Five behaviors of binding and catalysis, derived and placed

The full five-step comparison for each behavior, then every result placed at its parameter conditions, read backwards for design, and bounded by what the regime enumeration of a written mechanism guarantees.

CCBS · Lecture 9 · exposition · 7 October 2026

This chapter is the derivation companion to the core. The core shows five behaviors that cells build from binding and catalysis: the sign of regulation, a threshold, memory, adaptation and a rhythm. It explains each one by the allocation of a finite pool. Here each behavior is derived in full, with every model written out, every limit taken explicitly and every number traced to a calculation. Nothing below assumes an equation from the core.

The chapter has one method and one destination. The method is a five-step comparison between a binding-catalysis model and the effective rate law usually written in its place. The destination is a table that puts each example, and each published result about the same behavior, at the parameter conditions where it holds. A reader who finishes should be able to take a mechanism this page never mentions, run the five steps, and place a result on that table.

Parts 2 to 6 run the comparison on the five behaviors, one part each. Part 7 collects them. It shows that every boundary on the placement table has been derived, reads the table backwards for design, and states what the enumeration of regimes guarantees about a written mechanism and what it leaves to the closed dynamics. The lineage tells how the field reached each verdict. This page cites the literature for authority only.

1A behavior is a property of the closed loop

A response curve describes one experiment on a binding layer. A behavior belongs to the vector field on the totals, once production and removal close the loop.

The object of comparison is a vector field on totals, so we build one concretely first. Take a protein X that binds one partner Y. Binding is the elementary pair

where $k_+$ is the association rate constant, $k_-$ the dissociation rate constant, and $K_d$ their ratio, the dissociation constant, a concentration. Binding moves molecules between states without changing how many X there are. The totals count every state:

Binding is fast against everything else, so at every instant the pair sits at equilibrium, $K_dC_{XY}=XY$. Given the totals, eqs. (1) and (2) fix every free species and complex. That is Lecture 8's binding layer.

Production and removal change the totals, and that closes the loop. Suppose X is made by a composite step whose rate law $v_\mathrm{prod}$ depends on the binding layer, for instance on how much of a promoter is free. Its total then obeys an addition-removal balance:

Here q collects the totals and b the binding constants and activities. The binding layer computes the production law from the totals, and the balance moves the totals. A behavior is what this closed loop does.

The removal term has a rule, and the rule is part of the model. If free X is removed at frequency $\lambda_\mathrm{f}$ and bound X at $\lambda_\mathrm{b}$, the loss is $\lambda_\mathrm{f}X+\lambda_\mathrm{b}C_{XY}$. We write $\theta=\lambda_\mathrm{b}/\lambda_\mathrm{f}$ for the bound-to-free removal ratio. Growth dilution removes every state alike, so $\theta=1$ and the loss is $\lambda q_X$, a function of the total alone. Removal of free X only, for example by a protease that cannot reach bound X, has $\theta=0$, and binding then protects X. These are the two variants Lecture 8's coda handed over. Part 6 shows that the whole range of $\theta$ matters.

A behavior is one of four properties of that vector field. It can be how a steady state depends on a total, how many attractors exist, how far the output returns after a sustained step of input, or whether a stable cycle exists. The first is a static input-output behavior and the other three are dynamical. A steep response curve of the binding layer alone, measured at fixed totals, is a static relation of that layer, and a count of binding states is a property of the layer too. The loop can turn either into different closed-loop behaviors or into none.

Each comparison in this chapter runs five steps, and each step removes one ambiguity.

1. Model. Write the elementary binding steps, the composite production and removal steps with their rate laws, the totals, the removal rule and the observable.
2. Limit. Derive the effective rate law as a limit of that model, and name the assumptions that make it exact.
3. Match. Choose parameters so that both give the same response at one reference condition.
4. Move a total. Change what a cell changes: an input total, a regulator total, a site number, a removal rate.
5. Close the loop. Read the behavior from the closed dynamics: steady states, their stability, the error after a step, or a cycle.

Two of the five steps exist to keep the comparison fair, and two to make it informative. Step 2 makes the effective law a correct answer to a narrower question, not a strawman. Step 3 prevents a difference of scale from posing as a difference of mechanism. Step 4 is where the two models can part. Step 5 decides whether the parting is a difference in behavior.

State a behavior as a property of the closed vector field on the totals, with the removal rule named by its bound-to-free ratio $\theta$, and plan every comparison as model, limit, match, move a total, close the loop.

2One binding pair supplies the elementary calculation

Most effective laws in this chapter are the binding pair of eq. (1) evaluated in one of its regimes, and the multisite and multimer laws of Parts 3 and 4 are assembled from the same calculation. Drawn on the pair's map, a law is a cut that stays on one side of a boundary.

The pair has an exact solution, and the effective laws are its limits. Write the pair generically as $E+S\xrightleftharpoons[k_-]{k_+}C_{ES}$, with totals $q_E=E+C_{ES}$ and $q_S=S+C_{ES}$. Substituting $E=q_E-C_{ES}$ and $S=q_S-C_{ES}$ into $K_dC_{ES}=ES$ gives a quadratic, $C_{ES}^2-TC_{ES}+q_Eq_S=0$ with $T=q_E+q_S+K_d$. Its physical root is the smaller one, written without a cancelling difference:

The root lies between zero and the smaller total, as a complex must.

Lecture 8 sorts the pair into four regimes by which term wins each total. A regime names the winner of $q_E$ first, then the winner of $q_S$. Keeping only the winners and the binding relation gives one monomial law per regime. Each law is accurate to leading order deep inside its sector, where the stated inequalities hold with a wide margin:

The last regime is singular, and for a structural reason. The complex wins both totals, so the two winner equations, $C_{ES}\approx q_E$ and $C_{ES}\approx q_S$, select the same column and leave E and S undetermined. Its law needs one more balance. The difference of the totals is exact, $q_E-q_S=E-S$, so near the ray $E\approx S$, and binding gives $ES=K_dC_{ES}\approx K_dq_E$. A small free pool of size $\sqrt{K_dq_E}$ survives where nearly everything is bound. The condition $E\approx S$ needs the difference of the totals to be small on that scale, not merely the totals to look close on a log plot.

The sector lines are leading-order boundaries, not exact ties. An exact tie is an equal contribution of two terms to one total. Free and bound E contribute equally where $C_{ES}=q_E/2$, and binding then forces $S=K_d$, so the tie lies on the curve $q_S=K_d+q_E/2$. Free and bound S tie on $q_E=K_d+q_S/2$. The leading-order map draws the first of these as $q_S=\max(K_d,q_E)$, which differs from the exact curve by at most a factor of 2. Lecture 8 guaranteed agreement within such bounded factors and no more. A tie marks a boundary between two regimes. It is a different thing from the singular regime above, where the winners are dependent.

The Michaelis-Menten law is the pair wherever the free substrate wins its own total. If $S\approx q_S$, then $E=q_E-C_{ES}$ and $K_dC_{ES}=(q_E-C_{ES})q_S$ give

The condition $S\approx q_S$ holds in the regimes $(E,S)$ and $(C_{ES},S)$, that is when $q_S\gg q_E$ or $K_d\gg q_E$. This validity condition is the one Liu, Ren and Xiao derived for the law.¹ Outside it the substrate is held in the complex, and eq. (5) has no way to say so.

The Hill law with coefficient 1 is the free partner on a weak cut. Hold $q_E$ far below $K_d$, the partner in trace, and sweep $q_S$. The free species S then wins its total, so $C_{ES}/E=S/K_d\approx q_S/K_d$, and the free fraction of the partner is

This is a repression law of Hill coefficient 1 in the total $q_S$. Its log-slope, $-d\ln(E/q_E)/d\ln q_S=q_S/(K_d+q_S)$, never reaches 1. A promoter in trace read by a repressor is this cut.

A titration switch is the same partner on a tight cut. Hold $q_E$ far above $K_d$. Below the ray, where $q_E-q_S\gg\sqrt{K_dq_E}$, the complex holds nearly all of S, and the free partner falls only as the sites fill, $E\approx q_E-q_S$. Above it, where $q_S-q_E\gg\sqrt{K_dq_E}$, the complex holds nearly all of E, the free S is $q_S-q_E$, and $E\approx K_dq_E/(q_S-q_E)$. The $(C_{ES},S)$ law $E\approx K_dq_E/q_S$ needs the further excess $q_S\gg q_E$. Between these the free partner drops by a factor of order $q_E/K_d$, passing through $\sqrt{K_dq_E}$ on the ray. A promoter pool read by a repressor whose total passes the pool is this cut, and Parts 4 and 6 use it.

Only the winners of the conservation sums are a dominance regime. A regime is a choice of one winning term per total, nothing more. The observable, the controlled input and the removal rule are separate model choices. A different observable reads the same allocation differently, as free and total output do in Part 5. A different removal rule changes the vector field without changing any allocation. A different architecture changes the conservation sums themselves. All four must be named to interpret a result, and only the first is a regime.

A cut's height decides the kind of response, and the height is a ratio of totals to $K_d$. Below the corner a cut breaks once, at $q_S=K_d$, where one total changes winner, and the response is graded. Above it a cut crosses the singular ray, where in the leading-order map both totals change winner together, and the response switches. At finite binding the two exact ties stay distinct: on a cut at fixed $q_E\gg K_d$ they sit near $q_S=q_E/2$ and $q_S=2q_E$. Between them lies a broad intermediate band, and the free partner's collapse is concentrated in a much narrower rounding layer of width about $\sqrt{K_dq_E}$ around the ray. Lecture 11 develops the general geometry of such maps for networks with many pairs. This chapter needs only the one-pair map, applied once per example.

Locate an effective law as a cut through a binding pair's map: name the regime it stays in, write the condition on the totals that keeps it there, and predict a switch wherever a path crosses the singular ray.

3Matching fixes one scale and leaves the totals free

Matching two models at a reference condition sets one constant for each quantity matched. It does not make the models equivalent, because it says nothing about how either one stores a total.

A match sets as many constants as it equates. To match a floor, a ceiling and a midpoint of two response curves, one chooses three parameters. The choice is a calibration, a fact about the parameters, not about the mechanisms. The binding-catalysis supplement of the lecturer's group makes the same point about a matched pair of induction circuits: the equality of the fitted curves is a calibration choice, not an identity of the full equilibrium.²

A match leaves three things free. It leaves the inventories: how much of each total sits in complexes rather than free. It leaves the joint states, such as which promoter states coexist. And it leaves the response to a total the match did not touch. Step 4 moves exactly such a total.

A match can move the two models to opposite sides of a regime boundary. Part 2 matches an induced activator to an induced repressor in free inducer. The match forces the activator to bind its inducer about $10^4$ times more weakly than the repressor does. On the map of Figure 1, it puts the activator's inducer pair near the corner and the repressor's far above it. The two circuits then answer a change of inducer total in opposite ways, and the match itself predicted it.

A contrast is stated with its conditions, and only a counterexample under those conditions refutes it. Each contrast below names its model, its observable, its controlled input, its removal rule and the one parameter it moves. A reproduced counterexample under exactly those statements would refute that contrast. A second successful fit of the effective law refutes nothing. The law should keep working wherever its regime holds, and it can agree by accident outside it.

After matching two models, list what the match fixed and which totals it left free, check where each model's pairs sit on the map, and state every contrast with the conditions a counterexample would have to meet.

4The repressor circuit reduces to one equation

A repressor circuit has five binding species, but its equilibrium is one monotone equation for the free repressor, and its output reads that free repressor.

1 · Model, repressor. The job is to switch on a gene, the output protein G, when an inducer I appears. The repressor R binds the promoter D and blocks expression. The inducer binds the repressor, and the complex $C_{RI}$ cannot bind the promoter. The binding steps are elementary, and expression and removal are composite:

Here a is the expression activity of a free promoter and $\delta$ the removal frequency of the output. The two dissociation constants are $K_{RI}=k_{-RI}/k_{+RI}$ and $K_{DR}=k_{-DR}/k_{+DR}$. The letter pair in a subscript names the partners. At equilibrium $K_{RI}C_{RI}=RI$ and $K_{DR}C_{DR}=DR$, and the three totals are

Every species follows from the free repressor. The promoter total and its binding relation give $D=q_DK_{DR}/(K_{DR}+R)$ and $C_{DR}=q_DR/(K_{DR}+R)$. The inducer total gives, in the same way, $C_{RI}=q_IR/(K_{RI}+R)$. Substituting both into the repressor total leaves one equation for R:

The left side rises strictly with R, from zero at $R=0$ to at least $q_R$ at $R=q_R$. So eq. (9) has exactly one root in $(0,q_R]$, for every set of totals.

The output reads the free repressor. Production is $v=aD$, so the production per promoter, in units of the free-promoter activity, is

This is the observable throughout Part 2. Because it reads a free species, it inherits the titration switch of §2 whenever the inducer total drives the free repressor across a ray.

The illustrative parameters put the repressor far above its operator constant. In nanomolar, $q_R=100$, $K_{DR}=1$, $K_{RI}=0.01$ and $q_D=0.1$, a trace promoter. The excess of the repressor over its operator constant is $\rho=q_R/K_{DR}=100$. These are model parameters chosen for clarity, not measurements.

Reduce a one-promoter repressor circuit to the single monotone equation (9) for the free repressor, and read its output as the free-promoter fraction (10).

5The activator circuit reduces to one equation, and its order is capped

The activator circuit also reduces to one equation, for the inducer-bound activator. Its output is a complex, and a complex holding one inducer can never rise faster than the inducer total.

1 · Model, activator. The activator A binds the inducer, and only the complex $C_{AI}$ binds the promoter, as $C_{DAI}$. The free activator cannot bind the promoter. Both promoter states express:

The constants $a_0$ and $a_1$ are the activities of the free and the loaded promoter. At equilibrium $K_{AI}C_{AI}=AI$ and $K_{DA}C_{DAI}=DC_{AI}$, with totals

Every species follows from the inducer-bound activator. Write $c=C_{AI}$. The promoter gives $C_{DAI}=q_Dc/(K_{DA}+c)$. Let $s(c)=c+C_{DAI}$ be what the complexes hold of both the activator and the inducer. Then $A=q_A-s(c)$ and $I=q_I-s(c)$, and the binding relation of the pair becomes

The left side falls and the right side rises as c grows from zero to the point where $s(c)=\min(q_A,q_I)$. So eq. (13) has one root. The production is $v=a_0D+a_1C_{DAI}=a_0q_D+(a_1-a_0)C_{DAI}$, and the rise above the floor is proportional to the active complex.

The active complex never rises faster than the inducer total, and the proof takes two observations. First, the share of the inducer that sits in the active complex depends only on the free activator and the free promoter. Second, both fall when inducer is added. Take the first. Every species is a product of free constituents, $C_{AI}=AI/K_{AI}$ and $C_{DAI}=DAI/(K_{AI}K_{DA})$, so the free inducer cancels from the share:

This share rises with A and with D.

The second observation follows from the Jacobian of the totals, which is positive definite. Differentiate the totals with respect to the logarithms of the free constituents A, I and D. Each species contributes its concentration once for every constituent it contains, so the Jacobian is the symmetric matrix

It is positive definite, because for any vector w the form $w^{\mathsf T}Mw$ is a sum of species concentrations times squares. Adding inducer at fixed $q_A$ and $q_D$ changes the free logarithms by $M^{-1}$ times $(0,dq_I,0)$. The two entries of $M^{-1}$ that matter are cofactors:

The first numerator is nonnegative because $q_D\ge C_{DAI}$. So free activator and free promoter both fall as the inducer total rises, the share (14) falls, and

The lower bound holds because $C_{DAI}=q_D-D$ rises as D falls. This is exact for every parameter set of the circuit (11), promoter loading included.

An order of at most 1 caps the Hill coefficient at 2. Measure steepness by $n_H=\ln81/\ln(\mathrm{EC}_{90}/\mathrm{EC}_{10})$, where $\mathrm{EC}_{10}$ and $\mathrm{EC}_{90}$ are the inputs giving 10% and 90% of the rise. The normalized rise is $C_{DAI}/C_{DAI}^{\max}$, and by eq. (17) its log-slope is at most 1. A ninefold rise of the response, from 0.1 to 0.9, then needs at least a ninefold rise of input. So $\mathrm{EC}_{90}/\mathrm{EC}_{10}\ge9$ and $n_H\le\ln81/\ln9=2$, whatever the parameters. The repressor's output (10) reads a free species, which carries no such bound. The lecturer's thesis states the general form of this dichotomy: an activating regulator acts through a complex, whose order in a total is bounded between 0 and 1, while a repressing regulator acts through a free species, whose order can become arbitrarily negative under tight binding.³

Reduce the activator circuit to eq. (13), and bound its response without computing it: the active complex has order at most 1 in the inducer total, so its Hill coefficient in that total is at most 2.

6Three assumptions give one free-inducer law

With binding at equilibrium, the free inducer clamped and the promoter in trace, both circuits become one Michaelis-Menten law with a floor. Matching them costs one affinity ratio of about $10^4$.

2 · Limit. Three assumptions give the usual effective law. Binding is at equilibrium. The free inducer I is the controlled input, as when a large medium holds it fixed. The transcription factor far exceeds the promoter, $q_D\ll q_R$ and $q_D\ll q_A$, so loading the promoter does not deplete the factor. The fast subsystem removed is the binding layer, justified by binding equilibrium. The promoter in trace is a dominance choice: the factor wins its own total.

The repressor circuit becomes Michaelis-Menten in free inducer, with a floor. With the promoter in trace, the free repressor is the share of its total that holds no inducer, $R=q_RK_{RI}/(K_{RI}+I)$. Substituting into eq. (10) and rearranging,

The floor $1/(1+\rho)$ is the leak of a repressed promoter, and the rise above it is half complete at the free inducer $K_\mathrm{eff}$.

The activator circuit becomes the same form. With the promoter in trace, the inducer-bound activator is $c=q_AI/(K_{AI}+I)$, and the loaded fraction of the promoter is $c/(K_{DA}+c)$. With $\rho_A=q_A/K_{DA}$ that fraction rearranges to $\tfrac{\rho_A}{1+\rho_A}\,I/(K_{AI}/(1+\rho_A)+I)$. The production per promoter is $(a_0/a)$ times the free fraction plus $(a_1/a)$ times the loaded fraction. Its half-rise point is $K_{AI}/(1+\rho_A)$.

3 · Match. Three choices make the two laws identical. Give both factors the same excess, $\rho_A=\rho=100$. Set the activities so that the floors and ceilings agree: the floor needs $a_0=a/(1+\rho)$, and a ceiling of 1 then needs $a_1=a(2+\rho)/(1+\rho)$. Equal half-rise points need

At the illustrative parameters $K_{AI}=102.01\ \mathrm{nM}$ and $K_\mathrm{eff}=1.01\ \mathrm{nM}$ in both circuits. The exact circuits then agree with each other and with eq. (18) within 0.1% (numerical, Figure 2, left).

The match moves the two inducer pairs to opposite sides of the corner. Eq. (19) makes the repressor bind its inducer $(1+\rho)^2\approx10^4$ times more tightly than the activator does. Excess activator lowers the half-rise point, because a few induced activators already fill the promoter. Excess repressor raises it, because nearly all of the repressor must be inactivated. On the map of Figure 1, the activator and its inducer sit at $q_A/K_{AI}\approx0.98$, at the corner, and the repressor and its inducer at $q_R/K_{RI}=10^4$, far above it. The first cut is graded and the second switches, and §7 confirms both.

In this regime the two designs differ only in the sign of one sensitivity. Savageau compared activator and repressor control in systems that were otherwise identical, and found that the only inherent difference was their response to regulatory mutations (p. 5647).⁴ In eq. (18) and its activator twin, a change of the regulator's own total is such a mutation. §8 derives the two responses and finds them equal and opposite. That verdict is exact in the regime of the three assumptions.

Before trusting a matched induction law, name its three assumptions, write the affinity ratio the match forces, here $(1+\rho)^2$, and locate each circuit's inducer pair on the map.

7In the inducer total, the repressor switches where its store runs out

When the cell controls the inducer total, each circuit must first fill its own store of bound inducer. The repressor's store is its whole total, so it switches there. The activator's store is small, so it stays graded.

4 · Move a total. Let the cell set the inducer total $q_I$, as it does for a metabolite or a second messenger it makes itself. The inducer then splits between free and bound, and only the free part enters the law (18).

Counting the stored inducer at half rise gives both half-rise totals. At half rise both circuits hold free inducer $K_\mathrm{eff}$, by eq. (18). Take the repressor first. Setting eq. (10) to the half level $(2+\rho)/(2(1+\rho))$ gives $R=q_R/(2+\rho)$. All other repressor holds an inducer, apart from at most $q_D$ on the promoter, so the stored inducer is $q_R(1+\rho)/(2+\rho)=99.0\ \mathrm{nM}$. For the activator, the bound inducer at $I=K_{AI}/(1+\rho)$ is $q_AI/(K_{AI}+I)=q_A/(2+\rho)=0.98\ \mathrm{nM}$. Adding free and stored inducer,

The exact half-rise totals are 99.98 and 2.04 nM (numerical). The activator's estimate is 2.4% low, mostly because the loaded promoter also stores inducer, about $0.05\ \mathrm{nM}$ at half rise.

The shapes differ as much as the positions. In the inducer total the activator stays graded, with $n_H=1.02$, inside the cap of eq. (17). The repressor switches, with $n_H=25.7$ (both numerical).

In the tight-inducer limit the two responses are mirror images, and their Hill coefficients are related exactly. Let the inducer binding become tight, $K_{RI}\to0$, with $K_{AI}$ rematched by eq. (19), and write $x=q_I/q_R$. Then every inducer is bound. The free repressor is $q_R-q_I$ while $x<1$, so eq. (10) becomes $1/(1+\rho(1-x))$, and the normalized rises of the two circuits are

Solving each for its 10% and 90% points gives

at $\rho=100$, exact in the limit. For large $\rho$, $n_R\approx\rho/2$. In this limit the repressor computes a comparison of two totals, $v/(aq_D)\approx K_{DR}/(K_{DR}+\max(0,q_R-q_I))$, the titration switch of §2 with the repressor as the sink for inducer. The activator reads the smaller of $q_I$ and $q_A$ through one saturating promoter.

At finite inducer affinity the steepness has a closed form, and it locates the tightness condition. With the promoter in trace, the response depends on the free inducer alone, so the 10% and 90% points in free inducer are $K_\mathrm{eff}/9$ and $9K_\mathrm{eff}$. The inducer total adds the stored inducer, $C_{RI}=q_RI/(K_{RI}+I)$, so $\mathrm{EC}_x=I_x\big(1+q_R/(K_{RI}+I_x)\big)$. With $\sigma=q_R/K_{RI}$, the ratio of the two points gives

As $\sigma\to0$ the ratio is 81 and $n_H=1$. As $\sigma\to\infty$ it becomes the tight limit (22). The switch appears when the stored inducer outweighs the free inducer at both points, which needs $\sigma\gg10+9\rho$, roughly $K_{RI}\ll K_{DR}/9$: the repressor must bind its inducer sufficiently more tightly than its operator. Eq. (23) is continuous, so there is no universal threshold. At $\rho=100$ and the $n_H$ criterion used here, a ratio of 0.1 already gives 6.1 and a ratio of 1 gives 1.93, which is why the factor of ten is quoted as a working rule for these parameters. Eq. (23) gives 25.8, 6.11 and 1.93 at $K_{RI}/K_{DR}=0.01$, 0.1 and 1, against exact values of 25.7, 6.10 and 1.93 with promoter loading (numerical, Figure 3).

Predict a half-rise total by adding the free inducer the law needs to the inducer each regulator stores, and read from eq. (23) how steep the repressor's switch at $q_I\approx q_R$ is for its affinity ratio: at $q_R=100K_{DR}$ a ratio of 0.1 gives $n_H\approx6$, and the switch sharpens continuously as the ratio falls.

8The regulator and promoter totals separate the circuits again

The regulator's own total moves the repressor's threshold one for one and barely moves the activator's. The promoter total, which the effective law does not contain, separates them a third time.

The repressor's threshold tracks its own total. By eq. (20) the repressor's half-rise total is about $q_R$. Sweeping the factor's total over 50, 100 and 200 nM moves it to 49.5, 100.0 and 201.0 nM. The activator's half-rise total moves only from 3.01 to 2.04 to 1.55 nM (numerical). The activator's store, $q_A/(2+\rho_A)$, stays near $K_{DA}$ whatever $q_A$ is. What moves is its free-inducer half point, $K_{AI}/(1+\rho_A)$, which falls as $q_A$ rises at fixed $K_{AI}$.

In free inducer the two sensitivities to the regulator total are equal and opposite. Hold the free inducer at the half-rise point $I=K_\mathrm{eff}$ and change the factor's total. For the repressor, eq. (10) with $R=q_RK_{RI}/(K_{RI}+I)$ gives $d\ln v/d\ln q_R=-(R/K_{DR})/(1+R/K_{DR})$, and at the half point $R/K_{DR}=\rho/(2+\rho)$. For the activator, the loaded fraction $\vartheta=c/(K_{DA}+c)$ has $d\vartheta/d\ln q_A=\vartheta(1-\vartheta)$, and at the half point $\vartheta=\rho/(2(1+\rho))$. Working both through,

This is exact within the reduced model, and it is Savageau's difference of response to regulatory mutations made quantitative.⁴

In the inducer total the repressor's sensitivity is about fifty times its free-inducer value. Holding the inducer total at each circuit's half-rise point, the sensitivity is $-24.7$ for the repressor, about 50 times the $-0.495$ of eq. (24), and $+0.25$ for the activator, about half its eq. (24) value (numerical). The two circuits now differ about a hundredfold in size of sensitivity. Near its threshold the repressor turns a 1% change in its own total into about a 25% change in output. This is a static sensitivity: it predicts how a change of the factor's total slower than $1/\delta$ shows up in the output. No noise was simulated.

The promoter total separates the circuits a third time. Raise $q_D$ past the factor's total. With tight promoter binding, an uninduced repressor occupies one site each, so the free promoter is about $\max(0,q_D-q_R)$ and the production per promoter about $(q_D-q_R)/q_D$ once $q_D>q_R$. With saturating inducer, the activator loads one site each, so the loaded promoter is about $\min(q_D,q_A)$ and the production per promoter is about $a_0/a+(a_1-a_0)/a\cdot\min(1,q_A/q_D)$. Both estimates give 0.50 at $q_D=200\ \mathrm{nM}$, against exact values of 0.505 (numerical). At 1000 nM the repressor circuit reaches 0.90 and the activator circuit falls to 0.11. The effective law (18) contains no $q_D$, so it predicts a production per promoter independent of copy number in both.

Use the regulator total and the promoter total as two further inputs: in free inducer the sensitivities to the regulator total are $\mp\rho/(2(1+\rho))$, and in the inducer total a repressor near its threshold amplifies changes in its own total many fold.

9Closing the inducer loop: removal decides what the steady state reads

Closing the output loop adds nothing while the inducer total is fixed. Once the inducer is made and removed, the removal rule decides whether the steady state reads the free inducer or the total.

5 · Close the loop. With the inducer total fixed, the output obeys $\dot q_G=v-\delta q_G$, which filters v with the same time $1/\delta$ in both circuits. Closing that loop changes no comparison. Let the inducer total itself be dynamic, supplied by $\varnothing\rightsquigarrow I$ with rate law u, and dilute the output at $\delta=\lambda$. Time is in units of $1/\lambda$.

If only free inducer is removed, both circuits reach the same steady state, at different times. The inducer obeys $\dot q_I=u-\delta_II$, so at steady state $I^*=u/\delta_I$ exactly, whatever the binding. Both circuits settle at the output the free-inducer law (18) gives, 0.505 when $I^*=K_\mathrm{eff}$, that is at $u=0.101\,\lambda q_R$ with $\delta_I=10\lambda$. The repressor must first fill a store of about $q_R$, which takes about $q_R/u\approx10/\lambda$. It reaches half its final output at $\lambda t=10.6$, against 0.85 for the activator (numerical). Binding protects the stored inducer from removal.

If every inducer-containing state is diluted, the steady state reads the total. Uniform dilution gives $\dot q_I=u-\lambda q_I$, so

in both circuits, exactly, whatever the binding. At the same supply the steady total is 10.1 nM, and the activator reaches 0.84 while the repressor stays at 0.011 (numerical). The repressor turns on only when the steady total passes its half-rise total, about $q_R+K_\mathrm{eff}$, that is when $u>\lambda(q_R+K_\mathrm{eff})$. At $u=2\lambda q_R$, eq. (25) crosses $q_R$ when $e^{-\lambda t}=1/2$, at $\lambda t=\ln2$, and the production per promoter switches there. The output protein follows through its own removal filter, with time $1/\delta$. The repressor total is itself held by synthesis at $\lambda q_R$, so the threshold sits where the two synthesis rates match, the in vivo titration threshold of Buchler and Louis (pp. 1106 to 1110).⁶

Close an inducer loop with a named removal rule: removal of free inducer only makes the steady state read the free inducer and delays the repressor by about $q_R/u$, while uniform dilution makes it read the total and switches the repressor where the inducer supply passes $\lambda q_R$.

10The modification cycle with every complex kept

The canonical switch is a covalent-modification cycle. Written with both enzyme complexes kept in the substrate total, it is two binding pairs and one slow balance with exactly one stable steady state.

Two measures of steepness are used, and a threshold is a claim about the second. The Hill coefficient $n_H=\ln81/\ln(\mathrm{EC}_{90}/\mathrm{EC}_{10})$ of §5 summarizes the whole rise from 10% to 90%. The local order $d\ln(\text{output})/d\ln(\text{input})$ is the slope on log axes at one point, Lecture 8's gain. A hard saturation corner can raise $n_H$ without any point of high local order, so a threshold is checked with the local order.

1 · Model. A kinase E converts the substrate X into its modified form Y, and a phosphatase F converts it back. Each enzyme binds its substrate form in an elementary step and acts in a composite catalytic step, with co-substrates held fixed and the reverse activity neglected:

The totals are $q_E=E+C_{EX}$, $q_F=F+C_{FY}$ and the substrate total $q_S=X+Y+C_{EX}+C_{FY}$, which the cycle conserves. The modified total, free or bound to the phosphatase, is $q_Y=Y+C_{FY}$, and the unmodified total is $q_S-q_Y=X+C_{EX}$.

Each enzyme forms one pair with its own substrate form, so eq. (4) gives both complexes. The kinase pairs with the unmodified total and the phosphatase with the modified total: $C_{EX}=C(q_E,q_S-q_Y;K_E)$ and $C_{FY}=C(q_F,q_Y;K_F)$, where $C(\cdot)$ is the root (4) and $K_E$, $K_F$ are the two dissociation constants. The slow variable obeys

The root (4) rises with each of its totals. So the first term falls and the second rises as $q_Y$ grows, and eq. (27) has exactly one steady state, which is stable. The cycle cannot hold two states by itself.

The experiment varies the ratio of the enzyme totals. The input is $r=q_E/q_F$ with equal catalytic constants, $k_E=k_F$, so r is the ratio of the two maximal rates. The readout is the modified fraction $y=q_Y/q_S$. Both enzymes share one dissociation constant K, which is also their Michaelis constant when catalysis is slow against unbinding.

Write a modification cycle as two binding pairs plus the slow balance (27), and conclude from monotonicity alone that it has one stable steady state at every input.

11Saturation is the route to steepness, and a large phosphatase share caps it

In this cycle the steepness comes from saturating both enzymes. One saturated enzyme gives at most $n_H=2$. For the modified fraction read against the enzyme ratio, a phosphatase holding more than a small share of the substrate caps the steepness however saturated the enzymes are.

2 · Limit. Goldbeter and Koshland took both enzymes in trace, so that the complexes are negligible in the substrate total and each free substrate form wins its own total.⁷ Each complex then follows eq. (5). With $\bar K=K/q_S$, setting eq. (27) to zero gives their closed form:

These are their Eqs. 7 to 10 (pp. 6841 to 6842) with equal constants for the two enzymes. 3 · Match. In the trace limit the exact cycle (27) reproduces eq. (28) to $6\times10^{-6}$ (numerical).

Saturation sets the steepness. For a saturating substrate, $\bar K\ll1$, expand the logarithm. Since $\ln81+2\ln(0.1/0.9)=0$, only the first-order terms survive: $\ln R_V\approx2\bar K(10-10/9)=(160/9)\bar K$. Therefore

At $q_S=100K$ the rise from 10% to 90% spans r from 0.919 to 1.088, an 18% change of the enzyme ratio. At $q_S=K$ the switch is barely there, $n_H=1.33$ (both from eq. 28).

One saturated enzyme gives at most 2, so a sharper switch needs both. Let the phosphatase work in its first-order range, $\bar K_F\to\infty$, in the two-constant form of Eq. 10. The response ratio becomes $R_V=81(\bar K_E+0.1)/(\bar K_E+0.9)$, which falls to 9 as the kinase saturates, so $n_H\to2$ and stops (exact). Goldbeter and Koshland's condition is that "one or more" of the enzymes work in the zero-order region (p. 6840). One saturated enzyme already makes the response ultrasensitive, up to 2. A switch steeper than that needs both. Saturation is the route to steepness in this cycle. It is not the only route in general, and §13 reaches a threshold through sequestration instead.

4 · Move a total. An abundant phosphatase caps the switch, because it holds the modified substrate. Raise the phosphatase total to a fraction $f=q_F/q_S$ of the substrate and take binding tight. In that limit each complex is the smaller of its two totals, and flux balance at equal catalytic constants sets $C_{EX}=C_{FY}$. There are two cases, $f<1/2$ and $f\ge1/2$. Take $f<1/2$ first. Below the balance point $r=1$ the phosphatase out-runs the kinase, so every modified substrate sits on the phosphatase: $C_{FY}=q_E$, and $y=rf$. Above it the phosphatase is full, $C_{FY}=q_F$, the kinase holds the whole unmodified total $q_F$, and $y=1-f$. The normalized rise reaches 10% at $r=0.1(1-f)/f$ on the linear part. It reaches 90% on the linear part only if $0.9(1-f)\le f$, and otherwise at the jump, $r=1$.

For $f\ge1/2$ the phosphatase can never be filled, because the plateau $1-f$ would leave it less modified substrate than its own total. Once the kinase total passes half the substrate, at $r=1/(2f)$, each enzyme holds the whole of its substrate form, and $C_{EX}=C_{FY}$ forces $y=1/2$. The response is $y=rf$ up to $r=1/(2f)$ and then the plateau $1/2$, with no jump at $r=1$. At $q_S=1$, $f=0.8$, $r=100$ and $K=10^{-8}q_S$ the exact cycle gives $y=0.5000000083$ (numerical). Normalized by its plateau the rise is linear from start to end, so $n_H=\ln81/\ln9=2$. The two cases combine into

with the plateau $1-f$ below $f=1/2$ and $1/2$ above it. The cap is 41.7 at $f=0.1$ and 3.0 at $f=0.3$. Unbounded steepness of this readout needs $f\le1/11$, a phosphatase scarcer than $q_S/11$. That condition is exact in the tight limit for this normalized readout, the total modified fraction against the ratio of enzyme totals. It is not a requirement for steepness in general, and not a necessary condition for every catalytic switch. The exact cycle approaches these caps as the substrate saturates (numerical, Figure 4). Goldbeter and Koshland remarked that enzyme ratios of 0.1 or less change little at their parameters (p. 6842). That agrees, because at $q_S/K=10$ the cap of 42 is far above the trace value of 3.7.

5 · Close the loop. The cycle has one stable state at every input, by §10, so its steepness is a static input-output behavior: the steady state as a function of the enzyme ratio. It is not an autonomous dynamical behavior such as memory or a cycle. A steep static response becomes one only when a loop feeds the output back, which Part 4 does with a titration layer instead.

Read a catalytic switch's steepness from eq. (29), about $0.25\,q_S/K$, with both enzymes saturated, since one alone gives at most $n_H=2$. For the modified fraction against the enzyme ratio, read the cap of eq. (30) from the phosphatase share: unbounded only below $f=1/11$, and 2 from $f=9/19$ on.

12In free ligand, negative cooperativity flattens

A receptor with two sites, only the doubly bound state active, can make a threshold with one binding layer. Measured against free ligand, though, negative cooperativity makes its response flatter, not steeper.

1 · Model. A receptor R binds ligand L at two identical sites, first with the site constant $K_1$ and then, with one site filled, with $K_2$. An empty receptor offers two sites, so its association runs at $2k_{+1}RL$. A doubly bound receptor can lose either ligand, so its dissociation runs at $2k_{-2}C_{RL_2}$. Writing those factors into the elementary arrows,

Equilibrium of each step then gives

with totals $q_R=R+C_{RL}+C_{RL_2}$ and $q_L=L+C_{RL}+2C_{RL_2}$. The cooperativity is $c=K_1/K_2$. It is negative when $c<1$, so that the second ligand binds more weakly than the first. The observable is the active fraction $C_{RL_2}/q_R$. The illustrative constants are $K_1=10^{-4}\ \mu\mathrm M$ and $K_2=10^{-2}\ \mu\mathrm M$, so $c=0.01$.

The exact equilibrium is one monotone equation for the free ligand. Substituting eq. (32) into the ligand total,

The left side rises strictly with L, so there is one root, and every species follows from it.

2 · Limit. The classical treatment takes the ligand in excess, so that the free ligand equals its total. The fast subsystem removed is the binding layer, and the dominance choice is that free ligand wins its own total. The active fraction is then a function of free ligand alone:

Its Hill coefficient lies between 1 and 2. For strong negative cooperativity, once $L\gg K_1$ the first site is full and eq. (34) becomes $L/(2K_2+L)$, a Michaelis-Menten law, so $n_H\to1$. For strong positive cooperativity it becomes $L^2/(K_1K_2+L^2)$, so $n_H\to2$. The computed values are 1.000 at $c=10^{-4}$, 1.005 at $c=0.01$, 1.19 at $c=1$ and 1.98 at $c=10^4$ (numerical, from eq. 34). Ha and Ferrell describe the same trend (p. 990).⁸

3 · Match. Eq. (34) is exact when the receptor cannot deplete the ligand, that is when $q_R\ll K_2$. There the response to the ligand total rises half-way near $q_L=2K_2$, with $n_H=1.02$ at $q_R=10^{-4}\ \mu\mathrm M$ (numerical), as the free-ligand law does.

Write a multisite receptor with its statistical factors in the elementary arrows, solve it by the monotone equation (33), and expect a free-ligand response with $n_H$ between 1 and 2 that negative cooperativity pushes toward 1.

13In the ligand total, the first sites buffer the input

When the receptor total exceeds its second dissociation constant and the input is the ligand total, the singly bound receptors absorb the first ligands one each. The response stays flat up to the receptor total and then rises straight to full.

4 · Move a total. Raise the receptor total past $K_2$, to $q_R=1\ \mu\mathrm M$, and sweep the ligand total. Count before computing. The first site binds a hundred times more tightly than the second, so the first ligands go to empty receptors. Until $q_L$ reaches $q_R$, nearly every receptor holds at most one ligand, and the active state stays empty. Past that point the two totals read $C_{RL}+C_{RL_2}\approx q_R$ and $C_{RL}+2C_{RL_2}\approx q_L$, because nearly every receptor and every ligand is bound. Subtracting,