From stationary dynamics to equilibrium weights

Time-scale separation removes fast transients. Detailed balance supplies something more: compatible weights that let us calculate molecular occupancies without specifying every kinetic step. The distinction turns an enzyme cycle into a test of equilibrium, then turns equilibrium into a modelling tool for repression and allostery.

CCBS · EST 5053 · Fall 2026 · Lecture 7 · core (full) · 22 September 2026 · 10:40–12:15 · lecturer-led, with student contributions

Lecture 6 gave us a smaller problem, not always an easier algebraic one. Its fast binding layer follows the changing totals. But when an enzyme binds both substrate and product, finding that fast allocation already requires coupled equations. Must we solve the detailed kinetic model every time we change the molecular state space?

Equilibrium physics offers another route. We will find the condition under which transition ratios define consistent state weights. Then we will use those weights to calculate gene regulation. The method applies to a suitable fast subsystem inside a driven cell. It does not require the whole cell to be at equilibrium.

0–5 min

Course business.

5–11 min

1. The remaining problem. Fast stationarity versus equilibrium.

11–18 min

2. State. What information predicts the future?

18–26 min

3. Jumps. Molecular configurations and molecule counts.

26–34 min

4. Generator. Probability conservation and stationarity.

34–46 min

5. Circulation. Matched rings and the enzyme wheel.

46–55 min

6. Detailed balance. Pairwise cancellation and the cycle test.

55–65 min

7. Weights. Path independence and reversible stationary movies.

65–77 min

8. Repression. Binding entropy, concentration and activity.

77–85 min

9. Allostery. Eight states, one active fraction.

85–90 min

10. Closure. Fast allocation and slow total dynamics.

90–95 min

Lecturer addendum and the question carried to Lecture 8.

Selection if time runs short. The expandable derivations and playground are outside this clock. First omit the count-chain walkthrough in §3. Then show §9's state table and result without its limiting calculations. If necessary, defer the rest of §9. Keep the enzyme-current comparison, pairwise balance, path independence and the concentration factor in §8. Those steps establish the argument.

1What fast stationarity leaves unsolved

A fast subsystem may settle quickly without becoming simple to solve. Equilibrium is additional physical structure, not another name for fast.

Keep one enzyme in view throughout the lecture. Substrate $S$ enters, binds free enzyme $E$, is converted while bound, and leaves as product $P$. Product can bind too. The enzyme occupies three states: free, substrate-bound $C_{ES}$, or product-bound $C_{EP}$. Here is the mechanism we are simplifying:

The straight arrows resolve the binding and catalytic steps of this model. The squiggly arrows denote composite supply and removal, with fluxes $v_{\mathrm{in}}$ and $v_{\mathrm{out}}$. Binding preserves the three totals. Slow conversion changes the substrate and product totals. Figure 6 will show how the same three enzyme states can carry a steady current.

Choose the observation clock before simplifying a network. Let $t_{\mathrm{bind}}$ be a binding relaxation time, $t_{\mathrm{obs}}$ the time over which the dynamics of interest changes, and $t_{\mathrm{env}}$ a still slower environmental time. When the first is much shorter than the second, binding can track its conditional stationary state. When the third is much longer, those environmental variables can be treated as parameters during observation.

The stationary constraint still needs a solution. In Lecture 6, the fast rate function was called $g$. Setting its leading fast equation to zero produced a reconstruction $x^{\mathrm{fast}}=h(x^{\mathrm{slow}})$. That calculation can involve many coupled balances. Even when a stationary solution is unique, the expression for it need not be transparent.

The physical question is whether the fast transitions can balance individually. A subsystem exchanging with equilibrium-compatible surroundings can relax to thermodynamic equilibrium. Its probabilities are constrained by free energies and the ways molecules can be arranged. A subsystem exposed to maintained chemical driving can instead keep circulating even after its probabilities stop changing. We need a language that can display both cases.^{1, 2}

Separate the clock test from the equilibrium test. Fast attraction licenses a stationary reduction. Balanced physical transitions supply the extra simplification we will derive.

2State replaces the relevant past

A state contains the information a model needs to predict its future, given the model's parameters and future inputs.

For an autonomous deterministic model, the state selects a trajectory. Lecture 3 wrote the canonical form $\dot x=f(x)$, with $x$ a vector of state variables. Once we know $x(t_0)$ at the present time $t_0$, existence and uniqueness give the future trajectory. We do not need to replay the earlier history to compute the next step.

The spring shows why a visible variable need not be a complete state. Let $q$ be displacement, $m$ mass, $k$ the spring constant and $\zeta$ the damping coefficient. The second-order equation becomes a first-order state system by including velocity $v=\dot q$:

Two springs can have the same displacement and opposite velocities. They immediately move in different directions. Position alone loses predictive information. The pair $(q,v)$ restores it. Here $q$ is a mechanical position. Later, a species-indexed $q_X$ will again denote a molecular total.

For a stochastic model, the state determines a conditional probability law. Knowing a promoter is ON need not tell us exactly when it will switch OFF. It can tell us the probabilities of switching over future intervals. The state is sufficient if older history adds no predictive information once that present state and the external conditions are specified.

Molecular configurations give us useful discrete states. A promoter can be free or repressor-bound. An enzyme can be empty, substrate-bound or product-bound. A channel can occupy open and closed conformations. These are choices of model resolution. They work when unresolved motions can be represented by the proposed transition rules. Discreteness alone does not establish that the chosen states contain all relevant memory.

Test a proposed state by asking whether two systems with that same state can have different future laws because of omitted history. If they can, add the missing variables or use a model with memory.

3Molecular states jump with conditional rates

A continuous-time Markov chain describes a system that occupies a discrete state and makes random transitions whose hazards depend on its present state.

The ON/OFF promoter makes a transition hazard concrete. Write its random state as $\xi(t)$, encoding OFF as zero and ON as one. Let $a>0$ be the OFF-to-ON hazard and $b>0$ the ON-to-OFF hazard. In a short interval $\Delta t$, conditional on being OFF, the probability of turning ON is

The remainder $o(\Delta t)$ becomes negligible compared with $\Delta t$ as the interval shrinks. The hazard has units of inverse time. It is not a probability by itself. A large hazard means a short typical wait in the source state.

Markov and time-homogeneous are separate assumptions. The Markov property says the present state screens off the earlier history. Time homogeneity says the transition rules do not also depend explicitly on the clock time. Here we hold the surroundings fixed and use constant hazards within each state. A changing input could produce a time-dependent Markov model, or require the input to be included in the state.

The addition/removal model is another Markov state space. For composite production and first-order removal, use the chemical accounting $\varnothing\rightsquigarrow X$ and $X\rightsquigarrow\varnothing$. Their count propensities are $b_0$ and $dn_X$, respectively. The state is the integer count $N_X(t)$, not the chemical species symbol $X$. Its possible values form the chain $0,1,2,\ldots$. This nearest-neighbour count process is called a birth-death process.

The count and concentration laws use different rate units. Here $b_0$ is an event frequency, producing one molecule per event, while $d$ is an inverse-time constant multiplying the eligible molecule count. At fixed volume, Lecture 5's conversion $\Omega=N_A V$ gives concentration input $b_0/\Omega$. The count mean obeys $d\langle N_X\rangle/dt=b_0-d\langle N_X\rangle$. No concentration flux should be inserted directly as a transition hazard without this conversion.

Turn a molecular configuration or count into a Markov model by specifying its conditional event hazards, their units and the external conditions held fixed.

4Stationarity balances each node

The generator collects the transition hazards into one probability equation. A stationary solution balances arrivals and departures at each state.

Build the ON/OFF equation by counting probability flow. Let $p_{\mathrm{off}}(t)$ and $p_{\mathrm{on}}(t)$ be state probabilities, in that order. The ON probability gains $a p_{\mathrm{off}}$ per unit time and loses $b p_{\mathrm{on}}$. The OFF probability has the opposite balance:

Use the same convention for a general finite state space. The entry $q_{ij}$ is the hazard from state $j$ to state $i$, for distinct states. A negative diagonal records the total exit hazard. With probabilities in a column,

The indices are destination, source. In particular, $\sum_iq_{ij}=0$ for each source $j$. If $1$ denotes the column of ones, then $1^\top Q=0$ and $d(1^\top p)/dt=0$. A doubly indexed $q_{ij}$ is a generator entry. A species-indexed $q_X$ remains a concentration total.

This is the chemical master equation on a state graph. A reaction with count change $\gamma_r$ and propensity $a_r(n)$ adds that hazard to the entry from count state $n$ to $n+\gamma_r$. Distinct reaction channels can contribute to the same matrix entry. Lecture 3's $\dot x=\Gamma v(x)$ balances concentrations across reactions. Here $\dot p=Qp$ balances probabilities across states. The two equations describe different quantities.

At stationarity the probability vector stops changing. Write $p^*$ for a nonnegative, normalized stationary distribution:

A finite irreducible chain has one such distribution. Irreducible means every state can reach every other through positive-rate paths. For an infinite count space, existence and normalization need separate checks. Stationarity says nothing yet about balancing each forward transition against its reverse.

The two-state example can be solved without matrix algebra. Setting $a p_{\mathrm{off}}^*=b p_{\mathrm{on}}^*$ and using normalization gives $p_{\mathrm{on}}^*=a/(a+b)$. Substituting $p_{\mathrm{off}}=1-p_{\mathrm{on}}$ into the time-dependent equation also gives

Thus the stationary probability depends on a ratio of hazards, while the relaxation time is $1/(a+b)$. Multiplying both hazards by ten preserves the occupancy but accelerates the dynamics tenfold. We will need exactly this distinction when equilibrium weights replace rate constants.

Construct a column generator and solve $Qp^*=0$ with normalization. Interpret that solution as node balance, not as absence of transitions or proof of thermodynamic equilibrium.

5Fixed occupancies can hide an enzyme current

A stationary system can keep moving around a cycle. An enzyme processing maintained substrate and product reservoirs gives that probability current a physical meaning.

Compare two rings with the same three states. In both rings, counterclockwise hazards are one per second. Set clockwise hazards to one per second in the first ring and two per second in the second. Rotation symmetry gives $p_1^*=p_2^*=p_3^*=1/3$ in both. Direct substitution into the node balances verifies that result.

Directional traffic distinguishes the rings. Define one-way probability flux $\phi_{i\to j}=q_{ji}p_i$ and net edge current $J_{i\to j}$ by subtracting its reverse:

The balanced ring has $J=(1)(1/3)-(1)(1/3)=0$ on every edge. The driven ring has $J=(2)(1/3)-(1)(1/3)=1/3\,\mathrm{s}^{-1}$ clockwise on every edge. At each node, one net current enters and the same current leaves. The occupancy stays fixed.

In a single ring, stationarity makes the oriented edge currents equal. For the orientation $1\to2\to3\to1$, node 2 has $\dot p_2=J_{1\to2}-J_{2\to3}$. Setting that derivative and the other node derivatives to zero gives $J_{1\to2}=J_{2\to3}=J_{3\to1}$. Call the common value $J_{\mathrm{cycle}}$. Adding the three values would count transitions along the cycle, not three independent cycles.

Now resolve a familiar enzyme into three molecular states. The enzyme binds substrate, catalyzes a reversible conversion while occupied, and releases product. Treat the following as resolved elementary steps of the model:

A bare species symbol denotes its free concentration. A complex is named by its constituents. Thus $C_{ES}$ is substrate-bound enzyme, while $ES$ is the product of two free concentrations. The enzyme total is $q_E=E+C_{ES}+C_{EP}$.

Maintained free substrate and product turn this into a linear state model for one enzyme. In state order $(E,C_{ES},C_{EP})$, the clockwise hazards are $k_{+S}S$, $k_{\mathrm{cat}}^+$ and $k_{-P}$. The reverse hazards are $k_{-S}$, $k_{\mathrm{cat}}^-$ and $k_{+P}P$. A bimolecular binding constant has concentration-inverse time-inverse units. Multiplication by the fixed free ligand produces the required inverse-time hazard.

The reservoirs provide the free energy for sustained conversion. Define the dimensionless cycle affinity as the logarithm of the clockwise/reverse hazard-product ratio:

For a thermodynamically consistent enzyme coupled only to this substrate-to-product conversion, $\mathcal A=(\mu_S-\mu_P)/(k_BT)$. Here $\mu_S,\mu_P$ are chemical potentials per molecule, $k_B$ is Boltzmann's constant and $T$ is absolute temperature. Internal enzyme free-energy changes cancel around a complete turn. The net reservoir change remains.

A finite substrate stock can drive a transient, but maintained throughput needs maintained conditions. As a closed stock converts, substrate falls and product accumulates. The chemical driving changes. Maintaining a nonzero cycle current indefinitely instead requires continued supply and removal, or another specified source of free energy. The picture is like flowing water turning a mill: the rotating wheel is not itself the source of the flow.

The count and concentration readings agree after multiplication by the enzyme total. For an ensemble of equivalent enzymes under the same fixed reservoirs, the state probabilities equal the corresponding enzyme fractions. The net concentration throughput is $q_EJ_{\mathrm{cycle}}$. The current alone has units of inverse time. Multiplying by $q_E$ gives concentration per time, the flux used in Lecture 3. A driven cycle dissipates free energy even when no useful external work is extracted.¹

Look for stationary currents as well as occupancies. For an enzyme, identify the reservoir conversion behind the cycle and convert its per-enzyme current to a concentration flux with $q_E$.

6Detailed balance cancels each edge current

Detailed balance requires every resolved transition to have the same probability flux as its physical reverse. It is stronger than stationarity.

Balance each pair before summing at a node. Write $p^{\mathrm{eq}}$ for an equilibrium distribution. Detailed balance requires

Each net current in (8) is then zero. Summing those pairwise cancellations at any node gives $Qp^{\mathrm{eq}}=0$. Thus detailed balance implies stationarity. The right-hand ring in Figure 5 supplies the counterexample to the converse.

Equilibrium does not stop the microscopic jumps. In the balanced ring, each one-way edge flux is still $1/3\,\mathrm{s}^{-1}$. Forward and reverse transitions continue. Detailed balance removes their directional imbalance, not their traffic. A single trajectory at equilibrium is not a molecule frozen in one state.

The three-state cycle supplies a rate-only equilibrium test. Rearrange (11) to express each neighbouring probability ratio. Multiplication around the cycle makes every probability appear once above and once below:

The forward and reverse rate products must match. For the balanced ring the ratio is one. For the driven ring it is $2^3/1^3=8$, so its affinity is $\ln8$. No equilibrium distribution can balance all three of those driven edge pairs simultaneously. The problem is not failure to guess the right probabilities.

Keep the earlier use of “equilibrium” in view. In dynamical-systems language, an ODE equilibrium is simply a fixed point. Lecture 4's stability test asks whether nearby trajectories return to it. Here thermodynamic equilibrium imposes physical balance on resolved transitions. Neither a stable fixed point nor a stationary probability vector proves that stronger property.

Test a proposed equilibrium by comparing each forward/reverse flux or the rate products around cycles. State which physical channels have been resolved before drawing a thermodynamic conclusion.

7Equilibrium weights are path-independent

Balanced cycle products let us assign one consistent weight to every state. The occupancy calculation no longer needs every individual kinetic rate.

A probability ratio can be built along any path. Detailed balance gives $p_j^{\mathrm{eq}}/p_i^{\mathrm{eq}}=q_{ji}/q_{ij}$. Suppose the ratio from state 1 to 2 is two, and from 2 to 3 is two. Then the indirect path assigns state 3 four times the weight of state 1. A direct edge from 1 to 3 must give the same ratio.

Compatible cycles make the assignment independent of the chosen route. Select state 1 as a root with weight one. Multiply rate ratios along a path to each other state. If two paths gave different answers, following one forward and the other backward would give a cycle-product ratio different from one. Therefore the cycle criterion is exactly what makes this construction consistent.

Logarithms turn multiplicative weights into additive state potentials. Define $\beta=1/(k_BT)$. Assign a potential $G_i$ so the ratio across each edge is

The constant $Z$, called the partition function, normalizes the weights. Adding the same constant to every $G_i$ changes neither ratios nor probabilities. In the compatible example, choose $G_1=0$. Then $G_2=-k_BT\ln2$ and $G_3=-k_BT\ln4$.

The weight construction proves the equivalence of the first three statements. Pairwise balance implies the cycle criterion by (12). The cycle criterion makes the root-to-state construction path-independent. Normalizing those weights gives a positive distribution that balances every edge and therefore is stationary. No general stationary linear solve was required.¹

Time reversal states the same balance as a movie test. At equilibrium, a stationary movie and its backward replay have the same probability law. Individual enzymes still jump. But neither a jump nor an entire sequence of jumps has a statistically preferred direction. A driven enzyme wheel does: over a long movie, more complete conversions run from substrate to product than backward.

A state potential needs a physical interpretation before it is a thermodynamic free energy. Equation (13) first establishes a mathematical compatibility property. In a molecular model, $G_i$ includes the chosen reservoir contributions and any internal multiplicities absorbed into the state description. The kinetic ratios must describe physical reverse channels with compatible energy exchange. Merely defining $G_i=-k_BT\ln p_i^*$ for an arbitrary stationary vector does not pass the edge-ratio test. The driven ring has a perfectly positive stationary vector and fails it.

The simplification concerns occupancies, not every dynamical question. Multiplying every hazard by the same factor leaves all ratios and equilibrium probabilities unchanged. Relaxation and first-passage times change. We can dispense with many kinetic constants when predicting equilibrium allocation, but not when predicting speed. The time-reversal statement here concerns molecular configurations and counts. A mechanical state containing velocity requires velocity reversal as well.

Construct equilibrium weights from compatible edge ratios, then normalize. Use the resulting probabilities for allocation while retaining kinetic information for response times and checking the physical meaning of the potential.

8State weights predict repression

A two-state gene recovers the familiar repression law by counting configurations and assigning activities. The concentration factor comes from molecular multiplicity.

Specify the subsystem before assigning its weights. Consider one gene that is free, $G$, or bound by a repressor, $C_{GR}$. Hold free repressor concentration $R$ fixed. This is justified by a large reservoir or negligible depletion from binding the gene. Assume the binding configurations equilibrate rapidly relative to the readout and that their resolved transitions are not themselves fuel-driven.

Binding competes with the number of ways a repressor can remain free. Imagine $N$ indistinguishable repressors distributed among $M$ background lattice sites, with at most one per site. If the gene is free, there are $\binom{M}{N}$ arrangements. If one repressor is bound to the gene, the remaining repressors have $\binom{M}{N-1}$ arrangements. Thus

Binding may be energetically favourable, but it removes the freedom to occupy many solution locations. More free repressors increase the number of bound arrangements relative to free-gene arrangements. This is why the binding weight depends on concentration, not only on one binding-energy value.

The lattice count becomes a concentration ratio with an explicit volume conversion. Let each background lattice cell have volume $v_{\mathrm{site}}$, so $M=V/v_{\mathrm{site}}$ in solution volume $V$. In the free-gene state, molar repressor concentration is $R=N/(N_A V)$. The corresponding standard concentration $c^\circ=1/(N_Av_{\mathrm{site}})$ then gives

The reservoir assumption also requires that binding one repressor does not appreciably change the free pool. In this counting picture, that means many available repressors. The lattice cell is a bookkeeping reference, not a measured physical compartment inside the cell.

A standard binding free energy packages the remaining state preference into a measurable affinity. Let $\Delta G_{\mathrm{bind}}^\circ$ be the binding free energy associated with the chosen standard concentration. Multiplying the concentration factor by the binding Boltzmann factor gives

The free energy sets the measurable binding affinity. A more favourable binding free energy lowers $K_d$. The concentration factor is dimensionless. It also follows from the dilute-solution chemical potential, whose concentration-dependent term is $k_BT\ln(R/c^\circ)$.⁴

Bookkeeping conventions cannot change the physical weight ratio. Changing the standard concentration changes the numerical standard free energy consistently, leaving $K_d$ unchanged. Multiplicity can be counted explicitly or absorbed into a coarse-state free energy, but not counted twice.

Normalize the two weights to obtain the occupancy. Choose the free-state weight as one. Then the state sum and bound probability are

At $R=K_d$, the states have equal weights and the bound probability is one-half. At low free repressor the bound fraction is approximately $R/K_d$. At high free repressor it approaches one. This is the Langmuir binding isotherm. Its algebraic resemblance to Michaelis–Menten does not identify binding $K_d$ with catalytic-QSSA $K_M$.

Occupancy becomes regulation only after we assign activities. Suppose the free gene initiates transcription at frequency $r_0$, while the bound gene is silent. The mean initiation frequency per gene is the state-probability-weighted activity:

We have recovered the effective repression law from Lecture 3. We specified allowed states, a binding free energy and state activities. We did not specify every microscopic association pathway or solve its transient kinetics. A leaky bound state would instead contribute its nonzero activity times $p_{\mathrm{bound}}$.

Keep the probability and concentration interpretations distinct. One gene is either bound or not bound at a given moment. In an ensemble of equivalent genes under the same reservoir conditions, the expected bound fraction is $p_{\mathrm{bound}}$. In the deterministic ensemble description, $q_G=G+C_{GR}$ and $C_{GR}/q_G=p_{\mathrm{bound}}$. Multiplying (17) by $q_G$ gives a concentration production flux. Subsequent RNA and protein removal still affect the measured expression level.

Derive a repression law by enumerating states, retaining the concentration factor in their weights, and averaging declared state activities. Identify free concentration and the readout assumptions before comparing the result with expression data.

9Allostery shifts the active population

The same states-and-weights method handles a protein whose ligand-binding affinity depends on its conformation. Binding can regulate catalysis by shifting the probability of an active state.

Choose a two-site protein with two concerted conformations. Call them active $A$ and inactive $I$. In the Monod–Wyman–Changeux construction, the whole protein changes conformation together. Within a fixed conformation, the two sites are equivalent and bind independently. Two sites have four occupancy patterns: empty, first bound, second bound and both bound. With two conformations, there are eight molecular states.⁵

Each conformation has its own affinity and unliganded weight. Let $c$ denote free ligand concentration. Let $K_A$ and $K_I$ be dissociation constants within the active and inactive conformations. Choose unliganded active weight one and unliganded inactive weight $L=e^{-\beta(G_I^0-G_A^0)}$, where the superscript zero denotes the unliganded state. Summing the four weights in each row of Figure 9 gives

The factor two counts the two distinct single-bound patterns. There is no direct interaction between sites within one conformation. Their occupancies become coupled in the full ensemble because both sites favour the same protein-wide conformational shift.

Preferential ligand binding can overcome an inactive conformational bias. For a simple numerical calculation, set $L=100$, $K_A=1\,\mathrm{nM}$ and $K_I=100\,\mathrm{nM}$. With no ligand, the active probability is $1/(1+100)=1/101$. At $c=10\,\mathrm{nM}$, the active weight is $(1+10)^2=121$, while the inactive weight is $100(1+0.1)^2=121$. Half the population is active. The plot uses $L=10$ instead, so the difference between activity and binding is easy to see. At $c=1\,\mathrm{nM}$ in that plot, about 28% of proteins are active but only 15% of sites are occupied.

The high-ligand limit need not be fully active. At large $c$, the leading quadratic terms give $p_A\to1/[1+L(K_A/K_I)^2]=100/101$ for these choices. A ligand binding more strongly to the inactive conformation would instead favour inactivity. The allosteric response reflects relative affinities and conformational weights, not merely the existence of two sites.

An active fraction is useful only for a stated activity readout. If an active protein catalyzes a slow reaction at frequency $r_A$ and an inactive one has frequency $r_I$, the mean frequency is $\bar r=r_Ap_A+r_I(1-p_A)$. This assumes the readout does not invalidate the fast equilibrium allocation. The response can resemble a Hill curve, but it is not generally an exact Hill law, and a fitted exponent need not equal the number of sites.

Build a richer regulatory law by listing joint molecular states and summing their weights. Keep ligand occupancy, conformational activity and catalytic output as separate readouts of that same distribution.

10Equilibrate the fast layer, evolve the totals

The equilibrium shortcut applies to a selected fast subsystem. Its changing conditions and slow activities still describe a driven biological system.

Return to the enzyme wheel and separate its clocks. If binding relaxes much faster than catalysis and supply, the leading fast problem contains the two binding pairs but not the slow catalytic edge. At fixed free substrate and product, that binding state graph is a tree. Its equilibrium ratios give relative enzyme weights $1$, $S/K_{d,S}$ and $P/K_{d,P}$, where $K_{d,S}=k_{-S}/k_{+S}$ and $K_{d,P}=k_{-P}/k_{+P}$.

Weighting those enzyme states by forward and reverse catalytic frequencies gives the signed net flux

Equation (20) is a rapid-binding approximation expressed in free concentrations. Its net flux can be nonzero. The fast allocation is equilibrated to leading order, while slow catalysis changes the surrounding composition. Small departures from limiting binding balance carry the finite throughput, just as Lecture 6's moving slow manifold has nonzero motion on the slow clock.

Finite pools require conservation alongside the weights. The totals $q_S=S+C_{ES}$ and $q_P=P+C_{EP}$ are conserved by binding, not by catalysis. With substrate input $v_{\mathrm{in}}$ and product output $v_{\mathrm{out}}$, all measured as concentration fluxes in a fixed volume, their balances are

These total balances are exact for the declared mechanism and boundary processes. The equilibrium constraints used to reconstruct the complexes are approximations. The free concentrations in (19) must be solved together with the total relations when binding appreciably depletes the pools. Equilibrium structure simplifies the closure without promising that every finite-pool calculation has a short explicit formula.

The modeller now has two distinct closure choices. A fast attracting subsystem can justify a stationary kinetic calculation. A fast subsystem with compatible equilibrium structure can instead use physical state weights. In either case, the chosen states and reservoirs must be specified, and state-dependent activities convert the fast distribution into slow output. A fast driven promoter that fails detailed balance needs its stationary currents retained rather than hidden inside equilibrium weights.

Use equilibrium to allocate a justified fast layer, then use its activities to evolve the totals. Keep conservation exact, label the equilibrium approximation, and leave driven channels outside it unless their stationary kinetics are explicitly retained.

Epilogue: equilibrium inside a living cell

Optional reading, outside the lecture clock. If a living cell must consume free energy, why have we spent a lecture on equilibrium?

A claim about a part is not a claim about the whole cell. A growing bacterium imports nutrients, exports waste and maintains chemical differences that would disappear if metabolism stopped. Its overall state is not thermodynamic equilibrium. Yet a repressor can bind and unbind many times before its gene makes another transcript. That fast occupancy can be close to equilibrium even while transcription, translation and growth consume free energy.²

Draw the boundary around the process you want to approximate. Inside a passive promoter-binding model are the free and bound gene states. Outside are the maintained repressor pool and the machinery that reads the promoter. The binding calculation predicts which fraction of the time the promoter is available. It does not claim that making RNA is reversible, or that maintaining the repressor pool is free. Slow changes in that pool move the equilibrium allocation that binding tracks.

The enzyme makes the boundary especially clear. Remove the slow catalytic edge from Figure 6. The fast binding graph has two branches, one for substrate and one for product. Each branch can balance with its own reservoir. Restore catalysis and those branches close a cycle. Now one completed turn converts substrate to product. Maintained substrate and product chemical potentials can drive that turn even though each binding branch remains very close to balance.

Close to balance does not mean exactly zero throughput. With very fast binding, large forward and reverse binding traffic differ by a small fraction. Their difference carries the same finite current as the slower catalytic step. The leading equilibrium weights discard that small relative imbalance, not the slow conversion. Increasing the binding speed in the playground makes the exact occupancies approach those weights while product formation continues.

A fast subsystem can also fail the equilibrium test. An ATP-consuming remodeler inside the promoter boundary may sustain a cycle. Accelerating all its transitions makes it relax faster but does not remove its chemical drive. Its stationary kinetics, rather than equilibrium weights, determine its occupancy. The distinction is physical: inspect the reactions and reservoirs inside the boundary, not only the timescale.

Equilibrium is useful because it can describe allocation without describing every motion. We retain the states, free energies, reservoir conditions and state activities that determine the observable. We set aside fast kinetic details only where the approximation permits it. A driven cell can therefore contain locally equilibrated binding layers whose changing activities organize its much slower dynamics. That is the modelling opportunity carried into Lecture 8.

Questions to develop further

The classroom story ends at a modelling capability. The following questions use that capability, but are not additional points on the 95-minute route. They can become focused student contributions or companion calculations.

• When can a static response reveal driving? For a specified four-state promoter with at most one transcription factor per state, equilibrium occupancy can be fractional-linear in free factor concentration and therefore monotone. Examine which kinetic changes permit an interior maximum, and why the conclusion depends on the chosen state model.
• What do finite molecule pools change? Replace the fixed repressor reservoir by conserved copy numbers. Derive the binding distribution and ask when the deterministic equilibrium fraction is an adequate approximation.
• What can energy buy beyond an occupancy curve? Compare systems with the same equilibrium probabilities but different relaxation times. Then examine driven proofreading, switching or adaptation with a stated output, speed and fuel budget.
• What does a receptor trajectory tell us about concentration? Choose an estimator and count the independent information in binding and unbinding events. A sensing limit is a statement about a measurement protocol as well as a physical receptor.

The exposition develops the calculations, the lineage traces their conceptual and experimental origins, and the worked extension compares equilibrium and driven promoter models. These companion chapters are released separately. They are not prerequisites for following the core.

Playground: an enzyme's current and energy budget

Watch a glycolytic conversion through the enzyme's states. Triosephosphate isomerase interconverts dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAP). GAP feeds the downstream reactions of glycolysis. The three-state model below uses $S=\mathrm{DHAP}$ and $P=\mathrm{GAP}$. It is a teaching model of binding, conversion and release, not a fitted microscopic mechanism of this enzyme.

The energy benchmark comes from a pathway calculation. Flamholz and colleagues found an optimized driving force of about $2.9\,\mathrm{kJ/mol}$ for bottleneck reactions of the classical glycolytic pathway, including this isomerase. Their calculation constrained metabolite concentrations to $1\,\mu\mathrm M$–$10\,\mathrm{mM}$, at pH 7.5 and ionic strength 0.2 M. This is a model-derived pathway benchmark, not a measured universal intracellular value. We use it as the default free-energy change, $\Delta_rG=-2.9\,\mathrm{kJ/mol}$. The kinetic constants and enzyme concentration below are illustrative.⁶

The complete cycle's energy constrains its rate ratios. For fixed free substrate and product, define the dimensionless cycle affinity $\mathcal A=-\Delta_rG/(RT)$. Let $J$ be signed net conversions per enzyme per second. Under local detailed balance, with no additional fuel channel or useful-work load, the stationary entropy production and dissipated power per enzyme are

At the teaching temperature $T=298.15\,\mathrm K$, $RT=2.479\,\mathrm{kJ/mol}$. The benchmark gives $\mathcal A=1.170$, or $1.170\,k_BT$ per net conversion. Multiplying by the actual current gives a power, not another energy. At equilibrium the net current vanishes although opposing traffic continues. With reversed drive both $J$ and $\mathcal A$ change sign, so entropy production remains nonnegative.