Constructing and analyzing a binding-catalysis model

Every derivation behind the core's three tools, and the five-step pipeline run on every example.CCBS · Lecture 8 · exposition · lecturer-written · revised 24 September 2026 This chapter is the reader's companion to the core. The core states three tools: the binding-catalysis model, leading-order dominance, and regulation read from the regime map. Here every step behind them is shown, with units, proofs and worked answers. The aim is transfer. A reader who finishes should be able to take an unfamiliar complex, write its four lines, map its regimes and read its responses. The chapter follows the core's four parts. Part 1 makes the core's two observations exact. Part 2 builds the model and proves that its binding layer has one solution. Part 3 turns dominance into a procedure and runs it on one binding pair. Part 4 justifies the core's five-step pipeline and then runs it on every regulatory example, each ending in a regime-sequence table. An optional part runs the pipeline on further mechanisms. The lineage traces where these ideas came from.

Conventions"Core §N" always means a section of the core. A bare §N means a section of this page. Concentrations are pure numbers in one canonical unit,

1\,\mu\mathrm M

, as in core §4, so no bar or reference concentration is carried. A regime is named by its winner tuple, never by a letter. Every curve is computed from stated illustrative parameters. None is a measurement.

How this page follows the core

Core section	What the core states	Where this page derives it
§1	the model-class question	the lineage
§2	binding moves constituents, catalysis changes totals	§1
§3	separated clocks put binding at equilibrium	§2
§4	the four-line model	§3, with every matrix of an enzyme in §9
§5	the counting matrix and its graph	§4
§6	one constant per complex, and one allocation	§5 to §9, with the uniqueness proof in §6
§7	select, solve, check	§11
§8	one pair: three sectors and a ray	§10 and §12 to §14
§9	the five-step pipeline	§15 and §16
§10	four responses along two paths	§17 to §19
§11	a competitor adds a candidate	§20
§12	joint states decide combination	§21 and §22
Coda	close the dynamics	§23 to §25
A1	the winner sets the Hill degree	§26
A2	a sponge shifts the input	§27
A3	one extra state makes a window	§28 to §30
beyond the core	series binding, modification cycles, transfer	§31 to §33

Part 1

A model class for biology

Core §2 and §3 narrow the class of cell models with two observations. This part makes both exact: binding conserves every total, and binding settles on a clock of its own.

1An intervention changes a total, and binding only moves it between species

Begin with the intervention to be predicted. Supplying a protein changes its total concentration across all the forms that contain it. A reporter may respond to the free form, while a catalytic assay measures a flux from a complex. So the model's coordinates must connect the total that is controlled with the species that is read. Core §2 sorts every step by what it changes. This section makes that sorting exact for the smallest case.

The notation follows the core. A bare letter such as E or S is a free concentration. A named complex is written $C_{ES}$ . The total of the enzyme constituent is $q_E$ , the sum over every species that contains the enzyme. A homodimer contributes two units to its monomer's total, and a ternary complex contributes to all three totals it contains.

Units come next. A dissociation constant $K_d$ of a one-to-one pair is a concentration. An association rate constant such as $k_{+S}$ has units $\mu\mathrm M^{-1}\mathrm s^{-1}$ . A dissociation rate constant $k_{-S}$ and a unimolecular conversion have units $\mathrm s^{-1}$ . Each rate law, a rate constant times the right concentrations, then has units of concentration per time. Fixing the canonical unit of core §4 turns every concentration into a pure number, so logarithms are defined and logarithmic derivatives are dimensionless.

One binding pair shows the accounting. Order the species as $x=(E,S,C_{ES})$ . The association $E+S\xrightleftharpoons[k_{-S}]{k_{+S}}C_{ES}$ has the stoichiometric column $\Gamma_{\mathrm{bind}}$ , the change in each species per event. The counting matrix L records how many copies of each constituent each species holds:

\begin{gathered}\Gamma_{\mathrm{bind}}=\begin{pmatrix}-1\\-1\\1\end{pmatrix},\qquad L=\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix},\\ q=Lx=\begin{pmatrix}E+C_{ES}\\S+C_{ES}\end{pmatrix}.\end{gathered}

The product $L\Gamma_{\mathrm{bind}}=0$ says that one association event moves one enzyme and one substrate constituent into the complex and changes neither total. Dissociation flips the sign of the column and leaves the conclusion unchanged. Neither statement needs equilibrium, so the cancellation holds during a binding transient too.

Scroll inside the figure to see it at reading size.

Figure 1. Constituent accounting for one binding pair. Binding moves one enzyme and one substrate constituent into the complex, so

L\Gamma_{\mathrm{bind}}=0

and neither total changes. The cancellation is exact at every instant, not only at equilibrium.

The same cancellation holds for any model. Write the species rate equation as $\dot x=\Gamma_{\mathrm{bind}}v_{\mathrm{bind}}+\Gamma_{\mathrm{cat}}v_{\mathrm{cat}}+b(x,t)$ , as in core §4. Here $v_{\mathrm{bind}}$ and $v_{\mathrm{cat}}$ are the binding and catalytic fluxes, and $b(x,t)$ collects boundary processes such as supply, export and dilution. Any constant L with $L\Gamma_{\mathrm{bind}}=0$ removes the binding fluxes:

\dot q=L\Gamma_{\mathrm{cat}}v_{\mathrm{cat}}+L\,b(x,t).

Uniform dilution at rate λ is one such boundary process. It removes every species in proportion, so it contributes $-\lambda x$ to b and $-\lambda q$ to the totals.

The rows of L must be complete. They must include every independent total that fast binding conserves, or the totals cannot label the binding state. Elemental mass alone can be too coarse. Substrate and product forms are separately conserved by fast binding, even though slow catalysis converts one into the other. Conversely, a total need not be conserved by the whole model to be a useful slow coordinate. It only has to be conserved by binding.

Name the controlled total, the molecular readout and the units first, and then check that binding leaves every total unchanged.

2Affinity fixes where binding settles, not how fast

An equilibrium constant says where binding settles. It says nothing about how quickly binding gets there. This section separates the two, because core §3 needs both: equilibrium to write the binding layer algebraically, and a fast clock to justify doing so.

Equilibrium comes first. For an isolated elementary pair the net association flux is $k_{+S}ES-k_{-S}C_{ES}$ . Setting it to zero gives the binding relation, with the dissociation constant as the ratio of the two rate constants:

K_dC_{ES}=ES,\qquad K_d=\frac{k_{-S}}{k_{+S}}.

In a binding network with several reversible steps, every step balances at once, which is Lecture 7's detailed balance.

A catalytic escape changes the relation. For a substrate complex that also turns over at $k_{\mathrm{cat}}$ , the complex obeys $\dot C_{ES}=k_{+S}ES-(k_{-S}+k_{\mathrm{cat}})C_{ES}$ . Setting this derivative to zero gives $K_MC_{ES}=ES$ with $K_M=(k_{-S}+k_{\mathrm{cat}})/k_{+S}$ , Briggs and Haldane's constant.¹ A small derivative and an equilibrium ratio are different assumptions. They agree when $k_{\mathrm{cat}}\ll k_{-S}$ , the condition stated in the note that closes core §6.

Fast exchange need not mean equilibrium either. A strongly connected state graph with a driven cycle has a unique stationary distribution and nonzero currents around the cycle. Gunawardena's linear framework eliminates such a graph without detailed balance.² Lecture 7's compatibility audit decides which subgraph may be replaced by equilibrium weights, and core §3 assigns every driven step to catalysis.

Speed is a separate calculation. For one pair at fixed totals, substituting $E=q_E-C_{ES}$ and $S=q_S-C_{ES}$ gives one equation for the complex:

\dot C_{ES}=k_{+S}\bigl[(q_E-C_{ES})(q_S-C_{ES})-K_dC_{ES}\bigr].

Linearizing around the physical root gives the relaxation frequency $k_{+S}(E+S+K_d)$ , evaluated at that root. So the local binding time is

t_{\mathrm{bind}}=\frac{1}{k_{+S}(E+S+K_d)}.

Multiplying $k_{+S}$ and $k_{-S}$ by one common factor changes $t_{\mathrm{bind}}$ and leaves $K_d$ , L and the equilibrium untouched. The lecturer's thesis writes the binding time of a complex as $(k_{-S}+k_{\mathrm{cat}}+k_{+S}S)^{-1}$ . That is this expression when free enzyme is scarce, with $k_{\mathrm{cat}}$ added when the complex also turns over.³

Scroll inside the figure to see it at reading size.

Figure 2. Equal affinities, different binding clocks. Three exact transients of one closed pair, each starting with no complex. Illustrative parameters:

q_E=q_S=1\,\mu\mathrm M

and

K_d=0.3\,\mu\mathrm M

. Only the common factor

k_{+S}

changes, with

k_{-S}=K_dk_{+S}

. The final allocation is the same, and the relaxation times differ tenfold per step.

A fast clock is not automatic. The same thesis estimates that an association constant near $10^8\,\mathrm M^{-1}\mathrm s^{-1}$ with a 10 nM partner gives an association frequency of only about $1\,\mathrm s^{-1}$ .³ The physical reason for the usual separation is chemical. Binding makes and breaks weak bonds and relaxes fast, while catalysis rearranges covalent bonds and is slower.⁴ In metabolism, binding equilibrates within milliseconds to seconds, while metabolite levels change over tens of seconds.⁵

So compare the relaxation time with the clocks of the experiment. In a network with many complexes, use the slowest binding mode at fixed totals, not one chosen association rate. The reduction needs that mode to stay fast over every state the experiment visits. A start away from equilibrium adds a brief initial transient.

The reduced model describes the state after that transient. It still lets the complex change, because the equilibrium allocation follows the totals, $\dot x^{\mathrm{eq}}=D_qx^{\mathrm{eq}}\,\dot q$ (§7). At a finite binding speed, a small departure from equilibrium supplies the binding flux that this motion needs. The equilibrium equation therefore does not claim that the complex has stopped changing.

Check the binding relaxation against the experiment's slow clocks, separately from computing where binding settles.

Part 2

Tool 1: the binding-catalysis model

Core §§4 to 6 write the model in four lines and state that its binding layer has one solution. This part derives the four lines, builds each matrix, proves the uniqueness and computes the solution.

3Projecting the rate equation onto the totals gives four lines

The four lines of core §4 come from one projection and one closure. The projection is exact. The closure is the fast-equilibrium approximation of §2. This section derives both and fixes every dimension.

Start with the sizes. Let n be the number of species, d the number of independent totals, m the number of catalytic steps and p the number of external fluxes. Then the concentrations are $x\in\mathbb R_{>0}^n$ and the totals $q\in\mathbb R_{>0}^d$ . The counting matrix is $L\in\mathbb R^{d\times n}$ , the catalytic stoichiometry on species is $\Gamma_{\mathrm{cat}}\in\mathbb R^{n\times m}$ , and the activity matrix is $K^{\mathrm{cat}}\in\mathbb R_{\ge0}^{m\times n}$ .

The projection multiplies the species equation of §1 by L. Binding drops out, and what remains is catalysis and exchange on the totals:

\dot q=Sv+S^{\mathrm{ext}}v^{\mathrm{ext}},\qquad S=L\Gamma_{\mathrm{cat}},\qquad S^{\mathrm{ext}}v^{\mathrm{ext}}=L\,b(x,t).

Here v is the vector of catalytic fluxes, and $S\in\mathbb R^{d\times m}$ says how each catalytic step changes each total. The external fluxes $v^{\mathrm{ext}}\in\mathbb R^{p}$ are supply, export and dilution, and $S^{\mathrm{ext}}\in\mathbb R^{d\times p}$ says which totals each one changes. Writing exchange as a separate stoichiometric term follows the lecturer's metabolic work.⁵ This line is exact. It holds whether or not binding is at equilibrium.

The activity line reads each catalytic flux from the species that performs it. Row j of $K^{\mathrm{cat}}$ holds the catalytic constant of step j in the column of its enzyme state, so $v=K^{\mathrm{cat}}x$ . The entries are frequencies, so v has units of concentration per time. For a resolved first-order catalytic step this law is exact. For a composite production process, such as transcription from a promoter state, one frequency per state is an assumption. A time-resolved question may need more internal states.

The closure replaces binding dynamics by binding equilibrium. It adds one logarithmic relation per complex, collected as $N\log x=\log k$ , where N has $n-d$ rows and k holds the equilibrium constants (§5). With the conservation line, the model is

\begin{aligned} q&=Lx &&\text{conservation}\\ N\log x&=\log k &&\text{equilibrium}\\ v&=K^{\mathrm{cat}}x &&\text{activity}\\ \dot q&=Sv+S^{\mathrm{ext}}v^{\mathrm{ext}} &&\text{accumulation.} \end{aligned}

The lecturer's flux-species sampling paper writes the same four lines, without the external term.⁶ Conservation and accumulation are exact accounting. Equilibrium is the closure, and activity is exact only for resolved first-order steps.

Scroll inside the figure to see it at reading size.

Figure 3. The binding-catalysis model links two operations. At each slow time, solve conservation and equilibrium for the species, read the fluxes from the active states, and update the totals. The next allocation uses the updated totals. The arrows connect calculations. They are not reaction steps.

Checkpoint: dimensions and units

A model has five species, three totals and two catalytic steps. Find the size of L, N, $\Gamma_{\mathrm{cat}}$ , S and $K^{\mathrm{cat}}$ . Why does $K^{\mathrm{cat}}$ carry units of inverse time while S is a pure count?

Worked answer

L is 3×5 and N is 2×5, since $n-d=2$ . $\Gamma_{\mathrm{cat}}$ is 5×2, S is 3×2 and $K^{\mathrm{cat}}$ is 2×5. The product $SK^{\mathrm{cat}}x$ maps five concentrations to two fluxes and then to three rates of change. Each entry of S counts a change per event. Each entry of $K^{\mathrm{cat}}$ sets how often one molecule of the acting state performs the event.

Project the species equation with L to get the accumulation line, and keep track of which of the four lines is exact accounting and which is the fast-equilibrium closure.

4The complex list fixes the counting matrix

The first line needs L, and L is fixed by the complex list. This section builds L column by column, as core §5 does, and adds the rank condition that makes it complete.

Order the species with the d free atomic species first and the r other species after them, so $n=d+r$ . "Atomic" means a reference constituent of the binding model, such as an enzyme, a substrate or a protein monomer. It need not be a chemical element. The entry $L_{ij}$ counts the copies of atomic species i in species j, so

q=Lx,\qquad L=[\,I_d\mid L^{(c)}\,],\qquad q_i=x_i^{(a)}+\sum_{j=1}^{r}L^{(c)}_{ij}x_j^{(c)}.

Here $x^{(a)}$ are the free atomic species, $x^{(c)}$ the other species, and $L^{(c)}$ their compositions.³ A complex $C_{ES}$ contributes once to each of two totals. A dimer $C_{RR}$ contributes twice to the monomer total. An isomer $E^\ast$ contributes once to the enzyme total, exactly as E does.

The same matrix is a graph. Draw one square per total and one circle per species, and join total i to species j by an edge of weight $L_{ij}$ when it is positive. This bipartite composition graph has L as its bi-adjacency matrix.⁷ Its edges mean "contains". They prescribe no assembly pathway.

Scroll inside the figure to see it at reading size.

Figure 4. Count multiplicities and keep distinct conformations. A weighted edge counts copies, so a dimer contributes two to its total. An isomer repeats the composition column of its reference state, yet keeps its own affinity and activity. Both drawings are composition graphs, not reaction graphs.

Two definitions name what L may contain. The thesis calls L isomer-atomic when it has the identity block, nonnegative integer entries and no zero column. Repeated columns are then allowed.³ The author manuscript calls L faithful-atomic when its columns are also distinct.⁷ Repeated columns are what let two conformations carry the same material and still differ in activity. These are definitions of the present formulation, not terms from the original enzyme papers.

L must also be complete. If $\Gamma_{\mathrm{bind}}$ holds the binding columns, require

L\Gamma_{\mathrm{bind}}=0,\qquad \operatorname{rank}\Gamma_{\mathrm{bind}}=n-d.

The first condition says that binding conserves every row. The second says that the rows capture every conservation law of binding. Counting elemental mass alone can miss an independent protein-form total, and a missing total leaves the allocation underdetermined.

Build L column by column from each complex's composition, keep multiplicities and distinct conformations, and check that its rows span every conservation law of binding.

5Equilibrium needs one constant per complex

Composition fixes the form of the equilibrium equations, and one constant per complex fixes the allocation. This section derives the monomials of core §6, their units, and the rule for changing the basis of N.

At equilibrium each non-free species is a monomial in the free atomic species. Its exponents are its composition column, and its coefficient is its association constant $b_j$ :

x_j^{(c)}=b_j\prod_{i=1}^{d}\bigl(x_i^{(a)}\bigr)^{L^{(c)}_{ij}}.

One constant suffices because equilibrium is path independent (Lecture 7). Every assembly route reaches the same weight, so the product of step constants along any route is the same $b_j$ .

The units follow from the count. Let $s_j=\sum_iL^{(c)}_{ij}$ be the number of constituents in species j. Then $b_j$ has units of concentration to the power $1-s_j$ . A binary complex has $s_j=2$ and $b_j=1/K_d$ , with $K_d$ a concentration. A three-constituent complex has a constant in inverse concentration squared. An isomer ratio is dimensionless. In the canonical unit each $b_j$ becomes a pure number, and these units say how it rescales if the unit changes. A catalytic $K_M$ is a different, kinetic quantity (§2), not another name for $K_d$ .

Taking logarithms turns the monomials into linear equations. Collect them in the association-affinity matrix $N^{\mathrm{aa}}$ :

N^{\mathrm{aa}}\log x=\log b,\qquad N^{\mathrm{aa}}=[\,-(L^{(c)})^{\mathsf T}\mid I_r\,].

Row j subtracts the logarithms of the constituents of species j and adds the logarithm of the species. The matrices L and $N^{\mathrm{aa}}$ have ranks d and r, which sum to n, and $N^{\mathrm{aa}}L^{\mathsf T}=0$ . Such a pair is a Gale pair, also called a pair of dual configurations.³ L counts what binding preserves. N states how equilibrium shares it out.

Any full-rank N with $NL^{\mathsf T}=0$ describes the same states, provided its constants transform with it. Split $N=[\,N_1\mid N_2\,]$ into its first d columns and its last r. Orthogonality gives $N_1=-N_2(L^{(c)})^{\mathsf T}$ , and full rank makes the square block $N_2$ invertible. So

N=N_2N^{\mathrm{aa}},\qquad (L^{(c)})^{\mathsf T}=-N_2^{-1}N_1,\qquad \log k=N_2\log b.

The equilibrium set $\{x>0:N\log x=\log k\}$ is unchanged by this change of basis. Changing N while keeping $\log k$ fixed would change the physical model.

One pair in two conventions

For $x=(E,S,C_{ES})$ , the association row is $N^{\mathrm{aa}}=(-1,-1,1)$ with $b=1/K_d$ . Reversing it gives the dissociation row $N=(1,1,-1)$ with $k=K_d$ . Both equations say $ES=K_dC_{ES}$ . The thesis writes N in the dissociation direction, so its N and $\log k$ carry the opposite sign.³

Worked change of coordinates

Take the promoter of core A3. A regulator R binds gene G twice in sequence, $G+R\xrightleftharpoons[k_{-1}]{k_{+1}}C_{GR}$ and $C_{GR}+R\xrightleftharpoons[k_{-2}]{k_{+2}}C_{GRR}$ , with stepwise dissociation constants $K_1=k_{-1}/k_{+1}$ and $K_2=k_{-2}/k_{+2}$ . In the order $x=(G,R,C_{GR},C_{GRR})$ , counting gives

L=\begin{pmatrix}1&0&1&1\\0&1&1&2\end{pmatrix},\qquad N^{\mathrm{aa}}=\begin{pmatrix}-1&-1&1&0\\-1&-2&0&1\end{pmatrix},\qquad b=\begin{pmatrix}1/K_1\\1/(K_1K_2)\end{pmatrix}.

The second constant has units of inverse concentration squared, because $C_{GRR}$ holds three constituents. The pathway basis instead uses the two binary steps. Its matrix and constants are

N_{\mathrm{path}}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}N^{\mathrm{aa}}=\begin{pmatrix}-1&-1&1&0\\0&-1&-1&1\end{pmatrix},\qquad \log k_{\mathrm{path}}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}\log b=\begin{pmatrix}-\log K_1\\-\log K_2\end{pmatrix}.

The second row subtracts the weight of the binary complex from that of the ternary one. Exponentiating it gives $C_{GRR}=C_{GR}R/K_2$ . A canonical row states an equilibrium relation. It does not assert a simultaneous three-molecule encounter.

The economy of the equilibrium model is now precise. The complex list, the compositions and one constant per complex fix the allocation. A kinetic model would also need the binding pathways and their rates, so the same equilibrium can carry different transients (§2).

Write one monomial per complex from its composition column, and transform the constants together with N whenever the basis changes.

6A strictly convex potential makes the allocation unique

To use the totals as the state, every positive total must have exactly one positive allocation. Core §6 states this. This section proves it in two parts: uniqueness from strict convexity, and existence from a lower bound on the same potential.

Set up the hypotheses. The d free atomic species are present, so $L=[\,I_d\mid L^{(c)}\,]$ , and every other column has nonnegative counts. Every concentration is a pure number in the canonical unit. Extend the association constants to the free columns by $b_i=1$ . Then equilibrium and conservation read

x_j=b_j\exp\bigl(L_j^{\mathsf T}z\bigr),\qquad q=Lx,\qquad z_i=\log x_i^{(a)},

where $L_j$ is column j of L and z holds the logarithms of the free atomic species.

Define the potential

\Phi(z)=\sum_{j=1}^{n}b_j\exp\bigl(L_j^{\mathsf T}z\bigr)-q^{\mathsf T}z.

Its gradient is $Lx-q$ , the conservation residual, so a stationary point is exactly an allocation with the right totals. Its Hessian is

\nabla^2\Phi=L\operatorname{diag}(x)L^{\mathsf T}.

Uniqueness follows from convexity. For any nonzero u the quadratic form is $\sum_jx_j(L_j^{\mathsf T}u)^2$ . The free columns contribute $\sum_ix_i^{(a)}u_i^2>0$ , so the form is strictly positive. A strictly convex function has at most one stationary point.

Existence follows from a lower bound. Dropping the nonnegative complex terms gives $\Phi(z)\ge\sum_i(e^{z_i}-q_iz_i)$ . For positive $q_i$ , each term is bounded below and grows without bound as $z_i\to\pm\infty$ . Along any unbounded sequence at least one term grows without bound, so Φ is coercive. It therefore attains its minimum at a finite z, and that minimum is the stationary point.

So every positive total has exactly one positive allocation. The invertible Hessian also makes the allocation a smooth function of the totals and constants (§7). Horn and Jackson's theory of balanced networks is the general setting for uniqueness within a conserved class.⁸ The lecturer's thesis proves the same statement another way: the map $\log x\mapsto(\log q,\log k)$ is a global change of coordinates.³

What this theorem covers

The identity block supplies both the strict convexity and the coercivity. Positive constants alone do not replace it. Repeated composition columns keep the proof intact, because they only add positive exponential terms. If a total is zero, every species that carries that constituent is zero. Remove those species and solve on the remaining positive face, since the logarithmic coordinates exclude zero.

The result selects one allocation in each class of totals. It says nothing about how fast binding relaxes, and nothing about steady states once production and removal are added. Those are statements about different equations, and Lecture 9 meets circuits with several steady states.

Prove uniqueness by strict convexity and existence by coercivity, and state separately what happens at zero totals and once slow dynamics are added.

7The allocation moves smoothly with the totals

The positive-definite Hessian of §6 does more than prove uniqueness. It gives the derivative of the allocation, which the gains of Part 3 and the Jacobian of §25 both need.

At fixed constants, differentiating the monomials gives $dx=\operatorname{diag}(x)L^{\mathsf T}dz$ , and so $dq=L\,dx=M\,dz$ with

M=L\operatorname{diag}(x)L^{\mathsf T}.

M is positive definite by §6, so dz can be solved for. The derivative of the allocation, and its logarithmic form H, are then

D_qx^{\mathrm{eq}}=\operatorname{diag}(x)L^{\mathsf T}M^{-1},\qquad H=\frac{\partial\log x}{\partial\log q}=L^{\mathsf T}M^{-1}\operatorname{diag}(q).

Check the dimensions. L is d×n and M is d×d, so both $D_qx^{\mathrm{eq}}$ and H are n×d. The first maps a change of totals to a change of species. H maps a fractional change of totals to a fractional change of species. Its entries are the gains. At leading order they become the exponents of a regime's monomial laws (§11).

When the totals move, the allocation follows, $\dot x^{\mathrm{eq}}=D_qx^{\mathrm{eq}}\,\dot q$ . "Equilibrium" here names the fast state at the current totals, not a state in which nothing changes. If a constant also changes in time, add the derivative with respect to that constant.

Why the algebraic equations are locally nonsingular

The Jacobian of conservation and equilibrium with respect to x is the square matrix

\mathcal J_x=\begin{pmatrix}L\\N\operatorname{diag}(x)^{-1}\end{pmatrix}.

Suppose $\mathcal J_x\delta x=0$ . The second block gives $\operatorname{diag}(x)^{-1}\delta x\in\ker N=\operatorname{im}L^{\mathsf T}$ , so $\delta x=\operatorname{diag}(x)L^{\mathsf T}w$ for some w. The first block then gives $Mw=0$ . Positive definiteness forces $w=0$ , hence $\delta x=0$ , and $\mathcal J_x$ is invertible.

So one differentiation of the algebraic equations fixes the species derivative from the total derivative. This is the local index-one property of the model on the positive orthant. It says nothing about the speed of the kinetic binding layer.

Differentiate the allocation through $M=L\operatorname{diag}(x)L^{\mathsf T}$ , read the gains from H, and keep algebraic nonsingularity separate from kinetic speed.

8Newton's method in log coordinates computes the allocation

The convexity proof also gives an algorithm. Minimize Φ by Newton's method in the log coordinates z. Each species then stays positive by construction, and the Hessian M is the Newton matrix.

A reproducible Newton step

At a trial z, compute $x_j=b_j\exp(L_j^{\mathsf T}z)$ , the residual $r=Lx-q$ and $M=L\operatorname{diag}(x)L^{\mathsf T}$ . Solve $M\,\delta z=-r$ . Try $z+\alpha\,\delta z$ with $\alpha=1$ , and halve α until Φ decreases. Stop only when the scaled residual $\max_i|r_i|/q_i$ meets the requested tolerance.

Very small positive totals make that relative test demanding. When the task has a detection floor, add an explicit absolute tolerance. Never accept a small step as evidence that every total balances.

A multivalent example checks the method. Take three atomic species A, B and D, and three complexes $C_{AB}$ , $C_{ABD}$ and $C_{AA}$ . Illustrative constants give the monomials

C_{AB}=10AB,\qquad C_{ABD}=50ABD,\qquad C_{AA}=3A^2,

and illustrative totals $q=(1.2,\,0.7,\,0.5)$ for A, B and D. Every total counts its complexes with their copy numbers, so $q_A=A+C_{AB}+C_{ABD}+2C_{AA}$ . The verified implementation compares its matrix of gains against centered finite differences in log coordinates. The largest discrepancy is $4.8\times10^{-11}$ . Independent scalar solutions check the competitive and series-binding states of §20 and §31.

A fast driven cycle needs a different solver. Solve the normalized stationary linear equations of its state graph and keep the currents. Strong connectivity can give a unique stationary state without detailed balance.² The extension shows how such a model can still contain subgraphs that do satisfy the equilibrium assumptions.

Solve in log free concentrations, damp the Newton step, and report a conservation residual scaled to the task.

9The reversible enzyme, matrix by matrix

Core §6 writes the reversible enzyme as a complete model. This section builds every matrix explicitly, so that the pattern can be copied for a new system.

Order the five species as $x=(E,S,P,C_{ES},C_{EP})$ . The two binding pairs and the catalytic step are elementary, so each arrow carries its rate constants:

E+S\xrightleftharpoons[k_{-S}]{k_{+S}}C_{ES},\qquad C_{ES}\xrightleftharpoons[k_{\mathrm{cat}}^{-}]{k_{\mathrm{cat}}^{+}}C_{EP},\qquad C_{EP}\xrightleftharpoons[k_{+P}]{k_{-P}}E+P.

Here $k_{\mathrm{cat}}^{+}$ and $k_{\mathrm{cat}}^{-}$ are the forward and reverse catalytic constants. The dissociation constants are $K_{d,S}=k_{-S}/k_{+S}$ and $K_{d,P}=k_{-P}/k_{+P}$ .

Scroll inside the figure to see it at reading size.

Figure 5. One enzyme gives every matrix a physical meaning. The rows of L count the enzyme, substrate and product constituents. The columns of

\Gamma_{\mathrm{cat}}

convert

C_{ES}

into

C_{EP}

and back. Their projection

S=L\Gamma_{\mathrm{cat}}

moves one unit between the substrate and product totals. The rows of

K^{\mathrm{cat}}

name the acting state of each step.

Conservation and equilibrium come first. The counting matrix has one row per total, and the association rows follow from its complex block:

L=\begin{pmatrix}1&0&0&1&1\\0&1&0&1&0\\0&0&1&0&1\end{pmatrix},\qquad N^{\mathrm{aa}}=\begin{pmatrix}-1&-1&0&1&0\\-1&0&-1&0&1\end{pmatrix},\qquad \log b=\begin{pmatrix}-\log K_{d,S}\\-\log K_{d,P}\end{pmatrix}.

The first row of the equilibrium line reads $\log C_{ES}-\log E-\log S=-\log K_{d,S}$ . Exponentiating gives $ES=K_{d,S}C_{ES}$ , a relation between concentrations in the canonical unit.

Activity and accumulation come next. The catalytic columns change only the two complexes, and projecting them with L gives S:

\Gamma_{\mathrm{cat}}=\begin{pmatrix}0&0\\0&0\\0&0\\-1&1\\1&-1\end{pmatrix},\qquad S=L\Gamma_{\mathrm{cat}}=\begin{pmatrix}0&0\\-1&1\\1&-1\end{pmatrix},\qquad K^{\mathrm{cat}}=\begin{pmatrix}0&0&0&k_{\mathrm{cat}}^{+}&0\\0&0&0&0&k_{\mathrm{cat}}^{-}\end{pmatrix}.

The columns of $\Gamma_{\mathrm{cat}}$ and S, and the rows of $K^{\mathrm{cat}}$ , follow the flux order $v=(v_+,v_-)$ , forward then reverse. Supply and free-product export are composite boundary processes, $\varnothing\rightsquigarrow S$ and $P\rightsquigarrow\varnothing$ . Their rate laws are $v_{\mathrm{in}}$ and $v_{\mathrm{out}}=k_{\mathrm{out}}P$ , so

v^{\mathrm{ext}}=\begin{pmatrix}v_{\mathrm{in}}\\k_{\mathrm{out}}P\end{pmatrix},\qquad S^{\mathrm{ext}}=\begin{pmatrix}0&0\\1&0\\0&-1\end{pmatrix}.

Uniform dilution would add $-\lambda q$ . The accumulation line then reads, component by component,

\dot q_E=0,\qquad \dot q_S=v_{\mathrm{in}}-(v_+-v_-),\qquad \dot q_P=(v_+-v_-)-k_{\mathrm{out}}P.

The binding layer reduces to one equation in free enzyme. Each free ligand follows from its total, $S=q_SK_{d,S}/(K_{d,S}+E)$ and $P=q_PK_{d,P}/(K_{d,P}+E)$ , and the enzyme total then reads

E+\frac{q_SE}{K_{d,S}+E}+\frac{q_PE}{K_{d,P}+E}=q_E.

The left side increases strictly from zero at $E=0$ to at least $q_E$ at $E=q_E$ , so exactly one root is physical. This is the scalar form of the uniqueness of §6.

Checkpoint: the conserved quantities

Does catalysis conserve the substrate total? Does it conserve the sum of the substrate and product totals?

Worked answer

One forward event changes $(q_S,q_P)$ by $(-1,+1)$ , so the substrate total is not conserved. The sum is conserved by both catalytic steps, as the columns of S show. Supply and export change the sum. Binding conserves each total separately, which is why both are coordinates of the model.

Write every matrix of a new model explicitly, then check each catalytic column against the totals it should change.

Part 3

Tool 2: leading-order dominance

Core §7 states the method and core §8 runs it on one pair. This part solves the pair exactly, writes select, solve and check as equations, bounds the error of every leading-order law, and resolves the singular ray that the leading order leaves open.

10One pair has one physical root

Every leading-order law in this part is checked against the exact equilibrium of one pair. This section computes that reference and shows how to evaluate it without losing digits. It is the calculation behind the optional box of core §8.

Substitution gives one quadratic. With $E=q_E-C_{ES}$ and $S=q_S-C_{ES}$ , the binding relation $K_dC_{ES}=ES$ becomes

C_{ES}^2-T\,C_{ES}+q_Eq_S=0,\qquad T=q_E+q_S+K_d.

Its discriminant is positive, because it can be written as a sum of squares and positive terms:

T^2-4q_Eq_S=(q_E-q_S)^2+2K_d(q_E+q_S)+K_d^2.

Exactly one root is physical. Call the left side $p(C_{ES})$ , and let $m=\min(q_E,q_S)$ . Then $p(0)=q_Eq_S>0$ , while $p(m)=-K_dm<0$ , because $(m-q_E)(m-q_S)=0$ . So the smaller root lies strictly between zero and the smaller total, and the larger root lies beyond it. Only the smaller root leaves both free species positive.

Rationalizing the smaller root avoids cancellation. The textbook form subtracts two nearly equal numbers when binding is weak. Multiplying by the conjugate gives the form the core uses:

C_{ES}=\frac{2q_Eq_S}{T+\sqrt{T^2-4q_Eq_S}}.

Near complete sequestration the free enzyme needs its own formula. There $E=q_E-C_{ES}$ is again a difference of nearly equal numbers. Eliminating the complex instead gives a quadratic for E itself, with $h=K_d+q_S-q_E$ :

E^2+hE-K_dq_E=0.

Use $E=2K_dq_E/\bigl(h+\sqrt{h^2+4K_dq_E}\bigr)$ when $h\ge0$ , and $E=\bigl(\sqrt{h^2+4K_dq_E}-h\bigr)/2$ when $h<0$ . The two expressions are algebraically equal. Each avoids cancellation on its own side of $h=0$ , so together they are accurate everywhere.

Any computed root should pass two checks before it is used. Both totals must be recovered, and $K_dC_{ES}/(ES)$ must equal one to the working precision. The companion figures and the verifier of this chapter apply both checks to every exact curve they draw.

Scroll inside the figure to see it at reading size.

Figure 6. The smaller total limits the complex. The exact complex

C_{ES}/q_E

against added substrate

q_S/q_E

, at

q_E=1\,\mu\mathrm M

and three illustrative values of

K_d

. The dashed line is the tight limit

C_{ES}=\min(q_E,q_S)

. As

K_d

falls the exact root approaches it, except near

q_S=q_E

, where the free pools of §13 remain.

Evaluate the physical root in a cancellation-free form, and confirm both totals and the binding relation before trusting it.

11Select, solve, check: dominance as a procedure

Core §7 states dominance in three steps: select a winner for each total, solve for monomial laws, and check the discarded terms. This section writes each step as an equation, so that the method runs the same way in every hand. It then proves two facts that the core uses. A leading-order law errs by a bounded factor, and its exponents are the limits of the exact gains.

The method starts from an exact rewriting of each total. A total is a sum of positive terms, $q_i=\sum_jL_{ij}x_j$ , where $L_{ij}$ counts the copies of constituent i in species j. Choose one term $L_{ij_i}x_{j_i}$ as the winner of total i. The rest of the total, measured against the winner, is the residual fraction:

\eta_i=\frac{\sum_{j\ne j_i}L_{ij}x_j}{L_{ij_i}x_{j_i}},\qquad q_i=L_{ij_i}x_{j_i}\,(1+\eta_i).

The second relation holds for every positive state, whichever winner is chosen. Leading order sets every $\eta_i$ to zero. The same comparison, normalized by the total, is the dominance profile of the flux-species sampling paper, $D_{ij}=L_{ij}x_j/q_i$ .⁶ Row i of D holds the shares of total i, so its entries are nonnegative and sum to one. The winner's share is $D_{ij_i}=1/(1+\eta_i)$ .

Selection has a graph test. On the bipartite composition graph a choice of winners is an orientation, and it is realizable exactly when the orientation is acyclic.⁷ Collect the winners' composition columns into the winner matrix $L_\star$ , whose row i is $L_{j_i}^{\mathsf T}$ . A regime is regular when $L_\star$ is invertible. It is singular otherwise, as when one species wins two totals and its column appears twice.

Solving is linear algebra in log coordinates. At equilibrium every species is a monomial, $x_j=b_j\exp(L_j^{\mathsf T}z)$ , where z holds the logarithms of the free atomic species and $b_j$ is the association constant of species j (§6). Taking the logarithm of each winner's term gives d equations, one per total:

L_\star z=\log q-c_\star-\log(1+\eta),\qquad (c_\star)_i=\log\bigl(L_{ij_i}b_{j_i}\bigr).

Here log acts on each entry, and $c_\star$ collects the counts and constants of the winners. The equation is exact. Leading order drops $\log(1+\eta)$ , and a regular regime then has the unique solution $z^{\mathrm{lo}}=L_\star^{-1}(\log q-c_\star)$ . Every species becomes a monomial in the totals and the affinities:

\log x_j^{\mathrm{lo}}=\log b_j+\rho_j^{\mathsf T}(\log q-c_\star),\qquad \rho_j^{\mathsf T}=L_j^{\mathsf T}L_\star^{-1}.

The row vector $\rho_j^{\mathsf T}$ holds the exponents of species j in the regime. On log axes each law is a straight line with these slopes, which is how saturation regimes appear in the geometric approach of Marken, Xiao and Murray.⁹ The exponents are rational, because $L_\star$ has integer entries.

Checking turns each discarded term into an inequality. The regime holds where every discarded term is smaller than its total's winner, $L_{ij}x_j<L_{ij_i}x_{j_i}$ . Substituting the monomial laws makes each inequality linear in $\log q$ and the log affinities. So each regime occupies a polyhedron in log parameter space, and its inequalities are its consistency conditions.⁹ The regime-index paper runs the same loop over every choice of winners: enumerate, solve, substitute into the inequalities, and keep the consistent regimes.¹⁰

The exact equation also measures the error of every law. Subtracting the leading-order solution gives $z-z^{\mathrm{lo}}=-L_\star^{-1}\log(1+\eta)$ , and therefore

\frac{x_j}{x_j^{\mathrm{lo}}}=\prod_{i=1}^{d}(1+\eta_i)^{-\rho_{ji}}.

This identity is exact. Where each winner is the largest term of its total, $1\le1+\eta_i\le n_i$ , with $n_i$ the number of terms in total i. The error is then a bounded factor, fixed by the gains and the term counts and independent of the size of the totals. A positive gain makes the law an overestimate by at most $n_i^{\rho_{ji}}$ from that total, and a negative gain an underestimate. Deep inside the region every $\eta_i$ vanishes, and the law becomes exact.

One pair shows the size of the bound. At $q_E=q_S=K_d$ the exact state has $E=S=0.618K_d$ and $C_{ES}=0.382K_d$ , so the free species still win and the regime is $(E,S)$ . The complex has gains $(1,1)$ there, and both residual fractions equal 0.618. The law $q_Eq_S/K_d=K_d$ therefore overestimates the exact complex by $1.618^2=2.62$ , inside the bound $2^2=4$ .

The exact gains come from the same shares. Differentiating $q_i=\sum_jL_{ij}x_j$ shows that the log change of a total is the share-weighted average of the log changes of its species:

d\log q=D\,d\log x.

When one share in a row is near one, the total moves with its winner.³ At equilibrium $d\log x=L^{\mathsf T}dz$ . Eliminating dz gives the gain matrix H of §7 in terms of the shares alone:

H=L^{\mathsf T}\bigl(DL^{\mathsf T}\bigr)^{-1}.

This agrees with $H=L^{\mathsf T}M^{-1}\operatorname{diag}(q)$ from §7, because $M=\operatorname{diag}(q)\,DL^{\mathsf T}$ . Leading order replaces each row of D by the indicator of its winner. Then $DL^{\mathsf T}$ becomes $L_\star$ , and H becomes $L^{\mathsf T}L_\star^{-1}$ , whose rows are the exponents $\rho_j^{\mathsf T}$ . The exponents of a regime are the limits of the exact gains.

A singular regime needs one more equation. When a species j wins two totals i and k, its column appears twice in $L_\star$ . Combine the two totals so that the shared winner cancels:

L_{kj}\,q_i-L_{ij}\,q_k=\sum_{l\ne j}\bigl(L_{kj}L_{il}-L_{ij}L_{kl}\bigr)x_l.

Every term on the right is a discarded term, so this balance survives at leading order. It holds exactly. For one pair the complex wins both totals, and the balance is $q_E-q_S=E-S$ . §13 solves it.

Scroll inside the figure to see it at reading size.

Figure 7. Select, solve, check, for the four regimes of one pair. Each panel selects one winner per total, solves the binding relation, and states the conditions that the discarded terms impose. The three regular panels end in the gains of the complex, the exponents

\rho^{\mathsf T}

of its law. The singular panel selects the complex twice, so it keeps the exact balance

q_E-q_S=E-S

instead.

Run dominance as select, solve and check, and read the error of each leading-order law from the residual fractions and the gains.

12One pair has three sectors and a ray

Running the procedure on one pair produces the regime map of core §8. This section derives each check from its discarded fraction, locates the exact boundaries, and measures how far they sit from the leading-order rays.

Two exact ratios do all the checking. The binding relation $K_dC_{ES}=ES$ gives

\frac{C_{ES}}{E}=\frac{S}{K_d},\qquad \frac{C_{ES}}{S}=\frac{E}{K_d}.

Within the enzyme total, the complex outweighs the free enzyme exactly when free substrate exceeds $K_d$ . Within the substrate total, the complex outweighs the free substrate exactly when free enzyme exceeds $K_d$ . Every check below is one of these two comparisons, evaluated with the regime's laws.

Take $(E,C_{ES})$ as the worked case. Select $q_E\approx E$ and $q_S\approx C_{ES}$ . Solve: binding gives $S\approx K_dq_S/q_E$ . Check: the discarded fractions are the residual fractions of §11, $C_{ES}/E\approx q_S/q_E$ and $S/C_{ES}\approx K_d/q_E$ . Both must be below one. So the regime needs $q_S<q_E$ and $K_d<q_E$ . That is two conditions, where a guess of "enzyme in excess" would supply only one.

Regime	Select	Discarded fractions at leading order	Holds when
$(E,S)$	$q_E\approx E,\ q_S\approx S$	$C_{ES}/E\approx q_S/K_d$ , $C_{ES}/S\approx q_E/K_d$	$q_E<K_d$ and $q_S<K_d$
$(E,C_{ES})$	$q_E\approx E,\ q_S\approx C_{ES}$	$C_{ES}/E\approx q_S/q_E$ , $S/C_{ES}\approx K_d/q_E$	$q_S<q_E$ and $K_d<q_E$
$(C_{ES},S)$	$q_E\approx C_{ES},\ q_S\approx S$	$E/C_{ES}\approx K_d/q_S$ , $C_{ES}/S\approx q_E/q_S$	$q_E<q_S$ and $K_d<q_S$
$(C_{ES},C_{ES})$	$q_E\approx C_{ES}\approx q_S$	$E/C_{ES}\approx S/C_{ES}\approx\sqrt{K_d/q_E}$	the ray $q_E\approx q_S>K_d$

The three regular regions share one corner. Their boundaries are the rays $q_E=K_d$ , $q_S=K_d$ and $q_E=q_S$ , all starting at $(K_d,K_d)$ . Together the three laws say that the complex is the smallest of three candidates, $C_{ES}\approx\min(q_E,q_S,q_Eq_S/K_d)$ , which is the form in the lecturer's thesis.³ The thesis derives the same three regimes of the complex. The lecturer's paper with Khammash and Doyle reads them as the regimes of Michaelis-Menten kinetics, including the enzyme-excess regime, where the complex holds the substrate total, that a substrate-excess assumption misses.^{3, 4}

The exact boundaries are curves, and each lies within a factor of two of the rays. The complex equals the free enzyme when $S=K_d$ , which gives $q_S=K_d+q_E/2$ . It equals the free substrate when $E=K_d$ , which gives $q_E=K_d+q_S/2$ . Take the first curve. Its leading-order counterpart is $q_S=\max(K_d,q_E)$ , and the ratio $(K_d+q_E/2)/\max(K_d,q_E)$ stays between 1/2 and 3/2. The second curve is its mirror image.

Between the two exact curves the complex outweighs both free species. At large totals this wedge runs from $q_S=q_E/2$ to $q_S=2q_E$ , a band of constant width $\log4$ on log axes. At leading order it collapses onto the singular ray, which is why the ray has no area in totals. The regime-index paper gives the four regimes of one pair the shares 1/3, 1/3, 1/3 and 0 of log parameter space, while in free concentrations the singular regime is a full quadrant.¹⁰

The error of each law stays bounded up to its boundary. By §11 each free species errs by at most a factor of two in the regular regimes. The complex errs by at most a factor of two in $(E,C_{ES})$ and $(C_{ES},S)$ , and by at most four in $(E,S)$ , where both of its gains are one.

Scroll inside the figure to see it at reading size.

Figure 8. The leading-order regime map of one pair, in totals. Three rays from

(K_d,K_d)

bound the regular sectors

(E,S)

(E,C_{ES})

and

(C_{ES},S)

, each labelled with its law. The diagonal beyond the corner is the singular ray

(C_{ES},C_{ES})

. The exact equal-contribution curves, not drawn, lie within a factor of two of the rays. The dashed cuts at

q_E=0.01K_d

and

q_E=100K_d

are the two paths that §17 reads as responses. This is the core's Figure 12.

Derive each regime of a pair from its two discarded fractions, and read its map as leading-order rays, exact curves within a factor of two.

13On the singular ray a small free pool survives

On the singular ray the complex wins both totals, yet the free pools are not zero. Their size sets the depth of every titration switch in this chapter, from the tight cut of §17 to the sponge of §27. This section solves them exactly and bounds the error of the two usual approximations.

Ray coordinates reduce the problem to one quadratic. Let $Q=(q_E+q_S)/2$ be the mean total and $\Delta=q_E-q_S$ the imbalance. The complex cancels from the imbalance, so $\Delta=E-S$ exactly. Let $w=E+S$ be the free sum. Then $C_{ES}=Q-w/2$ and $ES=(w^2-\Delta^2)/4$ , and $K_dC_{ES}=ES$ becomes a quadratic in w. Its positive root gives both free pools:

w=\sqrt{K_d^2+\Delta^2+4K_dQ}-K_d,\qquad E=\frac{w+\Delta}{2},\qquad S=\frac{w-\Delta}{2}.

These formulas are exact for every pair of totals, not only near the ray.

Leading order on the ray keeps the exact balance and approximates the complex. Replace $C_{ES}$ by the mean total Q in the binding relation, and use $S=E-\Delta$ . This gives the ray balance $E(E-\Delta)=K_dQ$ , so

E\approx\frac{\Delta+\sqrt{\Delta^2+4K_dQ}}{2}.

At matched totals, $\Delta=0$ , this is the core's $E\approx S\approx\sqrt{K_dq_E}$ . Away from matching, for $|\Delta|\gg\sqrt{K_dQ}$ , it approaches $E\approx\Delta$ on the enzyme-excess side and the small tail $E\approx K_dQ/|\Delta|$ on the substrate-excess side.

The ray has a width, and the width shrinks as binding tightens. The free pools cross over where $|\Delta|\sim\sqrt{K_dQ}$ . Relative to the totals that is $|\Delta|/Q\sim\sqrt{K_d/Q}$ , which vanishes as $K_d/Q\to0$ . So at leading order the singular regime is a ray.

Two bounds control the smaller free pool. Write $m=\min(q_E,q_S)$ and $x_{\min}=\min(E,S)$ . The larger free pool is $|\Delta|+x_{\min}$ and the complex is $m-x_{\min}$ , so binding gives

x_{\min}\bigl(|\Delta|+x_{\min}\bigr)=K_d\bigl(m-x_{\min}\bigr).

Dropping positive terms gives $x_{\min}\le\sqrt{K_dm}$ and, when $\Delta\ne0$ , $x_{\min}\le K_dm/|\Delta|$ . The first bound holds everywhere and is sharp at matching. The second is sharper far from matching.

The tight limit is a rectifier, and its error is exactly the smaller pool. As $K_d\to0$ the free enzyme tends to $\max(q_E-q_S,0)$ . The exact free enzyme exceeds this by $x_{\min}$ , so the absolute error is at most $\sqrt{K_dm}$ . At the tight-cut ray of §17, with $q_E=q_S=100K_d$ , the rectifier predicts no free enzyme, the ray balance predicts $0.1q_E$ , and the exact value is $0.0951q_E$ .

The relative error depends on the readout. For the complex it is $x_{\min}/(m-x_{\min})$ , which is small throughout. For the free enzyme below threshold the rectifier predicts zero, so its relative error is one, however tight the binding. A task that needs an output below a detection limit compares $x_{\min}$ with that limit. A task that uses the free enzyme's activity needs the ray balance.

Scroll inside the figure to see it at reading size.

Figure 9. A small complex error can hide a large free-pool error. Left: free enzyme across the ray, scaled by

\sqrt{K_dQ}

, exact, from the ray balance, and from the tight limit. Right: the relative error of the tight limit, for the complex readout and for the free-enzyme readout. Illustrative parameters:

Q=1\,\mu\mathrm M

and

K_d=10^{-4}\,\mu\mathrm M

. At the same operating point the two errors differ by orders of magnitude.

Near matching the free enzyme is also hypersensitive to the totals. §14 shows that its gains grow like $\tfrac12\sqrt{Q/K_d}$ on the ray, about 5 at $Q=100K_d$ . The lecturer's thesis derives the same hypersensitivity of the free enzyme near matched totals.³

Keep the exact difference of the totals on the singular ray, and judge any approximation there by its error on the output that the biology reads.

14The gains of the complex fill a triangle

A gain is a logarithmic derivative of a species with respect to a total. For the complex of one pair, its two gains are confined to a triangle whose corners are the three regular regimes. This section derives the triangle, shows that every interior point occurs, and places the two paths of §17 on it.

Implicit differentiation gives the gains. Differentiate $K_dC_{ES}=ES$ at fixed $K_d$ , with $dE=dq_E-dC_{ES}$ and $dS=dq_S-dC_{ES}$ . This gives $(K_d+E+S)\,dC_{ES}=S\,dq_E+E\,dq_S$ . Multiply each partial derivative by its total and divide by $C_{ES}$ , using $q_E/C_{ES}=(K_d+S)/S$ and $q_S/C_{ES}=(K_d+E)/E$ :

\rho_{C_{ES},q_E}=\frac{K_d+S}{W},\qquad \rho_{C_{ES},q_S}=\frac{K_d+E}{W},\qquad W=K_d+E+S.

The gain pair is a convex combination of three corners. Written as a vector,

\bigl(\rho_{C_{ES},q_E},\rho_{C_{ES},q_S}\bigr)=\frac{K_d}{W}(1,1)+\frac{E}{W}(0,1)+\frac{S}{W}(1,0).

The weights are positive and sum to one, so the pair lies strictly inside the triangle with corners $(1,1)$ , $(0,1)$ and $(1,0)$ . The lecturer's thesis derives the same convex combination.³

The corners are the three regular regimes. The corner $(1,1)$ is $(E,S)$ , where the complex is proportional to both totals. The corner $(0,1)$ is $(E,C_{ES})$ , where it follows the substrate total alone, and $(1,0)$ is $(C_{ES},S)$ . Dominance within W decides where the pair sits. It sits near $(1,1)$ when $K_d$ is the largest of the three weights, and near $(0,1)$ or $(1,0)$ when E or S is. When E and S tie far above $K_d$ , it sits at the midpoint $(1/2,1/2)$ of the hypotenuse, which is the singular ray.

Every interior point is attained. Given a point in the open triangle, set $W=K_d/(\rho_{C_{ES},q_E}+\rho_{C_{ES},q_S}-1)$ , $E=W(1-\rho_{C_{ES},q_E})$ and $S=W(1-\rho_{C_{ES},q_S})$ . These are positive free concentrations. The totals built from them, $q_E=E+ES/K_d$ and $q_S=S+ES/K_d$ , realize the point. So the open triangle is the full attainable set of the complex's gains.

The two paths of §17 trace two edges. The weak cut $q_E=0.01K_d$ keeps $E\le0.01K_d$ , so the first gain stays near one while the path runs down the right edge, from $(E,S)$ to $(C_{ES},S)$ . The tight cut $q_E=100K_d$ keeps $E+S\gg K_d$ , so the gains sum to nearly one while the path follows the hypotenuse, from $(E,C_{ES})$ through the ray to $(C_{ES},S)$ . At its ray point the exact gains are 0.525 each.

Scroll inside the figure to see it at reading size.

Figure 10. The gains of the complex fill a triangle. Each corner is a regular regime, and the singular ray sits at the midpoint of the hypotenuse. Each curve is one exact path, at fixed

q_E

with

q_S

swept, for the illustrative

K_d=1\,\mu\mathrm M

. The weak cut runs down the right edge. The tight cut follows the hypotenuse.

Free-species gains are not confined to the triangle. The same differentiation gives

\rho_{E,q_E}=\frac{(K_d+E)(K_d+S)}{K_dW},\qquad \rho_{E,q_S}=-\frac{S(K_d+E)}{K_dW}.

On the ray, where $E\approx S\approx\sqrt{K_dQ}\gg K_d$ , both have magnitude near $\tfrac12\sqrt{Q/K_d}$ . At the tight-cut ray point they are 5.52 and −4.99. This is the steepness of the titration switch, read as a gain.

Gains need not be integers. For a homodimer with $q_R=R+2C_{RR}$ and $C_{RR}=R^2/K_{\mathrm{dim}}$ , the regime in which the dimer wins gives $R\approx(K_{\mathrm{dim}}q_R/2)^{1/2}$ , a gain of one half. In general the exponents $\rho_j^{\mathsf T}=L_j^{\mathsf T}L_\star^{-1}$ of §11 are rational, and a two-copy column puts a two in $L_\star$ . The lecturer's thesis finds the same drop of the dimer's order.³ §26 runs the full dimer example.

Place the complex's two gains in the triangle whose corners are the regular regimes, and compute free-species gains separately, because they grow without bound near the ray.

Part 4

Tool 3: regulation read from the regime map

Core §9 to §12 and the coda read responses off the regime map with a five-step pipeline. This part first gives the reason behind each step. It then runs the pipeline on every regulatory example of the core, and each run ends in a regime-sequence table.

15The five-step pipeline, with a reason for each step

Core §9 turns a regime map into a response in five steps. This section states the steps once more and proves that each one does what the core says. Every proof applies a fact from Part 3 to a path through the map.

The analysis pipeline

Model. Write the complex list, the binding relations and the activities. The output is $v=K^{\mathrm{cat}}x$ .
Experiment. Name the swept totals $q_{\mathrm I}$ , the held totals $q_{\mathrm P}$ and the readout. Hold the affinities b and the activities fixed. The response is the map
$q_{\mathrm I}\ \longmapsto\ v=K^{\mathrm{cat}}x^{\mathrm{eq}}(q_{\mathrm I},q_{\mathrm P};b).$
Candidates. List the species that contain each total. Settle every total held in excess or in trace. The totals left over can change winner.
Regime sequence. Walk the path. In each regime select the winners, solve for monomial laws and check the discarded terms (§11). The first check to fail is a breakpoint.
Response. Join the pieces in a regime-sequence table: regime, input range, leading-order output and gain.

Five facts justify the steps. A sweep is a straight line in log coordinates. Excess and trace settle a total for the whole path. A regime holds on one interval of the path, and its first failing check ends it. A regular crossing joins the pieces, while a singular crossing makes a free species jump. Every piece holds within a bounded factor, and the exact curve checks it. The subsections take the facts in this order.

A sweep is a straight line in log coordinates

A sweep of one total moves log q along a straight line. Write the swept total as $q_{\mathrm I}=q_{\mathrm I}^0e^t$ , with t the log of the sweep factor. Then $\log q=\log q^0+t\,e_{\mathrm I}$ , where $e_{\mathrm I}$ is the unit vector of the swept total. Sweeping several totals in fixed proportion, as along a diagonal, gives another straight line.

Along such a line every leading-order law is a straight line on log axes. In a regime the law of species j is $\log x_j^{\mathrm{lo}}=\log b_j+\rho_j^{\mathsf T}(\log q-c_\star)$ (§11). Substituting the path gives $\log x_j^{\mathrm{lo}}=\mathrm{const}+t\,\rho_{j\mathrm I}$ . The slope $\rho_{j\mathrm I}$ is the gain of species j on the input. A readout $v_\alpha=\sum_jk_{\alpha j}x_j$ is, at leading order, its own largest term, so it is piecewise linear on log axes as well.

The affinities are held because they are parameters of the map, not coordinates of the path. They enter the checks through $c_\star$ , so changing one moves the boundaries and changes which regimes a path meets. Which reduced laws count as valid therefore depends on which variables the experiment can tune.¹⁰

Excess and trace settle a total

Two counting bounds settle a total, and neither needs an affinity. The first bounds a ligand held in excess. Every complex that contains the ligand also contains a partner, so bound ligand cannot exceed the partner total. For one pair, $S=q_S-C_{ES}\ge q_S-q_E$ , and therefore

\frac{S}{q_S}\ \ge\ 1-\frac{q_E}{q_S}.

A ligand far in excess of its partners wins its own total along the whole path, at every affinity.

Weak binding settles the same total by a second route. The reservoir law replaces free S by its total in the binding relation, which gives $C^{\mathrm{res}}=q_Eq_S/(K_d+q_S)$ . Its relative error is exact:

\frac{C^{\mathrm{res}}-C_{ES}}{C_{ES}}=\frac{E}{K_d+q_S}\ \le\ \min\Bigl(\frac{q_E}{K_d},\ \frac{q_E}{q_S}\Bigr).

To derive it, write $(C^{\mathrm{res}}-C_{ES})(K_d+q_S)=q_Eq_S-K_dC_{ES}-q_SC_{ES}$ . Replace $K_dC_{ES}$ by ES, with $E=q_E-C_{ES}$ and $S=q_S-C_{ES}$ . The right side collapses to $C_{ES}E$ . The bound follows from $E\le q_E$ . So the substrate behaves as a reservoir whenever the enzyme total is small against either $K_d$ or $q_S$ .

The second bound concerns a species held in trace. A complex holding a promoter G cannot exceed the promoter total, $C_{GR}\le q_G$ . So $C_{GR}/q_R\le q_G/q_R$ , and when $q_G\ll q_R$ the complex never wins the regulator total. A trace promoter reads its regulator without loading it.

Settling is what keeps step 4 short. A settled total keeps one winner along the path and contributes no breakpoint. The walk then runs over the one or two totals that can change winner.

A regime holds on one interval, and its first failing check ends it

Each check is linear along the path. A check compares a discarded term with its total's winner, $L_{ij}x_j^{\mathrm{lo}}<L_{ij_i}x_{j_i}^{\mathrm{lo}}$ . The log of this ratio is affine in log q (§11), and so it is affine in t. An affine function changes sign at most once. The checks of a regime therefore hold on an intersection of half-lines, which is one interval of the path, possibly empty.

Two consequences follow. A regime appears at most once along a straight path. And it ends where its first check fails, because that is where one discarded term catches up with its winner.

The failing term names the successor. Swap it in as the new winner of its total, then select, solve and check again, because the new laws change every other comparison. If the swapped-in species already wins another total, it now wins two. Then $L_\star$ repeats a column, and the crossing is singular. Past the singular ray the second total passes to its own next candidate.

A regular crossing joins the pieces

A regular crossing keeps the leading-order response continuous. Suppose only total i changes winner, from species $j_i$ to species j, and both winner matrices $L_\star$ and $L_{\star'}$ are invertible. On the boundary the two terms are equal, $L_{ij}x_j=L_{ij_i}x_{j_i}$ . So the laws of the old regime also satisfy the new equation for total i, $q_i=L_{ij}x_j$ . The equations for the other totals are shared. The new regime's equations have exactly one solution, because $L_{\star'}$ is invertible. The two laws therefore agree on the boundary, and the pieces meet.

The gain changes at a regular crossing, and the change is a rational number. The gains are rows of $L^{\mathsf T}L_\star^{-1}$ , built from integer matrices. For one pair every gain is 1, 0 or −1 (§12), so the change is an integer. In general it is not. When the dimer takes the repressor total in core A1, free monomer passes from $R\approx q_R$ to $R\approx(K_{\mathrm{dim}}q_R/2)^{1/2}$ , a gain change from 1 to 1/2 (§26). On log axes a regular crossing is a kink, and the response is graded.

A singular crossing makes a free species jump

A singular crossing changes two winners at once, and the free species that loses both totals jumps. On the tight cut of one pair, the enzyme total passes from E to $C_{ES}$ and the substrate total from $C_{ES}$ to S at the same point. Free enzyme follows $E\approx q_E$ before the ray and $E\approx K_dq_E/q_S$ after it. At $q_S=q_E$ the two laws give $q_E$ and $K_d$ . The pieces miss each other by the factor $q_E/K_d$ .

The singular balance of §11 places the value in the gap. For one pair it reads $q_E-q_S=E-S$ , and §13 solved it: on the ray both free pools equal $\sqrt{K_dq_E}$ . That is the geometric mean of the one-sided limits $q_E$ and $K_d$ , so on log axes the ray value sits halfway down the jump. A singular crossing is a switch, and its size grows with $q_E/K_d$ .

Every piece holds within a bounded factor, and the exact curve checks it

Leading order is the answer because its error is bounded. The identity of §11, $x_j/x_j^{\mathrm{lo}}=\prod_i(1+\eta_i)^{-\rho_{ji}}$ , holds exactly. Where the selected winners are the exact largest terms, each factor $1+\eta_i$ lies between 1 and the term count $n_i$ . The piece for species j is then within $\prod_in_i^{|\rho_{ji}|}$ of the exact value, however large the totals are.

The bound weakens only near boundaries. For one pair the rule of §12 is leading-order rays, exact curves within a factor of two, so the loss is confined to a band of fixed width on log axes. Near a regular boundary the neighbouring regime's bound takes over. Near a singular ray the jumping species is not bounded by either side, and the balance of §13 replaces its monomial law.

The exact curve is the check. The monotone scalar solves of §8 and §10 give it at every point of a path. Each table below lists exact values at sample points beside the leading-order laws. A check that passes by a small margin is a warning sign. When the candidates of a total stay within a factor of two of one another across a whole regime, leading order can name a winner that the exact solve does not confirm, and §23 meets such a case.

Settle totals by excess and trace, end each regime at its first failing check, and read a regular crossing as a kink and a singular crossing as a jump. Trust each piece within its bounded factor, and check it against the exact curve.

16Activities fix the sign of a response

The binding layer says where each constituent sits. The activities of step 1 say what each place produces. The same allocation can therefore activate, repress or do nothing, depending only on the activities.

Take a machine with a free state E and a bound state $C_{ES}$ , with activities $k_0$ and $k_1$ . Its output is

v=k_0E+k_1C_{ES}=k_0q_E+(k_1-k_0)\,C_{ES}.

Per machine, the output is $v/q_E=k_0(1-\theta)+k_1\theta$ , with occupancy $\theta=C_{ES}/q_E$ . Its derivative in θ is $k_1-k_0$ . So occupancy activates when $k_1>k_0$ , represses when $k_1<k_0$ , and changes nothing when the two agree.³ Affinity decides how θ follows the totals, and activity decides what a change in θ produces.

Several outputs collect into one matrix. Write each activity as $v_\alpha=\sum_jk_{\alpha j}x_j$ , and collect the coefficients into $K^{\mathrm{cat}}$ , one row per process. A coefficient can be a resolved unimolecular rate constant. It can also be the effective activity of a state for a composite process, such as transcription $C_{GR}\rightsquigarrow C_{GR}+X$ with rate law $k_1C_{GR}$ . The mechanism decides which. If a co-substrate varies during the experiment, its level or its state must enter the model rather than hide inside a constant coefficient.

At leading order the activity of the winner sets the output. Where E wins the machine total, $v\approx k_0q_E$ unless $k_0=0$ . Where $C_{ES}$ wins, $v\approx k_1q_E$ . A path that crosses from one regime to the other moves the output from one plateau to the other. The pieces of the regime tables below are these winner activities, multiplied by the monomial laws.

Scroll inside the figure to see it at reading size.

Figure 11. One binding curve, three regulatory effects. The figure is drawn for a promoter G that binds a regulator R, with transcription from G and from

C_{GR}

as composite processes. Only the activities differ between the three curves. A faster bound state activates, a slower one represses, and equal activities leave the output flat. The machine E with bound state

C_{ES}

is the same calculation.

Assign an activity to every state before reading a response, and read the sign of the response from $k_1-k_0$ .

17Four responses along two paths

Core §10 runs the pipeline on one binding pair and reads four responses from one table. This section derives every entry of that table and places an exact value beside each law.

1 · Model. The pair of §10 to §14, with $q_E=E+C_{ES}$ , $q_S=S+C_{ES}$ and $K_dC_{ES}=ES$ . Two activity assignments give two readouts: the complex active, read as $C_{ES}/q_E$ , and the free enzyme active, read as $E/q_E$ .

2 · Experiment. Sweep $q_S$ at fixed $q_E$ and $K_d$ . Take two paths, the weak cut $q_E=0.01K_d$ and the tight cut $q_E=100K_d$ . They are the two dashed horizontal lines of Figure 8.

3 · Candidates. The weak cut settles the substrate total. There $S/C_{ES}=K_d/E\ge K_d/q_E=100$ , so S wins $q_S$ along the whole path, and only the enzyme total can change winner. The tight cut settles nothing. Both totals keep two candidates each.

4 · Regime sequence. The weak cut starts in $(E,S)$ , with laws $E\approx q_E$ , $S\approx q_S$ and $C_{ES}\approx q_Eq_S/K_d$ . Its check $C_{ES}<E$ reads $q_S<K_d$ and fails first. Swapping $C_{ES}$ in for the enzyme total gives $(C_{ES},S)$ , a regular crossing. Its laws are $C_{ES}\approx q_E$ and $E\approx K_dq_E/q_S$ , and its checks hold for all $q_S>K_d$ .

The tight cut starts in $(E,C_{ES})$ , with laws $E\approx q_E$ , $C_{ES}\approx q_S$ and $S\approx K_dq_S/q_E$ . The check $S<C_{ES}$ reads $K_d<q_E$ and always holds. The check $C_{ES}<E$ reads $q_S<q_E$ and fails at $q_S=q_E$ . Swapping $C_{ES}$ in makes it win both totals, so the path crosses the singular ray $(C_{ES},C_{ES})$ . Past the ray the substrate total passes to S, and the path is in $(C_{ES},S)$ .

5 · Response. Dividing each law by $q_E$ gives the table. The exact column solves the pair at one sample point per regime.

Path	Regime	Range of $q_S$	$C_{ES}/q_E$	$E/q_E$	Exact at a sample point
weak cut, $q_E=0.01K_d$	$(E,S)$	$q_S<K_d$	$q_S/K_d$ , gain 1	1, gain 0	at $q_S=0.1K_d$ : 0.0902 and 0.910
weak cut, $q_E=0.01K_d$	$(C_{ES},S)$	$q_S>K_d$	1, gain 0	$K_d/q_S$ , gain −1	at $q_S=10K_d$ : 0.909 and 0.0910
tight cut, $q_E=100K_d$	$(E,C_{ES})$	$q_S<q_E$	$q_S/q_E$ , gain 1	1, gain 0	at $q_S=0.5q_E$ : 0.490 and 0.510
	$(C_{ES},C_{ES})$	the ray, $q_S\approx q_E$	about 1	$\sqrt{K_d/q_E}=0.1$	at $q_S=q_E$ : 0.905 and 0.0951
	$(C_{ES},S)$	$q_S>q_E$	1, gain 0	$K_d/q_S$ , gain −1	at $q_S=2q_E$ : 0.990 and 0.00981

The exact values sit where §15 says they should. Deep in a regime they approach the laws. At the weak cut's breakpoint $q_S=K_d$ , both pieces give 1 while the exact values are 0.499 and 0.501. That factor of two is the worst case, and it occurs at a boundary. On the tight cut the free-enzyme values at $q_S=0.5q_E$ and $2q_E$ are off by two for another reason: they are close to the ray, where the balance $q_E-q_S=E-S$ gives the better values $E\approx q_E-q_S$ and $E\approx K_dq_E/(q_S-q_E)$ .

The table holds four archetypes. On the weak cut the complex saturates, which is the Michaelis-Menten shape at leading order, one edge of the triangle of §14.⁴ The free enzyme shows graded repression. On the tight cut the complex shows a limiting pool, and the free enzyme shows a titration switch. The lecturer's thesis finds it beside the three vertex regimes of a pair, as the extreme case where both totals are mostly bound and nearly equal, and there the regime is a ray (Eq. 3.45).³ The switch needs matched totals, so among all totals it occupies only a thin strip along the ray.⁶

Scroll inside the figure to see it at reading size.

Figure 12. One binding pair, two paths, four responses. Top row: the weak cut

q_E=0.01K_d

. Bottom row: the tight cut

q_E=100K_d

. One panel in each row reads the complex, the other the free enzyme. Thick pieces are leading-order laws, colored by regime, and thin gray curves are the exact equilibrium. The dot marks the singular ray.

Swapping the input gives the rising threshold, and it is a second run of the same five steps. Hold a sequestering total $q_S=100K_d$ , sweep $q_E$ , and read free E. The path is a vertical line on the map of Figure 8. It runs from $(C_{ES},S)$ across the singular ray at $q_E=q_S$ into $(E,C_{ES})$ .

Regime	Range of $q_E$	Free E	Gain	Exact at a sample point
$(C_{ES},S)$	$q_E<q_S$	$K_dq_E/q_S$	1	at $q_E=0.5q_S$ : $0.962K_d$ , law $0.5K_d$
$(C_{ES},C_{ES})$	the ray, $q_E\approx q_S$	$\sqrt{K_dq_S}=10K_d$	jump	at $q_E=q_S$ : $9.51K_d$
$(E,C_{ES})$	$q_E>q_S$	$q_E$	1	at $q_E=2q_S$ : $101K_d$ , law $200K_d$

Free enzyme stays near $K_d$ until the sequestering pool fills, and then nearly every added enzyme is free. At leading order it jumps a hundredfold at $q_E=q_S$ . The two sample points beside the ray are again off by two, and the balance gives $E\approx K_dq_E/(q_S-q_E)$ and $E\approx q_E-q_S$ there. This threshold of molecular titration needs no cooperativity.¹¹ Bacteria use it to time replication. In the model of Fu, Xiao and Jun, about 300 chromosomal DnaA boxes bind the initiator DnaA with dissociation constants near 1 nM. About ten weaker sites at the origin, near 100 nM, trigger initiation only once the boxes are saturated.¹²

The low-occupancy end of the weak cut recovers Lecture 2's effective rate law. In $(E,S)$ , with the complex active at $k_{\mathrm{cat}}$ , the flux is $v=k_{\mathrm{cat}}C_{ES}\approx(k_{\mathrm{cat}}/K_d)\,q_Eq_S$ . The coefficient $k_{\mathrm{cat}}/K_d$ has units $\mathrm M^{-1}\mathrm s^{-1}$ . It is the second-order constant of the whole enzyme in this equilibrium reduction, and it holds only in $(E,S)$ .

Derive each entry of a regime table from its winners, and expect the exact value within a factor of two at a boundary and at the balance value on a singular ray.

18A reporter reads the switch

A titration switch becomes a measurable response when a reporter reads the free species. The pipeline shows where the reporter flips, and the exact inventory gives its midpoint and slope.

1 · Model. The pair of §17, read as an activator E and its sequestering partner S. A one-site reporter binds free E with dissociation constant $K_{\mathrm{rep}}$ , and its occupied fraction is

\theta=\frac{E}{K_{\mathrm{rep}}+E},\qquad \log\frac{\theta}{1-\theta}=\log\frac{E}{K_{\mathrm{rep}}}.

The second relation is the logit of θ. It is exact and shows that the reporter reads log E directly.¹¹

2 · Experiment. Sweep the activator total $q_E$ at a fixed sequestering total $q_S$ , and read θ. The illustrative values are $q_S=10\,\mu\mathrm M$ , $K_d=0.01\,\mu\mathrm M$ and $K_{\mathrm{rep}}=1\,\mu\mathrm M$ .

3 · Candidates. The reporter is held in trace, so its complex never wins the activator total (§15). The activator and sequestering totals can both change winner. The reporter's own total has two candidates, free and bound, which tie at $E=K_{\mathrm{rep}}$ .

4 · Regime sequence. The pair is tight, $q_S=1000K_d$ , so the path is the swapped run of §17. It runs through $(C_{ES},S)$ , crosses the singular ray at $q_E=q_S$ , and enters $(E,C_{ES})$ . Free E jumps from about $K_d$ to about $q_S$ there. The reporter's scale lies inside that jump, $K_d<K_{\mathrm{rep}}<q_S$ , so the reporter flips at the ray. The balance $q_E-q_S=E-S$ locates the flip: just past the ray S is small, $E\approx q_E-q_S$ , and half occupancy needs $q_E\approx q_S+K_{\mathrm{rep}}$ .

5 · Response.

Regime	Range of $q_E$	Free E	Reporter θ	Gain of θ	Exact at a sample point
$(C_{ES},S)$ , reporter free	$q_E<q_S$	$K_dq_E/q_S$	$K_dq_E/(K_{\mathrm{rep}}q_S)$	1	at $q_E=5$ : 0.0099, law 0.005
singular ray	$q_E\approx q_S$	from $q_E-q_S=E-S$	rises from $K_d/K_{\mathrm{rep}}$ to 1, half at $q_E\approx q_S+K_{\mathrm{rep}}$	switch	at $q_E=10$ : 0.237
$(E,C_{ES})$ , reporter bound	$q_E>q_S+K_{\mathrm{rep}}$	$q_E$	1	0	at $q_E=15$ : 0.834

The exact inventory refines the midpoint. Eliminating S with its total gives $C_{ES}=q_SE/(K_d+E)$ , so the activator total as a function of free E is

q_E=E+\frac{q_SE}{K_d+E},\qquad \frac{dq_E}{dE}=1+\frac{q_SK_d}{(K_d+E)^2}.

Half occupancy means $E=K_{\mathrm{rep}}$ . The logit relation makes the local Hill slope $n_H$ , the slope of the logit against $\log q_E$ , equal to the log gain of free E at that point:

q_{E,1/2}=K_{\mathrm{rep}}+\frac{q_SK_{\mathrm{rep}}}{K_d+K_{\mathrm{rep}}},\qquad n_H=\frac{q_{E,1/2}}{K_{\mathrm{rep}}\bigl[1+q_SK_d/(K_d+K_{\mathrm{rep}})^2\bigr]}.

With the illustrative values the midpoint is $10.901\,\mu\mathrm M$ , close to the balance estimate $q_S+K_{\mathrm{rep}}=11$ , and the local Hill slope is 9.93. The steepness has a simple reading. When $q_SK_d\ll K_{\mathrm{rep}}^2$ the bracket is near one, and $n_H\approx q_{E,1/2}/K_{\mathrm{rep}}$ . The reporter reads a small difference of two large totals, so its slope is the threshold divided by its own scale.

Two other slopes are easy to confuse with this one. The ordinary elasticity of θ is smaller by the factor $1-\theta$ , so it is half as large at the midpoint. And a Hill coefficient fitted to a whole noisy curve need not equal either local slope. One reporter site suffices because the nonlinearity sits upstream, in the finite pool. The yeast sequestration experiment of Buchler and Cross realized this mechanism, with fitted coefficients and experimental corrections distinct from this ideal calculation.¹³ In the lab below, "activator" is E and "inhibitor" is its sequestering partner S.

Choose the active state

Vary activator, inhibitor and affinity. Compare the free activator with the one-site reporter it drives.

Activator total (µM) 10Inhibitor total (µM) 10log₁₀ dissociation scale (µM) -2

The reporter is a trace probe with dissociation scale 1 µM, so its own binding does not deplete activator.

Place a reporter's scale inside a titration jump, locate its midpoint from the ray's balance, and compute the exact midpoint and local slope from the inventory written in free E.

19Effective laws are pieces of a regime sequence

The familiar effective laws of enzymology and gene regulation are pieces of regime sequences. Each piece holds in its own regime, and each regime has conditions that the effective law does not display. Three laws show this: the Michaelis-Menten law, the specificity constant and the Hill law.

The Michaelis-Menten law comes first. With a catalytic escape, the balance of the complex gives $K_M=(k_{-S}+k_{\mathrm{cat}})/k_{+S}$ rather than $K_d=k_{-S}/k_{+S}$ (§2).¹ The two agree when $k_{\mathrm{cat}}\ll k_{-S}$ . The curve in free substrate, $v/q_E=k_{\mathrm{cat}}S/(K_M+S)$ , has two pieces: $v\approx(k_{\mathrm{cat}}/K_M)q_ES$ where E wins the enzyme total, and $v\approx k_{\mathrm{cat}}q_E$ where $C_{ES}$ wins.

The specificity constant belongs to the first piece only. The slope of the curve at zero substrate is $k_{\mathrm{cat}}/K_M$ , with units $\mathrm M^{-1}\mathrm s^{-1}$ . Writing the flux in the substrate total also needs the substrate to win its own total, which the reservoir bound of §15 controls.

A Hill law needs two regime conditions. The ligand must win its own total, which is the abundance assumption behind Hill laws in total input.³ And no partly occupied state may win the receptor total. Take n sites with stepwise dissociation constants $K_1,\dots,K_n$ , so that the state with k ligands has weight $R^k/(K_1\cdots K_k)$ relative to the empty state. When only the empty and full states ever win, the full-state fraction has one corner, at $K_H=(K_1\cdots K_n)^{1/n}$ , where its gain drops from n to 0. That two-piece sequence is the Hill law.

Independent equivalent sites break the second condition. Their weights are binomial, $\binom nk(R/K_d)^k$ , so every state wins in turn. State k overtakes state $k-1$ at $R/K_d=k/(n-k+1)$ . The winners therefore step through all n+1 states between $R=K_d/n$ and $R=nK_d$ , and the full-state fraction $[R/(K_d+R)]^n$ descends a staircase of gains.

Mechanism	Winner of the receptor total	Range of R	Full-state fraction	Gain
only empty and full states win	empty	$R<K_H$	$(R/K_H)^n$	n
only empty and full states win	full	$R>K_H$	1	0
independent equivalent sites	the state with k ligands	$\frac{k}{n-k+1}K_d<R<\frac{k+1}{n-k}K_d$	$(R/K_d)^{n-k}/\binom nk$	$n-k$

A Hill exponent therefore needs a mechanism in which the partly occupied states never win, or it must be declared an empirical fit. Selecting only the full state in the binomial sum gives $[R/(K_d+R)]^n$ , a different curve from the Hill function with the same n. The regime-index paper finds Hill laws for sequential binding fragile for exactly this reason.¹⁰

Before using an effective law, name the regime it is a piece of, and check that regime's conditions on the path of the experiment.

20A competitor adds a candidate for the enzyme total

Core §11 adds an inhibitor I that binds the enzyme as $C_{EI}$ , and runs the pipeline with both ligands in excess. This section derives that run, then reruns the pipeline twice with step 3 changed. In the second run the enzyme titrates a tight inhibitor. In the third the inhibitor is a reservoir and the substrate is tight.

All three runs read one exact equation. The totals are $q_E=E+C_{ES}+C_{EI}$ , $q_S=S+C_{ES}$ and $q_I=I+C_{EI}$ , with $K_{d,S}C_{ES}=ES$ and $K_{d,I}C_{EI}=EI$ . Eliminating each ligand with its own total gives $C_{ES}=q_SE/(K_{d,S}+E)$ and $C_{EI}=q_IE/(K_{d,I}+E)$ , and so

E+\frac{q_SE}{K_{d,S}+E}+\frac{q_IE}{K_{d,I}+E}=q_E.

The left side increases from 0 to above $q_E$ as E runs from 0 to $q_E$ . So the equation has one root in that interval, and bisection finds it. Only $C_{ES}$ is active, $v=k_{\mathrm{cat}}C_{ES}$ .

Run 1: both ligands in excess

1 · Model. As above. 2 · Experiment. Sweep the substrate at fixed inhibitor levels, with both ligands far in excess of the enzyme. 3 · Candidates. Excess settles both ligand totals (§15): $S\approx q_S$ and $I\approx q_I$ . The enzyme total keeps three candidates, with $C_{ES}/E=s$ and $C_{EI}/E=i$ , where $s=S/K_{d,S}$ and $i=I/K_{d,I}$ . The winner is the largest of 1, s and i.

4 · Regime sequence. With $i<1$ the path starts in $(E,S,I)$ and crosses into $(C_{ES},S,I)$ at $s=1$ . With $i>1$ it starts in $(C_{EI},S,I)$ and crosses at $s=i$ . Each crossing is regular, because only the enzyme total changes winner. 5 · Response. Exactly, $q_E=E(1+s+i)$ , so $C_{ES}/q_E=s/(1+s+i)$ . The pieces keep the largest term of the denominator.

Path	Regime	Range of s	$C_{ES}/q_E$	Gain in s	Exact at a sample point
$i=0.1$	$(E,S,I)$	$s<1$	s	1	at $s=0.1$ : 0.0833
$i=0.1$	$(C_{ES},S,I)$	$s>1$	1	0	at $s=10$ : 0.901
$i=10$	$(C_{EI},S,I)$	$s<10$	$s/i$	1	at $s=1$ : 0.0833
$i=10$	$(C_{ES},S,I)$	$s>10$	1	0	at $s=100$ : 0.901

The competitor moves the saturation corner without touching the substrate affinity. The exact response is half saturated at $s=1+i$ , so the apparent substrate constant is $K_{d,S}(1+I/K_{d,I})$ . Its initial slope in s falls by the same factor, from 1 to $1/(1+i)$ . Leading order keeps the larger term, $K_{d,S}\max(1,i)$ . The boundaries separate leading-order laws. The exact response is smooth across all of them.

Scroll inside the figure to see it at reading size.

Figure 13. A competitor adds a candidate for the enzyme total. Left: the winner of the enzyme total with both ligands in excess, on log axes

S/K_{d,S}

and

I/K_{d,I}

, with dashed paths at

I/K_{d,I}=0.1

, 10 and 100. Right: the same paths as responses, leading-order pieces thick and the exact curves thin. The saturation corner moves to

K_{d,S}\max(1,I/K_{d,I})

Run 2: the enzyme titrates a tight inhibitor

1 · Model. As above, with finite ligands. 2 · Experiment. Sweep $q_I$ at fixed $q_E$ and $q_S=4q_E$ , and read $C_{ES}/q_E$ . The illustrative values are those of core Figure 15: $q_E=K_{d,S}=1\,\mu\mathrm M$ and $K_{d,I}=0.01\,\mu\mathrm M$ .

3 · Candidates. The substrate is in excess, $S\ge q_S-q_E=3q_E$ , so S wins its total. The enzyme and inhibitor totals can change winner. Name a regime by the winners of $(q_E,q_S,q_I)$ .

4 · Regime sequence. At low $q_I$ the path is in $(C_{ES},S,C_{EI})$ . The substrate complex wins the enzyme total because $S/K_{d,S}\approx4$ , and the tight inhibitor is almost all bound. The check $C_{EI}<C_{ES}$ reads $q_I<q_E$ and fails first. Swapping $C_{EI}$ in makes it win both the enzyme and the inhibitor totals, so the path crosses a singular ray at $q_I=q_E$ . Past it the inhibitor total passes to free I, and the path is in $(C_{EI},S,I)$ .

5 · Response. In the last regime $E\approx K_{d,I}q_E/q_I$ , and so $C_{ES}=ES/K_{d,S}\approx(K_{d,I}/K_{d,S})\,q_Eq_S/q_I$ .

Regime	Range of $q_I$	$C_{ES}/q_E$	Gain in $q_I$	Exact at a sample point
$(C_{ES},S,C_{EI})$	$q_I<q_E$	1, refined by the balance to $(1-q_I/q_E)\,S/(K_{d,S}+S)$	0	at $q_I=0.5q_E$ : 0.421, refined 0.391
singular ray	$q_I\approx q_E$	from $q_E-q_I=E+C_{ES}-I$ : about 0.17	switch	at $q_I=q_E$ : 0.156
$(C_{EI},S,I)$	$q_I>q_E$	$\dfrac{K_{d,I}}{K_{d,S}}\,\dfrac{q_S}{q_I}$	−1	at $q_I=4q_E$ : 0.0130, law 0.0100

The balance explains the first row. Subtracting the inhibitor total from the enzyme total cancels $C_{EI}$ : $q_E-q_I=E+C_{ES}-I$ . Before the ray, free I is tiny, so the enzyme that the inhibitor leaves, $q_E-q_I$ , is shared between E and $C_{ES}$ in the ratio $K_{d,S}:S$ . Each inhibitor takes one enzyme from the pool that substrate can reach. On the ray the balance and $EI\approx K_{d,I}q_E$ give $E^2(1+S/K_{d,S})\approx K_{d,I}q_E$ , hence $C_{ES}\approx0.17q_E$ . The same titration by a tight inhibitor appears in the lecturer's thesis.³

Run 3: a reservoir inhibitor and a tight substrate

1 · Model. As above. 2 · Experiment. Sweep $q_S$ at fixed $q_E$ and $q_I$ , and read free E. The illustrative values are $q_E=1$ , $K_{d,S}=0.001$ , $K_{d,I}=10$ and $q_I=100\,\mu\mathrm M$ .

3 · Candidates. The inhibitor is a reservoir: $C_{EI}/I=E/K_{d,I}\le q_E/K_{d,I}=0.1$ , so I wins its total. The inhibitor-bound and free enzyme then move together, $C_{EI}=(a-1)E$ with $a=1+q_I/K_{d,I}=11$ . Write $U=aE$ for this pool. The enzyme total becomes $q_E=U+C_{ES}$ , and $K_{d,S}C_{ES}=ES$ becomes

aK_{d,S}\,C_{ES}=U\,S.

So U and S form one pair with the effective constant $K'=aK_{d,S}=0.011$ . The competitor weakens the substrate's apparent affinity by the factor a, and a small $K_{d,S}$ alone no longer makes the pair tight.

4 · Regime sequence. The effective pair is tight, $q_E=91K'$ , so its path is a tight cut. Within the pool U the inhibitor complex outweighs free E by the factor $a-1=10$ , so $C_{EI}$ is the winner where U wins. The path runs from $(C_{EI},C_{ES},I)$ across the effective pair's singular ray at $q_S=q_E$ into $(C_{ES},S,I)$ .

5 · Response.

Regime	Range of $q_S$	Free E	Gain	Exact at a sample point
$(C_{EI},C_{ES},I)$	$q_S<q_E$	$K_{d,I}q_E/q_I=0.1$ , refined by the balance to $(q_E-q_S)/a$	0	at $q_S=0.1$ : 0.0825, refined 0.0818
singular ray	$q_S\approx q_E$	$\sqrt{K'q_E}/a=\sqrt{K_{d,S}q_E/a}=0.0095$	switch	at $q_S=q_E$ : 0.0091
$(C_{ES},S,I)$	$q_S>q_E$	$K_{d,S}q_E/q_S$	−1	at $q_S=10$ : 0.000111, law 0.000100

The ray row comes from §13 applied to the effective pair. There the deficit of the complex is about $\sqrt{K'q_E}$ , and it is shared by the pool U, of which free E is the fraction $1/a$ . Near the ray the matching must be avoided on the scale of the residual, not of the constant. With the gap $g=|q_E-q_S|$ , the smaller total $m=\min(q_E,q_S)$ and the missing complex δ, §13's balance reads $\delta(g+\delta)=K'(m-\delta)$ . The rectifier laws $E\approx(q_E-q_S)/a$ and $E\approx K_{d,S}q_E/(q_S-q_E)$ need both $K'\ll g$ and $K'm\ll g^2$ .

Read every competition from one scalar equation. Settle each ligand total by excess, titration or reservoir first, and a reservoir competitor then scales the substrate's effective constant by $1+q_I/K_{d,I}$ .

21Joint states decide how regulators combine

Core §12 reads combined regulation as a regime map of joint promoter states. This section derives the weights and the exact outputs, runs the pipeline along both paths, and then restores finite regulators.

1 · Model. Free regulators $R_1$ and $R_2$ bind a promoter with total $q_G$ . The joint states are empty, bound by one regulator, bound by the other, and bound by both. Relative to the empty promoter their weights are

w_0=1,\qquad w_1=\frac{R_1}{K_1},\qquad w_2=\frac{R_2}{K_2},\qquad w_{12}=\omega\,w_1w_2,\qquad Z=w_0+w_1+w_2+w_{12}.

Here $K_1$ and $K_2$ are the dissociation constants of each regulator from the empty promoter, and the positive factor ω is the interaction of the two bound regulators. Independent binding has $\omega=1$ , and mutual exclusion removes the double state. Each state holds $q_Gw_j/Z$ . Assign the states the activities $a_0$ , $a_1$ , $a_2$ and $a_{12}$ . Transcription is composite at this resolution, so its rate law is stated separately:

v_{\mathrm{out}}=q_G\,\frac{a_0+a_1w_1+a_2w_2+a_{12}\,\omega w_1w_2}{Z}.

Thermodynamic models of transcription enumerate promoter states in the same way. Turning an occupancy into a transcription rate is a separate assumption. It needs each state's own initiation kinetics, not only the probability that a polymerase is bound.¹⁴

Scroll inside the figure to see it at reading size.

Figure 14. A promoter response needs states and activities. The denominator sums the weights of the allowed states. The numerator weights the same states by activity. Independent binding sets

\omega=1

, and mutual exclusion removes the double state.

Checkpoint: an equilibrium interaction factor

The first regulator binds the empty promoter with dissociation constant $K_1$ . What is its dissociation constant when the second regulator is already bound, if the double state has weight $\omega w_1w_2$ ?

Worked answer

The ratio of the double-state weight to the second-regulator weight is $\omega w_1=R_1/(K_1/\omega)$ . So the conditional constant is $K_1/\omega$ , and the second regulator's is $K_2/\omega$ . Both assembly paths give the same double-state weight, which is equilibrium cycle compatibility.

2 · Experiment. Control both free regulators, as when each is in excess of the promoter. Take two paths across the plane $(w_1,w_2)$ : one regulator alone, sweeping $w_1$ with $w_2\ll1$ , and both together along the diagonal $w_1=w_2=x$ .

3 · Candidates. Each regulator wins its own total, so only the promoter total changes winner. Its candidates are the allowed joint states, and at leading order $v_{\mathrm{out}}\approx a_{\mathrm{win}}q_G$ , the activity of the winning state.

4 · Regime sequence. One regulator alone hands the promoter from G to $C_{GR_1}$ at $w_1=1$ in every architecture. On the diagonal the weights are 1, x, x and $\omega x^2$ . With independent sites and $\omega=1$ , the double state overtakes G at $x=1$ , where all four weights tie. With exclusive sites the two single states take over together at $x=1$ , along their tie ray. The interaction moves the diagonal's breakpoints. With $\omega>1$ the double state overtakes G first, at $x=\omega^{-1/2}$ . With $\omega<1$ the single states win on $1<x<1/\omega$ before the double state takes over.

5 · Response. Every active state has activity a, and $\omega=1$ for independent sites.

Path	Regime: winner of $q_G$	Range	OR	Mutual exclusion	Inactive double state
one regulator alone	G	$w_1<1$	0	0	0
one regulator alone	$C_{GR_1}$	$w_1>1$	a	a	a
diagonal $w_1=w_2=x$	G	$x<1$	0	0	0
	$C_{GR_1}$ tied with $C_{GR_2}$	$x>1$ , exclusive sites	none	a	none
	$C_{GR_1R_2}$	$x>1$ , independent sites	a	none	0

The exact outputs interpolate between these values, and at the corner $x=1$ they are furthest apart from them. With independent sites the weights factor, $Z=(1+w_1)(1+w_2)$ , and each activity assignment gives a closed form:

AND activates only the double state: $v_{\mathrm{out}}/(aq_G)=\frac{w_1}{1+w_1}\,\frac{w_2}{1+w_2}$ .
Addition sets $a_0=0$ and $a_{12}=a_1+a_2$ : $v_{\mathrm{out}}/q_G=\frac{a_1w_1}{1+w_1}+\frac{a_2w_2}{1+w_2}$ .
OR gives every bound state the activity a, so double occupation counts once: $v_{\mathrm{out}}/(aq_G)=1-\frac{1}{(1+w_1)(1+w_2)}$ .
Mutual exclusion has $Z=1+w_1+w_2$ : $v_{\mathrm{out}}/(aq_G)=\frac{w_1+w_2}{1+w_1+w_2}$ .

At the corner OR, mutual exclusion and the inactive double state give $3/4$ , $2/3$ and $1/2$ . With one regulator alone all three give the same curve, $w_1/(1+w_1)$ . The inactive double state reads 0, a, a and 0 in its four quadrants, so at leading order it computes exclusive-or. Its off quadrants approach zero only far from the corner, because the losing active states keep a share of the promoter.

Scroll inside the figure to see it at reading size.

Figure 15. Single-input curves do not determine the mixed response. Left: one input alone, where the three mechanisms coincide. Right: both inputs rising together, where they separate. The differences come from the state list and the double state's activity. Every single-input curve is identical.

A discriminating experiment therefore varies both regulators. A two-dimensional dose matrix tests the interaction directly, and a measurement of joint occupancy separates a binding change from an activity change. A response surface alone can still leave several mechanisms consistent with the data.

Which joint state changes the answer?

Control the two free regulator inputs, then change the double-state weight or activity. The empty state is inactive. Either singly occupied state has unit activity. Inputs are free concentrations divided by their dissociation constants.

w_1=R_1/K_1

w_2=R_2/K_2

1 Binding interaction

\log_{10}\omega

0 Double-state activity Mutually exclusive binding (remove double state)

State	Probability	Share of promoter pool
Empty
Regulator 1
Regulator 2
Both regulators

Output per gene in units of the single-state activity: .

Set one input to zero. Every choice gives the same isolated response. Restore both inputs to see which hidden assumption changes their combination. Finite supplied regulator totals require the additional balances derived in the text.

Finite regulators change step 3, because their totals are no longer settled. Each regulator total must count the promoter states that hold it:

q_{R_1}=R_1+q_G\,\frac{w_1+\omega w_1w_2}{Z},\qquad q_{R_2}=R_2+q_G\,\frac{w_2+\omega w_1w_2}{Z}.

These two balances fix both free regulators together, and more targets add more bound terms to the same balances. The output law is then evaluated at the resulting free concentrations. No new composition rule is needed. A solution without a short radical form is still unique, computable and differentiable, by the convex solve of §6.

Scroll inside the figure to see it at reading size.

Figure 16. Promoters can deplete the regulators they read. Both regulators have the same finite total, and every bound state is active. A larger promoter pool depletes its regulators at fixed affinities and activities. The dashed curve substitutes totals for free inputs, which holds only while the loading is small.

Checkpoint: independent binding with finite supplies

Set $q_{R_1}=q_{R_2}=q_G=1\,\mu\mathrm M$ , $K_1=K_2=1\,\mu\mathrm M$ and $\omega=1$ . Compare the either-bound fraction with the value obtained by substituting totals for free inputs.

Worked answer

Independence factors the occupancies, so each balance reads $1=R+R/(1+R)$ in micromolar units. The positive root is $R=(\sqrt5-1)/2\approx0.618\,\mu\mathrm M$ . The either-bound fraction is $1-(1+R)^{-2}\approx0.618$ , while substituting totals gives 0.75. Independent binding does not remove finite-pool depletion.

Derive combined regulation from joint states and their activities, test mechanisms along the diagonal where isolated curves agree, and close the regulator totals whenever their supplies are finite.

22At low copy number the state is a distribution

At small copy numbers the equilibrium object is a probability distribution over molecule counts, not a concentration. The core's note on promoters in §12 relies on two facts about that distribution, and this section proves both.

The first fact is an identity for occupancy. Let m labelled sites be exchangeable, and let c count the occupied ones. With $I_i$ the indicator that site i is occupied, $c=\sum_iI_i$ , and linearity of expectation gives

\Pr(I_i=1)=\frac{\mathbb E[c]}{m}.

The identity needs exchangeability, not independence. A finite ligand supply can correlate the sites while keeping their marginal probabilities equal.

The second fact is that the distribution is exact and small. Take $n_E$ enzyme molecules, $n_S$ substrate molecules and c complexes in a volume V. The association and dissociation propensities are $k_{+S}(n_E-c)(n_S-c)/(N_AV)$ and $k_{-S}c$ , where $N_A$ converts moles to molecules. Define $n_{K_d}=K_dN_AV$ , the dissociation constant counted in molecules. Detailed balance gives

\frac{\pi(c+1)}{\pi(c)}=\frac{(n_E-c)(n_S-c)}{n_{K_d}(c+1)},\qquad \pi(c)\propto\frac{(n_E)_c\,(n_S)_c}{c!\,n_{K_d}^c},

where π is the stationary distribution, $(n)_c=n(n-1)\cdots(n-c+1)$ is the falling factorial, and $0\le c\le\min(n_E,n_S)$ .

A two-enzyme example shows the distinction. With $n_E=2$ , $n_S=1$ and $n_{K_d}=0.1$ , the complex is present with probability $20/21$ , and each enzyme is occupied with probability $10/21$ . Both enzymes are never occupied together, because only one substrate exists. The deterministic root of the same numbers gives 0.916 complexes, while the stochastic mean is 0.952. The concentration model is a macroscopic description, and averages at low counts must use the distribution.

A promoter in maintained regulator reservoirs is the case where the two agree. There the normalized weights of §21 are the probabilities of the discrete promoter states, and their activity-weighted sum is the mean production rate. Lecture 6's fast-switching limit justifies using that mean for slow dynamics. A finite regulator pool needs the conditional distribution at the relevant counts instead.

Use the exchangeability identity for mean occupancy, the exact count distribution for any low-copy average, and the concentration model only where reservoirs make the two coincide.

23The open enzyme closes the dynamics

The core's coda closes a binding-catalysis model with supply and removal on the same totals. This section does it for the reversible enzyme of §9, derives the steady state exactly, and runs the pipeline on a supply sweep.

1 · Model. Order the species as $x=(E,S,P,C_{ES},C_{EP})$ . The substrate and product bind the enzyme, and catalysis is one resolved reversible chemical step:

E+S\xrightleftharpoons[k_{-S}]{k_{+S}}C_{ES}\xrightleftharpoons[k_{\mathrm{cat}}^{-}]{k_{\mathrm{cat}}^{+}}C_{EP}\xrightleftharpoons[k_{+P}]{k_{-P}}E+P.

Supply and export are composite boundary processes, $\varnothing\rightsquigarrow S$ and $P\rightsquigarrow\varnothing$ , with rate laws $v_{\mathrm{in}}$ and $v_{\mathrm{out}}=k_{\mathrm{out}}P$ . Only free product is exported.

The accumulation line follows from the species equations. Let $v_{\mathrm{b}S}=k_{+S}ES-k_{-S}C_{ES}$ and $v_{\mathrm{b}P}=k_{+P}EP-k_{-P}C_{EP}$ be the net binding fluxes, and $v_+-v_-=k_{\mathrm{cat}}^{+}C_{ES}-k_{\mathrm{cat}}^{-}C_{EP}$ the net catalytic flux. Then

\begin{aligned}\dot E&=-v_{\mathrm{b}S}-v_{\mathrm{b}P},&\dot S&=v_{\mathrm{in}}-v_{\mathrm{b}S},&\dot P&=-v_{\mathrm{b}P}-k_{\mathrm{out}}P,\\\dot C_{ES}&=v_{\mathrm{b}S}-(v_+-v_-),&\dot C_{EP}&=v_{\mathrm{b}P}+(v_+-v_-).&&\end{aligned}

Adding rows with the counts of L cancels every binding flux. The enzyme total $q_E=E+C_{ES}+C_{EP}$ is constant, and the two form totals $q_S=S+C_{ES}$ and $q_P=P+C_{EP}$ obey

\dot q_S=v_{\mathrm{in}}-(v_+-v_-),\qquad \dot q_P=(v_+-v_-)-k_{\mathrm{out}}P.

This is $\dot q=S\,v+S^{\mathrm{ext}}v^{\mathrm{ext}}$ with the matrices of §9, where the external fluxes form their own stoichiometric term.⁵ Catalysis changes the two form totals, yet each is conserved by binding, so both are slow coordinates.

The binding layer solves by one scalar equation. With $K_{d,S}C_{ES}=ES$ and $K_{d,P}C_{EP}=EP$ , eliminating each ligand with its form total gives

C_{ES}=\frac{q_SE}{K_{d,S}+E},\qquad C_{EP}=\frac{q_PE}{K_{d,P}+E},\qquad E+\frac{q_SE}{K_{d,S}+E}+\frac{q_PE}{K_{d,P}+E}=q_E.

The last equation is monotone in E on $(0,q_E)$ . At each integration step it supplies the states that catalysis and selective export read.

Scroll inside the figure to see it at reading size.

Figure 17. Five molecular states, two changing totals. Binding moves constituents between free and bound forms. Catalysis moves the substrate form into the product form. Supply and export are composite boundary processes with separately stated rate laws. Export reads free product, so bound product stays inside the model.

2 · Experiment. Vary the supply $v_{\mathrm{in}}$ and read the steady state. At steady state $v_+-v_-=v_{\mathrm{in}}=k_{\mathrm{out}}P$ , so the supply fixes $P=v_{\mathrm{in}}/k_{\mathrm{out}}$ and the form totals become outputs. The binding relation gives $C_{EP}=EP/K_{d,P}$ , and the catalytic balance gives $C_{ES}=v_{\mathrm{in}}/k_{\mathrm{cat}}^{+}+(k_{\mathrm{cat}}^{-}/k_{\mathrm{cat}}^{+})\,C_{EP}$ . The enzyme total then yields the exact steady state:

E=\frac{q_E-v_{\mathrm{in}}/k_{\mathrm{cat}}^{+}}{1+\bigl(1+k_{\mathrm{cat}}^{-}/k_{\mathrm{cat}}^{+}\bigr)\,v_{\mathrm{in}}/(k_{\mathrm{out}}K_{d,P})},\qquad S=\frac{K_{d,S}\,C_{ES}}{E}.

Every concentration is positive when $0<v_{\mathrm{in}}<k_{\mathrm{cat}}^{+}q_E$ . The upper limit is the capacity. At or above it the reduced model has no finite steady substrate inventory. The illustrative values are $q_E=K_{d,S}=K_{d,P}=1\,\mu\mathrm M$ , $k_{\mathrm{cat}}^{+}=1\,\mathrm s^{-1}$ , $k_{\mathrm{cat}}^{-}=0.2\,\mathrm s^{-1}$ and $k_{\mathrm{out}}=0.3\,\mathrm s^{-1}$ .

3 · Candidates. The enzyme total is at most either affinity, so $C_{ES}/S=E/K_{d,S}<1$ and $C_{EP}/P=E/K_{d,P}<1$ . Free S and free P win their form totals at every supply. Only the enzyme total can change winner, among E, $C_{ES}$ and $C_{EP}$ .

4 · Regime sequence. At low supply the path is in $(E,S,P)$ , with $E\approx q_E$ . The ratio $C_{EP}/E=P/K_{d,P}=v_{\mathrm{in}}/(k_{\mathrm{out}}K_{d,P})$ is exact, so the check $C_{EP}<E$ fails at $v_{\mathrm{in}}=k_{\mathrm{out}}K_{d,P}=0.3$ . The check $C_{ES}<E$ would fail later, at 0.6. The walk therefore predicts a product-bound regime $(C_{EP},S,P)$ next. Its checks hold for $k_{\mathrm{out}}K_{d,P}<v_{\mathrm{in}}<(k_{\mathrm{cat}}^{+}-k_{\mathrm{cat}}^{-})q_E$ , which is 0.3 to 0.8. Near capacity the path is in $(C_{ES},S,P)$ .

The exact curve overrules the middle prediction. The exact winners are E below $v_{\mathrm{in}}=0.278$ , the root of $10v^2+8v-3=0$ , and $C_{ES}$ above it. At the crossing $E=C_{ES}=0.342$ and $C_{EP}=0.317$ , the closest that $C_{EP}$ ever comes to winning. The predicted band is only a factor 2.7 wide, and its checks pass by margins under two. This is the warning case of §15. All three candidates stay within a factor of two of one another, so leading order names a winner that the exact solve does not confirm. A real product-inhibition regime needs $k_{\mathrm{out}}K_{d,P}\ll(k_{\mathrm{cat}}^{+}-k_{\mathrm{cat}}^{-})q_E$ , through slow export or tight product binding.

5 · Response. The substrate inventory is the readout. In $(E,S,P)$ , $S\approx K_{d,S}C_{ES}/q_E$ with $C_{ES}\approx v_{\mathrm{in}}(1+k_{\mathrm{cat}}^{-}q_E/(k_{\mathrm{out}}K_{d,P}))/k_{\mathrm{cat}}^{+}$ . Near capacity, the exact E is proportional to the capacity gap $q_E-v_{\mathrm{in}}/k_{\mathrm{cat}}^{+}$ , and S is inversely proportional to it.

Regime	Range of $v_{\mathrm{in}}$	Free E	Free S	Gain of S	Exact at a sample point
$(E,S,P)$	$v_{\mathrm{in}}<0.28$	$q_E$	$(5/3)\,v_{\mathrm{in}}$	1 in $v_{\mathrm{in}}$	at 0.05: $S=0.097$ , law 0.083
$(C_{EP},S,P)$ , predicted	0.3 to 0.8	not realized: $C_{EP}$ never exceeds 0.93 of the winner			at 0.5: $C_{EP}=0.28$ , $C_{ES}=0.56$
$(C_{ES},S,P)$	$0.28<v_{\mathrm{in}}<1$	$(1-v_{\mathrm{in}})/5$	$5/(1-v_{\mathrm{in}})$	−1 in the gap $1-v_{\mathrm{in}}$	at 0.95: $S=91.8$ , law 100

A supply step moves the state between two steady states. At $v_{\mathrm{in}}=0.2\,\mu\mathrm M\,\mathrm s^{-1}$ the free state is $(E,S,P)=(0.4444,0.5833,0.6667)\,\mu\mathrm M$ , with complexes $C_{ES}=0.2593$ and $C_{EP}=0.2963\,\mu\mathrm M$ . At $0.4\,\mu\mathrm M\,\mathrm s^{-1}$ it is $(0.2308,2.0000,1.3333)\,\mu\mathrm M$ , with complexes 0.4615 and $0.3077\,\mu\mathrm M$ . The form totals move from $(q_S,q_P)=(0.8426,0.9630)$ to $(2.4615,1.6410)\,\mu\mathrm M$ . The trajectory between them follows from the two accumulation equations and the scalar solve.

Scroll inside the figure to see it at reading size.

Figure 18. A supply step moves the equilibrium states. Supply changes from 0.2 to

0.4\,\mu\mathrm M\,\mathrm s^{-1}

at time zero, and the enzyme total stays at

1\,\mu\mathrm M

. Free and total quantities correspond to different assays, so compare each with the matching readout.

Supply, export and enzyme capacity

Change supply or export and compare the substrate and product inventories required for steady operation.

Substrate supply (µM/s) 0.4Free-product export rate (s⁻¹) 0.3Enzyme total (µM) 1

Both binding scales are 1 µM. Forward and reverse catalytic rates are 1 and 0.2 s⁻¹. Only free product is exported.

Close a binding-catalysis model by deriving the accumulation of each form total, solve its steady state exactly, and read the regimes of the enzyme total along the supply, trusting leading order only where its checks pass by a wide margin.

24Test a reduction on the output it must predict

A reduction is tested by comparing it with the resolved dynamics on the output of interest. The test keeps the equilibrium constants fixed and speeds up binding, so that only the clock separation changes.

The protocol has four parts. First, set $k_{+S}=k_{+P}=\varepsilon^{-1}\,\mu\mathrm M^{-1}\mathrm s^{-1}$ , and choose $k_{-S}$ and $k_{-P}$ to keep $K_{d,S}$ and $K_{d,P}$ fixed. The catalytic and export rates are unchanged, so ε measures binding speed against the fixed catalytic clock. Second, start the resolved and reduced models at the same state, the reduced steady state at supply $0.2\,\mu\mathrm M\,\mathrm s^{-1}$ , and step the supply to 0.4 at time zero. Third, compare the two form totals on 301 samples over 60 s, and record the largest absolute error in either. Fourth, control the numerics. The stiff resolved equations use an implicit Radau method with relative tolerance $2\times10^{-10}$ and absolute tolerance $2\times10^{-12}\,\mu\mathrm M$ . The reduced system is integrated independently by DOP853 and by fixed-step RK4, which agree to within $10^{-9}\,\mu\mathrm M$ . The resolved enzyme total stays constant to numerical precision.

ε	Largest total error (µM)	Largest species error (µM)
0.1	0.3598	0.3578
0.01	0.03397	0.03374
0.001	0.003377	0.003354

The error falls in proportion to ε across the tested range. The control speeds up only substrate binding and leaves product binding on its original clock. The fully equilibrated reduction then keeps a total error near $3.21\,\mu\mathrm M$ , however fast substrate binding becomes. A valid partial reduction would keep product binding dynamic rather than eliminate it.

Scroll inside the figure to see it at reading size.

Figure 19. The reduction improves only for the fast states it eliminates. Largest error in the form totals over 0 to 60 s against the five-species model. Both binding pairs fast: the error falls with ε. Only substrate binding fast: the error stays near

3.21\,\mu\mathrm M

. The comparison supports the reduction on the tested range, not a global error theorem.

Test a reduction against resolved dynamics with its equilibrium held fixed, and measure the error on the totals or outputs the model must predict.

25Gains enter the slow Jacobian

The gains of Part 3 are the binding layer's contribution to slow dynamics. The Jacobian of the closed model factors into stoichiometry, flux size, pool size and these gains, and each factor can be read separately.

Start with the log sensitivities of the activities. For a positive activity $v_\alpha=\sum_jk_{\alpha j}x_j$ with nonnegative coefficients, define $F_{\alpha i}=\partial\log v_\alpha/\partial\log q_i$ . The chain rule through the gain matrix $H=\partial\log x/\partial\log q$ of §7 gives

F_{\alpha i}=\sum_j\frac{k_{\alpha j}x_j}{v_\alpha}\,H_{ji},\qquad D_qv=\operatorname{diag}(v)\,F\,\operatorname{diag}(q)^{-1}.

The weights $k_{\alpha j}x_j/v_\alpha$ are the shares of the activity carried by each state, and they sum to one. So the sensitivity of a process is the share-weighted average of its states' gains. At leading order one state carries the activity, and its row of $L^{\mathsf T}L_\star^{-1}$ becomes the process's sensitivity (§11).

The Jacobian of the slow model follows. For $\dot q=S\,v+S^{\mathrm{ext}}v^{\mathrm{ext}}$ , with $S=L\Gamma_{\mathrm{cat}}$ ,

J_q=S\,\operatorname{diag}(v)\,F\,\operatorname{diag}(q)^{-1}+D_q\bigl(S^{\mathrm{ext}}v^{\mathrm{ext}}\bigr).

Stoichiometry enters through S, flux size through $\operatorname{diag}(v)$ , pool size through $\operatorname{diag}(q)^{-1}$ , and binding through F. Within a regime F is a matrix of rational gains, so the Jacobian inherits the regime's structure. Lecture 9 uses this structure for stability, and Lecture 11 develops the geometry of the possible gains.

Directional activities must stay separate. Write $v_+$ and $v_-$ as two positive activities before taking logarithms. A net flux can vanish at equilibrium or at a stall, where its logarithm is undefined. The derivative of the net flux is still finite, as the difference of two well-defined terms. Keeping the two directions preserves the physics near balance and avoids spurious infinite sensitivities.

Build the slow Jacobian from stoichiometry, fluxes, pools and binding gains, keeping forward and reverse activities as separate positive terms.

Optional · further runs of the pipeline

The pipeline on further mechanisms

Core A1 to A3 run the pipeline three more times. This part derives those runs and adds four mechanisms the core leaves out: negative cooperativity, hidden conformations, series binding and a modification cycle. It ends with the pipeline stated for a system that no lecture has met.

26The winner of the repressor total sets the Hill degree

Core A1 shows that a dimerizing repressor gives a Hill law of degree two in total repressor only in one regime. This section derives the exact inventory, the exact midpoint and the table, with exact values beside the laws.

1 · Model. Two free repressor monomers R bind as the dimer $C_{RR}$ , and the dimer binds the promoter G as $C_{GRR}$ :

2R\xrightleftharpoons[k_{-\mathrm{dim}}]{k_{+\mathrm{dim}}}C_{RR},\qquad G+C_{RR}\xrightleftharpoons[k_{-\mathrm{prom}}]{k_{+\mathrm{prom}}}C_{GRR}.

The dimer association flux is defined as $k_{+\mathrm{dim}}R^2$ , so any combinatorial factor sits inside that rate constant. The dissociation constants are $K_{\mathrm{dim}}=k_{-\mathrm{dim}}/k_{+\mathrm{dim}}$ and $K_{\mathrm{prom}}=k_{-\mathrm{prom}}/k_{+\mathrm{prom}}$ , both concentrations. Only free G produces, through the composite process $G\rightsquigarrow G+X$ with rate law $v=k_0G$ . The binding relations and totals are

C_{RR}=\frac{R^2}{K_{\mathrm{dim}}},\qquad C_{GRR}=\frac{G\,C_{RR}}{K_{\mathrm{prom}}},\qquad q_G=G+C_{GRR},\qquad q_R=R+2C_{RR}+2C_{GRR}.

The promoter total eliminates G, and the repressor total becomes one scalar equation for free monomer:

\frac{G}{q_G}=\frac{1}{1+R^2/K_H^2},\qquad q_R=R+\frac{2R^2}{K_{\mathrm{dim}}}+\frac{2q_GR^2}{K_H^2+R^2},\qquad K_H=\sqrt{K_{\mathrm{dim}}K_{\mathrm{prom}}}.

The right side of the second equation increases from zero to infinity with R, so every repressor total has one root. The solve keeps both dimer storage and promoter loading.

2 · Experiment. Sweep the repressor total $q_R$ , and read the free promoter fraction $G/q_G$ .

3 · Candidates. The promoter is in trace, $2q_G\ll q_R$ , so $C_{GRR}$ never wins the repressor total (§15). The repressor total has candidates R and $2C_{RR}$ , which tie at $R=K_{\mathrm{dim}}/2$ . The promoter total has candidates G and $C_{GRR}$ , which tie at $R=K_H$ . A regime is named by the winners of $(q_R,q_G)$ .

4 · Regime sequence. The order of the two ties decides the sequence. When $K_H<K_{\mathrm{dim}}/2$ , the promoter switches while the monomer still wins, and the path runs through $(R,G)$ , $(R,C_{GRR})$ and $(C_{RR},C_{GRR})$ . When $K_H>K_{\mathrm{dim}}/2$ , the dimer takes the repressor total first, and the path runs through $(R,G)$ , $(C_{RR},G)$ and $(C_{RR},C_{GRR})$ . Every crossing is regular.

5 · Response. Where the monomer wins, $R\approx q_R$ . Where the dimer wins, $C_{RR}\approx q_R/2$ , and the binding relation gives $R\approx(K_{\mathrm{dim}}q_R/2)^{1/2}$ . The monomer's gain drops from 1 to 1/2 at that crossing, a rational change (§15). The exact column uses the two panels of Figure 20, both with $K_H=1\,\mu\mathrm M$ and $q_G=10^{-4}\,\mu\mathrm M$ .

Regime	Range of $q_R$	Free monomer R	$G/q_G$	Gain	Exact at a sample point
$(R,G)$	$q_R<\min(K_H,K_{\mathrm{dim}}/2)$	$q_R$	1	0	left, $q_R=0.1$ : 0.990
$(R,C_{GRR})$	$K_H<q_R<K_{\mathrm{dim}}/2$	$q_R$	$(K_H/q_R)^2$	−2	left, $q_R=10$ : 0.0135, law 0.01
$(C_{RR},G)$	$K_{\mathrm{dim}}/2<q_R<2K_{\mathrm{prom}}$	$(K_{\mathrm{dim}}q_R/2)^{1/2}$	1	0	right, $q_R=10$ : 0.953
$(C_{RR},C_{GRR})$	$q_R>\max(K_{\mathrm{dim}}/2,\,2K_{\mathrm{prom}})$	$(K_{\mathrm{dim}}q_R/2)^{1/2}$	$2K_{\mathrm{prom}}/q_R$	−1	left, $q_R=1000$ : $2.5\times10^{-5}$ , law $2\times10^{-5}$ . Right: 0.167, law 0.2

So a squared dependence on free monomer becomes a squared dependence on total repressor only in $(R,C_{GRR})$ , where the monomer wins the repressor total. The lecturer's thesis derives these dimer orders, and the regime-index paper gives the same regime sequence.^{3, 10}

The exact midpoint tests which regime a measured curve is in. At half activity $R=K_H$ , so $C_{RR}=K_{\mathrm{prom}}$ and $C_{GRR}=q_G/2$ . The repressor total there is

q_{R,1/2}=K_H+2K_{\mathrm{prom}}+q_G.

The three terms count free monomer, free-dimer constituents and promoter-bound constituents at the midpoint. A Hill-2 law in total repressor needs the first term to exceed both others. In the left panel the terms are 1, 0.02 and 0.0001, so the midpoint is $1.0201\,\mu\mathrm M$ and the monomer wins. In the right panel they are 1, 200 and 0.0001, so the dimer wins, and the midpoint sits at $2K_{\mathrm{prom}}$ .

Scroll inside the figure to see it at reading size.

Figure 20. The winner of the repressor total sets the Hill degree. Left:

K_{\mathrm{dim}}=100\,\mu\mathrm M

and

K_{\mathrm{prom}}=0.01\,\mu\mathrm M

, so the path runs through

(R,G)

and

(R,C_{GRR})

, and the slope is −2. Right: the two values swapped, so the path runs through

(C_{RR},G)

and

(C_{RR},C_{GRR})

, and the slope is −1. Thick pieces are leading order. Thin gray curves solve the finite-pool model with promoter total

10^{-4}\,\mu\mathrm M

. This is the core's Figure 19.

Self-repression closes this model, and the closure needs a removal law. Let the product X be the repressor itself, so production is $G\rightsquigarrow G+R$ . The repressor total then obeys $\dot q_R=k_0G-v_{\mathrm{out}}$ , where $v_{\mathrm{out}}$ must say which protein states are removed. Removing every form at one frequency $\delta_R$ gives $v_{\mathrm{out}}=\delta_Rq_R$ . Removing only free monomer gives $v_{\mathrm{out}}=\delta_RR$ . The two choices give different slow dynamics from the same equilibrium.

Count every protein copy, find the winner of the repressor total, and only then assign a Hill degree to a response plotted against total protein.

27A sponge moves the threshold

Core A2 shows that a tight sponge moves a promoter's threshold to the sponge total. This section derives the exact inventory, runs the pipeline twice, and then shows that the same stored material slows the free signal.

1 · Model. Free repressor R binds the promoter G or the sponge T:

G+R\xrightleftharpoons[k_{-G}]{k_{+G}}C_{GR},\qquad T+R\xrightleftharpoons[k_{-T}]{k_{+T}}C_{RT},\qquad K_{GR}=\frac{GR}{C_{GR}},\qquad K_{RT}=\frac{RT}{C_{RT}}.

Only free G produces. The totals are $q_G=G+C_{GR}$ , $q_T=T+C_{RT}$ and $q_R=R+C_{GR}+C_{RT}$ . Eliminating G and T with their totals gives one monotone equation for free repressor:

q_R=R+\frac{q_GR}{K_{GR}+R}+\frac{q_TR}{K_{RT}+R},\qquad \frac{G}{q_G}=\frac{K_{GR}}{K_{GR}+R}.

Half repression needs $R=K_{GR}$ , whatever the sponge does. There the promoter holds $q_G/2$ and the sponge holds $q_TK_{GR}/(K_{RT}+K_{GR})$ , so the exact midpoint is

q_{R,1/2}=K_{GR}+\frac{q_G}{2}+\frac{q_TK_{GR}}{K_{RT}+K_{GR}}.

2 · Experiment. Sweep $q_R$ at a fixed sponge total $q_T$ , and read $G/q_G$ .

Run 1: a trace promoter and a tight sponge

3 · Candidates. The promoter is in trace, so $C_{GR}$ never wins the repressor total. The sponge and repressor totals form the one-pair map of §12, with T in the enzyme's role and R in the substrate's. The promoter total has candidates G and $C_{GR}$ , which tie at $R=K_{GR}$ . A regime is named by the winners of $(q_T,q_R,q_G)$ .

4 · Regime sequence. The sponge is tight, $K_{RT}\ll K_{GR}<q_T$ . While $q_R<q_T$ the path is in $(T,C_{RT},G)$ , and the free remainder $R\approx K_{RT}q_R/q_T$ stays below $K_{RT}$ . At $q_R=q_T$ the path crosses the sponge pair's singular ray. Past it, free repressor wins its total, and the path is in $(C_{RT},R,C_{GR})$ .

5 · Response. The exact column uses the parameters of the core's Figure 20: $K_{GR}=0.2$ , $K_{RT}=0.005$ , $q_G=0.02$ and $q_T=1\,\mu\mathrm M$ .

Regime	Range of $q_R$	Free repressor R	$G/q_G$	Gain	Exact at a sample point
$(T,C_{RT},G)$	$q_R<q_T$	$K_{RT}q_R/q_T$	1	0	at $q_R=0.5$ : 0.976
singular ray	$q_R\approx q_T$	from $q_R-q_T=R-T$	falls from 1 to $K_{GR}/q_T$	switch	at $q_R=1$ : 0.753
$(C_{RT},R,C_{GR})$	$q_R>q_T$	$q_R$ , refined to $q_R-q_T$	$K_{GR}/q_R$ , refined to $K_{GR}/(K_{GR}+q_R-q_T)$	−1	at $q_R=2$ : 0.168, law 0.1, refined 0.167

The refined column comes from the ray's balance. With the promoter in trace, $q_R-q_T=R-T$ . Past the ray the free sponge is small, so $R\approx q_R-q_T$ , and half repression needs $q_R\approx q_T+K_{GR}$ . The exact midpoints for sponge totals 0, 1 and $2\,\mu\mathrm M$ are 0.21, 1.19 and $2.16\,\mu\mathrm M$ . The DnaA boxes of §17 act as such a sponge.¹²

The shifted law needs its own conditions, and they are stricter than the midpoint's. The balance shows that replacing R by $q_R-q_T$ needs $C_{GR}\ll R$ and $T\ll R$ . Sufficient conditions are $q_G\ll K_{GR}$ , $R\gg K_{RT}$ and $q_TK_{RT}\ll R^2$ . The last one bounds how close to the ray the shifted law can be used. A nearly full sponge can still leave enough free sites to violate it, so use the exact equation whenever a small difference matters.

Scroll inside the figure to see it at reading size.

Figure 21. An added binding partner changes the available signal. Exact solutions at sponge totals 0, 4 and

8\,\mu\mathrm M

, with

K_{GR}=1\,\mu\mathrm M

K_{RT}=0.01\,\mu\mathrm M

and

q_G=0.1\,\mu\mathrm M

. The midpoints fall at 1.05, 5.01 and

8.97\,\mu\mathrm M

, as the exact midpoint formula gives. One scalar solve covers the unrepressed range, the crossover and the repressed range.

Run 2: a tight promoter makes a rectifier

3 · Candidates. A tight promoter changes this step, because the promoter total is no longer in trace. The complex $C_{GR}$ now becomes a candidate of the repressor total. Take $K_{RT}\ll K_{GR}\ll q_G$ , so the sponge binds first and the promoter binds next. The illustrative values are $K_{GR}=0.001$ , $K_{RT}=10^{-5}$ , $q_G=1$ and $q_T=2\,\mu\mathrm M$ .

4 · Regime sequence. Past the sponge's ray the balance holds over an interval, not just on the ray. The complex $C_{RT}$ wins both the sponge and the repressor totals from $q_R=q_T$ onward. Their difference is then a total of its own, with an exact balance:

q_R-q_T=R+C_{GR}-T.

Its candidates are walked like any others. While the promoter has room, $C_{GR}$ carries the balance, so $G\approx q_G-(q_R-q_T)$ . Once the promoter is full, free repressor carries it, $R\approx q_R-q_T-q_G$ , and $G\approx K_{GR}q_G/R$ .

5 · Response. Dividing each regime's law for G by $q_G$ gives the table.

Regime	Range of $q_R$	Carrier of $q_R-q_T$	$G/q_G$	Exact at a sample point
$(T,C_{RT},G)$	$q_R<q_T$	none	1	at $q_R=1$ : 0.990
$C_{RT}$ wins $q_T$ and $q_R$	$q_T<q_R<q_T+q_G$	$C_{GR}$	$1-(q_R-q_T)/q_G$	at $q_R=2.5$ : 0.483, law 0.5
$C_{RT}$ wins $q_T$ and $q_R$ , $C_{GR}$ wins $q_G$	$q_R>q_T+q_G$	R	$K_{GR}/(q_R-q_T-q_G)$	at $q_R=4$ : 0.0010, law 0.0010

The free promoter therefore follows $G\approx\max\bigl(q_G-\max(q_R-q_T,0),0\bigr)$ , a rectifier with a linear ramp of width $q_G$ . The balance also supplies an exact error diagnostic. Combining the totals gives

G-\bigl(q_G-(q_R-q_T)\bigr)=R-T.

The ramp is accurate where the free pools are small against the signal, $R+T\ll\min(q_R-q_T,\;q_G-q_R+q_T)$ . At $q_R=2.5\,\mu\mathrm M$ the diagnostic is $-0.017\,\mu\mathrm M$ , which is the whole gap between 0.483 and 0.5. The condition compares omitted pools with the intended signal. An affinity hierarchy alone does not guarantee it near either end of the ramp.

A sponge also slows the free signal

The stored material that moves the threshold also buffers changes in free R. Take supply and removal as composite processes, $\varnothing\rightsquigarrow R$ with rate law u and $R\rightsquigarrow\varnothing$ with rate law $\delta_RR$ , so only free repressor is removed. The total obeys $\dot q_R=u-\delta_RR$ . Differentiating the equilibrium inventory in R gives

\frac{dq_R}{dR}=\beta(R)=1+\frac{q_GK_{GR}}{(K_{GR}+R)^2}+\frac{q_TK_{RT}}{(K_{RT}+R)^2},\qquad \dot R=\frac{u-\delta_RR}{\beta(R)}.

Here β is the buffering factor, the change in total repressor per unit change in free repressor. Because $\beta\ge1$ , loading slows every change in free R. Near a steady state the relaxation time is $\beta(R^*)/\delta_R$ , where $R^*=u/\delta_R$ is the steady free repressor. The steady free level does not depend on the sponge at fixed supply. The total and the transient do. At fixed total, by contrast, adding a sponge lowers free R directly. Retroactivity theory makes exactly this distinction between a module in isolation and a module under load.¹⁵

Add a finite sponge

Increase the sponge amount and see how much total repressor is needed to repress the same gene.

Repressor total (µM) 5Sponge total (µM) 4log₁₀ sponge dissociation scale (µM) -2

Gene total 0.1 µM; gene–repressor dissociation scale 1 µM. Both complexes count toward total repressor.

Solve static sequestration and dynamic buffering from one shared inventory, and use the ray's balance to place the threshold and to test any shifted law.

28Extra states create windows and conformational switches

Core A3 shows that one inactive extra state turns a rising response into a window. This section derives the window's exact peak and then runs the pipeline on a conformational ensemble, the MWC model.

Run 1: a window from an inactive double state

1 · Model. A regulator binds a promoter twice in sequence, as $C_{GR}$ and then $C_{GRR}$ , with stepwise dissociation constants $K_1$ and $K_2$ . The state weights are 1, $R/K_1$ and $R^2/(K_1K_2)$ . Only the singly bound state produces, at activity a:

\frac{v}{a\,q_G}=\frac{C_{GR}}{q_G}=\frac{R/K_1}{1+R/K_1+R^2/(K_1K_2)}.

2 · Experiment. Sweep a controlled free regulator R, held in excess of the promoter. 3 · Candidates. Only the promoter total changes winner, among G, $C_{GR}$ and $C_{GRR}$ .

4 · Regime sequence. When $K_1<K_2$ , the three states win in turn, with breakpoints at $R=K_1$ and $R=K_2$ . When $K_1>K_2$ , the singly bound state never wins, and G hands the promoter directly to $C_{GRR}$ at $R=\sqrt{K_1K_2}$ .

5 · Response. The illustrative values are those of the core's Figure 21: $K_1=0.1$ and $K_2=10\,\mu\mathrm M$ for the open window, swapped for the closed one.

Window	Regime	Range of R	$C_{GR}/q_G$	Gain	Exact at a sample point
open, $K_1<K_2$	G	$R<K_1$	$R/K_1$	1	at $R=0.01$ : 0.0909, law 0.1
	$C_{GR}$	$K_1<R<K_2$	1	0	at $R=1$ : 0.833
	$C_{GRR}$	$R>K_2$	$K_2/R$	−1	at $R=100$ : 0.0909, law 0.1
closed, $K_1>K_2$	G	$R<\sqrt{K_1K_2}$	$R/K_1$	1	peak at $R=1$ : 0.0476, law 0.1
closed, $K_1>K_2$	$C_{GRR}$	$R>\sqrt{K_1K_2}$	$K_2/R$	−1	peak at $R=1$ : 0.0476, law 0.1

The exact peak confirms the table. Differentiating the fraction in R gives a numerator proportional to $1-R^2/(K_1K_2)$ , so the maximum is at $R=\sqrt{K_1K_2}$ in both cases. With $s=\sqrt{K_2/K_1}$ , the weights there are 1, s and 1, and the peak is

\max_R\frac{C_{GR}}{q_G}=\frac{s}{2+s}.

For the open window $s=10$ and the peak is 0.833. For the closed window $s=0.1$ and the peak is 0.0476, while leading order gives s = 0.1. The ratio stays above 1/3, the bound for a total with three terms (§15).

Checkpoint: activate the double state too

Core A3 asks what changes when the double state also produces at activity a. Which end of the curve changes, and which state wins there?

Worked answer

The output becomes $(R/K_1+R^2/(K_1K_2))/Z$ , one minus the empty fraction. The low end is unchanged, because G still wins below the first breakpoint. The high end changes. There $C_{GRR}$ wins the promoter total, and it is now active, so the output saturates at a instead of falling as $K_2/R$ . The window becomes a monotone saturation.

Run 2: a conformational switch

1 · Model. An allosteric protein with n equivalent sites takes two concerted conformations. MWC's historical letters label them R, for relaxed, which binds ligand tightly, and T, for tense, which binds it weakly. Lecture 7 called them active and inactive. Let L be the free ligand, $K_R$ and $K_T$ the site dissociation constants of the two conformations, and Λ the ratio of unliganded T to unliganded R. The state weights sum to

Z(L)=\Bigl(1+\frac{L}{K_R}\Bigr)^n+\Lambda\Bigl(1+\frac{L}{K_T}\Bigr)^n.

The binomial expansion of each power counts the occupancy states of one conformation. The relaxed fraction is $f_R=(1+L/K_R)^n/Z$ . The mean number of bound ligands per protein is $\bar m=L\,d\log Z/dL$ . These are different readouts of one ensemble, and neither gives a catalytic rate without the activities of the states.¹⁶

2 · Experiment. Sweep free ligand in reservoir, and read $f_R$ . The illustrative values are $\Lambda=1000$ , $n=4$ , $K_R=1$ and $K_T=100\,\mu\mathrm M$ .

3 · Candidates. The protein total has two pooled candidates, the relaxed and the tense ensembles. At leading order each factor $1+L/K$ is replaced by its larger term, $\max(1,L/K)$ . That costs at most a factor of two per site, so at most $2^n=16$ per ensemble.

4 · Regime sequence. The tense ensemble wins at low ligand, because its weight is Λ against 1. The relaxed ensemble's weight grows as $(L/K_R)^n$ once $L>K_R$ , while the tense weight stays near Λ below $K_T$ . The relaxed ensemble therefore takes over at $L^\star=K_R\Lambda^{1/n}=5.62\,\mu\mathrm M$ , which lies below $K_T$ as the step assumed.

5 · Response.

Winning ensemble	Range of L	$f_R$	Gain	Exact at a sample point
tense, relaxed sites empty	$L<K_R$	$1/\Lambda$	0	at $L=0.1$ : 0.00146, law 0.001
tense, relaxed sites filling	$K_R<L<L^\star$	$(L/K_R)^n/\Lambda$	n	at $L=2$ : 0.0696, law 0.016
relaxed	$L>L^\star$	1	0	at $L=20$ : 0.990

The errors sit where §15 puts them. At the corner $L=K_R$ the exact value is 15 times the law, just inside the bound of 16. At the tie $L=L^\star$ the exact fraction is 0.607. The pooled table is coarser than the true per-state sequence. Resolving each occupancy state as its own candidate gives the finer staircase of §19.

A finite ligand total closes the calculation with one more balance, $q_L=L+q_{\mathrm{prot}}\,\bar m(L)$ , where $q_{\mathrm{prot}}$ is the protein total. In the full state representation each conformation and occupancy is its own species, with its own column of the counting matrix. States with equal composition may carry different weights, and a homodimer's column counts two copies of its monomer.

List every conformation and occupancy as a candidate, read windows and switches from the order in which they win, and restore finite ligand accounting before reading an ensemble curve as a response to total input.

29Negative cooperativity makes a buffer and a threshold

A two-site receptor separates binding cooperativity from response sharpness. With negative cooperativity, the singly occupied state absorbs the first ligand, and a threshold appears in total ligand. The pipeline shows where the threshold sits and why it needs finite supply.

1 · Model. Let $q_R$ be the receptor total, L the free ligand, and $K_1$ , $K_2$ the microscopic dissociation constants of the first and second occupation. The two singly occupied states are equivalent, so their pooled weight carries a factor of two:

w_0=1,\qquad w_1=\frac{2L}{K_1},\qquad w_2=\frac{L^2}{K_1K_2},\qquad Z=w_0+w_1+w_2,\qquad p_i=\frac{w_i}{Z}.

Here $p_i$ is the fraction of receptors with i ligands, so $C_{RL}=q_Rp_1$ and $C_{RL_2}=q_Rp_2$ . The cooperativity is $c=K_1/K_2$ . Negative cooperativity means $c<1$ : the first ligand binds more readily than the second. Only the doubly occupied receptor is active, so the output is $p_2$ . The ligand total closes the model:

q_L=L+q_R\,(p_1+2p_2).

2 · Experiment. Sweep $q_L$ at $q_R=1\,\mu\mathrm M$ , and read $p_2$ . The illustrative affinities are $(K_1,K_2)=(10^{-4},10^{-2})\,\mu\mathrm M$ , so $c=0.01$ .

3 · Candidates. Both totals can change winner. The receptor total has candidates R, $C_{RL}$ and $C_{RL_2}$ . The ligand total has candidates L, $C_{RL}$ and $2C_{RL_2}$ . A regime is named by the winners of $(q_R,q_L)$ .

4 · Regime sequence. The path starts in $(R,C_{RL})$ : each ligand fills a first site. The binding relations then give $L\approx K_1q_L/(2q_R)$ and $C_{RL_2}\approx(c/4)\,q_L^2/q_R$ . The check $C_{RL}<R$ fails at $q_L=q_R$ , where $C_{RL}$ wins both totals, a singular ray. Past it the path is in $(C_{RL},C_{RL_2})$ . A second singular ray at $q_L=2q_R$ hands both totals to $C_{RL_2}$ , and the path ends in $(C_{RL_2},L)$ .

The two rays carry balances, and with tight binding they hold across the whole middle interval. Subtracting totals cancels the complex that wins both:

q_L-q_R=L+C_{RL_2}-R,\qquad q_L-2q_R=L-C_{RL}-2R.

Between the rays the free pools L and R are small, so the first balance gives $p_2\approx q_L/q_R-1$ . This is sharper than the regular law $q_L/(2q_R)$ , which holds only within its factor of two.

5 · Response.

Regime	Range of $q_L$	$p_2$	Gain	Exact at a sample point
$(R,C_{RL})$	$q_L<q_R$	$(c/4)(q_L/q_R)^2$	2	at $q_L=0.5$ : 0.0012, law 0.0006
singular ray	$q_L\approx q_R$	from $q_L-q_R=L+C_{RL_2}-R$	threshold	at $q_L=1$ : 0.045
$(C_{RL},C_{RL_2})$	$q_R<q_L<2q_R$	$q_L/(2q_R)$ , balance $q_L/q_R-1$	1	at $q_L=1.5$ : 0.483, law 0.75, balance 0.5
singular ray	$q_L\approx2q_R$	from $q_L-2q_R=L-C_{RL}-2R$	corner	at $q_L=2$ : 0.868
$(C_{RL_2},L)$	$q_L>2q_R$	1	0	at $q_L=3$ : 0.981
positive cooperativity, $c=100$ : $(R,C_{RL_2})$	$q_L<2q_R$	$q_L/(2q_R)$	1	at $q_L=1$ : 0.454, law 0.5

The table shows a buffer and then a threshold. With $c=0.01$ the singly occupied state soaks up the first $q_R$ of ligand and produces nothing. Output then rises along the ramp $q_L/q_R-1$ . With positive cooperativity, $c=100$ , the singly occupied state never wins, and the output rises from the start with no threshold. Finite affinity rounds the corners.

Scroll inside the figure to see it at reading size.

Figure 22. Negative cooperativity can sharpen a total-ligand response. The symmetric two-site model with receptor total

1\,\mu\mathrm M

. The affinity pairs

(K_1,K_2)

are

(10^{-4},10^{-2})

(10^{-3},10^{-3})

and

(10^{-2},10^{-4})\,\mu\mathrm M

, giving

c=0.01

, 1 and 100. The output is double occupancy. Left: against controlled free ligand. Right: against total ligand. The panels use different input scales. They are model calculations, not digitized data.

Ha and Ferrell tested the effect with DNA oligonucleotides. Their receptor strand bound two distinct ligand strands, and overlapping, separated or adjacent binding sequences changed the cooperativity. They measured the doubly bound product while varying the supplied ligands. The symmetric identical-ligand calculation above is their conceptual model. It is not a claim that their assay used identical ligands.¹⁷

The activity assignment is essential. If each bound ligand contributes independently, the output is proportional to $p_1+2p_2$ , and the singly occupied receptors already produce half the full output. The buffering interval then vanishes in the tight limit. Negative cooperativity alone does not make a sharp regulatory response.

Read cooperativity from which occupied state wins, and read sharpness from the chosen activity and the ligand supply.

30Hidden conformations pool into one effective pair

Two conformations with the same composition test the whole construction. The pipeline runs on an effective pair, and its table shows which parameters a bulk response can identify.

1 · Model. Enzyme E binds substrate S into an open complex $C_{ES}^{\mathrm{o}}$ , which interconverts quickly with a closed complex $C_{ES}^{\mathrm{c}}$ . The fast mechanism is resolved as

E+S\xrightleftharpoons[k_{\mathrm{off}}]{k_{\mathrm{on}}}C_{ES}^{\mathrm{o}},\qquad C_{ES}^{\mathrm{o}}\xrightleftharpoons[k_{\mathrm{co}}]{k_{\mathrm{oc}}}C_{ES}^{\mathrm{c}}.

Only the closed state is productive. Its slow conversion $C_{ES}^{\mathrm{c}}\rightsquigarrow E+P$ is composite at this resolution, with rate law $v=k_{\mathrm c}C_{ES}^{\mathrm{c}}$ . With species ordered $(E,S,C_{ES}^{\mathrm{o}},C_{ES}^{\mathrm{c}})$ , the fast totals are $q_E=E+C_{ES}^{\mathrm{o}}+C_{ES}^{\mathrm{c}}$ and $q_S=S+C_{ES}^{\mathrm{o}}+C_{ES}^{\mathrm{c}}$ , and

L=\begin{pmatrix}1&0&1&1\\0&1&1&1\end{pmatrix},\qquad N^{\mathrm{aa}}=\begin{pmatrix}-1&-1&1&0\\-1&-1&0&1\end{pmatrix}.

This L repeats a column, so it is isomer-atomic but not faithful-atomic (§4). Define $K_{\mathrm o}=k_{\mathrm{off}}/k_{\mathrm{on}}$ , a concentration, and $\rho=k_{\mathrm{oc}}/k_{\mathrm{co}}$ , a pure number. Binding gives $C_{ES}^{\mathrm{o}}=ES/K_{\mathrm o}$ and $C_{ES}^{\mathrm{c}}=\rho\,ES/K_{\mathrm o}$ . With $C_{\mathrm{bound}}=C_{ES}^{\mathrm{o}}+C_{ES}^{\mathrm{c}}$ , the finite problem reduces exactly to one pair:

K_{\mathrm{eff}}C_{\mathrm{bound}}=ES,\qquad K_{\mathrm{eff}}=\frac{K_{\mathrm o}}{1+\rho},\qquad v=k_{\mathrm{eff}}C_{\mathrm{bound}},\qquad k_{\mathrm{eff}}=\frac{\rho\,k_{\mathrm c}}{1+\rho}.

For a closed substrate-to-product experiment the slow balances are $\dot q_E=0$ , $\dot q_S=-v$ and $\dot q_P=v$ . Product does not bind here, so $q_P=P$ , and including it adds one identity row and one free column to L.

2 · Experiment. Sweep $q_S$ at fixed $q_E$ , and read the flux v. 3 · Candidates. The two conformations stay in the fixed ratio ρ at equilibrium, so they enter as one pooled candidate, $C_{\mathrm{bound}}$ . The pooling rests on that equilibrium ratio, not on equal composition alone. If the conformational step were slow, both states would stay dynamic.

4 · Regime sequence. The one-pair map of §12 applies with $K_{\mathrm{eff}}$ in place of $K_d$ . On the weak cut, $q_E\ll K_{\mathrm{eff}}$ , the path runs from $(E,S)$ into $(C_{\mathrm{bound}},S)$ at $q_S=K_{\mathrm{eff}}$ .

5 · Response.

Regime	Range of $q_S$	Flux v	Gain	Parameters it depends on
$(E,S)$	$q_S<K_{\mathrm{eff}}$	$(k_{\mathrm{eff}}/K_{\mathrm{eff}})\,q_Eq_S=(\rho k_{\mathrm c}/K_{\mathrm o})\,q_Eq_S$	1	$\rho k_{\mathrm c}/K_{\mathrm o}$ only
$(C_{\mathrm{bound}},S)$	$q_S>K_{\mathrm{eff}}$	$k_{\mathrm{eff}}q_E$	0	$\rho k_{\mathrm c}/(1+\rho)$ only

The last column exposes an identification limit. Every piece, and the exact curve, depends on the parameters only through $K_{\mathrm{eff}}$ and $k_{\mathrm{eff}}$ . For any positive ρ, setting $K_{\mathrm o}=(1+\rho)K_{\mathrm{eff}}$ and $k_{\mathrm c}=(1+\rho)k_{\mathrm{eff}}/\rho$ gives exactly the same flux curve at every substrate total. With $K_{\mathrm{eff}}=1\,\mu\mathrm M$ and $k_{\mathrm{eff}}=1\,\mathrm s^{-1}$ , ρ of 1 or 9 makes one half or nine tenths of the bound enzyme closed. A closed-state measurement separates mechanisms that total binding and bulk flux cannot.

Scroll inside the figure to see it at reading size.

Figure 23. Identical bulk responses can hide different active states. The two models share

K_{\mathrm{eff}}=1\,\mu\mathrm M

k_{\mathrm{eff}}=1\,\mathrm s^{-1}

and

q_E=1\,\mu\mathrm M

, so their flux curves coincide at every substrate total. Their conformational allocation differs. The parameters are constructed to show an exact ambiguity, not fitted to data.

Pool conformations that share a composition only through their equilibrium ratio, and separate the parameters a bulk response identifies from those that need a state-resolved measurement.

31Series binding tunes amplitude or threshold

A series architecture uses the first complex as the second binding partner. The downstream complex then enters the upstream totals, so the exact problem is coupled. The pipeline shows two hierarchies, one that tunes the output amplitude and one that tunes the input threshold.

1 · Model. Ligand L binds receptor R, and the complex binds a partner P:

L+R\xrightleftharpoons[k_{1,-}]{k_{1,+}}C_{LR},\qquad C_{LR}+P\xrightleftharpoons[k_{2,-}]{k_{2,+}}C_{LRP}.

The dissociation constants are $K_1=k_{1,-}/k_{1,+}$ and $K_2=k_{2,-}/k_{2,+}$ . The active output is $C_{LRP}$ . The totals are

q_L=L+C_{LR}+C_{LRP},\qquad q_R=R+C_{LR}+C_{LRP},\qquad q_P=P+C_{LRP}.

One scalar solve gives the exact state. Let $B=C_{LR}+C_{LRP}$ be the ligand-bound receptor. Then $C_{LR}=(q_L-B)(q_R-B)/K_1$ and $C_{LRP}=q_PC_{LR}/(K_2+C_{LR})$ , so B is the root of

F(B)=B-C_{LR}(B)-\frac{q_P\,C_{LR}(B)}{K_2+C_{LR}(B)}=0,\qquad 0\le B\le\min(q_L,q_R).

On that interval $C_{LR}(B)$ decreases, so F increases. F is nonpositive at the left end and nonnegative at the right end. Bisection therefore finds the unique physical state, with no reservoir assumption.

The downstream partner changes the upstream occupancy. Binding P lowers the free binary complex needed to hold a given B, so it draws more ligand and receptor into the bound population. Adding P can raise the first layer's occupancy with every binding constant fixed.

Scroll inside the figure to see it at reading size.

Figure 24. A downstream partner changes an upstream occupancy. Exact solutions of the series network with

q_R=1\,\mu\mathrm M

K_1=0.5\,\mu\mathrm M

and

K_2=0.1\,\mu\mathrm M

, at partner totals 0, 0.3 and

3\,\mu\mathrm M

. The plotted readout counts both binary and ternary complexes.

2 · Experiment. Sweep $q_L$ , and read $C_{LRP}$ . A regime is named by the winners of $(q_L,q_R,q_P)$ .

Run 1: the amplitude hierarchy

3 · Candidates. Take $q_R\ll K_1$ , $q_R\ll K_2$ and $q_P\ll K_2$ . The reservoir bound of §15 settles two totals. Bound ligand is at most $q_R$ , and $C_{LR}/L=R/K_1\le q_R/K_1$ , so L wins its total. $C_{LRP}/P=C_{LR}/K_2\le q_R/K_2$ , so P wins its total. Only the receptor total can change winner. 4 · Regime sequence. R hands the receptor to $C_{LR}$ at $q_L=K_1$ . The ternary complex never wins, because $C_{LRP}/C_{LR}=P/K_2\le q_P/K_2$ .

5 · Response. The illustrative values are $q_R=q_P=0.1$ and $K_1=K_2=10\,\mu\mathrm M$ .

Regime	Range of $q_L$	$C_{LRP}$	Gain	Exact at a sample point
$(L,R,P)$	$q_L<K_1$	$q_Pq_Rq_L/(K_1K_2)$	1	at $q_L=1$ : $9.0\times10^{-5}$ , law $1.0\times10^{-4}$
$(L,C_{LR},P)$	$q_L>K_1$	$q_Pq_R/K_2$	0	at $q_L=100$ : $8.9\times10^{-4}$ , law $1.0\times10^{-3}$

The plateau $q_Pq_R/K_2$ is set by the receptor. Changing the receptor total therefore tunes the achievable output without changing the partner, which is useful when the partner's abundance is constrained by other functions. The plateau stays below $q_P$ , because $q_R\ll K_2$ .

Run 2: the threshold hierarchy

3 · Candidates. Take $q_P\ll K_2\ll q_R\ll K_1$ , and keep $q_L\ll K_1$ . Then L wins its total and R wins its total, since $C_{LR}/R=L/K_1\ll1$ . So $C_{LR}\approx q_Lq_R/K_1$ , and only the partner total can change winner. 4 · Regime sequence. P hands the partner to $C_{LRP}$ where $C_{LR}=K_2$ , at

q_L=K_{\mathrm{eff}}=\frac{K_1K_2}{q_R}.

5 · Response. The illustrative values are $q_R=1$ , $q_P=0.001$ , $K_1=1000$ and $K_2=0.01\,\mu\mathrm M$ , so $K_{\mathrm{eff}}=10\,\mu\mathrm M$ .

Regime	Range of $q_L$	$C_{LRP}/q_P$	Gain	Exact at a sample point
$(L,R,P)$	$q_L<K_{\mathrm{eff}}$	$q_L/K_{\mathrm{eff}}$	1	at $q_L=1$ : 0.0907, law 0.1
$(L,R,C_{LRP})$	$K_{\mathrm{eff}}<q_L\ll K_1$	1	0	at $q_L=100$ : 0.901

Here the receptor sets the threshold. Lowering the receptor total raises $K_{\mathrm{eff}}$ , so it can suppress activity without removing any ligand. The exact values follow the hyperbola $q_L/(K_{\mathrm{eff}}+q_L)$ to within one percent across the tested range.

Direct binding is the comparison. If L bound P directly, with dissociation constant $K_{LP}$ and $q_P\ll K_{LP}$ , the path would run from $(L,P)$ into $(L,C_{LP})$ at $q_L=K_{LP}$ . The series architecture adds one control knob in each hierarchy. The amplitude hierarchy adds the receptor factor to the plateau, and the threshold hierarchy adds it to the breakpoint. Each holds only under its own settling conditions, and the exact root covers the transitions between them.

Keep every downstream complex in the upstream totals, settle the totals the hierarchy allows, and read from the table which knob sets the amplitude and which sets the threshold.

32A modification cycle switches where both enzymes saturate

A modification cycle is the classic source of zero-order ultrasensitivity. The pipeline shows that its switch is a singular regime of the two enzyme totals. The finite model then proves separately that the steady state is unique and stable.

1 · Model. Enzyme E binds the unmodified target X with dissociation constant $K_E$ , and enzyme F binds the modified target Y with $K_F$ . Catalysis is composite at this resolution: $C_{EX}\rightsquigarrow E+Y$ with rate law $k_EC_{EX}$ , and $C_{FY}\rightsquigarrow F+X$ with rate law $k_FC_{FY}$ . For a phosphorylation cycle these laws assume maintained co-substrates and negligible reverse activities. Binding conserves $q_E$ , $q_F$ , $q_X=X+C_{EX}$ and $q_Y=Y+C_{FY}$ . Catalysis conserves the target total $q_{\mathrm{target}}=q_X+q_Y$ .

The slow dynamics reduce to one equation. Write $C(q_1,q_2,K)$ for the exact complex of one pair with totals $q_1$ , $q_2$ and dissociation constant K, the root of §10:

C(q_1,q_2,K)=\frac{2q_1q_2}{T+\sqrt{T^2-4q_1q_2}},\qquad T=q_1+q_2+K.

Then the modified total obeys

\dot q_Y=f(q_Y)=k_E\,C(q_E,\,q_{\mathrm{target}}-q_Y,\,K_E)-k_F\,C(q_F,\,q_Y,\,K_F).

The pair's complex increases with either total, because both of its gains are positive (§14). So $f'(q_Y)<0$ on the interior. Also $f(0)>0$ and $f(q_{\mathrm{target}})<0$ . The cycle therefore has exactly one steady state, and it is stable. A scalar root between zero and the target total computes it.

2 · Experiment. Vary the capacity ratio $r=V_E/V_F$ , with capacities $V_E=k_Eq_E$ and $V_F=k_Fq_F$ , and read the modified fraction $y=q_Y/q_{\mathrm{target}}$ . Take trace enzymes, so complexes are negligible in the target totals, and write $\bar K_E=K_E/q_{\mathrm{target}}$ and $\bar K_F=K_F/q_{\mathrm{target}}$ . The steady balance $V_EC_{EX}/q_E=V_FC_{FY}/q_F$ becomes

r\,\frac{1-y}{\bar K_E+1-y}=\frac{y}{\bar K_F+y}\quad\Longleftrightarrow\quad r=\frac{1+\bar K_E/(1-y)}{1+\bar K_F/y}.

The second form is exact, and each factor is an enzyme total over its complex. In trace, $E/C_{EX}=K_E/X=\bar K_E/(1-y)$ , so the numerator is $q_E/C_{EX}$ . Likewise the denominator is $q_F/C_{FY}$ .

3 · Candidates. The target totals are settled, since the enzymes are in trace. The enzyme totals change winner. E's candidates are E and $C_{EX}$ , and $C_{EX}$ wins when $\bar K_E<1-y$ . F's candidates are F and $C_{FY}$ , and $C_{FY}$ wins when $\bar K_F<y$ . A regime is named by the winners of $(q_E,q_F)$ .

4 · Regime sequence. Keeping the larger term of each factor gives the regimes. With $\bar K_E,\bar K_F\ll1$ , the path runs from $(C_{EX},F)$ , where $r\approx y/\bar K_F$ , into $(C_{EX},C_{FY})$ , where both factors are near one and $r\approx1$ . It ends in $(E,C_{FY})$ , where $r\approx\bar K_E/(1-y)$ . The middle regime is singular. Its leading-order law fixes r rather than y, so y jumps from about $\bar K_F$ to about $1-\bar K_E$ across $r=1$ . The exact form above is the balance that places y inside the jump.

5 · Response. The exact column uses $\bar K_E=\bar K_F=0.01$ .

Regime	Range of r	Modified fraction y	Gain	Exact at a sample point
$(C_{EX},F)$	$r<1$	$r\bar K_F$	1	at $r=0.1$ : 0.0011, law 0.001
$(C_{EX},C_{FY})$	$r\approx1$	from the exact balance	switch	0.1 at $r=0.919$ , 0.5 at $r=1$ , 0.9 at $r=1.088$
$(E,C_{FY})$	$r>1$	$1-\bar K_E/r$	0	at $r=10$ : 0.9989, law 0.999
$(E,F)$ , when $\bar K_E,\bar K_F\gg1$	all r	$y/(1-y)\approx r\bar K_F/\bar K_E$	1 in the logit	graded, no switch

The switch is steep because the jump is singular. At $r=1$ the slope of $\log(y/(1-y))$ against $\log r$ is $(1+2\bar K)/(2\bar K)=51$ , for $\bar K_E=\bar K_F=\bar K=0.01$ . y moves from 0.1 to 0.9 while r changes by 18 percent. When both enzymes stay unsaturated, the cycle sits in $(E,F)$ and the response is graded. Goldbeter and Koshland derived this zero-order amplification and also examined appreciable complexes and nonproductive binding.¹⁸ The finite model above keeps the productive complexes exactly, and its readout counts free and bound modified target.

Scroll inside the figure to see it at reading size.

Figure 25. Opposed saturated enzymes amplify a capacity imbalance. Total modified fraction in the exact two-arm model, with target total

1\,\mu\mathrm M

K=K_E=K_F

and

q_{\mathrm{enz}}=q_E=q_F

. Unsaturated enzymes,

K=1

, give a graded response. Saturated trace enzymes,

K=0.01

and

q_{\mathrm{enz}}=0.01

, give the switch. Appreciable enzyme pools,

q_{\mathrm{enz}}=0.3

, change the response without creating a second steady state.

Prove uniqueness from the monotone pair complex, and read the switch from the regime where both enzyme totals are won by their complexes.

33The pipeline carries over to a new system

A new binding-catalysis system goes through the same three tools in a fixed order. Each tool has a check that the examples of this page have exercised.

Tool 1 builds the model. State the intervention and the readout first. List the molecular states, count their constituents into L, and check that the totals are complete (§1, §4). Assign one affinity per complex, with units, and one activity per state (§5, §16). Solve the positive allocation before simplifying anything (§6, §8).

Tool 2 maps the regimes. For each total, list its candidates and select, solve and check (§11). Keep the exact solve available at every boundary, and use the ray's balance wherever a singular crossing or a small difference carries the output (§13).

Tool 3 reads responses. Run the five steps of §15 along the path of the experiment, and end in a regime-sequence table with exact values beside the laws. Then test the dynamical closure on its own terms. Affinity data fix the allocation but not the relaxation, so compare clocks (§2) and check the reduced model against the resolved one on the output it must predict (§24). The extension runs the same three tools on a transporter that carries a physical current.

Transfer exercise with answer

In the conformational example of §30, suppose both bound states produce, at frequencies $k_{\mathrm o}^{\mathrm{cat}}$ and $k_{\mathrm c}^{\mathrm{cat}}$ . What changes in the pooled model and in its regime table?

Worked answer

The equilibrium inventory and $K_{\mathrm{eff}}=K_{\mathrm o}/(1+\rho)$ are unchanged. The effective activity becomes the population-weighted average

k_{\mathrm{eff}}=\frac{k_{\mathrm o}^{\mathrm{cat}}+\rho\,k_{\mathrm c}^{\mathrm{cat}}}{1+\rho}.

The regime table keeps its breakpoint at $q_S=K_{\mathrm{eff}}$ , and both pieces take the new $k_{\mathrm{eff}}$ . If the two frequencies match, moving bound enzyme between conformations changes no activity per bound enzyme. Changing the conformational weight can still move the breakpoint through $K_{\mathrm{eff}}$ .

Build, map and read a new model in separate steps, and tie every approximation to the output it must predict.

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