The cell as a biomachine

Binding and catalysis provide a model class in which molecular mechanisms determine how regulation combines.

Lecture 8 · core · 95-minute session · Fall 2026

We can now analyze a specified biochemical model. The next question is which models to write. The first seven lectures supplied ways to choose relevant scales, describe reaction dynamics and justify reductions. Here we use those capabilities to construct a class of models with explicit molecular states, physical parameters and calculable regulatory functions.

We already have a working example. Lecture 6's supplied enzyme used exact totals, a fast reconstruction of the complexes and slow input-output balances. Its self-repressor used the same construction to derive a regulatory law. Lecture 7 added the equilibrium-compatibility test. We now turn these examples into one model class and use it to answer a question an isolated Hill curve leaves open: how do two regulators combine?

The exposition gives full derivations and worked checkpoints. The lineage traces why these distinctions became necessary. The extension tests the construction on a transporter with a physical current.

If time runs short, skip numerical sweeps and the exhibition first. Read the enzyme matrices from the figure without walking through every entry. In §8 keep competition and leave the sponge as a worked example. In §9 keep the independent-versus-exclusive comparison. Preserve the canonical equations, the finite-total root and the joint-state derivation.

1A useful model class joins biological questions to analysis

A biological question does not select a reaction network by itself. The observable, intervention, spatial scale, timescale and molecule counts determine which mechanisms a model must resolve. Once the model is specified, rate equations, stochastic trajectories, stability analysis and justified reductions can be used to understand it. A useful model class makes these two activities meet.

Binding and state-dependent catalysis provide such a class for a broad range of biochemical regulation. Binding allocates molecular constituents among complexes and conformations. Activities of those states change chemical forms or produce outputs. The allocation and activity problems have distinct physical parameters, and their composition retains the molecular pools shared between modules.

The model class fills a practical gap between general reaction lists and ready-made response curves. A general reaction network permits many mechanisms but does not itself select the physical inventories or a tractable binding subproblem. An isolated Hill law gives a convenient response but leaves its molecular composition unspecified. Binding and catalysis retain that composition while exposing a state-allocation problem that can be solved separately from the slow dynamics.

The class is a structural choice within chemical reaction modeling. It does not imply that every cellular process is at equilibrium or that arbitrary biological dynamics are automatically tractable. Spatial transport, driven state cycles and slow conformations remain explicit when their scales matter. The equilibrium layer earns its place by simplifying a substantial part of the model without imposing an abundance hierarchy among its constituents.

The destination is concrete: two transcription factors regulate one gene, so how do their effects combine? A state-based construction will determine the answer and supply the production laws needed for the next lecture.

Choose a model class by the molecular information it preserves and the analytical structure it exposes for the biological question.

2Binding allocates states and catalysis acts from those states

Define binding reactions as association-dissociation steps and compatible transitions between internal states. They preserve the constituents assigned to the binding layer. For example, an enzyme and a substrate can occupy separate species or the same complex without changing the amount of enzyme or substrate-form constituent.

A resolved catalytic channel acts from an identified molecular state. In the class considered here, a channel with reactant state concentration $x_j$ has flux $k x_j$. Its frequency k has inverse-time units. A reversible conversion uses two opposed channels, each with a nonnegative frequency. Bimolecular encounters are represented in the binding mechanism when the physical bound intermediate is resolved.

This is Lecture 6's reversible, product-binding enzyme. Here $k_f=k_{\mathrm{cat}}^+$ and $k_r=k_{\mathrm{cat}}^-$ are shorter labels for the same two catalytic frequencies. Neither reversibility nor binding affinity by itself establishes that the binding layer relaxes fast enough.

The first two pairs allocate enzyme between substrate and product complexes. The last pair changes the chemical form inside the complex. Binding preserves substrate-form and product-form totals separately. Catalysis transfers material between them while preserving the enzyme total.

A promoter can instead have a composite production activity. At that resolution, write $G_i\rightsquigarrow G_i+X$ and state its assumed rate $a_iG_i$ separately. This representation includes the state-dependent output in the same accounting framework, while acknowledging that transcription or translation has unresolved internal steps. The linear activity law is then a model assumption, not a consequence of an elementary reaction arrow.

Classify each process by what it conserves, what it changes and which state determines its activity.

3Constituent totals remove binding flux by exact accounting

A constituent total counts every modeled state containing that constituent. Consider the elementary binding pair $A+B\xrightleftharpoons[k_{\mathrm{off}}]{k_{\mathrm{on}}}C_{AB}$. Bare A and B denote free concentrations, $C_{AB}$ denotes the complex concentration, and the controllable totals are $q_A=A+C_{AB}$ and $q_B=B+C_{AB}$. Each complex uses one unit from each pool.

Let x be the vector of all modeled species concentrations. Separate a chosen fast binding stoichiometry $\Gamma_{\mathrm{bind}}$ from the remaining reaction stoichiometry $\Gamma_{\mathrm{cat}}$. Their flux vectors are $v_{\mathrm{bind}}$ and $v_{\mathrm{cat}}$. Boundary supply and removal contribute a species-rate vector r, and uniform volume growth contributes dilution at rate λ:

These are concentration balances, as in Lecture 3. Reaction fluxes have concentration-per-time units. Here $r-\lambda x$ is the boundary vector called $b(x,t)$ in Lecture 6, with uniform growth dilution displayed separately. Growth changes concentration without destroying molecules.

Choose the rows of L as a complete independent set of the fast network's constituent conservation laws. Then $q=Lx$ and $L\Gamma_{\mathrm{bind}}=0$. Multiplication gives an exact identity, even while binding is far from its eventual equilibrium:

The canceled binding fluxes still matter: they determine x and therefore the catalytic activities. But they cannot directly change their own constituent totals. Uniform dilution projects to $-\lambda q$; selective degradation of only free species must instead be evaluated on those free species. Neither fitted first-order removal nor growth dilution becomes an elementary reaction merely because its rate is linear.

Build the physical totals and project the species equations before choosing a binding approximation.

4The canonical model has a binding layer and an activity layer

Group slow activity channels by the change they produce in the binding totals. Put each distinct change in a column of $\Gamma_q$. Let $K_{\mathrm{cat}}$ be the nonnegative matrix that sums the activities of the molecular states contributing to each channel. Its entries have inverse-time units. Then $\Gamma_qK_{\mathrm{cat}}x$ is the corresponding rate of change of totals.

To close that exact projection, impose the compatible rapid-binding equations. Let $c^\circ$ be a fixed reference concentration. Let N contain independent binding stoichiometric rows, oriented consistently with dimensionless equilibrium ratios κ. Completeness means $\ker N=\operatorname{im}L^{\top}$. The canonical differential-algebraic system is

The logarithmic equation is the familiar binding constraint written row by row. For the elementary pair $A+B\xrightleftharpoons[k_{\mathrm{off}}]{k_{\mathrm{on}}}C_{AB}$, one row of N gives

Every logarithm is dimensionless. This representation applies to positive concentrations. If a total is zero, the species containing that constituent vanish. Use the original algebraic balances or their continuous limit at that boundary, not the logarithm of zero. Compatibility means that additional binding paths impose no contradictory equilibrium ratios, the cycle test from Lecture 7.

The differential equation changes the available totals. The algebraic equations allocate those totals among molecular states. The boundary term $r_q$ contains projected supply, removal and growth. It can depend on reconstructed free species, so the formula does not assume every boundary rate is a function of totals alone.

There are two independent conditions behind this form. The admitted activities must be linear in their explicit source states, or be modeled that way at the chosen resolution. The binding layer must track a compatible equilibrium on the timescale of the outputs. Multiplying both the association rate constant and the dissociation rate constant by the same factor leaves equilibrium affinities fixed but changes whether that tracking is accurate.

The lecturer's current supplement develops this canonical form from binding mechanisms with physical constituent totals. The exposition gives the full construction, uniqueness proof and derivative formula.¹

A useful uniqueness result holds for an ideal compatible binding layer with positive affinities, a complete nonnegative constituent inventory and an explicit free species for each independent constituent. Each positive set of totals selects one positive equilibrium state $x^{\mathrm{eq}}(q)$. A strictly convex potential proves this result.

The map $x^{\mathrm{eq}}(q)$ reconstructs the full species vector. Its complex components play the role of Lecture 6's $h_{C_{ES}}(q)$. The algebraic state moves when the totals move. Binding equilibrium therefore does not make every complex constant. A unique binding state at fixed totals also does not imply a unique or stable steady state of the slow dynamics.

Construct the total-state-activity model while separating exact projection, equilibrium reconstruction and its timescale justification.

5The enzyme parameters have separate physical meanings

Use the resolved reversible enzyme introduced above, with species ordered as $x=(E,S,P,C_{ES},C_{EP})^{\mathsf T}$. Its totals are $q_E=E+C_{ES}+C_{EP}$, $q_S=S+C_{ES}$ and $q_P=P+C_{EP}$. The matrices in the canonical equation are

The two activity rows select the substrate-bound and product-bound enzyme. The two stoichiometric columns say that forward conversion consumes one substrate-form unit and produces one product-form unit, while reverse conversion does the opposite. Neither conversion changes the enzyme inventory.

The affinities $K_S=k_{S,-}/k_{S,+}$ and $K_P=k_{P,-}/k_{P,+}$ allocate enzyme through $K_SC_{ES}=ES$ and $K_PC_{EP}=EP$. The catalytic frequencies determine the activity of those occupied states. Changing enzyme expression changes $q_E$. Changing binding speed at fixed affinity changes the relaxation clock. These interventions act on different parts of the model.

Keep Lecture 6's names for substrate supply $v_{\mathrm{in}}$ and net catalytic flux $v_{\mathrm{cat}}$. For free-product export coefficient $k_{\mathrm{out}}$, the projected boundary vector is $r_q=(0,v_{\mathrm{in}},-k_{\mathrm{out}}P)^{\mathsf T}$. Supply has concentration-per-time units. Export depends on free product P, not total product $q_P$. This fixed-volume example has no growth dilution.

The enzyme model is an executable construction. Define $v_{\mathrm{cat}}=k_fC_{ES}-k_rC_{EP}$. Its totals obey $\dot q_S=v_{\mathrm{in}}-v_{\mathrm{cat}}$ and $\dot q_P=v_{\mathrm{cat}}-k_{\mathrm{out}}P$, while $\dot q_E=0$. Eliminating substrate and product with their inventories gives the same increasing free-enzyme equation worked in Lecture 6:

Its root lies between zero and $q_E$. The two fractions are the complexes. Solve for E, recover the complexes, evaluate $v_{\mathrm{cat}}$ and advance the totals. This algorithm does not replace free substrate or free product by their totals.

Translate a molecular mechanism into inventories, affinities, activities and boundaries that can be perturbed separately.

6One binding pair generates several regulatory regimes

The equilibrium equation and two inventories determine the complex. Define the dissociation constant $K_d=k_{\mathrm{off}}/k_{\mathrm{on}}$, which has concentration units. Equilibrium gives $K_dC_{AB}=AB$. Replacing the free concentrations using the two totals yields

The physical bounds select one root of the quadratic. The complex must satisfy $0\le C_{AB}\le\min(q_A,q_B)$. With $T=q_A+q_B+K_d$, the acceptable root is

The other root requires more complex than the pools contain. Reject it using the smaller-total bound, then recover free A and B from their inventories. Both total amounts matter whenever binding appreciably depletes either partner.

The rationalized expression is useful for numerical calculation. In weak binding, the first form subtracts nearly equal numbers. The second form evaluates the same physical root without that cancellation.

The equilibrium constant carries a specific kinetic assumption. The catalytic complex balance from Lecture 2 gives $K_M=(k_{\mathrm{off}}+k_{\mathrm{cat}})/k_{\mathrm{on}}$ when catalytic escape contributes to complex removal. Our $K_d$ belongs to binding equilibrium. The two agree approximately when catalytic escape is negligible compared with dissociation.²

Saturation follows when binding barely depletes the input partner. If $q_A\ll K_d$, then $C_{AB}/B=A/K_d\le q_A/K_d\ll1$. Thus $B\approx q_B$ throughout the input range, giving

The non-saturating end of this curve is effectively mass action. For $q_B\ll K_d$ as well, $C_{AB}\approx q_Aq_B/K_d$. If the complex acts at frequency $k_{\mathrm{cat}}$, the flux is approximately $(k_{\mathrm{cat}}/K_d)q_Aq_B$. This is the same effective-rate interpretation developed in Lecture 2, with the binding-equilibrium constant appropriate to the present reduction.

Tight binding creates a smaller-pool bottleneck when the complex is active. As $K_d$ becomes small, the complex approaches $\min(q_A,q_B)$. Adding more of the already abundant partner then does little. The limiting constituent sets the amount of active complex.

The same tight binding creates a threshold when free A is active. Accounting gives $A\approx\max(q_A-q_B,0)$. Below the matching point, B sequesters added A. Above it, additional A remains free. This is the molecular-titration mechanism of ultrasensitive regulation.³

Finite affinity rounds the threshold on a calculable concentration scale. At equal totals $q_A=q_B=q$, the free species satisfy $A=B$ and $A^2=K_d(q-A)$. For tight binding, this gives $A\approx\sqrt{K_dq}$. A finite-affinity binding pair therefore approximates a rectifier with a rounded neighborhood near equivalence.

Use the finite binding root to identify saturation, a bottleneck or a threshold for the molecular state the experiment actually reads.

7Binding regulates catalysis through state-specific activities

Occupancy alone does not specify a production rate. Suppose a gene exchanges between free $G$ and bound $C_{GR}$, with total $q_G$. Let the two states produce transcript T at effective frequencies $k_0$ and $k_1$. Their composite transcription processes and rate law are

A binding change regulates the output according to the relative activities of the states. If the bound fraction is $\theta=C_{GR}/q_G$, then $v_{\mathrm{tx}}/q_G=k_0(1-\theta)+k_1\theta$. Increasing binding activates when $k_1>k_0$, represses when $k_1<k_0$, and has no effect on this output when the two activities match.

The same construction extends to many conformations and complexes. Each state has an equilibrium weight and an activity for each modeled process. The total activity is the sum $v_\alpha=\sum_j k_{\alpha j}x_j$. Allosteric ensembles and equilibrium promoter models supply important examples, with their own kinetic conditions for relating occupancy to output.⁴⁵

The Michaelis form is recovered by choosing a bound enzyme as the active state. In rapid equilibrium, $v/q_E=k_{\mathrm{cat}}S/(K_d+S)$, with free substrate S as the input. The distinct catalytic quasi-steady approximation uses $K_M=(k_{\mathrm{off}}+k_{\mathrm{cat}})/k_{\mathrm{on}}$. Substituting supplied total for free substrate needs a depletion check.²

The non-saturating slope has a direct interpretation. For $S\ll K_M$, $v\approx(k_{\mathrm{cat}}/K_M)q_ES$. The coefficient $k_{\mathrm{cat}}/K_M$ has concentration-inverse time-inverse units. It is the effective second-order coefficient of the composite conversion within that approximation.

A Hill form also needs a state mechanism. If an n-site receptor has only two appreciably populated states, empty and fully occupied, with weights 1 and $(R/K)^n$, its active fraction is $R^n/(K^n+R^n)$. This requires suppression of intermediate states. Independent sites with only the fully occupied state active instead give $[R/(K_d+R)]^n$. Neither description licenses an elementary n-body association arrow.

Lecture 6 gave another explicit route to degree two: proteins first dimerize, then the dimer binds the promoter. Under its fast constraints, promoter repression depends on free monomer as $1/[1+(R/K_H)^2]$. Rewriting that expression in total protein additionally requires monomers to dominate the total. A Hill exponent and a choice of concentration variable are two separate modeling decisions.

Derive a response from its populated states and activities, then state the assumptions behind any Michaelis or Hill simplification.

8A second binding reaction can tune the effective affinity

Competition modifies allocation without changing the substrate's microscopic affinity. Let enzyme E bind substrate S or inhibitor I in mutually exclusive complexes, with dissociation constants $K_S,K_I$. At controlled free S and I, the shared enzyme inventory gives

The effective binding scale is $K_{S,\mathrm{eff}}=K_S(1+I/K_I)$. The factor comes from the enzyme state occupied by inhibitor. It is not a new microscopic constant.

The same relation can retain finite substrate if free I is controlled. Add $q_S=S+C_{ES}$ and use the finite-pair root with $K_{S,\mathrm{eff}}$. If total inhibitor $q_I$ is controlled instead, add $q_I=I+C_{EI}$ and solve the shared pool. Replacing I by its total without this check can change the prediction.

A second example: a finite sponge shifts a threshold

Adding a binding partner changes the amount of signal available to an existing module. Let repressor R bind either a gene G or a sponge T, whose total is $q_{\mathrm{sponge}}$. The gene responds to free R, while both complexes draw from total repressor. With dissociation constants $K_{GR}$ and $K_{RT}$, the three balances give

The shared balance determines the whole response without fitting a new repression curve. Its right side increases strictly with R, so one scalar solve gives the free signal for any three totals. At half repression, $R=K_{GR}$, which immediately gives

A tight sponge shifts the threshold by approximately its own total amount. When $K_{RT}\ll K_{GR}$, nearly every sponge molecule is occupied at the repression midpoint. The added repressor required is therefore approximately $q_{\mathrm{sponge}}$. Below the threshold, the exact balance automatically prevents a negative available-repressor concentration.

Compose binding reactions through shared pools and identify when they produce a tunable effective affinity or threshold.

9Joint promoter states determine how two regulators combine

Let free regulators $R_1,R_2$ bind a promoter with total $q_G$. Define dimensionless inputs $u=R_1/K_1$ and $v=R_2/K_2$. The allowed promoter states are empty, bound by regulator 1, bound by regulator 2, and bound by both. Their weights are $1,u,v,\omega uv$, where positive dimensionless ω describes the equilibrium interaction between binding events.

Assign production frequencies $a_0,a_1,a_2,a_{12}$ to the four states. Transcription is a composite process at this resolution. The separate state-weighted output law is

For one promoter in a well-maintained regulator reservoir, normalized weights are probabilities of discrete gene states. Their activity-weighted average is the fast-switching operation from Lecture 6, not a fractional gene molecule. The finite-inventory concentration equations below describe a deterministic binding layer. At low counts with appreciable shared-pool fluctuations, Lecture 5's count model and a conditional stationary distribution are needed instead. A deterministic root is not generally the exact mean of that stochastic model.

A product of one-input Hill functions follows under specific assumptions. If binding is independent, $\omega=1$, the denominator factorizes. If only the double state acts with frequency a, then $v_{\mathrm{out}}/(q_Ga)=u/(1+u)\,v/(1+v)$. This is an AND response derived from states.

An additive law follows from a different activity assignment on the same independent binding architecture. Set $a_0=0$ and $a_{12}=a_1+a_2$. Then $v_{\mathrm{out}}/q_G=a_1u/(1+u)+a_2v/(1+v)$.

An OR response uses the same activity a for either single state and the double state. Set $a_0=0$ and $a_1=a_2=a_{12}=a$. Double occupation then contributes once, giving $v_{\mathrm{out}}/(q_Ga)=1-1/[(1+u)(1+v)]$.

Mutual exclusion changes the state list. Without a double state, the denominator is $1+u+v$. If both singly occupied states have activity a and the empty state is inactive, the normalized response is $(u+v)/(1+u+v)$. A continued fraction may arise from sequential elimination in a particular architecture, but algebraic appearance alone does not identify that architecture.

Consider three mechanisms with equal singly occupied activities and zero empty-state activity. The first has independent sites and an equally active double state. The second excludes double occupation. The third has independent sites but an inactive double state. With the second regulator absent, every mechanism gives $u/(1+u)$. The same agreement holds on the other input axis.

The mechanisms disagree when both inputs equal their dissociation constants. At $u=v=1$, their normalized outputs are respectively $3/4$, $2/3$, and $1/2$. The third mechanism eventually loses output as joint occupation dominates. No improvement in the precision of the isolated curves can recover this missing mixed-state information.

A minimal discriminating experiment therefore varies both regulators. A two-dimensional dose matrix tests interaction directly. Measuring joint occupancy can further separate altered binding from altered activity. A response surface alone may still leave several mechanisms consistent with the observations.

The promoter response above is expressed in free regulators. If those regulators are supplied in finite amounts, their totals must also count bound promoter states. With $Z=1+u+v+\omega uv$, the two balances are

These balances and the weights determine both free concentrations together. Additional targets contribute additional bound terms to the same balances. The activity law is then evaluated at the resulting free concentrations. No new empirical composition rule is needed.

Derive combined regulation from joint states and activities, then close finite regulator inventories when the experiment controls totals.

10The construction supplies the dynamics for Lecture 9

A complete model now has an architecture, physical parameters, a state reconstruction and an output law. For a produced protein X, let $V(q)$ be the production flux obtained from that construction. If every modeled X-containing state undergoes uniform removal at rate δ, the total balance is

The removal assumption matters. Selective removal of free X instead gives a term proportional to X. Binding redistributes the substrate of removal as well as the states that produce output. Addition and removal must use the same molecular accounting.

Uniform growth dilution contributes $-\lambda q_X$, directly from Lecture 3's count-to-concentration conversion. A degradation term proportional only to free X is a different mechanism, as in Lecture 6's self-repressor. Once the reduced vector field is specified, its steady states require zero total rates. Lecture 4's Jacobian or Lyapunov analysis then tests their stability. Neither conclusion follows from the binding constraint alone.

Lecture 9 asks what happens when these laws are connected into feedback and other dynamical motifs. It will compare binding-catalysis models with effective-law reductions through simulations, operating points and stability analysis. MultiFate supplies a concrete example: competitive dimerization determines active transcription-factor states, while production and removal determine which expression states are stable.⁶

Later lectures develop different consequences. Lecture 11 studies the response orders available to an architecture. Lecture 12 uses binding-generated nonlinearities for molecular computation. Lecture 16 asks how architectures and parameters can be inferred. The present construction gives each of them a molecular model to work with.

Close the derived regulatory functions with explicit addition and removal, then analyze the resulting cellular dynamics.

An exhibition to explore after class

Each example extends a different part of the construction. The companion essays retain the full equations, source context and numerical comparisons.

• Negative cooperativity with finite ligand. An inactive singly occupied receptor can buffer ligand before double occupancy produces output. The readout and inventory determine the threshold.⁷
• BMP ligand mixtures. Competing receptor complexes have different affinities and activities. Changing receptor expression can change which feature of a ligand mixture the cell reads.⁸
• MultiFate. Homo- and heterodimer competition redistributes finite transcription-factor pools. Active homodimers enter an addition-removal circuit.⁶
• Antithetic control. Shared sequestration terms can cancel from a controller difference. Production, removal and feedback determine whether that difference integrates output error.⁹
• Growth limitation. A catalytic process can be limited by its machine pool or its substrate supply. A growth model must identify which resource each pool represents.¹⁰
• Protein-carried substrates. The glycine cleavage system transfers a chemical group on H protein’s lipoyl arm. Carrier abundance and its allocation among chemical and bound states can therefore matter to a substrate model.¹¹
• Opposed modification and coupled transport. The exposition develops a modification cycle and the one-pair response-order triangle. The extension retains driven transport steps while equilibrating compatible binding subgraphs.

References