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Elementary modes

Set of the simplest pathways in a biochemical network

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Through Biochemistry, Enzyme and Stoichiometry

The set of elementary modes for a given network is unique (up to an arbitrary scaling factor). Given the fundamental nature of elementary modes in relation to uniqueness and non-decomposability, the term 'pathway' can be defined as an elementary mode. Note that the set of elementary modes will change as the set of expressed enzymes change during transitions from one cell state to another. Mathematically, the set of elementary modes is defined as the set of flux vectors v that satisfy the steady state condition

Nv(x,p)=0

where N is the stoichiometry matrix, v is the vector of rates, x the vector of steady state floating (or internal) species and p, the vector of system parameters.

An important condition is that the rate of each irreversible reaction must be non-negative, virr≥0.

A more formal definition is given by:

An elementary mode, vi, is defined as a vector of fluxes, v1,v2,…, such that the three conditions listed in the following criteria are satisfied.

  1. The vi vector must satisfy: Nvi=0, that is: the steady state condition.
  2. For all irreversible reactions: vi≥0. This means that all flow patterns must use reactions that proceed in their most natural direction. This makes the pathway described by the elementary mode a thermodynamically feasible pathway.
  3. The vector vi must be elementary. That is, it should not be possible to generate vi by combining two other vectors that satisfy the first and second requirements using the same set of enzymes that appear as non-zero entries in vi. In other words, it should not be possible to decompose vi into two other pathways that can themselves sustain a steady state. This is called elementarity. A more formal test is that the null space of the submatrix of N that only involves the reactions of vi is of dimension one and has no zero entries.

Example

Consider a simple branched pathway with all three steps irreversible. Such a pathway will admit two elementary modes which are indicated in thicked (or red) reaction lines.

Because both v2 and v3 are irreversible, and elementary mode lying on both these reactions is not possible since it would mean one reactions going against its thermodynamic direction. Each mode in this system satisfies the three conditions described above. The first condition is steady state, that is for each mode vi, it has to be true that Nvi=0.

Algebraically the two modes are given by:

v1=[110]v2=[101]

By substituting each of these vectors into Nvi=0, it is easy to show that condition one is satisfied. For condition two we must ensure that all reactions that are irreversible have positive entries in the corresponding elements of the elementary modes. Since all three reactions in the branch are irreversible and all entries in the elementary modes are positive, condition two is satisfied.

Finally, to satisfy condition three, we must ask whether we can decompose the two elementary modes into other paths that can sustain a steady state while using the same non-zero entries in the elementary mode. In this example, it is impossible to decompose the elementary modes any further without disrupting the ability to sustain a steady state. Therefore, with all three conditions satisfied, we can conclude that the two vectors shown above are elementary modes.

All possible flows through a network can be constructed from linear combinations of the elementary modes, that is:

v=∑λiviwhereλ≥0

such that the entire space of flows through a network can be described. λi must be greater than or equal to zero to ensure that irreversible steps aren't inadvertently made to go in the reverse direction. For example, the following is a possible steady-state flow in the branched pathway.

v=2.5[110]+0.5[101]=[3.02.50.5]

If one of the outflow steps in the simple branched pathway is made reversible, an additional elementary mode becomes available, representing the flow between the two outflow branches. An additional mode emerges because, with only the first two modes, it is impossible to represent a flow between the two branches because the scaling factor, λi, cannot be negative (which would be required to reverse the flow).

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