From a reaction list to a model you can defend

A worked guide to state choice, stoichiometric accounting, effective rate laws, conserved pools, and the difference between a balance and a dynamical explanation.

CCBS · EST 5053 · Fall 2026 · lecture 3 · exposition · study after the core lecture

The core page is the argument delivered in the room. This page is for the slower task that follows: taking an unfamiliar biochemical diagram and turning it into equations whose symbols, assumptions, units, and boundary can all be explained. The lineage essay asks where this language came from and what its history warns us not to overclaim. Here the test is practical. At the end, you should be able to build and audit the model yourself.

1Choose the boundary and the state

A model becomes dynamical only when its present state contains enough information to determine its next velocity.

Begin with the question, not the pathway map. If the question is how a substrate pool responds over seconds, enzyme binding states may need to be explicit. If the question is how a protein level responds over generations, those same binding states may be replaceable by an effective law, while cell volume and growth cannot be ignored. A state is not a list of everything that exists. It is the smallest list that makes the chosen future depend on the present rather than on an unrecorded past.

For a deterministic concentration model, fix an ordered state vector

Write down four declarations beside it:

1. Boundary. Which species and processes are inside? Is nutrient supply imposed, or is the environment itself dynamic?
2. Scale. Are the variables molecule counts, concentrations, fractions, or held totals? Do not mix these in one equation without a conversion.
3. Clock. Which processes are fast enough to be eliminated, and what numerical comparison licenses that move?
4. Input. Is a signal a fixed parameter, a prescribed function of time, or another state variable?

If all parameters and boundary conditions are fixed and the right-hand side depends only on $x$, the model is autonomous: $\dot x=F(x)$. A scheduled nutrient pulse gives $\dot x=F(x,t)$, which is nonautonomous. It can become autonomous again only by adding a state that generates the pulse. This distinction is not cosmetic. A trajectory in state space is unambiguous only when one state has one velocity.

2Translate one arrow without guessing

Reactants determine what must meet. Products minus reactants determine what one event changes.

For reaction $i$, use one fixed species order and write

The bare squiggle makes only a net accounting claim. The reactant vector $\alpha_i$, product vector $\beta_i$, and change vector $\gamma_i$ answer different questions. For $2A+B\rightsquigarrow3B$, ordered as $(A,B,C)$,

If a mechanism declares this to be an elementary, well-mixed event, write $2A+B\xrightarrow{k_i}3B$; its rate is then $v_i=k_iA^2B$. The exponents come from $\alpha_i$, because two $A$ molecules and one $B$ molecule must be present. The contribution to the $B$ equation is $+2v_i$, because one event changes $B$ by two. Using the net vector as a kinetic exponent would ask for a negative concentration power of $A$ and would miss the consumed $B$ altogether.

3Assemble the autonomous system

Stack one change vector per reaction and one rate per reaction. The matrix multiplication performs all sign bookkeeping.

For $m$ reactions, form the $n\times m$ stoichiometric matrix

Species index the rows; reactions index the columns. This dimensional check catches many transcription errors: $\Gamma$ maps a vector of $m$ reaction fluxes into $n$ species velocities.

For an elementary reaction, the deterministic mass-action rate is

Equation (4) is not attached automatically to every arrow. An arrow for transcription, transport through several conformations, or a complete enzyme turnover is composite. Its column still accounts correctly for net material change, but its rate must be derived from a more detailed mechanism, measured, or declared as a phenomenological law. This separation between stoichiometry and kinetics is what lets the same network be tested under several rate models.

Use the reversible binding and catalysis mechanism

With state $(E,S,C_{ES},P)^T$,

The first row gives $\dot E=-k_+ES+(k_-+k_{\mathrm{cat}})C_{ES}$; the other three follow without a new sign decision. Check the units: each component of $v$ has concentration per time, while the entries of $S$ are molecules changed per event and therefore dimensionless in the concentration balance.

4Read constraints before solving

The left null space finds conserved quantities. The right null space finds flux combinations that can coexist at steady state.

If a row vector $\ell^T$ satisfies $\ell^T\Gamma=0$, then

Thus $\ell^Tx$ is conserved by every kinetic law placed on the same arrows. For (6), $\ell^T=(1,0,1,0)$ gives $q_E=E+C_{ES}$. A second vector, $(0,1,1,1)$, gives $S+C_{ES}+P$. Each initial condition is confined to the intersection of the nonnegative region with these conservation planes. Chemical reaction network theory calls that set a stoichiometric compatibility class.¹

The right null space answers a different question. At a steady state,

The flux vector need not vanish. It only needs to lie in the right null space of $\Gamma$. This is why a metabolic steady state is not chemical equilibrium. Carbon can flow into, through, and out of every reaction while internal concentrations remain fixed. Equilibrium is a stronger statement involving the cancellation of forward and reverse currents; Lecture 7 develops that distinction.

5Derive growth dilution from counts

A stable concentration in a growing bacterium need not mean the molecules are being destroyed.

Let $N_X(t)$ be the number of molecules of $X$, $V(t)$ the cell volume, and $x=N_X/V$ the concentration. Under balanced exponential growth, $\dot V=\lambda V$. The quotient rule gives

The term $-\lambda x$ is effective removal by dilution. No molecule is chemically degraded. If molecules are produced at rate $a$ per unit volume, then $\dot N_X=aV$ and

At one doubling time, $T_d=\ln2/\lambda$, the displacement from the steady concentration has halved. The molecule count and volume can both rise while concentration approaches a constant. For a fixed molecular machine making $k$ molecules per time instead, the source is $k/V(t)$, not a constant. A constant source in concentration units already assumes production capacity scales with volume.

Growth changes the language of conservation. If chemistry alone conserves $q=\ell^Tx$, then dilution gives $\dot q=-\lambda q$. A living cell replenishes the pool, so a more realistic coarse equation is $\dot q=\beta-\lambda q$ with $q^*=\beta/\lambda$. On a fast signaling timescale the pool is approximately conserved. Across generations it is a held total whose level the cell controls by expression.

6Pick the view that exposes the question

Stoichiometry-flux and addition-removal are two readings of the same balance, not two rival theories.

Read row $j$ of (3) and collect its positive and negative terms:

The matrix view is reaction-centric. It makes coupled conversions, conservation laws, and flux routes visible. The addition-removal view is species-centric. It makes the balance of one measured output visible. Neither loses information if every term is retained.

A metabolic system

Lump glycolysis into an investment step and a payoff step:

With rows ordered as $(G,I,\mathrm{ATP},\mathrm{ADP},W)$,

The dense columns are the biology. One pass consumes ATP before it produces more, and $\mathrm{ATP}+\mathrm{ADP}$ is unchanged by both lumped steps. Writing five separate addition-removal equations is possible, but it hides the shared reaction event that changes several metabolites at once.

A regulated gene

Let repressor $R$ bind promoter $G$, let free promoter produce protein $P$, and include actual production and removal:

Most columns change one recorded species. After fast promoter binding is reduced, the protein equation becomes

Here addition-removal is the clearer view because regulation changes the shape of the addition term while the removal term stays first order. Calling (11) production-loss would be misleading: conversion can add one species without producing matter, and dilution removes concentration without degrading a molecule. Use the generic name addition-removal. Reserve production, degradation, dilution, import, and conversion for the actual mechanisms.

7Write the driven enzyme cycle

Supply and removal turn a one-pass conversion into an open system with sustained flux and a genuine steady state.

Add substrate at constant concentration flux $v_{\mathrm{in}}$, let enzyme bind and convert it, and remove product first order:

For the ordered state $x=(E,S,C_{ES},P)^T$, the mixed-resolution system has three elementary enzyme steps and two composite boundary processes:

Multiplying the rows gives

Every term has an auditable origin. The supply appears only in the free substrate row. Binding removes one free enzyme and one free substrate and adds one complex. Unbinding reverses that change. Catalysis removes the complex and returns free enzyme while adding product. Product removal touches only the product row.

8Transform to pools exactly

Eliminating internal binding motion by changing variables is exact. Closing the resulting equations is a separate step.

The vectors left unchanged by binding suggest the coordinates

Differentiate these definitions and substitute (18):

No approximation occurred. Association and dissociation disappeared because they only redistribute material within each total. Equation (20) is lower-dimensional accounting, but it is not yet a closed dynamical system: the right-hand side still contains $C_{ES}$, which is not among the three displayed state variables.

This distinction is worth naming:

9Close the composite model

The quasi-steady complex is a selected root of an algebraic equation, not a new elementary reaction.

The complex equation in (18) can be written

If complex relaxation is fast compared with motion of the substrate pool, set the bracket approximately to zero. In free variables, use $E=q_E-C_{ES}$ but leave $S$ free:

Then $v=k_{\mathrm{cat}}C_{ES}$ gives the familiar Briggs-Haldane form

This is not rapid equilibrium unless $k_-\gg k_{\mathrm{cat}}$. Under rapid equilibrium, $K_M\simeq k_-/k_+=K_{d}$. Under the standard quasi-steady argument, catalysis contributes to the denominator through $K_M=(k_-+k_{\mathrm{cat}})/k_+$. Briggs and Haldane introduced this steady-complex logic in 1925; Segel and Slemrod later made the small parameter explicit.²³

An experiment usually controls total substrate, not free substrate. Substitute both conservation relations into the same closure:

The physical root is

Choose the smaller root because $0\le C_{ES}\le\min(q_E,q_S)$. For numerical work, avoid subtracting nearly equal large numbers and use the equivalent form

The closed composite model is therefore

Three regimes of the same root

Equation (25) is exact and unreadable. What makes it usable is that it collapses to a hyperbola in most of the parameter plane, and the collapse has to be earned rather than assumed. Write $b=q_E+q_S+K_M$ and pull it out of the radical:

Nothing is approximated yet. Since $q_S\le b$ we have $u\le 4q_E/b$, so $u$ is small whenever the enzyme total is small against $b$. Expanding $\sqrt{1-u}=1-u/2+O(u^{2})$ and then dropping $q_E$ from $b$ gives

Both steps need the same thing, $q_E\ll K_M+q_S$. That is a sum, so it has two independent sufficient conditions, and they are different regimes with different physical content.

• Trace enzyme, $q_E\ll K_M$. The textbook assumption. It makes $u$ small for every substrate concentration, so (32) holds on both arms of the curve. At $q_E=K_M/100$ the worst relative error of (32) against the exact root is 0.98% over eight decades of $q_S$.
• Substrate excess, $q_E\ll q_S$. Apply $q_S\le b$ once more, now in the denominator, and $u\le 4q_E/b\le 4q_E/q_S$. The expansion runs through with $K_M$ never mentioned. At $q_E=100K_M$, where the first condition fails by four orders of magnitude, (32) is still accurate to 0.011% for every $q_S>10q_E$.
• Titration, $K_M\ll q_E$. Here $u$ is not small and (32) is not available. Because (24) is symmetric in $q_E$ and $q_S$, the same statement with the totals exchanged does the same work, so the regime is $K_M$ small against the totals. Setting $K_M=0$ in (24) gives $b^{2}-4q_Eq_S=(q_E-q_S)^{2}$ exactly, hence

The complex reads the scarcer partner and the free species are rectified at the crossing. At $q_E=100K_M$ that law is right to 9.5% across six decades, and its one soft point is $q_S=q_E$, where expanding (25) about the crossing gives

Here that predicts a 10% shortfall against a measured 9.5%, and the corner sharpens as $\sqrt{K_M/q_E}$. That square root is the width of the threshold, and it is what makes titration a device rather than a defect.

10Check the reduction with numbers

A reduction becomes persuasive when its steady state, transient mismatch, and timescale ratio can all be calculated.

Use one reproducible parameter set:

First recover the microscopic unbinding rate from (21), taking care with units:

Thus $k_-/k_{\mathrm{cat}}=999$: the complex typically dissociates many times per catalytic event. For a run beginning at $q_S=20\,\mu\mathrm{M}$, $\varepsilon=1/(100+20)=8.3\times10^{-3}$. Both comparisons predict a rapid initial adjustment.

At steady state, flux balance gives the solution before any root finder is used:

Then $E^*=q_E-C_{ES}^*=0.5\,\mu\mathrm{M}$. Equation (24) gives $S^*=100\,\mu\mathrm{M}$ and $q_S^*=100.5\,\mu\mathrm{M}$. The input flux is below $V_{\max}=10\,\mu\mathrm{M}\,\mathrm{s}^{-1}$, so a finite steady substrate is possible. If $v_{\mathrm{in}}\ge V_{\max}$, no finite substrate concentration can make catalysis keep up; the pool grows without bound in this open model. That failure is a model prediction, not a numerical instability.

Quote both as times, and leave $\lambda$ to the growth rate it already names in §5. The complex's own equation is $k_+$ times the left side of (24), so $\dot C_{ES}=k_+(C_{ES}-C_-)(C_{ES}-C_+)$ with $C_\pm$ its two roots, and the fast time constant is exactly $\tau_{\mathrm{fast}}=[k_+(C_+-C_-)]^{-1}$, here $50\,\mu\mathrm{s}$. The pools relax with $\tau_{\mathrm{slow}}=[k_{\mathrm{cat}}(dC_{ES}/dq_S)]^{-1}=40\,\mathrm{s}$. The separation is $\tau_{\mathrm{slow}}/\tau_{\mathrm{fast}}=8\times10^5$. The reduced trajectory should therefore miss the initial boundary layer but track the slow motion afterward.

11Open the central-dogma network

A regulated protein equation is the last line of a multi-player network, not a primitive Hill curve.

Write a minimal transcription-and-translation mechanism before reducing it:

Growth adds dilution removal $\lambda$ to every concentration. The exact state could contain all seven free and complexed species in (34), plus whatever produces $R$. If promoter and ribosome binding relax rapidly relative to mRNA and protein levels, those complexes can be replaced by effective addition functions:

A further regime with unsaturated translation gives $f_{\mathrm{tl}}=k_PM$. If promoter regulation is summarized by a Hill repression law,

Equation (36) is useful precisely because it keeps the two slow filters explicit. The mRNA response time is $(\delta_M+\lambda)^{-1}$; the protein response time is $(\delta_P+\lambda)^{-1}$. In growing E. coli, measured mRNA half-lives are commonly minutes, while dilution can dominate removal of stable proteins.⁶⁷ A delayed protein rise can therefore emerge from two first-order state equations in series even though no explicit time delay appears.

12Derive activation and repression

A Hill-shaped addition term is an occupancy model after states have been removed, not mass action applied to a composite arrow.

Start with one rapid binding step between promoter $D$ and activator $A$:

Substitute $D=q_D-C_{DA}$ into the equilibrium relation:

If the bound promoter initiates at rate $r_1$ and the free promoter at $r_0$, average over the two rapidly interconverting states:

If binding blocks initiation, the productive fraction is free instead:

Cooperative or concerted binding models can yield the effective forms

The exponent $n$ is not automatically the number of molecules in one elementary collision. Hill introduced his equation as an empirical description of oxygen binding, and allosteric state models later showed how several coupled conformations can produce a compact sigmoidal occupancy law.⁴⁵ A fitted Hill exponent is a slope descriptor unless a mechanism establishes a literal stoichiometry.

13Distinguish free from total

The textbook hyperbola is written in the free ligand. The cell and the experiment usually control a total.

Equation (38) treats free activator $A$ as the independent variable. But if activator is sequestered on promoters or decoys, the controlled variable is $q_A=A+C_{DA}$. Together with $q_D=D+C_{DA}$, fast binding gives

This is the same quadratic as the enzyme closure (24). It has two especially useful regimes.

The crossover is rounded over a scale of order $\sqrt{K_{d}q_D}$. Increasing the held binding-partner total can therefore sharpen the free-output response without adding a cooperative binding site. Buchler and Cross engineered this molecular-titration mechanism in yeast and found that both threshold and apparent sharpness tracked inhibitor abundance.¹³ Ha and Ferrell showed the complementary lesson: a response that looks graded against free ligand can be sharply thresholded against total ligand.¹⁴

Several system-level computations follow by composing these regimes:

1. Threshold. A stoichiometric decoy holds free input near zero until the total input exceeds the held decoy total.
2. Threshold plus saturation. Feed the leftover input into a trace-enzyme step. The first module sets where the response begins; the second sets its upper limit.
3. Ratio sensing. Let two ligands compete for one scarce receptor pool. In the low-occupancy limit, the fraction of receptor assigned to ligand 1 is $(q_1/K_1)/(q_1/K_1+q_2/K_2)$. Scaling both ligand totals together leaves the fraction unchanged.

14Follow shared pools through metabolism

Allostery changes a flux law. Stoichiometry and conservation determine how that local regulation propagates through the rest of the network.

Phosphofructokinase catalyzes a committed glycolytic step that consumes fructose-6-phosphate and ATP. Adenine nucleotides also bind regulatory sites. In mammalian muscle PFK, distinct sites mediate activating and inhibitory effects of AMP, ADP, and ATP, and ATP is simultaneously a substrate.⁹ A useful model must therefore keep two statements separate:

1. Local mechanism: occupancy of catalytic and allosteric sites changes the flux $v_{\mathrm{PFK}}(x)$.
2. System coupling: ATP, ADP, and AMP participate in other reactions and are linked by an approximately held adenylate pool and rapid adenylate kinase.

A common one-number summary is the adenylate energy charge

Chapman, Fall, and Atkinson measured values near 0.8 in growing E. coli and a marked decline during starvation.¹⁰ Within a sufficiently closed, rapidly interconverting pool, high AMP and high ATP report opposed energetic regimes. That does not make AMP activation the same molecular action as ATP inhibition. The binding sites are distinct, the regulatory effects are organism-specific, and synthesis, consumption, or compartment exchange can break the simple pool relation.