A computational approach to persistence, permanence, and endotacticity of biochemical reaction systems

Abstract

We introduce a mixed-integer linear programming (MILP) framework capable of determining whether a chemical reaction network possesses the property of being endotactic or strongly endotactic. The network property of being strongly endotactic is known to lead to persistence and permanence of chemical species under genetic kinetic assumptions, while the same result is conjectured but as yet unproved for general endotactic networks. The algorithms we present are the first capable of verifying endotacticity of chemical reaction networks for systems with greater than two constituent species. We implement the algorithms in the open-source online package CoNtRol and apply them to a large sample of networks from the European Bioinformatics Institute’s BioModels Database. We use strong endotacticity to establish for the first time the permanence of a well-studied circadian clock mechanism.

Appendices

Appendix A (Proof of Lemma 1)

We will consider the endotactic and strongly endotactic cases separately. The lower endotactic case follows with a trivial modification to the endotactic case.

(Not endotactic $\Longrightarrow$) Suppose that the network is not endotactic. We construct the sets ${{\mathcal {R}}}_-$, ${{\mathcal {R}}}_0$, and ${{\mathcal {R}}}_+$ in the following way: (1) ${{\mathcal {R}}}_-$ consists of those $R_i$ for which ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)}) < 0$ and ${{\mathbf {w}}} \cdot (y_{\rho (i)} - y_{\rho (j)}) \le 0$ for all other $R_j \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) < 0$; (2) ${{\mathcal {R}}}_0$ consists of those $R_i$ for which ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)}) = 0$; and (3) ${{\mathcal {R}}}_+$ consists of everything else. We have that E1 and E2 are satisfied by construction and that E4 is satisfied by the assumption that the network is not endotactic. To show that E3 is satisfied, we suppose that there is a pair $R_i \in {{\mathcal {R}}}_-$ and $R_j \in {{\mathcal {R}}}_+$ such that ${{\mathbf {w}}} \cdot (y_{\rho (i)} - y_{\rho (j)}) > 0$. By the construction of ${{\mathcal {R}}}_-$, we have ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)}) < 0$, and by the construction of ${{\mathcal {R}}}_-$ and ${{\mathcal {R}}}_0$, it follows that ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) > 0$. It follows by Definition 4 that the network is endotactic, which contradicts our assumption. It follows that E3 is satisfied.

(Not endotactic $\Longleftarrow$) Suppose there is a ${{\mathbf {w}}} \in {{\mathbb {R}}}^m$ which satisfies E1 - E4. It follows from E2 and E4 that there is a $R_i \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)} ) < 0$. Furthermore, from E1 and E3 we have that for every $R_j \in {{\mathcal {R}}}$, either ${{\mathbf {w}}} \cdot (y_{\rho (i)} - y_{\rho (j)}) \le 0$ (if $R_j \in {{\mathcal {R}}}_+$) or ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) = 0$ (if $R_j \in {{\mathcal {R}}}_0 \cup {{\mathcal {R}}}_-$). It follows that the network is not endotactic by the contrapositive of Definition 4. The cases for lower endotactic networks follows similarly.

(Not strongly endotactic $\Longrightarrow$) Suppose that the network is not strongly endotactic. We construct the sets ${{\mathcal {R}}}_-$, ${{\mathcal {R}}}_0$, and ${{\mathcal {R}}}_+$ in the following way: (1) ${{\mathcal {R}}}_-$ consists of those $R_i$ for which ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)}) < 0$ and ${{\mathbf {w}}} \cdot (y_{\rho (i)} - y_{\rho (j)}) \le 0$ for all other $R_j \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) < 0$; (2) ${{\mathcal {R}}}_0$ consists of those $R_j$ for which ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) = 0$ and ${{\mathbf {w}}} \cdot (y_{\rho (j)} - y_{\rho (i)}) < 0$ for all $R_i \in {{\mathcal {R}}}_-$; and (3) ${{\mathcal {R}}}_+$ consists of everything else. The condition E1, E2, and E4 are satisfied by the constructions of ${{\mathcal {R}}}_-$ and ${{\mathcal {R}}}_0$.

Now suppose that ${{\mathcal {R}}}_0 \not = \emptyset$ and there is a pair $R_i \in {{\mathcal {R}}}_0$ and $R_j \in {{\mathcal {R}}}_- \cup {{\mathcal {R}}}_+$ such that ${{\mathbf {w}}} \cdot (y_{\rho (i)} - y_{\rho (j)}) \ge 0$. $R_j \in {{\mathcal {R}}}_-$ violates the construction of ${{\mathcal {R}}}_0$ so that we must have $R_j \in {{\mathcal {R}}}_+$. By the construction of ${{\mathcal {R}}}_-$ and ${{\mathcal {R}}}_0$ we must have ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) > 0$ and ${{\mathbf {w}}} \cdot (y_{\rho (j)}-y_{\rho (i)}) < 0$ so that the network is endotactic by Definition 4. Now suppose that ${{\mathcal {R}}}_0 = \emptyset$ and there is a pair $R_i \in {{\mathcal {R}}}_-$ and $R_j \in {{\mathcal {R}}}_+$ such that ${{\mathbf {w}}} \cdot (y_{\rho (i)} - y_{\rho (j)}) > 0$. By the construction of ${{\mathcal {R}}}_-$ and ${{\mathcal {R}}}_0$ (which is empty), it follows that ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) > 0$ so that the network is endotactic by Definition 4. Since these results contradict our assumption, it follows that one of SE3(a) or SE(b) is satisfied, and we are done.

(Not strongly endotactic $\Longleftarrow$) Suppose there is a ${{\mathbf {w}}} \in {{\mathbb {R}}}^m$ which satisfies E1, E2, SE3(a), and E4. It follows from E2 and E4 that there is a $R_i \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)} ) < 0$. It follows from E1 and SE3(a) that there is an $R_i \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)}) = 0$ and ${{\mathbf {w}}} \cdot (y_{\rho (i)} - y_{\rho (j)}) < 0$ for every $R_j \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)}) < 0$. It follows by Definition 4 that the network is not strongly endotactic.

Now suppose there is a ${{\mathbf {w}}} \in {{\mathbb {R}}}^m$ which satisfies E1, E2, SE3(b), and E4. It follows from E2 and E4 that there is a $R_i \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(i)} - y_{\rho (i)} ) < 0$. It follows from E1 and SE3(b) that ${{\mathbf {w}}} \cdot (y_{\rho (j)} - y_{\rho (i)} \le 0$ for every $R_j \in {{\mathcal {R}}}$ such that ${{\mathbf {w}}} \cdot (y_{\rho '(j)} - y_{\rho (j)}) > 0$ so that the network is not endotactic by Definition 4 and therefore not strongly endotactic. This completes the cases, so we are done.

Appendix B (Proof of Lemma 2)

We first establish the general form of the standard quasi-steady state approximation (QSSA) of (9) and (10). Under the QSSA, we assume that the intermediate compounds (i.e. the species $S_iE$ and $S_iF$) equilibriate on a sufficiently short time-scale that their concentrations may be assumed to be constant. To accommodate this assumption, we set their derivatives equal to zero, so that we have

$$\begin{aligned} \dot{[S_iE]}&= 0, \quad i=0, \ldots , n-1 \nonumber \\ \dot{[S_iF]}&= 0, \quad i=1, \ldots , n. \end{aligned}$$

(19)

We may use (19) in conjunction with the conservation laws

$$\begin{aligned} E_T = [E] + \sum _{i=0}^{n-1} [ES_i], \quad \text{ and } \quad F_T = [F] + \sum _{i=1}^n [FS_i] \end{aligned}$$

(20)

to determine algebraic expressions for $[S_iE]$ and $[S_iF]$. The rate of each phosphorylation or dephosphorylation step in (9) or (10) may then be simplified to a single step which depends upon the equilibriated value of the corresponding step’s intermediate compound.

The form of the QSSA rates for (9) and (10) are well-known for the cases $n=0$ (one-step chain) and $n=1$ (two-step chain). To the best of the authors’ knowledge, the derivation contained here is the first to determine the QSSA rate form for the general $n$-site chains (9) and (10). We have the following results which we prove by direct substitution. Note that we present the QSSA form for the forward chains in (9) and (10) only. The form for the backward chains follow directly.

Lemma 5

The forward chain of the processive phosphorylation network (9) has the QSSA solution

$$\begin{aligned}{}[ES_{i}] = \displaystyle {\frac{\left( \prod _{j=0}^iK_j \right) E_T [S_0]}{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j\right) [S_0]}}, \quad i=0, \ldots , n-1 \end{aligned}$$

(21)

where $K_i = k_{2i+1}/(k_{2i+2}+k_{2i+3})$, $i=0, \ldots , n-1$.

Proof

We need to solve the system of equations

$$\begin{aligned} \dot{[S_0E]}= & {} k_1[S_0][E] - (k_2+k_3)[ES_0] = 0 \end{aligned}$$

(22)

$$\begin{aligned} \dot{[S_iE]}= & {} k_{2i+1}[ES_{i-1}] - (k_{2i+2}+k_{2i+3})[ES_i] = 0, \quad i=1, \ldots , n-1 \end{aligned}$$

(23)

$$\begin{aligned} E_T= & {} [E] + \sum _{i=0}^{n-1} [ES_i]. \end{aligned}$$

(24)

We substitute (24) into (22) and (23) and substitute $K_i = k_{2i+1}/(k_{2i+2}+k_{2i+3})$ for $i=0, \ldots , n-1$, to get the new system of equations

$$\begin{aligned}&K_0[S_0]\left( E_T - \sum _{i=0}^{n-1} [ES_i] \right) - [ES_0] = 0 \end{aligned}$$

(25)

$$\begin{aligned}&K_{i}[ES_{i-1}] - [ES_i] = 0, \quad i=1, \ldots , n-1 \end{aligned}$$

(26)

Substituting (21) into the left-hand size of (25) gives

$$\begin{aligned}&K_0[S_0] \left( E_T \!-\! \sum _{i=0}^{n-1} \left( \frac{\left( \prod _{j=0}^{i} K_{j} \right) E_T[S_0]}{1 \!+\! \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j \right) [S_0]}\right) \right) - \frac{K_0 E_T [S_0]}{1 \!+\! \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j \right) [S_0]}\\&\quad = K_0 E_T [S_0] \left( \frac{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j \right) [S_0] - \left( \sum _{i=0}^{n-1} \prod _{j=0}^{i} K_{j+1} \right) [S_0]}{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j \right) [S_0]}\right) \\&\qquad -\frac{K_0 E_T [S_0]}{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j \right) [S_0]}\\&\quad = \frac{K_0 E_T [S_0]}{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j \right) [S_0]} - \frac{K_0 E_T [S_0]}{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^k K_j \right) [S_0]} = 0. \end{aligned}$$

Substituting (21) into the left-hand side of (26) gives

$$\begin{aligned} K_{i} \displaystyle {\frac{\left( \prod _{j=0}^{i-1}K_j \right) E_T [S_0]}{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_j\right) [S_0]}} - \displaystyle {\frac{\left( \prod _{j=0}^{i}K_j \right) E_T [S_0]}{1 + \left( \sum _{k=0}^{n-1} \prod _{j=0}^{k} K_{j}\right) [S_0]}} = 0 \end{aligned}$$

and we are done. $\square$

Lemma 6

The forward chain of the distributive phosphorylation network (10) has the QSSA solution

$$\begin{aligned}{}[ES_{i}] = \displaystyle {\frac{K_{i}E_T[S_{i}]}{1 + \sum _{j=0}^{n-1} K_j[S_{j}]}}, \quad i=0,\ldots , n-1 \end{aligned}$$

(27)

where $K_i = k_{3i+1}/(k_{3i+2}+k_{3i+3})$ for $i=0, \ldots , n-1$.

Proof

We need to solve the system of equations

$$\begin{aligned} \dot{[S_iE]}= & {} k_{3i+1} [S_i][E] - (k_{3i+2} + k_{3i+3})[ES_i] = 0, \quad i=0, \ldots , n-1 \end{aligned}$$

(28)

$$\begin{aligned} E_T= & {} [E] + \sum _{i=0}^{n-1} [ES_i]. \end{aligned}$$

(29)

We substitute (29) into (28) and use the substitution $K_i = k_{3i+1}/(k_{3i+2}+k_{3i+3})$ to obtain the new system of equations

$$\begin{aligned} K_{i} [S_i] \left( E_T - \sum _{i=0}^{n-1} [ES_i]\right) - [ES_i] = 0, \quad i=0, \ldots , n-1. \end{aligned}$$

(30)

Substituting (27) into the left-hand side of (30) gives

$$\begin{aligned}&K_{i} [S_i] \left( E_T - \sum _{i=0}^{n-1} \frac{K_{i}E_T[S_{i}]}{1 + \sum _{j=0}^{n-1} K_j[S_{j}]}\right) - \frac{K_{i}E_T[S_{i}]}{1 + \sum _{j=0}^{n-1} K_j[S_{j}]} \\&\quad = K_i E_T [S_i] \left( \frac{1 + \sum _{j=0}^{n-1} K_j[S_{j}] - \sum _{i=0}^{n-1} K_i[S_{i}]}{1 + \sum _{j=0}^{n-1} K_j[S_{j}]} \right) - \frac{K_{i}E_T[S_{i}]}{1 + \sum _{j=0}^{n-1} K_j[S_{j}]} \\&\quad = \frac{K_{i}E_T[S_{i}]}{1 + \sum _{j=0}^{n-1} K_j[S_{j}]} - \frac{K_{i}E_T[S_{i}]}{1 + \sum _{j=0}^{n-1} K_j[S_{j}]} = 0 \end{aligned}$$

and we are done. $\square$

We now prove Lemma 2.

Proof

The form (14) follows by applying Lemma 5 independently to the phosphorylation and dephosphorylation chains in (9), then rearranging the constants in the denominator. The form (16) follows similarly by applying Lemma 6 to the chains in (10), and we are done. $\square$

Appendix C (Proof of Lemma 4)

The following is the full model for the PER (period) and TIM (timeless) protein system in Drosophila presented by Leloup and Goldbeter in Leloup and Goldbeter (1999):

$$\begin{aligned} \frac{dM_P}{dt}= & {} \nu _{sP} \frac{K_{IP}^n}{K_{IP}^n + C_N^n} - \nu _{mP} \frac{M_P}{K_{mP}+M_P} - k_d M_P \end{aligned}$$

(31)

$$\begin{aligned} \frac{dP_0}{dt}= & {} k_{sP} M_P - V_{1P} \frac{P_0}{K_{1P}+P_0} + V_{2P} \frac{P_1}{K_{2P}+P_1}-k_d P_0 \end{aligned}$$

(32)

$$\begin{aligned} \frac{dP_1}{dt}= & {} V_{1P} \frac{P_0}{K_{1P}+P_0} - V_{2P} \frac{P_1}{K_{2P}\!+\!P_1} - V_{3P} \frac{P_1}{K_{3P}\!+\!P_1} \!+\! V_{4P} \frac{P_2}{K_{4P}\!+\!P_2} - k_d P_1 \end{aligned}$$

(33)

$$\begin{aligned} \frac{dP_2}{dt}= & {} V_{3P} \frac{P_1}{K_{3P}\!+\!P_1} - V_{4P} \frac{P_2}{K_{4P}\!+\!P_2} \!-\!k_3P_2T_2 + k_4C - \nu _{dP} \frac{P_2}{K_{dP}+P_2}-k_d P_2 \end{aligned}$$

(34)

$$\begin{aligned} \frac{dM_T}{dt}= & {} \nu _{sT} \frac{K_{IT}^n}{K_{IT}^n + C_N^n} - \nu _{mT} \frac{M_T}{K_{mT}+M_T} - k_d M_T \end{aligned}$$

(35)

$$\begin{aligned} \frac{dT_0}{dt}= & {} k_{sT}M_T - V_{1T} \frac{T_0}{K_{1T}+T_0} + V_{2T} \frac{T_1}{K_{2T} + T_1}-k_dT_0 \end{aligned}$$

(36)

$$\begin{aligned} \frac{dT_1}{dt}= & {} V_{1T} \frac{T_0}{K_{1T}+T_0} - V_{2T} \frac{T_1}{K_{2T} + T_1} - V_{3T} \frac{T_1}{K_{3T} + T_1} + V_{4T} \frac{T_2}{K_{4T} + T_2} - k_dT_1 \end{aligned}$$

(37)

$$\begin{aligned} \frac{dT_2}{dt}= & {} V_{3T} \frac{T_1}{K_{3T} + T_1} - V_{4T} \frac{T_2}{K_{4T}+T_2} - k_3P_2T_2 + k_4 C - \nu _{dT} \frac{T_2}{k_{dT}+T_2} - k_d T_2 \end{aligned}$$

(38)

$$\begin{aligned} \frac{dC}{dt}= & {} k_3 P_2 T_2 -k_4C - k_1C + k_2C_N -k_{dC}C \end{aligned}$$

(39)

$$\begin{aligned} \frac{dC_n}{dt}= & {} k_1C -k_2 C_N -k_{dN}C_N. \end{aligned}$$

(40)

We now prove the following result, which proves Lemma 4 as a direct corollary.

Lemma 7

The system of equations (31)–(40) can be represented by the following network with $\kappa$-variable mass action kinetics:

where each species also undergoes linear degradation (i.e. $P_i \rightarrow 0$, $T_i \rightarrow 0$, $C_i \rightarrow 0$).

Proof

We split the proof into the following steps:

• 1. We prove that all trajectories are eventually bounded by a common bound. Our method is to construct a linear functional in all of the species which decreases along trajectories. This allows us to bound the Michaelis-Menten terms in (32)–(34) and (36)–(38).

• 2. We bound concentrations of $M_P$ and $M_T$ from below in the limit as $t \rightarrow \infty$ so that, after a finite time, we may incorporate the concentrations of $M_P$ and $M_T$ into the rate constants for the creation of $P_0$ and $T_0$. The bounds guarantee that this can be done with a $\kappa$-variable reaction rate.

• 3. We show that this resulting system corresponds to a $\kappa$-variable mass-action system with underlying network structure given by the reduced clock network.

1. Trajectories are bounded We define the linear functional

$$\begin{aligned}&H(M_P,P_0,P_1,P_2,M_T,T_0,T_1,T_2,C,C_N) \nonumber \\&\quad = (k_{sP}+1) M_p + (k_{sT}+1) M_T + k_d (P_0 + P_1 + P_2\nonumber \\&\qquad + \,T_0 + T_1 + T_2) + 2k_d (C + C_N). \end{aligned}$$

(41)

and introduce the bound values $M_P^* = M_T^* = k^*/k_d$, $P_0^* = P_1^* = P_2^* = T_0^* = T_1^* = T_2^* = k^* / k_d^2$, $C^* = k^* / (2k_d k_{dC}),$ and $C_N^* = k^* / (2k_d k_{dN})$ where $k^* = (k_{sP} + 1) \nu _{sP} + (k_{sT} + 1) \nu _{sT}$. We now show that, along trajectories of (31)–(40), there is a finite time $\tau > 0$ such that $H(t) \le H^*$ for $t \ge \tau$ where

$$\begin{aligned} H^* = (k_{sP}+1) M_p^* + (k_{sT}+1) M_T^* + k_d (P_0^* + P_1^* + P_2^* + T_0^* + T_1^* + T_2^*) + 2k_d (C^* + C_N^*). \end{aligned}$$

(42)

Consider the time derivative of (41) along trajectories of (31)–(40). After simplification, we have

$$\begin{aligned} \frac{dH(t)}{dt}= & {} (k_{sP} + 1) \nu _{sP} \frac{K_{IP}^n}{K_{IP}^n + C_N^n} + (k_{sT}+1) \nu _{sT} \frac{K_{IT}^n}{K_{IT}^n + C_N^n}\nonumber \\&- (k_{sP}+1)\nu _{mP} \frac{M_P}{K_{mP}+M_P}-(k_{sT}+1) \nu _{mT} \frac{M_T}{K_{mT} + M_T}\nonumber \\&- k_d \nu _{dP} \frac{P_2}{K_{dP}+P_2} - k_d \nu _{dT} \frac{T_2}{K_{dT}+T_2} \nonumber \\&-k_d (M_P + M_T) - k_d^2 (P_0 + P_1 + P_2 + T_0 + T_1 + T_2)\nonumber \\&- 2k_d ( k_{dC}C + k_{dN}C_N). \end{aligned}$$

(43)

Now suppose that $H(t) \ge H^*$. It follows by construction that at least one of the reactants obtains or exceeds its corresponding bound value so that the negative values in (43) exceed $k^*$ in magnitude. Noting that

$$\begin{aligned}&(k_{sP} + 1) \nu _{sP} \frac{K_{IP}^n}{K_{IP}^n + C_N^n} + (k_{sT}+1) \nu _{sT} \frac{K_{IT}^n}{K_{IT}^n + C_N^n}\\&\quad \le (k_{sP} + 1) \nu _{sP} +(k_{sT} + 1) \nu _{sT} = k^* \end{aligned}$$

we have that (43) is nonpositive whenever $H(t) \ge H^*$. The strict negativity of the remaining negative terms (43) ensures that (43) is bound strictly, so that

$$\begin{aligned} \frac{dH}{dt} < -\epsilon \end{aligned}$$

(44)

for some $0 < \epsilon$. Integrating (44) shows that there is a $\tau > 0$ such that $H(t) \le H^*$ for all $t \ge \tau$. Since all of the coefficients of the reactants in $H(t)$ are positive, it follows that every reactant is eventually bounded by the common bound $H^*$.

2. $M_P$ and $M_T$ are bounded away from zero We first define

$$\begin{aligned} M_P^*= & {} \frac{\nu _{sP} K_{IP}^n}{K_{IP}^n + (C_N^*)^n} \frac{k_{mP}}{\nu _{mP} + k_d k_{mP}}\nonumber \\ M_T^*= & {} \frac{\nu _{sT} K_{IT}^n}{K_{IT}^n + (C_N^*)^n} \frac{k_{mT}}{\nu _{mT} + k_d k_{mT}}. \end{aligned}$$

(45)

Now consider the equation (31) for the concentration of $M_P(t)$. Since all trajectories of the system (31)–(40) converge to a bounded set, we have that there is a $C_N^* > 0$ and a $\tau \ge 0$ such that $C_N(t) \le C_N^*$ for all $t \ge \tau$. It follows from this and that fact that $M_P(t) \ge 0$ that, for $t \ge \tau$, we have

$$\begin{aligned} \frac{dM_P(t)}{dt}= & {} \nu _{sP} \frac{K_{IP}^n}{K_{IP}^n + C_N^n} - \nu _{mP} \frac{M_P}{K_{mP}+M_P} - k_d M_P\nonumber \\\ge & {} \nu _{sP} \frac{K_{IP}^n}{K_{IP}^n + (C_N^*)^n} - \left( \frac{\nu _{mP}}{K_{mP}} + k_d \right) M_P. \end{aligned}$$

(46)

It follows from (46) that

$$\begin{aligned} M_P(t) \ge M_P^* + ( M_P(0) - M_P^*) e^{- \frac{\nu _{mP} + k_d K_{mP}}{k_{mP}} t}. \end{aligned}$$

It follows that $\displaystyle \liminf _{t \rightarrow \infty } MP(t) \ge M_P^*$. A similar argument shows that $\displaystyle \liminf _{t \rightarrow \infty } MT(t) \ge M_T^*$, and we are done.

3.Construction of network We consider equations (32)–(34) and (36)–(40). We have shown that all variables are bounded, so that Lemma 6 implies any Michaelis-Menten form in these variables may be written in $\kappa$-variable form. We also have shown that there is are $M_P^{*} \ge 0, M_P^{**} \ge 0, M_T^{*} \ge 0,$ $M_T^{**} \ge 0,$ and $\tau \ge 0$ such that

$$\begin{aligned} 0 < M_P^{*} \le M_P(t) \le M_P^{**} \quad \text{ and } \quad 0 < M_T^{*} \le M_T(t) \le M_T^{**} \end{aligned}$$

for $t \ge \tau$. We may therefore incorporate these variables into a $\kappa$-variable influx rate for $P_0$ and $T_0$ respectively. The remaining terms correspond to linear degradation terms which are already in $k$-variable form.

This yields a network which is $\kappa$-variable form. For example, we may write (32) as

$$\begin{aligned} \frac{dP_0}{dt} = \kappa _{sP}(t) - (\kappa _{1P}(t) + \kappa _d(t)) P_0 + \kappa _{2P}(t) P_1 \end{aligned}$$

where

$$\begin{aligned}&\kappa _{sP}(t) \in [ M_P^*, M_P^{**} ], \kappa _{1P}(t) \in \left[ \frac{V_{1P}}{K_{1P}+P_0^*}, \frac{V_{1P}}{K_{1P}} \right] ,\\&\kappa _{2P} \in \left[ \frac{V_{2P}}{K_{2P}+P_1^*}, \frac{V_{2P}}{K_{2P}} \right] , \text{ and } \quad \kappa _d(t) = k_d. \end{aligned}$$

Since the resulting network has the form of the reduced clock network, we are done.