1
Molecular networks
Life is based on an amazing number of biochemical reactions occurring at the cellular
level. These reactions occur in very complex networks where the product of one
reaction is then used as a component of another one, or possibly as a substance
increasing (an ‘enzyme’) or decreasing (an ‘inhibitor’) the rate at which another
reaction occurs. Parts of these networks are presented (and often simplified) in those
schemes, such as Krebs’ cycle, so prominent in textbooks of biochemistry.
Similarly to what we have done for models in ecology, in this book we do not
attempt to model a realistic network that may involve hundreds of chemical species
and reactions, but limit ourselves to describe the main principles and ideas developed
for modelling the building blocks of such complex networks. Namely, in Section 1.1 we
will model simple enzymatic reactions, in Section ?? we will describe the qualitative
properties of some small molecular networks, whose behaviour may mimic that of
relevant molecular processes.
The best introductory reference for a more extensive treatment of the topic is the
book by Keener and Sneyd “Mathematical Physiology”.
1.1
Modelling enzymatic reactions
A fundamental property of biochemical reactions in living cells is the fact that the
rate at which they occur is regulated by the presence of other molecules. An enzyme
is a protein that makes it possible that a reaction (that by itself would be extremely
slow) occurs over a short time-scale, as necessary for the cell functioning, but is
apparently not involved in the reaction itself, in the sense that its concentration at
the end is the same as at the beginning.
An enzymatic reaction is often presented as:
k1
→
S + E −−
−2→ P + E
k−1 SE −
←−−
k
In the diagram S refers to the ‘substrate’ the molecule that is converted into P , the
‘product’ of the reaction, while E refers to enzyme molecules. Looking only at the
left and right end-points of the diagram, one sees that a molecule of S is converted
into P , while one is left with same molecule of E present in the beginning. However,
one sees that E is involved in intermediate reactions, necessary to understand why
the presence of the enzyme is necessary for the reaction to occur.
1
The letters over each arrow in the diagram represent the rate constants at which
each reaction (represented by an arrow) occurs. We assume that all reactions occur
following the law of mass action: when a reaction involves two substances reacting,
the rate at which the reaction occurs is proportional to the product of the concentrations of the substances involved; the rate constant is the proportionality constant.
On the other hand, when a reaction involves a single substance, the reaction rate is
the product of the rate constant times the concentrations of that substance.
It is easy transforming the diagram into a set of differential equations, using the
principle that the concentration of each substance will increase at the sum of the rates
of all reactions producing it, and decrease at those destroying it. Namely setting
s(t) = concentration of substrate S at time t
e(t) = concentration of enzyme E at time t
we obtain
c(t) = concentration of complex SE at time t
p(t) = concentration of product P at time t
c0 (t) = k1 s(t)e(t) [input] − (k−1 c(t) + k2 c(t)) [output] .
Analogously
s0 (t) = k−1 c(t) − k1 s(t)e(t)
e0 (t) = k−1 c(t) + k2 c(t)) − k1 s(t)e(t)
p0 (t) = k2 c(t).
This appears to be a system of 4 differential equations with 4 unknowns s(t), e(t), c(t),
p(t). Typical initial conditions will be s(0) = s0 > 0, e(0) = e0 > 0, c(0) = p(0) = 0
(at time 0, start of the reaction, there is no product, nor complex SE).
One can immediately see that the first 3 equations do not depend on p(t); we can
then solve the first 3 equations, then compute p(t) (the quantity presumably more
relevant) as
Z
t
p(t) = k2
c(τ ) dτ.
0
Moreover, one may note that
d
[c(t) + e(t) = 0] =⇒ c(t) + e(t) = c(0) + e(0) = e0 =⇒ e(t) = e0 − c(t).
dt
One can then examine the system of 2 equations for s(t), c(t), using e(t) = e0 − c(t).
Finally the system to be studied is
s0 (t) = k−1 c(t) − k1 s(t)(e0 − c(t))
(1)
c0 (t) = k1 s(t)(e0 − c(t)) − (k−1 + k2 )c(t)
with initial values s(0) = s0 > 0, c(0) = 0.
It is certainly possible analyse the behaviour of system (1) as it is. However,
it is more customary and insightful simplifying it to a single equation, through the
2
application of singular perturbation, on the basis that the parameter e0 /s0 = ε is very
small. To perform this, we first choose to rescale the equations by the transformations
x=
s
s0
y=
c
e0
k−1 t.
(2)
Using standard steps (left to readers) and letting ˙ denote d/dτ , one obtains
k1 s0
,
k−1
with
k2
β=
k−1
α=
ẋ = ε(y − αx(1 − y))
ẏ = αx(1 − y) − (1 + β)y
(3)
If we let ε go to 0, the first equation becomes ẋ = 0 so that x(τ ) is a constant x,
while y satisfies a simple linear equation such that y(τ ) converges to the equilibrium
αx
.
1 + β + αx
Moving now to the slower time scale t0 = ετ , we can use the quasi-equilibrium
approximation in which the fast variable y is set at the equilibrium value obtained in
the fast time scale. In this time scale, we may see this procedure as an application
of Tikhonov’s Theorem (see below ). We then obtain
αx
αβx
dx
= y − αx(1 − y) = −αx + (1 + αx)
=−
.
0
dt
1 + β + αx
1 + β + αx
(4)
This is a single equation, from which one could obtain (by separation of variables)
an implicit equation for x(t0 ) at any value t0 .
It is customary to exploit the quasi-equilibrium approximation to obtain an approximate equation for the growth of the product p0 (t). From the original equation,
we have
k2 e0 αx(t)
ms(t)
αx(t)
0
=
=
,
p (t) = k2 c(t) = k2 e0 y(t)) = k2 e0
1 + β + αx(t)
1 + β + αx(t)
sh + s(t)
(5)
where in the rightmost expression (known as Michaelis-Menten equation) we have
returned to the original (dimensional) quantities.
The quantity m = k2 e0 represents the maximum rate (which is proportional to
enzyme concentration) at which the reaction may occur (if substrate were infinitely
k−1 + k2
abundant) while sh =
, known as the semi-saturation constant, represents
k1
the substrate concentration at which the reaction speed is equal to half the theoretical
maximum.
1.2
Modelling Inhibitory activities
A very important aspect in the regulation of molecular networks comes from the
fact that the presence of an inhibitory molecule may decrease the rate at which an
3
enzymatic reaction occurs. Very schematically, we can think of two main mechanism:
competitive inhibition, or allosteric inhibition. In the first case the inhibiting molecule
binds to the enzyme in the same site where it binds to the substrate, thus preventing
the formation of S-E complexes; in the latter case, the inhibiting molecule binds
to the enzyme in a different site, but this changes the conformation of the enzyme
reducing its affinity to the substrate.
We present now very simple models of the two cases. Let us start from the first
case, that, letting I represent the inhibitory molecule, can be drawn as
k
1
k
→
S + E −−
−−2→ P + E
k−1 SE
←−−
k3
→
I + E −−
k−3 IE
←−−
Then we obtain the system
s0 (t) = k−1 c1 (t) − k1 s(t)e(t)
e0 (t) = k−1 c1 (t) + k2 c1 (t) + k−3 c2 (t) − k1 s(t)e(t) − k3 i(t)e(t)
c01 (t) = k1 s(t)e(t) − k−1 c1 (t) − k2 c1 (t)
i0 (t) = k−3 c2 (t) − k3 i(t)e(t)
c02 (t) = k3 i(t)e(t) − k−3 c2 (t).
where i(t) now represents the concentration of free inhibitory molecules, c1 (t) that of
SE , c2 that of I − E complexes.
Conservation laws allow for a reduction of the dimension of the system. Indeed
d
(e(t) + c1 (t) + c2 (t)) = 0
dt
d
(i(t) + c2 (t)) = 0
dt
Using the second of these identities, we obtain e(t) + c1 (t) + c2 (t) ≡ e0 . Applying
then variable transformations similar to (2), namely
x=
s
s0
y1 =
c1
e0
y2 =
c2
e0
z=
i
i0
τ = k−1 t.
and letting ˙ denote d/dτ , one obtains
ẋ = ε(y1 − α1 x(1 − y1 − y2 ))
ẏ1 = α1 x(1 − y1 − y2 ) − (1 + β1 )y1
(6)
ẏ2 = α2 z(1 − y1 − y2 ) − β2 y2
ż = η(β2 y2 − α2 z(1 − y1 − y2 ))
with
α1 =
k1 s0
k3 i0
k2
k−3
, α2 =
β1 =
, β2 =
k−1
k−1
k−1
k−1
4
while ε = e0 /s0 and η = e0 /i0 are taken as small parameters.
One can then see that the equations for x and z are ‘slow’ equations, while those
for y1 and y2 are ‘fast’. Following, as above, the approach of quasi-stationary approximations, x and z will stay fixed while y1 and y2 will converge to the equilibrium of
the system of equations for the pair (y1 , y2 ). By letting ẏ1 = ẏ2 = 0, one obtains
α2 (1 + β1 )z
α1 β2 x + α2 (1 + β1 )z + β2 (1 + β1 )
(7)
Now the rate at which the product p(t) is produced is (in the original units) k2 c1 (t),
i.e.
y1 =
α1 β2 x
α1 β2 x + α2 (1 + β1 )z + β2 (1 + β1 )
p0 (t) ≈ k2 e0
y2 =
ms(t)
α1 β2 x(t)
=
α1 β2 x(t) + α2 (1 + β1 )z(t) + β2 (1 + β1 )
sh + s(t) + si i(t)
(8)
where
k−1 + k2
k3 (k−1 + k2 )i0
si =
.
k1
k−3 k1
One can say that the presence of the inhibitor changes the half-saturation constant of
a Michaelis-Menten function from sh to sh + si i(t), thus decreasing the reaction rate
over low-to-medium substrate concentration. The half-saturation constant becomes
larger as the concentration i(t) of inhibitor or its affinity k3 to the enzyme is higher,
or the degradation of EI complexes is lower.
Finally, one may note that, since i(t) + c2 (t) ≡ i0 , once y2 is known from (7), then
z = 1 − ηy2 is also known. As the right-hand side of (7) includes z, the situation may
look confusing. Indeed, one can plug the expression for y2 of (7) into z = 1 − ηy2 , and
obtain a 2nd-degree equation in z (Problem x); in th problem, it is asked to prove
that the resulting equation has a unique solution in (0, 1). Thus the quasi-stationary
approximation actually yields a unique value for z, hence for y1 and y2 for each value
of x. One can then obtain a 1-dimensional equation describing the evolution of x(t)
on the slow time scale.
However, if η is small, the variation of z because of this reaction is actually
negligible. The use of (8) is to show how the rate of an enzymatic reaction changes
in presence of a competitive inhibitor, whose density is mainly regulated by other
factors.
m = k2 e0
1.3
sh =
Cooperativity
Suppose that an enzyme can bind up to two substrate molecules. We can model the
reaction scheme as follows
k
1
k
−−
→
−−2→ P + E
k−1 C1
←−−
k3
k
→
C1 + S −−
−−4→ P + C1
k−3 C2
←−−
S+E
5
It often occurs that binding the first molecule changes the enzyme configuration,
so as to change the rate at which further binding occurs. An extreme degree of
cooperativity would correspond to the property that the rate k1 at which the enzyme
binds a first molecule is very low, but k3 , the rate for a second molecule, is very high.
On the other hand, independence of the site would mean that k1 = 2k3 (when the
enzyme is free there are two possible binding sites, while only one is left when the
enzyme is in the configuration C1 ) and, vice versa, that k−3 = 2k−1 (in C2 there are
two sites from which the substrate can detach, while only one site when the enzyme
is in C1 ). Finally, independence implies that k4 = 2k2 .
After these preliminary consideration, we sketch a model for this reaction scheme;
lete(t) the concentration of free enzyme, c1 (t) the concentration of enzyme bound to
one molecule, c2 (t) the concentration of enzyme bound to two substate molecules.
Conservation laws imply that e(t)+c1 (t)+c2 (t) ≡ e0 , the initial enzyme concentration;
hence we do not need to write down an equation for e0 (t). The previous scheme
translates into the system
s0 (t) = k−1 c1 (t) − k1 s(t)e(t) + k−3 c2 (t) − k3 s(t)c1 (t)
c01 (t) = k1 s(t)e(t) − k−1 c1 (t) − k2 c1 (t) − k3 s(t)c1 (t) + k−3 c2 (t) + k4 c2 (t)
c02 (t) = k3 s(t)c1 (t) − k−3 c2 (t) − k4 c2 (t)
p0 (t) = k2 c1 (t) + k4 c2 (t).
(9)
Following steps similar to those leading to Michaelis-Menten equation, again under
the assumption ε = e0 /s0 1, one sees that on a fast-time scale τ , s(τ ) remains
fixed at a value s, while c1 and c2 satisfy the linear system of equation
ċ1 = k1 s(1 − c1 − c2 ) − k−1 c1 − k2 c1 − k3 sc1 + k−3 c2 + k4 c2
ċ2 = k3 sc1 − k−3 c2 − k4 c2
or in vector notation
d c1
k1 s
c
=
+A 1
c
0
c2
dτ
2
−(k1 + k3 )s − (k−1 + k2 ) −k1 s + k−3 + k4
with A =
.
k3 s
−(k−3 + k4 )
(10)
As tr(A) < 0 and det(A) > 0, both eigenvalues of A have negative real part; hence
the solutions of the linear system (10) converge to its equilbrium, that can be easily
computed as
L2 e0 s
L1 L2 + L2 s + s2
e 0 s2
c2 =
L1 L2 + L2 s + s2
c1 =
where
L1 =
k−1 + k2
k1
L2 =
6
k4 + k−3
.
k3
(11)
(11) constitutes a quasi-stationary approximation for the variables c1 and c2 that
can be used in the “slow” equations for s and p in (9).
In particular, substituting then (11) into the expression for p0 (t) one arrives at
the law
(k2 L2 + k4 s)e0 s
p0 (t) =
(12)
L1 L2 + L2 s + s2
Depending on the values of the constants, the expression (12) can correspond to
rather different shapes of the function relating substrate concentration to reaction
rate.
In the case of independent sites (k1 = 2k3 , k−3 = 2k−1 , k4 = 2k2 as discussed
above), one arrives at the relation
p0 (t) = 2
k2 e0 s
sh + s
i.e. a Michalis-Menten with all constants doubled, as sh = 2(k−1 + k2 )/k1 .
Consider instead the limiting case of extreme cooperativity, i.e. let k1 → 0 while
k3 → +∞ with k1 k3 bounded. It follows that L1 → ∞, L2 → 0 and L1 L2 → K > 0.
We can then approximate the reaction rate as
p0 (t) =
k4 e0 s2
.
K + s2
This is an example of sigmoidal rate equation. More extreme example of sigmoidal
responses, in which the rate is close to 0 for s below a threshold, and maximal above
a threshold, can be obtained by substituting the exponent 2 with a large number n,
and obtaining
me0 sn
.
(13)
p0 (t) =
K + sn
This is the so-called Hill equation; it can be derived by assuming that an enzyme
can bind n substrate molecules, and that binding becomes efficient only after n − 1
molecules have been bound. It can also be used phenomenologically, without knowing
the molecular mechanisms behind it, to model a sigmoidal response.
2
Positive and negative feedback loops
As briefly discussed at the beginning of the chapter, actual molecular networks consist
of a large number of molecular species with very complex patterns of interaction
among them.
Wishing to go beyond the simple reactions analyzed in the previous Section, the
natural next step is to consider reaction loops, i.e. networks with two molecular
species, each acting on the other’s rates, either of formation or of decay. Since the
overall action may be positive (favouring the formation or inhibiting the decay) or
negative (favouring the decay or inhibiting the formation), in principle there are 3
7
possible cases: each molecular species is positive for the other; each is negative for
the other; one is positive for the other, which is negative for the first one).
We distinguish two kinds of loops: positive (+-+ or −-−) and negative (+-−).
Each can be realized in several different reaction schemes and each reaction may be
modeled as linear or Michaelis-Menten-type or cooperative. The dynamical properties will then depend on the details of reactions; we will show however (by way of
examples) that there are some common properties of positive vs. negative feedback
loops.
We thus present some specific examples, inspired to actual reaction networks, but
clearly extremely simplified.
A
Tikhonov’s theorem
1-IKHONOV THEOREM
Let z: z(t) and1t: y(t) denoten- and nt-dintensional
uectorsrespecriL)et,,
and the (autottontous)systembe
,tz,.v.p),
r*:
dv
rl
\
f:l( Z ,) ', p),
z(tu): ,0,
/
.r'(ro):.rb
(2.?a )
Ql b)
I'et z:ó(.v)
be a root of the equaÍiortF(t,.v,o):0
defined in sonte <'losed
artd boutuled donruin D c R"'. Consider the degenerures],srenl
4t :
l( +( v), Y,o)'
o,
and denote by y(t)
(a) z:
ó(l)
adjoined system
Y(ro): -t'o
'
(2' 8)
its solution. If:
rs an isoluted root in D, posiriuel.ystable with respect ro the
it
o:-,
;:F(2,y,0),
tt
(2.e)
unifo rmlyiny eD;
(b) the initial point (zo, yù lieswithin the donnin of influenceof e(y);
(c) rhesolution, : ,(f ) ol (2.8) belongsro D for all t e [to,T],
8
then the solution (z(t,tr),y(t,lL))
solution(Z(t):0(t(/)),
degenerate
T oe Uo, T),
of the ooerallsystem (2.7) tends ro the
r(t)) as p40, in thesenserhat, for an.y
J .imz^( t , p ) : z( t )
( 2. 10 a)
t, p) : ;v(r)
u r ,a-u(
(2 lob)
fo rto</<4 <T,and
f ort n4 t 4 T. <7 .
B
Exercises and problems
1. A molecule X is synthetised from a substrate S through an enzymatic reaction
of Michaelis-Menten kinetics, proportionally to the concentration of the enzyme
E in active configuration. The enzyme E moves from bound to active configuration at rate k1 , while the presence of X helps the opposite transition that
occurs at rate k−1 X, where X is the concentration of X. These reactions can be
summarised in the scheme to the right:
Finally the substrate is produced at constant rate, and X dissociates proportionally to its concentrations.
(a) Can we say that there is a positive or negative feedback of X on its synthesis?
(b) Write down a system of differential equations corresponding to the assumptions.
(c) Compute the (unique) equilibrium of the system, finding conditions for its
feasibility? [Hint: consider the equation for the sum S + X]
(d) Suggest what might be the behaviour of the system, when the conditions
for the feasibility of the equilibrium are not satisfied.
2. [Positive feedback on gene transcription (Griffith 1968).] Consider the simple
case of a protein that activates transcription of its own gene, as in the following
figure:
9
This mechanism is described by a pair of ordinary differential equations:
d[M ]
ε2 + ([P ]/Kl )2
− k2 [M ],
= ν1
dt
1 + ([P ]/Kl )2
d[P ]
= k3 [M ] − k4 [P ].
dt
(a) How must the variables be scaled to write the ODEs in dimensionless form:
ε2 + y 2
dx
=
− x,
dτ
1 + y2
dy
= κ(σx − y) ?
dτ
(b) Assume ε = 0.2, κ = 1. Draw phase plane portraits for several values of
σ.
(c) Show graphically how the number of equilibria of the system changes with
σ.
(d) Plot the equilibrium value of y as a function of σ. [it is enough a picture
showing the qualitative pattern.]
3. In the derivation of Michaelis-Menten equation, it is assumed that the backk−2
reaction SE ←− P + E is neglibly slow. Consider now the full system of
reactions:
k
k
1
−→
2
−→
k−1
k−2
S + E ←− SE ←− P + E
(a) Write down the system of equations corresponding to this scheme. Assume
that at the start, there is no complex SE nor product P .
(b) Note that there are two quantities that are conserved: [E] + [SE] and
[S] + [SE] + [P ]. Use this to decrease the number of equations.
(c) Make the equations non-dimensional, and in so doing introduce the paramete ε = E0 /S0 , ratio of initial values of enzyme and substrate.
(d) Go to the limit ε → 0+ which amounts to make the ‘quasi-equilibrium’
approximation.
(e) Consider the resulting differential equation for S(t) and show that it will
converge to an equilibrium S ∗ where the concentrations of substrate and
product will satisfy Haldane’s relation
P∗
k1 k2
=
.
∗
S
k−1 k−2
10