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Lecture - 13

Introduction to Enzymes and its Kinetics

We have just discussed the different techniques that are used to separate the proteins and
before that, we have also discussed different techniques that are used to identify the structure
of a protein. Now, let us do one problem to clarify any doubt to further reinforce our concept.
(Refer Slide Time: 00:47)

Now, there is a problem that, you are trying to determine the sequence of a protein that you
know is pure. I will give the most likely explanation for each of the following experimental
observations. Now, what are the observations?
The Sanger’s reagent (2,4-dinitro fluorobenzene or FDNB) identifies two N terminal
residues, alanine and leucine in roughly equal amounts. So, what does it indicate? It indicates
that the protein at least has two chains where one chain has got alanine as the N terminal
amino acid; and the other chain there has lecine as the N terminal amino acid. This is your X
chain and that is Y chain.

And we will come back to the topic that how these are tied together as per the problem itself.
The next information is that your protein has an apparent molecular weight of 80,000 as
determined by SDS-polyacrylamide gel electrophoresis. After treatment of the protein with
performic acid reveals two proteins of molecular weight 35,000 and 45,000. Formic acid
which is further oxidized is known as performic acid (HCO3H). We must remember that SDS
basically denatures the protein.
And then after denaturation, you do the gel electrophoresis and then you also run in a
molecular weight ladder. And from the molecular weight ladder, you can say that what is the
molecular weight of your protein. So, SDS PAGE says that the molecular weight is 80,000.
On the other hand, now you treat the protein with performic acid. The point we have not said
anything yet that is if there are two chains of proteins and if they are connected by disulfide
linkages; that means, there is a cysteine here.
So, that is usually the way of connecting two chains together through disulfide bonds. Now,
the question is, if you treat this one with performic acid(HCO3H) then this disulfide bond
breaks. And, this becomes a sulphonic acid and the other chain also becomes a sulphonic
acid; that means, the chains are now separated that is the most important thing.
So, what happens when you did this SDS PAGE? At that time, you get only one band 80,000
that corresponds to the denatured protein. But, when you treated with performic acid, then
you see that this is actually made up of two chains having molecular weights 45,000 and
35,000. That means, suppose the X is 45,000 then Y is 35,000 or vice versa.
So, now you have a rough estimate that after performic acid treatment, you get these two
bands; that means, these 45,000 and the 35,000 molecular weight chains are connected by at
least one disulfide linkage. There may be more than one disulfide; the HCO3H treatment does
not say how many disulfides are there. But, at least the minimum structure that you can write
here is that these are the two chains and this is alanine and this is leucine. And suppose this
is the 45,000 and this is the 35,000, but I again repeat this could be the other way around.
This could be 35,000 and that could be 45,000, because the information is not given that what
is the N terminal sequence of the 35,000 chain or what is the N terminal sequence of the
45,000 chain; that is not shown. So, we have to write this, or the other alternative
combination. The last one is size exclusion chromatography, that is called gel filtration. we
have learnt that. Size exclusion chromatography indicates that the native protein has an

apparent molecular weight of 160,000. Now, size exclusion chromatography does not disrupt
any structure of the protein. Whatever protein structure is there in the native state, that is
retained in size exclusion chromatography.
So, that means, if the protein is multimeric in nature, then that multimer will remain as one
unit. And when you do size exclusion chromatography depending on the rate of elusion and
the cut value; it is revealed that the native structure of the protein has a molecular weight of
160,000.I told you that the gel has a cut point that up to this molecular weight it will allow to
be retained; or this will be allowed to be passed through. So, you get 160,000 as the
molecular weight of the native protein. This immediately tells you that the protein must be
composed of two sets of this alanine and leucine structure. So, the actual protein must be
having a chain with alanine and another chain with leucine as their N terminal amino acids;
then these chains are connected by this disulfide linkage. And then these two proteins are
having some interaction to make a dimeric ensemble.
So, basically this is a tetramer, but dimer of hetero dimers;this is a heterodimer and then this
heterodimer is again dimerized. So, actual protein is a dimer of a heterodimer; that means,
this is a tetramer. So, the actual protein is a tetramer. This is how the development of
different techniques ultimately helped us to determine the structure of the protein. Suppose if
you only did the size exclusion chromatography, you would not have been able to know that
this protein existed in tetrameric structure. So, you have to do SDS PAGE, you have to also
do the performic acid treatment, if there are disulfide bonds that you need to break. And then
your Sanger’s method actually told you initially that there were two strands. So, that
completes our separation techniques and characterization techniques of proteins.

(Refer Slide Time: 08:01)

Now, we will go to our next topic that is the functional aspect of proteins. We have
completed the structural aspect, now we will go to the functional aspect of proteins. Proteins
have different functions in our body: one of the major function is that they act as the catalysts
and they catalyze various reactions. Such proteins are called enzymes. Some proteins act only
as receptors. Receptor means that you have a protein like this, and some molecule binds here
and then that sends a signal after binding. Receptors do not really catalyze any reaction, it is
only a site for binding and that binding is associated with generation of a signal.
Like what is shown here is rhodopsin, rhodopsin is the protein that is present in our eyes;
when light falls on it, it induces a change in the structure of rhodopsin and that creates the
signal. And finally, the signal is processed in the brain and we see the image.
Proteins are present in the muscles; even the silk that makes the beautiful sarees are also
made from proteins. Proteins are also present in storage materials ferritin; andtransport
proteins which transport (like hemoglobin transports oxygen to the cells). Proteins give us
immunity against the invading organisms that affect our body. These are called the antibodies
or immunoglobulins. Then they can be also hormones like insulin which metabolize the
glucose. There are structural proteins present in muscles; they are made up of proteins like
collagen. We will right now discuss the enzymes which are basically proteins that are acting
as catalysts. Later on, we will come back during our medicinal chemistry section to the
receptor chemistry.

The similarity between the the receptor and the enzyme is that in both, there are sites which
are called active sites pockets, where small molecules bind. But, in receptor there is no
reaction, there is a signal that is generated and in enzymes there is some reaction that takes
place so, that is the difference.
(Refer Slide Time: 11:27)

Now, let us go to the next slide. Enzymes are macro molecular biological catalysts. They
catalyze the reaction; catalyze means accelerating the reaction. The enzymes are made up of
proteins that already I have told, act like ,catalysts; enzymes increase the reaction rate by
lowering the activation energy. That means, we know that in any reaction if you want to
speed up the reaction, a catalyst only speeds of the reaction both ways: the forward and the
backward reaction. Ultimately, allowing the reaction to come to the equilibrium at a much
faster rate than the uncatalyzed reaction.
But, a catalyst cannot give higher yield that is a mis-concept that we have; catalysts only
lowers activation barrier which is directly related to the rate processes of the forward and the
backward reactions. This is the energy diagram. For the reactant, in case of biochemistry,
generally we do not write R. Instead of R, generally that is called substrate, the reactant is
called the substrate and the product is shown by P. So, if it is uncatalyzed, suppose this is
your activation energy; this is the pathway that the reaction takes. That means, now the
activation energy for the forward reaction is this.

When you have the catalyzed reaction, then what happens? It takes another path where the
activation energy is much lower. So, the enzymes acting as the catalyst, lower the activation
barrier for the reaction and so, that the reaction becomes faster.
(Refer Slide Time: 13:37)

There are still some interesting details here. The mechanism of enzyme catalysis will be
discussed now. Suppose enzyme has this type of shape. And this is what is called the active
site, where the substrate molecule comes and binds. And this ES complex is formed; then the
ES complex breaks into product. And then the enzyme is regenerated, to form the ES
complex, you have E plus the substrate. So, the substrate goes and binds to form the ES
complex. For simplicity, we are not considering the reversibility in this step, that could be
also reversible.
The initial analysis of enzyme catalyzed reaction was done by Michaelis and Menten and for
simplicity, they did not consider the reversibility of that step. First it is the enzyme plus
substrate to form the enzyme substrate complex; that is a reversible step, k1 and and k-1 are
the rate constants for the forward and backward reactions respectively; k2 is the rate constant
for the step involving the decomposition of the ES complex to yield the product. So, this k2
will ultimately give the product, the rate process for the formation of the breakdown of the
ES complex into the product. Now, because its catalysis, we know that the activation barrier
is lowered. But, interestingly when substrate binds to the enzyme, the energy of the substrate
also is decreased because the substrate is stabilized.

It is like when a baby is alone, the baby is crying with more energy, but as soon as the mother
comes and takes the baby into her lap, the baby stops crying; so baby is stabilized now. So,
the enzyme is like the mother, substrate is like the baby. Earlier the substrate energy was this
and now as soon as its complexes with the enzyme, it forms this ES complex.
And that energy has to be lower than this because it is stabilizing the whole process as there
are stabilizing interactions. Why the substrate will go here? Because, there are weak
interactions that stabilizes this complex like hydrogen bonding, or a salt bridge formation, or
electrostatic attactions, or hydrophobic interactions.
So, all these weak interactions play a part in this binding. But that has to cross an activation
barrier because every process needs an activation energy. So, there is an activation barrier
like this, but that is a very small activation barrier to form the ES complex. That means, your
original energy level now is being lowered. Then that goes into the ES complex which is the
transition state; where the bond starts breaking and new bonds are being made. So, this is
basically the activated complex or the transition state.
And then that goes to the product; at that point the product is still bound to the active site of
the enzyme and ultimately the product is released. So, this also lowers the actual energy of
this product that is released. Now what if I ask that between this substrate, product and the
transition state, which one is most stabilized by the enzyme? Suppose this is the substrate,so
the substrate binds and then it goes into the transition state. This is the transition state; where
something is happening on the substrate. So, now this will be TS and the transition state is
stabilized by the enzyme and then the transition state will first go into E and P. There will be
some bonding, definitely some stability; and then P goes out from the active site of the
enzyme and this makes the active site free for the next molecule which comes and binds.
Now, out of these three: S, then TS and the product P, which one is most stabilized by the
enzyme? The answer is that it is the transition state that is most stabilized by the enzyme,
because this goes down a little bit and this also goes down slightly.
They are also stabilized, but to have the lowering of the activation barrier, you need much
more lowering of the energy of the transition state as compared to the uncatalyzed reaction.
So, that is ultimately what I am trying to say that the enzyme active site must be most
complimentary to the transition state followed by the substrate, and the last is the product;
because product is released very easily from the enzyme.

So, earlier usually people used to think that the enzyme stabilizes the substrate, substrate goes
and binds. Yes it stabilizes, but then when it goes to the transition state, that is the highest
extent of stabilization of this. We will go to this aspect later on when we deal with synthetic
biology; that is why I spent so, much time discussing this one.
(Refer Slide Time: 19:51)

(Refer Slide Time: 19:59)

For enzyme, some structure is given here. The enzymes are proteins; this is a complex 3-D
structure of an enzyme.

(Refer Slide Time: 20:07)

But, let us go to the different classes of enzymes. It is like different organic reactions which
are classified according to their pathways and reagents. Like we have nucleophilic addition;
we have electrophilic addition; then substitution reaction; elimination reaction. Similarly,
based on the type of reaction, the enzymes are classified into six groups. So, whatever
enzymes we have in the living world, they will do one of these reactions. The first class is
what is known as oxidoreductase, the second one what is transferases. Oxidoreductase carry
out redox processes; like lactate dehydrogenasecarries out the conversion of lactic acid to
pyruvic acid.
What is lactic acid? Lactic acid is this, and that is getting converted to pyruvic acid by lactate
dehydrogenase. Then the second one is transferase; that means, they transfer a chemical
group from one place to another like an amine transferase; the amine group of one amino acid
is transferred to another substrate which lacks the amine and puts the amine there. So, they
are called transferases. Then comes the hydrolase. You already have some idea like the
protease which hydrolyze the peptide bond. There could be other types of hydrolase. So, that
means, when your bonds are broken by water, then those enzymes are called hydrolases.
So, a specific group is created for that; and there are many enzymes. Lysozyme is one of
them; we have chymotrypsin, trypsin, pepsin so, many. Why is lysozyme written here? It is
quite interesting that the lysozyme is present in our tears; lysozyme is an enzyme which
breaks the peptide bonds of proteins. And this lysozyme gives us a very good antibacterial

activity in the tear which is the water that is constantly washing our eyes. Thus hydrolases are
the ones which break a bond by water.
Then you have lysase. Lysases are enzymes responsible for non-hydrolytic bond cleavage.
So, if you have a bond cleavage which is done by something other than water, then you have
lysase type of enzymes. Then you have isomerase; isomerase is involved in some
isomerizations like say L to D conversion; an amino acid going from L to D, that is an
isomerization reaction. And then finally, there is a ligase; ligase is basically joining of two
things together (like synthesis); that means, your synthesis of new covalent bonds between
substrates using ATP hydrolysis; because, for any synthesis you need some energy and that
energy is provided by this adenosine triphosphate.
Now, it is important point to note that if I ask you that what are the different types of
enzymes? If you change the order of these classifications, then there is a problem. Because,
the International Union of Biochemistry says that if you have an enzyme which is
oxidoreductase, that should be given a specific number like our IUPAC nomenclature in
organic chemistry.
So, there is a nomenclature system for enzymes. So, any enzyme which is oxidoreductase,
that will get an initial number 1. And then the species from where it is isolated, and also how
many are already isolated from that species. And then the number is put like that; say see if I
say 3.1.1, the number 3 means that it belongs to hydrolase.
So that means, you cannot change the order that is written here, that is very important. So, if
some something is said at 6.1.2 so that means, it is a ligase.

(Refer Slide Time: 24:55)

Now, come back to the active site and what is the mechanism of this enzyme catalyzed
reaction? What is the mathematical expression of an enzyme catalyzed reaction? So, for
enzymes, as I said, this is , one type of active site and this is your substrate. See the substrate
has a geometry which is complementary to the active site of the enzyme. But again I warn
you that this shows as if the substrate is the best fit, but that is not so; it is a transition state
which has the best fit with the active site of the enzyme.
(Refer Slide Time: 25:25)

This is called the lock and key hypothesis. The enzyme mechanism was earlier described like
this that as if you have this active site which is a kind of a lock and the substrate is like a key.
As the key goes and fits into the lock, the substrate goes and binds to the enzyme. As you
rotate the key, the lock opens up. So, similar to this lock and key complex, this is the enzyme
and substrate complex.
If you want to open the lock, you have to spend some energy on ths lock and key; similarly if
you want to get a product you have to do something on this substrate. So, that is the the
homology between this lock and key complex and the enzyme-substrate complex .
(Refer Slide Time: 26:17)

Now, sometimes or actually frequently it is found that initially this active site may not be
perfectly complementary to the substrate. So, what happens? Like we have people of
different body weights and different dimensions. So, if we have a chair, initially the chair
may not fit properly according to your size. But what we do? We ultimately try to feel
comfortable by pushing our self inside the chair, so that ultimately there is a best fit between
me and the chair. The same thing happens here; the active site may not really have an exact
complementarity with the substrate. But, as the substrate goes inside the active site, the
enzyme adapts to fit the substrate maintaining the perfect complementarity.
So, this is called induced fit model of enzyme and substrate. So, it is not that the enzyme is
already having a perfectly tuned active site. It is like going to a shopping mall and buying a

shirt. The shirt may not fit that well. So, what we do? We go to the tailor and he cuts little bit
here and there and then it fits to the body very well. So, the same thing happens here, here the
enzyme itself is a kind of a tailor. So, it itself adjusts and then tries to fit the substrate into it.
(Refer Slide Time: 27:55)

Now, let us go to the Michaelis Menten equation. Before we go into that, it should be known
that some of the enzymes (of the six classes that have been mentioned) require a small
molecule for their activity to be shown. And those small molecules are called he cofactors.
So, cofactor is a non-protein, chemical compound that is bound either tightly or loosely; both
are possible for an enzyme; and they are required for catalysis. See that means, it is like you
have a big house, everything you have inside; there is refrigerator, TV everything; you have
gone outside and then you forgot or you lost your key.
So, if you lose your key, you cannot enter into the house. So, these cofactors are like the entry
point. The protein is made by the living system and if the cofactor is not there, the protein is
basically useless. Sometimes very simple things like a screw can be the most vital component
for some machine to be operational. Cofactors are small molecules which some proteins need
to show their catalytic activity. There are two types of cofactors: coenzymes and prosthetic
groups. If the cofactor is organic in nature, then that is called coenzyme.
On the other hand, prosthetic group is basically the cofactor which is tightly bound to the
protein.

(Refer Slide Time: 29:51)

Coenzyme is the non-protein component, loosely bound to the apoenzyme; apoenzyme means
the enzyme devoid of the cofactor. See we have said that there is an enzyme like this, there is
the active site here, but to do the catalysis, you need this small molecule (cofactor).
Now, there are many cases where cofactor is actually outside the enzyme. But, at the time of
reaction that comes and binds to the enzyme and then the substrate comes and the reaction
takes place. In some cases, the cofactor may already be bound to the enzyme through a
covalent bond. The prosthetic groups are the non-protein components, tightly bound to the
apoenzyme by covalent bonds. So, when will the cofactor be called a prosthetic group? When
it is tightly bound (attached through a covalent bond) to the enzyme.
Now, if you can strip this cofactor off, then this enzyme without the cofactor is called
apoenzyme. And when this whole thing is there; that means, the cofactor is attached to the
enzyme, then that is called the holoenzyme. If the cofactor is organic then that is called also
called coenzyme. Along with organic molecules, cofactor also includes metal ions; because
sometimes metal ions can also be very critical for some enzymes to show their activity.

(Refer Slide Time: 31:37)

Now, let us go to this famous equation called the Michaelis Menten equation. So, the reaction
that is there, that E plus S reversibly gives ES and then this ES going to the E plus product
(P). The enzyme becomes free and that can again bind with a new substrate molecule. So, this
is the kind of equation that we use.
(Refer Slide Time: 32:13)

And then, if you solve this, the first step is E plus S forming the ES complex. So, you have
two rate constants k1 and k-1, k-1 is the dissociation of the ES complex and k1 is the association

of E and S. k2 corresponds to the breakage of this ES complex. Usually we put it in a third
bracket because this is a transitory; it does not ultimately remain in the medium. It is
produced and then broken down either to product or again or it again reverts back to the
substrate and the enzyme.
So, we may solve this equation by assuming the steady state approximation. What is the
steady state approximation? It states that initially that substrate will go and bind to the
enzyme and then it will also dissociate, this is a reversible process and then after some time if
state is reached where the rate of dissociation of ES into E plus S and the rate of association
of E and S to form the ES will be same. If that be the case; that means, the change of
concentration of ES with time is equal to 0 because, if something is constant; that means, the
differentiation is 0. To be more precise, we can say that the rate of dissociation of ES into E
plus S and its dissociation to E and P is equal to the rate of association of E and S to form the
ES complex.
(Refer Slide Time: 33:41)

So, you apply this steady state and finally, you come to an equation which is the one shown
here. Initially you get this V0
.
In enzyme kinetics, as soon as you add the substrate, the
reaction starts. So, V0 is the initial rate; and that is given by gradient of the initial linear line
or initial straight line..

So, this V0 is the initial velocity. ou cannot have a velocity at time 0, it is very difficult to
measure that. So, what you can do is that you determine the velocity at some points and say T
= 1 minute then 2 minute then 3 minute. And as long as it is linear you can determine the
gradient and get the V0. So, if the V0 is there, then you can solve this equation. I am not going
into the solution; you can do it yourself or refer to any standard test book of this. V0
ultimately becomes k2(where k2 represents the breakage of the ES into E plus product)
multiplied with S0, (where S0 is the initial substrate concentration) and E0 (which is the initial
enzyme concentration, how much enzyme you have taken) divided by S0 plus Km .
Before that ,the equation is like this; initially you get V0 equal to k1k2 S0 multiplied with E0
divided by k1 multiplied to S0 plus k-1 and k2. And finally, if you divide the whole expression
by k1, you get this k2S0E0S0 plus k-1 plus k2 divided by k1. And this is what is called the
Michaelis Menten constant (Km). And so, Km is basically k-1 (that means, the rate constant
corresponding to breakage of the ES complex to E ans S) plus k2 (that means, the rate of
formation of the product from ES) divided by k1 (that is the rate constant for the association
of E and S to give ES).
This Km is the Michaelis Menten constant and this equation is what is called Michaelis
Menten equation.
(Refer Slide Time: 36:31)

Remember this is also an interesting issue that Km is basically the ratio of rate constants. But,
some of these rate constants are unimolecular rate constants, some are bimolecular rate
constants. So, overall it gets a unit which is a unit of concentration because, you have to
ultimately add a concentration to this part; Km. So, Km has to be in concentration unit
otherwise how can you add one concentration which is S and the Km.
So, you can say in both ways that, because the equation takes this form: S plus Km in the
denominator. So, Km must be having the unit of concentration. You can also do it by plugging
in the rate constants (with their respective units like litre per mole inverse) in the expression
for Km; this will give a concentration unit. You can actually simplify this equation for V0 if S
(which is the concentration of substrate) is much larger than Km then what happens? So, the
equation that earlier you had was V0 equal to k2 multiplied to S0 and E0 divided by S0 plus Km,
Now, when S is much larger than Km, then what happens? You can now neglect the Km part of
it. If you do that, you will see that V0 is becoming equal to k2 multiplied with E0 and that is
what is called the Vmax. The maximum velocity that you can get if you keep the enzyme
concentration constant is the Vmax.
Now, you can simplify the equation; earlier V0 was given by k2S0E0 divided by S0 plus Km;
now you just write that it will be k2 multiplied with E0 is Vmax. So, you can write Vmax
multiplied with S0 divided by S0 plus Km equals to V0. So, this is the final form of the
equation; some interesting observations are that when your V0 equals to half of Vmax; if you
put this value as half of Vmax then you will see that Km is equal to S.
I think that in the next session, we will do the further analysis of these parameters. So, there
are two important parameters that we learnt: Firstly Km (Michaelis Menten constant) is
characteristic of the enzyme that you have taken. And there is another parameter which is
Vmax; because this will ultimately give you that what is the efficiency of the enzyme. If Vmax is
very high; that means, the enzyme appears to be very efficient; maybe in the next session we
will start discussing on the Km.
Thank you.