The International Union of Biochemistry and Molecular Biology (IUBMB) has developed a classification scheme under which enzymes are classified into six major categories. They are summarized in this table, developed from descriptions in Horton's textbook, ch. 5:
|EC #||Name||Reactions Catalyzed||Sample Reaction||Comments|
|oxidation-reduction||lactate + NAD →
pyruvate + NADH + H+
|often involves NAD or FMN|
|2||transferases||transfer a group larger than water||alanine + α-ketoglutarate →
pyruvate + glutamine
|3||hydrolases||hydrolysis (elimination or transfer of H2O)||pyrophosphate + H2O →
phosphate + phosphate
|4||lyases||lysis making a double bond;
(converse: addition across double bond)
|pyruvate + H+ →
acetaldehyde + CO2
|often called synthases when reverse reaction is emphasized|
|5||isomerases||unimolecular reactions||L-alanine → D-alanine||some are mutases|
|6||ligases||joining of 2 substrates||glutamate +ATP +NH4+ → glutamine + ADP + Pi||generally require high-energy phosphate compounds like ATP|
Individual enzymes are numbered in a four-component system known as the "EC" system because the organization that originally developed the systm was called the Enzyme Commission. There are periods separating the levels of organization. Thus pancreatic elastase, a protease (enzyme that cleaves peptide bonds) is a hydrolase (category 3) and its full Enzyme Commission designation is 184.108.40.206:
If it helps you understand the system, think of the EC numbers as being like Internet Protocol (IP) addresses; the same logic applies. If you want to see the details of the EC numbering system, consult http://www.chem.qmw.ac.uk/iubmb/enzyme/.
The remainder of these notes concentrate on the kinetics of enzymatic reactions.
Kinetics is the study of reaction rates and the ways that they depend on concentrations of substrates, products, inhibitors, catalysts, and other effectors. If we consider the simple situation in which a reactant A is being converted to a product B under the influence of a catalyst C, such that at time t=0, [A] = A0, [B] = 0, then the rate or velocity of the reaction is expressed as d[B]/dt. In most situations more product will be produced per unit time if A0 is large than if it is small, and in fact the rate will be linear with the concentration at any given time:
d[B]/dt = v = k[A]
where v is the velocity of the reaction and k is a constant known as the forward rate constant. In this instance, since [A] has units of concentration and d[B]/dt has units of concentration / time, the units of k will be those of inverse time, e.g. sec-1.
Kinetics can become much more complicated than this if
more than one reactant is involved or if a catalyst whose concentration
influences the production of species B is present.
If more than one reactant is required for production of B,
then usually the reaction will be linear in the concentration
of the scarcest reactant and nearly independent of the concentration
of the more plentiful reactants.
If in the reaction
A + D → B
the initial concentrations of [A] and [D] are comparable, then the reaction rate will be linear in both [A] and [D]:
d[B]/dt = v = k[A][D] = k[A]1[D]1
This reaction is said to be first-order (rate directly proportional to concentration) in both [A] and [D]. The sum of all the exponents that appear in this equation is 1+1=2, so the reaction overall is said to be second-order. This time the units of k are L/(sec-1mol-1) if [A], [B], and [D] are expressed in moles/L.The rate in which the reverse reaction occurs may not be the same as the rate at which the forward reaction occurs. If the forward reaction rate of reaction 1 is designated as k1, the backward rate typically designated as k-1. In complex reactions, we may need to keep track of rates in the forward and reverse directions of multiple reactions. Thus in the conversion
A very common situation is one in which for some portion of the
time in which a reaction is being monitored, the concentration of the
enzyme-substrate complex is nearly constant. Thus in the general reaction
E + S ↔ ES ↔ E + P
where E is the enzyme, S is the substrate, ES is the enzyme-substrate complex (or "enzyme-intermediate complex"), and P is the product, we find that [ES] is nearly constant for a considerable stretch. This means that the rate at which new ES molecules are being produced in the first forward reaction is equal to the rate at which ES molecules are being converted to E and P. This is one of five assumptions we will need to make to complete this derivation of the Michaelis-Menten kinetics. The full list of assumptions is:
The rate of formation of ES is, by assumption (b),
vf = k1[E]avail[S] = k1([E]tot - [ES])[S]
because of assumption (c), which allows us to write the forward reaction as above, and assumption (d), which allows us to ignore production of [ES] from the right. Meanwhile, the rate of disappearance of ES on the right and left is a function of the [ES] concentration, and it is related to the conversion of ES leftward to E and S, which is k-1[ES], and the conversion rightward to E and P, which is k2[ES]. Thus
vd = k-1[ES] + k2[ES]
= (k-1 + k2)[ES].
Based on assumption (a), vf = vd. Therefore k1[ES] = (k-1 + k2)[ES]
which means if we define Km ≡ (k-1 + k2) / k1,
then [E]tot[S]/(Km + [S]) = [ES]
Now we can employ assumption (e), which tells us that the overall reaction rate, v0 = v2, That is, the overall velocity is equal to the velocity of the last step, the conversion of ES to E and P. But the latter velocity is v2 = k2[ES], so
v0 = k2[E]tot[S] / (Km + [S])
Under what circumstances, given a fixed [E]tot,
will the reaction operate at its maximal velocity?
It is logical to say that the highest possible rate we can achieve would
occur when the substrate concentration is very high,
i.e. [S] >> [E]tot.
If that is true, then v0 will be as large as it possibly
can be for a given [E]tot.
In that case, since [S] is very large, it will be much larger than
Km, so the denominator Km + [S]
≈[S], so for that case
v0 = k2[E]tot[S] / [S] = k2[E]
We describe the reaction velocity under these saturating conditions as Vmax, so we can write the general equation as
v0 = Vmax [S] / (Km + [S])
This is the form of the Michaelis-Menten equation with which we will ordinarily work.
We often are interested in the basic properties of an enzyme.
Km is an actual property of the enzyme;
Vmax is not,
because it will be much larger if we increase the enzyme concentration.
So we normalize it to the enzyme concentration and define
kcat ≡ Vmax / [E]tot
Now for our simple Michaelis-Menten system you can readily see that, kcat = k2; in more complex system kcat may have a more complex relationship with the underlying kinetic parameters.
The kinetic constants, Km, kcat, and Vmax provide information about the properties of an enzyme with respect to its substrates. The Michaelis constant Km measures how stable the enzyme-substrate complex is; kcat tells us the first-order rate constant for converting the enzyme-substrate complex, ES, to the product and the recovered enzyme; it describes the rate of the reaction when the substrate is not limiting. Vmax shows us the fastest possible rate at which the reaction will proceed given the current set of conditions. More briefly, Kcat characterizes the binding of the substrate to the enzyme; kcat characterizes the effectiveness of the enzyme in turnover, i.e. converting reactant to product or vice versa.
It's typical for the Km of an enzyme for a substrate to be close to the cellular concentration of that metabolite. This enables the enzyme to respond proportionally to changes in the substrate concentration; the enzyme then can act as a kind of buffer of the concentration of that substrate as external influences modify that concentration.
An important parameter is the ratio of kcat to Km. As Horton says, this measures the apparent second-order rate constant for the formation of enzyme and product from enzyme and substrate when the overall reaction is limited by the encounter of S with E. Some particularly effective enzymes deliver a value of kcat / Km in the range of 108 to 109 M-1s-1. Under these conditions the reaction is in fact diffusion-limited; that is, the reaction is proceeding as fast as it can given the ability of the two solutes to approach one another.
This ratio, kcat / Km, can be described as a measure of the specificity of the enzyme. Horton provides the details, but the general point is this: given two different substrates, the one with the higher kcat / Km ratio is the one for which the product can be produced the most rapidly.