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  Derivation of M-M Equation

Derivation of M-M Equation:

Leonor Michaelis and Mand L. Menton in the year 1913 planned a successful description for the influences of substrate concentration on the enzyme activity.

As per to them the enzyme E, and the substrate S combines quickly to make a complex, the enzyme substrate complex ES. After that this complex breaks down comparatively and slowly to make the product P of the reaction .These sequence of reactions can be presented in the following equations.

267_mm equation.jpg


 K1 & k2 are denotes the rate constants of the forward and backward reactions in the step 1 K3 & k4 are the rate constants of the forward and backward reactions correspondingly in the step2. This is right only for the enzyme reactions that fulfil the following conditions:

i. Only a single substrate and a single product are included.
Ii. The reaction proceeds necessarily to completion.
Iii. The concentration of the substrate is much greater as compared to that of the enzyme in the system.
Iv. An intermediate enzyme substrate complex is made.
V. The decomposition’s rate of the substrate is proportional to the concentration of the enzyme substrate complex.   

It is supposed that the concentration of S is much greater as compared to that of E and that only primary velocities are measured, in which only a small fraction of S has been converted. Within these conditions, concentration of P → ES may be ignored.

By applying law of mass action to the first step of the reaction where k1 and k2 are the rate constants for the forward and backward reaction correspondingly,

Forward reaction rate   = k1 [E] [S]                (1)
Backward reaction rate = k2 [ES]                   (2)

By applying law of mass action to the second step of the reaction where k3 and k4 are the rate constants for the forward and backward reaction correspondingly,

The rate of forward reaction = k3 [ES]             (3)

The rate of backward reaction can be ignored.
The total enzyme in the system can be presented like,

[Et] = [E] + [ES]                         (4)

In which [E] is the uncombined free enzyme concentration, [ES] the enzyme substate concentration and [Et] the total enzyme concentration.

The velocity of the complete reaction is

V = k3 [ES]                        (5)

The above reaction is the actual rate equation for the complete reaction but it is not helpful since neither k3 nor [ES] can be calculated directly. It is supposed that the reaction carries on at steady state in which the rate of formation of [ES] equivalent to the rate of degradation of [ES].

The formation rate of ES, Vf is proportional to E and S like in any second order reaction.

Vf = k1 [E] [S]
     = k1 ([ET] – [ES]) [S]                         (6)
The disappearance rate of (ES), Vd is
Vd = k2 [ES] + k3 [ES]
      = k2 + k3 [ES]                         (7)

Because in the steady state Vd = Vf

k1 ([Et] – [ES]) [S] = K2 + k3 [ES]                 (8)

Rearranging this equation provides
[S] ([Et] – [ES])      K2 + k3
——————— = ———     = Km                (9)
[ES]                                                k1
In which Km is the Michaelis – Menton constant, a helpful parameter characteristic of every enzyme and a substrate.
Rearranging this equation via solving for [ES]
[S] [Et] – [S] [ES]
--------------------------     =    Km                              (10)
            [ES]

km [ES] = [S] [Et] – [S] [ES]
km [ES] + [S] [ES] = [S] [Et]
[ES] ( Km + [S] ) = [S] [Et]
            [Et][S]
[ES] = ------------                                       (11)
             Km + [S]

As per to the previous equation (5), V =k3 [ES]. Substituting the value of [ES] (11) in this equation we acquire,
        k3 [Et] [S]
V = —————                         (12)
           Km + [S]
The maximal velocity Vmax is equivalent to
Vmax = k3 [Et]                         (13)
Substituting the value of k3 [Et] in the equation (12), the final Michaelis
– Menton rate equation becomes,
          Vmax [S]
V =  —————                         (14)
           Km + [S]

Here, while V is equal to half of the maximum velocity, that is V = Vmax / 2
So,   Vmax /2 =   Vmax [S] / Km + [S]                           (15)
Rearranging,
Km + [S] = 2 [S]
Hence
Km = [S]                                               (16)

 

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