From chemical balances to an adaptation claim

Start from the reaction list with the rate law of every composite step, the input, the measured output and the model boundary. The six steps are worked on the core's models, then run on a network the core does not treat.

CCBS · Lecture 10 · exposition · 13 October 2026

An adaptation mechanism must connect a specified function to feasible chemistry. The core chooses the adaptation biomachine, then follows binding and catalysis to the reaction orders that restore its output. This chapter teaches how to check that connection yourself. Begin with the full balances. Derive the limiting law, identify the measured quantity, and test the conditions under which the response returns.

Six questions organize the calculation. What identity constrains the output? Which observable does it constrain? Which dominance regime justifies the law? Is the proposed rest feasible? Does it attract nearby trajectories? Which changes preserve the result? The examples develop these questions in order. The final worked problem transfers them to a protein whose inactive form converts its active form.

The lineage explains where these questions came from. The research essay uses them to choose a perturbation that distinguishes candidate mechanisms. This chapter supplies the calculations those comparisons require.

1The audit starts from rate laws, an input, an output and a boundary

An adaptation claim says that a measured quantity returns to its old value after a persistent step of an input, for a class of inputs, and keeps doing so under a class of changes. Checking it needs more than a reaction list: the rate law of every composite step, the input, the output and the model boundary. With those written down, six questions taken in a fixed order test the claim and find its limits.

An adaptation claim names three things

An adaptation claim names an observable, a class of inputs and a class of changes. Hold the input at one value until the system rests, step it to another value, and hold it there. The claim is that a named observable then returns exactly to its old value (core Part 1 experiment). "Exactly" is a statement about every input in a class, not about one trace. "Robustly" is a statement about a class of changes to the network itself, such as a changed rate constant, an added load on the output or growth dilution (core A16).

A reaction list alone does not fix the dynamics, because a composite step carries no rate law. An elementary step is drawn with a straight arrow and its rate constant, and its mass-action rate law follows from the arrow. A composite step is drawn with a bare squiggle, and its rate law must be stated beside it. The core's incoherent feedforward loop shows why that statement matters. Its resolved binding-catalysis mechanism approaches a reduced model in which the degrader $X_2$ removes the output $X_1$ through a composite step. With the rate law $\delta x_1x_2$ the loop adapts exactly. With $\delta x_1x_2^f$ and $f\neq1$ it does not (§7). The composite reaction list is the same in both cases. The rate law differs. The resolved chemistry determines whether either law is a valid approximation (§4).

The audit's starting data has five parts. The reaction list, with every arrow classified as elementary or composite. The rate law of every composite step. The input: what is stepped, and in which units. The output: which species, complex or pool is measured. The model boundary: which processes the model contains, for instance whether growth dilutes its species. This page calls these five together the starting data.

The running example, written as starting data

The core's sequestering pair, written as starting data, is the running example. A reference process makes the actuator $Z_1$, the output $X$ makes the sensor $Z_2$, the two bind into the complex $C_{Z_1Z_2}$, which is removed, and the actuator makes the output, which is removed too (core §8):

Here $\mu$ is the reference rate, in c/T, $\theta$ the sensing constant, $k_+$ and $k_-$ the binding and unbinding constants, $\delta_C$ the removal frequency of the complex, $k$ the actuation constant and $\gamma$ the removal frequency of the output. The input is a change of $\gamma$, a first-order load $\ell$ on $X$, or a constant inflow $d_\mathrm{in}$ of $X$. The output is the free $X$. The boundary has no growth dilution, the actuation is linear in $Z_1$, and $X$ is made in one stage.

Six questions in a fixed order

Six questions, asked in order, audit the claim.

1. Identity. Find a relation among the rates, exact for the stated rate laws, whose zero at a rest constrains the output, and say what it stores, in which units (§2).
2. Observable. Find which quantity the identity pins, and how far each measured pool follows it (§3).
3. Regime. Decide whether the identity is exact or a limiting law, and if it is a limit, find the regime of the totals in which it holds (§4).
4. Positivity and reachability. Check that a rest with the pinned value exists with nonnegative amounts, and that the system can reach it (§5).
5. Stability. Test whether that rest attracts, with every conservation law kept in the linearization (§6).
6. Class of changes. Name the changes the identity survives and those it does not (§7).

The six questions organize the audit, while the chemistry sets the derivation order. Write exact species and total balances before imposing a closure or a dominance regime. An error identity suggested in step 1 remains provisional until step 3 establishes its assumptions. Only then can the feasibility and stability tests establish a returning response. The last step asks which changes preserve that result.

The audit records outcomes. It neither ranks models nor proves that a model cannot adapt. Each of the core's models illustrates one step's question in Part 2, and §8 gives every model's outcome at every step. A model can lack one kind of identity and still be input invariant. The core's subtractive compensator returns by a matched constant and has no identity at all. Its incoherent feedforward loop returns because its reaction orders match, and its identity is weighted rather than error-only (§2, core A9 and §9). So failing to find an identity of one kind does not show that none exists, and it does not show that the output cannot return.

Write the starting data of an adaptation claim (the classified reaction list, the rate law of each composite step, the input, the output and the boundary), and plan its audit as six questions in order: identity, observable, regime, positivity and reachability, stability with conservation kept, and the class of changes.

2Rate identities reveal what can pin the output

An identity is a relation among the rates that holds exactly for the stated rate laws. Three kinds appear among the core's models: an exact mass balance, an error-only identity and a weighted error identity. Each pins the output under its own condition. Cancelling, in a combination of rate equations, every step that does not read the output is a useful way to find one, though not the only way, and what an identity stores has units of its own.

Three kinds of identity

An exact mass balance is a combination of totals whose rate reduces, by stoichiometry, to a few fluxes. In the two-cycle feedback circuit the modified-B total $q_{B^*}$ changes only by two catalytic steps, $\dot q_{B^*}=k_{B1}C_3-k_{B2}C_4$ (§3). At a rest it pins a relation between two complexes, $k_{B1}C_3=k_{B2}C_4$. That relation is not yet a statement about a measured output. Turning it into one needs further assumptions, which is the work of step 2.

An error-only identity is a state whose rate depends on the system only through the output. Write it as $\dot z=g(y)$, where $y$ is the output and the function $g$ vanishes only at the set point $y^*$. Then every rest has $y=y^*$, whatever the other states and whatever the plant's constants. The core's integral loop has the simplest one, $\dot z=x$. Its chemotaxis teaching model has another, $\dot m=V_R-V_Ba/(K_B+a)$, whose right side depends on the state only through the receptor activity $a$ (core §5).

A weighted error identity carries a multiplier that depends on the state. Write it as $\dot z=h\,(y^*-y)$, where the multiplier $h$ depends on the state. A rest pins $y=y^*$ only where $h$ is nonzero. The core's incoherent feedforward loop has $z=x_1-(k_1/k_2)x_2$, with $\dot z=\delta x_2\,(x_1^*-x_1)$ (core §10). Its multiplier $\delta x_2$ is positive at every rest with a positive input, because there $x_2^*=k_2w/\gamma$. At zero input the multiplier vanishes and the rest pins nothing, which is why the core states its claim for $w>0$.

Finding an identity in the starting data

One useful route to an identity combines rate equations so that every step that does not read the output cancels. For mass-action steps this is linear algebra on the stoichiometry: look for a combination of species whose net change is zero in every step except those whose rate laws read the output. In the running example (1), binding and unbinding move a molecule between its free and bound forms, so they leave the totals $q_{Z_1}=Z_1+C_{Z_1Z_2}$ and $q_{Z_2}=Z_2+C_{Z_1Z_2}$ unchanged. Removal of the complex lowers both totals by one, so it cancels in their difference. Only the reference and sensing steps survive:

This is an exact mass balance that is also error-only. Its right side depends on the state only through $X$, and it vanishes only at $X=\mu/\theta$. So every rest has $X=\mu/\theta$, whatever $\gamma$, $k$ and the binding constants (exact).

The route is useful, not guaranteed. A linear combination of rate equations finds identities that are linear in the species. The identity of §9 is a logarithm, which no such combination produces. So this route can miss an identity in a nonlinear coordinate.

Specify the sampling clock before calculating the resting error

The sampled controller separates three operations within each time step. First calculate the action from the current output and memory. Next advance the output. Finally add the newly observed output to the memory. Let $t=0,1,2,\ldots$ label these updates and let $\Delta t$ be their fixed physical interval. Physical elapsed time is $t\Delta t$. A bracketed argument names an update, rather than a continuous-time derivative.

Physical units determine the gains in the sampled rule. Let output error $y[t]$ have units c, persistent disturbance $w[t]$ and opposing action $u[t]$ have units c/T, and accumulated error $z[t]$ have units c T. Define the sampled model by

Here $k_P$ has units 1/T and $k_I$ has units $1/\mathrm T^2$. The action in the output update uses $z[t]$. It does not use $z[t+1]$. This rule is a specified discrete model. It is not an exact sampled solution of the continuous differential equations below.

The core’s experiment uses dimensionless versions of these variables. Choose a fixed concentration unit $q_\mathrm{ref}$. For an active integral gain define

The normalized memory $\bar z$ measures its contribution to one step’s action. The normalized equations are

The core writes these normalized variables without bars. Integral-only control has $\bar k_P=0,\bar k_I=1$. Combined proportional and integral control has $\bar k_P=1,\bar k_I=1/4$. For proportional-only control the memory is absent and $\bar u=\bar y$.

A resting active memory forces zero output error. For a held $\bar w$ and $\bar k_I>0$, the last equation at a rest gives $0=\bar k_I\bar y^*$. Thus $\bar y^*=0$. The output balance then gives $\bar u^*=\bar w$, so $\bar z^*=\bar w$. The action remains nonzero after the error vanishes. Proportional-only control instead requires $\bar y^*=\bar w$. These calculations locate the rest. The stability calculation in §6 decides whether the updates reach it.

The continuous-time calculation uses a different clock. The calculation below develops the core’s continuous-time comparison. A convergence result for one clock does not automatically apply to the other.

The core’s control comparison can be calculated with physical units. Let $y$ be a concentration deviation, and let the disturbance $w$ and opposing action $u$ be concentration fluxes. The plant is $\dot y=w-u$. With $u=k_Py+k_Iz$ and $\dot z=y$, the memory has units c·T, $k_P$ has units $1/\mathrm T$, and $k_I$ has units $1/\mathrm T^2$.

A constant disturbance separates equilibrium from convergence. Proportional control alone gives $y^*=w/k_P$. With integral action, any rest instead has $y^*=0$ and $z^*=w/k_I$. Differentiate the plant equation after the disturbance has become constant:

Positive $k_P$ and $k_I$ put both characteristic roots in the left half-plane. With no proportional action, the roots are purely imaginary and the output oscillates. In the older comparison $\dot x=-\gamma x+d-k_Iz$, $\dot z=x$, the plant itself supplies damping through $\gamma$. The two comparisons therefore make the same point: the integral identity and the source of damping must be identified separately.

What the identity stores, in units

The stored quantity, the action it supplies and the disturbance it cancels are three objects with their own units. The core's integral loop shows them apart once its output carries units. Let $x$ be a concentration deviation from the set point, so that in $\dot x=-\gamma x+d-k_Iz$, $\dot z=x$ the disturbance $d$ is in c/T, $\gamma$ in 1/T, the memory $z$ in c·T and the gain $k_I$ in $1/\mathrm{T}^2$. At rest $z^*=d/k_I$, in c·T. The action $k_Iz^*=d$, in c/T, equals the disturbance in value and in units. The memory differs from both by the gain: at $k_I=2$ and $d=1$ it holds 0.5 (exact, core A9).

In the sequestering pair the three objects are a concentration, a flux and a frequency. The error integral $q_{Z_1}-q_{Z_2}$ is a concentration. The action is the actuator's flux $kZ_1^*$, in c/T, which at rest balances the output's removal, $kZ_1^*=\gamma\mu/\theta$. The disturbance is the removal frequency $\gamma$ itself, in 1/T. When $\gamma$ doubles, the action doubles: the actuator moves from 1 to 2, and the error integral from 0.98 to 1.99, in units of $\mu/\theta$ (numerical, core A14). The error integral is neither the disturbance nor the action.

The table compares the three objects across the core's models. In each row the memory, the action and the disturbance are different quantities, with different units.

A chemical concentration needs a positive implementation of signed memory

A physical engineering memory can increase or decrease through zero. A capacitor’s signed charge $Q$ obeys $\dot Q=I$. The sign records which plate has excess charge. A single chemical concentration $Z$ must instead satisfy $Z\ge0$. At $Z=0$, any physically admissible chemical balance has $\dot Z\ge0$. Therefore a global rule $\dot Z=\mu-\theta X$ is impossible when $\theta X>\mu$ on that boundary.

Three constructions avoid this conflict in different ways. A saturated removal process gives an approximate signed error rate inside a positive concentration regime. A difference of two nonnegative pools gives an exact signed coordinate. Multiplicative chemistry gives an error rate in the logarithm of a positive species. In the conversion example, $U$ is an active protein form and $D$ its inactive form. Each construction still needs an output, actuation, a feasible rest and a stability test.

The logarithmic construction follows by division, rather than cancellation of fluxes. In the mass-action conversion module of §9, $\dot D=D(k_\mathrm{conv}U-k_\mathrm{rec})$. Divide by $D>0$ to obtain the rate of $\ln(D/D_\mathrm{ref})$, where $D_\mathrm{ref}>0$ is a fixed concentration used only to make the logarithm dimensionless. Its stored coordinate is dimensionless. Its derivative has units 1/T. The same positive concentration can represent either sign of this coordinate by lying above or below $D_\mathrm{ref}$.

Two models with no error-integrating state

Two comparisons have no state that integrates the output's error, and each behaves for its own reason. Lecture 9's high-gain loop has the output as its only slow state, so the output itself holds the input, as an error of about $1/(1+H)$ of a small relative change, where $H$ is the loop gain (lineage §2). The subtractive compensator $u=-c\,d$ stores nothing at all, yet its output never moves when $c=1$, because the constant matches the plant (core A9). That is input invariance by a matched constant. Step 6 tests it, and a change of the plant's input gain breaks it (§7). Neither case says anything about networks in general.

Classify an identity as an exact mass balance, an error-only identity or a weighted error identity, look for it by cancelling every step that does not read the output, without reading a failed search as proof that none exists, say when its zero pins the output, and give the units of the memory, the action and the disturbance separately.

3The identity pins one quantity, and a pool adapts as far as that quantity dominates it

An identity pins something at rest, but not necessarily what an experiment measures. In the two-cycle feedback circuit the exact mass balance ties one complex to the bound enzyme. Three measurable quantities contain that complex to different degrees. In the feedforward mechanism, flux balance pins free output while total output acquires an input-dependent complex. Each calculation decomposes the measured quantity before judging adaptation.

The two-cycle feedback circuit, written as starting data

The design has twelve elementary steps on ten species. An input $I$ modifies a protein $A$ to $A^*$, the modified $A^*$ modifies a protein $B$ to $B^*$, the modified $B^*$ demodifies $A^*$, and a fixed enzyme $E$ demodifies $B^*$ (core §7). It is the two-node network that Jeynes-Smith and Araujo reduce from Ma and colleagues' three-node design, with its four complexes written out (Fig. 2, p. 3).¹ Each binding pair is followed by its catalytic step:

Here $k_{+i}$ and $k_{-i}$ are the binding and unbinding constants of pair $i$, and $k_{A1}$, $k_{A2}$, $k_{B1}$ and $k_{B2}$ the catalytic constants, named as in the core. The input is the input total $q_I$. The output is a pool of modified A, still to be chosen. The boundary is closed: no synthesis, no removal and no growth.

Six totals are tied by four conservation laws, and two independent totals move. The totals are $q_I=I+C_1$, $q_A=A+C_1$, $q_B=B+C_3$, $q_{A^*}=A^*+C_2+C_3$, $q_{B^*}=B^*+C_2+C_4$ and $q_E=E+C_4$. Only $q_I$ and $q_E$ are conserved individually. The A-protein total $A_0=q_A+q_{A^*}$ and the B-protein total $B_0=q_B+q_{B^*}$ are conserved too. That makes four conservation laws for ten species, since the twelve steps have rank 6, so six coordinates still move (exact). Two of them are independent totals, $q_{A^*}$ and $q_{B^*}$, with $q_A$ and $q_B$ moving opposite them. The other four are the amounts in the complexes $C_1$ to $C_4$, which also move. The totals determine those amounts only once a binding closure is imposed (§4). The two moving totals obey exact mass balances:

Every binding and unbinding step cancels from (4), because it moves a modified molecule between its free and bound forms without changing the total that counts it.

Two kinds of constant

A complex's binding relation holds with a dissociation constant at binding equilibrium. At binding equilibrium, $K_{B1}C_3=A^*B$ with $K_{B1}=k_{-3}/k_{+3}$, a dissociation constant. The same holds for the other three pairs.

At the steady state of the resolved model the same relation holds with a Michaelis constant. In (3) the complex $C_3$ forms at the rate $k_{+3}A^*B$ and leaves by unbinding and by catalysis, so at steady state $(k_{-3}+k_{B1})\,C_3=k_{+3}A^*B$. That is the binding relation with the Michaelis constant $K_{B1}=(k_{-3}+k_{B1})/k_{+3}$. The steady states plotted below in Figures 1 and 2 are exact for both models, with each $K$ read as the constant of the model in use. The core's parameter set PT, introduced below, fixes the Michaelis constants. Its time courses set each unbinding constant to 100 times its catalytic constant, so each dissociation constant is 100/101 of its Michaelis constant (core A7).

The sequestering pair's complex has a third kind of constant. It leaves by unbinding and by removal, so at steady state $Z_1Z_2=K\,C_{Z_1Z_2}$ with $K=(k_-+\delta_C)/k_+$. Identity (2) needs none of these constants, because it holds at every binding allocation.

What the balance pins

At any rest the second balance fixes the ratio of two complexes, and it puts $C_3$ near an input-independent value only while the enzyme is saturated. Setting $\dot q_{B^*}=0$ gives $k_{B1}C_3=k_{B2}C_4$, which ties $C_3$ to the bound enzyme $C_4$. Since $C_4=q_E-E$,

at every rest of (3), whatever the binding constants (exact). So $C_3$ lies a fraction $E/q_E$ below $\sigma$, and it is close to $\sigma$ only while the enzyme $E$ is nearly saturated. Jeynes-Smith and Araujo obtain the same set point in the limit of saturated B-cycle enzymes (p. 3).¹ Relation (5) gives it as an exact rest relation with an explicit correction.

Each measurable pool adds its own terms to $C_3$. Three quantities can be measured: free $A^*$, the partial pool $A^*+C_3$ and the full modified-A total $A^*+C_2+C_3$. Dividing by $\sigma$ and using (5),

exactly at every rest. The partial pool's error has two terms: the enzyme's free fraction and free $A^*$. The full modified-A total adds $C_2/\sigma$, the complex that holds most of the memory $q_{B^*}$, and the memory grows with the input. Jeynes-Smith and Araujo identify the partial pool as the quantity that tracks the set point when the B-cycle enzymes saturate (p. 5).¹

At parameter set PT the terms decide each pool's range. Set PT is the core's illustrative choice: $\sigma=1$, $A_0=2$, $B_0=50$ and $q_E=10^{-4}$, catalytic constants $k_{A1}=k_{A2}=25$, $k_{B1}=0.01$ and $k_{B2}=100$, and the Michaelis constants $K_{A1}=100$, $K_{A2}=10^{-4}$, $K_{B1}=0.2$ and $K_{B2}=10^{-6}$, with concentrations in units of $\sigma$ (core A7). Figure 1 plots the three terms against the input total. At low input the enzyme leaves saturation, and its free fraction $E/q_E$ grows. At high input $C_2/\sigma$ grows. Free $A^*/\sigma$ stays near 0.0041. So the partial pool stays within 5% of $\sigma$ from $q_I=0.065$ to at least $10^4$, the largest input computed, while the full modified-A total stays within 5% only from 0.064 to 5.23 (numerical).

Free $A^*$ is a third quantity, fixed by a binding relation. At binding equilibrium $A^*=K_{B1}C_3/B$ exactly. Since $C_3$ is near $\sigma$ and free $B$ is far larger than every complex that draws on it, free $A^*$ is nearly constant: $0.00407\sigma$ at $q_I=1$, against the leading-order estimate $K_{B1}\sigma/(B_0-\sigma-q_{B^*})=0.00408\sigma$ (numerical). It is nearly input invariant, and it sits 99.6% below $\sigma$. Not tracking $\sigma$ and not being input invariant are different failures.

The pinned quantity in the sequestering pair

In the sequestering pair the identity pins the free output, because the sensing step reads the free output. The rate law $\theta X$ in (1) is a function of free $X$, so (2) pins free $X$. If part of $X$ were bound by a partner that the sensing step cannot read, the identity would still pin free $X$, and the total of $X$ would be free to move.

Flux balance separates free feedforward output from total output

The feedforward mechanism makes a second observable distinction calculable. The input concentration $w>0$ drives output $X_1$ and degrader $X_2$ production at rates $k_1w$ and $k_2w$. Their complex $C_{12}$ catalytically removes output at rate $k_\mathrm{cat}C_{12}$. Free and bound degrader are removed at frequency $\gamma$. Removal of bound degrader releases intact output. The measured output may be free $X_1$ or total $q_1=X_1+C_{12}$. The boundary contains no separate output load or growth dilution.

The two total balances fix how much output is bound. With $q_2=X_2+C_{12}$, a rest requires $k_2w=\gamma q_2^*$ and $k_1w=k_\mathrm{cat}C_{12}^*$. Dividing these relations gives a bound fraction independent of the input:

The complex balance then pins the free output. The resolved steady-state relation is $X_1^*X_2^*=K_MC_{12}^*$, with $K_M=(k_-+k_\mathrm{cat}+\gamma)/k_+$. Since $X_2^*=(1-r)q_2^*$, cancellation of the positive $q_2^*$ gives

A finite positive rest requires $0<r<1$. Free output is input invariant. Total output grows with the input because the complex stores more output. For a change $\Delta w$, $\Delta X_1^*=0$ and $\Delta q_1^*=(rk_2/\gamma)\Delta w$, exactly within this model.

Turn an identity into an observable: write the rest relation it imposes, decompose each measured pool exactly into the pinned quantity plus its other terms, and read each pool's range from those terms, taking each binding relation at equilibrium or at a resolved steady state as the model requires.

4A limiting law holds only in its regime, and the regime is not the range

A chemical mechanism supplies an effective controller only under specified conditions. Start with a saturated memory-removal enzyme, then apply the same calculation to modification feedback and chemotaxis. Finally resolve the feedforward mechanism. Each example separates an exact balance from its limiting law and the input range over which a named observable stays within tolerance.

Derive zero-order memory removal from a catalytic complex

The simplest chemical example isolates the approximation that creates an integrator. Output $X$ produces memory $Z$ at rate $\theta X$. An enzyme removes the memory through $E+Z\xrightleftharpoons[k_-]{k_+}C_{EZ}\xrightarrow{k_\mathrm{cat}}E+P$. Production is composite. Binding and catalysis are elementary in this model. With fixed $q_E=E+C_{EZ}$, the two independent species balances are

Adding the balances identifies the memory before any approximation. Define $q_Z=Z+C_{EZ}$. Then $\dot q_Z=\theta X-k_\mathrm{cat}C_{EZ}$ exactly. Setting $X^*=k_\mathrm{cat}q_E/\theta$ rewrites the same equation as

The free enzyme is the exact defect in the proposed error identity. A rest obeys $X/X^*=1-E/q_E$.

An explicit plant supplies the missing feedback sign. Consider a measured free output $X$ supplied by the input flux $p_\mathrm{in}$ and removed by a memory-dependent composite process:

The removal law is part of this illustrative model’s starting data. Here $p_\mathrm{in}$ has units c/T, $\gamma$ has units 1/T, and $\kappa$ has units $1/(\mathrm{c\,T})$. Increasing the stored total $q_Z$ increases the output’s removal frequency. The boundary keeps $q_E$ fixed and has no memory dilution. The input is a held supply flux, not a change of the reference constants.

The saturated limit gives a candidate rest with a feasibility condition. Neglecting the free-enzyme defect gives $X=X^*=k_\mathrm{cat}q_E/\theta$ and $q_Z^*=(p_\mathrm{in}/X^*-\gamma)/\kappa$. A positive candidate requires $p_\mathrm{in}>\gamma X^*$. Positivity alone does not keep the enzyme saturated. The free $Z=q_Z-C_{EZ}$ must also pass the concentration test below. §6 checks stability of this chosen plant in the same limit.

A binding closure converts the defect into a concentration condition. Under rapid reversible binding, use $K_d=k_-/k_+$ and $E/q_E=K_d/(K_d+Z)$. To keep the fractional defect at most a chosen $\varepsilon\in(0,1)$, require $Z\ge K_d(1-\varepsilon)/\varepsilon$. For 5% precision, the free memory concentration must reach at least nineteen dissociation constants: $Z\ge19K_d$.

The same concentration test is exact at a resolved steady state with a Michaelis constant. Replace $K_d$ by $K_M=(k_-+k_\mathrm{cat})/k_+$. Using that relation during a transient needs a separate timescale argument.

A saturated modification cycle supplies high gain before feedback is closed

A modification fraction can control metabolism or transmit a signal. Glycogen phosphorylase’s phosphorylated and dephosphorylated forms motivate Goldbeter and Koshland’s covalent-modification model (p. 6840). Their calculation explains how saturated opposing enzymes can amplify a change in catalytic capacity.⁸ Here the input is the ratio of modifying to demodifying capacities. The output is the modified fraction. The substrate inventory is fixed, and feedback has not yet been closed.

A measured kinase-cascade switch supplies context for the gain calculation. Huang and Ferrell varied purified Mos in Xenopus oocyte extracts and measured a steep steady response of p42 MAPK (Fig. 3, p. 10083). Their analysis separates enzyme saturation from the additional steepness supplied by two successive phosphorylation encounters (pp. 10081–10082).⁹ That experiment does not identify every cascade with a single zero-order cycle. The calculation below isolates the single-cycle contribution.

Opposing saturated fluxes explain the switch used in the two-cycle feedback circuit. In the core’s single modification cycle, $E$ modifies $S$ and $F$ demodifies $S^*$. The substrate inventory is $S_0=S+S^*+C_{ES}+C_{FS^*}$. The modified total is $q_{S^*}=S^*+C_{FS^*}$.

The balance of catalytic capacities determines the modified fraction. Write $q_E=E+C_{ES}$ and $q_F=F+C_{FS^*}$ for the enzyme totals, and $k_E,k_F$ for their catalytic frequencies. At a resolved steady state, $K_E,K_F$ are the corresponding Michaelis constants. With enzymes scarce relative to $S_0$, set $f=q_{S^*}/S_0\approx S^*/S_0$. Balancing the two reduced fluxes gives

When $K_E/S_0$ and $K_F/S_0$ are small, this ratio stays close to one across a broad interval of $f$ away from zero and one. A small change of capacity ratio can therefore move the modified fraction far. This is zero-order ultrasensitivity. The lineage reconstructs Goldbeter and Koshland’s calculation with appreciable enzyme complexes.

A steep steady response does not by itself establish adaptation. The next construction makes one capacity depend on the controlled output and closes negative feedback. Its balances then identify the condition for approximate integration.

Binding allocation reduces the two-cycle feedback circuit

The two-cycle feedback circuit makes the second protein’s modified pool the memory. Its input is the held total $q_I$. Its candidate output is the modified-A total $q_{A^*}$, whose measured components were separated in §3. Increasing $q_{A^*}$ increases modification of B. Increasing $q_{B^*}$ increases demodification of A. The inventories $A_0$, $B_0$ and $q_E$ remain fixed. Those two influences close negative feedback.

Protein-modification feedback can buffer biological signaling, without establishing this circuit’s exact assumptions. Fritsche-Guenther and colleagues measured weak dependence of phosphorylated Erk on total Erk in some cellular backgrounds. They linked the buffering to post-translational feedback from Erk to Raf (Results and Discussion).¹⁰ This supplies a biological reason to study modification feedback. It does not prove the four-complex circuit’s joint dominance regime or perfect adaptation to a persistent input.

Reducing the six totals to two equations needs a binding closure first. The exact balances (4) are written in complexes, and the totals determine the complexes only through binding relations such as $C_1=IA/K_{A1}$. That relation holds at binding equilibrium when $K_{A1}$ is a dissociation constant. With the Michaelis constant it holds exactly at every rest of the resolved model (3). Along a trajectory it is a quasi-steady-state assumption, which needs binding to be fast against catalysis and against changes of the input. At set PT the full model's slow eigenvalues approach the leading ones as unbinding speeds up (§6). The exact balances (4) need no closure, and the leading-order model below rests on one.

Lecture 8's method gives each total one winner, and each winner turns one term of (4) into a function of totals. The core's joint regime keeps the B cycle zero order: $I$ wins $q_I$, $A$ wins $q_A$, $B$ wins $q_B$, $C_3$ wins $q_{A^*}$, $C_2$ wins $q_{B^*}$ and $C_4$ wins $q_E$. Each assumption enters (4) at one place:

• $I$ and $A$ win their totals, so $C_1=IA/K_{A1}\approx q_Iq_A/K_{A1}$.
• $C_2$ wins $q_{B^*}$, so $C_2\approx q_{B^*}$.
• $C_3$ wins $q_{A^*}$, so $C_3\approx q_{A^*}$.
• $C_4$ wins $q_E$, so $C_4\approx q_E$.
• $B$ wins $q_B$, so free $B$ is known from totals, $B\approx q_B=B_0-q_{B^*}$. It enters no term of (4) directly. It turns the other winner conditions into conditions on totals, and it fixes free $A^*=K_{B1}C_3/B$ (§3).
• Conservation of the A protein gives $q_A=A_0-q_{A^*}$, exactly.

Substituting into (4) gives the core's leading-order model:

Each assumption has an exact error term. Subtract (7) from (4), with every free species written through the totals. With the closure for $C_1$, the exact right side of the A balance differs from that of (7) by

This difference holds wherever the closure does. The exact right side of the B balance differs from that of (7) by

exactly, along every trajectory, because it uses only $q_{A^*}-C_3=A^*+C_2$ and $q_E-C_4=E$. So the A balance needs $C_1$ small against $I$ and against $A$, and $B^*+C_4$ small against $C_2$. The B balance needs $A^*+C_2$ small against $C_3$, and $E$ small against $C_4$. Those are five of the six winner ratios. The sixth, $C_3/B$, keeps $B$ the winner of its total.

The leading-order rest sets the memory in proportion to the input. Setting (7) to zero,

which at set PT is $q_{B^*}=q_I/100$ (leading order).

Where the regime ends

At set PT all six ratios stay below one fifth for input totals from 0.052 to 20, and two of them set the ends. Figure 2 plots each losing form over its total's winner, at the exact steady states. At the lower end the memory's complex stops dominating its own total against the enzyme's complex: $(B^*+C_4)/C_2=0.198$ at $q_I=0.052$. At the upper end the input starts to occupy $A$: $C_1/A=0.199$ at $q_I=20$, the condition that $A$ wins its total. The allocation ratio $(A^*+C_2)/C_3$ is 0.146 there, and it does not bind (numerical).

The leading-order model explains both ends, and only the full sweep establishes them. With (9), $C_2\approx q_{B^*}=q_I/100$ and $C_4\approx q_E=10^{-4}$, so $(B^*+C_4)/C_2\approx10^{-2}/q_I$, which reaches one fifth at $q_I=0.05$. And $C_1/A=I/K_{A1}\approx q_I/100$, which reaches one fifth at $q_I=20$ (leading order). Near the upper end the leading-order rest drifts from the full model: the full memory $q_{B^*}$ is 0.01005 at $q_I=1$ against the leading 0.01, and 0.145 at $q_I=20$ against the leading 0.20 (numerical). So the estimates read the ends correctly here, and the check is still the sweep of all six ratios.

The regime justifies the leading-order model, and it is not any observable's range. The partial pool stays within 5% of $\sigma$ up to at least $q_I=10^4$, far past the regime's upper end, because its error (6) contains neither $C_1$ nor $C_2$. The full modified-A total stays within 5% only up to 5.23, inside the regime, because its error adds $C_2/\sigma$. The regime and the two ranges are three different intervals (numerical).

Chemotaxis requires more than enzyme scarcity

The exact methylation balance counts modification events. Let $q_T$ be a fixed receptor concentration, $q_\mathrm{methyl}$ the concentration of receptor-bound methyl groups, and $m=q_\mathrm{methyl}/q_T$. A complex $C_{R,j}$ adds one group at rate $k_{R,j}C_{R,j}$. A complex $C_{B,j}$ removes one at rate $k_{B,j}C_{B,j}$. Hence

The activity-only teaching law imposes further assumptions on these fluxes. Bound CheR must remain approximately constant. Demethylation must depend on activity without a separate ligand dependence at fixed activity. The relevant catalytic ratios and binding constants must also lose their dependence on stored modification. Yi and colleagues identify the corresponding assumptions in their model, including negligible activity of unmethylated receptors. Their integral coordinate approximates total methylation. It is not automatically the literal mean $m$ used in the teaching model below.⁴

The chemotaxis teaching model places its boundary between rapid sensing and slower memory change. Its input is attractant concentration $L$. Its measured output is receptor activity $a$. Mean methylation $m$ is the changing memory. Enzyme abundances and effective methylation and demethylation capacities are held fixed. Receptor switching is represented by an equilibrium law, rather than by separate switching states. The reduced law does not resolve every CheB phosphorylation or recruitment step.

The teaching model treats the receptor activity as a fast equilibrium. A cluster of $N$ receptors with mean methylation $m$, in methyl groups per receptor, and attractant $L$ is active with probability

where $F$ is the free energy of the active state less that of the inactive state, a dimensionless difference with every energy divided by the thermal energy. Here $K_I$ and $K_A$ are the attractant's dissociation constants for an inactive and an active receptor, and $\epsilon_0$ and $\epsilon_1$ set the activity's offset and its rise per methyl group (core A12, model details). The cluster is assumed to switch much faster than methylation changes, so $a$ follows $m$ and $L$ at every instant. Methylation then obeys $\dot m=V_R-V_Ba/(K_B+a)$, with CheR methylating at the rate $V_R$ and CheB demethylating at most at $V_B$, half-maximally at the activity $K_B$.

The teaching model's identity is the limit of a bounded model in which CheR's rate does not depend on methylation. In the core's teaching model CheR methylates at the constant rate $V_R$, whatever the methylation $m$. Its bounded version multiplies CheR's rate by $s_R=(M-m)/(\kappa_R+M-m)$ and CheB's by $s_B=m/(\kappa_B+m)$, where $M$ is the number of sites per receptor (core A12, model details). Then $\dot m$ depends on $m$ directly, not only through the activity, and the activity-only identity is recovered as $\kappa_R$ and $\kappa_B$ go to zero while $m$ stays away from both site boundaries. The limit is not uniform at $m=0$ or $m=M$. With both constants equal to $\kappa$, the precision error after an attractant step is −2.3%, −0.23% and −0.023% at $\kappa=0.01$, 0.001 and $10^{-4}$ (numerical, the Lecture 10 analysis).

What the identity needs is a methylation rate that depends on the state only through the activity. A CheR out of saturation breaks it: the precision error is −7.2% at $\kappa_R=0.1$ and −35% at $\kappa_R=1$ (numerical, the Lecture 10 analysis). CheB acting on inactive receptors through the activity alone does not break it. So the condition is independence from $m$ and from the attractant at fixed activity, not saturation of both enzymes (core A12).

Resolve feedforward removal before testing its reaction orders

The feedforward calculation needs a convention for degrader turnover. Input $w$ drives production of $X_1$ and $X_2$ at rates $k_1w$ and $k_2w$. Their elementary binding-catalysis mechanism is $X_1+X_2\xrightleftharpoons[k_-]{k_+}C_{12}\xrightarrow{k_\mathrm{cat}}X_2+P$. As in the core, removal acts on free and bound degrader at frequency $\gamma$. Removing bound degrader releases intact output. With $q_1=X_1+C_{12}$ and $q_2=X_2+C_{12}$, the exact equations can be written as

Rapid reversible binding reconstructs the complex from both totals. At binding equilibrium, $(q_1-C_{12})(q_2-C_{12})=K_dC_{12}$, where $K_d=k_-/k_+$. The physical root is

When $q_1,q_2\ll K_d$, both totals are mainly free and $C_{12}\approx q_1q_2/K_d$. Thus $k_\mathrm{cat}C_{12}\approx\delta q_1q_2$ with $\delta=k_\mathrm{cat}/K_d$. The reduced totals are the $x_1,x_2$ of §2 and §7. Their bilinear law is exact as a reduced-model definition and approximate as a consequence of this chemistry.

The observable calculation in §3 supplies the exact resolved rest. It gives $r=k_1\gamma/(k_\mathrm{cat}k_2)$, $C_{12}^*=r q_2^*$ and $X_1^*=K_Mr/(1-r)$, with $K_M=(k_-+k_\mathrm{cat}+\gamma)/k_+$. A finite positive rest requires $w>0$ and $0<r<1$. The same calculation for the fast-binding model replaces $K_M$ by $K_d$. The next test asks when that model’s total output agrees with its bilinear approximation, rather than merely whether its free output is input invariant.

Two small ratios recover the core’s bilinear prediction for total output. Require $r\ll1$ and $q_2^*/K_d\ll1$. The fast-binding model gives the exact relative discrepancy from the leading level $K_dr$: