192
Chapter 6
that the enzyme could exist in two forms
E
= free enzyme, and
E S
= an enzyme complex with the
substrate. The conversion of substrate S to product
P
proceeds in two steps
*
‘ »
k
E + S
ES —
E + P
(6 2)
The first reaction is reversible, the second irreversible. Both reactions are assumed to be
elementary reactions with rate proportional to the concentration of reactants. Furthermore Briggs
and Haldane assumed that the concentration
(es)
of
ES
was constant in time, i.e.
ES
is in a
“pseudo-steady state”.
-
o =
k^e ■
s -
(es) —
k2 (es)
(6.3)
dt
The total enzyme “concentration” e0 was assumed to be constant and representing the “activity”
in eq. (6.1)
e0 = e
+ (es)
(6.4)
From (6.3)
(es) =
V o *
k^s
+ (A:_,
+ k 2)
e 0s
s +
k_x + k 2
k\
(6.5)
The rate of the reaction is determined by the decomposition of
ES
(i.e. the first reaction is much
faster in both directions than the second), and if
E S
decomposes by a first order reaction then
r - k 2 (es)
=
k2e0s
k_
| +
k2
k
2
e0s
s + K „
(
6
.
6
)
If the first reaction is infinitely fast and the substrate concentration is much higher than the
enzyme concentration
e0
then the concentration of the enzyme-substrate complex is
/
\
e ' s
(«)= —
K - e q
(6.7)
where
K
e, is the equilibrium constant for the dissociation of
ES
to
E
and
S.
Inserting (6.4) in (6.7)
and again assuming first order decomposition of
E S
by the second, rate determining reaction:
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