Having discussed some of the characteristic features of enzymes in the light of their
chemical nature, today we will discuss the functional nature of the enzymes and as most
of you are familiar, catalysis is the universal property of enzymes.
The essential characteristic of a catalyst is to accelerate the rate of reaction that
it is supposed to catalyze, but in the process it is not used up in itself and in an ideal
situation it must be possible to recover the catalyst at the end of the reaction. As a
biocatalyst in the living systems two very significant characteristic properties of enzymes
are noted that is the catalytic efficiency. As you will notice, as we go further, that
the enzymes provide a much more efficient catalytic system than their chemical counter
part. There are basic intrinsic mechanisms and the confirmation of the chemical nature
of enzymes is the responsible factor for providing the catalytic efficiency. We look into the
basic factors that provide the catalytic efficiency.
When I say catalytic efficiency, I normally mean the magnitude by which it is able to
increase the rate of reaction compared to an uncatalyzed reaction or the turn over number
of the enzyme that means number of moles of substrate converted to product per unit time
per mole of the catalyst. The other feature which is very important is their specificity
which is again not attributed to chemical catalyst and the enzymes provide a very specific
catalytic system specific to the extent that we can recognize even the isomers they can
distinguish. Therefore it will be interesting to note how they catalyze the reactions such
that these properties are built in, in the system.
Another feature which I think must be important to note is that an enzyme like any other chemical
catalyst does not influence the thermodynamic status of the reaction. That means whether
it is the free energy change or the equilibrium constant of the reaction, it remains unaltered
during the process. It only accelerates the reaction rate by an order of magnitude. Now
just as an example if you look at isomerization of glucose to fructose, the reaction can be
catalyzed by an alkali say NaOH.
You can get isomerization. You can also carry out isomerization by enzymatic route using
the glucose isomerase. In both the cases the equilibrium constant of the reaction is approximately
equal to one. So whether you carry out the reaction by alkali or by the presence of glucose
isomerase, the reaction will end up under equilibrium conditions at an equimolar mixture
of glucose and fructose and that remains unchanged. The only difference, lies by the virtue of
the presence of a catalyst or an enzyme in this case, will be that you can reach to the
equilibrium status much faster than in the case of an uncatalyzed reaction. Besides the
equilibrium conversion or equilibrium constant of the reaction, the catalysis or the enzyme
catalysis also follows all thermodynamic concepts that are applied to chemical reactions. Some
times it may look in the case of living systems that certain reactions that are catalyzed
are in totality have a positive free energy change. An example could be cited of conversion
of carboxylation of Acetyl coenzyme A to Melonyl coenzyme A. The enzyme carboxylase catalyses
the reaction. The reaction has a free energy change of +4.5 kilo calories per mole.
The reaction must be non-spontaneous, should not take place as per the normal thermodynamic
concepts. But we all know that this conversion does take place in the living cells. Apparently
it may look as if enzyme is doing something which is not allowed by a chemical thermodynamics
but in practice it is not so because the reaction is catalyzed in association with another hydrolytic
reaction which is energy yielding, which has a much higher negative free energy change
and that is hydrolysis of ATP to ADP and inorganic phosphate. The free energy change of this
is highly negative 8.9 kilo calories per mole. While the reaction takes place as a
coupled reaction of the two systems, the net reaction is a coupled reaction where the acetyl
coenzyme carboxylation is coupled to hydrolysis of ATP.
But in practice the reaction does take place in a living cell as shown in the second half.
That is the enzyme which uses biotin as a cofactor activates the carbon dioxide at the
expense of energy of hydrolysis of ATP and you get activated carbon dioxide with a negative
free energy change which is a spontaneous reaction and this activated carbon dioxide
then carboxylates the acetyl coenzyme A to melonyl coenzyme A, an enzyme along with the
cofactor biotin is released free.
The second reaction is also a negative free energy change and therefore it is also carried
out and the net reaction is the sum total of the two reactions, what we have shown here.
Therefore another feature is that the enzymes often show certain behavior which can be explained
on the basis of coupling of more than one enzyme reactions where by we are able to produce
certain net results which may look apparently to be non feasible but otherwise they follow
all the laws of chemical reaction kinetics as well as thermodynamics.
The catalytic behavior of enzymes can be probably best illustrated by chemical reaction rate
theory. One of the most accepted theoretical interpretation of chemical catalyzes can be
considered from point of view of chemical reaction rate theory which illustrates that
molecules can react only if they come in contact with each other. That means the molecules
do undergo certain collisions. Therefore for different molecules of reactants to react
collisions must take place. Any factor or any parameter which increases the frequency
of these collisions between the molecules will tend to increase the rate of reaction.
Of the two of the most sort after parameters, one is temperature. We know that if we increase
the temperature of a reaction system the frequency of collisions will increase and there by the
rate of reaction. Another is if you increase the reactant concentration again there also
the frequency of collisions will increase and therefore rate of reaction will also increase.
However while collisions among the molecules might lead to product formation as per this
chemical reaction rate theory, it is also understood by virtue of some of the studies
on the distribution of molecules which undergo reaction that not all the molecules that collide
will result into the product. Only certain molecules which have sufficient energy to
undergo reaction will result into the product formation and it will be only a fraction of
the total molecules that participate in the collisions process are able to yield the product.
If you consider two molecules in a binary reaction, bimolecular reaction that is A and
B and both of them are in their ground state keeping in view all their electronic, vibrational
and rotational ground state but possess translational energy which makes them to move for the collisions
to take place. If they are to react when they collide the rearrangement of the electronic
confirmation in each molecule must take place. Consider two molecules say for example A and
B which are supposed to react in the presence of a catalyst or without a catalyst. If the
reaction has to take place, then they must start colliding with each other. Initially
they will be at a ground state of their electronic, vibrational and rotational ground states but
they will posses the translational energy which can make them move and collide. They
can only react when the rearrangement of the electronic distribution in each molecule perhaps
by transfer of atoms from one molecule to the other takes place. It is clearly pointed
out by Eyring in his transition state theory that every chemical reaction proceeds via
formation of the unstable intermediate between reactants and the products.
Take the example of hydrolysis of an ester, a very common reaction even occurring in the
case of enzymatic reaction. To the ester molecule a water molecule is added. This addition of
water molecule results into an unstable transition state compound, which has partial bonds, partial
negative and positive charges at the molecule. This is a stable molecule which has much higher
free energy change than the reactants and will break down to give the corresponding
acid and the alcohol.
Eyrings transition state theory states that for each reaction to occur it must pass
through a transition state which is an unstable state of the molecules or unstable state some
where in between the reactants and the products and which has much higher energy than possessed
by the reactants of the ground state. This energy requirement is considered as the energy
of activation. That means the free energy required for the reactants to reach the level
of transition state compound is considered as the energy of activation. The transition
state compound will ultimately break down to give you the product or in other word the
energy of activation acts a barrier. Unless the reactant molecules cross the barrier they
cannot get converted to the product. From the energy diagram, consider the first stage
which is an uncatalyzed reaction. In the case of this reaction you see the reactants are
at a ground state.
The X-axis denotes the extent of reaction the Y-axis is the energy content and then
the molecules have to undergo an increase in energy till the point where a transition
state is reached and then this transition state will break down, will undergo reduction
in energy and lead to the products. These are the reactants and these are the products.
The difference between the ground state and the maximum level of energy at which the transition
state is generated is called as energy of activation. If you look at the energy content
of the molecules it will show you a Gaussian distribution. That means energy content per
molecule and the number of molecules in a system will have a Gaussian distribution and
only those molecules that possess sufficient energy which are showed here by a shaded portion,
only that fraction of molecules will undergo chemical transformation. That means the transition
state will be able to break down and form the products. Rest of the molecules which
have much lower energy content will not be able to lead to product formation.On the other
hand when we consider the second area where we use a catalyst the main role of a catalyst
as we will also further confirm is to reduce this energy of activation by some means.
Whether it is chemical catalysis or enzymatic catalysis the primary function or the primary
route by which a catalyst functions is by reduction in this activation energy. You see
that the energy of activation from here is reduced to this level. The number of molecules,
the fraction of total number of molecules has increased, which possess sufficient energy
for the transition state molecule to break down into product, and large number of molecules
can then take part in the reaction and result ultimately into the product. In conclusion
by lowering the energy of activation for the reaction, a catalyst makes it possible for
substrate molecules with a smaller internal energy to react. Even at a smaller internal
energy level the product formation can take place. You can see the related concentration
of the shaded portion in the two cases without the catalyst and in the presence of catalyst.
That will illustrate the point. Here it is the low energy of activation as a result of
catalyst and this indicates the high energy of activation in the absence of any catalyst.
If we look deeper into the transition state, the formation of transition state can be described
as a very easily studied, as an ordinary chemical equilibrium process. As I mentioned earlier
the reactant molecule has to pass through a transition state. The formation of transition
state starting from A and B can be considered as a simple chemical equilibrium process.
That means A and B combining to give you an activated complex AB* with an equilibrium
constant
of Keq or you can put it as say K*.
K* = [AB*]/ [A] [B]
[AB*] = K* [A] [B] (1) This chemical equilibrium follows the
other relationships. That is the free energy of activation which is required in this whole
process can also be given by
E* = -RT ln Keq
(2) Rate of reaction between A and B mind that we have not undergone the whole reaction
process. We have only come to the transition state complex and the rate of reaction will
depend upon the concentration of AB*.
The larger this concentration of the transition state complex, the higher will be the rate
of reaction and this rate will depend upon the frequency at which the activated complex
will decompose and this rate of decomposition will depend upon the vibrational frequency
of the bond that is breaking. From transition state some bonds have to break some has to
form. Ultimately the transition state complex is somewhere intermediate between the reactant
and the product. Some bonds are partially formed, they will be completely formed. Some
bonds are still in the formation stage and therefore the vibrational frequency of the
bond that is breaking will determine the rate of the reaction. This frequency can be arrived
at from the theoretical treatment from two routes, by quantum mechanics as well as from
classical mechanics. Consider the frequency of energy of excited
oscillator by quantum mechanics you get
E = h? and in classical mechanics you get
E = kt
h is the plancks constant, E is the energy, the ? is the frequency. Similarly here the
k is the Boltzmann constant and T is absolute temperature. Therefore
? = kt/h = RT/Nh
You are familiar with all these terms. That is R is a gas constant, T is absolute temperature,
N is the Avogadros number and h is the Plancks constant. The frequency at which
bonds in a transition state complex will break will be determined by these factors which
can be obtained by equating the energy of an oscillator by two treatments. We had as
we mentioned earlier the rate of the reaction r
r = ? [AB*] (3)
The rate of reaction will depend upon the concentration of the transition state complex
and the frequency of the break down of this transition state complex can be written as
r = RT/Nh K* [A] [B] (4)
The expression we got for the concentration of the transition state complex from the chemical
equilibrium process. This term RT/Nh x K* is the second order rate constant. Let us
say we define k2 as second order rate constant for the reaction of A and B. It becomes
K2[A] [B] (5)
or in other words the second order rate constant will be nothing else but
k2 = RT/Nh K* (6)
Again you notice a significant fact here that the rate constant is directly linked. All
these are constants. Temperature is a constant. If the temperature is constant these are all
constants. So the rate constant is dependent upon the equilibrium constant of the transition
state process, the conversion of the A and B into the transition state complex and this
also gives you a relationship between the rate constant and the K*, equilibrium constant
of the formation of transition state complex.
It will imply or it will also be a very logical conclusion that the faster reactions, if the
reaction has to go very fast rate, then they must have larger value of K*. That means at
a very fast rate the reactants must go into the transition state or they must have higher
concentration of AB*. This equation will also allow you to calculate the value of K* and
hence the value of activation energy during your earlier equation that is your E*
E* = -RTlnK*
If you use the equation number 6 and 2 you can calculate the value of energy of activation
for a given system. You can also conclude that the faster reactions will have a smaller
value of E*. No. Energy of activation is not equal to KT. This is the energy of the oscillator,
energy term for the oscillator based on the classical mechanics.
This is not energy of activation. E* is energy of activation. Energy of activation can be
calculated from here which will be related to K* and also can be related to k2 and the
two conclusions which we can make from make this analysis is that the fast reaction will
have a higher value of K*, equilibrium constant for the conversion of reactants into the transition
state or in other words the fast reactions will have higher concentration of AB* at equilibrium.
The second is that the first reaction will have smaller energy of activation. If you
combine equations 6 and 2 energy of activation for faster reaction if the k2 is high that
means lower energy of activation.These two conclusions do arrive from the theory of reaction
rate which is now well accepted world over. If you look at what we talked about
KT = RT/nh x K* (6) you can also write k2 as
k2 = RT/nh x exp (-E*/RT) (7)
Now this as all you know we call it as frequency factor which is often written as A.
The frequency factor A* you can say because A is the reactant,
A*exp (-E*/RT)
which is a very common expression what we all know as Arrhenius expression. The term
E* can again be separated into both enthalpic and entropic contributions
E* = ?H* - T?S* from second law of thermodynamics.
That means the energy of activation can also be considered both in terms of entropic contribution
as well as enthalpic contributions and if you separate into it then you can write
k2 = RT/nh . exp (?S*/R) . exp (-?H*/RT)
The final conclusion is that for a catalyst or an enzyme in our case as we are talking
about to be able to include the value of k2, it must be able to reduce the energy of activation
or increases the K*value that is equilibrium constant. This conclusion will apply commonly
to a chemically catalyzed reaction or an enzymatic catalytic reaction.
If you look at the performance of some of the enzymatic catalyst as compared to chemical
catalyst, you will notice that enzymes provide very efficient catalyst. Simple example where
I can compare the reaction which can take place both by chemical catalyst as well as
by enzymes is hydrolysis of hydrogen peroxide that is break down of hydrogen peroxide into
water and oxygen, a very common reaction that takes place in living cells by the enzyme
catalase.
It can also be carried out by colloidal platinum. The reaction is also feasible under normal
conditions without any catalyst. That means not a very non spontaneous reaction. It is
possible to carry out the reaction. The corresponding energy of activation for this reaction without
any catalyst is 18,000 kcal.mole-1 as against in the presence of colloidal platinum as catalyst
it is 11,700 and for the catalase it is 5500. This is just an illustrative example but a
similar example can be sited from a variety of reactions that can be carried out both
by chemical catalysis as well as by enzymatic catalysis and the conclusions are almost same.
The enzymes are able to bring down or reduce the energy of activation to a very large extent
compared to chemical catalyst and we will again see how this is done.
Another interesting reaction in the living systems is on oxidation of methane. The methane
mono oxygenase enzyme oxidize methane in to ethanol spontaneously in the living cells
particularly those organisms which can survive or consume methane as a carbon source. This
is the very first reaction which they carry out when the methane is used as carbon source.
This reaction can also be carried out chemically using a chemical catalyst, like zinc catalyst
at 400c and high pressure, almost few hundred atmospheres, and the turn over number is of
the order of 2 x 104 s-1, that means number of moles of methane oxidized per second per
mole of the catalyst.
The same reaction if you carry out at ambient conditions, 25c and ambient atmospheric
pressures the turn over number is as low as 1.5x10-7, almost negligible. In the presence
of enzyme, methane mono oxygenase at 25c and obviously at ambient pressure the turn
over number is 2.2 x 103, a much higher value although it is lower than the one which is
obtained at 400c and few hundred atmospheres. But compared to ambient conditions and the
same catalyst, the catalyst here is zinc and the value is much higher.
Very often a comparison of the first and the third case is not logical because the reaction
conditions are different and comparing the turn over number may not be a very reasonable
proposition. But if you compare under the same conditions the enzyme as a catalyst will
provide you much more efficient system and one should look at how the enzyme provides
this kind of efficiency compared to a chemical catalyst because the ultimate goal in both
cases is the same, to reduce the energy of activation. But the enzyme is able to reduce
the energy of activation to a large extent as compared to chemical catalyst and also
its turn over number is much higher.
Let us see what happens when we consider an enzymatic catalysis. Now in the case of enzymatic
catalysis we must bear clearly in our mind that one of the characteristic features is
that enzymes has a certain ligand binding sites and more specifically substrate binding
sites. That means an enzyme can bind to substrate forming enzyme substrate complex. This enzyme
substrate complex then again provides a system which will undergo the energy of activation,
undergo the transition state before the product is formed.
Consider an enzyme catalyzed system and an uncatalyzed system. If you consider let us
say E+A+B, same as what we are talking about reaction between A and B. Consider two routes.
One is E, the enzyme does not take part in reaction, an uncatalyzed reaction and it goes
to E plus P. In the second case we can also have that enzymes binds the substrate A and
B and forms a enzyme substrate complex EAB. It also undergoes a transition state EAB*
and then it goes to enzyme and product. Let us consider the rate constants k here ke here.
The lower route is catalyzed one, enzyme catalyzed. The upper root is uncatalyzed. That means
the enzyme has no role to play here. The enzyme binds to the substrate and then undergoes
transition state and then forms the product. In the middle case from the enzyme and transition
state AB you can also form the EAB activated complex.
If you consider the equilibrium constant of all these equilibrium reactions let us say
this is K*, this is KS, KE* and K*s. Thermodynamically we can notice that ultimately we are interested
in this. As a catalyzed reaction, we are interested in the EAB*, which is the transition state
complex which will give you a product at much faster rate.
Ke is much, much higher than K, the lower being a catalyzed reaction. Catalyzed reaction
will have a much higher reaction rate and so to really get the reaction at a high rate
we want a larger concentration of EAB*. If you look thermodynamically, this product EAB*
can be obtained by either of the routes. Either it can go this way and come here or it can
also come this way and reach to this side. From either way under equilibrium conditions,
it is possible to arrive at this complex and if you consider the two routes then your
KS . KE* = K*. KS*
Multiplication of equilibrium constant of these stages should be equal. As I said earlier
KE is greater than K and also
K = RT/nh . K*
Therefore the KE* >> K* and it will imply that KS*, equilibrium constant is much, much
greater than KS. This means that the binding of the enzyme to the AB complex, uncatalyzed
AB complex is much, much larger than the KS, the binding of the enzyme to A and B.
This implies that the enzyme binds AB* more tightly. The higher value of KS*means the
enzyme binds AB* more tightly than the substrate molecules. The general concept that the enzymes
have certain substrate binding sites on molecule but conformationally it is more complementary
to the substrate transition state and that was the contribution which was stated in very
clear terms by Linus Pauling that enzymes are molecules that are complementary in structure
to the activated complexes of the reactants they catalyse.
The attraction of the enzyme for the activated complex would thus lead to a decrease in its
energy and hence decrease the energy of activation for reaction and to an increase in rate of
reaction. The enzyme is in structure confirmationally which is very much complimentary to the transition
state of the substrates and by binding that it stabilizes those transition complex. Stabilization
of transition state means that the enzyme because of its confirmation which is complimentary
to the transition state, binds and stabilizes the transition state complex. So at any given
equilibrium the concentration of the EAB* that is the product of this binding is much,
much higher than as if enzyme does not take part.
If suppose this enzyme was not a bio catalyst. Suppose this E would have been a chemical
catalyst this binding would not have been feasible because there would have been no
complimentary catalytic function. The catalyst might be able to perform the catalytic function
by any other mechanism: by acid base catalysis or covalent catalysis which we will discuss
later but the contribution made by stabilizing the transition state of the substrate will
not be possible to be provided by the chemical catalyst which enzyme is able to provide and
there by it increases the stability of EAB* and thereby reduces the energy of activation
and leads to the formation of the product. That was a major contribution in the understanding
of the enzyme functions made by Linus Pauling.
For any reaction whether it is unimolecular or bimolecular the substrate or the reactant
has to undergo a transition state. The product cannot be formed without undergoing a transition
state. As a matter of fact practically there are hardly any reactions, very few reactions,
which are single molecular. Apparently hydrolysis may be unimolecular reactions but it is not
so. Even in the earlier example we discussed the hydrolysis of esters, it is an addition
of water molecule to the ester which leads to transition state. So most of the reactions
undergo or involve either cofactor or water molecule, excepting certain isomerases which
are truly unimolecular. But even in that case the molecule has to undergo a transition state
otherwise the catalyst reaction is not feasible according to reaction rate theory.
Coming back to the role of the energy of activation on the reaction rate constant, if you take
arbitrary value of energy of activation, say from 25kcal mol-1 downwards to 5kcal mol-1,
a decrease of activation energy results in a very significant increase in the reaction
rate.
The magnitude at every step of reduction of energy of activation by 5kcal mol-1 you will
see that starting from 3.16 x 10-6 at a energy of activation of 5kcal mol-1 you will go up
to 1.33 x 109. That means an increase of magnitude by 1015 fold of the rate constant. It is a
very, very significant increase in the reaction rate constant and it will give the rate of
reaction under any given set of conditions. That means if you maintain same temperature
and pressure and substrate concentration increase in the reaction rate constant by a factor
of 4.58 x 103 fold can be achieved for every drop of 5kcal mol-1 which is a very, very
large. So therefore even a decrease of the energy of activation by 5kcal mol-1 can lead
to an increase of the reaction rate by a factor of 4.5 x 103. Very significant number and
in case of most of the enzyme the decrease is of the order of about 15-20 kcal mol-1
in the case of energy of activation. Therefore one can achieve a very high catalytic efficiency
in the case of enzymatic catalysis and as you notice that this efficiency is primarily
achieved by two distinct factors as we saw today which are unique compared to the chemical
catalyst.
One is that enzymes have binding sites for substrate molecule. The second is the enzyme
conformation itself is complementary to the enzyme transition state complex thereby it
can stabilize the transition state and increase the concentration of transition state in any
given reaction at equilibrium. Now if you look at the whole picture of the enzymatic
catalyses again, in the energy diagram we notice the bold lines which go here from the
ground state E+A+B uncatalyzed reaction that is the non enzymatic reaction is this one.
The molecules have to reach to this state of the energy so that the transition state
can form and then it can be broken down to form the enzyme and the product.
What happens in the presence of an enzyme molecule? The enzyme binds to the substrates
A and B, forms EAB and this process also undergoes energy of activation. That means some amount
of energy is released by this binding and the enzyme substrate complex instead of starting
from the level of ground state of A and B now has reached to a state little higher than
ground state by binding process and from there it goes to the bound state.
Even the binding process will undergo the transition state and for this binding process
EAB is the product. Then from here EAB has to go to transition state EAB* as I showed
in the earlier comparative figure and this EAB* is much, much smaller than the energy
level of AB* for uncatalyzed reaction. So you have an advantage. This EAB* is reduced
by virtue of the complementary nature of binding of the enzyme molecule to the AB*. The concentration
of EAB* is increased by significant amount and at lower energy level this equilibrium
can be achieved. Two factors which contribute to the reduction in activation energy or the
efficiency of the enzymatic catalyst are: one is binding of the substrate and other
is complementary of the enzyme molecule to that of the transition state. The same process
follows on the reverse of reaction that means EAB* under goes break down into products whereby
instead of product directly being formed it forms EP, the enzyme bound to the product,
the enzyme product complex and then this EP again breaks down to the next level of your
product which is same as of the uncatalyzed reaction. Therefore one is by binding process
the enzyme processes started at much high energy level then the ground state as applicable
in the case of uncatalyzed reaction and the other is the complementarity of the enzyme
conformation to that of the transition state complex. These two factors put together give
you much higher catalytic efficiency as we noted in the case of one of the examples which
I have illustrated to carry out the conversion of reactants into product. The dotted line
here indicates the process. In fact I have shown here only one step just for sake of
clarity and removing the congestion whereas this process might undergo more than one humps.
For example binding of E to A, the EA further binds to B in different sequences depending
on the nature of catalysis. But there can be more than one processes undergoing here
and thereby the binding process can provide you a large fraction of the activated energy
to be compensated during binding process and reducing the energy of activation. Such a
process therefore leads to a very efficient catalytic system.
In my first presentation I mentioned about the specificity. Coming back to the specificity
the presence of ligand binding sites on the enzyme molecule are responsible for specificity.
The so called ligand binding sites and when we consider ligand binding site along with
the catalytic sites what we consider as active sites on the molecule they constitute for
the specificity and we will discuss that in our subsequent lecture. So today if you recall
we have considered in general the basic theoretical considerations behind the catalysis as it
applies to any chemical process. Secondly how enzyme under the same background is able
to provide much larger reduction in activation energy and provide a more efficient catalyst
compared to chemical catalyst. With this we will stop.
