From reactions to systems: how biochemical behavior emerges

Stoichiometry turns molecular events into dynamics. One driven enzyme cycle, written twice, shows what a lumped arrow costs. Gene expression, metabolism, signaling, sequestration, and growth then reveal what the same equation means in different biological settings.

CCBS · EST 5053 · Fall 2026 · lecture 3 · core page (full) · 2026-09-08 (Tue) 10:40–12:15 · lecturer-led

This core is the 95-minute spoken argument. Use the exposition to reconstruct the derivations and work the transfer problems, and the lineage essay to see how the field's separate traditions became this systems language.

Lectures 1 and 2 gave us credible parts. We estimated molecular sizes and copy numbers, collision and catalytic times, and the rate laws associated with elementary binding and catalytic steps. A parts list still does not explain a phenotype. This lecture changes scale. It asks how coupled reactions become an evolving system, and how levels, timescales, saturation, competition, and signal processing emerge from that system.¹

0–6 min

Parts to systems. Change the question from one reaction's rate to a network's behavior.

6–14 min

One event. Read reactant, product, and net-change vectors from an arrow.

14–23 min

The network. Assemble $\Gamma$, attach elementary mass action, fix the notation, and close the autonomous ODE.

23–30 min

Growth. Move from molecule count to concentration and expose dilution as a reaction.

30–42 min

Two views. Read the same accounting as $\Gamma v$ or as addition minus removal, on two real networks, declaring every composite rate law.

42–59 min

One cycle, twice. Drive an enzyme cycle, then lump it, watch the rate law absorb the difference, and find the two independent conditions that let it be a hyperbola.

59–64 min

Gene expression. Resolve §5's lumped arrow and watch its declared constant stop being one.

64–73 min

Fast binding. Pay for the closure with lecture 2's ledger, derive the saturating laws, then ask which variable was swept.

73–79 min

Metabolism. Read PFK allostery through the coupled adenylate pool.

79–91 min

Conservation. Compare phosphorylation and BMP competition, then compute in totals.

91–95 min

Cliffhanger. See one restoring balance and state the stability problem.

All eleven is 95 minutes with no slack. If it runs long, §6's open-question box comes off first (2 min): §10 asks the same question at minute 79 and then answers it. Next is §9 (6 min), which nothing downstream consumes, though put Figure 9 up for thirty seconds and say what a shared pool does to a rate law. Never drop §6's two boxes on the regimes of (21), because §8's $\varepsilon$ and §10's two regimes are both read back to them. Dropping the first leaves 93 and both leave 87.

1We know the parts; behavior belongs to the network

A single reaction can have a rate. Only a connected state can have a trajectory.

Imagine measuring every enzyme in a pathway and every one of its isolated rates. That still does not tell us whether an intermediate accumulates, a signal is filtered, or a protein concentration returns after a perturbation. Each reaction changes the concentrations used by other reactions. The output of one arrow becomes the input or regulator of another. Behavior is therefore a property of the coupled vector field, not an adjective attached to one arrow.

One grammar covers most of what follows, and it is worth stating before we use it. Molecules bind and unbind. Binding redistributes them among states. The productive transitions, meaning catalysis, transport and fuel coupling, then run at rates that depend on which state a molecule is in. That is a modeling claim and not a claim about all of biology. Spontaneous chemistry, diffusion, assembly and physical deformation also matter. What the claim buys is that one accounting scheme carries the whole lecture.

Refuse to answer a behaviour question from a parts list, and say what you need instead. Ask which species each rate reads and which each rate changes. Two cells holding identical enzymes at identical concentrations can differ in every balance they reach.

2One reaction becomes one change vector

The left side determines what must be present. The difference between right and left determines what one event changes.

Fix an order for $n$ species, $x=(x_1,\ldots,x_n)^T$. Write reaction $i$ as

The reactant vector is $\alpha_i=(\alpha_{1i},\ldots,\alpha_{ni})^T$. The product vector is $\beta_i$. One firing changes the state by the stoichiometric vector

For the net reaction $2A+B\rightsquigarrow3B$, ordered as $(A,B,C)$, the three vectors are

That bare squiggle makes only a net accounting claim. If a mechanism declares the whole event elementary, rewrite it as $2A+B\xrightarrow{k_i}3B$; then mass action uses $A^2B$. Yet one event changes $B$ by $+2$. Stoichiometry is a difference. Kinetics counts the reactants that must meet.

Read three vectors off any reaction arrow, and never let two of them merge. $\alpha_i$ is what must be present, $\beta_i$ is what is left, and $\gamma_i=\beta_i-\alpha_i$ is what one firing changes. The state moves by $\gamma$, and a mass-action exponent comes from $\alpha$.

3A reaction list becomes an autonomous system

Stack change vectors into a matrix, stack event rates into a vector, and the arrows can run forward in time.

For $m$ reactions, assemble the real $n\times m$ stoichiometric matrix $\Gamma$ by placing $\gamma_i$ in column $i$. If $v_i(x)$ is the rate per volume of reaction $i$, then bookkeeping gives

For an elementary reaction in a well-mixed concentration model, mass action closes its rate:

The exponent comes from $\alpha_i$, not $\gamma_i$. A composite arrow such as transcription or a whole enzyme turnover can still occupy a column of $\Gamma$. What it cannot do is take (4) with it. Its rate law has to be derived, measured, or explicitly assumed. Stoichiometric accounting does not grant mass action, and §5 pays that bill in front of the room.

One property of (3) is worth naming while it is still small. With fixed parameters and a fixed environment it is autonomous: the right-hand side depends on the current state and not explicitly on time. A timed nutrient pulse would break that, unless the environment is promoted to a state variable and the pulse becomes part of the network.

Turn any reaction list into $\dot x=\Gamma v(x)$, and then say out loud which half of it you got for free. The arrows give every column of $\Gamma$. Only an elementary arrow also gives its own rate law, so every composite column is a rate you still owe, by derivation, measurement, or open declaration.

4Growth is the reaction every species runs

Molecule number and concentration are different state variables in a growing cell, and the difference is one more column in $\Gamma$.

Let $N(t)$ be molecule number, $V(t)$ cell volume, and $x=N/V$ concentration. Balanced exponential growth at rate $\lambda$ gives $\dot V=\lambda V$. Differentiate the quotient:

Read the second term as chemistry and it is the composite boundary process $X_j\rightsquigarrow\varnothing$, with effective rate $\lambda x_j$. It is a strange one. It runs for every species at once, so it contributes $n$ columns rather than one, and every one of those columns carries the same effective coefficient. That coefficient belongs to the cell and not to the molecule. No bond is broken. Collecting the columns,

A zero-order source in a concentration model means production is constant per unit volume: $\dot N=aV$. Then

Starting from zero, $x(t)=x^*(1-e^{-\lambda t})$. After one doubling time $T_d=\ln2/\lambda$, concentration has reached half of its steady value even though both volume and molecule number increased. If one fixed molecular machine instead makes $k$ molecules per minute, then $\dot x=k/V(t)-\lambda x$. The dilution term survives, but the concentration-level source is not constant.

Two payments for introducing this now. First, every example that follows can be given a real steady state instead of an end state. Second, it settles what a conserved total means in a cell that is growing.

Give every species in a growing cell one more removal column, all of them at the same rate $\lambda$, and read the steady level straight off: a species made at $k$ per unit volume sits at $k/\lambda$. That rate belongs to the cell rather than to the molecule, and it breaks no bond.

5One system, two accounting views

The matrix form emphasizes coupling. The addition-removal split emphasizes the balance of one species. How you write it follows what you are asking.

Read row $i$ of (3) and collect positive and negative contributions:

This is an identity, not a modeling choice. Two networks show why the choice of form still matters.

Metabolism: few arrows, dense columns

Lump glycolysis into an investment and a payoff, with $I$ for the triose pool and $W$ for pyruvate:

Order the species $(G,I,\mathrm{ATP},\mathrm{ADP},W)$ and the two columns are forced:

Three things are visible in the matrix and in nothing else. The net of one pass is $G+2\,\mathrm{ADP}\rightsquigarrow2\,\mathrm{ATP}+2W$, which is the ATP yield of glycolysis. The adenylate pool $q_{\mathrm{ade}}=\mathrm{ATP}+\mathrm{ADP}$ is untouched by both reactions, so it couples them however far apart they are drawn. And ATP is a reactant of the pathway that produces it: read the ATP row alone, $\dot{[\mathrm{ATP}]}=4v_2-2v_1$, and that fact is gone. Metabolism keeps the matrix because the coupling is the coefficients.

Gene expression: many arrows, sparse columns

Now one repressor $R$ that binds the promoter $D$ of the next gene and blocks it, with transcription and translation lumped into the single arrow $r_4$. The list is deliberately mixed resolution: binding is elementary and carries its rate constants, the other four arrows are composite and carry none.

By §3 those arrows fix the columns and nothing else, so the four composite rates have to be declared. Declare both productions constant and both removals first order. Only $r_3$ gets its rate for free, because mass action applies to it. Order the species $(R,D,C_{DR},P)$:

The promoter is conserved: $\ell^T=(0,1,1,0)$ gives $\ell^T\Gamma=0$, so $q_D=D+C_{DR}$ is the gene dose rather than a dynamical variable. Four of the five columns hold a single entry. Every species is made by one process and removed by one process, and only binding couples two rows. So splitting into addition and removal discards nothing, and with the dilution term of (6):

Every constant in (13) was read off a row of (12), and four of the five rows were declarations rather than consequences. That is what a composite arrow costs, and it is §3's warning made concrete: the arrows alone would have given the columns and left the right-hand side empty. Two of those declarations are provisional. §7 resolves $r_4$ and finds that $k_P$ is not a constant, and §8 replaces $D$ by $q_DK_{d}/(K_{d}+R)$, after which the second row is still addition minus removal with every regulatory decision inside the addition term. The difficulty has moved out of the stoichiometry and into one scalar function, which is precisely where the addition-removal form puts the reader's attention.

Pick the accounting form by where the difficulty sits, since the two are the same equation. Keep $\Gamma v$ when the coefficients carry the coupling, as in the metabolic pair. Split into addition minus removal when each species has one source and one sink, because then all the biology has moved into a single term and you can point at it.

6Work one driven cycle, detailed and composite

A closed cycle runs down. Give it a source and a sink and it has a steady state, and then the same system can be written at two resolutions. Comparing them takes four moves: write both, solve the closure that connects them, take its limits, and price the lump.

Take the enzyme cycle and drive it. Substrate is supplied at a constant rate, product is removed first order, and the middle is unchanged:

Without the two boundary arrows the substrate is consumed once and the product piles up, and the system has an end state rather than a steady one. The supply and the drain hold it away from equilibrium, which is where a cell lives. Lecture 7 asks what that costs. The drain here can be read as export, as consumption by the next enzyme, or as the dilution of §4, in which case $k_{\mathrm{out}}=\lambda$.

Order the species $(E,S,C_{ES},P)$. The five columns comprise three elementary enzyme steps and two composite boundary processes; their five rates are

Multiplication gives all four ODEs without inventing a single sign, and it shows at once that $E+C_{ES}$ is untouched by every column.

The same system in pools

Change variables to the combinations that binding does not move. Binding shifts $E$, $S$ and $C_{ES}$ together, and the combinations it leaves alone are

The product is not an exception to the rule. Its pool is the rule evaluated on a species that binds nothing. In these variables the system reads

This is exact. No reduction has been made, no timescale has been assumed, and $k_{+}$ and $k_{-}$ have disappeared from the equations. Their disappearance is not an approximation: association and dissociation move the state along $(-1,-1,1,0)$, and (16) lists three vectors orthogonal to it. It is also not closed. The middle rate names $C_{ES}$, which is a species and not a pool.

The composite reaction, and what it costs

Read (17) as a reaction list rather than as three equations, and it is three arrows:

Substrate is produced, substrate becomes product, product is removed. Two dynamic variables instead of four, three columns instead of five, and no complex anywhere. The enzyme has left the state vector and become a parameter of the middle arrow. This is as simple as a reaction system gets.

The bill arrives in the rate law, which is no longer a monomial. Closing (17) takes one algebraic equation, and §8 says when we are allowed to write it. Set $\dot C_{ES}\simeq0$ and the quasi-steady complex satisfies $ES=K_MC_{ES}$ with $K_M=(k_-+k_{\mathrm{cat}})/k_+$. Substituting $E=q_E-C_{ES}$ and $S=q_S-C_{ES}$:

whose smaller root is the physical one, since $C_{ES}$ can exceed neither total. Solve it:

Take the minus sign. The plus sign is the other root, and it exceeds both totals, so it would put more enzyme in complex than the cell owns.

That is the missing rate law, $v(q_S)=k_{\mathrm{cat}}C_{ES}$, in closed form, with nothing assumed beyond the quasi-steady step. Look at it for a moment, because it is not a hyperbola. The square root is where the difficulty went. A network of monomials has closed into an algebraic function, and that function is what the composite arrow of (18) carries.

Keep (19) and (20) in view, because two later sections are about this one equation. §8 writes the same closure in the free substrate, where the square root disappears entirely. §10 meets the same quadratic again with $q_E$ renamed, and takes both of its limits instead of one.

Now expand (20). When $q_E$ is small against $K_M+q_S$, the square root flattens and $C_{ES}\simeq q_Eq_S/(K_M+q_S)$, so the composite rate law is

Michaelis–Menten, written in the total substrate. That the familiar law survives being rewritten in the controlled variable is a statement about a ratio, not a law of nature. The next two boxes reach (21) twice, because that ratio can be made small in two independent ways, and a cell does not always get the first one.

What the numbers do

Take lecture 2's medians and let the rest be forced. $K_M=100\,\mu\mathrm{M}$ and $k_{\mathrm{cat}}=10\,\mathrm{s^{-1}}$ are the medians of 5194 and 1942 measured pairs. Choose a typical rather than diffusion-limited on-rate, $k_{+}=10^{8}\,\mathrm{M^{-1}s^{-1}}$, and then $k_{-}=k_{+}K_M-k_{\mathrm{cat}}=9990\,\mathrm{s^{-1}}$ is not a choice. Set $q_E=1\,\mu\mathrm{M}$, $v_{\mathrm{in}}=5\,\mu\mathrm{M\,s^{-1}}$ and $k_{\mathrm{out}}=0.05\,\mathrm{s^{-1}}$. The steady state then needs no equation solved:

Note first where those medians put the example. A median $K_M$ of $100\,\mu\mathrm{M}$ is three decades above the floor two boxes up. So this worked enzyme is in the trace-enzyme regime and that reading is safe for it. A fast enzyme would need substrate excess instead.

Here that is $C_{ES}^{*}=0.5$, $E^{*}=0.5$, $S^{*}=100$, $q_S^{*}=100.5$ and $P^{*}=100$, all in $\mu\mathrm{M}$.¹⁵ Two consequences follow from the first expression alone.

Figure 6 is the lecture's one honest measurement of what lumping costs. After the first millisecond the composite model tracks the detailed one to $6\times10^{-3}\,\mu\mathrm{M}$ in $q_S$, which is 0.006% of the steady value. Inside that first millisecond it is wrong by as much as $0.16\,\mu\mathrm{M}$ in the complex, a third of the steady complex, because it starts the complex already full. That is the whole content of a lumped arrow: right everywhere except in the layer it was built to skip.

This move recurs for the rest of the course, and it is worth naming once while the example is small enough to check. Lecture 6 asks when the closure is legitimate. Lecture 8 asks what the closure does to reaction order. Lecture 11 gives the geometry of all of its limits. Lecture 13 works only in the composite form. Every time the arrow count drops, the rate law absorbs the difference.

Lump any mechanism you like, then price the lump before you use it. The arrows that disappear come back inside the rate law as the exact algebraic solution of the states you removed, which is why the rate law stops being a monomial. A lumped model is right everywhere except in the layer it was built to skip, and here that layer is the first millisecond.

7Gene expression makes regulation a production law

Promoter state controls production. Messenger degradation and growth control memory.

§5 lumped transcription and translation into the single arrow $r_4$, and declared its rate to be $k_PD$. Resolve it. RNA polymerase makes messenger $M$ from a promoter that is free, ribosomes translate $M$ into $P$, RNases remove $M$, and intrinsic proteolysis plus the dilution of §4 remove $P$. Same repressor, same promoter, same conserved $q_D$, one species added in the middle:

Compare with (13) and the lumped arrow gives up its secret. One species became two, so the protein equation now waits on a pool that (13) did not have. Hold $M$ at its quasi-steady value and $\dot P=k_{\mathrm{tl}}f_{\mathrm{tx}}(R)/(\delta_M+\lambda)-(\delta_P+\lambda)P$. Line it up against (13) and $k_PD=k_{\mathrm{tl}}f_{\mathrm{tx}}(R)/(\delta_M+\lambda)$. So §5's constant was never a constant. It is a regulated function divided by a messenger lifetime, and it is constant only while $R$ is fixed and $M$ is fast. A declared rate law is a promise that some later section has to keep.

At steady state, solve in causal order:

The same denominators set response times. A genome-wide E. coli study found about 80% of measured mRNA half-lives between 3 and 8 minutes.⁶ For many stable bacterial proteins, intrinsic degradation is small compared with dilution, so $\delta_P\approx0$ is useful. It is not a universal statement that bacterial proteins never degrade. Growth also changes gene dosage, polymerase and ribosome supply, and cell size, so dilution is the first whole-cell coupling, not the last.⁷

Treat every declared rate constant as a debt and go and find what it was made of. Resolving §5's lumped $k_PD$ turned it into $k_{\mathrm{tl}}f_{\mathrm{tx}}(R)/(\delta_M+\lambda)$, so it was never a constant. It holds only while the regulator is fixed and the messenger is fast.

8Fast binding leaves a saturating effective rate

Michaelis–Menten and Hill laws resemble one another because both remove fast binding states from a slower model. Lecture 2 is where the removal was paid for.

§6 set $\dot C_{ES}\simeq0$ and called it a closure. That is a claim about timescales, and it was measured last week. Lecture 2's ledger put a substrate finding an enzyme at 70 ns and one catalytic turnover at 0.1 s, and its identity $k_{-}/k_{\mathrm{cat}}=k_{+}/(k_{\mathrm{cat}}/K_M)-1$ turned that gap into a count.¹ Put §6's constants in: $10^{8}/10^{5}-1=999$. The complex forms and falls apart a thousand times per molecule of product made.

That count is not the licence, and two reductions get confused here, so separate them. Treating the bound fraction as an equilibrium is the one that needs the count to be large. Setting the complex derivative to zero needs something else. It needs Segel and Slemrod's small parameter $\varepsilon=q_E/(K_M+q_S)$, read at the substrate the run starts from.⁴ For §6's numbers that is $8\times10^{-3}$. Compute $\varepsilon$, not the count.

The enzyme reduction is §6's algebra in a different variable. The work is already done. Its closure (19) is $(q_E-C_{ES})(q_S-C_{ES})=K_MC_{ES}$, and the free substrate is $S=q_S-C_{ES}$. Substitute, and the quadratic term is gone:

One closure, two variables, two laws that do not look alike. In the free substrate it is linear in $C_{ES}$, so (25) is exact and carries no condition. In the total it is quadratic, and (21) needed a limit. This is the Briggs–Haldane quasi-steady argument.³ It is not rapid equilibrium, and $K_M$ is not generally a binding dissociation constant. Lecture 6 asks when the reduction is accurate.

The two laws differ by one number. Equation (21) is written in $q_S$ and (25) in $S$. Those two variables differ by $C_{ES}$. So the two laws say the same thing whenever the complex holds only a small share of the substrate. That share is $q_E/(K_M+S+q_E)$, and the analysis script checks it over five decades of free substrate.¹⁵

Now compare that share with $\varepsilon$ above. Same numerator, and denominators that agree once the complex is small. So one number is doing two jobs here. It licenses the quasi-steady step. It also makes the free and the total substrate interchangeable, which is what lets the textbook hyperbola stand in for the cell's. §6 found two independent ways to make that number small, and either one is enough.

Promoter reduction, and it is $r_3$ of (11). Let that binding equilibrate, so $v_{\mathrm{b}}\simeq0$ and $K_{d}=k_{\mathrm{off}}/k_{\mathrm{on}}=RD/C_{DR}$. With the promoter conservation $q_D=D+C_{DR}$ that §5 read off the matrix, this is two equations in two unknowns:

That is the promise §5 made, kept. Its declared $k_PD$ becomes $k_Pq_DK_{d}/(K_{d}+R)$, a falling function of the repressor and of nothing else, so the promoter leaves the state vector exactly as the enzyme complex left it in (18). A repressor reads $p_{\mathrm{free}}$ and an activator reads $p_{\mathrm{bound}}$, and either way $f_{\mathrm{tx}}=r_0+(r_1-r_0)p$. A concerted or effective cooperative model replaces $R$ by $R^n$ and $K_{d}$ by $K^n$:

The exponent of a fitted Hill curve is an effective slope unless a binding mechanism justifies a literal stoichiometry. Binding equilibrium, time-scale separation, and promoter conservation are the reasons these production laws are not elementary mass-action monomials. Allosteric state models make the same warning concrete: a compact occupancy curve can follow from several coupled binding states rather than one literal $n$-body collision.⁵

Compute $\varepsilon=q_E/(K_M+q_S)$ before you quote a saturating law, and then say which variable the law is written in. That one number licenses the quasi-steady step and makes free and total interchangeable, and it has to do both jobs before the textbook curve is also the cell's. In the free concentration the hyperbola is exact. In the total it is a claim about $\varepsilon$.

9Metabolism exposes coupled fluxes and shared pools

Stoichiometry routes matter through the network. Allosteric binding changes a flux without changing its stoichiometric column.

Glycolytic carbon passes through branches while ATP, ADP, NADH, phosphate, and enzyme pools participate in several reactions, which is what the two dense columns of (10) were a caricature of. The form $\dot{x}=\Gamma v(x)$ prevents a local rate law from hiding those shared obligations. At a metabolic steady state, $\Gamma v(x^*)=0$. Individual fluxes need not be zero. Material can flow through every reaction while each internal concentration remains constant, which is why the same stoichiometric matrix also anchors constraint-based metabolic analysis.⁸

Phosphofructokinase couples the composite net reaction $F6P+ATP\rightsquigarrow F1,6BP+ADP$ near a committed glycolytic step. Fast binding at regulatory sites changes that catalytic flux. Mammalian muscle PFK has functionally opposed adenine-nucleotide sites: AMP and ADP can activate, while ATP and ADP can inhibit depending on the site and concentration.⁹ ATP is also a substrate. The rate law is therefore not “ATP is bad” or “AMP is good”. It is a state-dependent consequence of catalytic and allosteric occupancy.

All of that presumes one timescale, and it is worth naming. Adenylate kinase has to be fast enough to keep the three nucleotides in equilibrium with each other. When it is, write the pool and its energy coordinate as

Chapman, Fall, and Atkinson measured an energy charge near 0.8 in growing E. coli, with marked decline during starvation.¹⁰ The useful systems point is the constraint, not an alleged equivalence. AMP activation cannot generally be relabeled ATP inhibition. Only on the specified pool manifold do the two concentrations carry related information, and the PFK regulation itself remains organism-specific.

One thing to carry forward. Allostery changed a flux here without touching a single column of $\Gamma$, and it did so by occupying a finite number of sites. §10 takes that mechanism on its own, with the catalysis stripped away, and finds that occupancy of a finite pool is enough to compute with.

Separate the two ways one metabolite can change a flux, because only one of them shows up in $\Gamma$. Appearing in a column is stoichiometry and is fixed by the chemistry. Occupying a regulatory site changes the rate law and leaves every column alone. ATP does both at once in glycolysis, which is why “ATP is bad” is not a rate law.

10Conserved pools turn binding into computation

A conserved total limits the reachable state. Competition for that total couples reactions that never touch in a pathway cartoon.

If a row vector $\ell^T$ lies in the left null space of $\Gamma$, then

This invariant confines each initial condition to one stoichiometric compatibility class, the structural state space emphasized in chemical reaction network theory.² Both matrices of §5 carried one, and by §4 each is an invariant of the chemistry that growth then holds at a level the cell chooses.

A phosphorylation cycle redistributes a response regulator between $Y$ and $Y_P$. Without synthesis or degradation on the signaling timescale, $q_Y=Y+Y_P$, or that sum plus explicitly modeled enzyme complexes, is fixed. Kinase and phosphatase activities control the fraction phosphorylated. In the EnvZ/OmpR system, writing the finite EnvZ and OmpR pools explicitly is essential to the model's input-output behavior.¹¹

Binding conservation creates the same kind of coupling. The BMP system contains multiple ligand variants and multiple type-I and type-II receptor variants. Ligands compete for shared receptor pools, and the resulting complexes can phosphorylate Smad with different activities. Antebi and colleagues showed that this competitive architecture can generate additive, ratiometric, balance-detection, and imbalance-detection responses.¹² Sequestration is therefore not passive storage. Occupying one receptor changes every route that needs that receptor.

Two regimes of one binding reaction

§6 and §8 each left a question open, and one small network answers both. §6 asked what the closure does when neither of its two conditions holds. §8 asked what a binding law looks like in the totals a cell can actually set. Take one reaction and name the two species by their roles. $X_P$ is the partner the cell holds fixed, the enzyme or the receptor or the decoy, with total $q_P$, where the subscript is for parameter. $X_I$ is the input being swept, with total $q_I$. Fast binding plus the two conservation laws gives four equations,

Eliminate the two free concentrations and one equation is left,

whose smaller root is the physical one. This is (19) again, with $(q_E,q_S,K_M)$ renamed $(q_P,q_I,K)$. Both questions are therefore questions about one equation, and (30) is written in totals, so the second one is already answered.