Computing Weakly Reversible Deficiency Zero Network Translations Using Elementary Flux Modes

Abstract

We present a computational method for performing structural translation, which has been studied recently in the context of analyzing the steady states and dynamical behavior of mass-action systems derived from biochemical reaction networks. Our procedure involves solving a binary linear programming problem where the decision variables correspond to interactions between the reactions of the original network. We call the resulting network a reaction-to-reaction graph and formalize how such a construction relates to the original reaction network and the structural translation. We demonstrate the efficacy and efficiency of the algorithm by running it on 508 networks from the European Bioinformatics Institutes’ BioModels database. We also summarize how this work can be incorporated into recently proposed algorithms for establishing mono- and multistationarity in biochemical reaction systems.

A: Appendix—Parametrization Method

While two structural translations have the same stoichiometric matrices $\varGamma$ and $\varGamma '$, they may nevertheless have different mass-action systems (1) due to differences in $R({\mathbf {x}})$. In this Appendix, we outline the method by which a steady-state parametrization may be constructed from a structural parametrization as constructed by the algorithm presented in Sect. 3.3.

For ease of notation and continuity, rather than repeating the technical definitions and Theorems introduced by Müller and Regensberger (2012, 2014) and Johnston et al. (2018), we outline the parametrization procedure through examples.

1.1 A.1: Histidine Kinase Model

We use the histidine kinase network in Fig. 1a as a motivating example. Through application of the algorithm presented in Sect. 3.3, we were able to correspond the following CRN (left) with the indicated structural translation (right):

Although these two networks have the same reaction vectors (i.e., $\varGamma = \varGamma '$), the governing system of differential equations (1) does not coincide due to differences in $R({\mathbf {x}})$. Specifically, the source complex of $r_1$ and $r_4$ differs in Network 1 from Network 2. To accommodate this difference, we map the source complexes from Network 1 into a secondary set of complexes known as kinetic-order complexes in Network 2. We can represent this with the following network:

Network () is an example of a generalized chemical reaction network (GCRN) (Müller and Regensburger 2012, 2014). In a GCRN, each vertex is assigned two complexes: a stoichiometric complex (unbracketed) and a kinetic-order complex (bracketed). In the corresponding generalized mass-action system

$$\begin{aligned} \frac{d{\mathbf {x}}}{dt} = \varGamma \tilde{R}({\mathbf {x}}) \end{aligned}$$

(17)

the reaction vectors forming $\varGamma$ are determined by the differences of the stoichiometric complexes, while the monomials in $\tilde{R}({\mathbf {x}})$ are determined by the kinetic-order complexes. Denote the ith kinetic-order complex by $\tilde{y}_i$, we have that $\tilde{R}({\mathbf {x}})$ has entries $\tilde{R}_i({\mathbf {x}}) = \prod _{j=1}^m x_j^{[\tilde{y}_{s(i)}]_j}$. For example, the term in $\tilde{R}({\mathbf {x}})$ corresponding to $r_1$ in () is $k_1 x$ rather than $k_1 x y$. It can be easily checked that dynamical equations (17) corresponding to () coincide with dynamical equations (1) corresponding to Network 1.

Note that when converting from Network 2 to (), we split the vertex $X+Y_p$. Consequently, the stoichiometric complex $X+Y_p$ is repeated at vertices 3 and 4 in () (indicted with $\dag$). This is allowed by Johnston et al. (2018) and, in fact, required since $r_3$ and $r_4$ would otherwise have multiple kinetic-order complexes at a single vertex [although, with some supplemental conditions, this is allowed by Johnston (2015) and Tonello and Johnston (2018)]. In order to regain weak reversibility, Johnston et al. (2018) introduce a new set of edges (called “phantom reactions") which connect stoichiometrically identical complexes. Notice that introducing such reactions introduces zero columns in $\varGamma$ and therefore does not alter the corresponding system of differential equations (17).

For technical reasons, Johnston et al. (2018) imposed further rules upon the splitting of stoichiometric complexes and the introduction of phantom reactions. They define equivalence classes of stoichiometrically identical complexes and select from within each such class a distinguished vertex (indicated with a $\star$). The set of phantom reactions is then introduced such that:

• 1. All “true reactions” (i.e., from the set $\mathcal {R}$) which have their product at any vertex in this equivalence class have the distinguished vertex as its product.

• 2. The phantom reactions between vertices on this equivalence class consist only of reactions with the distinguished complex as its source and the remaining complexes as the product.

We may interpret the distinguished vertices as hubs through which are all paths through an equivalence class of stoichiometrically identical complexes must pass. Such a construction produces a $V^{\star }$-directed GCRN which is important in the construction of positive parametrizations.

For network (), we select vertex 3 as the distinguish vertex (indicated with $\star$) and label the phantom edge with a free parameter $\sigma$:

Notice that only $r_2$ has a product in the equivalence class of vertices $\{3, 4\}$ and its product is the distinguished complex 3 (condition 1), and the only reaction on vertices $\{3, 4 \}$ goes from the distinguished vertex 3 to the remaining vertex 4 (condition 2). GCRN () is therefore $V^{\star }$-directed. Notice that the rate constant $\sigma$ corresponding to the phantom edge $3 \rightarrow 4$ joins stoichiometrically identical complexes and consequently does not appear in the system of differential equations (1).

Johnston et al. (2018) showed that, if the deficiency of the structural translation (called the effective deficiency) is zero and the corresponding GCRN is $V^{\star }$-directed, then the positive steady-state set of original dynamical system (1) can be characterized by the complex-balanced steady states of dynamical system (17), namely, the equation

$$\begin{aligned} A_k \tilde{R}({\mathbf {x}}) = 0 \end{aligned}$$

(19)

where $A_k \in {\mathbb {R}}^{n \times n}$ is the Laplacian of the reaction graph of the GCRN. For network (), this corresponds to the system:

$$\begin{aligned} \left[ \begin{array}{cccc} -k_1 &{} 0 &{} 0 &{} k_4 \\ k_1 &{} - k_2 &{} k_3 &{} 0 \\ 0 &{} k_2 &{} -k_3 - \sigma &{} 0 \\ 0 &{} 0 &{} \sigma &{} - k_4 \end{array} \right] \left[ \begin{array}{c} x \\ x_py \\ xy_p \\ y_p \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array} \right] \end{aligned}$$

Relationships between $\text{ ker }(A_k)$ and the steady-state set of mass-action systems have been studied extensively in recent years. It is known that, for weakly reversible networks, $\text{ ker }(A_k)$ can be characterized by algebraic combinations of the rate constants of a network known as “tree constants” (Johnston 2014; Johnston et al. 2018) which we summarize in “Appendix A.2.” For this network, we can directly compute that $\text{ ker }(A_k) = \text{ span } \{( K_1, K_2, K_3, K_4 )\}$ where

$$\begin{aligned} \begin{aligned} K_1&= k_2k_4\sigma ,&K_2&= k_1(k_3 + \sigma )k_4,&K_3&= k_1k_2k_4,&K_4&= k_1k_2 \sigma \end{aligned} \end{aligned}$$

are the tree constants. The steady-state condition $(x, x_py,xy_p,y_p) \in \text{ ker }(A_k)$ gives the implicit equations

$$\begin{aligned} \frac{x}{k_2k_4\sigma } = \frac{x_py}{k_1(k_3+\sigma )k_4} = \frac{xy_p}{k_1k_2k_4} = \frac{y_p}{k_1k_2\sigma }. \end{aligned}$$

Taking pairwise differences, this gives the following log-linear system of equations:

$$\begin{aligned} \left[ \begin{array}{cccc} -\,1 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 \\ -\,1 &{} 0 &{} 0 &{} 1 \end{array} \right] \left[ \begin{array}{c} \ln (x) \\ \ln (x_p) \\ \ln (y) \\ \ln (y_p) \end{array} \right] = \left[ \begin{array}{c} \frac{k_1(k_3 + \sigma )}{k_2\sigma } \\ \frac{k_1}{\sigma } \\ \frac{k_1}{k_4} \end{array} \right] \end{aligned}$$

(20)

Surprisingly, the solvability of system (20) depends on the deficiency of network () taken with only the kinetic-order complexes:

The deficiency of network () is known as the kinetic-order deficiency (Müller and Regensburger 2012, 2014). We can compute that the deficiency of () is zero so that the kinetic-order deficiency of () is zero. Consequently, log-linear system (20) is guaranteed to be consistent and therefore have a solution for all values of rate constants (including $\sigma$) Johnston et al. (2018). For this example, (20) can be solved for the log concentrations, which can then be exponentiated to give the following parametrization:

$$\begin{aligned}\left\{ \begin{aligned} x&= \frac{k_4}{\sigma } ,&y&= \tau , \\ x_p&= \frac{k_1(k_3+\sigma )k_4}{k_2\sigma ^2\tau }, \; \; \; \; \;&y_p&= \frac{k_1}{\sigma }, \end{aligned} \right. \end{aligned}$$

where $\sigma , \tau > 0$ are positive parameters. Notice that the parameter $\tau$ has arisen from parametrizing the null space of the coefficient matrix in (20), which is the span of the vector $(0,-\,1,1,0)$.

1.2 A.2: General Procedure for Parametrizations

For a given GCRN, we let $\tilde{y}_i$ denote the kinetic-order complex at the vertex labeled i and define ${\mathcal {T}} \subseteq \mathcal {R}$ to be the set of all trees which span the linkage class containing the vertex i and have vertex i as a unique sink. The tree constants $K_i$ corresponding to the vertex labeled i are given by

$$\begin{aligned} K_i = \sum _{T \in {\mathcal {T}}} \prod _{r_j \in T} k_j. \end{aligned}$$

(22)

By Lemma 12 of Johnston et al. (2018), if the GCRN has an effective deficiency of zero, we have the following representations of the steady-state set of the corresponding generalized mass-action system:

$$\begin{aligned} \frac{{\mathbf {x}}^{\tilde{y}_i}}{K_i} = \frac{{\mathbf {x}}^{\tilde{y}_j}}{K_j} \; \; \; \Longleftrightarrow \; \; \; (\tilde{y}_j - \tilde{y}_i)^{\mathrm{T}} \ln ({\mathbf {x}}) = \ln \left( \frac{K_j}{K_i} \right) \end{aligned}$$

(23)

for all vertices i and j belonging to the same linkage class. We can use the log-linear equation on the right of (23) to construct a linear system in the log concentrations. We define a matrix M such columns $M_{\cdot , k} = \tilde{y}_j - \tilde{y_i}$ and a vector b with entries $b_k = \ln (K_j/K_i)$ where the pairs (i, j) are chosen to be a maximal set of colorredvertices such that the resulting set spans the vertices of the underlying GCRN and does not have any nontrivial cycles. This process produces the following log-linear system

$$\begin{aligned} M^{\mathrm{T}} \ln ({\mathbf {x}}) = b. \end{aligned}$$

(24)

An effective deficiency of zero guarantees all steady states can be found by solving (24) (Lemma 12, Johnston et al. 2018). A kinetic-order deficiency of zero guarantees the solvability of this system for all values of the rate constants (Theorem 14 part 1, Johnston et al. 2018). A GCRN with a nonzero kinetic-order deficiency, however, may still produce solvable system (24) provided certain supplemental conditions on the rate parameters are satisfied (Theorem 14 part 2, Johnston et al. 2018).

The example in “Appendix A.1” suggests the following general procedure for determining a positive steady-state parametrization for mass-action systems (1):

• Step 1: Construct a weakly reversible, deficiency zero structural translation by the algorithm presented in Sect. 3.3.

• Step 2: Transfer source complexes from the original CRN as kinetic-order complexes in the network GCRN, splitting stoichiometric complexes as necessary.

• Step 3: Within each equivalence class of stoichiometrically identical complexes, select distinguished vertices and phantom edges so that the resulting GCRN is $V^{\star }$-directed. Note that the choice of distinguished vertices may be made arbitrarily.

• Step 4: Compute the kinetic-order deficiency. (The deficiency of the network with only the kinetic-order complexes from the $V^{\star }$-directed network found in Step 3.) If the kinetic-order deficiency is zero, skip to Step 5; otherwise proceed to Step 4*.

• Step 4*: Determine a basis $\{ c_1, \ldots , c_{\tilde{\delta }} \}$ of $\text{ ker }(M)$ and for every vector $c_i$ attempt to solve the system $c_i^{\mathrm{T}} b = 0$ for the phantom edge parameters $\sigma _j$ according to Theorem 15 of Johnston et al. (2018). If these conditions cannot be satisfied, the procedure fails. Otherwise, substitute the solved parameters $\sigma _j$ into the GCRN constructed in Step 3 and proceed to Step 5.

• Step 5: Compute the “tree constants” at each vertex of this $V^{\star }$-directed GCRN.

• Step 6: Set up and solve log-linear system (24) for the concentrations.

1.3 A.3: ZigZag Model Example

Reconsider the zigzag model of plant–pathogen interactions (). We now outline how the steps described in “Appendix A.2” apply to this network.

Step 1: We were able to use the algorithm described in Sect. 3.3 to determine the following structural translation:

As expected by the algorithm, this network is weakly reversible and deficiency zero. It follows from Lemma 12 of Johnston et al. (2018) that all of the steady states can be found by setting up and solving log-linear system (24).

Steps 2 & 3: Notice that the complexes $X_3 + X_4$, $X_9 + X_{10} + X_{11}$, and $X_9 + X_{11}$ have multiple source complexes which are translated to them from (). We therefore split these vertices in () when assigning kinetic-order complexes. We also need to select distinguish complexes and add phantom edges to satisfy the conditions of being $V^\star$-directed given in “Appendix A.1.” This can be accomplished by the following network, where the phantom edges are labeled with $\sigma _i$, $i=1, \ldots , 4$, the equivalence classes of stoichiometrically identical complexes are indicated with the symbols $\S$, $\dag$, and $\ddag$, and the distinguished vertices are indicated with $\star$.

Step 4: The kinetic-order deficiency is the deficiency of the CRN produced by considering only the kinetic-order (bracketed) complexes in (). It can be quickly computed that the deficiency is $\delta = n - \ell - s = 17-4-13 = 0$. It follows from Theorem 14 of Johnston et al. (2018) that the remainder of the steps may be performed to yield a steady-state parametrization.

Step 5: From (), we compute the following tree constants:

$$\begin{aligned} \begin{aligned} K_1&= k_2k_5(k_4+\sigma _1)&K_{10}&= k_8k_{10}k_{11}k_{12}k_{14}k_{15}k_{16}(k_{18}+k_{19})\\ K_2&= k_1(k_4+\sigma _1)k_5&K_{11}&= k_8k_{11}k_{12}k_{14}k_{15}k_{16}(k_{17}k_{19}+(k_{18}+k_{19})\sigma _2)\\ K_3&= k_1k_3k_5&K_{12}&= k_{11}k_{12}(\sigma _3+\sigma _4+k_{13})k_{14}k_{15}k_{16}((\sigma _2+k_{17})k_{19}+k_{18}\sigma _2)\\ K_4&= k_1k_3\sigma _1&K_{13}&= k_8k_{10}k_{11}k_{12}k_{14}k_{15}k_{16}k_{17}\\ K_5&= k_7&K_{14}&= k_8k_{10}k_{11}k_{14}k_{15}k_{16}k_{17}k_{19}\\ K_6&= k_6&K_{15}&= k_8k_9k_{12}k_{14}k_{15}k_{16}(k_{17}k_{19}+(k_{18}+k_{19})\sigma _2)\\ K_7&= k_8k_{10}k_{11}k_{12}k_{15}k_{16}(k_{18}+k_{19})\sigma _2&K_{16}&= k_{21}\\ K_8&= k_8k_{11}k_{12}k_{14}k_{16}((\sigma _2+k_{17})k_{19}+k_{18}\sigma _2)\sigma _3&K_{17}&= k_{20}\\ K_9&= k_8k_{11}k_{12}k_{14}k_{15}((\sigma _2+k_{17})k_{19}+k_{18}\sigma _2)\sigma _4\\ \end{aligned} \end{aligned}$$

Step 6: Log-linear system (24) can be set up for any maximal set of pairs of vertices lying in the same linkage class. We take the pairs

$$\begin{aligned}&\{ 1, 2\}, \{1, 3\}, \{1, 4\}, \{5, 6\}, \{7, 8\}, \{7, 9\}, \{7, 10\}, \{7, 11\},\\&\quad \{7, 12\}, \{7, 13\}, \{7, 14\}, \{7, 15\}, \{16, 17\}. \end{aligned}$$

This gives the following linear system in log concentrations (24):

$$\begin{aligned} \left[ \begin{array}{ccccccccccccc}-\,1&{}-\,1&{}1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ -\,1&{}-\,1&{}0&{}1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ -\,1&{}-\,1&{}0&{}1&{}1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}-\,1&{}-\,1&{}1&{}0&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}1&{}0&{}1&{}-\,1&{}0&{}0&{}0\\ 0&{}0&{}0&{}1&{}0&{}0&{}0&{}0&{}1&{}-\,1&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}1&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}1&{}-\,1&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}1&{}0&{}-\,1&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}-\,1&{}0&{}1&{}0\\ 0&{}0&{}0&{}0&{}1&{}0&{}0&{}0&{}0&{}-\,1&{}0&{}0&{}0\\ 1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}-\,1&{}0&{}0&{}0\\ 0&{}0&{}0&{}-\,1&{}0&{}0&{}0&{}0&{}0&{}0&{}-\,1&{}0&{}1\end{array} \right] \left[ \begin{array}{c} \ln (x_1) \\ \ln (x_2) \\ \ln (x_3) \\ \ln (x_4) \\ \ln (x_5) \\ \ln (x_6) \\ \ln (x_7) \\ \ln (x_8) \\ \ln (x_9) \\ \ln (x_{10}) \\ \ln (x_{11}) \\ \ln (x_{12}) \\ \ln (x_{13}) \end{array} \right] = \left[ \begin{array}{c} \ln \left( K_2/K_1 \right) \\ \ln \left( K_3/K_1 \right) \\ \ln \left( K_4/K_1 \right) \\ \ln \left( K_6/K_5 \right) \\ \ln \left( K_8/K_7 \right) \\ \ln \left( K_9/K_7 \right) \\ \ln \left( K_{10}/K_7 \right) \\ \ln \left( K_{11}/K_7 \right) \\ \ln \left( K_{12}/K_7 \right) \\ \ln \left( K_{13}/K_7 \right) \\ \ln \left( K_{14}/K_7 \right) \\ \ln \left( K_{15}/K_7 \right) \\ \ln \left( K_{17}/K_{16} \right) \end{array} \right] \end{aligned}$$

Since the kinetic-order deficiency is zero, this is a consistent system and therefore guaranteed to have a solution for all rate constants (Theorem 14, Johnston et al. 2018). Solving the system for $\ln (x_i)$ and then exponentiating gives the following solution, which is a rational parametrization of the steady-state set of mass-action system (1) corresponding to () in the parameters $\sigma _1, \sigma _2, \sigma _3, \sigma _4 \in {\mathbb {R}}y_{> 0}$:

$$\begin{aligned} \begin{aligned} x_1&= \frac{ k_9 k_{12}(k_{17} k_{19}+(k_{18}+k_{19})\sigma _2)\sigma _1}{k_5k_8k_{10}k_{17}k_{19}}&x_8&= \frac{k_{12}(k_{13}+\sigma _3+\sigma _4)((k_{17}+\sigma _2)k_{19}+k_{18}\sigma _2) \sigma _1}{ k_5k_{10}k_{11} k_{17} k_{19}}\\ x_2&= \frac{k_2 (k_4+\sigma _1)k_5 k_{10} k_{11} k_{17}k_{19} \sigma _4 }{ k_1 k_3 k_9k_{12}k_{16} (k_{17} k_{19}+(k_{18} + k_{19}) \sigma _2) \sigma _1}&x_9&= \frac{ k_{12}(k_{17} k_{19}+(k_{18}+k_{19}) \sigma _2)\sigma _1 }{k_5 k_{10} k_{17} k_{19}}\\ x_3&= \frac{(k_4+\sigma _1)\sigma _4 }{ k_3k_{16}}&x_{10}&= \frac{k_{12} (k_{18}+k_{19})\sigma _1 \sigma _2 }{k_5 k_{14}k_{17} k_{19}}\\ x_4&= \frac{\sigma _4}{k_{16}}&x_{11}&= \frac{k_{14}}{\sigma _2}\\ x_5&= \frac{\sigma _1}{k_5}&x_{12}&= \frac{ k_{12}\sigma _1}{k_5 k_{19}}\\ x_6&= \frac{ k_5 k_7\sigma _3}{k_6k_{15} \sigma _1 }&x_{13}&= \frac{ k_{14} k_{20}\sigma _4}{k_{16} k_{21}\sigma _2}\\ x_7&= \frac{\sigma _3}{k_{15}}\\ \end{aligned} \end{aligned}$$

(27)

Notice that this parametrization does not guarantee that for a given initial condition ${\mathbf {x}}(0) \in {\mathbb {R}}_{\ge 0}^m$ the parametrization intersects the relevant compatibility class $({\mathbf {x}}(0) + S) \cap {\mathbb {R}}_{\ge 0}^m$. For this example, we can observe that $x_8$ experiences no stoichiometric change in any of the system’s interactions and therefore we have $x_8(t)=x_8(0)$ for all $t \ge 0$. This requirement combined with (27) imposes further conditions on the rate constants which must be satisfied for a positive steady state to exist.

1.4 A.4: MAPK Model Example

Reconsider MAPK model ().

Step 1: We were able to use the algorithm described in Sect. 3.3 to determine the following structural translation:

Steps 2 & 3: The complexes $X+K+M$ and $X_p + K +M$ are both assigned multiple kinetic complexes and therefore must be split. Setting 1 and 3 as the distinguished complexes and introducing phantom edges gives the following $V^{\star }$-directed GCRN:

where $\sigma _1$ and $\sigma _2$ indicate the phantom edges, $\dag$ and $\ddag$ indicate equivalence classes of stoichiometrically identical complexes, and $\star$ indicates the distinguished vertex within each class.

Step 4: We can compute that the kinetic-order deficiency is one. We therefore have one condition of the form $c^{\mathrm{T}} b = 0$ where $c \in \text{ ker }(M)$ to satisfy on the rate constants in order to apply the method prescribed by Theorem 14 of Johnston et al. (2018). We suspend discussion of the construction of the matrix M to Step 6, but note that the required condition is

$$\begin{aligned} \frac{(k_{11}+k_{12})\sigma _2}{k_{16}\sigma _1} = 1 \; \Longrightarrow \; \sigma _2 = \frac{k_{11}+k_{12}}{k_{16}} \sigma _1. \end{aligned}$$

(29)

That is, we eliminate one of our free parameters to satisfy the condition. Since this result is positive, we may proceed.

Step 5: After substituting (29) into (), we can compute the following tree constants

$$\begin{aligned} \begin{aligned} K_1&= (k_2+k_3)(k_5+k_6)k_7k_{9}k_{10}(k_{11}+k_{12})k_{12}k_{14}k_{15}\sigma _1\\ K_2&= k_1(k_5+k_6)k_7k_{9}k_{10}(k_{11}+k_{12})k_{12}k_{14}k_{15}\sigma _1 \\ K_3&= k_1k_3 (k_5 + k_6) k_7 k_{9} k_{10}(k_{11} + k_{12}) (k_{13} + k_{14}) k_{15} k_{16} \\ K_4&= k_1k_3k_4k_7k_{9}k_{10}(k_{11}k_{13}+k_{11}k_{14}+k_{12}k_{13}+k_{12}k_{14})k_{15}k_{16}\\ K_5&= k_1 k_3 k_4 k_6(k_8 + k_{9}) k_{10} (k_{11} k_{13} + k_{11} k_{14} + k_{12} k_{13}+ k_{12} k_{14})k_{15} k_{16} \\ K_6&= k_1k_3k_4k_6k_7k_{10}(k_{11}k_{13}+k_{11}k_{14}+k_{12}k_{13}+k_{12}k_{14})k_{15}k_{16}\\ K_7&= k_1k_3(\sigma _1(k_5+k_6)k_{11}+k_{16}k_4k_6)k_7k_9(k_{11}+k_{12})(k_{13}+k_{14})k_{15}\\ K_8&= k_1k_3(k_5 + k_6)k_7 k_{9} k_{10} (k_{11}+k_{12})(k_{13} + k_{14}) k_{15} \sigma _1 \\ K_9&= k_1k_3(k_5+k_6)k_7k_{9}k_{10}(k_{11}+k_{12})k_{12}k_{15}\sigma _1\\ K_{10}&= (k_2\sigma _1+k_3\sigma _1+k_1k_3)(k_5+k_6)k_7k_{9}k_{10}(k_{11}+k_{12})k_{12}k_{14}\sigma _1\\ K_{11}&= (k_2k_5+k_2k_6+k_3k_5+k_3k_6)k_7k_{9}k_{10}(k_{11}+k_{12})k_{12}k_{14}k_{15}\sigma _1^2/k_{16} \end{aligned} \end{aligned}$$

Step 6: Log-linear system (24) can be set up for any maximal set of pairs of vertices lying in the same linkage class. We take the pairs

$$\begin{aligned} \{ 1, 2\}, \{1, 3\}, \{1, 4\}, \{1, 5\}, \{1, 6\}, \{1, 7\}, \{1, 8\}, \{1, 9\}, \{1, 10\}, \{1, 11\}. \end{aligned}$$

This gives the following log-linear system (24):

$$\begin{aligned} \left[ \begin{array}{ccccccccccc}-\,1&{}0&{}0&{}0&{}-\,1&{}1&{}0&{}0&{}0&{}0&{}0\\ -\,1&{}1&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0&{}0\\ -\,1&{}0&{}0&{}0&{}-\,1&{}0&{}1&{}0&{}0&{}0&{}0\\ -\,1&{}0&{}1&{}1&{}-\,1&{}0&{}0&{}0&{}0&{}0&{}0\\ -\,1&{}0&{}0&{}0&{}-\,1&{}0&{}0&{}1&{}0&{}0&{}0\\ -\,1&{}0&{}0&{}0&{}-\,1&{}0&{}0&{}0&{}1&{}0&{}0\\ -\,1&{}1&{}0&{}1&{}-\,1&{}0&{}0&{}0&{}0&{}0&{}0\\ -\,1&{}0&{}0&{}0&{}-\,1&{}0&{}0&{}0&{}0&{}1&{}0\\ -\,1&{}0&{}0&{}0&{}-\,1&{}0&{}0&{}0&{}0&{}0&{}1\\ 0&{}0&{}0&{}1&{}-\,1&{}0&{}0&{}0&{}0&{}0&{}0\end{array} \right] \left[ \begin{array}{c} \ln (X) \\ \ln (X_p) \\ \ln (X_{pp}) \\ \ln (M) \\ \ln (K) \\ \ln (XK) \\ \ln (X_pK) \\ \ln (X_{pp}M) \\ \ln (X_pM) \\ \ln (X^*_pM) \\ \ln (XM) \end{array} \right] = \left[ \begin{array}{c} \ln \left( K_2/K_1 \right) \\ \ln \left( K_3/K_1 \right) \\ \ln \left( K_4/K_1 \right) \\ \ln \left( K_5/K_1 \right) \\ \ln \left( K_6/K_1 \right) \\ \ln \left( K_7/K_1 \right) \\ \ln \left( K_8/K_1 \right) \\ \ln \left( K_9/K_1 \right) \\ \ln \left( K_{10}/K_1 \right) \\ \ln \left( K_{11}/K_1 \right) \end{array} \right] \end{aligned}$$

(30)

Note that, in Step 4, we used the left kernel vector $c = (0,1,0,0,0,0,-\,1,0,0,1)$ of the coefficient matrix $M^{\mathrm{T}}$ of (30). Since we have satisfied the condition $c^{\mathrm{T}} b = 0$ with (29), this is a consistent system. We can solve this system and exponentiate to obtain the following steady-state parametrization:

$$\begin{aligned} \begin{aligned} X&= \frac{(k_2+k_3)k_{15}\tau _1}{(k_2+k_3)\sigma _1+k_1k_3}&X_pK&= \frac{k_1k_3k_4(k_{13}+k_{14})k_{15}k_{16}\tau _1\tau _2}{((k_2+k_3)\sigma _1+k_1k_3)(k_5+k_6)k_{12}k_{14}\sigma _1}\\ X_p&= \frac{k_1k_3(k_{13}+k_{14})k_{15}k_{16}\tau _1}{((k_2+k_3)\sigma _1+k_1k_3)k_{12}k_{14}\sigma _1}&X_{pp}K&= \frac{k_1k_3k_4k_6(k_{13}+k_{14})k_{15}k_{16}\tau _1\tau _2}{((k_2+k_3)\sigma _1+k_1k_3)(k_5+k_6)k_9k_{12}k_{14}\sigma _1}\\ X_{pp}&= \frac{k_1k_3k_4k_6(k_8+k_9)(k_{13}+k_{14})k_{15}k_{16}^2\tau _1}{((k_2+k_3)\sigma _1+k_1k_3)(k_5+k_6)k_7k_9k_{12}k_{14}\sigma _1^2}&X_pM&= \frac{k_1k_3(\sigma _1(k_5+k_6)k_{11}+k_4k_6k_{16})(k_{13}+k_{14})k_{15}\tau _1\tau _2}{((k_2+k_3)\sigma _1+k_1k_3)(k_5+k_6)k_{10}k_{12}k_{14}\sigma _1}\\ M&= \frac{\sigma _1\tau _2}{k_{16}}&X^*_pM&= \frac{k_1k_3k_{15}\tau _1\tau _2}{((k_2+k_3)\sigma _1+k_1k_3)k_{14}}\\ K&= \tau _2&XM&= \tau _1\tau _2\\ XK&= \frac{k_1k_{15}\tau _1\tau _2}{(k_2+k_3)\sigma _1+k_1k_3}&\end{aligned} \end{aligned}$$

(31)

in the parameters $\sigma _1, \tau _1, \tau _2 \in {\mathbb {R}}_{>0}$.

Parametrization (31) is quite useful in the context of determining the capacity for mono- and multistationarity within stoichiometric compatibility classes of mass-action system (1) corresponding to MAPK network (). The steady states are not toric so that the results of Pérez Millán et al. (2012) and Müller et al. (2016) cannot be applied. We can, however, apply the computational procedure of Corollary 2 of Conradi et al. (2017). To satisfy the assumptions, we note that the network has the following conservation laws and is therefore dissipative:

$$\begin{aligned} X_{\mathrm{tot}}= & {} X + X_p + X_{pp} + XK + X_pK + XM + X_pM + X_p^*M + X_{pp}M\\ K_{\mathrm{tot}}= & {} K + XK + X_pK \\ M_{\mathrm{tot}}= & {} M + XM + X_pM + X_p^*M + X_{pp}M. \end{aligned}$$

It also has no critical siphons so that there are no boundary equilibria Angeli et al. (2007). Computing the determinant of the Jacobian $a({\hat{x}})$ evaluated along parametrization (31) yields a rational function in the three parameters $\sigma _1, \tau _1,$ and $\tau _2$ with a strictly positive denominator. It can be checked that, in the numerator of $a({\hat{x}})$, $\tau _1^2 \tau _2^2$ and $\sigma _1^2 \tau _1 \tau _2^3$ are extremal with respect to the corresponding Newton polytope and that the coefficients have mixed sign in $a({\hat{x}})$. It follows that the mechanism exhibits multistationarity for some choices of rate constants and initial conditions. It should be noted that parametrization (31) reduces the dimension of the system from 11 variables to 3 which allows significantly faster computation and analysis of $a({\hat{x}})$.