Practice English Speaking&Listening with: Lecture - 3 Enzymes as Biocatalysts

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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


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


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


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


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.

The Description of Lecture - 3 Enzymes as Biocatalysts