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Chemical kinetics for various
processes (I):
(1) Enzyme Reactions  The MichaelisMenten mechanism
 an example of consecutive elementary reactions
(1) Enzyme Reactions  The MichaelisMenten mechanism  an example
of consecutive elementary reactions
The MichaelisMenten mechanism is an example of a reaction involving
an intermediate. We shall consider a general case where a substrate S is
converted into products P (catalysed by the enzyme E) through the following
mechanism:
where ES denotes the substrate/enzyme intermediate, see fig. 1.
Fig. 1: The basis of the MichaelisMenten
mechanism of enzyme action. Only a fragment of the large enzyme molecule
E is shown.
This means that the rate of product formation is given by:
where is we assume a preequilibrium and apply the steady state approximation
we get:
i.e.:
Also, because the enzyme is not being consumed in the reaction, we may
conclude that the total concentration of the enzyme is given by (assuming
that the reaction volume remains constant):
i.e.:
i.e.:
i.e.:
i.e.:
i.e. the rate of formation of the product may be expressed in terms
of the effective rate constant, k, as:
where (see figure 2a):
where K_{M} is the Michaelis constant, given by:
According to these equations:

The rate of enzymolysis varies linearly with the initial enzyme concentration;

The rate of enzymolysis varies in a more complicated manner with the substrate
concentration (see plot of k/k_{b}vs. [S]
/ K_{M })
The values of K_{M} and k_{b} may be obtained by inverting
the equation:
to give the LineweaverBurk equation:
A plot of 1/k vs. 1/[S] (LineweaverBurk plot) which gives
1/k_{b} as intercept, and K_{M}/k_{b} as gradient.
(see fig. 2b.)
(a) 
(b) 
Fig. 2: (a)
The variation of the effective rate constant k with
substrate concentration according to the MichaelisMenten mechanism.(b)
A LineweaverBurk plot for the analysis of an enzymolysis
that proceeds by a MichaelisMenten mechanism, and the significance of
the intercepts and the slope.
>>> Let is now consider the
scenarios …
(A) When [S] >> K_{M} :
The rate of enzymolysis is given by:
This means that:
The rate becomes zeroorder in S.
Under these conditions, the rate is constant (i.e. There is so much S present,
that it remains at effectively the same concentration, even though products
are forming.
The rate of formation is at a maximum, and (1) k_{b} / [E]_{0}
is known as the maximum velocity, and
k_{b} is known
as the maximum turnover number.
(B) When, [S] is small, i.e. [S] << K_{M}
In such cases, we make use of:
and assume that K_{M} + [S] is approximately K_{M}, i.e.:
i.e. the rate is proportional to both [S] and [E]_{0}.

