Answer all parts Substrate A reacts with enzyme E to form an activated complex A
ID: 717522 • Letter: A
Question
Answer all parts
Substrate A reacts with enzyme E to form an activated complex AE that gives product P in the following reaction scheme: -l (E)o is the initial concentration of total enzyme sites and (A)o is the initial concentration of A. 1. Write the differential equations that describe the change in concentration of A and AE as a function of time, and the enzyme site balance. Non-dimensionalize the ODEs using the following variables, which are non-dimensionalized to be of order 1 (O(1): (A)o (E)o Show the derivation to write the non-dimensionalized ODEs in the following form, and define the constants , and u' in terms of dimensional variables: 2) =-yz + "(1-2) dz In the parlance of singular perturbation theory, e is the "small" parameter that cannot be set to zero to obtain an accurate solution to z(T) for initial values of t. Using non- dimensional variables, write the inequality that must hold for the pseudo-steady-state hypothesis (PSSH) to accurately describe this system and write the equality that is a consequence of applying the PSSH. 2. 3. Rewrite this inequality and equality in dimensional form. Explain why this inequality satisfies the two requirements needed for the PSSH on AE to be a valid one.Explanation / Answer
(1)
A basic enzyme reaction model by Michaelis and Menten can be represented with the following
diagram:
S + E
k1
k1
SE
k2
P + E (1)
where S is the substrate, E is the enzyme, SE is the substrate-enzyme complex, P is the product,
and k1, k1, and k2 are constants [3]. This diagram states that one molecule of the enzyme
combines with one molecule of the substrate to form a molecule of the substrate-enzyme complex.
The substrate-enzyme complex could decompose back into enzyme and substrate or may give rise
to one molecule of the product and one molecule of the enzyme. Let s = [S], e = [E], c = [SE],and p = [P], where [ ] denotes the concentration of a substance. The Law of Mass Action states
that “the rate of a reaction is proportional to the product of the concentration of the reactants”
[3]. We use this law to derive the following equations for the enzyme reaction described above:
ds
dt = k1es + k1c,
de
dt = k1es + (k1 + k2)c,
dc
dt = k1es (k1 + k2)c,
dp
dt = k2c,
s(0) = s0, e(0) = e0, c(0) = 0, p(0) = 0.
(2)
where k1, k1, k2, s0, and e0 are positive constants [3].
Notice that
dc
dt +
de
dt = 0 c(t) + e(t) = e0.
Hence, we can write
e(t) = e0 c(t). (3)
Moreover,
dp
dt = k2c p(t) = k2
t
0
c( )d.
Thus, we only need the equations for s and c. Plugging equation (3) into the first three equations
of system (2) we obtain the following reduced system
ds
dt = k1e0s + (k1s + k1)c,
dc
dt = k1e0s (k1s + k1 + k2)c,
s(0) = s0, c(0) = 0.
(4)
For enzyme-substrate reactions, it is usually assumed that the initial stage of enzyme-substrate
complex formation occurs very fast, and subsequently the reaction goes to a quasi-equilibrium or
quasi-steady state, in which the concentration of the complex barely changes ( dc
dt 0) [3]. In this
case, solving the second equation of system (4) for c c(t) = k1e0s(t)
k1s(t) + k1 + k2
=
e0s(t)
s(t) + k1+k2
k1
=
e0s(t)
s(t) + Km
(5)
where Km denotes the Michaeli’s constant.
Plugging the above result for c(t) into the first equation in system (4) we obtain the following:
ds
dt = k1e0s(t) + (k1s(t) + k1)
e0s(t)
s(t) + Km
=
k1e0s(t)(s(t) + Km) + (k1s(t) + k1)e0s(t)
s(t) + Km
=
(k1 k1Km)e0s(t)
s(t) + Km
=
k2e0s(t)
s(t) + Km
(6)
Now, we can solve for s(t) by using separation of variables and integrating both sides of equation
(6) from 0 to t to obtain
s(t) + Km ln(s(t)) = s0 + Km ln(s0) k2e0t. (7)
Notice that the solution we obtained here does not satisfy the initial conditions in system (4); for,
if c(0) = 0, then either s0 = s(0) = 0 or e0 = 0, by equation (5). This result would imply that
either there is no substrate or no enzyme (or neither) in the beginning of the process. However,
this quasi-steady state solution might be “a reasonable approximation for most of the time” [3].
(2)
As done by Murray in [3], we use the substitutions
= k1e0t, u( ) = s(t)
s0
, v( ) = c(t)
e0
=
k2
k1s0
, K =
k1 + k2
k1s0
=
Km
s0
, =
e0
s0
(8)
to rewrite system (4) in the form
du
d = u + (u + K )v,
dv
d = u (u + K)v
u(0) = 1, v(0) = 0.
(9)
We will focus our discussion on this non-dimensional system. The equations in system (9) are not
easy to solve analytically. In our analysis, we must keep in mind that is a very small, positive
parameter. Notice that is multiplied by dv
d . As we saw before, we cannot expect to find a
uniformly valid approximate solution of the system by simply setting = 0, because that would
reduce the order of the equation. This is, in fact, what we did before when we derived the
approximate solution given by equations (5) and (7), which do not satisfy the initial conditions in
system (4). This situation is characteristic of a singular perturbation problem.
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