A remark on the number of steady states in a multiple futile cycle
Liming Wang and Eduardo D. Sontag
Department of Mathematics
Rutgers University, New Brunswick, NJ, USA
Abstract
The multisite phosphorylation-dephosphorylation cycle is a motif repeatedly used in cell signaling.
This motif itself can generate a variety of dynamic behaviors like bistability and ultrasensitivity without
direct positive feedbacks. In this paper, we study the number of positive steady states of a general
multisite phosphorylation-dephosphorylation cycle, and how the number of positive steady states varies
by changing the biological parameters. We show analytically that (1) for some parameter ranges, there
are at least n + 1 (if n is even) or n (if n is odd) steady states; (2) there never are more than 2n − 1
steady states (in particular, this implies that for n = 2, including single levels of MAPK cascades, there
are at most three steady states); (3) for parameters near the standard Michaelis-Menten quasi-steady
state conditions, there are at most n + 1 steady states; and (4) for parameters far from the standard
Michaelis-Menten quasi-steady state conditions, there is at most one steady state.
Keywords: futile cycles, bistability, signaling pathways, biomolecular networks, steady states
1 Introduction
A promising approach to handling the complexity of cell signaling pathways is to decompose pathways into
small motifs, and analyze the individual motifs. One particular motif that has attracted much attention in
recent years is the cycle formed by two or more inter-convertible forms of one protein. The protein, denoted
here by S0, is ultimately converted into a product, denoted here by Sn, through a cascade of “activation”
reactions triggered or facilitated by an enzyme E; conversely, Sn is transformed back (or “deactivated”)
into the original S0, helped on by the action of a second enzyme F . See Figure 1.
S S0 2S1
SSS nn−1n−2
Figure 1: A futile cycle of size n.
Such structures, often called “futile cycles” (also called substrate cycles, enzymatic cycles, or enzymatic
inter-conversions, see [1]), serve as basic blocks in cellular signaling pathways and have pivotal impact on
the signaling dynamics. Futile cycles underlie signaling processes such as GTPase cycles [2], bacterial
two-component systems and phosphorelays [3, 4] actin treadmilling [5]), and glucose mobilization [6], as
well as metabolic control [7] and cell division and apoptosis [8] and cell-cycle checkpoint control [9]. One
very important instance is that of Mitogen-Activated Protein Kinase (“MAPK”) cascades, which regulate
http://arxiv.org/abs/0704.0036v2
primary cellular activities such as proliferation, differentiation, and apoptosis [10–13] in eukaryotes from
yeast to humans.
MAPK cascades usually consist of three tiers of similar structures with multiple feedbacks [14–16].
Each individual level of the MAPK cascades is a futile cycle as depicted in Figure 1 with n = 2. Markevich
et al.’s paper [17] was the first to demonstrate the possibility of multistationarity at a single cascade
level, and motivated the need for analytical studies of the number of steady states. Conradi et al. studied
the existence of multistationarity in their paper [19], employing algorithms based on Feinberg’s chemical
reaction network theory (CRNT). (For more details on CRNT, see [31,32].) The CRNT algorithm confirms
multistationarity in a single level of MAPK cascades, and provides a set of kinetic constants which can
give rise to multistationarity. However, the CRNT algorithm only tests for the existence of multiple steady
states, and does not provide information regarding the precise number of steady states.
In [18], Gunawardena proposed a novel approach to the study of steady states of futile cycles. His
approach, which was focused in the question of determining the proportion of maximally phosphorylated
substrate, was developed under the simplifying quasi-steady state assumption that substrate is in excess.
Nonetheless, our study of multistationarity uses in a key manner the basic formalism in [18], even for the
case when substrate is not in excess.
In Section 2, we state our basic assumptions regarding the model. The basic formalism and background
for the approach is provided in Section 3. The main focus of this paper is on Section 4, where we derive
various bounds on the number of steady states of futile cycles of size n. The first result is a the lower
bound for the number of steady states. Currently available results on lower bounds, as in [29], can only
handle the case when quasi-steady state assumptions are valid; we substantially extend these results to the
fully general case by means of a perturbation argument which allows one to get around these restricted
assumptions. Another novel feature of our results in this paper is the derivation of an upper bound
of 2n − 1, valid for all kinetic constants. Models in molecular cell biology are characterized by a high
degree of uncertainly in parameters, hence such results valid over the entire parameter space are of special
significance. However, when more information of the parameters are available, sharper upper bounds can
obtained, see Theorems 4 and 5. We finally conclude our paper in Section 5 with a conjecture of an n+ 1
upper bound.
We remark that the results given here complement our work dealing with the dynamical behavior
of futile cycles. For the case n = 2, [25] showed that the model exhibits generic convergence to steady
states but no more complicated behavior, at least within restricted parameter ranges, while [27] showed a
persistence property (no species tends to be eliminated) for any possible parameter values. These papers
did not address the question of estimating the number of steady states. (An exception is the case n = 1,
for which uniqueness of steady states can be proved in several ways, and for which global convergence to
these unique equilibria holds [27].)
2 Model assumptions
Before presenting mathematical details, let us first discuss the basic biochemical assumptions that go into
the model. In general, phosphorylation and dephosphorylation can follow either distributive or processive
mechanism. In the processive mechanism, the kinase (phosphatase) facilitates two or more phosphorylations
(dephosphorylations) before the final product is released, whereas in the distributive mechanism, the kinase
(phosphatase) facilitates at most one phosphorylation (dephosphorylation) in each molecular encounter.
In the case of n = 2, a futile cycle that follows the processive mechanism can be represented by reactions
as follows:
S0 + E ←→ ES0 ←→ ES1 −→ S2 + E
S2 + F ←→ FS2 ←→ FS1 −→ S0 + F ;
and the distributive mechanism can be represented by reactions:
S0 + E ←→ ES0 −→ S1 + E ←→ ES1 −→ S2 + E
S2 + F ←→ FS2 −→ S1 + F ←→ FS1 −→ S0 + F.
Biological experiments have demonstrated that both dual phosphorylation and dephosphorylation in MAPK
are distributive, see [14–16]. In their paper [19], Conradi et al. showed mathematically that if either phos-
phorylation or dephosphorylation follows a processive mechanism, the steady state will be unique, which,
it is argued in [19], contradicts experimental observations. So, to get more interesting results, we assume
that both phosphorylations and dephosphorylations in the futile cycles follow the distributive mechanism.
Our structure of futile cycles in Figure 1 also implicitly assumes a sequential instead of a random
mechanism. By a sequential mechanism, we mean that the kinase phosphorylates the substrates in a
specific order, and the phosphatase works in the reversed order. This assumption dramatically reduces the
number of different phospho-forms and simplifies our analysis. In a special case when the kinetic constants
of each phosphorylation are the same and the kinetic constants of each dephosphorylation are the same,
the random mechanism can be easily included in the sequential case. Biologically, there are systems, for
instance the auto-phosphorylation of FGF-receptor-1, that have been experimentally shown to follow a
sequential mechanism [33].
To model the reactions, we assume mass action kinetics, which is standard in mathematical modeling
of molecular events in biology.
3 Mathematical formalism
In this section, we set up a mathematical framework for studying the steady states of futile cycles. Let us
first write down all the elementary chemical reactions in Figure 1:
S0 + E
koff0
kcat0
→ S1 + E
Sn−1 + E
konn−1
koffn−1
kcatn−1
→ Sn + E
S1 + F
loff0
lcat0
→ S0 + F
Sn + F
lonn−1
loffn−1
lcatn−1
→ Sn−1 + F
where kon0 , etc., are kinetic parameters for binding and unbinding, ES0 denotes the complex consisting of
the enzyme E and the substrate S0, and so forth. These reactions can be modeled by 3n + 3 differential-
algebraic equations according to mass action kinetics:
= −kon0s0e+ koff0c0 + lcat0d1
= −konisie+ koffi
ci + kcati−1ci−1 − loni−1sif + loffi−1
di + lcatidi+1, i = 1, . . . , n− 1
= konjsje− (koffj
+ kcatj )cj , j = 0, . . . , n− 1 (1)
= lonk−1skf − (loffk−1
+ lcatk−1)dk, k = 1, . . . , n,
together with the algebraic “conservation equations”:
Etot = e+
Ftot = f +
di, (2)
Stot =
The variables s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f stand for the concentrations of
S0, . . . , Sn, ES0, . . . , ESn−1, FS1, . . . , FSn, E, F
respectively. For each positive vector
κ =(kon0 , . . . , konn−1 , koff0
, . . . , koffn−1
, kcat0 , . . . , kcatn−1 ,
lon0 , . . . , lonn−1 , loff0
, . . . , loffn−1
, lcat0 , . . . , lcatn−1) ∈ R
(of “kinetic constants”) and each positive triple C = (Etot, Ftot, Stot), we have a different system Σ(κ, C).
Let us write the coordinates of a vector x ∈ R3n+3+ as:
x = (s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f),
and define a mapping
Φ : R3n+3+ × R
+ × R
+ −→ R
with components Φ1, . . . ,Φ3n+3 where the first 3n components are
Φ1(x, κ, C) = −kon0s0e+ koff0c0 + lcat0d1,
and so forth, listing the right hand sides of the equations (1), Φ3n+1 is
ci − Etot,
and similarly for Φ3n+2 and Φ3n+3, we use the remaining equations in (2).
For each κ, C, let us define a set
Z(κ, C) = {x |Φ(x, κ, C) = 0}.
Observe that, by definition, given x ∈ R3n+3+ , x is a positive steady state of Σ(κ, C) if and only if x ∈ Z(κ, C).
So, the mathematical statement of the central problem in this paper is to count the number of elements
in Z(κ, C). Our analysis will be greatly simplified by a preprocessing. Let us introduce a function
Ψ : R3n+3+ × R
+ × R
+ −→ R
with components Ψ1, . . . ,Ψ3n+3 defined as
Ψ1 = Φ1 +Φn+1
Ψi = Φi +Φn+i +Φ2n+i−1 +Ψi−1, i = 2, . . . , n
Ψj = Φj, j = n+ 1, . . . , 3n + 3.
It is easy to see that
Z(κ, C) = {x |Ψ(x, κ, C) = 0},
but now the first 3n equations are:
Ψi = lcati−1di − kcati−1ci−1 = 0, i = 1, . . . , n,
Ψn+1+j = konjsje− (koffj
+ kcatj )cj = 0, j = 0, . . . , n− 1
Ψ2n+k = lonk−1skf − (loffk−1
+ lcatk−1)dk = 0, k = 1, . . . , n,
and can be easily solved as:
si+1 = λi(e/f)si, (3)
di+1 =
fsi+1
, (5)
where
kcatiLMi
KMi lcati
, KMi =
kcati + koffi
, LMi =
lcati + loffi
, i = 0, . . . , n− 1. (6)
We may now express
0 si,
0 ci and
1 di in terms of s0, κ, e and f :
si = s0
1 + λ0
+ λ0λ1
+ · · ·+ λ0 · · ·λn−1
:= s0ϕ
ci = es0
+ · · ·+
λ0 · · ·λn−2
KMn−1
:= es0ϕ
, (7)
di = fs0
+ · · · +
λ0 · · ·λn−1
LMn−1
:= fs0ϕ
Although the equation Ψ = 0 represents 3n+3 equations with 3n+3 unknowns, next we will show that
it can be reduced to two equations with two unknowns, which have the same number of positive solutions
as Ψ = 0. Let us first define a set
S(κ, C) = {(u, v) ∈ R+ × R+ |G
1 (u, v) = 0, G
2 (u, v) = 0},
where G
1 , G
2 : R
+ −→ R are given by
1 (u, v) = v (uϕ
1(u)− ϕ
2(u)Etot/Ftot)− Etot/Ftot + u,
2 (u, v) = ϕ
0(u)ϕ
2 (u)v
2 + (ϕκ0 (u)− Stotϕ
2 (u) + Ftotuϕ
1 (u) + Ftotϕ
2 (u)) v − Stot.
The precise statement is as follows:
Lemma 1 There exists a mapping δ : R3n+3 −→ R2 such that, for each κ, C, the map δ restricted to
Z(κ, C) is a bijection between the sets Z(κ, C) and S(κ, C).
Proof. Let us define the mapping δ : R3n+3 −→ R2 as δ(x) = (e/f, s0), where
x = (s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f).
If we can show that δ induces a bijection between Z(κ, C) and S(κ, C), we are done.
First, we claim that δ(Z(κ, C)) ⊆ S(κ, C). Pick any x ∈ Z(κ, C), we have that x satisfies (3)-(5).
Moreover, Φ3n+2(x, κ, C) = 0 yields
Etot = e+ es0ϕ
and thus
1 + s0ϕ
1(e/f)
. (8)
Using Φ3n+1(x, κ, C) = 0 and Φ3n+2(x, κ, C) = 0, we get:
e(1 + s0ϕ
1(e/f))
f(1 + s0ϕ
2(e/f))
, (9)
which is G
1 (e/f, s0) = 0 after multiplying by 1 + s0ϕ
2(e/f) and rearranging terms.
To check that G
2 (e/f, s0) = 0, we start with Φ3n+3(x, κ, C) = 0, i.e.
Stot =
Using (7) and (8), this expression becomes
Stot = s0ϕ
Etots0ϕ
1(e/f)
1 + s0ϕ
1 (e/f)
Ftots0ϕ
2(e/f)
1 + s0ϕ
2(e/f)
= s0ϕ
eFtots0ϕ
1(e/f)
f(1 + s0ϕ
2(e/f))
Ftots0ϕ
2(e/f)
1 + s0ϕ
2(e/f)
where the last equality comes from (9).
After multiplying by 1 + s0ϕ
2 (e/f), and simplifying, we get
ϕκ0 (
)ϕκ2 (
)s20 +
)− Stotϕ
Ftotϕ
) + Ftotϕ
s0 − Stot = 0,
that is, G
2 (e/f, s0) = 0. since both G
1 (e/f, s0) and G
2 (e/f, s0) are zero, δ(x) ∈ S(κ, C).
Next, we will show that S(κ, C) ⊆ δ(Z(κ, C)). For any y = (u, v) ∈ S(κ, C), let the coordinates of x be
defined as:
s0 = v
si+1 = λiusi
1 + s0ϕ
1 (u)
di+1 =
fsi+1
for i = 0, . . . , n − 1. It is easy to see that the vector x = (s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f) satisfies
Φ1(x, κ, C) = 0, . . . ,Φ3n+1(x, κ, C) = 0. If Φ3n+2(x, κ, C) and Φ3n+3(x, κ, C) are also zero, then x is an
element of Z(κ, C) with δ(x) = y. Given the condition that G
i (u, v) = 0 (i = 1, 2) and u = e/f, v = s0,
we have G
1 (e/f, s0) = 0, and therefore (9) holds. Since
1 + s0ϕ
1(e/f)
in our construction, we have
Ftot = f(1 + s0ϕ
2(e/f)) = f +
To check Φ3n+3(x, κ, C) = 0, we use
2 (e/f, s0)
1 + s0ϕ
2(e/f)
2 (e/f, s0) = 0 and 1 + s0ϕ
2(e/f) > 0. Applying (7)-(9), we have
di = s0ϕ
0(e/f) +
eFtots0ϕ
1(e/f)
f(1 + s0ϕ
2 (e/f))
Ftots0ϕ
2(e/f)
1 + s0ϕ
2(e/f)
= Stot.
It remains for us to show that the map δ is one to one on Z(κ, C). Suppose that δ(x1) = δ(x2) = (u, v),
where
xi = (si0, . . . , s
0, . . . , c
n−1, d
1, . . . , d
i, f i), i = 1, 2.
By the definition of δ, we know that s10 = s
0 and e
1/f1 = e2/f2. Therefore, s1i = s
i for i = 0, . . . , n.
Equation (8) gives
1 + vϕκ1 (u)
= e2.
Thus, f1 = f2, and c1i = c
i , d
i+1 = d
i+1 for i = 0, . . . , n − 1 because of (3)-(5). Therefore, x
1 = x2, and δ
is one to one.
The above lemma ensures that the two sets Z(κ, C) and S(κ, C) have the same number of elements. From
now on, we will focus on S(κ, C), the set of positive solutions of equations G
1 (u, v) = 0, G
2 (u, v) = 0,
1 (u, v) = v (uϕ
1(u)− ϕ
2(u)Etot/Ftot)− Etot/Ftot + u = 0, (10)
2 (u, v) = ϕ
0(u)ϕ
2 (u)v
2 + (ϕκ0(u)− Stotϕ
2 (u) + Ftotuϕ
1 (u) + Ftotϕ
2(u)) v − Stot = 0. (11)
4 Number of positive steady states
4.1 Lower bound on the number of positive steady states
One approach to solving (10)-(11) is to view G
2 (u, v) as a quadratic polynomial in v. Since G
2 (u, 0) < 0,
equation (11) has a unique positive root, namely
−Hκ,C(u) +
Hκ,C(u)2 + 4Stotϕ
0(u)ϕ
2 (u)
2ϕκ0 (u)ϕ
2 (u)
, (12)
where
Hκ,C(u) = ϕκ0(u)− Stotϕ
2(u) + Ftotuϕ
1(u) + Ftotϕ
2(u). (13)
Substituting this expression for v into (10), and multiplying by ϕκ0 (u), we get
F κ,C(u) :=
−H̃κ,C(u) +
H̃κ,C(u)2 + 4Stotϕ
0 (u)ϕ
2 (u)
2ϕκ2 (u)
uϕκ1(u)−
ϕκ2(u)
ϕκ0(u)+uϕ
0 (u) = 0.
So, any (u, v) ∈ S(κ, C) should satisfy (12) and (14). On the other hand, any positive solution u of (14)
(notice that ϕκ0(u) > 0) and v given by (12) (always positive) provide a positive a solution of (10)-(11),
that is, (u, v) is an element in S(κ, C). Therefore, the number of positive solutions of (10)-(11) is the same
as the number of positive solutions of (12) and (14). But v is uniquely determined by u in (12), which
further simplifies the problem to one equation (14) with one unknown u. Based on this observation, we
have:
Theorem 1 For each positive numbers Stot, γ, there exist ε0 > 0 and κ ∈ R
+ such that the following
property holds. Pick any Etot, Ftot such that
Ftot = Etot/γ < ε0Stot/γ, (15)
then the system Σ(κ, C) with C = (Etot, Ftot, Stot) has at least n + 1 (n) positive steady states when n is
even (odd).
Proof. For each κ, γ, Stot, let us define two functions R+ × R+ −→ R as follows:
κ,γ,Stot(ε, u) = H
κ,(εStot,εStot/γ,Stot)(u) (16)
= ϕκ0(u)− Stotϕ
2(u) + ε
uϕκ1(u) + ε
ϕκ2(u),
κ,γ,Stot(ε, u) = F
κ,(εStot,εStot/γ,Stot)(u) (17)
κ,γ,Stot(ε, u) +
κ,γ,Stot(ε, u)2 + 4Stotϕ
0 (u)ϕ
2 (u)
2ϕκ2 (u)
(uϕκ1(u)− γϕ
2 (u))
− γϕκ0(u) + uϕ
0(u).
By Lemma 1 and the argument before this theorem, it is enough to show that there exist ε0 > 0 and κ ∈
+ such that for all ε ∈ (0, ε0), the equation F̃
κ,γ,Stot(ε, u) = 0 has at least n+1 (n) positive solutions
when n is even (odd). (Then, given Stot, γ, Etot, and Ftot satisfying (15), we let ε = Etot/Stot < ε0,
and apply the result.)
A straightforward computation shows that when ε = 0,
κ,γ,Stot(0, u) = Stot (uϕ
1(u)− γϕ
2(u))− γϕ
0 (u) + uϕ
= λ0 · · ·λn−1u
n+1 + λ0 · · ·λn−2
KMn−1
(1− γβn−1)− γλn−1
+ · · ·+ λ0 · · ·λi−2
KMi−1
(1− γβi−1)− γλi−1
ui + · · · (18)
(1− γβ0)− γλ0
u− γ,
where the λi’s and KMi ’s are defined as in (6), and βi = kcati/lcati . The polynomial F̃
κ,γ,Stot(0, u)
is of degree n + 1, so there are at most n + 1 positive roots. Notice that u = 0 is not a root because
κ,γ,Stot(0, u) = −γ < 0, which also implies that when n is odd, there can not be n + 1 positive roots.
Now fix any Stot and γ. We will construct a vector κ such that F̃
κ,γ,Stot(0, u) has n+ 1 distinct positive
roots when n is even.
Let us pick any n+ 1 positive real numbers u1 < · · · < un+1, such that their product is γ, and assume
(u− u1) · · · (u− un+1) = u
n+1 + anu
n + · · · + a1u+ a0, (19)
where a0 = −γ < 0 keeping in mind that ai’s are given. Our goal is to find a vector κ ∈ R
+ such that
(18) and (19) are the same. For each i = 0, . . . , n − 1, we pick λi = 1. Comparing the coefficients of u
in (18) and (19), we have:
(1 + a0βi) = ai+1 − a0 − 1. (20)
Let us pick KMi > 0 such that
(ai+1 − a0 − 1)− 1 < 0, then take
(ai+1 − a0 − 1)− 1
in order to satisfy (20). From the given
λ0, . . . , λn−1,KM0 , . . . ,KMn−1 , β0, . . . , βn−1,
we will find a vector
κ =(kon0 , . . . , konn−1 , koff0
, . . . , koffn−1
, kcat0 , . . . , kcatn−1 ,
lon0 , . . . , lonn−1 , loff0
, . . . , loffn−1
, lcat0 , . . . , lcatn−1) ∈ R
such that βi = kcati/lcati , i = 0, . . . , n− 1, and (6) holds. This vector κ will guarantee that F̃
κ,γ,Stot(0, u)
has n + 1 positive distinct roots. When n is odd, a similar construction will give a vector κ such that
κ,γ,Stot(0, u) has n positive roots and one negative root.
One construction of κ (given λi,KMi , βi, i = 0, . . . , n − 1) is as follows. For each i = 0, . . . , n − 1, we
start by defining:
LMi =
λiKMi
consistently with the definitions in (6). Then, we take
koni = 1, loni = 1,
koffi
= αiKMi , kcati = (1− αi)KMi , lcati =
1− αi
KMi , loffi
= LMi − lcati ,
where αi ∈ (0, 1) is chosen such that
loffi
= LMi −
1− αi
KMi > 0.
This κ satisfies βi = kcati/lcati , i = 0, . . . , n− 1, and (6).
In order to apply the Implicit Function Theorem, we now view the functions defined by formulas in
(16) and (17) as defined also for ε ≤ 0, i.e. as functions R×R+ −→ R. It is easy to see that F̃
κ,γ,Stot(ε, u)
is C1 on R × R+ because the polynomial under the square root sign in F̃
κ,γ,Stot(ε, u) is never zero. On
the other hand, since F̃
κ,γ,Stot(0, u) is a polynomial in u with distinct roots, ∂F̃
κ,γ,Stot
(0, ui) 6= 0. By the
Implicit Function Theorem, for each i = 1, . . . , n+ 1, there exist open intervals Ei containing 0, and open
intervals Ui containing ui, and a differentiable function
αi : Ei → Ui
such that αi(0) = ui, F̃
κ,γ,Stot(ε, αi(ε)) = 0 for all ε ∈ Ei, and the images αi(Ei)’s are non-overlapping. If
we take
(0, ε0) :=
(0,+∞),
then for any ε ∈ (0, ε0), we have {αi(ε)} as n+ 1 distinct positive roots of F̃
κ,γ,Stot(ε, u). The case when
n is odd can be proved similarly.
The above theorem shows that when Etot/Ftot is sufficiently small, it is always possible for the futile
cycle to have n + 1 (n) steady states when n is even (odd), by choosing appropriate kinetic constants κ.
We should notice that for arbitrary κ, the derivative of F̃ at each positive root may become zero, which
breaks down the perturbation argument. Here is an example to show that more conditions are needed:
n = 2, λ0 = 1, λ1 = 3, γ = 6, β0 = β1 = 1/12, K0 = 1/8, K1 = 1/2, Stot = 5,
we have that
κ,γ,Stot(0, u) = 3u3 − 12u2 + 15u− 6 = 3(u− 1)2(u− 2)
has a double root at u = 1. In this case, even for ε = 0.01, there is only one positive root of F̃
κ,γ,Stot(ε, u),
see Figure 2.
1 2 3
Figure 2: The plot of the function F̃
κ,γ,Stot(0.01, u) on [0, 3]. There is a unique positive real solution
around u = 2.14, the double root u = 1 of F̃
κ,γ,Stot(0, u) bifurcates to two complex roots with non-zero
imaginary parts.
However, the following lemma provides a sufficient condition for ∂F
κ,γ,Stot
(0, ū) 6= 0, for any positive
ū such that F̃
κ,γ,Stot(0, ū) = 0.
Lemma 2 For each positive numbers Stot, γ, and vector κ ∈ R
+ , if
1− γβj
holds for all j = 1, · · · , n − 1, then ∂F̃
κ,γ,Stot
(0, ū) 6= 0.
See Appendix for the proof.
Theorem 2 For each positive numbers Stot, γ, and vector κ ∈ R
+ satisfying condition (21), there exists
ε1 > 0 such that for any Ftot, Etot satisfying Ftot = Etot/γ < ε1Stot/γ, the number of positive steady
states of system Σ(κ, C) is greater or equal to the number of (positive) roots of F̃
κ,γ,Stot(0, u).
Proof. Suppose that F̃
κ,γ,Stot(0, u) has m roots: ū1, . . . , ūm. Applying Lemma 2, we have
κ,γ,Stot
(0, ūk) 6= 0, k = 1, . . . ,m.
By the perturbation arguments as in Theorem 1, we have that there exists ε1 > 0 such that F̃
κ,γ,Stot(ε, u)
has at least m roots for all 0 < ε < ε1.
The above result depends heavily on a perturbation argument, which only works when Etot/Ftot is
sufficiently small. In the next section, we will give an upper bound of the number of steady states with no
restrictions on Etot/Ftot, and independent of κ and C.
4.2 Upper bound on the number of steady states
Theorem 3 For each κ, C, the system Σ(κ, C) has at most 2n− 1 positive steady states.
Proof. An alternative approach to solving (10)-(11) is to first eliminate v from (10) instead of from (11),
Etot/Ftot − u
uϕκ1(u)− (Etot/Ftot)ϕ
, (22)
when uϕκ1(u) − (Etot/Ftot)ϕ
2 (u) 6= 0. Then, we substitute (22) into (11), and multiply by (uϕ
1(u) −
(Etot/Ftot)ϕ
2 (u))
2 to get:
P κ,C(u) := ϕκ0ϕ
+ (ϕκ0 − Stotϕ
2 + Ftotuϕ
1 + Ftotϕ
uϕκ1 −
− Stot
uϕκ1 −
= 0. (23)
Therefore, if uϕκ1(u) − (Etot/Ftot)ϕ
2 (u) 6= 0, the number of positive solutions of (10)-(11) is no greater
than the number of positive roots of P κ,C(u).
In the special case when uϕκ1(u) − (Etot/Ftot)ϕ
2(u) = 0, by (10), we must have u = Etot/Ftot, and
thus ϕκ1 (Etot/Ftot) = ϕ
2 (Etot/Ftot). Substituting into (11), we get a unique v defined as in (12) with
u = Etot/Ftot. But notice that in this case u = Etot/Ftot is also a root of P
κ,C(u), so also in this case
the number of positive solutions to (10)-(11) is no greater than the number of positive roots of P κ,C(u).
It is easy to see that P κ,C(u) is divisible by u. Consider the polynomial Qκ,C(u) := P κ,C(u)/u of
degree 2n + 1. We will first show that Qκ,C(u) has no more than 2n positive roots, then we will prove by
contradiction that 2n distinct positive roots can not be achieved.
It is easy to see that the coefficient of u2n+1 is
(λ0 · · · λn−1)
LMn−1
and the constant term is
FtotKM0
So the polynomial Qκ,C(u) has at least one negative root, and thus has no more than 2n positive roots.
Suppose that S(κ, C) has cardinality 2n, then Qκ,C(u) must have 2n distinct positive roots, and each of
them has multiplicity one. Let us denote the roots as u1, . . . , u2n in ascending order. We claim that none
of them equals Etot/Ftot. If so, we would have ϕ
1(Etot/Ftot) = ϕ
2 (Etot/Ftot), and Etot/Ftot would
be a double root of Qκ,C(u), contradiction.
Since Qκ,C(0) > 0, Qκ,C(u) is positive on intervals
I0 = (0, u1), I1 = (u2, u3), . . . , In−1 = (u2n−2, u2n−1), In = (u2n,∞),
and negative on intervals
J1 = (u1, u2), . . . , Jn = (u2n−1, u2n).
As remarked earlier, ϕκ1 (Etot/Ftot) 6= ϕ
2 (Etot/Ftot), the polynomial Q
κ,C(u) evaluated at Etot/Ftot
is negative, and therefore, Etot/Ftot belongs to one of the J intervals, say Js = (u2s−1, u2s), for some
s ∈ {1, . . . , n} .
On the other hand, the denominator of v in (22), denoted as B(u), is a polynomial of degree n and
divisible by u. If B(u) has no positive root, then it does not change sign on the positive axis of u. But v
changes sign when u passes Etot/Ftot, thus v2s−1 and v2s have opposite signs, and one of (u2s−1, v2s−1)
and (u2s, v2s) is not a solution to (10)-(11), which contradicts the fact that both are in S(κ, C).
Otherwise, there exists a positive root ū of B(u) such that there is no other positive root of B(u)
between ū and Etot/Ftot. Plugging ū into Q
κ,C(u), we see that Qκ,C(ū) is always positive, therefore, ū
belongs to one of the I intervals, say It = (u2t, u2t+1) for some t ∈ {0, . . . , n}. There are two cases:
1. Etot/Ftot < ū. We have
u2s−1 < Etot/Ftot < u2t < ū.
Notice that v changes sign when u passes Etot/Ftot, so the corresponding v2s−1 and v2t have different
signs, and either (u2s−1, v2s−1) /∈ S(κ, C) or (u2t, v2t) /∈ S(κ, C), contradiction.
2. Etot/Ftot > ū. We have
ū < u2t+1 < Etot/Ftot < u2s.
Since v changes sign when u passes Etot/Ftot, so the corresponding v2t+1 and v2s have different
signs, and either (u2t+1, v2t+1) /∈ S(κ, C) or (u2s, v2s) /∈ S(κ, C), contradiction.
Therefore, Σ(κ, C) has at most 2n− 1 steady states.
4.3 Fine-tuned upper bounds
In the previous section, we have seen that any (u, v) ∈ S(κ, C), u 6= Etot/Ftot must satisfy (22)-(23), but
not all solutions of (22)-(23) are elements in S(κ, C). Suppose that (u, v) is a solution of (22)-(23), it is
in S(κ, C) if and only if u, v > 0. In some special cases, for example, when the enzyme is in excess, or the
substrate is in excess, we could count the number of solutions of (22)-(23) which are not in S(κ, C) to get
a better upper bound.
The following is a standard result on continuity of roots; see for instance Lemma A.4.1 in [30]:
Lemma 3 Let g(z) = zn + a1z
n−1 + · · ·+ an be a polynomial of degree n and complex coefficients having
distinct roots
λ1, . . . , λq,
with multiplicities
n1 + · · ·+ nq = n,
respectively. Given any small enough δ > 0 there exists a ε > 0 so that, if
h(z) = zn + b1z
n−1 + · · ·+ bn, |ai − bi| < ε for i = 1, . . . , n,
then h has precisely ni roots in Bδ(λi) for each i = 1, . . . , q.
Theorem 4 For each γ > 0 and κ ∈ R6n−6+ such that ϕ
1 (γ) 6= ϕ
2 (γ), and each Stot > 0, there exists
ε2 > 0 such that for all positive numbers Etot, Ftot satisfying Ftot = Etot/γ < ε2Stot/γ, the system
Σ(κ, C) has at most n+ 1 positive steady states.
Proof. Let us define a function R+ × C −→ C as follows:
κ,γ,Stot(ε, u) = Q
κ,(εStot,εStot/γ,Stot)(u),
and a set B
κ,γ,Stot(ε) consisting of the roots of Q̃
κ,γ,Stot(ε, u) which are not positive or the corresponding
v’s determined by u’s as in (22) are not positive, Since Q̃
κ,γ,Stot(ε, u) is a polynomial of degree 2n + 1,
if we can show that there exists ε2 > 0 such that for any ε ∈ (0, ε2), Q̃
κ,γ,Stot(ε, u) has at least n roots
counting multiplicities that are in B
κ,γ,Stot(ε), then we are done.
In order to apply Lemma 3, we regard the function Q̃
κ,γ,Stot as defined on R× C. At ε = 0:
κ,γ,Stot(0, u) = [ϕκ0ϕ
2(γ − u)
2 + (ϕκ0 − Stotϕ
2 )(uϕ
1 − γϕ
2)(γ − u)− Stot(uϕ
1 − γϕ
= [ϕκ0ϕ
2(γ − u)
2 + ϕκ0(uϕ
1 − γϕ
2)(γ − u)− Stotϕ
1 − γϕ
2)(γ − u)− Stot(uϕ
1 − γϕ
= [ϕκ0 (γ − u)u(ϕ
1 − ϕ
2) + Stotu(uϕ
1 − γϕ
2 )(ϕ
2 − ϕ
1)]/u
= (ϕκ2 − ϕ
1)(uϕ
0 + Stot(uϕ
1 − γϕ
2 )− γϕ
= (ϕκ2 − ϕ
κ,γ,Stot(0, u)
Let us denote the distinct roots of Q̃
κ,γ,Stot(0, u)/u as
u1, . . . , uq,
with multiplicities
n1 + · · ·+ nq = 2n+ 1,
and the roots of ϕκ1 − ϕ
u1, . . . , up, p ≤ q,
with multiplicities
m1 + · · ·+mp = n, ni ≥ mi, for i = 1, . . . , p.
For each i = 1, . . . , p, if ui is real and positive, then there are two cases (ui 6= γ as ϕ
1(γ) 6= ϕ
2(γ)):
1. ui > γ. We have
1(ui)− γϕ
2 (ui) > γ(ϕ
1 (ui)− ϕ
2(ui)) = 0.
2. ui < γ. We have
1(ui)− γϕ
2 (ui) < γ(ϕ
1 (ui)− ϕ
2(ui)) = 0.
In both cases, uiϕ
1(ui)− γϕ
2 (ui) and γ − ui have opposite signs, i.e.
1 (ui)− γϕ
2(ui))(γ − ui) < 0.
Let us pick δ > 0 small enough such that the following conditions hold:
1. For all i = 1, . . . , p, if ui is not real, then Bδ(ui) has no intersection with the real axis.
2. For all i = 1, . . . , p, if ui is real and positive, the following inequality holds for any real u ∈ Bδ(ui):
(uϕκ1(u)− γϕ
2 (u))(γ − u) < 0. (24)
3. For all i = 1, . . . , p, if ui is real and negative, then Bδ(ui) has no intersection with the imaginary
axis.
4. Bδ(uj)
Bδ(uk) = ∅ for all j 6= k = 1, . . . , q.
By Lemma 3, there exists ε3 > 0 such that for all ε ∈ (0, ε3), the polynomial Q̃
κ,γ,Stot(ε, u)/u has exactly
nj roots in each Bδ(uj), j = 1, . . . , q, denoted by u
j (ε), k = 1, . . . , nj .
We pick one such ε, and we claim that none of the roots in Bδ(ui), i = 1, . . . , p with the v defined as
in (22) will be an element in S. If so, we are done, since there are
1 ni ≥
1 mi = n such roots, of
κ,γ,Stot(ε, u) which are in B
κ,γ,Stot(ε).
For each i = 1, . . . , p, there are two cases:
1. ui is not real. Then condition 1 guarantees that u
i (ε) is not real for each k = 1, . . . , ni, and thus is
κ,γ,Stot(ε).
2. ui is real and positive. Pick any root u
i (ε) ∈ Bδ(ui), k = 1, . . . , ni, the corresponding v
i (ε) equals
γ − uki (ε)
uki (ε)ϕ
i (ε)) − γϕ
i (ε))
) < 0
followed from (24). So (uki (ε), v
i (ε)) /∈ S(κ, C), and u
i (ε) ∈ B
κ,γ,Stot(ε).
3. ui is real and negative. By condition 1 and 3, u
i (ε) is not positive for all k = 1, . . . , ni.
The next theorem considers the case when enzyme is in excess:
Theorem 5 For each γ > 0, κ ∈ R6n−6+ such that ϕ
1 (γ) 6= ϕ
2(γ), and each Etot > 0, there exists ε3 > 0
such that for all positive numbers Ftot, Stot satisfying Ftot = Etot/γ > Stot/(ε3γ), the system Σ(κ, C) has
at most one positive steady state.
Proof. For each γ > 0, κ ∈ R6n−6+ such that ϕ
1 (γ) 6= ϕ
2 (γ), and each Etot > 0, we define a function
R+ × C −→ C as follows:
κ,γ,Etot(ε, u) = Q
κ,(Etot,Etot/γ,εEtot)(u).
Let us define the set C
κ,γ,Etot(ε) as the set of roots of Q̄
κ,γ,Etot(ε, u) which are not positive or the
corresponding v’s determined by u’s as in (22) are not positive. If we can show that there exists ε3 > 0
such that for any ε ∈ (0, ε3) there is at most one positive root of Q̄
κ,γ,Etot(ε, u) that is not in C
κ,γ,Etot(ε),
we are done.
In order to apply Lemma 3, we now view the function Q̄
κ,γ,Etot as defined on R× C. At ε = 0:
κ,γ,Etot(0, u) = (γ − u)
(γ − u)ϕκ0ϕ
ϕκ0 +
uϕκ1 +
(uϕκ1 − γϕ
:= (γ − u)R
κ,γ,Etot(u).
Let us denote the distinct roots of Q̄
κ,γ,Etot(0, u)/u as
u1(= γ), u2, . . . , uq,
with multiplicities
n1 + · · ·+ nq = 2n+ 1,
and u2, . . . , uq are the roots of R
κ,γ,Etot(u) other than γ.
Since ϕκ1(γ) 6= ϕ
2(γ), R
κ,γ,Etot(u) is not divisible by u− γ, and thus n1 = 1.
For each i = 2, . . . , q, we have
(γ − ui)ϕ
0(ui)ϕ
2 (ui) = −
ϕκ0(ui) +
1(ui) +
ϕκ2(ui)
1 (ui)− γϕ
2(ui)) .
If ui > 0, then ϕ
0(ui)ϕ
2 (ui) and ϕ
0 (ui) +
1 (ui) +
ϕκ2 (ui) are both positive. Since uiϕ
1(ui)−
γϕκ2(ui) and γ − ui are non zero, uiϕ
1 (ui)− γϕ
2(ui) and γ − ui must have opposite signs, that is
1 (ui)− γϕ
2(ui))(γ − ui) < 0.
Let us pick δ > 0 small enough such that the following conditions hold for all i = 2, . . . , q:
1. If ui is not real, then Bδ(ui) has no intersection with the real axis.
2. If ui is real and positive, then for any real u ∈ Bδ(ui), the following inequality holds:
(uϕκ1(u)− γϕ
2 (u))(γ − u) < 0. (25)
3. If ui is real and negative, then Bδ(ui) has no intersection with the imaginary axis.
4. Bδ(uj)
Bδ(uk) = ∅ for all i 6= k = 2, . . . , q.
By Lemma 3, there exists ε3 > 0 such that for all ε ∈ (0, ε3), the polynomial Q̄
κ,γ,Etot(ε, u) has exactly
nj roots in each Bδ(uj), j = 1, . . . , q, denoted by u
j (ε), k = 1, . . . , nj .
We pick one such ε, and if we can show that all of the roots in Bδ(ui), i = 2, . . . , q are in C
κ,γ,Etot(ε),
then we are done, since the only roots that may not be in C
κ,γ,Etot(ε) are the roots in Bδ(u1), and there
is one root in Bδ(u1).
For each i = 2, . . . , p, there are three cases:
1. ui is not real. Then condition 1 guarantees that u
i (ε) is not real for all k = 1, . . . , ni.
2. ui is real and positive. Pick any root u
i (ε), k = 1, . . . , ni, the corresponding v
i (ε) equals
γ − uki (ε)
uki (ε)ϕ
i (ε))− γϕ
i (ε))
So, uki (ε) is in C
κ,γ,Etot(ε).
3. ui is real and negative. By conditions 1 and 3, u
i (ε) is not positive for all k = 1, . . . , ni.
5 Conclusions and discussions
Here we have set up a mathematical model for multisite phosphorylation-dephosphorylation cycles of size
n, and studied the number of positive steady states based on this model. We reformulated the question
of number of positive steady states to question of the number of positive roots of certain polynomials,
through which we also applied perturbation techniques. Our theoretical results depend on the assumption
of mass action kinetics and distributive sequential mechanism, which are customary in the study of multisite
phosphorylation and dephosphorylation.
An upper bound of 2n−1 steady states is obtained for arbitrary parameter combinations. Biologically,
when the substrate concentration greatly exceeds that of the enzyme, there are at most n + 1 (n) steady
states if n is even (odd). And this upper bound can be achieved under proper kinetic conditions, see
Theorem 1 for the construction. On the other extreme, when the enzyme is in excess, there is a unique
steady state.
As a special case of n = 2, which can be applied to a single level of MAPK cascades. Our results
guarantees that there are no more than three steady states, consistent with numerical simulations in [17].
We notice that there is an apparent gap between the upper bound 2n−1 and the upper bound of n+1
(n) if n is even (odd) when the substrate is in excess. If we think the ratio Etot/Ftot as a parameter ε,
then when ε≪ 1, there are at most n+1 (n) steady states when n is even (odd), which coincides with the
largest possible lower bound. When ε ≫ 1, there is a unique steady state. If the number of steady states
changes “continuously” with respect to ε, then we do not expect the number of steady states to exceed
n + 1 (n) if n is even (odd). So a natural conjecture would be that the number of steady states never
exceed n+ 1 under any conditions.
6 Acknowledgment
We thank Jeremy Gunawardena for very helpful discussions.
7 Appendix
proof of Lemma 2: Recall that (dropping the u’s in ϕκi , i = 0, 1, 2)
κ,γ,Stot(0, u) = uϕκ0 + Stot(uϕ
1 − γϕ
2)− γϕ
κ,γ,Stot
(0, u) = ϕκ0 + Stot(uϕ
1 − γϕ
′ − (γ − u)(ϕκ0 )
Since F̃
κ,γ,Stot(0, ū) = 0,
Stot(ūϕ
1 − γϕ
2 ) = (γ − ū)ϕ
that is,
γ − ū =
Stot(ūϕ
1 − γϕ
Therefore,
κ,γ,Stot
(0, ū) = ϕκ0 + Stot(uϕ
1 − γϕ
Stot(ūϕ
1 − γϕ
(ϕκ0)
= ϕκ0 +
ϕκ0(uϕ
1 − γϕ
′ − (ūϕκ1 − γϕ
= ϕκ0 +
((1 + λ0ū+ λ0λ1ū
2 + · · · + λ0 · · ·λn−1ū
(1− γβ0) + 2
(1− γβ1)ū+ · · ·+ n
λ0 · · · λn−2
KMn−1
(1− γβn−1)ū
λ0 + 2λ0λ1ū+ · · · + nλ0 · · ·λn−1ū
(1− γβ0)ū+
(1− γβ1)ū
2 + · · ·+
λ0 · · ·λn−2
KMn−1
(1− γβn−1)ū
= ϕκ0 +
λ0 · · ·λi−1ū
(j + 1− i)
λ0 · · ·λj−1
(1− γβj)ū
λ0 · · ·λi−1ū
λ0 · · ·λj−1ū
+ Stot
λ0 · · ·λi−1ū
(j + 1− i)
λ0 · · ·λj−1
(1− γβj)ū
λ0 · · ·λi−1ū
λ0 · · ·λn−1ū
λ0 · · ·λj−1ū
1 + Stot(j + 1− i)
1 − γβj
where the product λ0 · · ·λ−1 is defined to be 1 for the convenience of notation.
Because of (21),
(j + 1− i)
1− γβj
so we have ∂F̃
κ,γ,Stot
(0, ū) > 0.
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http://arxiv.org/abs/math/0701575
	Introduction
	Model assumptions
	Mathematical formalism
	Number of positive steady states
	Lower bound on the number of positive steady states
	Upper bound on the number of steady states
	Fine-tuned upper bounds
	Conclusions and discussions
	Acknowledgment
	Appendix

  The multisite phosphorylation-dephosphorylation cycle is a motif repeatedly
used in cell signaling. This motif itself can generate a variety of dynamic
behaviors like bistability and ultrasensitivity without direct positive
feedbacks. In this paper, we study the number of positive steady states of a
general multisite phosphorylation-dephosphorylation cycle, and how the number
of positive steady states varies by changing the biological parameters. We show
analytically that (1) for some parameter ranges, there are at least n+1 (if n
is even) or n (if n is odd) steady states; (2) there never are more than 2n-1
steady states (in particular, this implies that for n=2, including single
levels of MAPK cascades, there are at most three steady states); (3) for
parameters near the standard Michaelis-Menten quasi-steady state conditions,
there are at most n+1 steady states; and (4) for parameters far from the
standard Michaelis-Menten quasi-steady state conditions, there is at most one
steady state.

Introduction
A promising approach to handling the complexity of cell signaling pathways is to decompose pathways into
small motifs, and analyze the individual motifs. One particular motif that has attracted much attention in
recent years is the cycle formed by two or more inter-convertible forms of one protein. The protein, denoted
here by S0, is ultimately converted into a product, denoted here by Sn, through a cascade of “activation”
reactions triggered or facilitated by an enzyme E; conversely, Sn is transformed back (or “deactivated”)
into the original S0, helped on by the action of a second enzyme F . See Figure 1.
S S0 2S1
SSS nn−1n−2
Figure 1: A futile cycle of size n.
Such structures, often called “futile cycles” (also called substrate cycles, enzymatic cycles, or enzymatic
inter-conversions, see [1]), serve as basic blocks in cellular signaling pathways and have pivotal impact on
the signaling dynamics. Futile cycles underlie signaling processes such as GTPase cycles [2], bacterial
two-component systems and phosphorelays [3, 4] actin treadmilling [5]), and glucose mobilization [6], as
well as metabolic control [7] and cell division and apoptosis [8] and cell-cycle checkpoint control [9]. One
very important instance is that of Mitogen-Activated Protein Kinase (“MAPK”) cascades, which regulate
http://arxiv.org/abs/0704.0036v2
primary cellular activities such as proliferation, differentiation, and apoptosis [10–13] in eukaryotes from
yeast to humans.
MAPK cascades usually consist of three tiers of similar structures with multiple feedbacks [14–16].
Each individual level of the MAPK cascades is a futile cycle as depicted in Figure 1 with n = 2. Markevich
et al.’s paper [17] was the first to demonstrate the possibility of multistationarity at a single cascade
level, and motivated the need for analytical studies of the number of steady states. Conradi et al. studied
the existence of multistationarity in their paper [19], employing algorithms based on Feinberg’s chemical
reaction network theory (CRNT). (For more details on CRNT, see [31,32].) The CRNT algorithm confirms
multistationarity in a single level of MAPK cascades, and provides a set of kinetic constants which can
give rise to multistationarity. However, the CRNT algorithm only tests for the existence of multiple steady
states, and does not provide information regarding the precise number of steady states.
In [18], Gunawardena proposed a novel approach to the study of steady states of futile cycles. His
approach, which was focused in the question of determining the proportion of maximally phosphorylated
substrate, was developed under the simplifying quasi-steady state assumption that substrate is in excess.
Nonetheless, our study of multistationarity uses in a key manner the basic formalism in [18], even for the
case when substrate is not in excess.
In Section 2, we state our basic assumptions regarding the model. The basic formalism and background
for the approach is provided in Section 3. The main focus of this paper is on Section 4, where we derive
various bounds on the number of steady states of futile cycles of size n. The first result is a the lower
bound for the number of steady states. Currently available results on lower bounds, as in [29], can only
handle the case when quasi-steady state assumptions are valid; we substantially extend these results to the
fully general case by means of a perturbation argument which allows one to get around these restricted
assumptions. Another novel feature of our results in this paper is the derivation of an upper bound
of 2n − 1, valid for all kinetic constants. Models in molecular cell biology are characterized by a high
degree of uncertainly in parameters, hence such results valid over the entire parameter space are of special
significance. However, when more information of the parameters are available, sharper upper bounds can
obtained, see Theorems 4 and 5. We finally conclude our paper in Section 5 with a conjecture of an n+ 1
upper bound.
We remark that the results given here complement our work dealing with the dynamical behavior
of futile cycles. For the case n = 2, [25] showed that the model exhibits generic convergence to steady
states but no more complicated behavior, at least within restricted parameter ranges, while [27] showed a
persistence property (no species tends to be eliminated) for any possible parameter values. These papers
did not address the question of estimating the number of steady states. (An exception is the case n = 1,
for which uniqueness of steady states can be proved in several ways, and for which global convergence to
these unique equilibria holds [27].)
2 Model assumptions
Before presenting mathematical details, let us first discuss the basic biochemical assumptions that go into
the model. In general, phosphorylation and dephosphorylation can follow either distributive or processive
mechanism. In the processive mechanism, the kinase (phosphatase) facilitates two or more phosphorylations
(dephosphorylations) before the final product is released, whereas in the distributive mechanism, the kinase
(phosphatase) facilitates at most one phosphorylation (dephosphorylation) in each molecular encounter.
In the case of n = 2, a futile cycle that follows the processive mechanism can be represented by reactions
as follows:
S0 + E ←→ ES0 ←→ ES1 −→ S2 + E
S2 + F ←→ FS2 ←→ FS1 −→ S0 + F ;
and the distributive mechanism can be represented by reactions:
S0 + E ←→ ES0 −→ S1 + E ←→ ES1 −→ S2 + E
S2 + F ←→ FS2 −→ S1 + F ←→ FS1 −→ S0 + F.
Biological experiments have demonstrated that both dual phosphorylation and dephosphorylation in MAPK
are distributive, see [14–16]. In their paper [19], Conradi et al. showed mathematically that if either phos-
phorylation or dephosphorylation follows a processive mechanism, the steady state will be unique, which,
it is argued in [19], contradicts experimental observations. So, to get more interesting results, we assume
that both phosphorylations and dephosphorylations in the futile cycles follow the distributive mechanism.
Our structure of futile cycles in Figure 1 also implicitly assumes a sequential instead of a random
mechanism. By a sequential mechanism, we mean that the kinase phosphorylates the substrates in a
specific order, and the phosphatase works in the reversed order. This assumption dramatically reduces the
number of different phospho-forms and simplifies our analysis. In a special case when the kinetic constants
of each phosphorylation are the same and the kinetic constants of each dephosphorylation are the same,
the random mechanism can be easily included in the sequential case. Biologically, there are systems, for
instance the auto-phosphorylation of FGF-receptor-1, that have been experimentally shown to follow a
sequential mechanism [33].
To model the reactions, we assume mass action kinetics, which is standard in mathematical modeling
of molecular events in biology.
3 Mathematical formalism
In this section, we set up a mathematical framework for studying the steady states of futile cycles. Let us
first write down all the elementary chemical reactions in Figure 1:
S0 + E
koff0
kcat0
→ S1 + E
Sn−1 + E
konn−1
koffn−1
kcatn−1
→ Sn + E
S1 + F
loff0
lcat0
→ S0 + F
Sn + F
lonn−1
loffn−1
lcatn−1
→ Sn−1 + F
where kon0 , etc., are kinetic parameters for binding and unbinding, ES0 denotes the complex consisting of
the enzyme E and the substrate S0, and so forth. These reactions can be modeled by 3n + 3 differential-
algebraic equations according to mass action kinetics:
= −kon0s0e+ koff0c0 + lcat0d1
= −konisie+ koffi
ci + kcati−1ci−1 − loni−1sif + loffi−1
di + lcatidi+1, i = 1, . . . , n− 1
= konjsje− (koffj
+ kcatj )cj , j = 0, . . . , n− 1 (1)
= lonk−1skf − (loffk−1
+ lcatk−1)dk, k = 1, . . . , n,
together with the algebraic “conservation equations”:
Etot = e+
Ftot = f +
di, (2)
Stot =
The variables s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f stand for the concentrations of
S0, . . . , Sn, ES0, . . . , ESn−1, FS1, . . . , FSn, E, F
respectively. For each positive vector
κ =(kon0 , . . . , konn−1 , koff0
, . . . , koffn−1
, kcat0 , . . . , kcatn−1 ,
lon0 , . . . , lonn−1 , loff0
, . . . , loffn−1
, lcat0 , . . . , lcatn−1) ∈ R
(of “kinetic constants”) and each positive triple C = (Etot, Ftot, Stot), we have a different system Σ(κ, C).
Let us write the coordinates of a vector x ∈ R3n+3+ as:
x = (s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f),
and define a mapping
Φ : R3n+3+ × R
+ × R
+ −→ R
with components Φ1, . . . ,Φ3n+3 where the first 3n components are
Φ1(x, κ, C) = −kon0s0e+ koff0c0 + lcat0d1,
and so forth, listing the right hand sides of the equations (1), Φ3n+1 is
ci − Etot,
and similarly for Φ3n+2 and Φ3n+3, we use the remaining equations in (2).
For each κ, C, let us define a set
Z(κ, C) = {x |Φ(x, κ, C) = 0}.
Observe that, by definition, given x ∈ R3n+3+ , x is a positive steady state of Σ(κ, C) if and only if x ∈ Z(κ, C).
So, the mathematical statement of the central problem in this paper is to count the number of elements
in Z(κ, C). Our analysis will be greatly simplified by a preprocessing. Let us introduce a function
Ψ : R3n+3+ × R
+ × R
+ −→ R
with components Ψ1, . . . ,Ψ3n+3 defined as
Ψ1 = Φ1 +Φn+1
Ψi = Φi +Φn+i +Φ2n+i−1 +Ψi−1, i = 2, . . . , n
Ψj = Φj, j = n+ 1, . . . , 3n + 3.
It is easy to see that
Z(κ, C) = {x |Ψ(x, κ, C) = 0},
but now the first 3n equations are:
Ψi = lcati−1di − kcati−1ci−1 = 0, i = 1, . . . , n,
Ψn+1+j = konjsje− (koffj
+ kcatj )cj = 0, j = 0, . . . , n− 1
Ψ2n+k = lonk−1skf − (loffk−1
+ lcatk−1)dk = 0, k = 1, . . . , n,
and can be easily solved as:
si+1 = λi(e/f)si, (3)
di+1 =
fsi+1
, (5)
where
kcatiLMi
KMi lcati
, KMi =
kcati + koffi
, LMi =
lcati + loffi
, i = 0, . . . , n− 1. (6)
We may now express
0 si,
0 ci and
1 di in terms of s0, κ, e and f :
si = s0
1 + λ0
+ λ0λ1
+ · · ·+ λ0 · · ·λn−1
:= s0ϕ
ci = es0
+ · · ·+
λ0 · · ·λn−2
KMn−1
:= es0ϕ
, (7)
di = fs0
+ · · · +
λ0 · · ·λn−1
LMn−1
:= fs0ϕ
Although the equation Ψ = 0 represents 3n+3 equations with 3n+3 unknowns, next we will show that
it can be reduced to two equations with two unknowns, which have the same number of positive solutions
as Ψ = 0. Let us first define a set
S(κ, C) = {(u, v) ∈ R+ × R+ |G
1 (u, v) = 0, G
2 (u, v) = 0},
where G
1 , G
2 : R
+ −→ R are given by
1 (u, v) = v (uϕ
1(u)− ϕ
2(u)Etot/Ftot)− Etot/Ftot + u,
2 (u, v) = ϕ
0(u)ϕ
2 (u)v
2 + (ϕκ0 (u)− Stotϕ
2 (u) + Ftotuϕ
1 (u) + Ftotϕ
2 (u)) v − Stot.
The precise statement is as follows:
Lemma 1 There exists a mapping δ : R3n+3 −→ R2 such that, for each κ, C, the map δ restricted to
Z(κ, C) is a bijection between the sets Z(κ, C) and S(κ, C).
Proof. Let us define the mapping δ : R3n+3 −→ R2 as δ(x) = (e/f, s0), where
x = (s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f).
If we can show that δ induces a bijection between Z(κ, C) and S(κ, C), we are done.
First, we claim that δ(Z(κ, C)) ⊆ S(κ, C). Pick any x ∈ Z(κ, C), we have that x satisfies (3)-(5).
Moreover, Φ3n+2(x, κ, C) = 0 yields
Etot = e+ es0ϕ
and thus
1 + s0ϕ
1(e/f)
. (8)
Using Φ3n+1(x, κ, C) = 0 and Φ3n+2(x, κ, C) = 0, we get:
e(1 + s0ϕ
1(e/f))
f(1 + s0ϕ
2(e/f))
, (9)
which is G
1 (e/f, s0) = 0 after multiplying by 1 + s0ϕ
2(e/f) and rearranging terms.
To check that G
2 (e/f, s0) = 0, we start with Φ3n+3(x, κ, C) = 0, i.e.
Stot =
Using (7) and (8), this expression becomes
Stot = s0ϕ
Etots0ϕ
1(e/f)
1 + s0ϕ
1 (e/f)
Ftots0ϕ
2(e/f)
1 + s0ϕ
2(e/f)
= s0ϕ
eFtots0ϕ
1(e/f)
f(1 + s0ϕ
2(e/f))
Ftots0ϕ
2(e/f)
1 + s0ϕ
2(e/f)
where the last equality comes from (9).
After multiplying by 1 + s0ϕ
2 (e/f), and simplifying, we get
ϕκ0 (
)ϕκ2 (
)s20 +
)− Stotϕ
Ftotϕ
) + Ftotϕ
s0 − Stot = 0,
that is, G
2 (e/f, s0) = 0. since both G
1 (e/f, s0) and G
2 (e/f, s0) are zero, δ(x) ∈ S(κ, C).
Next, we will show that S(κ, C) ⊆ δ(Z(κ, C)). For any y = (u, v) ∈ S(κ, C), let the coordinates of x be
defined as:
s0 = v
si+1 = λiusi
1 + s0ϕ
1 (u)
di+1 =
fsi+1
for i = 0, . . . , n − 1. It is easy to see that the vector x = (s0, . . . , sn, c0, . . . , cn−1, d1, . . . , dn, e, f) satisfies
Φ1(x, κ, C) = 0, . . . ,Φ3n+1(x, κ, C) = 0. If Φ3n+2(x, κ, C) and Φ3n+3(x, κ, C) are also zero, then x is an
element of Z(κ, C) with δ(x) = y. Given the condition that G
i (u, v) = 0 (i = 1, 2) and u = e/f, v = s0,
we have G
1 (e/f, s0) = 0, and therefore (9) holds. Since
1 + s0ϕ
1(e/f)
in our construction, we have
Ftot = f(1 + s0ϕ
2(e/f)) = f +
To check Φ3n+3(x, κ, C) = 0, we use
2 (e/f, s0)
1 + s0ϕ
2(e/f)
2 (e/f, s0) = 0 and 1 + s0ϕ
2(e/f) > 0. Applying (7)-(9), we have
di = s0ϕ
0(e/f) +
eFtots0ϕ
1(e/f)
f(1 + s0ϕ
2 (e/f))
Ftots0ϕ
2(e/f)
1 + s0ϕ
2(e/f)
= Stot.
It remains for us to show that the map δ is one to one on Z(κ, C). Suppose that δ(x1) = δ(x2) = (u, v),
where
xi = (si0, . . . , s
0, . . . , c
n−1, d
1, . . . , d
i, f i), i = 1, 2.
By the definition of δ, we know that s10 = s
0 and e
1/f1 = e2/f2. Therefore, s1i = s
i for i = 0, . . . , n.
Equation (8) gives
1 + vϕκ1 (u)
= e2.
Thus, f1 = f2, and c1i = c
i , d
i+1 = d
i+1 for i = 0, . . . , n − 1 because of (3)-(5). Therefore, x
1 = x2, and δ
is one to one.
The above lemma ensures that the two sets Z(κ, C) and S(κ, C) have the same number of elements. From
now on, we will focus on S(κ, C), the set of positive solutions of equations G
1 (u, v) = 0, G
2 (u, v) = 0,
1 (u, v) = v (uϕ
1(u)− ϕ
2(u)Etot/Ftot)− Etot/Ftot + u = 0, (10)
2 (u, v) = ϕ
0(u)ϕ
2 (u)v
2 + (ϕκ0(u)− Stotϕ
2 (u) + Ftotuϕ
1 (u) + Ftotϕ
2(u)) v − Stot = 0. (11)
4 Number of positive steady states
4.1 Lower bound on the number of positive steady states
One approach to solving (10)-(11) is to view G
2 (u, v) as a quadratic polynomial in v. Since G
2 (u, 0) < 0,
equation (11) has a unique positive root, namely
−Hκ,C(u) +
Hκ,C(u)2 + 4Stotϕ
0(u)ϕ
2 (u)
2ϕκ0 (u)ϕ
2 (u)
, (12)
where
Hκ,C(u) = ϕκ0(u)− Stotϕ
2(u) + Ftotuϕ
1(u) + Ftotϕ
2(u). (13)
Substituting this expression for v into (10), and multiplying by ϕκ0 (u), we get
F κ,C(u) :=
−H̃κ,C(u) +
H̃κ,C(u)2 + 4Stotϕ
0 (u)ϕ
2 (u)
2ϕκ2 (u)
uϕκ1(u)−
ϕκ2(u)
ϕκ0(u)+uϕ
0 (u) = 0.
So, any (u, v) ∈ S(κ, C) should satisfy (12) and (14). On the other hand, any positive solution u of (14)
(notice that ϕκ0(u) > 0) and v given by (12) (always positive) provide a positive a solution of (10)-(11),
that is, (u, v) is an element in S(κ, C). Therefore, the number of positive solutions of (10)-(11) is the same
as the number of positive solutions of (12) and (14). But v is uniquely determined by u in (12), which
further simplifies the problem to one equation (14) with one unknown u. Based on this observation, we
have:
Theorem 1 For each positive numbers Stot, γ, there exist ε0 > 0 and κ ∈ R
+ such that the following
property holds. Pick any Etot, Ftot such that
Ftot = Etot/γ < ε0Stot/γ, (15)
then the system Σ(κ, C) with C = (Etot, Ftot, Stot) has at least n + 1 (n) positive steady states when n is
even (odd).
Proof. For each κ, γ, Stot, let us define two functions R+ × R+ −→ R as follows:
κ,γ,Stot(ε, u) = H
κ,(εStot,εStot/γ,Stot)(u) (16)
= ϕκ0(u)− Stotϕ
2(u) + ε
uϕκ1(u) + ε
ϕκ2(u),
κ,γ,Stot(ε, u) = F
κ,(εStot,εStot/γ,Stot)(u) (17)
κ,γ,Stot(ε, u) +
κ,γ,Stot(ε, u)2 + 4Stotϕ
0 (u)ϕ
2 (u)
2ϕκ2 (u)
(uϕκ1(u)− γϕ
2 (u))
− γϕκ0(u) + uϕ
0(u).
By Lemma 1 and the argument before this theorem, it is enough to show that there exist ε0 > 0 and κ ∈
+ such that for all ε ∈ (0, ε0), the equation F̃
κ,γ,Stot(ε, u) = 0 has at least n+1 (n) positive solutions
when n is even (odd). (Then, given Stot, γ, Etot, and Ftot satisfying (15), we let ε = Etot/Stot < ε0,
and apply the result.)
A straightforward computation shows that when ε = 0,
κ,γ,Stot(0, u) = Stot (uϕ
1(u)− γϕ
2(u))− γϕ
0 (u) + uϕ
= λ0 · · ·λn−1u
n+1 + λ0 · · ·λn−2
KMn−1
(1− γβn−1)− γλn−1
+ · · ·+ λ0 · · ·λi−2
KMi−1
(1− γβi−1)− γλi−1
ui + · · · (18)
(1− γβ0)− γλ0
u− γ,
where the λi’s and KMi ’s are defined as in (6), and βi = kcati/lcati . The polynomial F̃
κ,γ,Stot(0, u)
is of degree n + 1, so there are at most n + 1 positive roots. Notice that u = 0 is not a root because
κ,γ,Stot(0, u) = −γ < 0, which also implies that when n is odd, there can not be n + 1 positive roots.
Now fix any Stot and γ. We will construct a vector κ such that F̃
κ,γ,Stot(0, u) has n+ 1 distinct positive
roots when n is even.
Let us pick any n+ 1 positive real numbers u1 < · · · < un+1, such that their product is γ, and assume
(u− u1) · · · (u− un+1) = u
n+1 + anu
n + · · · + a1u+ a0, (19)
where a0 = −γ < 0 keeping in mind that ai’s are given. Our goal is to find a vector κ ∈ R
+ such that
(18) and (19) are the same. For each i = 0, . . . , n − 1, we pick λi = 1. Comparing the coefficients of u
in (18) and (19), we have:
(1 + a0βi) = ai+1 − a0 − 1. (20)
Let us pick KMi > 0 such that
(ai+1 − a0 − 1)− 1 < 0, then take
(ai+1 − a0 − 1)− 1
in order to satisfy (20). From the given
λ0, . . . , λn−1,KM0 , . . . ,KMn−1 , β0, . . . , βn−1,
we will find a vector
κ =(kon0 , . . . , konn−1 , koff0
, . . . , koffn−1
, kcat0 , . . . , kcatn−1 ,
lon0 , . . . , lonn−1 , loff0
, . . . , loffn−1
, lcat0 , . . . , lcatn−1) ∈ R
such that βi = kcati/lcati , i = 0, . . . , n− 1, and (6) holds. This vector κ will guarantee that F̃
κ,γ,Stot(0, u)
has n + 1 positive distinct roots. When n is odd, a similar construction will give a vector κ such that
κ,γ,Stot(0, u) has n positive roots and one negative root.
One construction of κ (given λi,KMi , βi, i = 0, . . . , n − 1) is as follows. For each i = 0, . . . , n − 1, we
start by defining:
LMi =
λiKMi
consistently with the definitions in (6). Then, we take
koni = 1, loni = 1,
koffi
= αiKMi , kcati = (1− αi)KMi , lcati =
1− αi
KMi , loffi
= LMi − lcati ,
where αi ∈ (0, 1) is chosen such that
loffi
= LMi −
1− αi
KMi > 0.
This κ satisfies βi = kcati/lcati , i = 0, . . . , n− 1, and (6).
In order to apply the Implicit Function Theorem, we now view the functions defined by formulas in
(16) and (17) as defined also for ε ≤ 0, i.e. as functions R×R+ −→ R. It is easy to see that F̃
κ,γ,Stot(ε, u)
is C1 on R × R+ because the polynomial under the square root sign in F̃
κ,γ,Stot(ε, u) is never zero. On
the other hand, since F̃
κ,γ,Stot(0, u) is a polynomial in u with distinct roots, ∂F̃
κ,γ,Stot
(0, ui) 6= 0. By the
Implicit Function Theorem, for each i = 1, . . . , n+ 1, there exist open intervals Ei containing 0, and open
intervals Ui containing ui, and a differentiable function
αi : Ei → Ui
such that αi(0) = ui, F̃
κ,γ,Stot(ε, αi(ε)) = 0 for all ε ∈ Ei, and the images αi(Ei)’s are non-overlapping. If
we take
(0, ε0) :=
(0,+∞),
then for any ε ∈ (0, ε0), we have {αi(ε)} as n+ 1 distinct positive roots of F̃
κ,γ,Stot(ε, u). The case when
n is odd can be proved similarly.
The above theorem shows that when Etot/Ftot is sufficiently small, it is always possible for the futile
cycle to have n + 1 (n) steady states when n is even (odd), by choosing appropriate kinetic constants κ.
We should notice that for arbitrary κ, the derivative of F̃ at each positive root may become zero, which
breaks down the perturbation argument. Here is an example to show that more conditions are needed:
n = 2, λ0 = 1, λ1 = 3, γ = 6, β0 = β1 = 1/12, K0 = 1/8, K1 = 1/2, Stot = 5,
we have that
κ,γ,Stot(0, u) = 3u3 − 12u2 + 15u− 6 = 3(u− 1)2(u− 2)
has a double root at u = 1. In this case, even for ε = 0.01, there is only one positive root of F̃
κ,γ,Stot(ε, u),
see Figure 2.
1 2 3
Figure 2: The plot of the function F̃
κ,γ,Stot(0.01, u) on [0, 3]. There is a unique positive real solution
around u = 2.14, the double root u = 1 of F̃
κ,γ,Stot(0, u) bifurcates to two complex roots with non-zero
imaginary parts.
However, the following lemma provides a sufficient condition for ∂F
κ,γ,Stot
(0, ū) 6= 0, for any positive
ū such that F̃
κ,γ,Stot(0, ū) = 0.
Lemma 2 For each positive numbers Stot, γ, and vector κ ∈ R
+ , if
1− γβj
holds for all j = 1, · · · , n − 1, then ∂F̃
κ,γ,Stot
(0, ū) 6= 0.
See Appendix for the proof.
Theorem 2 For each positive numbers Stot, γ, and vector κ ∈ R
+ satisfying condition (21), there exists
ε1 > 0 such that for any Ftot, Etot satisfying Ftot = Etot/γ < ε1Stot/γ, the number of positive steady
states of system Σ(κ, C) is greater or equal to the number of (positive) roots of F̃
κ,γ,Stot(0, u).
Proof. Suppose that F̃
κ,γ,Stot(0, u) has m roots: ū1, . . . , ūm. Applying Lemma 2, we have
κ,γ,Stot
(0, ūk) 6= 0, k = 1, . . . ,m.
By the perturbation arguments as in Theorem 1, we have that there exists ε1 > 0 such that F̃
κ,γ,Stot(ε, u)
has at least m roots for all 0 < ε < ε1.
The above result depends heavily on a perturbation argument, which only works when Etot/Ftot is
sufficiently small. In the next section, we will give an upper bound of the number of steady states with no
restrictions on Etot/Ftot, and independent of κ and C.
4.2 Upper bound on the number of steady states
Theorem 3 For each κ, C, the system Σ(κ, C) has at most 2n− 1 positive steady states.
Proof. An alternative approach to solving (10)-(11) is to first eliminate v from (10) instead of from (11),
Etot/Ftot − u
uϕκ1(u)− (Etot/Ftot)ϕ
, (22)
when uϕκ1(u) − (Etot/Ftot)ϕ
2 (u) 6= 0. Then, we substitute (22) into (11), and multiply by (uϕ
1(u) −
(Etot/Ftot)ϕ
2 (u))
2 to get:
P κ,C(u) := ϕκ0ϕ
+ (ϕκ0 − Stotϕ
2 + Ftotuϕ
1 + Ftotϕ
uϕκ1 −
− Stot
uϕκ1 −
= 0. (23)
Therefore, if uϕκ1(u) − (Etot/Ftot)ϕ
2 (u) 6= 0, the number of positive solutions of (10)-(11) is no greater
than the number of positive roots of P κ,C(u).
In the special case when uϕκ1(u) − (Etot/Ftot)ϕ
2(u) = 0, by (10), we must have u = Etot/Ftot, and
thus ϕκ1 (Etot/Ftot) = ϕ
2 (Etot/Ftot). Substituting into (11), we get a unique v defined as in (12) with
u = Etot/Ftot. But notice that in this case u = Etot/Ftot is also a root of P
κ,C(u), so also in this case
the number of positive solutions to (10)-(11) is no greater than the number of positive roots of P κ,C(u).
It is easy to see that P κ,C(u) is divisible by u. Consider the polynomial Qκ,C(u) := P κ,C(u)/u of
degree 2n + 1. We will first show that Qκ,C(u) has no more than 2n positive roots, then we will prove by
contradiction that 2n distinct positive roots can not be achieved.
It is easy to see that the coefficient of u2n+1 is
(λ0 · · · λn−1)
LMn−1
and the constant term is
FtotKM0
So the polynomial Qκ,C(u) has at least one negative root, and thus has no more than 2n positive roots.
Suppose that S(κ, C) has cardinality 2n, then Qκ,C(u) must have 2n distinct positive roots, and each of
them has multiplicity one. Let us denote the roots as u1, . . . , u2n in ascending order. We claim that none
of them equals Etot/Ftot. If so, we would have ϕ
1(Etot/Ftot) = ϕ
2 (Etot/Ftot), and Etot/Ftot would
be a double root of Qκ,C(u), contradiction.
Since Qκ,C(0) > 0, Qκ,C(u) is positive on intervals
I0 = (0, u1), I1 = (u2, u3), . . . , In−1 = (u2n−2, u2n−1), In = (u2n,∞),
and negative on intervals
J1 = (u1, u2), . . . , Jn = (u2n−1, u2n).
As remarked earlier, ϕκ1 (Etot/Ftot) 6= ϕ
2 (Etot/Ftot), the polynomial Q
κ,C(u) evaluated at Etot/Ftot
is negative, and therefore, Etot/Ftot belongs to one of the J intervals, say Js = (u2s−1, u2s), for some
s ∈ {1, . . . , n} .
On the other hand, the denominator of v in (22), denoted as B(u), is a polynomial of degree n and
divisible by u. If B(u) has no positive root, then it does not change sign on the positive axis of u. But v
changes sign when u passes Etot/Ftot, thus v2s−1 and v2s have opposite signs, and one of (u2s−1, v2s−1)
and (u2s, v2s) is not a solution to (10)-(11), which contradicts the fact that both are in S(κ, C).
Otherwise, there exists a positive root ū of B(u) such that there is no other positive root of B(u)
between ū and Etot/Ftot. Plugging ū into Q
κ,C(u), we see that Qκ,C(ū) is always positive, therefore, ū
belongs to one of the I intervals, say It = (u2t, u2t+1) for some t ∈ {0, . . . , n}. There are two cases:
1. Etot/Ftot < ū. We have
u2s−1 < Etot/Ftot < u2t < ū.
Notice that v changes sign when u passes Etot/Ftot, so the corresponding v2s−1 and v2t have different
signs, and either (u2s−1, v2s−1) /∈ S(κ, C) or (u2t, v2t) /∈ S(κ, C), contradiction.
2. Etot/Ftot > ū. We have
ū < u2t+1 < Etot/Ftot < u2s.
Since v changes sign when u passes Etot/Ftot, so the corresponding v2t+1 and v2s have different
signs, and either (u2t+1, v2t+1) /∈ S(κ, C) or (u2s, v2s) /∈ S(κ, C), contradiction.
Therefore, Σ(κ, C) has at most 2n− 1 steady states.
4.3 Fine-tuned upper bounds
In the previous section, we have seen that any (u, v) ∈ S(κ, C), u 6= Etot/Ftot must satisfy (22)-(23), but
not all solutions of (22)-(23) are elements in S(κ, C). Suppose that (u, v) is a solution of (22)-(23), it is
in S(κ, C) if and only if u, v > 0. In some special cases, for example, when the enzyme is in excess, or the
substrate is in excess, we could count the number of solutions of (22)-(23) which are not in S(κ, C) to get
a better upper bound.
The following is a standard result on continuity of roots; see for instance Lemma A.4.1 in [30]:
Lemma 3 Let g(z) = zn + a1z
n−1 + · · ·+ an be a polynomial of degree n and complex coefficients having
distinct roots
λ1, . . . , λq,
with multiplicities
n1 + · · ·+ nq = n,
respectively. Given any small enough δ > 0 there exists a ε > 0 so that, if
h(z) = zn + b1z
n−1 + · · ·+ bn, |ai − bi| < ε for i = 1, . . . , n,
then h has precisely ni roots in Bδ(λi) for each i = 1, . . . , q.
Theorem 4 For each γ > 0 and κ ∈ R6n−6+ such that ϕ
1 (γ) 6= ϕ
2 (γ), and each Stot > 0, there exists
ε2 > 0 such that for all positive numbers Etot, Ftot satisfying Ftot = Etot/γ < ε2Stot/γ, the system
Σ(κ, C) has at most n+ 1 positive steady states.
Proof. Let us define a function R+ × C −→ C as follows:
κ,γ,Stot(ε, u) = Q
κ,(εStot,εStot/γ,Stot)(u),
and a set B
κ,γ,Stot(ε) consisting of the roots of Q̃
κ,γ,Stot(ε, u) which are not positive or the corresponding
v’s determined by u’s as in (22) are not positive, Since Q̃
κ,γ,Stot(ε, u) is a polynomial of degree 2n + 1,
if we can show that there exists ε2 > 0 such that for any ε ∈ (0, ε2), Q̃
κ,γ,Stot(ε, u) has at least n roots
counting multiplicities that are in B
κ,γ,Stot(ε), then we are done.
In order to apply Lemma 3, we regard the function Q̃
κ,γ,Stot as defined on R× C. At ε = 0:
κ,γ,Stot(0, u) = [ϕκ0ϕ
2(γ − u)
2 + (ϕκ0 − Stotϕ
2 )(uϕ
1 − γϕ
2)(γ − u)− Stot(uϕ
1 − γϕ
= [ϕκ0ϕ
2(γ − u)
2 + ϕκ0(uϕ
1 − γϕ
2)(γ − u)− Stotϕ
1 − γϕ
2)(γ − u)− Stot(uϕ
1 − γϕ
= [ϕκ0 (γ − u)u(ϕ
1 − ϕ
2) + Stotu(uϕ
1 − γϕ
2 )(ϕ
2 − ϕ
1)]/u
= (ϕκ2 − ϕ
1)(uϕ
0 + Stot(uϕ
1 − γϕ
2 )− γϕ
= (ϕκ2 − ϕ
κ,γ,Stot(0, u)
Let us denote the distinct roots of Q̃
κ,γ,Stot(0, u)/u as
u1, . . . , uq,
with multiplicities
n1 + · · ·+ nq = 2n+ 1,
and the roots of ϕκ1 − ϕ
u1, . . . , up, p ≤ q,
with multiplicities
m1 + · · ·+mp = n, ni ≥ mi, for i = 1, . . . , p.
For each i = 1, . . . , p, if ui is real and positive, then there are two cases (ui 6= γ as ϕ
1(γ) 6= ϕ
2(γ)):
1. ui > γ. We have
1(ui)− γϕ
2 (ui) > γ(ϕ
1 (ui)− ϕ
2(ui)) = 0.
2. ui < γ. We have
1(ui)− γϕ
2 (ui) < γ(ϕ
1 (ui)− ϕ
2(ui)) = 0.
In both cases, uiϕ
1(ui)− γϕ
2 (ui) and γ − ui have opposite signs, i.e.
1 (ui)− γϕ
2(ui))(γ − ui) < 0.
Let us pick δ > 0 small enough such that the following conditions hold:
1. For all i = 1, . . . , p, if ui is not real, then Bδ(ui) has no intersection with the real axis.
2. For all i = 1, . . . , p, if ui is real and positive, the following inequality holds for any real u ∈ Bδ(ui):
(uϕκ1(u)− γϕ
2 (u))(γ − u) < 0. (24)
3. For all i = 1, . . . , p, if ui is real and negative, then Bδ(ui) has no intersection with the imaginary
axis.
4. Bδ(uj)
Bδ(uk) = ∅ for all j 6= k = 1, . . . , q.
By Lemma 3, there exists ε3 > 0 such that for all ε ∈ (0, ε3), the polynomial Q̃
κ,γ,Stot(ε, u)/u has exactly
nj roots in each Bδ(uj), j = 1, . . . , q, denoted by u
j (ε), k = 1, . . . , nj .
We pick one such ε, and we claim that none of the roots in Bδ(ui), i = 1, . . . , p with the v defined as
in (22) will be an element in S. If so, we are done, since there are
1 ni ≥
1 mi = n such roots, of
κ,γ,Stot(ε, u) which are in B
κ,γ,Stot(ε).
For each i = 1, . . . , p, there are two cases:
1. ui is not real. Then condition 1 guarantees that u
i (ε) is not real for each k = 1, . . . , ni, and thus is
κ,γ,Stot(ε).
2. ui is real and positive. Pick any root u
i (ε) ∈ Bδ(ui), k = 1, . . . , ni, the corresponding v
i (ε) equals
γ − uki (ε)
uki (ε)ϕ
i (ε)) − γϕ
i (ε))
) < 0
followed from (24). So (uki (ε), v
i (ε)) /∈ S(κ, C), and u
i (ε) ∈ B
κ,γ,Stot(ε).
3. ui is real and negative. By condition 1 and 3, u
i (ε) is not positive for all k = 1, . . . , ni.
The next theorem considers the case when enzyme is in excess:
Theorem 5 For each γ > 0, κ ∈ R6n−6+ such that ϕ
1 (γ) 6= ϕ
2(γ), and each Etot > 0, there exists ε3 > 0
such that for all positive numbers Ftot, Stot satisfying Ftot = Etot/γ > Stot/(ε3γ), the system Σ(κ, C) has
at most one positive steady state.
Proof. For each γ > 0, κ ∈ R6n−6+ such that ϕ
1 (γ) 6= ϕ
2 (γ), and each Etot > 0, we define a function
R+ × C −→ C as follows:
κ,γ,Etot(ε, u) = Q
κ,(Etot,Etot/γ,εEtot)(u).
Let us define the set C
κ,γ,Etot(ε) as the set of roots of Q̄
κ,γ,Etot(ε, u) which are not positive or the
corresponding v’s determined by u’s as in (22) are not positive. If we can show that there exists ε3 > 0
such that for any ε ∈ (0, ε3) there is at most one positive root of Q̄
κ,γ,Etot(ε, u) that is not in C
κ,γ,Etot(ε),
we are done.
In order to apply Lemma 3, we now view the function Q̄
κ,γ,Etot as defined on R× C. At ε = 0:
κ,γ,Etot(0, u) = (γ − u)
(γ − u)ϕκ0ϕ
ϕκ0 +
uϕκ1 +
(uϕκ1 − γϕ
:= (γ − u)R
κ,γ,Etot(u).
Let us denote the distinct roots of Q̄
κ,γ,Etot(0, u)/u as
u1(= γ), u2, . . . , uq,
with multiplicities
n1 + · · ·+ nq = 2n+ 1,
and u2, . . . , uq are the roots of R
κ,γ,Etot(u) other than γ.
Since ϕκ1(γ) 6= ϕ
2(γ), R
κ,γ,Etot(u) is not divisible by u− γ, and thus n1 = 1.
For each i = 2, . . . , q, we have
(γ − ui)ϕ
0(ui)ϕ
2 (ui) = −
ϕκ0(ui) +
1(ui) +
ϕκ2(ui)
1 (ui)− γϕ
2(ui)) .
If ui > 0, then ϕ
0(ui)ϕ
2 (ui) and ϕ
0 (ui) +
1 (ui) +
ϕκ2 (ui) are both positive. Since uiϕ
1(ui)−
γϕκ2(ui) and γ − ui are non zero, uiϕ
1 (ui)− γϕ
2(ui) and γ − ui must have opposite signs, that is
1 (ui)− γϕ
2(ui))(γ − ui) < 0.
Let us pick δ > 0 small enough such that the following conditions hold for all i = 2, . . . , q:
1. If ui is not real, then Bδ(ui) has no intersection with the real axis.
2. If ui is real and positive, then for any real u ∈ Bδ(ui), the following inequality holds:
(uϕκ1(u)− γϕ
2 (u))(γ − u) < 0. (25)
3. If ui is real and negative, then Bδ(ui) has no intersection with the imaginary axis.
4. Bδ(uj)
Bδ(uk) = ∅ for all i 6= k = 2, . . . , q.
By Lemma 3, there exists ε3 > 0 such that for all ε ∈ (0, ε3), the polynomial Q̄
κ,γ,Etot(ε, u) has exactly
nj roots in each Bδ(uj), j = 1, . . . , q, denoted by u
j (ε), k = 1, . . . , nj .
We pick one such ε, and if we can show that all of the roots in Bδ(ui), i = 2, . . . , q are in C
κ,γ,Etot(ε),
then we are done, since the only roots that may not be in C
κ,γ,Etot(ε) are the roots in Bδ(u1), and there
is one root in Bδ(u1).
For each i = 2, . . . , p, there are three cases:
1. ui is not real. Then condition 1 guarantees that u
i (ε) is not real for all k = 1, . . . , ni.
2. ui is real and positive. Pick any root u
i (ε), k = 1, . . . , ni, the corresponding v
i (ε) equals
γ − uki (ε)
uki (ε)ϕ
i (ε))− γϕ
i (ε))
So, uki (ε) is in C
κ,γ,Etot(ε).
3. ui is real and negative. By conditions 1 and 3, u
i (ε) is not positive for all k = 1, . . . , ni.
5 Conclusions and discussions
Here we have set up a mathematical model for multisite phosphorylation-dephosphorylation cycles of size
n, and studied the number of positive steady states based on this model. We reformulated the question
of number of positive steady states to question of the number of positive roots of certain polynomials,
through which we also applied perturbation techniques. Our theoretical results depend on the assumption
of mass action kinetics and distributive sequential mechanism, which are customary in the study of multisite
phosphorylation and dephosphorylation.
An upper bound of 2n−1 steady states is obtained for arbitrary parameter combinations. Biologically,
when the substrate concentration greatly exceeds that of the enzyme, there are at most n + 1 (n) steady
states if n is even (odd). And this upper bound can be achieved under proper kinetic conditions, see
Theorem 1 for the construction. On the other extreme, when the enzyme is in excess, there is a unique
steady state.
As a special case of n = 2, which can be applied to a single level of MAPK cascades. Our results
guarantees that there are no more than three steady states, consistent with numerical simulations in [17].
We notice that there is an apparent gap between the upper bound 2n−1 and the upper bound of n+1
(n) if n is even (odd) when the substrate is in excess. If we think the ratio Etot/Ftot as a parameter ε,
then when ε≪ 1, there are at most n+1 (n) steady states when n is even (odd), which coincides with the
largest possible lower bound. When ε ≫ 1, there is a unique steady state. If the number of steady states
changes “continuously” with respect to ε, then we do not expect the number of steady states to exceed
n + 1 (n) if n is even (odd). So a natural conjecture would be that the number of steady states never
exceed n+ 1 under any conditions.
6 Acknowledgment
We thank Jeremy Gunawardena for very helpful discussions.
7 Appendix
proof of Lemma 2: Recall that (dropping the u’s in ϕκi , i = 0, 1, 2)
κ,γ,Stot(0, u) = uϕκ0 + Stot(uϕ
1 − γϕ
2)− γϕ
κ,γ,Stot
(0, u) = ϕκ0 + Stot(uϕ
1 − γϕ
′ − (γ − u)(ϕκ0 )
Since F̃
κ,γ,Stot(0, ū) = 0,
Stot(ūϕ
1 − γϕ
2 ) = (γ − ū)ϕ
that is,
γ − ū =
Stot(ūϕ
1 − γϕ
Therefore,
κ,γ,Stot
(0, ū) = ϕκ0 + Stot(uϕ
1 − γϕ
Stot(ūϕ
1 − γϕ
(ϕκ0)
= ϕκ0 +
ϕκ0(uϕ
1 − γϕ
′ − (ūϕκ1 − γϕ
= ϕκ0 +
((1 + λ0ū+ λ0λ1ū
2 + · · · + λ0 · · ·λn−1ū
(1− γβ0) + 2
(1− γβ1)ū+ · · ·+ n
λ0 · · · λn−2
KMn−1
(1− γβn−1)ū
λ0 + 2λ0λ1ū+ · · · + nλ0 · · ·λn−1ū
(1− γβ0)ū+
(1− γβ1)ū
2 + · · ·+
λ0 · · ·λn−2
KMn−1
(1− γβn−1)ū
= ϕκ0 +
λ0 · · ·λi−1ū
(j + 1− i)
λ0 · · ·λj−1
(1− γβj)ū
λ0 · · ·λi−1ū
λ0 · · ·λj−1ū
+ Stot
λ0 · · ·λi−1ū
(j + 1− i)
λ0 · · ·λj−1
(1− γβj)ū
λ0 · · ·λi−1ū
λ0 · · ·λn−1ū
λ0 · · ·λj−1ū
1 + Stot(j + 1− i)
1 − γβj
where the product λ0 · · ·λ−1 is defined to be 1 for the convenience of notation.
Because of (21),
(j + 1− i)
1− γβj
so we have ∂F̃
κ,γ,Stot
(0, ū) > 0.
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http://arxiv.org/abs/math/0701575
	Introduction
	Model assumptions
	Mathematical formalism
	Number of positive steady states
	Lower bound on the number of positive steady states
	Upper bound on the number of steady states
	Fine-tuned upper bounds
	Conclusions and discussions
	Acknowledgment
	Appendix
