Enzymes

Although some of the stuff we have been covering so far is interesting, the topic we have to cover now, enzymes and enzyme mechanism, is facinating. After learning about enzymes and how they work, you will really be able to realize the level of complexity and organization that chemistry in living systems has.

We have mentioned some definitions and properties of enzymes in the past, but we will clean the slate and go over all of them again.

- First and foremost, enzymes are biochemical catalysts that are crucial for life. Enzymes catalyze chemical reactions, i.e., they make the reactions go a lot faster than what they would if we did not had them around. Without them, processes such as protein synthesis would take hundreds or thousands of years. In some cases, these processes would not even happen! If not all, 99.99% of the chemistry going on in cells is catalyzed by enzymes: Protein synthesis, lipid synthesis, conversion of fuels (glucose, light) into energy and synthetic precursors, etc.. Without them non of this would be possible, and living organisms could never reach the level of structural diversity and organization needed for life and self-perpetuation.

- As opposed to chemical catalysts, enzymes catalyze reactions in very mild conditions. Most enzymes work in water, at pHs around 7, normal pressure, and 37^oC. Compare to Pd-C (a catalyst used to reduce double bonds), where we have to use an H₂ atmosphere, usually high pressure, relatively high temepratures, and organic solvents.

- Except for catalytic RNA molecules that also catalyze certain reactions, enzymes are proteins. All the concepts that we have discussed for proteins (amino acid components, non-covalent interactions, primary, secondary, tertiary, and quaternary structure, folding, etc., etc.) apply straight forwardly to enzymes.

- Many enzymes do their bussines alone, using exclusively the amino acid residues (backbone and most oftenly side chain groups) that they have. However, many enzymes require other non-amino acid groups to function. These are called cofactors, and can be one of many metals (Fe²⁺, Mg²⁺, Zn²⁺, Mn²⁺), or more complex organic molecules. In these latter case, these groups are referred to as coenzymes. Most vitamins are coenzymes (vitamin B12 is also referred to as coenzyme B12). When the cofactor or coenzyme is bound to the enzyme, it is, as we know, its prosthetic group.

- An enzyme, complete with its cofactor or coenzyme, is referred to as an holoenzyme. When the prosthetic group is removed, and we have only the amino acid component of the enzyme, we refer to it as an apoenzyme or an apoprotein. The apoenzyme or apoprotein is therefore inactive.

- Certain enzymes can be modified by covalent attachment of additional chemical groups as a way of regulating their activity. For example, phosphorilation or glycosidation of certain residues will make an enzyme to turn on or off (to be active or inactive). These chemical modifications are not to be confused with cofactors.

We won't spend a lot of time here, as we will usually use common or trivial names for enzymes. Generally, enzymes are named according to the reaction they catalyze, by adding the termination '-ase' to the name of their substrate or to a short list of words describing the reaction they catalyze. Thus, urease catalyzes the hydrolisis of urea, DNA polymerase the polymerization of DNA monomers, HIV protease catalyzes the hydrolisis of certain proteins (peptide bonds), etc., etc..

There are, however, formal names which are a bit more explicit. The enzyme that catalyzes the following reaction:

is ATP:glucose phosphotransferase, because it transfers a phospahte unit (PO₄^3-) from ATP to glucose, making in the process ADP and glucose-6-phosphate. The trivial name for this enzyme is hexokinase, which without knowing what it does may be confusing.

In general, we will see enzymes classified in 6 main families according to the type of reaction they catalyze:

- Oxidoreductases: Oxidation-reduction reaction

- Tranferases: Transfer of functional groups.

- Hydrolases: Hydrolysis reactions.

- Lyases: Group elimination reactions. Formation of double bonds.

- Isomerases: Isomerization reactions.

- Ligases: Bond formation coupled with ATP hydrolysis.

What enzymes do

By now, we more or less know what they do, but we do not know it yet in detail. We will examine what is that ernzymes affect (or provide) for a reaction that would otherwise take eons to happen (or not happen at all), happen in milliseconds.

For any chemical reaction to occur, a number of different conditions must be met. First, if the reaction involves the condensation between two reactants to form a product, the two molecules will have to come close together in space, and orient themselves in a particular orientation with respect to each other. Second, the electron clouds of cetrain atoms or group of atoms in both molecules have to change form to allow them to interact with one another. Finally, the certain atomic or bonding orbitals in both molecules are converted in bonding orbitals between the two molecules to form the final bonding structure of the product.

Each of these steps requires some work, even if the reaction is one of the reactions we commonly know as happening spontaneously. First, time will pass before two molecules come together in space in the right orientation and actually 'hit' each other. In the lab, we can solve this problem by heating up the mixture: More heat means more molecular movement, and therefore, more chances of having collisions between molecules in the right orientation. Second, the two molecules will have to modify their electronic structure to form new bonds. Again, we need energy, which in the lab can be again mimicked by heating the system up.

In a biological system, we cannot 'heat up' the system for any of these processes to happen. Here is when enzymes come to the rescue. An enzyme will provide an environment in which a certain reaction is energetically more favourable. This environment will first bring the two molecules together, therefore increasing their probablility of 'seeing' each other, and it will also give the two molecules the energy needed to rearrange their electrons to form a bond. Although we still have to figure out how is that the enzyme gives this energy to the molecules, or substrates, we can define certain distinguishing features of an enzyme using what we just said:

The environment provided by the enzyme which favour a particular reaction is called its active site. We will soon see why the active site of a particular enzyme catalyses a specific reaction (how is that it 'gives' energy to the reaction to occur).

Now we move to a simpler reaction, the breaking of a bond in a substrate (S) to form a product (P). The reaction could be the hydrolysis of a peptide bond in a polypeptide, for example:

Here 'P' can be one or more products. This reaciton, which is an equilibrium, will have a characteristic equilibrium constant K, which tells us how much of each we have under certain conditions:

As you know from PhyChem, Principles, or GenChem, the equillibrium constant (or the ratio between products and substrates) is connected with the difference in free energy (DG) of the reaction:

The DG tells us the difference between the free energy of the products (G_p) and the free energy of the substrates (G_s), or DG = G_p - G_s. If the G_p is lower than G_s, it means that we have a negative DG, and therefore it is more favourable to be a product than a reactant (or substrate). Usually, the DG^o of a reaction is given in normal conditions of pressure, temperature, and concentration (1M for all components). However, it will be very hard, for example, to get [H⁺] = 1M, so in biochemistry we use DG^o', which refers to reactions happening at a pH of 7.0 ([H⁺] = 1 x 10^-7).

However, as you also probably recall from these chemistry courses, that a reaction is favourable (or spontaneous energetically) does not mean that the reaction will happen immediatelly. We have to remember that for the reaction to happen, we will have to rearrange the electronic structure of the substrate to form the transition state (i.e., a high energy specie that can go either way: towards reactants or towards products), and this takes a lot of energy. This energy is known as the activation energy (DG^R) of the reaction, and is this value that determines the rate at which the reaction will happen. The rate of a reaction is defined by the ammount of substrate that is consumed (or product that is created) per second, or rate law:

k_forward is the rate constant, and indicates how fast are substrates consumed (or products made). For a first order reaction like the one we are looking at here, k has units of s^-1. If ve have more substrates, S₁, S₂, S₃, the rate law may or may not depend one more than one of them, and we will have second order, third order, etc., rate laws. The activation energy (or barrier) is related to the rate of the reaction by the following expression:

Ais the Arrenhius constant. Note that it is similar but different to the one for the equilibrium of the reaction. We clearly see that the higher the activation energy (DG^R), the smaller the rate.

We can analyze a reaction like this by ploting its reaction coordinate diagram, in which we graph the free energy of the system versus its position along the reaction, or the reaction coordinate:

In this example, we can see that the free energy of the products (G_p) is smaller than the free energy of the substrate (G_s), and therefore the reaction is spontaneous thermodynamically; The equilibrium constant K_eq is large. Hoewever, the activation energy (DG^R) to get to the transition state (TS) is huge, and therefore the reaction will be slow; the rate constant k_forward is small. The reaction is not favourable kinetically, or in other words, the reaction is under kinetic control. This means that despite that it is a favourable reaction in thermodynamic terms, it won't happen readily due to its kinetics.

One thing in favour of the activation energy: As we see, it is what determines the speed of a reaction, and therefore if it will happen or not. We saw when we discussed proteins and peptide bonds that the hydrolysis of a peptide bond was thermodynamically favoured, but kinetically controlled. This meant that it had a large activation energy. If it wasn't for the DG^R, proteins would spontaneously hydrolyze, and there would not be a chance for life.

As we said before, in the lab we can give the necessary energy to overcome DG^R by pumping in thermal energy: Heat will give more energy to the molecules, and they will be able to overcome the energy barriers. In living systems, we cannot do this, and we therefore use enzymes to catalyze the reaction. In general, a catalyst is a molecule or molecular specie that combines with the substrates in such a way that it lowers the activation energy of the reaction. Usually, this means that it will combine and stabilize the transition state. Basically, everything depends on probabilities of things happening or not happening: the more time we have the TS around, the more chances we have for the reaction to go forward (or backwards), therefore we increase the reaction rate.

The reasoning is the same if the catalyst is an enzyme. The enzyme will combine with the substrate, and it will stabilize the transition state of the reaction (we'll see how later) so that the energy needed to get to it (the activation energy DG^R) is smaller. This will in turn increase the rate of the reaction. We re-write the equilibrium as:

We have more species, or intermediates, in the reaction: ES and EP. This should not be confused with the transition state! These are species that are long lived, and of relatively low energy. The reaction coordinate diagram is now:

The first, and perhaps most important, thing that we see is that the DG of the reaction is the same weather we use a catalyst or we don't. If the DG of the reaction is the same, K_eq, the equilibrium constant of the reaction, will also be the same. Ergo, a catalyst (enzyme) will not affect the position of the equilibrium in the reaction.

We also see that that after the enzyme combines to the substrate (ES), the energy required to get to the transition state is a lot smaller than without the enzyme. Smaller DG^R means larger k_forward, and a faster reaction. Again, we have to keep in mind that we get faster to the equilibrium, but we do not affect the equilibrium. Another important concept is that in the presence of the enzyme, the transition state may or may not be the same as in the nuncatalyzed reaction. The enzyme may decide to take the reaction through a different path which it can stabilize better than the path followed by the reaction in the absence of the enzyme.

Since we have now more that one specie taking place in the reaction towards products (S, E, ES and EP), we see that there are two new processes (reactions) that we have to take into account. These are binding of the substrate to the enzyme (E + S G ES) and release of the product from the enzyme (EP G E + P). Each of these processes is basically a new reaction, and will have its own DG and  DG^R. Therefore, we will have more equilibrium (K_eq) and rate (k_forward) constants. The reaction is not an unimolecular first order reaction, but higher in order. We will talk about this when we start discussing enzyme kinetics in detail.

Finally, the last leg of the diagram accounts for the release of product and 'recovery' of the enzyme. This enzyme can be in principle used over and over again to catalyze the same reaction. This is one of the rules that define a catalyst in general.

In sum, enzymes will make reactions that otherwise won't happen, happen. Going back to our peptide bond example, enzymes (guided by other signals that the organism generates) will lower the activation energy for the hydrolysis of certain peptide bonds in the degradation of a protein (proteases, chimotrypsin, trypsin, etc., etc.). In general, we can expect increases in the rates of reaction in the order of 10⁷ to 10¹⁴ fold (that is, 10 million to 100 trillion times faster!)

How do enzymes do what they do

Fine. Now we know in thermodynamic and kinetic terms why a reaction moves faster with the enzyme around. This explanations are macroscopical, i.e., they explain the 'whys', but not the 'hows'. To understand the 'hows' we need to look at the microscopic level. We said that the enzyme will stabilize the transition state of the reaction, which may be or may not be the same trnasition state that we see in the uncatalyzed reaction. How?

- First, the substrate (or substrates) will have to bind to the active sites of the enzyme. In this step of the reaction, all the concepts that we learned before about weak interactions (van der Waals, hydrophobic, h-bonding, electrostatic interactions) come back to haunt us. It is all these forces that will make an enzyme avid for its substrate. Also, at this stage, all the concepts of complementarity that we learned come to play. An enzyme that binds glucose will not bind benzene in its binding site, because benzene cannot complement the h-bonding sites present in the enzyme active site. Furthermore, it will not bind tightly to mannose (another hexose monosaccharide), because the stereochemistry in certain carbons of the substrate (the sugar) is different, and therefore there are going to be missmatches in the binding site. It is due to this (i.e., almost perfect substrate-active site complementarity) that enzymes are as specific for their substrate as they are.

- Second, all these weak interactions between the substrate and active site will contribute to the free energy of the ES complex. All these small contributions to the free energy of the ES complex are collectivelly named the binding energy of the ES complex. It is primarily from this reservoir of free energy that the enzyme will be able to take the necessary energy to overcome (or lower) the activation energy.

- Third, what if the enzyme binds to its substrate really really perfectly? The thing will be so stable that it would just sit there, and no reaction would happen, because what we need to get to is the transition state, which is structurally different from the substrate, and therefore would have less interactions with the enzyme. Not only do we need good complementarity between the substrate and the enzyme, but what the enzyme really shoots for is perfect complementarity with the transition state. In this way, the transition state becomes stabilized by weak interactions, and we increase the chance of it being converted to products. By having complementarity to the transition state, the combined energy contributions from weak interactions will lower the activation energy needed to reach the transition state.

Next time we will disect the binding energy into some of its components, and we will discuss types of catalysis done by enzymes, as  well as the functional groups that carry them out.

Prepared by Guillermo Moyna, 1999.