The Michaelis-Menten equation:

A distinctive feature of enzyme catalyzed reactions is that they reach a maximum, limiting rate of reaction as the substrate concentration is progressively increased. (This would not be the case with ordinary chemical reactions, see the previous topic).

When the rate (velocity) of an enzyme-catalyzed reaction is plotted against the concentration of the substrate, a rectangular hyperbola is obtained.

This saturation phenomenon led to the hypothesis of the formation of a reversible enzyme-substrate complex, and this implies much less enzyme than substrate.

Derivation of the L. Michaelis-M. Menten (1913) equation for an enzyme with single substrate (similar to the derivation of the Briggs-Haldane equation):

1. WRITE THE CHEMICAL EQUATION:

                 k1      k2
E  +  S  <===>  ES        ---> E +  P
                k-1

2. AT STEADY STATE: RATE OF "ES" FORMATION EQUALS RATE OF BREAKDOWN

k₁[Efree][S] = (k_-1 + k₂)[ES]

3. DEFINE K_m:

K_m-1 = k₁/ (k_-1 + k₂)

or K_m = (k_-1 + k₂)/ k₁

4. DIVIDE BOTH SIDES OF THE STEADY STATE EQUATION (above):
by (k_-1 + k₂), and substitute K_m for small k's:

[E]_free [S]/ K_m = [ES]

5. REPLACE [E]_free by [E]_total - [ES] (this must be done because the value of E_free cannot be determined, it must be expressed in other terms that we can measure.)

([E]_totalM - [ES])[S]/ K_m = [ES]

6. SOLVE FOR [ES]

[S][E_t] - [ES][S] = K_m [ES]

[S][E_t] = [ES]( K_m + [S])

[ES] = [S][E_t]/(K_m + [S])

At this point have [ES] in terms of the measurable quantities [E_t] and [S]

Rate of product formation (v) = k_cat[ES], and define v_max = k_cat[E_t] (when all enzyme has bound substrate) therefore:

k_cat[ES] = k_cat [Et] [S]/(K_m + [S])

v = v_max [S]/(K_m + [S]) This is the Michaelis-Menten equation!

K_m physically corresponds to the [S] that produces half the maximal enzyme activity. When [ES] = [E]_total, then v = v_max.

The Michaelis-Menten equation describes enzyme activity as a function of V_max, K_m and substrate concentration.

This general form of equation describes a rectangular hyperbola.

Making kinetic measurements:

Making kinetic measurements gives us valuable information such as: the concentration of substrate required for maximal enzyme activity, and an estimate of how tightly (strongly) the substrate interacts with the enzyme.

Kinetic measurements are made by varying the concentration of substrate and measuring the initial rate of product formation.

A mathematician needs 3 points to define a regular curve (in this case 0, K_m, and V_max), but this type of data is difficult to plot.

The double reciprocal plot (Lineweaver-Burk plot)

A more useful graph is obtained if the Michaelis Menten equation is transformed to the form y = mx + b, so a straight line is produced when the data are plotted graphically. This permits easier curve fitting using a least squares program. The Michaelis-Menton equation can be transformed into a linear form as follows:

v = v_max[S]/([S] + K_m) Take the reciprocal of each side of the equation

1/V = 1/V_max + Km\_m/V_max x 1/[S] Slope-intercept form

A straight line is obtained when 1/V is plotted against 1/[S]; K_m and V_max are constants

Measuring catalytic efficiency:

[E] + [S] <==> [ES] -> [P]

No matter how fast an enzyme can catalyze a reaction, there is a physical limit on the overall rate of the reaction.

This rate is determined by how quickly the enzyme and the substrate collide (i.e. it is diffusion controlled).

Bimolecular chemical reactions are limited to a maximum rate of 10⁸ to 10⁹ molar^-1sec^-1.

For enzyme-catalyzed reactions, we know that:

a. k_cat has the units of inverse time (sec^-1)

b. K_m (the concentration of [S] at half V_max has the units of molarity (M)

Taking the ratio of k_cat/K_m gives units of molar^-1 sec^-1 (the units of the rate of a bimolecular reaction)

k_cat/K_m = k_catk₁/(k_-1 + k_cat) but, for an efficient enzyme we would expect k_-1 would be very small,

therefore, the expression above tends to k_catk₁/k_cat => k₁ (as k_-1 approaches zero)

In other words, collision of the enzyme and its substrate limits the rate of the reaction (i.e., it is diffusion controlled!).

For acetylcholine esterase, carbonic anhydrase, triose phosphate isomerase this ratio is between 10⁸ and 10⁹ molar^-1 sec^-1, indicating that they have achieved near catalytic perfection. This is not bad for stumbling around in the evolutionary dark!

Turnover number:

The turnover number describes the number of moles of substrate converted to product per mole of enzyme per unit time (at enzyme saturation).

V_max = k_cat[E_T] = mmoles/min.

If we know the amount of enzyme responsible for this reaction, then

mmoles/min. x (mmoles enzyme)^-1 = min.^-1

For catalase this value is 10⁶ min^-1

It is difficult to envision any process that occurs one million times a minute, but many enzymes do it with ease!

It is a fundamental characteristic of the protein, the substrate molecule, temperature, and the composition of the aqueous solution. Temperature determines the amount of kinetic energy molecules have; the viscosity of the aqueous solution determines how fast molecules diffuse (and thereby the rate of collision between them).

Under physiological conditions, enzymes are usually not fully saturated (rather they are in the range of 0.05 -> 0.5 saturation).

This allows for control of the enzyme activity and a reserve capacity for catalysis.