AA3 – Networks



Flux Cones of Metabolic Networks

Project Heads

Alexander Bockmayr, Martin Henk

Project Members

Frederik Wieder

Project Duration

01.01.2021 − 31.12.2021

Located at

FU Berlin


The project aims at better understanding the structure and function of metabolic networks in biology by analyzing geometric, algebraic and combinatorial properties of the steady-state flux cone, a polyhedral cone defined by the network.

Metabolic network analysis is an important field in computational biology with numerous applications, e.g. in medicine and biotechnology. Metabolic networks can be seen as (weighted) hypergraphs where the nodes correspond to chemical species (called metabolites) and the hyperarcs to reversible or irreversible chemical reactions.

Formally, a metabolic network is given by a set \mathcal{M} of metabolites, a set \mathcal{R} = Irr \uplus Rev of irreversible and reversible reactions, and a stoichiometric matrix S\in\mathbb{R}^{\mathcal{M} \times \mathcal{R}}. We always assume that the metabolic network is in steady-state, i.e., we focus on the polyhedral cone \Gamma  = \{ v\in \mathbb{R}^\mathcal{R} \mid Sv = 0,\ v_{\Irr} \geq 0 \}, which is called the (steady-state) flux cone. An elementary flux mode (EFM) is a flux vector v \in \Gamma \setminus \{0\} with inclusion-minimal support \text{supp}(v) = \left\{ i \in \mathcal{R} \mid v_i \not= 0 \right\}. Studying EFMs is of major biological interest because they correspond to minimal functional units of the network. Up to scaling EFMs are uniquely determined by their support. Therefore, we can identify two EFMs e,e' \in \Gamma whenever \text{supp}(e) = \text{supp}(e'). The finite set \mathcal{E} of all EFMs defines a conic basis of the flux cone \Gamma, i.e., every v \in \Gamma can be written as a non-negative linear combination of EFMs.

The main goal of the project is to understand the structure of the set of all EFMs in large (genome-scale) metabolic networks with possibly hundreds or thousands of reactions. In general, the number of EFMs is exponential in the number of reactions. Therefore, computing all EFMs is often not feasible in practice. To address this problem, Röhl and Bockmayr (2019) introduced the notion of a minimum set of EFMs (MEMo), which is an inclusion-minimal set of EFMs generating the flux cone, and developed a method to compute such a set. For non-pointed cones, MEMos are not unique, and each MEMo captures only some of the biologically relevant metabolic pathways. Therefore, one is interested in computing more than one or possibly all MEMos.

In order to find all MEMos, we plan to proceed as follows:

  • Compute all minimal proper faces of the cone.
  • Determine the EFMs in each minimal proper face.
  • Compute all reversible EFMs in the lineality space.

Given this information, all possible MEMos can be generated by choosing one irreversible EFM from each minimal proper face together with a set of reversible EFMs forming a basis of the lineality space.

Related Publications

Röhl A, Bockmayr A: Finding MEMo: minimum sets of elementary flux modes. Journal of Mathematical Biology, 79(5), 1749-1777, 2019

Wieder F, Henk M, Bockmayr A: On the geometry of elementary flux modes. Journal of Mathematical Biology, 87:50, 2023

Selected Pictures

Example of a metabolic network.

A) Metabolic network involving two irreversible and one reversible reaction. The network has three EFMs: e^1 = (1,0,1),\ e^2 = (0,1,-1), and e^3 = (1,1,0).

B) The corresponding flux cone \Gamma = \{x \in \mathbb{R}^3 \mid x_1-x_2-x_3 = 0, x_1,x_2 \geq 0\}, shown in light grey, is spanned by the EFMs e^1 and e^2. The third EFM e^3 lies inside \Gamma and is the sum of e^1 and e^2. Projecting \Gamma in direction of the reversible reaction 3 results in the pointed cone proj_{\{1,2\}}(\Gamma ) = \{x \in \mathbb{R}^3 \mid x_3 = 0, x_1,x_2 \geq 0\}, shown in dark grey. It is generated by the extreme rays g^1=(1,0,0) and g^2=(0,1,0), which correspond to the minimal proper faces of \Gamma.

C) Removing reaction 3 results in the network in C), which has only one EFM e=(1,1).