Consider a chemical reaction in which molecule A is converted to equal amounts of B and C.

In chemical reactions, A is termed a reactant whereas B and C are products. The rate of formation of B is a product of the concentration of A (designated by [A]) and a rate constant (k) characteristic for the reaction.

Using this equation, one can describe in graphical from how the rate of the formation of product B will be proportional to the [A].  The slope of the graph will equal the rate constant for the reaction.

This is called a first order rate since it is proportional to the [A]¹.  An example of this sort of reaction would be the hydrolysis of Nutrasweet or Aspartame, the peptide artificial sweetener.  This molecule is a dipeptide of aspartate and phenylalanine with a methyl ester on the carboxy-terminal end.  When used in diet soft drinks, which have a pH of about 3, there is a gradual acid hydrolysis of the molecule, producing a non-sweet product.

 This gradual hydrolysis destroys the sweet-tasting property of the soft drink.  By knowing the rate of hydrolysis of Nutrasweet at the pH and normal storage conditions of the product, food scientists can predict a shelf life which will help ensure consumer satisfaction.

Things can sometimes be more complex.  For a reaction in which two molecules of A are reactants, the rate of product formation is proportional to the square of [A]. 

This is a second order rate since the formation of product is proportional to [A]².  Notice that a graph of product formation versus [A] is a curve, but when the rate of product formation is plotted versus [A]², the line on the graph becomes a straight line.

These graphs are useful because they allow actual experimental data to be plotted and easily determine n (order) for Aⁿ without having to use a computer program to solve the equations.

Part 2.  Simple Enzyme Kinetics:

In biochemistry, a reactant is called a substrate and the reaction is catalyzed by an enzyme.  Enzymes are catalysts (Chapter 1) and as a consequence enzymatic reactions behave in a fundamentally different manner from purely chemical reactions.  Because enzymes themselves are complex biomolecules, they are sensitive to environmental conditions such as pH, temperature, ionic strength and the presence of organic or metal cofactors.  When examining the kinetics of an enzyme in vitro (from the Latin 'in glass'), the conditions need to be defined.

Think of enzymes as small particles which bind substrates on their surface to catalyze reactions to form products.  There are two rate constants (k) for each reaction because reactions are always reversible.  The rate constant for the combination of E and S to form the ES complex is designated as k₁.  The dissociation of the ES complex to reform E and S is designated as k_-1.  The same convention is used for the second step in catalysis in which the ES complex reacts to form free E plus product (P).

If we look at what happens when substrate and enzyme are mixed, it can be divided into four periods  

General assumption [E] <<< [S]

In biological systems, enzymes are often present at concentrations of 10^-12 to 10^-9 M, whereas substrates are usually present at concentrations of 10^-6 to 10^-2 M.

1) Instantaneous time

The time it takes to bring ES into equilibrium with E + S
the reverse rate increases until it becomes equal to forward rate (k₁[S] = k_-1[ES])
[ES] increases and then becomes constant
[S] decreases so little that it is essentially constant
During this time [P] = 0

2) Initial Rate

[P] is negligible, but starts to increase
[ES] is constant and depends on [S], i.e. higher [S], higher [ES]

The rate of product formation is constant

3) Longer Times

[P] > 0; math becomes difficult because forward and back ratios change constantly

4) Equilibrium (very long times)

(k₁[S] = k_-1[ES])

(k₂[ES] = k_-2[P])

there is no net conversion of S to P

In studying the properties of enzymes, biochemists usually work under conditions of initial rate, because it is a simple, controlled system. It is easy to measure and the mathematics are relatively simple.

At low concentrations, the [ES] is proportional to the [S] and the rate of the reaction is proportional to [S] or first order. Biochemists also call the  function a velocity (v)

However, at higher [S], we reach a point at which all the enzyme molecules have bound a molecule of substrate and [E] = 0. At that point, adding more S causes no increase in [ES] and no increase in the reaction. This is called zero-order kinetics.

The maximal rate is called the maximal velocity or V_max and occurs when [substrate] is said to be "saturating".  In the terms of chemical kinetics, this lack of change in velocity of product formation is said to be zero order since the response is proportional to [S]⁰.

Connecting the linear portion of the graph at low [S] and the horizontal portion at high [S], we see a nonlinear relationship between velocity and [S] which is hyperbolic in nature.  Also, the velocity of product formation is usually designated v.

This type of kinetic response to [S] is termed Michaelis-Menten.  In this system, one can define the substrate concentration that causes a rate equal to half-maximal velocity (0.5 V_max).  Biochemist call this the K_m (Michaelis constant) and it is a measure of the affinity of the enzyme for the substrate molecule.

High K_m is low affinity of the enzyme for the substrate.
Low K_m is a high affinity of the enzyme for the substrate.

Does this have a practical meaning?  

Michaelis Equation

At low [S] the v is a small % of V_max

At high [S] the v is 99% of V_max

Things to notice:

1. Even at high [S], v < V_max

2. at v / V_max = 0.5, then [S] = K_m

In real terms, K_m is the [S] that causes half-maximal velocity

Consider the figure below of three enzymes with the same V_max value and different K_m values.  The V_max has arbitrarily been set at a value of 1.0.

Relative to Enzyme 2, Enzyme 1 has a lower K_m value and thus a greater affinity for the substrate.  Because Enzyme 1 has a greater affinity for the substrate at all concentrations, a greater proportion of the Enzyme will be in the ES form and therefore the velocity catalyzed by that enzyme will be greater.  Similarly, Enzyme 3 has a higher K_m value than Enzyme 2, indicating a lower affinity for the substrate.  At any given [S], less of Enzyme 3 will form an ES complex than for Enzyme 2.  This results in a lower velocity for enzyme 3 at all substrate concentrations.

The K_m is a characteristic for each enzyme. The V_max depends on how much enzyme is added to the assay (if the amount of enzyme is doubled, the V_max is doubled).

Notice that the K_m (or affinity) is equal for the two enzymes.  Doubling the amount of enzyme in the assay doubles the velocity at every substrate concentration.

What is the physiological significance of K_m?

Most Biochemical reactions occur as part of a pathway.  Below is a pathway with four intermediates, M, N, O and P.  Three enzymes, E1, E2, and E3 catalyze the reaction of each intermediate to form the next intermediate in the pathway.

There is a normal concentration of N in the cell. Too little slows down the whole pathway, too much may be toxic or wasteful.  For long term control, increasing or decreasing the amount of E2 may increase or decrease the production of O and P from N. However, in biological systems, short term controls are essential; increasing the activity of E2 by synthesizing more of this protein would probably take hours in eukaryotic systems.

For many biochemical systems, normal physiological substrate concentration, also called steady-state concentration, is near the K_m of the enzyme which utilizes it as a substrate.

As a consequence, any increase in cellular [N] causes an increase in E2 activity which causes N to be converted to O at a higher velocity and would tend to cause [N] to decrease. As [N] drops below the K_m, the activity of E2 decreases and N utilization decreases and the [N] starts to rise. This concentration around which [N] oscillates is called the "steady state concentration". This is not the equilibrium concentration. 

Competitive Inhibitors

These are molecules, often similar in structure to the substrate, which bind in the active site and displace some of the substrate.  Because they compete with the substrate for the active site, they cause inhibition by causing an increase in the apparent K_m (decrease in apparent affinity) for the substrate.

This inhibition can be overcome by increasing the [S] which, in turn, displaces the inhibitor. Since [I] is finite, the enzyme can still reach V_max at infinite [S]. However in the presence of the inhibitor, the apparent K_m for the substrate is increased.

Noncompetitive Inhibitors

These are molecules or metals which bind or react to inhibit, but do not interfere with the binding of substrate to the enzyme. The amount of activity is inhibited at all S concentrations. The K_m for substrate is unaffected, but the V_max is decreased.

For an example of inhibitors which mimic the molecular structure of a substrate, consider the new class of synthetic HIV protease inhibitors; see if you can determine which part of their structure resembles a peptide.

Crixivan^®-indinavir (Merck)

Invirase^®-saquinavir (Roche)

Norvir^®-ritonavir (Abbott)

Vlracept^®-nelfinivir (Agouron)

VX-478^®(Vertex)

• Enzymes and Chemical Reactions [local]
• MIT Biology Hypertextbook: Enzyme Biochemistry [local]
• Enzymes and Catalysis [local]
• Information about Enzyme Inhibitors [local](don't read the section about regulation yet).
• Review of enzyme kinetics [local]