From a molecule count to a biochemical rate law

A tutorial on constructing, interpreting and testing the quantitative models of Lecture 2. Counts determine concentrations, transport supplies encounters, and a mechanism determines how encounters become product.

CCBS · EST 5053 · Fall 2026 · Lecture 2 exposition

A rate constant answers a question about a particular event. A molecule can reach an enzyme, bind it, leave it, or become product. Each event has a different clock. To predict a cell's response, we must connect those clocks to the concentrations and pools that actually change.

The core develops that argument at lecture pace. This chapter supplies the intermediate calculations and worked decisions. The lineage follows the experiments and ideas that made the calculations possible. You need powers of ten, derivatives and simple integrals. Probability and differential-equation methods are introduced where used.

1Four calculations and one decision

Learn to connect a molecular census to the rate of a specified cellular change.

The calculation begins with an observable. Suppose we want the product flux after a substrate increase. We need the free substrate concentration, the amount of enzyme, and a law for product formation. A protein count alone does not supply that law, and a binding rate alone does not supply the product flux.

1. Count: convert copies and volume to concentration, keeping free and total pools distinct.
2. Time transport: compare diffusion across the relevant distance with the rate at which the relevant pool changes.
3. Resolve the mechanism: distinguish elementary events from composite conversions and boundary flows.
4. Calculate the output: derive the rate law and examine the concentration range in which it will be used.

The final decision is whether the model may omit a spatial gradient or a fast intermediate. That decision needs an error criterion and an observation timescale. A typical cellular number is a useful starting input, but it cannot establish the validity of every model of that cell.

Before selecting a rate constant, name the event and the observable whose rate you need.

2Estimate, calibrate and scale

Use the methods of Lecture 1 while keeping their assumptions attached.

A bound answers a one-sided question. If all of a cell's wet mass were protein, dividing that mass by an assumed mean protein mass would bound the number of molecules under that mass assumption. Estimating the actual count additionally requires the water and dry-mass composition.

A calibration transfers only where the same law and parameters apply. Measuring diffusion of one sphere can determine the combination of temperature and viscosity in the Stokes–Einstein relation. Moving to a different solvent or a crowded cytoplasm requires new inputs. A theoretical displacement calculated from that relation is a consistency check, rather than an independent calibration.

A scaling law transfers a ratio. At fixed diffusion coefficient, a tenfold longer exploration distance requires a hundredfold longer time. If the diffusion coefficient also changes, include that change explicitly. We will repeatedly write the relation, attach units to each factor, and then compare the result with an independently obtained value.

Label each input as an assumption, a measurement or a derived value before carrying it into the next estimate.

3A cubic micron is a femtolitre

Convert cell size into the denominator of every concentration.

A litre is $10^{-3}\,\mathrm{m}^3$, whereas a cubic micron is $(10^{-6}\,\mathrm m)^3=10^{-18}\,\mathrm m^3$. Therefore

Molarity, written M, means moles per litre. Avogadro's constant $N_A\approx6\times10^{23}\,\mathrm{mol}^{-1}$ converts molecule count $n$ to amount in moles. For a volume $V$ measured in litres,

The convenient mnemonic is one copy per bacterium of order one nanomolar. Keep 1.7 when doing arithmetic at exactly 1 fL. At 4 fL the same one copy is about 0.42 nM. Thus 50 copies represent about 83 nM in 1 fL and 21 nM in 4 fL. Neither number by itself tells us whether a binding site is occupied.

Use $c=n/(N_AV)$ to transfer a copy number between cell sizes without treating concentration as a property of the count alone.

4Estimate a protein mass and choose the right average

A log midpoint is an estimate of scale; a mass balance requires the mean appropriate to the molecules being counted.

Protein mass is length times residue mass. A dalton, Da, is a molecular mass unit of approximately $1.66\times10^{-24}\,\mathrm g$. Free amino acids bracket roughly 75–204 Da. The geometric midpoint is $\sqrt{75\times204}\approx124\,\mathrm{Da}$. Peptide formation removes one water per bond, so subtracting about 18 Da suggests a residue scale near 110 Da. The terminal correction is negligible for a chain of hundreds of residues.

A geometric midpoint gives the middle of a logarithmic bracket. For plausible endpoints $a,b$, the midpoint has logarithm $(\ln a+\ln b)/2$ and value $\sqrt{ab}$. If the target value is $m$, its relative error is exactly

The estimate is exact when the endpoints are equally far below and above the target on a log scale. That identity explains why an unrepresentative extreme can spoil the bracket. It does not identify an unknown median or mean from endpoints alone.

For example, 50–5000 residues has midpoint 500, whereas a bracket of commonly encountered globular proteins at 100–1000 gives 316. Taking 300 residues as a working scale gives

Use a log bracket to estimate a scale, then use an abundance-weighted mass when converting total protein mass to molecule count.

5Separate wet mass, dry mass and protein mass

Two composition fractions turn the mass of a reference cell into a protein count.

A 1 fL cell at an assumed density $\rho=1.1\,\mathrm{g/mL}$ weighs about 1.1 pg, because $1\,\mathrm{fL}=10^{-12}\,\mathrm{mL}$. Assume a dry fraction of 0.30 and a protein fraction of dry mass of 0.55. Define $c_p$ as protein mass per cell volume. Then

The protein molecule count is this protein mass divided by the mass per molecule:

The all-protein bound uses a different numerator. Dividing the full 1.1 pg by the same assumed mean mass gives $2.0\times10^7$ molecules. The composition factors reduce that answer by approximately six. The larger number remains a bound under the chosen mean-mass assumption, not an estimate of the actual proteome.

Transfer the formula before transferring the number. A 2000 fL mammalian cell at an assumed 0.20 g/mL protein concentration and a 400-residue mean has about $5.5\times10^9$ protein molecules. The volume multiplication is exact within the model, while the composition and mean mass remain organism-dependent inputs.

Compute protein count from $n_p=c_pV/m_p$, with water, dry-mass composition and the molecular-mass conversion all visible.

6Concentration becomes useful beside a binding scale

Compare the free concentration of the right ligand with the constant for its own interaction.

A proteome total does not specify an individual enzyme concentration. Dividing a few million proteins among a few thousand species gives a species-average abundance near a thousand copies, or micromolar in a 1 fL cell. Abundances are uneven, and some genes are not expressed. This quotient is a census scale, not a measurement of a typical selected protein.

A total metabolite pool is also different from a substrate pool. Bennett and colleagues measured 103 metabolites with a summed concentration near 300 mM in glucose-grown E. coli. Individual pools span much smaller values, with glutamate at 96 mM. An enzyme encounters its own substrate, so the sum cannot decide its saturation.³

To interpret binding, let $L$ and $B$ denote free ligand and free binding site. A bare species symbol will always mean a free concentration. Write the complex as $C_{BL}$ and the total site concentration as $q_B=B+C_{BL}$. For the elementary binding pair

equilibrium requires equal forward and reverse fluxes. Thus $k_{\mathrm{on}}BL=k_{\mathrm{off}}C_{BL}$, or $C_{BL}=BL/K_d$. Substitute this into the total:

At $L=K_d$ half the sites are occupied. At one tenth and ten times $K_d$, occupancy is approximately 9% and 91%. This comparison requires free ligand. If the sites bind an appreciable fraction of the supplied ligand, Appendix A1 solves the finite-total problem.

Enzymatic saturation uses a related but distinct concentration scale. In the Michaelis–Menten model, substrate saturation depends on $S/K_M$. Section 21 derives $K_M$ and shows why it need not equal the binding dissociation constant $K_d$.

Use free ligand versus its own $K_d$ for binding occupancy and free substrate versus its own $K_M$ for Michaelis–Menten saturation.

7Distinguish material volume from crowding

A mass-to-volume conversion supplies a size estimate, while crowding asks how one molecule restricts another.

Protein's partial specific volume is approximately $\bar v=0.73\,\mathrm{mL/g}$. Multiplying by the mass of one dalton converts it to $1.21\,\text{\AA}^3/\mathrm{Da}$, or $0.00121\,\mathrm{nm}^3/\mathrm{Da}$. Thus a 33 kDa protein has material volume near 40 nm³. A sphere of that volume has radius

This is an equivalent compact radius. Hydration, shape and internal cavities can make the hydrodynamic radius larger. Erickson explains how sedimentation, gel filtration and microscopy probe these different sizes.⁴

The protein material fraction can be calculated without counting individual proteins. Multiply mass concentration by partial specific volume. Our estimate gives $0.1815\times0.73\approx0.13$, or 13%. The 0.24 g/mL comparison gives 17.5%. Neither includes the surrounding space denied to another molecule's centre.

For comparison, a 4.6 Mbp DNA molecule at 0.34 nm per base pair has contour length about 1.6 mm. Treating the double helix as a 2 nm diameter cylinder gives

That is about 0.5% of the reference cell's volume. It is the material cylinder volume, rather than the volume occupied by the nucleoid and its conformations.

State whether a size or volume describes material, hydrodynamic motion, a conformation or exclusion of a specified probe.

8Checkpoint: count and interpret

Carry the volume and the free-pool assumption through each calculation.

1. A protein has 12 copies in a 1 fL cell. Estimate its concentration and a site's occupancy at $K_d=5\,\mathrm{nM}$, assuming negligible ligand depletion. Repeat at 4 fL.
2. Estimate the protein count in a 2000 fL cell at 0.20 g/mL protein and 400 residues per protein.
3. Compare the mass of a 300-residue protein with its 900-nucleotide coding RNA, using 110 Da per residue and 330 Da per nucleotide. Explain why this does not fix their hydrodynamic sizes.
4. A 2.5 MDa ribosome has a diameter of order 20 nm. Apply the compact-protein formula as a rough comparison and identify its extra assumption for this RNA–protein assembly.

Worked answers are in Solutions. Keep the calculation separate from any biological inference that needs additional information.

A correct concentration estimate specifies the cell volume; an occupancy estimate additionally specifies free ligand and the binding model.

9Derive the displacement statistic

Independent random steps produce a mean-square displacement proportional to time.

Consider a walker taking a step of length $\ell$ every $\Delta t$, with independent signs $s_i=\pm1$ of equal probability. After $n$ steps, $x_n=\ell\sum_i s_i$. Angle brackets denote an average over possible walks. The mean displacement is zero, while

The first sum is $n$, because each squared sign is one. The cross terms vanish because independent signs have zero mean. With $t=n\Delta t$, define the diffusion coefficient $D=\ell^2/(2\Delta t)$:

The square root is an ensemble displacement scale. Individual paths can move farther or less far, and can return. The time $L^2/(6D)$ at which the root-mean-square radial displacement equals $L$ is not generally a first-arrival time or a confined compartment's mixing time. Those questions need boundary conditions.

At long times in this ideal random-walk model, the one-dimensional displacement density is Gaussian:

The integral follows by substituting $u=x^2/(4Dt)$. It relates a measured mean absolute displacement to $D$, provided the displacement statistic and dimensionality match the formula.

Choose the displacement statistic first; use $L^2/D$ as an exploration scale rather than a universal exact arrival time.

10Connect diffusion to drag

Thermal equilibrium supplies the Einstein relation, and viscous flow supplies the sphere's mobility.

Mobility relates velocity to an applied force. Write $u=bF$, where $b$ is mobility. To find its relation to diffusion, place a dilute suspension in a potential $U(x)$. The force is $F=-U'(x)$. At thermal equilibrium the particle density follows $c(x)\propto e^{-U(x)/(k_BT)}$, where $k_B$ is Boltzmann's constant and $T$ is absolute temperature.

The current combines drift with diffusion: $j=bFc-Dc'$. Differentiating the equilibrium density gives $c'=Fc/(k_BT)$. Requiring zero net current yields

This is a relation between thermal fluctuations and the response to a weak force. Its use assumes an equilibrium thermal bath and the same linear friction law for both descriptions. Einstein's 1905 argument connected diffusion and osmotic equilibrium in this way.⁶

Viscosity determines the drag on a slowly moving sphere. Dynamic viscosity $\eta$ measures shear stress per velocity gradient. Around a sphere of radius $a$ moving at speed $u$, the gradient scales as $u/a$. Stress times area therefore scales as $(\eta u/a)a^2=\eta au$. The exact solution for a no-slip sphere in an unbounded incompressible Newtonian fluid at negligible inertia gives $F=6\pi\eta au$.⁷

Here $Da$ means the product of diffusion coefficient and radius, not the dalton unit. At fixed temperature and viscosity, doubling radius halves diffusion. For compact objects of the same density, radius scales as the cube root of mass, so doubling mass multiplies $D$ by $2^{-1/3}\approx0.79$.

Using $k_B\approx1.38\times10^{-23}\,\mathrm{J/K}$, $T\approx298\,\mathrm K$ and water viscosity $\eta\approx0.89\times10^{-3}\,\mathrm{Pa\,s}$ gives $Da\approx0.245\,\mu\mathrm m^3/\mathrm s$. A radius of 2.5 nm then gives $D\approx100\,\mu\mathrm m^2/\mathrm s$. Check the units: $\mathrm J/(\mathrm{Pa\,s})=\mathrm m^3/\mathrm s$.

Use measurements in cells when the model concerns cells. Elowitz and colleagues measured GFP diffusion at $7.7\pm2.5\,\mu\mathrm m^2/\mathrm s$ in E. coli, versus about 87 in buffer. The difference depends on the molecule and condition. For the examples below, 7.7 is a GFP-like protein input and 200 µm²/s is an assumed small-metabolite input.⁹

Predict diffusion from mobility under specified fluid assumptions, and replace water inputs with relevant cellular measurements before judging a cellular clock.

11Test mixing against pool change

A compartment can be well mixed for one process and spatially structured for another.

The exploration estimate is $\tau_D=L^2/D$. At $D=200\,\mu\mathrm m^2/\mathrm s$, a metabolite explores 1 µm in about 5 ms, 20 µm in 2 s and 1 mm in 5000 s. At a protein coefficient of 7.7 µm²/s, the corresponding times are 0.13 s, 52 s and about 36 h. These are calculations with fixed illustrative coefficients.

Mixing must be fast compared with the change the model predicts. If a pool concentration is $S$ and its consumption flux density is $J$, then $\tau_{mathrm{pool}}=S/J$. The ratio

compares spatial redistribution with consumption. A value much smaller than one supports a well-mixed approximation when sources, boundaries and binding do not create additional local structure. A single enzyme's turnover interval is not the pool-consumption time: one must also know how many enzymes are consuming how large a pool.

A simple spatial model shows the alternative. In a half-space $x\ge0$, maintain $S(0)=S_b$, let substrate diffuse with coefficient $D$, and consume it with a specified effective first-order rate $kS$. At steady state, $DS''-kS=0$. Trying $S\propto e^{rx}$ gives $r=\pm\sqrt{k/D}$. Boundedness far from the source selects

Across distance $L$, the governing ratio is $kL^2/D=(L/\ell_D)^2$. At $D=200\,\mu\mathrm m^2/\mathrm s$, $k=0.02\,\mathrm s^{-1}$ gives $\ell_D=100\,\mu\mathrm m$, whereas $k=2000\,\mathrm s^{-1}$ gives $\ell_D\approx0.32\,\mu\mathrm m$. Even a bacterial length can support a substantial gradient in the second model.

At a metre, the same protein coefficient gives an exploration time of order four thousand years. Diffusion alone cannot explain rapid end-to-end delivery over that distance. This comparison motivates additional transport or local production; identifying motors requires biological evidence beyond the scaling calculation.

Compare $L^2/D$ with the relevant pool-change time, and retain spatial structure when the model cannot justify removing it.

12Checkpoint: choose the transport clock

Practice deciding what the estimate does and does not establish.

5. A 1 µm compartment has substrate at 0.1 mM, consumption at 1 mM/s and $D=200\,\mu\mathrm m^2/\mathrm s$. Compare exploration and pool replacement. What additional information would test local gradients?
6. Estimate diffusion over 10 µm at $D=7.7\,\mu\mathrm m^2/\mathrm s$. Compare with a specified directed speed of 1 µm/s, and explain why similar arrival times do not identify the mechanism.
7. For the half-space example, find the distance at which concentration is half the boundary value.
8. Under the compact-sphere assumptions, what happens to $D$ when molecular mass doubles? Why might a measured fusion protein behave differently?

Transport calculations support a model decision only after distance, observable and competing timescale have been specified.

13Choose a biochemical description

Binding and catalysis organize many cellular mechanisms, with transport and physical variables included when the question needs them.

A biochemical model tracks molecular constituents and their transformations. Binding changes which molecules are associated. Catalytic mechanisms convert substrates into products while regenerating the enzyme. These operations provide a useful vocabulary for metabolism, regulation and gene expression.

The description may also need voltage, force, shape, position or fluid flow. A transporter responds to an electrochemical difference, and a motor couples a chemical cycle to displacement. Specifying a reaction network does not prove that concentrations alone determine those observables.

Catalysts affect kinetic accessibility. A favorable free-energy change establishes a thermodynamic direction under specified conditions, but does not set the waiting time. Measurements and extrapolations of uncatalyzed biological reactions span seconds to extremely long chemical times. Enzymes make many such transformations occur on physiological clocks.¹⁰¹¹

A catalyst does not change the equilibrium constant of the same overall reaction. Cells can additionally change substrate supply, product removal, enzyme state and energy coupling. The next sections make these choices explicit enough to assign rate laws.

List the chemical species and the additional physical variables required by the observable before choosing the model's resolution.

14Derive mass action from molecule counts

The volume and Avogadro factors explain both the concentration dependence and the units.

Let distinct species A and B have counts $n_A,n_B$ in a well-mixed volume $V$ measured in litres. Suppose one A has an effective encounter volume per time $\sigma$, measured in L/s. For a short interval $dt$, the expected number of B molecules in that encounter volume is $(n_B/V)\sigma dt$.

Multiplying by the A count gives the expected event rate $r=\sigma n_An_B/V$, in events/s. This argument assumes dilute, effectively independent encounters and a fixed probability of reaction per encounter, incorporated in $\sigma$. Write free molar concentrations $A=n_A/(N_AV)$ and $B=n_B/(N_AV)$. Dividing the event rate by $N_AV$ gives

Thus $k$ has units $\mathrm{L\,mol^{-1}s^{-1}}=\mathrm{M^{-1}s^{-1}}$, and $v$ has units M/s. For a unimolecular event, a per-molecule event rate $k$ directly has units s⁻¹, giving $v=kA$.

Repeated reactants require distinct molecular combinations. For two A molecules, the number of unordered pairs is $n_A(n_A-1)/2$, rather than $n_A^2$. In a large-count deterministic convention the pair factor is absorbed into the rate constant and the concentration dependence becomes quadratic. At low counts the exact combinatorial expression matters.

A mass-action concentration flux of total reactant order $m$ has coefficient units $\mathrm{M}^{1-m}\mathrm s^{-1}$. Reaction order describes concentration dependence. Molecularity describes the participating molecules in one elementary event. Their relation depends on the mechanism and on the ideal-mixing assumptions just used.

Convert event rates to concentration flux with $N_AV$, and check that every rate law ends in concentration per time.

15Keep reaction resolution visible

An effective mass-action form can describe a composite process without making it elementary.

An elementary chemical step uses a straight arrow with a microscopic rate constant. For example, $A+B\xrightarrow{k}C_{AB}$ represents the modeled single association event. A reversible elementary pair uses $A+B\xrightleftharpoons[k_-]{k_+}C_{AB}$, with separate forward and reverse fluxes.

A composite conversion uses a bare squiggly arrow. For $S\rightsquigarrow P$, state its rate law separately. It may be measured, phenomenological or derived from a resolved mechanism. Even if that law reduces to $v=k_{\mathrm{eff}}S$, the process remains composite.

Count transitions are a separate notation. One S-to-P event maps counts as $(n_S,n_P)\mapsto(n_S-1,n_P+1)$. This mathematical map reports a state change rather than a chemical mechanism. Similarly, a causal connection or a transport vector is not a chemical-reaction arrow.

A reversible arrow asserts that both directions are modeled. It does not establish equality of the two fluxes. Conversely, mathematical mass-action theorems depend on the specified equations and their structural hypotheses. A composite process that has a valid mass-action law can meet those hypotheses. The course's arrow convention records physical resolution; it is not itself a theorem about dynamics.

Classify each arrow separately, then justify its rate law; a first-order fitted or reduced law does not change a composite arrow into an elementary one.

16Write supply, expression and resource allocation

Three examples show what remains to be specified after the reaction accounting is correct.

An enzyme with a model boundary

For a substrate supply and product removal, write $\varnothing\rightsquigarrow S$ and $P\rightsquigarrow\varnothing$. The symbol $\varnothing$ means outside the retained system. Choose a supply flux $v_{\mathrm{in}}$ and a removal law, such as $v_{\mathrm{out}}=\delta P$, separately. These terms do not claim creation or destruction of matter in the complete physical system.

The internal minimal enzyme mechanism is

In this minimal model, each displayed straight arrow is treated as an elementary event. More detailed enzymes can have extra intermediates, reversible chemistry and distinct release steps. For those mechanisms the measured saturated turnover need not be the microscopic constant of any one step.

Gene expression and pseudo-first-order association

Write $G\rightsquigarrow G+M$, $M\rightsquigarrow M+P$, and removal of M and P with bare squiggles. These are composite transcription, translation and removal processes. A passive channel crossing is also composite when the pore mechanism is not resolved.

One can explicitly resolve a polymerase association $G+R\xrightleftharpoons[k_-]{k_+}C_{GR}$. If the free polymerase concentration $R$ is effectively fixed, association has flux $k_+RG=k_{\mathrm{obs}}G$, where $k_{\mathrm{obs}}=k_+R$ has units s⁻¹. This is pseudo-first-order association. Completed transcription still includes initiation, elongation and termination, so its flux is a different quantity.

The protein cost of a ribosome

A resource calculation can bound a composite production process. Take the core's protein budget of about 0.84 MDa per ribosome. At 110 Da per residue, this is approximately 7600 residues. At a translation rate $u=16.5\,\mathrm{residues/s}$, one active ribosome needs about $n/u\approx460\,\mathrm s$ of serial translation time to make that much protein. This excludes rRNA production, assembly and every other protein the cell needs.¹²

If each ribosome devotes a fraction $\phi$ of its total translation capacity to ribosomal proteins, the idealized production coefficient is $\phi u/n$. Let $n_R$ denote the ribosome count. With immediate assembly and no other limitation,

The ideal doubling time at $\phi=1$ is about 320 s, or five minutes. This is shorter than the serial translation time because newly made ribosomes also translate. At a 20 min doubling time, the same model requires $\phi\approx320/1200\approx0.27$. This is an allocation estimate under the specified inputs, rather than a universal bound on the active fraction. Ribosome allocation and the fraction actively translating depend on growth conditions.¹³¹²

Use mechanism to derive an internal rate, a boundary model to specify supply or removal, and resource accounting to bound composite production.

17Checkpoint: build a model with honest arrows

State what is retained, what is hidden and which law governs each flux.

9. Classify: a net combustion equation, a channel crossing with the pore omitted, a resolved one-step DNA–protein association, and completed transcription.
10. A study fits $v=k_{\mathrm{eff}}S$ to an enzyme-mediated conversion. Give a mechanistic regime where this is reasonable and a concentration change that could invalidate it.
11. Write two sequential elementary binding events for a promoter binding two repressors. Explain why writing two repressors on one arrow does not by itself establish a simultaneous elementary event or cooperative affinity.
12. Use the ribosome budget to calculate the idealized allocation fraction at a 40 min doubling time, and list the omitted costs.

A network model is reviewable when every retained species, hidden process and flux assumption can be identified explicitly.

18Solve the absorbing-sphere problem

Compute the volume per time associated with ideal diffusive capture.

Place a perfectly absorbing sphere of contact radius $a$ in an unbounded dilute solution. Let $c(r)$ be substrate number density, in molecules/m³, and let its value far away be $c_\infty$. Substrate diffuses with constant coefficient $D$. Assume spherical symmetry, a stationary profile and no force field.

Conservation makes the inward arrival rate independent of shell radius. The outward radial current is $j_r=-D\,dc/dr$. Because concentration increases outward, this current is negative. Define the positive inward arrival rate $J=-4\pi r^2j_r$. Thus

The boundaries determine the constants. The perfect sink imposes $c(a)=0$, and the reservoir imposes $c(\infty)=c_\infty$. Hence $B=c_\infty$ and $A=ac_\infty$. Differentiating gives the surface gradient:

Multiply the gradient by diffusion coefficient and area:

The superscript V marks a volume-rate coefficient, $k_{\mathrm{diff}}^{(V)}=4\pi Da$, in m³/s. Its linear radius dependence results from area growing as $a^2$ while the surface gradient falls as $1/a$. The profile has a long $1/r$ tail, rather than a depletion region with a sharp outer boundary.

When both partners diffuse, use relative motion. Independent displacements subtract to give a relative-coordinate variance proportional to $D_A+D_B$. Contact occurs when centre separation equals $a_A+a_B$. Therefore

For identical spheres this is $16\pi Da$, since both factors double. A population of identical particles also needs a pair-counting convention when translating encounters into particle loss. Smoluchowski's coagulation analysis treated the encounter problem and the evolving particle population as connected but distinct steps.¹⁴

Calculate an ideal encounter coefficient only after specifying relative diffusion, contact geometry and the absorbing boundary.

19Convert volume per time to molar units

The same physical capture process has different numerical coefficients when concentration units change.

A molar concentration $S$ contains $1000N_AS$ molecules per cubic metre. Substitute that number density into the single-target arrival rate from section 18:

Now $k_{\mathrm{diff}}$ has units M⁻¹s⁻¹ and $J$ is the per-target arrival frequency in s⁻¹. Multiplying by a molar target concentration would give a flux density in M/s.

For an illustrative pair with $D_A+D_B=770\,\mu\mathrm m^2/\mathrm s=7.7\times10^{-10}\,\mathrm m^2/\mathrm s$ and $a_A+a_B=2.4\,\mathrm{nm}=2.4\times10^{-9}\,\mathrm m$,

The swept-volume interpretation is statistical. About 23 fL/s does not mean that one target uniformly empties twenty-three physical cells each second. The derivation assumes a maintained dilute reservoir and independent capture.

Equal spheres provide a size-cancellation check. Under the same Stokes–Einstein assumptions for both partners,

The right side is still a volume per time. Applying the molar conversion gives approximately $7.4\times10^9\,\mathrm{M^{-1}s^{-1}}$ in water at 25 °C. Unequal radii leave a factor $(a_A+a_B)(a_A^{-1}+a_B^{-1})$; the cancellation does not produce a universal bound for every size ratio.

Always display the litre-per-cubic-metre and Avogadro factors when converting $4\pi Da$ to M⁻¹s⁻¹.

20Change the encounter assumptions

A useful benchmark identifies which physical assumptions would change its value.

The calculator below takes radii and diffusion coefficients directly. Radius inferred from molecular mass is only an equivalent-sphere approximation, and diffusion inside a cell should be supplied for the relevant molecule and condition. The tool's numerical output is an ideal encounter benchmark, not a prediction that every contact makes a bound complex.

Finite surface reactivity changes the absorbing boundary. Let $\kappa$, in m/s, relate the inward surface current to the concentration at contact: $D c'(a)=\kappa c(a)$. Keep $c(r)=c_\infty-A/r$. Substitution gives $DA/a^2=\kappa(c_\infty-A/a)$, and hence

The coefficient approaches the perfect sink when $\kappa a\gg D$, and the surface-reaction value $4\pi a^2\kappa$ when $\kappa a\ll D$. Collins and Kimball developed the finite-reactivity boundary description. It accounts for diffusive revisits rather than treating each macroscopic encounter as one independent orientation lottery.¹⁵

Orientation, conformational changes and electrostatic interactions can affect productive association. DNA binding may add nonspecific capture and movement along DNA. Such mechanisms change the target or the transport problem. A measured association rate above a force-free small-target estimate therefore calls for checking that model's assumptions.¹⁶¹⁷

Use an encounter benchmark to identify a missing physical assumption, and keep it distinct from the measured association coefficient.

21Derive the specificity constant as the non-saturating slope

An enzyme-mediated conversion becomes effectively mass-action-like at small free substrate relative to its Michaelis constant.

From the elementary mechanism to the rate curve

Return to the minimal mechanism of section 16. Let $q_E=E+C_{ES}$ be the conserved total enzyme concentration and let $v=k_{\mathrm{cat}}C_{ES}$ be product flux. Substrate association fills the complex, and dissociation plus catalysis empty it:

If free substrate is maintained at a fixed value, the complex approaches a stationary concentration. Setting the right side to zero and collecting the terms in $C_{ES}$ gives

For a closed reaction whose substrate declines, this expression is an approximation after a fast transient, with conditions developed in section 23. It describes an initial-rate assay when product effects and substrate depletion are negligible during the measurement. Briggs and Haldane's complex balance gives this interpretation without requiring binding equilibrium.¹⁸

Why the ratio matters

The low-substrate limit means non-saturating relative to this enzyme's own $K_M$. Rewrite the rate per enzyme as

The initial slope of this curve is the specificity constant:

It predicts how much additional flux each enzyme supplies per additional free substrate concentration. The composite conversion remains $S\rightsquigarrow P$, with the separate effective law $v\approx(k_{\mathrm{cat}}/K_M)q_ES$. If $q_E$ is fixed, this is first order in S with coefficient $(k_{\mathrm{cat}}/K_M)q_E$ in s⁻¹. If enzyme amount also varies, the effective law is bilinear.

Take $k_{\mathrm{cat}}=10\,\mathrm s^{-1}$ and $K_M=100\,\mu\mathrm M=10^{-4}\,\mathrm M$. Their ratio is $10^5\,\mathrm{M^{-1}s^{-1}}$. At $S=1\,\mu\mathrm M$, the linear approximation gives 0.1 product per enzyme per second. The exact curve gives $10/101\approx0.0990\,\mathrm s^{-1}$. At $q_E=1\,\mu\mathrm M$, those correspond to fluxes of 0.100 and 0.0990 µM/s. The relative overestimate of the linear law is exactly $S/K_M=1\%$.

The useful comparison changes at saturation. When $S\gg K_M$, the per-enzyme flux approaches $k_{\mathrm{cat}}$. Two enzymes with equal specificity constants can have different saturated capacities. For example, $(k_{\mathrm{cat}},K_M)=(10\,\mathrm s^{-1},100\,\mu\mathrm M)$ and $(100\,\mathrm s^{-1},1000\,\mu\mathrm M)$ both have specificity $10^5\,\mathrm{M^{-1}s^{-1}}$. They agree to leading order at low S and differ strongly as S increases.

From association to successful conversion

A formed complex has two competing exits with constant rates. Its survival probability until time t is $e^{-(k_{\mathrm{off}}+k_{\mathrm{cat}})t}$. Integrating the catalytic exit probability density yields

This identity refers to the actual association coefficient of the minimal mechanism. It is not obtained by replacing that coefficient with an ideal diffusion value from another pair. A ratio to an appropriate ideal encounter benchmark measures distance from that benchmark; a ratio to the measured $k_{\mathrm{on}}$ gives the conditional product probability in this mechanism.

Use $k_{\mathrm{cat}}/K_M$ to compare non-saturating product flux per enzyme, $k_{\mathrm{cat}}$ to compare saturated capacity, and measured $k_{\mathrm{on}}$ to interpret individual binding attempts.

22Checkpoint: compare the right coefficients

Units establish comparability, while the measured event determines the interpretation.

13. Evaluate the ideal molar encounter coefficient for two equal spheres in water at 25 °C. Identify both factors of two in the relative-coordinate calculation.
14. For the two enzymes in section 21, compute flux per enzyme at 1 µM and at 10 mM substrate. Which parameter predicts the ranking in each regime?
15. A minimal mechanism has $k_{\mathrm{on}}=10^6\,\mathrm{M^{-1}s^{-1}}$, $k_{\mathrm{off}}=90\,\mathrm s^{-1}$ and $k_{\mathrm{cat}}=10\,\mathrm s^{-1}$. Find $K_M$, specificity and product probability.
16. An efficiency is one hundredth of a computed ideal encounter coefficient. What additional measurement is needed before calling one in a hundred formed complexes productive?

A dimensionally valid comparison still needs matching molecular partners, assay conditions and event definitions.

23Derive waiting, relaxation and quasi-steady state

The same stationary rate curve can hide very different microscopic binding kinetics.

One complex and one complete catalytic cycle

A complex lifetime ends at either exit. Integrating its exponential survival gives $\tau_{\mathrm{dwell}}=1/(k_{\mathrm{off}}+k_{\mathrm{cat}})$. For binding without catalysis, residence is instead $1/k_{\mathrm{off}}$. Neither is the mean time from free enzyme to the next product.

Let $T_F$ be that product waiting time at fixed free S, starting with free enzyme. The first association takes mean time $1/(k_{\mathrm{on}}S)$. The ensuing bound episode lasts $1/(k_{\mathrm{off}}+k_{\mathrm{cat}})$. If dissociation occurs, the enzyme starts over, with probability $1-p_{\mathrm{product}}$. Therefore

This renewal calculation reproduces the stationary Michaelis–Menten rate. The mean number of failed bound episodes before product is $(1-p_{\mathrm{product}})/p_{\mathrm{product}}=k_{\mathrm{off}}/k_{\mathrm{cat}}$. Failed episodes are individual events; they are not repetitions of macroscopic equilibration.

Bar-Even and colleagues used the corresponding specificity-to-association comparison in their analysis of futile encounters.²⁰ Their broad initial-encounter model and our resolved bound-state model define the encounter differently. A success fraction must retain the event definition used in its denominator.

Population relaxation at fixed free substrate

At clamped S, the complex equation is linear. Define $\lambda=k_{\mathrm{on}}S+k_{\mathrm{off}}+k_{\mathrm{cat}}$ and its stationary value $C_*=q_ES/(K_M+S)$. Then $\dot C_{ES}=-\lambda(C_{ES}-C_*)$, so

For binding alone at fixed free ligand, omit $k_{\mathrm{cat}}$. For a finite binding-only pool, both free partners change as complex forms. With $f(C)=k_{\mathrm{on}}(q_E-C)(q_S-C)-k_{\mathrm{off}}C$, linearizing at equilibrium gives

The extra free-enzyme term is the response of the second depleted pool. The star denotes the equilibrium free concentrations, which must first be found from mass conservation.

Two mechanisms with the same rate curve

Both have $K_M=100\,\mu\mathrm M$ and specificity $10^5\,\mathrm{M^{-1}s^{-1}}$. In mechanism 1, dissociation dominates the exits, so $K_d=99.9\,\mu\mathrm M\approx K_M$. In mechanism 2, product formation dominates, and $K_d\approx0.99\,\mu\mathrm M$. Thus the same full stationary rate curve does not identify binding affinity or establish rapid binding equilibrium.

A slowly changing substrate pool

For a closed initial reaction, let $C_{ES}(0)=P(0)=0$, $S(0)=S_0$, and conserve $E+C_{ES}=q_E$ and $S+C_{ES}+P=S_0$. The exact balances include

During the initial complex transient, temporarily freeze S at $S_0$. Its characteristic duration is $t_c=[k_{\mathrm{on}}(K_M+S_0)]^{-1}$. The initial substrate-removal slope has magnitude $k_{\mathrm{on}}q_ES_0$, so the estimated fractional change during that interval is

When $\varepsilon\ll1$, freezing substrate for this fast phase is self-consistent. After the transient, the standard quasi-steady-state approximation follows the moving value $C_{ES}\approx q_ES/(K_M+S)$. The corresponding substrate-depletion scale is $t_s=(K_M+S_0)/(k_{\mathrm{cat}}q_E)$. This is the classical closed-reaction setting for the sufficient small-parameter criterion developed by Segel and Slemrod.²¹

Quasi-steady does not mean literally unchanging. As substrate decreases, the approximate complex concentration decreases too. Its small derivative is omitted in the leading balance between larger formation and removal fluxes. An approximation intended for product accumulation after the transient need not reproduce the initial complex rise. Other initial conditions, comparable binding totals or external driving can require another reduction or the full equations.²²

For either matched mechanism at $q_E=1\,\mu\mathrm M$ and $S_0=1\,\mathrm{mM}$, $\varepsilon\approx9.1\times10^{-4}$. Standard QSSA can be accurate for product even though mechanism 2 fails the condition $k_{\mathrm{cat}}\ll k_{\mathrm{off}}$ needed for the rapid-equilibrium interpretation. Being near thermodynamic equilibrium of the overall S-to-P reaction is a third question, involving reverse flux and chemical driving.

Derive the clock for the requested observable, and check rapid equilibrium and quasi-steady state as separate hypotheses.

24Compare pool replacement with enzyme adjustment

Rapid flux changes can use substrate and enzyme state before enzyme abundance has substantially adjusted.

A pool-to-flux calculation

Take glucose uptake of 8 mmol per gram dry weight per hour as a representative assumed flux. Our density and dry fraction imply that one gram dry mass occupies $1/(1.1\times0.30)=3.03\,\mathrm{mL}$ of cell volume. Therefore the volumetric glucose flux is

For a measured PEP pool of 0.18 mM and an assumed lower-glycolytic throughput of twice the glucose flux, $\tau_{\mathrm{pool}}=0.18/(2\times0.733)\approx0.12\,\mathrm s$. A 9.6 mM ATP pool with an illustrative turnover flux of thirty times glucose flux gives about 0.44 s. The stoichiometric or energetic flux multipliers are inputs to these estimates, rather than measurements supplied by the concentration survey.³