction we have here is a pre
equilibrium which means it has a forward rate
constant and a backward rate constant. If
we consider the forward rate constant to be
k1 and the reverse rate constant to be k ñ1
and this to be k2 then we have our enzyme
plus substrate.
So basically if we just draw what it looks
like? We have an Enzyme and we have Substrate
then they will form an Enzyme substrate complex.
Then what ever reaction is to occur or the
active site occurs then we get back the enzyme
but our product looks likes. Because now the
substrate has changed to the product but our
enzyme remains the same i.e., what is unique
about these catalyst.
You have to understand that these are all
chemical reactions that are going on in the
active site but it reverts back to where it
has to be after the reaction in the event
forming the product. Now we will see how we
can get to expression for Enzyme Kinetics.
So we have E + S in a k1 and a kñ1 going
to an ES complex forming E + P.
Now I am going to define a velocity for the
reaction. A velocity for the reaction is going
to be v it is going to ?, this is going to
be k2 and the velocity of the reaction is
going to be rate at which the products are
formed. So it is going to be equal to k2 [ES].
What is the rate at which products are formed?
It is d[P]/dt. The d[P]/dt is the rate at
which products are formed. So I have my d[P]/dt
which is actually nothing but the velocity.
Now what do I want? I want an expression for
ES, I have to consider d[ES]/dt to get an
expression for ES. So now if I have a steady
state which you have studied what it telling
is the rate of its destruction is equal to
the rate of its formation. So the production
and its destruction are at the same rate.
So if I apply that to the expression that
I have here, what can I write for d[ES]/dt?
I can write k1 enzyme concentration [E] substrate
concentration [S] ñ k2[ES] why minus is because
it is being destroyed in the backward reaction
so it is ñ k2[ES]. So I have an expression
that tells me that this d[ES]/dt is going
to be k1[ES] because it is formed by k1 it
is destroyed by kñ1 and k2.Now if I am to
apply the steady states then I have to equate
these two to zero.
Why because its rate of formation is equal
to the rate of destruction. So there is no
change of the concentration of the Enzyme
substrate complex with time. If there is no
change with time then this quantity becomes
zero because I am at a steady state.
Now if I am at a steady state therefore I
will get a new expression for [ES]. Now what
expression do I get for [ES]? It is going
to 
I have d[ES]/dt was k1[E][S] ñ kñ1[ES] ñ
k2[ES]. What did I do? I equated this two
zero so I can get a value for [ES] which is
going to be k1[E][S] / .
Now you have to remember that when you are
doing this reaction there are certain quantities
that you have measured, it is your experiment.
What do you know? You know the amount of the
substrate that you have added, you know the
amount of total Enzyme that you have added
but you do know at what time how much amount
Enzyme substrate is actually being formed.
So what you know is you know these quantities,
you know what S is and you also know what
the total enzyme concentration is. Now at
any time T the total enzyme concentration
is going to be the what ever free Enzyme is
left plus what ever has formed as the Enzyme
substrate complex because the enzyme cannot
go anywhere else, it either is free or it
has formed an Enzyme substrate complex. So
that is what our ET amounts to. So if I change
the expression, instead of writing the Enzyme
concentration I can write ET ñ [ES]. I can
write this for the free enzyme where do I
have this quantity in my expression here.
So this is where we are going to substitute
this. So if we do that what do we get 
is (kñ1+ k2) [ES] = k1 (ET ñ [ES]) [S] which
is what I have. So here I have another expression
for [ES] which I can take over on to the left
hand side. So if I do that I will amount on
the right hand side to k1 [ET] [S]. if take
this [ES] part over on to the left hand side
then what I will end up with on the right
hand side is a k1 [ET] [S] and what I will
have on the left hand side is a (kñ1 + k2
+ k1 [S]) / [ES] and this going to be equal
to k1 [ET][S].
Now here I am going to use some more information.
The velocity of the reaction where it was
k2 [ES] and here I am going to use another
term. If this is k2 [ES] then I can define
Vmax. The Vmax is the maximum velocity that
the enzyme can attain and that is the measure
of total Enzyme concentration because of all
the Enzyme want to react then it would attain
the maximum velocity possible. This is something
you should get very clear.
If you considering the maximal velocity it
is highest reaction rate that can be attained
because all of you enzyme is going to be saturated
with the substrate. So it is going to be k2
ET. So basically if we work out this whole
reaction by substituting ? as k2[ES] and putting
Vmax as k2 [ET] if we multiply both sides
by k2 because we are going to put ? instead
of k2[ES].
so I am going have on the left hand side (kñ1+
k1 [S]) [ES] + ? because we 
are 
looking at this and this together with the
S and the k2 and the [ES] form the ?. and
what do we have on the right hand side is
we still have k1 ET [S]. Now if I multiply
both sides by k2, I can put a ?max form here.
Eventually what we are going to get is we
are going to define another quantity that
is going to be a combination of the rate constants.
it is the Michaelis constant that is equal
to the term (kñ1+ k2) / k1. This is another
quantity where we are going to use. And the
final expression after you do all the algebra
is going to work out to ? = ?max[S] / (KM+[S]).
This is your final expression that you are
going to get after you do all the algebra.
Now if I want to have a physical interpretation
of what KM value actually is then what you
can get is the expression is ? = ?max[S] / (KM+[S]).
Now if I make this velocity ?max/2 then if
I just say that the velocity is half attained
the ?max is half attained then I have ?max/2
= ?max[S] / (KM+[S]) and if you work out the
algebra you will get this specific case as
KM = [S].
So what does this mean? It means the KM or
the Michaelis constant is the substrate concentration
when half the maximum velocity is attained.
So all you have to do is just do the algebra.
Now we look at our Enzyme Kinetics we work
through the whole set of expressions. What
do we have from the expressions? We get Michaelis-
Menten kinetics and this is the expression
that we got there and ? = ?max[S] / KM +[S].
What are the features of Enzymes Kinetics
or Michaelis -Menten kinetics? It assumes
the formation of an enzyme -substrate complex.
There is pre equilibrium where the enzyme
-substrate complex is in equilibrium with
the free enzyme.
Then the breakdown of the enzyme is slower
than the formation of ES and the breakdown
of ES to re-form E and S. So basically what
we are looking at is we are looking at a picture
like this which exactly what happens. We have
the velocity and we have the substrate concentration,
as you increase the substrate concentration.
You are going to have saturation. Why? You
are talking about an Enzyme which is a protein
so it has a minimum number of active sites,
you had a limited amount. So if even though
you increase the substrate concentration the
enzyme has a definite capacity to accept the
substrate. So at a time you are going to attain
saturation and at high substrate concentration
the velocity is going to be independent of
the substrate concentration where you have
zero ordered kinetics. Initially you have
first ordered kinetics at low substrate concentrations
where ? is proportional to [S]. And what is
KM? This is ?max, this is half of ?max, this
is KM. it is a substrate concentration at
which the maximum velocity is reached.
So what we did today was we understand what
enzyme kinetics is what enzymes actually do
and their specific classifications based on
the types of reactions that they catalyze.
Next class we will consider more of this Enzyme
kinetics and how we can inhibit the kinetics
of enzymes. Thank you!
