Mechanism and Kinetics (with reference to Catalytic Antibodies)

Dr. Poretz has asked me to talk about a couple of aspects of catalytic antibodies, or rather basic aspects of enzymes which are applicable to catalytic antibodies. I shall do this with less physical chemistry than in Proteins & Enzymes, but still a bit.

First, what do we mean by catalysis? Increase of rate of a reaction, brought about by something (the catalyst) which is not consumed in the reaction, may be reused. A reaction may be looked at as passing from one valley, representing stable reactants, over a mountain pass to another valley, the products. When the molecules are at the top of the pass, where they may with equal probability go ahead to products or back to reactants, they are said to be in the transition state. Energy must be put in to a reaction to raise the reactants to the top of the pass; this energy is called the free energy of activation, ∆G*, always positive because it represents energy which must be put in to reach the unlikely transition state. The role of the catalyst is to find a lower pass over the mountain range, a pathway with a lower activation energy.

The transition state may be described as a condition of the reactants - let's just use one, call it A* - in equilibrium with the ground state A, so that K*, the equilibrium constant of activation, = [A*]/[A]. An alternative way of looking at the role of a catalyst is that it stabilizes the transition state, increases the ratio [A*]/[A]. The rate constant of the overall reaction, k, by which [A] is multiplied to get the rate, is proportional to the amount of the reactant in the transition state, k = (kT/h)K*, where k is Boltzmann's constant and h is Planck's constant. The factor kT/h is the frequency of decomposition of the transition state, which is the same is the vibrational frequency n of the bond breaking. At 25° C n = 6.212 x 10¹² s^-1.

The ∆G* of activation is of course related to the equilibrium constant K* in the usual way, which I’m sure Dr. Kahn has described to you in General Biochemistry, ∆G* = -RTlnK* = -RTln([A]*/[A]) = -RT(ln[A]* - ln[A]), -(∆G*/RT) = ln[A]* - ln [A], -(∆G*/RT) + ln[A] = ln[A*]. Taking antilogs, [A]e^-∆G*/RT = [A*]. This relates the concentration of [A*] to the conc. of ground-state A and the difference in free energy between the ground state and the transition state. Since the exponent is -∆G*/RT, a negative number, e^-∆G*/RT is a number <1, a small fraction, and the larger ∆G* is, the smaller e^-∆G*/RT, and the smaller the fraction of A in the transition state. In our mountain pass metaphor, ∆G represents the energy which must be expended to reach the top of the pass, and is proportional to the vertical rise from the valley floor to the pass.

The ∆G* may further be separated into enthalpy and entropy, ∆G* = ∆H* - T∆S*, k = (kT/h)e^-∆H*/RTe^∆S*/R. If one can measure the rate of reaction k as a function of temperature, one can obtain (although not terribly accurately) values of ∆G^*, ∆H^* and ∆S* by various rearrangements of this equation, plotting k or lnk vs. various functions of t; but I spare you. The enthalpy term ∆H represents the heat change in the reaction, while the entropy term ∆S represents roughly the change of orderedness in a reaction. Entropy may be thought of as a measure of disorder, and one law of thermodynamics states that entropy tends always toward a maximum, energy must be put in to a reaction to reverse order, decrease S, have a negative value of T∆S. My classic example of this is that if I put out my garbage can the night before it is to be picked up, and a dog or a raccoon knocks the can over and scatters the garbage across the street, it takes me much more energy to collect the garbage into the can again than it did the dog or raccoon to knock it over; I am making the garbage more ordered by putting it into the can.

Any transition state is likely to be a very ordered state of the molecule or molecules undergoing reaction, a precise intermediate state between reactant and product, which means that entropy is lost compared to the ground state. If ∆S^*is negative, the term -T∆S^* is positive, and ∆G^* is increased, making the reaction more difficult. One way that an enzyme can accelerate a reaction is by selecting a particular conformation of the substrate, one which is rare but not unknown in solution, and binding it selectively. This conformation is already part way toward the transition state, a sort of base camp part way up the pass, and by selecting it the enzyme makes the final push to the pass a less arduous climb. In the paper by Campbell et al. which I have given you, one catalytic antibody selects the conformation 3 in Fig. 1, which is much less often found in solution than the diequatorial conformer 2, and binds it.

Furthermore, a reaction making more molecules out of fewer - an increase of disorder - will have a favorable overall entropy term, while one making fewer molecules will have an unfavorable entropy term. This implies that the formation of a transition state in the reaction of two or more molecules, once separate but brought together for reaction, has a very substantial negative entropy term - although in many cases the is partially balanced by the displacement of water molecules from a relatively ordered state around the reactants to the disordered state of bulk H₂O.

This is particularly true when two or more reactants must be brought together into a precise arrangement, having started from quite random positions in solution. The molecules must lose translational entropy – randomness of position in the solution - rotational entropy - freedom of the whole molecule & its parts to rotate - and some vibrational entropy, freedom of the parts of a molecule to vibrate with respect to each other. Picking a specific rare conformation of the molecule also constitutes a loss of entropy.

This describes the reaction of two molecules meeting each other in solution. But it also describes the formation of a complex between substrate and catalyst; a considerable entropy penalty must be paid to form the complex, the more so the more precisely the substrate fits the active site of the catalyst. To form the complex, favorable enthalpy of binding interactions between substrate and enzyme must be used to balance the loss of entropy.

In a sense, the enzyme has three roles in catalyzing a reaction. One is the minimization of the entropy penalty implicit in forming the transition state, both the bringing together of two molecules to react and the introduction of the third participant, the enzyme itself. One cannot escape the entropy penalty entirely, but with enzymes the entropy penalty is paid in the formation of the enzyme-substrate complex, in which energetically favorable binding interactions with negative ∆H - charge attraction, van der Waals contacts, hydrogen bonding - are formed which compensate for the unfavorable entropy change. Not only can two or more molecules react with each other without paying the entropy penalty in forming the transition state, but groups on the enzyme can be involved in chemical catalysis of the reaction, the second way enzymes catalyze. In catalysis of a reaction by small molecules these catalytic involvements must accelerate the reaction more than 1000-fold to overcome the entropy penalty and display net acceleration of the reaction. Binding to an enzyme allows a number of catalytic groups to act, each raising the rate another 100-1000 fold, without any entropy penalty for their involvement, until an otherwise unlikely reaction such as the rearrangement of succinate to methylmalonate can take place at a reasonable rate. Thus the decrease of activation entropy, thanks to binding, allows the expression of other mechanisms of catalysis. Of course a tremendous entropy penalty had to be paid in the synthesis of the large, highly ordered enzyme molecule, but only once, the enzyme molecule can be used for many, many reactions, while in small-molecule catalysis the transition state complex must be formed anew for each reaction. You could say that this is "the meaning of life" - the living system synthesizes catalysts which diminish the payment of entropy of activation by the reusability of the catalysts.

The third way enzymes catalyze, and the major way by which catalytic antibodies catalyze, is by forming binding interactions which stabilize the transition state in a structural way. Besides initially binding the substrate, binding interactions maximize binding complementarity in the transition state. More groups on the enzyme interact with the substrate in the transition state, or the interactions are closer and at better angles. The improvement of the binding interactions provides negative ∆G which balances the positive (unfavorable) ∆G of distorting the substrate physically and chemically to the transition state structure, as well as the positive T∆S of going to a specific transition state in which bonds are not vibrating. Indeed, it is advantageous for the K_m for the substrate to be relatively high, representing weak binding of the ground state of the substrate, in order to maximize the gain in binding interaction in going to the transition state. K_ms may be higher than the in vivo concentration of the substrate! In our metaphor of the mountain pass, the overall energy expenditure to reach the transition state is lessened if you don't have to walk far down valley before starting up the route to the lower pass. This effect is limited by the fact that K_m >[S] implies that less than half of the enzyme has substrate bound at any given moment; this means that more enzyme must be synthesized to keep the metabolic rate up.

This effect is studied by the use of transition state analogs: a structure for the transition state of the reaction is guessed at, and a compound with similar structure is synthesized and shown to bind much better than the actual substrates, because less of the binding energy is used to approximate the transition state, so that the net ∆G of binding is greater. An example is that lactones, what you get by oxidizing a carbohydrate to the acid level while maintaining the ring structure, have that end of the ring flat like the carbonium ion intermediate of glycoside hydrolysis, which is similar in structure to the transition state of the reaction; such lactones bind 100 to 4000 times better to the hydrolytic enzymes than their substrates. Transition state analogs often use different elements; for instance, the transition state analog for chymotryptic hydrolysis of a substrate like phenylethanecarboxylic acid esters is phenylethaneboronic acid, with -B(OH)₂ replacing -COOH.

From these conclusions the approach to catalytic antibodies should be obvious, at least in principle. The basic idea dates back to a paper by Linus Pauling in American Scientist in 1948, and was well stated by Jencks in 1969: "one way to synthesize an enzyme is to prepare an antibody to a haptenic group which resembles the transition state of a given reaction. The combining sites of such antibodies should be complementary to the transition state and should cause an acceleration by forcing bound substrates to resemble the transition state." Imagine the transition state for the desired reaction - this comes from study of the uncatalyzed reaction and/or of how the enzyme found in nature catalyzes; synthesize a molecule which is similar in structure to the supposed transition state, such as 4 in Fig. 1 of the Campbell et al. paper; and elicit antibodies to it. These should then bind the substrates, but favor the transition state of the reaction between them by binding it much better; and once you get the transition state you are half way to the products. Of course the idea could only be brought to fruition by the development of monoclonal antibodies, enabling examination of what a single species of protein could do.

How do you tell which of several catalytic antibodies is best? By using simple enzyme kinetics, which I'm sure Dr. Kahn told you about in General Biochemistry. You will remember that v = , where V_max = k_cat[E_]t, the rate constant of the reaction times the concentration of enzyme, and K_m is a sort of dissociation constant, defined as equal to the substrate concentration at which v = . In the simplest enzyme kinetic mechanism, E + S E·S Æ E + P, K_m = . Note that K_m is not just the dissociation constant of the enzyme-substrate complex, K_s = , but as k₂, the rate constant of the forward reaction, becomes larger, so also does K_m; the faster the enzyme kicks out product and is free to bind new substrate molecules, the harder it is to saturate.

If an enzyme will work on a variety of similar substrate molecules, we define the 'best' substrate as that which has the highest ratio of V_max to K_m, or k_cat/K_m. Similarly, if we have a variety of catalytic antibodies acting on the same substrate, we define the 'best' as the one with the highest k_cat/K_m ratio, combining high rate of catalysis at saturation - k_cat - with low K_m. By the argument given above K_m cannot be too low, or k_cat will be reduced because the activation energy to go from the E·S complex to the transition state will be greater. But it also cannot be too high, or uneconomically high substrate concentrations will be needed to get a good rate. Stewart and Benkovic have described efficiency of antibody catalysis in terms of the ratio between the dissociation constant K_s of the substrate and that of the transition state K , = ; the greater the ratio of the dissociation constants, the more free energy of binding is available to stabilize the transition state and thus accelerate the reaction.

I remind you only briefly of how to determine K_m and V_max. Normally one measures the rate of the reaction at different substrate concentrations (and a constant enzyme level). This gives a hyperbolic plot of v vs [S]. One rearranges the data into one of several linear forms, usually 1/v vs 1[S], the Lineweaver-Burk plot (though for statistical reasons I prefer the Woolf plot, [S]/v vs [S]), and determines the slope and intercept of the plot. In the L-B plot, the y intercept is 1/V_max, the slope K_m/V_max. This gives an approximate value; for best results you go to a computer and fit the data directly to the Michaelis-Menten equation by a non-linear least-squares treatment.

To get k_cat from V_max you also need to know the molar concentration of the enzyme or abzyme, which you get from the protein concentration and the molecular weight, when you have pure enzyme or abzyme. Note that you can determine K_m of the enzyme even in crude extract, as long as there are no other enzymes present catalyzing the reaction or binding a lot of the substrate, but a meaningful determination of V_max requires pure enzyme, because otherwise the protein concentration includes other proteins. If I have, say, V_max for one cloned version of an enzyme in crude extract of E. coli which is twice that of another version, is it because the first version is inherently better - higher k_cat - or because there is less enzyme present in the second extract, or even that there are more other proteins present in the second? I have to purify the enzyme to know which, to determine the molar concentration of enzyme - although for some enzymes there is a technique called enzyme titration, production of one mole product per mole enzyme, to do this.

The reason for wanting to know k_catis not just to determine which abzyme is best by k_cat/K_m; k_cat is the k you use in calculations of ∆G*, ∆H* and ∆S* mentioned earlier. Campbell et al. cite determinations of these quantities for the reaction catalyzed by two abzymes with chorismate mutase activity. One acted mainly by lowering the ∆S* of the reaction, the other acted by lowering ∆H*, the ∆S* was actually higher than for the spontaneous reaction. They then used an nmr technique called transferred nuclear Overhauser effect to show that substrate bound to the latter abzyme was in conformation 3, closer to the transition state of the reaction.

One point of this analysis is that it points out limitations of catalytic antibodies which cannot be overcome by the basic procedure of selecting an antibody which binds the transition state analog well. There is no selection for groups on the enzyme which will play a role in chemical catalysis of the reaction. And real enzymes typically change their own conformation in binding the transition state; this does not necessarily happen in an abzyme (in the one thoroughly studied case it doesn't). However, mutagenesis of the cloned abzyme, which I assume Dr. Poretz will talk about, can improve the abzyme along these lines, although introducing conformational changes, involving the whole protein, not just the active site, will be more difficult than inserting catalytic side chains at the active site.