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Biology · Systems & Quantitative Biology

Rates, balances, biological networks and dynamical change

Two invented pools contain twelve and eight amount units, twenty in total. Three amount units per time unit move from the first to the second, while one moves back. Each pool changes even though the combined total does not. A single arrow or total can hide that distinction.

Research Design, Data & Scientific Practice introduces observations, comparisons and uncertainty. Gene Regulation & Cell Signalling develops biological feedback, while Population Ecology & Life-History Patterns develops rates and population boundaries. Their English chapters are available in this working manuscript; publication remains pending. Start with arithmetic, averages and ordinary graph reading. A derivative is introduced as an instantaneous rate of change. Exponentials and the circular sine–cosine example are explained where used; a first reading can follow their values and arrows, then return to their formulas with further mathematical preparation.

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By the end, you can

Use a balance to distinguish internal transfers from boundary flows, and distinguish network structure from a complete rule for change.

01

Count what crosses the chosen boundary

A balance accounts for change as contributions with signs. For a quantity inside a boundary, inflows add and outflows subtract. An internal transfer can count as an outflow for one subpool and an inflow for another. When those subpools form a closed combined system, the internal terms cancel in its total balance.

Write the first amount as X and the second as Y. In the original fixed-rate case, dX/dt = -3 + 1 = -2 and dY/dt = 3 - 1 = 2. The notation dX/dt means the instantaneous rate at which X changes with time. Adding the two equations gives d (X + Y)/dt = 0. It states conservation of this assigned total, not constancy of each part.

A conserved quantity depends on what is being counted. Conservation of matter does not mean that the number of every molecular species stays fixed in a reaction. A balance for a particular species may include production and consumption terms. Units and stoichiometry must justify which combination is conserved; merely drawing a closed box is insufficient.

Internal changes cancel in the combined balance

Internal changes cancel in the combined balanceTwo pools retain a total of twenty while equal and opposite net changes redistribute their assigned amounts over one stated interval.First pool12Second pool831Amount units per time unitNo transfer crosses the combined boundaryStart128Combined amount: 20One time unit later1010Combined amount: 20dX/dt=-3+1; dY/dt=3-1; d(X+Y)/dt=0Net rate: -2 + 2 = 0
An amount-transfer arrow carries amount per time units. It does not serve as a generic activation arrow, and the fixed-rate interval is bounded.
Read the diagram in words

The first pool's net rate is minus two, and the second pool's is plus two amount units per time unit. After one time unit with the assigned rates fixed, both contain ten. The total is twenty at both times because the internal transfer terms cancel.

Pool
PoolAmount units
First pool12
Second pool8
Amount units per time unit
PoolPoolAmount units per time unit
First poolSecond pool3
Second poolFirst pool1
Combined amount
Time unitsFirst poolSecond poolCombined amount
012820
1101020
Net rate
PoolNet rate
First pool-2
Second pool2
Combined amount0

dX/dt=-3+1; dY/dt=3-1; d(X+Y)/dt=0

Two original pools with the same amount unit, no boundary inflow or outflow, and fixed internal transfer rates for the interval 0..1. No actual molecular reaction, organism, measured parameter or infinite-time continuation is supplied.

02

A rate is not already a finite change

A rate has quantity per time units. A change has quantity units. If the assigned net rate minus two stays constant for one time unit, X changes by minus two. Starting at twelve gives ten. Y starts at eight and rises to ten. This calculation uses the fixed-rate assumption over that whole interval.

If the rate depends on the current state, replacing the whole interval by the initial rate is generally an approximation. The state changes, so the rate may change too. Numerical simulation often updates a state through a sequence of steps; the step size and rule can affect the computed trajectory. Exact conservation of an algebraic total is a useful check but does not prove every simulated value is accurate.

A model for nonnegative amounts also needs an appropriate domain. Continuing a fixed negative rate indefinitely would eventually produce a negative amount. That would signal that the teaching rule has left a meaningful amount domain. Here the displayed interval ends with both pools at ten; no indefinitely valid biological transfer law is asserted.

03

A graph needs a meaning for every edge

A biological network can represent material transfers, regulatory influences, statistical associations or other relationships. The same arrow shape can therefore carry different meanings. An activation arrow indicates a specified positive influence under a model's conditions; it is not automatically a molecule moving between two locations.

The original three-node regulatory graph has a positive A-to-B influence, a negative B-to-C influence and a positive C-to-A influence. Following the closed path gives one negative sign overall. That is a statement about the assigned signs. It does not provide response strength, delay, initial state or an equation for any node.

A network becomes a dynamical model only when suitable state variables and change rules are supplied. The rules must say how contributions combine, what depends on time and what can be observed. Distinct kinetic choices can share a topology yet produce different behaviour. Finding a signed cycle therefore suggests a question about dynamics; it does not by itself demonstrate an oscillation or mechanism in a living system.

Loop sign leaves the dynamics open

Loop sign leaves the dynamics openThree abstract regulatory influences form a negative signed loop without supplying kinetic strengths, delays or a predicted oscillation.+−+Abstract component AAbstract component BAbstract component CAssigned positive influenceAssigned negative influenceProduct of the three signs: (+1) × (−1) × (+1) = -1Strength, delays and change rules are unspecifiedRegulatory influence, not amount flow
Signs distinguish the influence meanings. A closed path is a topological property, not a measured time series.
Read the diagram in words

The A-to-B and C-to-A influences are positive; B-to-C is negative. Their sign product is minus one. This negative loop does not define an oscillation, because response rules, strengths and timescales have not been supplied.

Regulatory influence, not amount flow
Regulatory influence, not amount flowRegulatory influence, not amount flowAssigned positive influence / Assigned negative influence
Abstract component AAbstract component B1
Abstract component BAbstract component C-1
Abstract component CAbstract component A1
Product of the three signs
Product of the three signs
-1

Strength, delays and change rules are unspecified

Original abstract signed regulatory network. These edges describe assigned influences, not material movement, measured correlations or verified causal relationships in a particular organism. Topology alone supplies no trajectory.

Worked example

Follow the internal transfer for one time unit

X loses three and gains one amount unit per time unit, so its net rate is minus two. Y gains three and loses one, so its net rate is plus two. The net rate of the sum is zero.

With both rates held constant, X (1) = 12 - 2 = 10 and Y (1) = 8 + 2 = 10. The initial and final totals are both twenty. There are internal changes without a boundary flow for the combined pair.

In the separate regulatory graph, the product of the signs is (+1) × (-1) × (+1) = -1. It is a negative regulatory loop. There is no amount unit on those influences and no supplied trajectory, so applying the transfer calculation to those edges would change their meaning.

Check your understanding

  1. If a closed combined system has a constant total, must every subpool stay constant?

    Show reasoning

    No. Opposite internal changes cancel in the combined balance. Here X decreases by two and Y increases by two over the stated interval, leaving the total twenty.

  2. When may the initial rate be multiplied by the whole interval as an exact change?

    Show reasoning

    When that rate is constant throughout the interval under the model. If it changes with state or time, the initial-rate calculation generally approximates rather than exactly determines the finite change.

  3. Does the negative sign of a regulatory loop establish oscillation?

    Show reasoning

    No. It identifies the signed topology only. Kinetics, strength, nonlinearity, delays and initial conditions can affect the dynamics; none is specified by the sign product alone.

What to carry forward

Keep quantities, rates, boundaries and edge meanings separate before inferring a trajectory from a graph.

References

Resources supporting this lesson’s explanations and examples.

  1. Computational systems biology (Nature 420, 206–210 (2002); doi:10.1038/nature01254)Hiroaki KitanoResearch abstractpubmed.ncbi.nlm.nih.govRead source
  2. Design principles of biochemical oscillators (Nature Reviews Molecular Cell Biology 9, 981–991 (2008); doi:10.1038/nrm2530)Béla Novák and John J. TysonResearch abstractnature.comRead source
  3. The systems biology markup language (SBML): a medium for representation and exchange of biochemical network models (Bioinformatics 19(4), 524–531 (2003); doi:10.1093/bioinformatics/btg015)M. Hucka and colleagues, with the SBML ForumResearch abstractpubmed.ncbi.nlm.nih.govRead source