How To Calculate The Catalytic Efficiency

Enzymes are proteins in biological systems that help speed along reactions that would otherwise take place far more slowly than without the aid of the enzyme. As such, they are a kind of catalyst. Other, non-biological catalysts play a role in industry and elsewhere (for example, chemical catalysts aid in the combustion of gasoline to enhance the capabilities of gas-powered engines). Enzymes, however, are unique in their mechanism of catalytic action. They work by lowering the activation energy of a reaction without changing the energy states of the reactants (the inputs of a chemical reaction) or the products (the outputs). Instead, they in effect create a smoother path from reactants to products by lowering the amount of energy that needs to be "invested" in order to receive a "return" in the form of products.

Given the role of enzymes and the fact that many of these naturally occurring proteins have been co-opted for human therapeutic use (one example being lactase, the enzyme that aids in the digestion of milk sugar that millions of people's bodies fail to produce), it is not surprising that biologists have come up with formal tools to assess how well specific enzymes do their jobs under given, known conditions – that is, determine their catalytic efficiency.

An important attribute of enzymes is their specificity. Enzymes, generally speaking, serve to catalyze only one of the hundreds of biochemical metabolic reactions that are unfold within the human body at all times. Thus a given enzyme may be thought of as a lock, and the specific compound on which it acts, called a substrate, can be likened to a key. The part of the enzyme with which a substrate interacts is known as the active site of the enzyme.

Enzymes, like all proteins, consist of long strings of amino acids, of which there are about 20 in human systems. The active sites of enzymes therefore usually consist of amino acid residues, or chemically incomplete chunks of a given amino acid, which may be "missing" a proton or other atom and carry a net electric charge as a result.

Enzymes, critically, are not changed in the reactions they catalyze – at least not after the reaction is over. But they do undergo temporary changes during the reaction itself, a necessary function in allowing the reaction at hand to proceed. To carry the lock-and-key analogy further, when an substrate "finds" the enzyme required for a given reaction and binds to the enzyme's active site (the "key insertion"), the enzyme-substrate complex undergoes changes ("key turning") that result in the release of a newly formed product.

The interaction of the substrate, enzyme, and product in a given reaction can be represented as follows:

\(E + S ⇌ ES → E + P\)

Here, E represents the enzyme, S is the substrate, and P is the product. Thus, you may envision the process as loosely akin to a lump of modeling clay (S) becoming a fully formed bowl (P) under the influence of a human craftsperson (E). The craftsperson's hands may be thought of as the active site of the "enzyme" this person embodies. When the lump up clay becomes "bound" to the person's hands, they form a "complex" for a time, during which the clay is molded into a different and predetermined shape by the action of the hand to which it is joined (ES). Then, when the bowl is fully shaped and no further work is needed, the hands (E) release the bowl (P), and the process is complete.

Now consider the arrows in the above diagram. You'll notice that the step between E + S and ES has arrows moving in both directions, implying that, just as enzyme and substrate can bind together to form an enzyme-substrate complex, this complex can dissociate in the other direction to release the enzyme and its substrate in their original forms.

The unidirectional arrow between ES and P, on the other hand, shows that the product P never spontaneously joins with the enzyme responsible for its creation. This makes sense in light of the previously noted specificity of enzymes: If an enzyme binds to a given substrate, then it does not also bind to the resulting product or else that enzyme would then be specific for two substrates and hence not specific at all. Also, from a common-sense standpoint, it would make no sense for a given enzyme to make a given reaction work more favorably in both directions; this would be like a car that rolls with both uphill and downhill with equal ease.

Think of the general reaction in the previous section as the sum of three different competing reactions, which are:

\(1) \; E + S → ES \
2) \; ES → E + S \
3) \; ES → E + P\)

Each of these individual reactions has its own rate constant, a measure of how quickly a given reaction proceeds. These constants are specific to particular reactions and have been experimentally determined and verified for a plethora of different substrate-plus-enzyme and enzyme-substrate complex-plus-product groupings. They can be written in a variety of ways, but generally, the rate constant for reaction 1) above is expressed as k₁, that of 2) as k_-1, and that of 3) as k₂ (this is sometimes written k_cat).

Without diving into the calculus needed to derive some of the equations that follow, you can probably see that the velocity at which product accumulates, v, is a function of the rate constant for this reaction, k₂, and the concentration of ES present, expressed as [ES]. The higher the rate constant and the more substrate-enzyme complex present, the more rapidly the ultimate product of the reaction accumulates. Therefore:

\(v = k_2[ES]\)

However, recall that two other reactions besides the one that creates the product P are occurring at the same time. One of these is the formation of ES from its components E and S, while the other is the same reaction in reverse. Taking all of this information together, and understanding that the rate of formation of ES must equal its rate of disappearance (by two opposing processes), you have

\(k_1[E][S] = k_2 [ES] + k_{-1}[ES]\)

Dividing both terms by k₁ yields

\([E][S] = {(k_2 + k_{-1}) \above{1pt} k_1} [ES]\)

Since all of the "k" terms in this equation are constants, they can be combined into a single constant, K_M:

\(K_M= {(k_2 + k_{-1}) \above{1pt} k_1}\)

This allows the equation above to be written

\([E][S] = K_M[ES]\)

K_M is known as the Michaelis constant. This can be regarded as a measure of how fast the enzyme-substrate complex disappears via the combination of becoming unbound and new product being formed.

Going all the way back to the equation for velocity of product formation, v = k₂[ES], substitution gives:

\(v = [E][S] \Bigg( {k_2 \above{1pt} K_M}\Bigg)\)

The expression in parentheses, k₂/K_M, is known as the specificity constant, also called the kinetic efficiency. After all of this pesky algebra, you finally have an expression that assesses the catalytic efficiency, or enzyme efficiency, of a given reaction. You can calculate the constant directly from the concentration of enzyme, the concentration of substrate and the velocity of product formation by re-arranging to:

\(\Bigg( {k_2 \above{1pt} K_M}\Bigg)= {v \above{1pt}[E][S]}\)