The Biochemistry 111 lecture course

Enzyme kinetics

This page explains enzyme kinetics in much more detail than you need to know . It is here for those of you who wish to extend your understanding of the subject.

Maude Menten

Though many biologists recognize Menten as the co-discoverer of one of the fundemental equations in Biochemistry, the Michaelis/Menten equation, they know little more. Maud Menten became one of the first women doctors in Canada in 1911. Two years later she and Leonor Michaelis published a paper describing the relationship between the rate of an enzyme-catalysed reaction and the concentration of the enzyme's substrate. Dr. Menten went on in 1944 to publish on what may be the first use of electrophoresis to separate proteins. She then went on to help develop an important technique in enzyme histochemistry. During her 50 year career she was involved in the publication of some 70 scholarly articles. Though under-appreciated, Maud Menten certainly stands as an important contributor to the fields of enzyme kinetics and histochemistry.

Assaying Enzymes

To study an enzyme, an assay is necessary. The assay is a measurement of a chemical reaction, which might involve measuring the formation of the product. For example, b-galactosidase catalyses the following reaction:

lactose ---> glucose + galactose

For this reaction, measuring the formation of glucose would constitute an assay. Because this is technically difficult, an easier way to follow the reaction is to use a substrate that gives a colored reaction product. b-galactosidase can catalyze many reactions of the following general type:

y-galactose --> y + galactose

in the 'normal' reaction, y-galactose = lactose = glucose-galactose (y=glucose)
Two alternatve substrates for b-galactosidase are:

ONPG = ONP-galactose (ONP = o-nitro-phenol)
X-gal = X-galactose (X = 4-chloro-3- bromo indole)

Both ONPG and X-gal are colorless, but when hydrolysed by b-galactosidase, they produce colored products:

ONPG --> galactose + ONP
(colorless) (colorless) (yellow)

X-gal --> galactose + 4-Cl-3-Br-indigo
(colorless) (colorless) (deep blue)

These colored reaction products are much easier to detect and can be used to assay (measure) enzyme activity more easily that the 'normal' reaction catalyzed by b-galactosidase.

Enzyme catalysis

Here is an example of an enzyme catalyzed reaction:

The enzyme converts S to P (substrate to product). Initially, [P] is small, so the majority reaction is S-->P. Later, as [P] grows, the back reaction rate increases, until equilibrium is reached (enzymes catalyze BOTH forward AND reverse reactions). To measure the kinetic properties of a given enzyme, you must perform many experiments like the one above. Keep the enzyme concentration constant and measure the initial rate of product formation (before the reaction is anywhere near equilibrium) at several different initial substrate concentrations. Then plot the initial velocity of the reaction: Vo, as a function of [S].

Some qualitative statements about this graph:

Velocity is dependent on [S]
The more enzyme added, the faster the reaction goes

Why isn't the graph linear? As [S] gets large, the enzyme becomes limiting - all enzyme molecules are 'busy' operating on the substrate and the rate of reaction depends on the amount of enzyme, not the amount of substrate.(And since the amount of enzyme used in this series of experiments is fixed, the rate asymtotically approaches a maximum.)

NOTE: the two preceeding graphs look very similar, but mean quite different things! It is important to understand the difference between an individual reaction: [P] vs. t (from which you get Vo) and a kinetic graph of many such reactions: Vo vs. [S].This kinetic behavior can be modelled mathematically. That is what we discuss next, after a short review of chemical kinetics.

Enzyme kinetics I

The symbols represent (in our example):
E enzyme (b-galactosidase)
S substrate (lactose)
ES enzyme-substrate complex
P product (glucose or galactose)

First consider the initial velocity of the reaction. In this case, there will be a negligibly small amount of product present ( [P]<5% of [S] is considered negligible). Under these conditions, the back reaction is negligible, that is, k-2[P]= 0 (approximately). The initial velocity is simply:

Vo = k2[ES] (1)

The problem with this equation is that the quantity [ES] cannot be measured. However,
[S] (the initial concentration of substrate) is known,
[P] (product produced) can be measured, and
[E]tot (the amount of enzyme added to the reaction) is known.

Now what we can do is use the rate equations plus a few other assumptions to derive an expression for [ES] (which we cannot measure) in terms of quantities which we can measure ([S], [P], and [E]tot). Okay, here we go . . .

Enzyme kinetics II

Enzyme kinetics III - example 1

Example1: [S] very large

Enzyme kinetics IV - Example 2

Example 2: [S] small

Enzyme kinetics V - Example 3

Example 3: [S]=Km

Enzyme kinetics VI

Enzyme kinetics VII

Enzyme kinetics VIII

Competitive Inhibition

The inhibitor binds to the active site and competes with the substrate for binding. In this case, at low [S], the inhibitor competes for the active site and effectively lowers the [S] at the active site. This lowers the rate at low [S]. This increases the apparent Km. At very high [S], the level of S overcomes the inhibition by mass action and Vo approaches Vmax. Vmax is unchanged because all of the enzyme molecules are active.

Competitive Inhibition

Competitive inhibition occurs when the substrate and a substance resembling the substrate are both added to the enzyme.
A theory called the "lock-key theory" of enzyme catalysts can be used to explain why inhibition occurs.The lock and key theory utilizes the concept of an "active site." The concept holds that one particular portion of the enzyme surface has a strong affinity for the substrate. The substrate is held in such a way that its conversion to the reaction products is more favorable. If we consider the enzyme as the lock and the substrate the key - the key is inserted in the lock, is turned, and the door is opened and the reaction proceeds. However, when an inhibitor which resembles the substrate is present, it will compete with the substrate for the position in the enzyme lock. When the inhibitor wins, it gains the lock position but is unable to open the lock. Hence, the observed reaction is slowed down because some of the available enzyme sites are occupied by the inhibitor. If a dissimilar substance which does not fit the site is present, the enzyme rejects it, accepts the substrate, and the reaction proceeds normally.

This diagram was cropped from

Non-Competitive Inhibition

The inhibitor binds to another site on the enzyme and inactivates the enzyme molecule. This effectively reduces the [E]tot available for catalysis. Since Vmax is proportional to [E]tot, Vmax is reduced. Since the remaining active enzyme molecules are unaltered, Km is unchanged.

The lock and key theory - non-competitive inhibition

Non-competitive inhibitors are considered to be substances which when added to the enzyme alter the enzyme in a way that it cannot accept the substrate.

This diagram was cropped from

These pages are brought to you by the Biochemistry Department, University of Otago 1999 for use by undergraduate BIOC 111 students. Material for these pages has been obtained from a variety of sources. Where possible these sources have been acknowledged. Enquiries or corrections should be addressed to the webmaster. Site designed by Jason Tagg, driven by a custom FileMaker Pro solution.