From order of magnitude in biology to binding and catalysis

Molecule counts give concentrations; diffusion gives encounter times. To turn those into biochemical fluxes, we need to know what happens after molecules meet. This lecture connects those estimates through binding and catalysis, then asks how their different clocks shape cellular regulation. CCBS · EST 5053 · Fall 2026 · lecture 2 · core page · 2026-09-03 (Thu) 10:40–12:15 · lecturer-taught Take a cell one micron across. How many molecules fit, how long do they take to meet, and how many meetings make product? We will carry the same reference cell through all three questions. Each answer supplies an input to the next, until we can compare the time available for metabolic regulation with the time needed to make more enzyme.

This page is the lecture note The twelve main sections follow the classroom argument. A1–A5 provide optional depth on affinity, residence time and the assumptions behind kinetic reductions. The exposition is the companion for extended derivations and practice; the lineage supplies historical context and sources. Measured inputs are cited where used, and illustrative assumptions are identified separately.

The 95 minutes, and what comes off

The spine, as timed blocks. Section 7 is the exercise that transfers, and section 8 is the anchor number of the course. The optional part is not in the clock.
Block	Section	min
Admin	what is due, what is coming	4
1	§1 a cubic micron is a femtolitre	5
2	§2 what is in it	9
3	§3 Brownian motion and D	9
4	§4 mixing times	5
5	§5 the core is chemistry	5
6	§6 the two arrows	9
7	§7 three systems, written both ways	11
8	§8 the diffusion limit	8
9	§9 what enzymes actually do	10
10	§10 the timescale ledger	6
11	§11 metabolism cannot wait	9
Addendum	§12 where this goes	3
Optional	A1 to A5, not lectured	0

The main blocks total 93 minutes, including four minutes of administration, leaving two minutes for questions and transitions. The measured fast-enzyme examples in §9 and the historical box in §11 are reading branches; their point is carried by the figures and main text. If the board work runs long, drop the gene-expression example in §7 first (3 min). Next, shorten §10 to a thirty-second reading of Figure 10 (saving 5.5 min). Keep the ribosome estimate, the effective rate law in §9 and the regulation argument in §11. These cuts leave 90 and 84.5 minutes respectively. A1–A5 are outside the classroom clock. They open the mechanism beneath the measured parameters and explain why rapid equilibrium and quasi-steady state require different checks. Lecture 6 develops those reductions in full. The main argument defines every quantity it needs before using it.

Part 1

What is in a cell

How much stuff a cell has, and what follows from there being so little of it.

1A cubic micron is a femtolitre

One identity, and the conversion that falls out of it. Memorise both. Start with the length ladder, which fixes the scale of everything after it.¹² Atom 1 Å, bacterium 1 µm, human 1 m.

Figure 1. The length ladder, with the objects of this course marked, and four estimates for the one-femtolitre reference cell. The third box converts one copy into a concentration at this volume.

\begin{aligned}1\,\mu\mathrm{m}^3&=10^{-18}\,\mathrm{m}^3=10^{-15}\,\mathrm{L}=\pf{1\,\mathrm{fL}},\\1\,\mathrm{fL}\text{ of water}&\text{ has mass }\pf{1\,\mathrm{pg}}.\end{aligned}

(1)

Use molar concentration: $1\,\mathrm{M}=1\,\mathrm{mol/L}$ , and Avogadro's constant $N_A\approx6.0\times10^{23}\,\mathrm{mol}^{-1}$ converts molecules to moles. One molecule in one femtolitre is

c = 1 / (N_{A} \cdot 10^{-15}\,\mathrm{L}) = 1.66\times 10^{-9}\,\mathrm{M} \approx \pf{1\,\mathrm{nM}}

(2)

The rule to carry is one molecule in a one-femtolitre cell is about one nanomolar. The more precise conversion is 1.66 nM per copy. Ten copies therefore give about 17 nM. A count alone does not tell us whether a binding site is occupied: we will need to compare that concentration with an affinity.

The reference volume is a convenient round number, not a fixed size for every bacterium. Volkmer and Heinemann measured E. coli volumes from about 1.5 to 4.4 fL across growth conditions.¹ At fixed copy number, a larger cell means a lower concentration; at fixed concentration, it means more copies.

Convert a molecule count to molar concentration using $c=n/(N_A V)$ , and state the cell volume before applying the one-copy-to-nanomolar rule.

2What is in it

A representative protein weighs about 33 kDa and a femtolitre holds a few million proteins. We can build both estimates from molecular-size brackets and cell composition, then check them against measurements.

To count the proteins in a cell you need two quantities: the mass of one protein, and the total mass of protein. Take them in that order. Neither is a number to memorise, and getting both from brackets is the point of the section.

The move: bracket it, then take the geometric mean

When you know a quantity's two extremes but not its middle, estimate it as $\sqrt{x_{\min}x_{\max}}$ . That is the arithmetic mean of the logarithms, so it lands in the middle of the range of magnitudes rather than being dragged towards the top by the larger endpoint. One rule for using it: bracket what is common, not what is possible. A rare extreme is still an extreme, and the mean cannot know it is rare. The exposition proves the rule: the estimate's error is exactly the bracket's log-asymmetry about the truth, and nothing else.

The mass of one protein. Two brackets, multiplied. The dalton measures molecular mass: $1\,\mathrm{Da}\approx1.66\times10^{-24}\,\mathrm{g}$ . Amino acids are small organic molecules; glycine is the bare backbone at 75 Da, tryptophan carries a double ring at 204 Da. A peptide bond releases one water, so a residue inside a chain weighs 18 Da less than the free acid.

m_{\mathrm{residue}} \approx \sqrt{75\times 204}\,\mathrm{Da} - 18\,\mathrm{Da} = \pf{106\,\mathrm{Da}}

(3)

Carry 110 Da per residue, which is the round number the field quotes. The bracket's job here was to tell us we can trust it, and checking a quoted constant against a bracket is the cheapest use this method has. Now the length. One folded domain is about 100 residues and a large multi-domain enzyme about 1000, so $\sqrt{100\times 1000} = 316$ , call it 300 residues. Bracket what is possible instead, 50 residues to 5000, and the same move returns 500, which is 1.7 times the 300-residue reference. Nothing is wrong with the arithmetic. The upper endpoint is simply rare, and that is the rule in the box earning its keep. Multiply the two brackets:

m_{\mathrm{protein}} \approx 300 \times 110\,\mathrm{Da} = \pf{33\,\mathrm{kDa}}

(4)

The total mass of protein. Equation (1) says one femtolitre of water weighs one picogram, and a cell is a little denser than water at 1.1 g/mL, so a 1 µm³ cell weighs about 1.1 pg wet. Two subtractions turn that into protein, and they are where this estimate is usually lost.

First the water. Biomass is about 70% water, as a rough composition estimate, and it leaves 0.33 pg of dry mass. Then the rest of the dry mass: take protein as about 55% of it, with RNA, DNA, lipid and the wall accounting for the rest. So a cell carries about 0.18 pg of protein, a protein mass concentration of $c_{p} \approx 0.2\,\mathrm{g/mL}$ . Divide by (4):

N = \frac{c_{p}V}{m_{\mathrm{protein}}} = \frac{0.18\times10^{-12}\,\mathrm{g}}{33000\times1.66\times10^{-24}\,\mathrm{g}} = \pf{3.3\times 10^{6}} \text{ per fL}

(5)

Both subtractions are worth doing out loud, because each one is a factor. Put the whole wet mass on top and you get $2\times 10^{7}$ , six times too many. Stop after the water and you get $6\times 10^{6}$ , which is wrong as an answer and useful as a bound: a cell cannot hold more protein than that.

Against the measurements

The measured protein mass concentration in E. coli is 0.24 g/mL, which gives 4.4 million using our assumed mean protein mass, about a third above (5). Milo's systematic pass over the census data puts it at 2 to 4 million per cubic micron, and finds comparable densities in the bacterial, yeast and mammalian examples considered.² That near-universality is why the number is worth carrying at all. The same pass finds several proteome-wide censuses reporting values 3 to 10 times lower, and argues they are mis-calibrated: order of magnitude used as lecture 1 advertised, as a benchmark that catches an experiment.

The comparisons support the scale of both brackets. The abundance-weighted mean residue mass is 110 Da, against 106 from (3). The abundance-weighted mean protein length is 300 residues in E. coli and 400 in yeast and human, against 316 from the common bracket and 500 from the wide one.

The concentration ladder

Three million proteins, but of how many kinds? Use the protein-length estimate again, now to estimate coding capacity. A 300-residue protein needs $3\times 300 = 900$ nucleotides of coding sequence, and a bacterial genome is dense, about 90% coding. E. coli's is 4.6 Mbp, so

n_{\mathrm{genes}} \approx \frac{0.9 \times 4.6\times 10^{6}\,\mathrm{nt}}{900\,\mathrm{nt}} = \pf{4600}

(6)

This is close to roughly 4,400 annotated protein-coding genes. The agreement checks the scale; the mean length across genes need not equal the abundance-weighted mean across protein molecules. If all 4,400 species shared our protein count, their arithmetic mean would be $3.3\times10^6/4400\approx750$ copies, or about 1 µM. Actual expression is highly uneven and some genes are silent. Abundant individual metabolites instead reach millimolar concentrations: Bennett et al. measured a total pool near 300 mM, including glutamate at 96 mM.³ The total pool and the concentration of any one species are different quantities.

Table 1. Representative concentration scales, with rounded copy numbers for a 1 fL reference cell. Individual proteins and growth conditions span wider ranges.
Species	copies per cell	concentration	regime
abundant metabolite (glutamate, ATP, FBP)	10⁶–10⁸	1–100 mM	many copies; mixing checked in §4
many enzymes	10²–10⁴	0.1–10 µM	the regulated layer (lectures 7, 8)
low-abundance transcription factor	10–10²	10–100 nM	counting starts to matter (lecture 5)
one molecule	1	1 nM	discreteness (lecture 5)

Concentrations become informative when compared with a binding constant. For one independent site at equilibrium, define $K_d$ as the free ligand concentration at half occupancy. The occupied fraction is $L/(K_d+L)$ , where $L$ is free ligand concentration. Below $K_d$ the site is mostly empty; above it, mostly occupied. Compare the ligand concentration with the site's affinity, not the receptor's abundance. Enzyme saturation uses a related but generally different concentration, $K_M$ , introduced in §9.

Two geometric estimates close the ledger. Treating the 4.6 Mbp genome as a 2 nm diameter DNA cylinder gives 1.6 mm of contour length but only 0.5% of a femtolitre in material volume. Our three million proteins occupy about 13%, using the specific volume of compact protein.⁴ These are material volumes, not the space occupied by a folded chromosome or hydrated, mutually excluding molecules. They motivate the next question: how freely do molecules move through this crowded interior?

Build a protein count from molecular mass and cell composition, then compare the concentration of a particular ligand with the binding constant relevant to its target.

Part 2

How fast things move

One formula turns a size into a diffusion coefficient. One scaling turns that into a time.

3Brownian motion, and the formula for D

Watch something jiggle, estimate how far it goes in a second, and lecture 1's scaling gives a diffusion coefficient.

In 1827 Brown watched granules released by pollen suspended in water. Think of a granule about 1 µm across, with radius $a\approx0.5\,\mu\mathrm{m}$ , wandering about $x\approx1\,\mu\mathrm{m}$ in one second. These are order-of-magnitude inputs for the example, not Brown's own quantitative measurements. The much larger pollen grain diffuses more slowly; its greater viscous drag, rather than weight alone, is the relevant distinction.

That is enough to estimate the diffusion coefficient $D$ , which measures squared displacement per unit time. Lecture 1 got $t \approx x^{2}/\kappa$ out of the heat equation by replacing every derivative with its variation over one characteristic scale. Diffusion of matter is that equation with $D$ in place of $\kappa$ , and here we read it the other way round: the displacement scale and the time are our inputs, so the constant is what falls out.

\frac{\partial c}{\partial t} = D\frac{\partial^{2}c}{\partial x^{2}} \,\Rightarrow\, \frac{c}{t} \approx \frac{Dc}{x^{2}} \,\Rightarrow\, D \approx \frac{x^{2}}{t} = \frac{(1\,\mu\mathrm{m})^{2}}{1\,\mathrm{s}} = \pf{1\,\mu\mathrm{m}^{2}/\mathrm{s}}

(7)

A displacement and a time give a transport coefficient. Section 4 runs the same line forwards, turning a size into a time.

A calibrated theoretical example

Pearle and colleagues reconstructed Brown's observations and analyzed what his microscope could resolve. Their Table I calculates an expected mean displacement of $0.74\,\mu\mathrm{m}$ in one second for a sphere of radius $0.5\,\mu\mathrm{m}$ in water at 20 °C. Inverting $\langle|x|\rangle\approx0.80\sqrt{2Dt}$ gives $D\approx0.43\,\mu\mathrm{m}^2/\mathrm{s}$ .⁵ This puts (7)'s rough scale in context; it is a theoretical comparison, not an independent experimental measurement of D. The lineage develops the historical observations.

Now go from a granule to a protein, which is a factor of a few hundred in radius. That needs to know how $D$ depends on size, and the answer is less obvious than it looks.

Write $\eta$ for the fluid viscosity. Diffusion reflects thermal motion opposed by viscous drag, so the question is the drag on a sphere of radius $a$ moving at speed $v$ . Two powers of $a$ are in play, and they pull opposite ways. A viscous stress is the viscosity times a velocity gradient, and the fluid has to go from $v$ at the sphere's surface to zero far away, over the only length in the problem. So the gradient is $v/a$ and the stress is $\eta v/a$ . That stress acts over the sphere's surface, which is $a^{2}$ . Multiply:

F \approx \frac{\eta v}{a}\cdot a^{2} = \pf{\eta a v}, \quad \text{one power of }a\text{, not two}

(8)

That near-cancellation is the whole answer to why 1/a, and it is worth saying out loud, because the naive guess is that drag goes as the area. A bigger sphere does present more surface, and it sits in a gentler gradient by the same factor, so the drag grows only linearly in size. This also fixes what $a$ means: it is the one length that sets both the area and the gradient, so it is the hydrodynamic radius, the size of the object plus whatever solvent it drags along. Stokes did the flow properly in 1851 and got the prefactor, $F = 6\pi \eta a v$ . Mobility is $b=v/F=1/(6\pi\eta a)$ . Einstein's relation $D=bk_BT$ , with Boltzmann constant $k_B$ and absolute temperature $T$ , closes it:

D = k_{B}T / (6\pi \eta a) \,\,\Rightarrow \,\, \pf{D \propto 1/a}

(9)

At fixed temperature and viscosity, $Da=k_BT/(6\pi\eta)$ is independent of particle size. In water at 25 °C, using $\eta=0.89\,\mathrm{mPa\,s}$ , it is $0.245\,\mu\mathrm{m}^3/\mathrm{s}$ . Our round inputs give $1\,\mu\mathrm{m}^2/\mathrm{s}\times0.5\,\mu\mathrm{m}=0.5\,\mu\mathrm{m}^3/\mathrm{s}$ , within a factor of two. The calculated 20 °C example in the box gives about $0.215\,\mu\mathrm{m}^3/\mathrm{s}$ ; its different temperature must be respected in a precise comparison. This is a temperature- and solvent-dependent quantity, not a universal constant.

Now use it. A protein is close-packed amino acids, so it weighs what packed organic matter weighs, about 1.4 g/mL. The 33 kDa of (4) then occupies 40 nm³ and has radius $(3V/4\pi)^{1/3} \approx 2\,\mathrm{nm}$ . That is the factor of a few hundred promised above. Divide the constant by it:

D_{\mathrm{protein}} \approx \frac{0.245\,\mu\mathrm{m}^{3}/\mathrm{s}}{2\,\mathrm{nm}} \approx \pf{100\,\mu\mathrm{m}^{2}/\mathrm{s}} \quad \text{in water}

(10)

Figure 2. Stokes-Einstein over five decades of size, computed and not fitted, with measured comparison points and a hollow marker for the calculated Brown-scale example at 20 °C. Glucose provides a comparison with the line. GFP's plotted hydrodynamic radius is inferred from its solution diffusion coefficient, so that point is a calibration, not an independent test. Water sits above it, correctly: the formula assumes the particle is large compared with the solvent. The dashed line is the same relation divided by 11, an illustration based on the GFP measurement, not a universal cytoplasmic correction.

The cytoplasm is not water

Elowitz and colleagues measured GFP inside E. coli at $D=7.7\pm2.5\,\mu\mathrm{m}^2/\mathrm{s}$ , compared with $87\,\mu\mathrm{m}^2/\mathrm{s}$ in dilute solution. Their larger fusion diffused at about $2.5\,\mu\mathrm{m}^2/\mathrm{s}$ .⁶ The elevenfold GFP slowdown is an empirical comparison for those conditions. For the next estimates use about $10\,\mu\mathrm{m}^2/\mathrm{s}$ for a protein, and an illustrative $200\,\mu\mathrm{m}^2/\mathrm{s}$ for a small metabolite. Transport depends on the molecule and its environment.

Estimate a diffusion coefficient from a displacement and time, or from hydrodynamic size, temperature and viscosity; distinguish a water estimate from an in-cell measurement.

4τ ≈ x²/D, and what it predicts

The same scaling as (7), run forwards. A null model earns its keep by being wrong in a specific place.

Equation (7) read that line backwards, taking a measured wander and returning $D$ . Read forwards it returns a time from a size, where $x$ is the distance to explore. This is a characteristic diffusion time; a precise mixing time also depends on geometry, boundaries and the required degree of uniformity:

\pf{\tau \approx x^{2}/D}

(11)

Figure 3. Mixing time against distance, from

\tau = x^{2}/D

with in-cytoplasm coefficients. The shaded millisecond-to-hour interval is a reference window for comparing cellular processes. The quadratic distance dependence makes diffusion increasingly costly over larger lengths; whether it is fast enough depends on the process being supplied.

Table 2. $\tau = x^{2}/D$ for a protein at the measured in-cell value $D = 7.7\,\mu\mathrm{m}^{2}/\mathrm{s}$ , and for an illustrative metabolite coefficient of $200\,\mu\mathrm{m}^2/\mathrm{s}$ .
Compartment	size	τ, protein	τ, metabolite	verdict
E. coli	1 µm	0.13 s	5 ms	test against the reaction clock
yeast	5 µm	3.2 s	0.12 s	test against the reaction clock
mammalian cell	20 µm	52 s	2 s	compare with the reaction clock
Xenopus egg	1.2 mm	2 days	2 h	slow diffusion permits gradients
squid giant axon	1 cm	150 days	6 days	too slow for rapid supply
human motor axon	1 m	4000 yr	160 yr	too slow for rapid supply

Read the last two rows as a prediction. Diffusion alone is far too slow for rapid delivery along a centimetre or metre of axon. Sustained transport over these distances requires another mechanism; motor-driven transport is one biological solution. The estimate identifies a transport problem before specifying how the cell solves it.

Now read the bacterial row conditionally. A metabolite explores a micron in about 5 ms. Compare that with the time in which the reaction changes a substantial fraction of the local pool: §11 will estimate about 0.12 s for PEP replacement, roughly twenty times longer. This makes a spatially uniform approximation plausible for that comparison. A single enzyme's turnover time is not by itself the relevant pool-consumption clock. Faster local reactions, binding to structures or localized sources can still maintain gradients, even in bacteria.

Compare $x^2/D$ with the reaction or response time under study before assuming a compartment is well mixed.

Part 3

The biochemical core

Concentrations and transport constrain reactions. Enzymes make many otherwise slow conversions available on cellular timescales.

5The core is chemistry

The counts and transport clocks become useful together when we ask how a cell converts one molecular species into another.

A cell builds, consumes and modifies molecules through biochemical reactions. It also moves, exerts forces and maintains electrical gradients. Reaction networks provide a framework for its biochemical transformations, coupled where needed to spatial transport, mechanics and electrical dynamics. Treating a region as isothermal and well mixed simplifies that framework; it does not prove that every cellular state is a chemical concentration.

Figure 4. The biochemical focus of this course, coupled to physical processes. The connecting lines indicate coupling, not chemical reactions or a literal cell boundary. Choosing a biochemical model requires deciding which physical variables can be held fixed and which must be modeled.

Begin with accounting. A composite conversion lists species $X_i$ with reactant coefficients $\alpha_i$ and product coefficients $\beta_i$ :

\alpha_1X_1+\dots+\alpha_nX_n\;\rightsquigarrow\;\beta_1X_1+\dots+\beta_nX_n

(12)

The list must respect chemical composition, including any species omitted because they belong to a reservoir. It specifies the net conversion, but neither the molecular route nor its speed. This is where a cell's catalysts matter. Many thermodynamically favourable reactions are extremely slow without them: Stockbridge et al. compile uncatalysed half-times at 25 °C ranging from 27 seconds to $1.1\times10^{12}$ years for selected reactions, with the longest inferred by temperature extrapolation.¹³

Enzymes lower activation barriers and make such conversions accessible on biological timescales. Cells regulate enzyme abundance and state, and also change substrates, products, cofactors and driving conditions. Binding and catalysis are therefore central recurring operations. The following minimal enzyme mechanism separates them: enzyme $E$ binds substrate $S$ to form complex $C_{ES}$ , which can dissociate or produce $P$ .

E+S\;\xrightleftharpoons[k_{-1}]{k_1}\;C_{ES}\;\xrightarrow{k_2}\;E+P

(13)

Here $k_1$ is the association coefficient, $k_{-1}$ the dissociation rate and $k_2$ a single conversion-and-release rate in the assumed mechanism. Real enzymes may require additional bound states and separate chemical and release steps. The straight arrows assert that those events are resolved as elementary in this minimal model; they do not establish that every enzyme has exactly this mechanism.

Two questions to keep separate

At fixed temperature and pressure, the Gibbs free-energy change determines the thermodynamic direction of a net reaction. Activation barriers determine its kinetics. A catalyst changes the route and its speed, but does not change the equilibrium of the overall reaction. A favourable conversion can remain negligibly slow.

The next section makes the difference between (12) and (13) explicit. It is the difference between stating what changes and supplying a mechanism from which a rate law can be built.

Separate net chemical conversion, thermodynamic driving and kinetic mechanism; use binding and catalysis as a model whose unresolved steps remain explicit.

Part 4

What a reaction is

One word, two objects. Different arrows keep mechanism and net accounting distinct when we assign rate laws.

6Two meanings of the word

An elementary step specifies a microscopic event. A composite reaction specifies a net change. The route from an arrow to a rate law depends on which one we mean.

Definition and notation

A reaction step, or elementary reaction (基元反应), represents one microscopic event with no intervening reaction intermediate. Use a straight arrow carrying its microscopic rate constant.

An overall reaction, or composite reaction (复合反应), combines unresolved steps into their net change. Use a bare squiggly arrow and state any derived or fitted rate law separately. A simple first-order law does not make an unresolved process elementary.

Figure 5. The arrow convention records model resolution. The composite arrow remains bare even when we know its effective rate law; §9 supplies an example.

Table 3. Mechanism and net conversion, following the distinction in the 2025 note.¹¹
Object	Notation	What it supplies
Elementary binding pair	$E+S\xrightleftharpoons[k_{-1}]{k_1}C_{ES}$	Resolved association and dissociation events
Composite enzyme conversion	$S\rightsquigarrow P$	Net substrate consumption and product formation
Composite transcription	$G\rightsquigarrow G+M$	Gene $G$ retained; message $M$ produced; resources omitted

A third notation records a state change rather than a chemical reaction. If $n_S,n_P$ are molecule counts, one net conversion applies $(n_S,n_P)\mapsto(n_S-1,n_P+1)$ . The map records bookkeeping and carries no mechanistic claim. Lecture 3 packages these changes into a stoichiometric matrix; lecture 5 applies them to discrete counts.

Concentration notation for rate laws

In a rate law, a bare species symbol denotes its free molar concentration. Thus $A B$ is multiplication, never the name of a complex. Write a bound complex as $C_{AB}$ , and a constituent total as $q_A$ , summed over all forms containing A. For (13), $q_E=E+C_{ES}$ . Multi-letter subscripts that name a process are upright, as in $k_{\mathrm{on}}$ and $v_{\mathrm{in}}$ .

Why an elementary step gives mass action

Consider distinct reactants A and B in a homogeneous dilute solution, with encounters sufficiently uncorrelated. Let $\sigma$ be the effective reactive volume sampled by one A per second, in litres per second. With $n_B/V$ B molecules per litre, one A reacts at rate $\sigma n_B/V$ . Multiply by $n_A$ , then divide by $N_A V$ to convert events per second into molar flux:

v=\frac{\sigma n_A n_B}{N_A V^2}=N_A\sigma\,AB\equiv kAB,\qquad A=\frac{n_A}{N_A V},\quad B=\frac{n_B}{N_A V}

(14)

This is the law of mass action. The bimolecular coefficient $k=N_A\sigma$ has units $\mathrm{L\,mol}^{-1}\mathrm{s}^{-1}=\mathrm{M}^{-1}\mathrm{s}^{-1}$ . Section 8 computes the ideal diffusive encounter version of $\sigma$ . The Avogadro conversion matters: a molecular volume per second and a molar rate coefficient are related, but have different units.

For an elementary step with reactant coefficients $\alpha_i$ , the deterministic dilute-solution form is $v=k\prod_i X_i^{\alpha_i}$ . The units of $k$ depend on total reaction order: first-order steps use $\mathrm{s}^{-1}$ , second-order steps use $\mathrm{M}^{-1}\mathrm{s}^{-1}$ . Tethering, membrane confinement and spatial correlations require their own concentration measures and encounter assumptions; §4 checks only one part of that modeling choice.

The habit to install

Ask which microscopic events justify a proposed rate law. For resolved elementary steps, mass action follows under the stated mixing and independence assumptions. For a composite reaction, derive the effective law from a mechanism or establish it experimentally. Its stoichiometry alone does not determine it.

Classify an arrow before assigning its rate law, and distinguish microscopic mass action from a composite law that still needs derivation or measurement.

7Three systems, written both ways

Opening an overall arrow adds mechanistic detail. Work through three examples and identify both the rate laws obtained and the processes still unresolved.

At the board, write the net conversion first, then name the intermediate species needed for the question being asked. Keep composite arrows visible wherever the mechanism remains compressed.

Figure 6. Three systems at different resolutions. Opening enzyme conversion supplies internal elementary rate laws; supply and removal remain composite. Opening transcription exposes polymerase binding but leaves message synthesis unresolved. Ribosome production is a resource-and-assembly process whose protein cost can be estimated without pretending it is one elementary event.

An enzyme, fed and drained. Resolving $S\rightsquigarrow P$ as (13) adds E and $C_{ES}$ . Four species replace two, and $q_E=E+C_{ES}$ is conserved if enzyme is neither made nor lost. The internal rates are $k_1ES$ , $k_{-1}C_{ES}$ and $k_2C_{ES}$ . Substrate supply and product removal need separate boundary laws, such as a fixed input flux and a fitted first-order removal law. Those choices do not follow from the enzyme mechanism.

Gene expression. Let G be a free gene, RNAP a free RNA polymerase and $C_{G,\mathrm{RNAP}}$ their bound complex. Polymerase association can be resolved as an elementary binding step. The arrow from that complex to completed message M still combines initiation, elongation and termination. If free polymerase is effectively constant, its elementary association flux $k_1G\,\mathrm{RNAP}$ becomes a pseudo-first-order expression $k_{\mathrm{eff}}G$ , with $k_{\mathrm{eff}}=k_1\mathrm{RNAP}$ . Identifying this with the rate of completed messages needs additional assumptions about the downstream steps.

Ribosome production. The bookkeeping shorthand $R\rightsquigarrow2R$ expresses autocatalytic production of ribosomes. Existing ribosomes translate ribosomal proteins, while transcription supplies rRNA and assembly makes new functional particles. This hides resources, parallel branches and intermediate states. Even without resolving them all, translation capacity gives a useful estimate.

A protein-production capacity bound

Take about 0.84 MDa of protein per ribosome. Using §2's 110 Da per residue gives $n\approx7600$ amino-acid residues. A translating ribosome adds about $u=16.5\,\mathrm{aa/s}$ in good growth conditions.²¹ If all ribosomes were active, all translation made ribosomal protein, and rRNA supply and assembly were immediate, the ideal growth law would be $dR/dt=(u/n)R$ , with

t_2=\frac{n\ln2}{u}\approx\frac{7600\times0.693\,\mathrm{aa}}{16.5\,\mathrm{aa/s}}\approx320\,\mathrm{s}\approx\pf{5\text{ minutes}}

(15)

If only a fraction $\phi$ of this total translation capacity makes ribosomal protein, the ideal doubling time becomes $n\ln2/(\phi u)$ . A 20-minute doubling time would require $\phi\approx0.27$ in this model. This is an allocation estimate, not a universal upper bound on the fraction of active translation. Inactive ribosomes, turnover and assembly alter the inference. Scott et al.'s measured relation between ribosome-affiliated protein fraction and growth rate supplies the empirical starting point for lecture 14.²²

Across these examples, stop resolving a mechanism when the remaining approximation is adequate for the question. Boundary processes such as import, export and dilution use composite arrows unless their intervening mechanism is explicitly resolved. For example, write $\varnothing\rightsquigarrow S$ and specify $v_{\mathrm{in}}$ separately. The arrow marks entry into the modeled species inventory, not necessarily transport through a particular protein.

Write a system at two resolutions, name the extra species and conserved totals, and identify which rate laws remain assumptions after a mechanism is opened.

Part 5

How fast a reaction can go

An ideal encounter calculation sets a reference scale. Product formation introduces another coefficient, and concentrations turn both into comparable clocks.

8The diffusion limit, in three lines

For a perfectly absorbing sphere in a three-dimensional fluid, dimensional reasoning gets the encounter coefficient almost all the way. Geometry supplies the prefactor.

The association step of (13) requires molecules to meet. Estimate that opportunity first, assuming diffusion in a uniform fluid, capture on the whole spherical surface, a maintained far-field concentration and no steering forces. Call the ideal encounter coefficient $k_{\mathrm{diff}}$ ; the actual association coefficient $k_{\mathrm{on}}$ can be smaller because contact need not make a complex.

Line one: the units. Before converting to molar units, write the encounter coefficient as $k_{\mathrm{diff}}^{(V)}$ , a volume per second. Diffusion coefficient $D$ and capture radius $a$ give the dimensional form

[D]=\mathrm{length}^2/\mathrm{time},\quad[a]=\mathrm{length}\quad\Longrightarrow\quad k_{\mathrm{diff}}^{(V)}\sim Da.

(16)

Line two: the gradient. A stationary sphere absorbs substrate at its surface, so the substrate number density is zero at radius $a$ and approaches $c_\infty$ far away. The only geometric length is $a$ , suggesting a surface gradient of order $c_\infty/a$ .

Line three: Fick's law times the area. Multiplying that gradient by D gives an inward number flux per unit area, $j$ . Multiplying by the sphere's area gives the capture count per second, $J$ :

j\approx D\frac{c_\infty}{a},\qquad J\approx4\pi a^2j=4\pi Da\,c_\infty\equiv k_{\mathrm{diff}}^{(V)}c_\infty.

(17)

The exact steady solution is $c(r)=c_\infty(1-a/r)$ , and its surface gradient is indeed $c_\infty/a$ . Thus the estimate gives the exact coefficient for this geometry. The profile has a long $1/r$ depletion tail; it does not end one radius from the surface.

Figure 7. Absorbing-sphere geometry and its normalised steady concentration profile. Inward arrows show diffusive transport direction. At

r=2a

, concentration is half the bulk value; it approaches bulk gradually farther away. The dashed tangent marks the surface gradient. The exposition develops the full boundary-value calculation.

Let both partners move. Use the relative diffusion coefficient $D_A+D_B$ and contact radius $a_A+a_B$ . With Stokes–Einstein for each sphere, only their size ratio $\rho=a_A/a_B$ survives:

k_{\mathrm{diff}}^{(V)}=4\pi(D_A+D_B)(a_A+a_B)=\frac{2k_BT}{3\eta}\left(2+\rho+\frac1\rho\right).

(18)

Convert the physical coefficient to molar units

Equation (18), with SI inputs, has units $\mathrm{m}^3/\mathrm{s}$ . Multiply by $10^3\,\mathrm{L/m}^3$ and by $N_A$ , as §6 requires. For equal spheres, the bracket is 4:

k_{\mathrm{diff}}=N_A(10^3\,\mathrm{L/m}^3)\frac{8k_BT}{3\eta}=\pf{7.4\times10^9\,\mathrm{M}^{-1}\mathrm{s}^{-1}}\quad\text{in water at }25\,^{\circ}\mathrm{C}.

(19)

The common particle size cancels for equal spheres. The size-ratio factor increases for unequal partners; it is not bounded across arbitrary sizes. For the enzyme–small-metabolite pair used below, the coefficient is about $1.4\times10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ . These are ideal geometric references, not universal bounds for every molecule, search geometry or force field.

The naive $Da$ estimate for one sphere in water, after molar conversion, is $1.5\times10^8\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ . The equal-pair answer is about 50 times larger: $4\pi$ from spherical geometry, a factor of two from relative diffusion and another two from contact radius. In cytoplasm use the appropriate diffusion coefficients of both partners; the GFP slowdown alone cannot determine the change for a protein–metabolite pair.

We have now counted ideal encounters. An enzyme's useful output is product, so the next step is to identify the rate coefficient for completing the whole conversion.

Compute an ideal encounter reference with the stated diffusion coefficients and capture radius, then convert volume per second into $\mathrm{M}^{-1}\mathrm{s}^{-1}$ before comparing it with molar kinetic data.

9What enzymes actually do

Section 8 counted encounters. How many of those encounters become product? To compare an enzyme with the diffusion limit, we need a rate coefficient for the whole journey.

Continue to distinguish §8's ideal encounter coefficient $k_{\mathrm{diff}}$ , and reserve $k_{\mathrm{on}}$ for the enzyme's actual association coefficient. A collision need not form a bound complex, and a bound substrate can leave before being converted. The coefficient for making product must include both losses.

Why divide turnover by a concentration?

Consider an initial-rate assay of the single-substrate enzyme in §7, with negligible product. Let $v$ be product concentration made per second, $S$ the free substrate concentration, and $q_E = E+C_{ES}$ the total enzyme concentration, free plus bound. Two measured parameters describe its Michaelis–Menten rate curve:

$k_{\mathrm{cat}}$ , in $\mathrm{s}^{-1}$ , is the turnover number: products made per enzyme per second when substrate is saturating.
$K_M$ , in $\mathrm{M}$ , is the half-saturation concentration: at $S=K_M$ , the rate is half its maximum. It need not equal the binding dissociation constant $K_d$ .

After the initial binding transient, the rate law is the following; lecture 6 derives when this approximation is valid.¹⁰

\frac{v}{q_E}=\frac{k_{\mathrm{cat}}S}{K_M+S}

(19a)

The non-saturating limit means $S/K_M\ll1$ : substrate is low relative to the half-saturation concentration, so most enzymes are waiting. In this regime, far from saturation, replace the denominator by $K_M$ :

S\ll K_M:\quad v\approx\underbrace{\frac{k_{\mathrm{cat}}}{K_M}}_{\mathrm{M}^{-1}\mathrm{s}^{-1}}q_E S;\qquad S\gg K_M:\quad v\approx k_{\mathrm{cat}}q_E

(19b)

$k_{\mathrm{cat}}/K_M$ is the slope of flux per enzyme against substrate concentration in this limit. At fixed $q_E$ , $\partial(v/q_E)/\partial S\approx k_{\mathrm{cat}}/K_M$ : each extra unit of substrate concentration adds this much flux per enzyme. It is called the specificity constant, or catalytic efficiency, and its units are $\mathrm{M}^{-1}\mathrm{s}^{-1}$ . At the same non-saturating substrate concentration and enzyme abundance, twice the ratio means twice the product flux. At saturation, compare $k_{\mathrm{cat}}$ instead.

The composite reaction acquires an effective mass-action law

This closes the question posed in §6. Compress binding, possible release, and conversion into the single composite reaction, and give its derived rate law separately:

S\rightsquigarrow P,\qquad v\approx\frac{k_{\mathrm{cat}}}{K_M}\,q_E S\qquad(S\ll K_M)

The rate is proportional to both enzyme and substrate concentration, so it has the mass-action form with effective coefficient $k_{\mathrm{cat}}/K_M$ . This coefficient describes the complete conversion from free substrate to product. $k_{\mathrm{on}}$ describes complex formation, while $k_{\mathrm{cat}}$ describes saturated turnover. The squiggly arrow stays bare because the mechanism remains composite even when its rate law becomes this simple.

If $q_E$ is held fixed, the rate is first order in substrate, with coefficient $(k_{\mathrm{cat}}/K_M)q_E$ in $\mathrm{s}^{-1}$ .

Give the number a job

Suppose $k_{\mathrm{cat}}=10\,\mathrm{s}^{-1}$ and $K_M=100\,\mu\mathrm{M}=10^{-4}\,\mathrm{M}$ . Their ratio is $10^5\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ . At $S=1\,\mu\mathrm{M}\ll K_M$ ,

\frac{v}{q_E}\approx (10^5\,\mathrm{M}^{-1}\mathrm{s}^{-1})(10^{-6}\,\mathrm{M})=\pf{0.1\,\mathrm{s}^{-1}}

(19c)

That is about one product per enzyme every 10 seconds, even though a saturated enzyme makes ten per second. The units tell you which multiplication turns the coefficient into a rate.

What Figure 8 compares

Diffusive encounters occur at a rate proportional to $k_{\mathrm{diff}}ES$ . In the low-substrate limit $E\approx q_E$ , so this has the same concentration factors as (19b). We can therefore compare $k_{\mathrm{cat}}/K_M$ with an encounter coefficient: their dimensionless ratio benchmarks product formation against the ideal encounter opportunity. It is a literal success fraction only when the encounter model and reaction mechanism describe the same pair under the same conditions. Both coefficients have units of $\mathrm{M}^{-1}\mathrm{s}^{-1}$ . A turnover number in $\mathrm{s}^{-1}$ cannot go on this axis.

Figure 8. Circles measure product formation,

k_{\mathrm{cat}}/K_M

: the survey median, EcoRI methyltransferase, β-lactamase and a SOD mutant. The square measures complex formation,

k_{\mathrm{on}}

, for barnase binding its inhibitor barstar. Diamonds are ideal encounter references,

k_{\mathrm{diff}}

, for two specified pairs in water, with

4\pi Da

converted to molar units. These share units, but count different events. The median is about five decades below the ideal enzyme–metabolite encounter coefficient; the selected fast enzymes show how much closer product formation can get.⁷⁹

Real enzymes must do more than collide: the substrate has to reach a small active-site patch in the right orientation, which can cost up to three orders of magnitude.⁷ Two ways to improve access are electrostatic steering and searching along DNA.

Measured examples of faster access

Steering. Superoxide dismutase acts on a tiny charged ion and funnels it electrostatically. An interface mutant reaches $k_{\mathrm{cat}}/K_M=1.7\times 10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ , reported as the fastest enzyme in Stroppolo et al.'s 2001 review. That is about 70% of our ideal-sphere reference for this pair.⁷

Getzoff et al.'s mutagenesis experiments show that the arrangement of charges matters. Neutralizing selected acidic residues increased the measured rate by roughly two- to threefold; a corresponding charge reversal was less effective. The authors linked this difference to the local network positioning the charged groups.³⁰

Salt also changes steering. a colicin nuclease and its immunity protein associate at $10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ at low salt and $10^{7}\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ at high.⁸

A lower-dimensional search. EcoRI methyltransferase's $k_{\mathrm{cat}}/K_M$ rises about fourfold when its DNA substrate is lengthened from 14 to 429 base pairs.⁷ DNA outside the recognition site helps the enzyme find it: association with DNA followed by motion along it can enlarge the effective target. The rate therefore depends on the search geometry. The ideal-sphere diamonds are references for their stated pairs, not a universal upper bound for every geometry or electrostatic field. Halford and Marko nevertheless find most measured protein–DNA association coefficients within a factor of three of $10^8\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ .⁸

Now ask how fast a loaded enzyme works

Figure 8 concerns the low-substrate coefficient. Figure 9 asks the other question in (19b): at saturation, how many products does one enzyme make per second? Bar-Even and colleagues compiled parameters for enzymes acting on their natural substrates. The median turnover number is about $10\,\mathrm{s}^{-1}$ .⁹

Figure 9. An illustrative log-normal curve chosen to match two reported statistics: median

k_{\mathrm{cat}}\approx10\,\mathrm{s}^{-1}

and about 60% between

1

and

100\,\mathrm{s}^{-1}

. It is not the measured histogram. The band is the theoretical turnover range quoted by Bar-Even et al.,

10^6\text{--}10^7\,\mathrm{s}^{-1}

; it is a separate estimate from the second-order diffusion references in Figure 8.

Table 4. Reported distributions for natural substrates; each row has its own sample size.⁹
Parameter	median	where about 60% sit	comparison
$k_{\mathrm{cat}}$ (N = 1942)	10 s⁻¹	1–100 s⁻¹	theoretical turnover estimate: 10⁶–10⁷ s⁻¹
$k_{\mathrm{cat}}/K_{M}$ (N = 1882)	10⁵ M⁻¹ s⁻¹	10³–10⁶ M⁻¹ s⁻¹	diffusion benchmark quoted in the paper: 10⁸–10⁹ M⁻¹ s⁻¹
$K_{M}$ (N = 5194)	100 µM	10–1000 µM	observed lower tail: >99% above 0.1 µM

The survey's median specificity constant is $10^5\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ . Relative to its quoted diffusion benchmark of $10^8\text{--}10^9\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ , the gap is three to four decades; relative to §8's ideal whole-surface encounter estimate near $10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ , it is about five. The size of the gap depends on the reference. Either comparison says that product formation uses only a small part of the encounter opportunity. It does not tell us whether the loss occurs before a complex forms or after it forms. A4 separates those possibilities.

Far from saturation, you can give the composite enzymatic conversion an effective mass-action law: $v\approx(k_{\mathrm{cat}}/K_M)q_E S$ . Its coefficient $k_{\mathrm{cat}}/K_M$ , in $\mathrm{M}^{-1}\mathrm{s}^{-1}$ , is the slope of flux per enzyme against substrate concentration and the quantity to compare with encounter coefficients. At saturation, use $k_{\mathrm{cat}}$ in $\mathrm{s}^{-1}$ .

10The ledger of reaction timescales

Table 1 said what a dissociation constant is read against. This is the same thing for a rate, and the span is ten and a half decades.

Turn a coefficient into a time before putting it on this ledger. For a free enzyme at fixed substrate, the ideal waiting time to an encounter is $1/(k_{\mathrm{diff}}S)$ . The saturated turnover time is $1/k_{\mathrm{cat}}$ . These are the first and fourth rows; the low-substrate wait for a product instead uses $1/[(k_{\mathrm{cat}}/K_M)S]$ , as (19c) showed. An encounter time is not the time for binding to equilibrate.

Figure 10. A timescale ledger combining reaction, transport and growth estimates. The 70 ns row is an ideal encounter at

S=1\,\mathrm{mM}

, using the enzyme–metabolite reference in water; actual association in a cell can be slower. The 0.1 s row is saturated catalytic turnover. Their shaded separation is a comparison of two clocks, not a measurement of binding equilibration. Shamir et al. survey cellular timescales for E. coli and HeLa.²⁰

Table 5. Illustrative clocks, not a single set of simultaneous measurements. The conformational-change scale is from Shamir et al.; transcription uses 45 nt/s within their 10–100 nt/s range. The mRNA value is a representative half-life, and the last two rows use a chosen 40-minute doubling time.²⁰
Reaction	written as	timescale	set by
an ideal substrate encounter	contact with a free enzyme	70 ns	$1/(k_{\mathrm{diff}}S)$ , at 1 mM in water
a protein changes shape	$C \rightsquigarrow C^{*}$	1 ms	ligand-induced change, source estimate
a metabolite crosses the cell	$S_{\mathrm{here}} \rightsquigarrow S_{\mathrm{there}}$	5 ms	(11)
one catalytic turnover	$C_{ES} \xrightarrow{k_{\mathrm{cat}}} E + P$	0.1 s	$1/k_{\mathrm{cat}}$ , Table 4
the PEP pool turns over	$\mathrm{PEP} \rightsquigarrow \mathrm{pyr}$	0.12 s	pool over flux, Table 6
one protein is translated	$M \rightsquigarrow M + P$	18 s	300 aa at 16.5 aa/s
one gene is transcribed	$G \rightsquigarrow G + M$	22 s	1 kbp at 45 nt/s
half an mRNA pool is degraded	$M \rightsquigarrow \varnothing$	5 min	representative mRNA half-life
a protein concentration is halved by dilution	$P \rightsquigarrow \varnothing$	40 min	(20)
a cell divides	$N \rightsquigarrow 2N$	40 min	the doubling time

Here G denotes a gene, M its message and P its encoded protein in the expression rows; $C^*$ is a changed conformation and N denotes cells. Each row identifies its own event or fraction completed: a turnover time, a transit estimate and a pool half-life need not be the same mathematical kind of time constant. The arrow identifies model resolution, but it never tells you the timescale by itself. A hypothetical fixed-step simulation resolving the shortest listed clock throughout one division would require at least order $10^{10}$ steps. Lectures 5 and 6 develop methods for handling disparate clocks without resolving every fluctuation that finely.

To estimate a waiting time, first identify the event: encounter, product formation at low substrate, or saturated turnover. Then invert the rate for that event, with concentration included where needed.

Part 6

Regulation on different clocks

Compare metabolite-pool turnover with enzyme-abundance adjustment to identify which mechanisms can contribute to a rapid response.

11Metabolism cannot wait for gene expression

Cells regulate metabolism through both enzyme abundance and existing enzyme activity. Their relative clocks tell us why rapid disturbances need the second route.

Start with the time represented by a metabolite pool. Divide its concentration by the flux consuming it. For an illustrative glucose uptake of $8\,\mathrm{mmol}/(\mathrm{g}_{\mathrm{DW}}\,\mathrm{h})$ and a cell volume of $3.03\,\mathrm{mL}/\mathrm{g}_{\mathrm{DW}}$ , where $\mathrm{g}_{\mathrm{DW}}$ denotes grams of dry biomass and the volume follows from §2's density and dry-mass fraction, the glucose-equivalent flux is $J=8/(3.03\times3600)\,\mathrm{mol/(L\,s)}\approx0.73\,\mathrm{mM/s}$ . Combine this assumed flux with the measured pools below.

Table 6. Pool replacement estimates. Concentrations are from glucose-fed *E. coli*.³ Flux factors are illustrative: two three-carbon units per glucose for PEP and 3PG, and an assumed 30 ATP turnovers per glucose for the ATP row. The 8.8 mM hexose-phosphate measurement combines glucose-6-phosphate, glucose-1-phosphate and fructose-6-phosphate. Fluxes here are not measured reaction fluxes in the concentration experiment.
Metabolite	pool	assumed consumption flux	pool / flux	40 min / turnover
phosphoenolpyruvate (PEP)	0.18 mM	$2J$	0.12 s	20 000
ATP	9.6 mM	$30J$	0.43 s	5 600
3-phosphoglycerate (3PG)	1.5 mM	$2J$	1.0 s	2 400
combined hexose-phosphate pool	8.8 mM	$J$	12 s	200
fructose-1,6-bisphosphate (FBP)	15 mM	$J$	20 s	120

At steady state these pools are replenished, so they do not empty. Pool divided by flux measures a replacement time. It also estimates depletion time if supply stops and consumption initially continues at its old rate; as concentration falls, the consumption rate may change. The small PEP pool can therefore change substantially on a subsecond timescale when incoming and outgoing fluxes become imbalanced.

Now compare an enzyme-abundance response. Let $q_E$ be the concentration of a stable enzyme, $\beta$ its synthesis flux and $\mu$ the exponential cell-growth rate. Assume no active degradation and a sustained step to constant synthesis $\beta_{\mathrm{new}}$ , at fixed $\mu$ . Then

\begin{aligned} \frac{dq_E}{dt}&=\beta_{\mathrm{new}}-\mu q_E,\qquad q_{E,\mathrm{new}}=\frac{\beta_{\mathrm{new}}}{\mu},\\ q_E(t)&=q_{E,\mathrm{new}}+(q_{E,\mathrm{old}}-q_{E,\mathrm{new}})e^{-\mu t},\\ t_{1/2,\mathrm{response}}&=\frac{\ln2}{\mu}=\pf{\text{cell doubling time}}. \end{aligned}

(20)

The half-time is the time to cover half the distance to the new steady concentration. It is not a minimum delay before new protein appears. A sufficiently large synthesis increase can rapidly change the concentration relative to its old value. Conversely, after synthesis is switched off, dilution alone removes half the stable protein concentration in one doubling time. Transcription, translation and maturation add their own clocks; most mRNA half-lives measured by Bernstein et al. under their conditions were 3–8 minutes.¹⁵

What the comparison establishes

For a 40-minute doubling time, the abundance-response half-time in (20) is about 20,000 PEP replacement times. This comparison does not forbid expression-based metabolic regulation over minutes or generations. It shows why a rapid flux imbalance cannot generally be corrected by waiting for a large adjustment of stable enzyme abundance. Existing enzymes and their substrates, products or regulators must carry the immediate response.

Section 9 already supplied one such route. At fixed enzyme abundance and non-saturating substrate, $v\approx(k_{\mathrm{cat}}/K_M)q_E S$ , so a substrate change changes flux directly. Near saturation that response weakens. Product concentrations and allosteric effectors provide further routes, depending on the mechanism. The need for a fast response does not by itself identify which route dominates.

Two experiments ask complementary questions

A transient experiment. Link, Kochanowski and Sauer switched E. coli between pyruvate and labelled glucose or fructose media every 30 seconds. Flux through glycolysis reversed rapidly, with substantial labelling of PEP within five seconds after glucose addition. Their short perturbations were designed to probe existing enzyme kinetics and metabolite regulation before substantial expression changes.¹⁶

A comparison across steady states. Hackett et al. measured enzymes, metabolites and fluxes in 25 yeast chemostat conditions. Their fitted models attributed about twice as much flux variation to metabolites collectively as to enzyme abundance, with substrates carrying the largest average contribution. Allosteric contributions were concentrated in strongly forward-driven reactions in this dataset.¹⁷ This is evidence about those conditions and models; it neither follows from the transient timescale argument nor establishes a universal division of metabolic control.

From feedback inhibition to an equilibrium model

Umbarger's 1956 experiments showed isoleucine inhibiting the first step of its biosynthetic pathway. Work on pyrimidine synthesis by Yates and Pardee, followed by Gerhart and Pardee's experiments on aspartate transcarbamylase, helped distinguish catalytic and regulatory interactions.¹⁸ Such distinctions require kinetic or structural evidence; chemical dissimilarity alone does not establish noncompetitive inhibition. Monod, Wyman and Changeux proposed an equilibrium model of allosteric transitions in 1965.¹⁹ Applying it requires checking that its states equilibrate on the clock of interest, a separate question developed in lecture 7.

Compare a pool replacement time with a specified enzyme-abundance response, then identify which mechanisms can change flux on the required timescale.

Addendum

Where this goes

The last two minutes. The two components, and the rest of the course as a map.

12Where this goes

Binding, catalysis and their coupling form a recurring framework for the rest of the course. Its approximations must be checked for each system.

Figure 11. A useful regime to test: reversible binding relaxes quickly enough to approximate by equilibrium, while slower catalysis carries a net flux. The example assumes

k_{\mathrm{on}}=10^8\,\mathrm{M}^{-1}\mathrm{s}^{-1}

k_{\mathrm{off}}=10^4\,\mathrm{s}^{-1}

S=1\,\mathrm{mM}

, and

k_{\mathrm{cat}}=10\,\mathrm{s}^{-1}

: binding relaxation is about 10 µs and saturated turnover 0.1 s. These assumptions illustrate the separation; the survey median alone does not establish it. The dashed arrow denotes regulation.

Table 7. What lecture 2 hands to each later lecture.
Handed on	To	Used for
net-change notation, mass action for reaction steps	lecture 3	the rate equation $dx/dt = \Gamma v$
copy-to-concentration conversion at a stated cell volume	lecture 5	when noise becomes physics
the enzyme of §7, and A4's $\varepsilon = q_{E}/(K_{M}+S_{0})$	lecture 6	the licence for quasi-steady state
rapid-equilibrium binding, when the microscopic rates and driving timescale permit it (A4)	lecture 7	equilibrium statistical mechanics of regulation
binding and catalysis, with boundary flows and justified closure	lecture 8	a reduced description of biochemical dynamics
saturation, and $K_{M}$ against pool size	lecture 11	reaction order as a log-derivative
flux, pools and turnover times	lecture 13	flux balance analysis
the ribosome estimate of §7	lecture 14	the growth laws
biochemical mechanisms, physical coupling and model resolution	lecture 16	what a foundational virtual cell has to be

Five things to leave with

Convert counts to concentrations at a stated volume: one molecule per femtolitre is about $1\,\mathrm{nM}$ . Compare $x^2/D$ with the process of interest before treating concentrations as spatially uniform.
Binding and catalysis provide a useful biochemical framework. Thermodynamic driving, kinetic barriers and physical transport answer different questions.
An elementary reaction uses a labelled straight arrow. A composite reaction uses a bare squiggle, even when a derived effective law has a mass-action form.
At non-saturating substrate, $k_{\mathrm{cat}}/K_M$ is the slope of flux per enzyme against substrate concentration. Its median near $10^5\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ lies about five decades below our ideal enzyme–metabolite encounter reference. At saturation, turnover is governed by $k_{\mathrm{cat}}$ instead.
Subsecond metabolic disturbances need responses through existing enzymes. Substrate sensitivity can supply one route. Treating regulatory binding as equilibrium additionally requires the microscopic-rate and driving-timescale checks in A4.

Use the exposition for extended calculations and exercises, and the lineage for historical development. The optional sections below answer a narrower question raised by the main argument: which extra measurements let us move from a rate curve to the microscopic kinetics beneath it?

Carry concentrations, event-specific clocks and the elementary/composite distinction into a network model, and state the additional conditions needed for each reduction.

Optional · not in the 95 minutes

The third rate constant

Where an affinity comes from, how an association rate turns it into a residence time, and when the specificity constant supports an equilibrium approximation. Section 9 supplied the measured parameters; this part opens the mechanism beneath them.

A1What an off-rate adds

Section 9 defined the parameters of the rate curve. The microscopic mechanism tells us which combinations of binding, release and conversion those parameters measure.

Return to the minimal mechanism (13). It resolves a reversible elementary binding pair and one elementary conversion-and-release channel, with three microscopic rate constants. Real multistep enzymes can require a more detailed mechanism.

E + S \,\xrightleftharpoons[\,k_{-1}\,]{\,k_{1}\,}\, C_{ES} \,\xrightarrow{\,k_{2}\,}\, E + P

(21)

Microscopic constants and observable combinations

$k_{1}\equiv k_{\mathrm{on}}$ , in $\mathrm{M}^{-1}\mathrm{s}^{-1}$ . Formation of the bound complex. Section 8 estimates an ideal encounter reference, not a measurement of this coefficient.
$k_{-1}\equiv k_{\mathrm{off}}$ , in $\mathrm{s}^{-1}$ . The complex coming apart without making product.
$k_{2}\equiv k_{\mathrm{cat}}$ , in $\mathrm{s}^{-1}$ . The complex going on to product in this minimal mechanism. In a multistep enzyme, measured turnover need not be one elementary rate constant.
$K_{d}\equiv k_{-1}/k_{1}$ , the dissociation constant. The concentration at which half the enzyme is bound, with catalysis switched off. It has concentration units and is read against the ligand concentrations in Table 1.
$K_{M}\equiv (k_{-1}+k_{2})/k_{1}$ , in $\mathrm{M}$ , the half-saturation concentration. It approaches $K_d$ when $k_2\ll k_{-1}$ ; A4 explains what evidence is needed to check that condition.

Now the arithmetic that makes the rest of this part short. $K_{d} = k_{\mathrm{off}}/k_{\mathrm{on}}$ holds for this elementary reversible binding pair. With catalysis switched off,

k_{\mathrm{off}} = k_{\mathrm{on}}K_{d}, \qquad \tau_{\mathrm{res}} = 1/k_{\mathrm{off}}

(22)

Once $k_{\mathrm{on}}$ and $K_d$ are known, their product fixes the dissociation rate. With catalysis active, the mean time spent in $C_{ES}$ before either exit is $1/(k_{\mathrm{off}}+k_{\mathrm{cat}})$ ; this differs from the dissociation-only residence time in (22). Measuring $k_{\mathrm{cat}}$ and $K_M$ alone leaves association and dissociation undetermined: many microscopic pairs give the same rate curve.

Distinguish dissociation-only residence from total complex dwell time, and identify which additional microscopic rate a measured Michaelis–Menten curve leaves unknown.

A2Where an affinity comes from: count the contacts

Affinity depends exponentially on binding free energy. A contact-count model makes that sensitivity visible, provided its energetic and geometric assumptions remain explicit.

For dilute ideal binding, the standard molar binding free energy $\Delta G^\circ$ and the dissociation constant satisfy

\Delta G^\circ=RT\ln(K_d/c^\circ),\qquad c^\circ=1\,\mathrm{M},\qquad RT\ln10\approx\pf{1.36\,\mathrm{kcal/mol}}\quad(298\,\mathrm{K}).

(23)

Here $R=N_Ak_B$ is the gas constant. A tenfold decrease in $K_d$ makes binding more favourable by $RT\ln10$ per mole, or $k_BT\ln10\approx2.3k_BT$ per molecular pair. The standard concentration makes the logarithm dimensionless.

A possible slope. Binding in water exchanges interactions with solvent for interactions across an interface. A hydrogen-bond energy in vacuum cannot simply be credited in full. For an illustrative additive model, assign each favourable contact a net free-energy gain of $\epsilon=2.5k_BT$ , after solvent effects. Then

\frac{\epsilon}{k_BT\ln10}=\frac{2.5}{\ln10}\approx\pf{1.09\text{ decades per assumed contact}}.

(24)

An opposing cost. Binding restricts relative translation and orientation. At the 1 M standard state, the available volume per molecule is $V^\circ=1/(N_Ac^\circ)=1.66\,\mathrm{nm}^3$ . Assume a bound translation volume of $10^{-3}\,\mathrm{nm}^3$ and an orientational volume of $0.15^3$ within the freely rotating measure $8\pi^2$ . The corresponding model cost is

\frac{\Delta G_{\mathrm{immob}}^\circ}{RT}\approx\ln\frac{1.66}{10^{-3}}+\ln\frac{8\pi^2}{0.15^3}\approx17.5,\qquad\frac{17.5}{\ln10}\approx7.6.

(25)

Combining the offset with n identical favourable contacts gives

K_d\approx(1\,\mathrm{M})\,10^{\,7.6-1.09n}.

(26)

In this chosen model, about seven contacts compensate the standard-state immobilisation cost, and each additional contact gains about a decade. That does not mean seven contacts suffice for occupancy at an arbitrary ligand concentration. Half occupancy at $1\,\mu\mathrm{M}$ requires $K_d\approx10^{-6}\,\mathrm{M}$ , or roughly thirteen model contacts. Nor is a model contact equivalent to a heavy atom or a universal hydrogen bond.

The same bookkeeping explains a possible benefit of avidity. After one site binds, a suitably tethered second site can make contacts without paying the full translational search cost again. The actual gain depends on effective local concentration, geometry and tether entropy, so it cannot be read from contact count alone.

What the literature does and does not check

Finkelstein and Janin estimated standard-state translational and rotational entropy losses corresponding to roughly $18k_BT$ for looser complexes and $24\text{--}27k_BT$ for tightly constrained ones. Their result shows how strongly residual motion affects the offset.²³ It supports the scale of an immobilisation cost, not one universal value.

Fersht et al.'s mutation experiments found typical energetic losses of 0.5–1.5 kcal/mol, about 0.8–2.5 thermal units, when deleting good neutral hydrogen-bonding groups in tyrosyl-tRNA synthetase.²⁹ Their energies were inferred from changes in $k_{\mathrm{cat}}/K_M$ and explicitly concern transition-state binding. They demonstrate solvent-dependent, variable increments; they do not directly measure a universal equilibrium-affinity slope. Equation (24)'s 2.5-unit contact remains an illustrative assumption near the upper end of that neutral range.

Kuntz et al.'s survey found an initial upper envelope of binding strength near 1.5 kcal/mol per non-hydrogen atom and a tendency to level off for larger ligands.²⁴ Atom count differs from contact count, so this is not an independent measurement of (24)'s assumed slope. The observed envelope, including its exceptions, is a dataset pattern rather than a thermodynamic ceiling on affinity.

Convert an affinity ratio to a binding free-energy difference, and state the standard-state, contact and confinement assumptions before estimating affinity from geometry.

A3The residence ladder, and its floor

Affinity fixes a ratio of rates. A residence time needs an absolute rate as well.

For the reversible pair in A1, holding $K_d$ fixed while changing $k_{\mathrm{on}}$ requires a proportional change in $k_{\mathrm{off}}$ . If an encounter calculation supplies an applicable upper reference for association, it supplies a conditional lower reference for dissociation-only residence:

\tau_{\mathrm{res}}=\frac1{k_{\mathrm{on}}K_d}\ge\frac1{k_{\mathrm{on}}^{\max}K_d},\qquad k_{\mathrm{on}}^{\max}\approx10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}.

(27)

The inequality assumes this association bound applies to the specified geometry and conditions. Section 9 showed why steering and alternative search geometries require revisiting it.

Figure 12. Illustrative residence times at five affinities. Hollow markers use the common reference

k_{\mathrm{on}}^{\max}=10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}

; filled markers use the stated example association rates. The first three coincide because their assumed association rate equals the reference. These are conditional calculations, not five measurements.

An enzyme. A4's example assumes $k_{\mathrm{on}}=1.4\times10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ , $K_M=100\,\mu\mathrm{M}$ and $k_{\mathrm{cat}}=10\,\mathrm{s}^{-1}$ . It implies $k_{\mathrm{off}}\approx1.4\times10^6\,\mathrm{s}^{-1}$ , or a dissociation time near 0.7 µs. Figure 12 uses a rounded $10^{10}$ association reference instead and places its $100\,\mu\mathrm{M}$ row at 1 µs. Both estimates require the stated on-rate; neither follows from affinity alone.

A regulatory binder. At $K_d=1\,\mathrm{nM}$ and $k_{\mathrm{on}}=10^9\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ , the dissociation-only residence is one second. Whether this represses a gene depends also on free repressor concentration, rebinding and competition with transcription. A short individual residence need not imply low time-averaged occupancy.

An antibody example. At $K_d=0.1\,\mathrm{nM}$ and association coefficients $10^5\text{--}10^6\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ , the implied residence is $10^4\text{--}10^5\,\mathrm{s}$ , roughly three to thirty hours. These inputs must be paired consistently; an association range quoted for a population does not determine the residence of every antibody in it. Foote and Eisen discuss kinetic and affinity constraints in antibody responses.²⁵

Whole-surface capture and productive association differ

Northrup and Erickson's Brownian-dynamics calculations compared freely diffusing spheres that bind on any contact with proteins requiring an aligned contact patch. The latter associated much more slowly, while repeated close encounters reduced the penalty relative to a single random-orientation estimate.²⁶ This explains why a diffusion-controlled encounter coefficient and a productive association coefficient need not coincide. Changes in conformational or energetic barriers can also alter the rates.

Use an association coefficient together with affinity to estimate residence; do not infer the lifetime of a complex from its equilibrium occupancy alone.

A4When low efficiency implies rapid equilibrium

Low efficiency can mean that encounters seldom bind, or that bound substrate usually escapes. Only the second explanation directly supports rapid-equilibrium binding.

Briggs and Haldane replaced Michaelis and Menten's equilibrium assumption with a steady state on the complex. Two pages, 1925, and the expression this part needs:¹⁰

\frac{k_{\mathrm{cat}}}{K_{M}} = \frac{k_{1}k_{2}}{k_{-1}+k_{2}} \,\,\Longrightarrow\,\, \frac{k_{\mathrm{cat}}/K_{M}}{k_{1}} = \frac{k_{2}}{k_{-1}+k_{2}}

(28)

The denominator on the left is the actual association coefficient $k_1$ . The right is the probability that a formed complex makes product before dissociating: two competing exponential exits, with rates $k_2$ and $k_{-1}$ . Thus the specificity constant is association multiplied by success probability. Invert the relation:

\frac{k_{-1}}{k_{2}} = \frac{k_{1}}{k_{\mathrm{cat}}/K_{M}} - 1 = \pf{\text{mean dissociations before one product}}

(29)

This counts unsuccessful binding attempts in the minimal mechanism, assuming the enzyme can bind again. It is also $(1/k_2)/(1/k_{-1})$ : the catalytic time divided by the dissociation time. It does not count macroscopic equilibration events. At fixed substrate, the binding-only relaxation time is $\tau_{\mathrm{bind}}=1/(k_1S+k_{-1})$ .

Bar-Even et al. used the corresponding specificity-to-association ratio to analyze futile encounters.³¹ Their broader initial-encounter model includes processes preceding a productively bound active-site complex. Its success fraction and our bound-complex success fraction must be interpreted with their respective event definitions.

The same specificity constant, two different mechanisms

Take $k_{\mathrm{cat}}=10\,\mathrm{s}^{-1}$ and $K_M=100\,\mu\mathrm{M}$ , so $k_{\mathrm{cat}}/K_M=10^5\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ . If $k_1=1.4\times10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ , (29) gives about $1.4\times10^5$ dissociations per product. Binding reverses much faster than catalysis, consistent with a rapid-equilibrium approximation.

But choose $k_1=2\times10^5\,\mathrm{M}^{-1}\mathrm{s}^{-1}$ and $k_{-1}=10\,\mathrm{s}^{-1}$ . The same $k_{\mathrm{cat}}$ and $K_M$ result, while half the formed complexes make product. Dissociation is now as slow as catalysis. The measured rate curve and the gap to the ideal diffusion reference cannot distinguish these mechanisms.

Two reductions need different checks. Rapid equilibrium is supported by $k_{-1}\gg k_2$ and binding relaxation faster than changes in the substrate or other driving inputs. Then $K_M\approx K_d$ . A standard quasi-steady-state reduction instead approximates the complex by its instantaneous stationary value after an initial transient; in a standard closed assay starting with free enzyme and substrate, $\varepsilon=q_E/(K_M+S_0)\ll1$ is a sufficient smallness condition, where $S_0$ is initial free substrate.¹⁴ That reduction does not require binding equilibrium.

Reversible metabolic flux also needs a mechanism that includes the reverse reaction. Its thermodynamic driving force is a separate question from how quickly binding relaxes. The rapid-equilibrium approximation, the quasi-steady-state approximation, and near-equilibrium net flux describe different properties.

Lecture 6 derives the reductions and their errors, including the initial transient and the choice of retained variables. Here the important distinction is already available: a stationary complex concentration need not be a binding equilibrium.

Divide $k_{\mathrm{cat}}/K_M$ by the measured $k_{\mathrm{on}}$ to infer product probability per formed complex in (21). Dividing by an ideal diffusion reference cannot replace that measurement.

A5How tight binding can limit an enzyme

Tighter binding can improve capture, slow release or change the catalytic barrier. Its effect depends on which state is stabilised and which rate limits the cycle.

First, distinguish a model condition from a performance limit. For A4's rapid-equilibrium approximation, require $k_{\mathrm{off}}\gg k_{\mathrm{cat}}$ . Using $k_{\mathrm{off}}=k_{\mathrm{on}}K_d$ gives

K_d\gg\frac{k_{\mathrm{cat}}}{k_{\mathrm{on}}}=\frac{10\,\mathrm{s}^{-1}}{1.4\times10^{10}\,\mathrm{M}^{-1}\mathrm{s}^{-1}}\approx\pf{0.7\,\mathrm{nM}}.

(30)

This is a condition for that approximation at the assumed rates. An enzyme can violate it and still function: bound substrate may be converted rather than frequently released. It is not a universal lower limit on enzyme affinity.

Product release can limit turnover. If product must dissociate before another cycle, its mean release time contributes to the total cycle time. A slow product off-rate can therefore cap turnover. For an elementary product-binding pair, $k_{\mathrm{off},P}=k_{\mathrm{on},P}K_{d,P}$ . Product affinity is a separate quantity from substrate affinity; chemical resemblance does not force them to be equal. Albery and Knowles analyzed how stabilising bound intermediates can eventually reduce turnover by trapping the enzyme.²⁷

Substrate and transition-state stabilisation have different effects. In a transition-state-theory comparison of the same chemical transformation, with comparable kinetic prefactors, the acceleration of the intrinsic chemical step can be written

\frac{k_{\mathrm{chem}}}{k_{\mathrm{uncat}}}\approx\frac{K_{d,S}}{K_{d,\mathrm{TS}}}.

(31)

Here $K_{d,S}$ is substrate dissociation affinity and $K_{d,\mathrm{TS}}$ is the formal affinity for a transition-state configuration. It is a thermodynamic-cycle construction, not an equilibrium measurement on a stable transition-state species. Stabilising bound substrate alone, while keeping the bound transition-state energy fixed, raises the barrier from the substrate well and slows that chemical step. Measured $k_{\mathrm{cat}}$ need not equal $k_{\mathrm{chem}}$ if another step limits turnover. Wolfenden and Snider use such comparisons to quantify catalytic power.²⁸

Saturation also changes responsiveness. Return to the measured rate law in §9. At fixed $q_E$ , $k_{\mathrm{cat}}$ and $K_M$ , its substrate elasticity, the fractional rate change per fractional substrate change, is

\frac{\partial\ln v}{\partial\ln S}=\frac{K_M}{K_M+S}=\frac1{1+S/K_M}.

(32)

At $S\ll K_M$ this approaches one, the effective mass-action limit. At $S=K_M$ it is one half; at $S=100K_M$ it is about 0.01. Increasing substrate from $K_M$ to $100K_M$ nearly doubles flux per enzyme but reduces fractional substrate sensitivity about fiftyfold. This is a local property of one rate law, not the control coefficient of a whole metabolic network.

Likewise, reducing $K_M$ at fixed substrate and turnover raises saturation while lowering substrate sensitivity. Actual molecular changes can alter several parameters together. Bennett et al. found substrate concentrations above $K_M$ for many measured pairs, with lower-glycolytic intermediates near their consuming enzymes' $K_M$ values.³ This motivates questions about capacity and responsiveness; it does not prove that all enzymes are optimised to have $K_M$ equal to their pool concentration.

Judge binding by the job it serves

A binder may need sustained occupancy; a regulator may need occupancy that tracks a changing signal; an enzyme must complete and repeat a catalytic cycle. Compare ligand concentration, residence, release and conversion rates with the relevant task. Affinity alone is not a measure of performance.

Ask which bound state a change stabilises, then assess its effects on capture, chemical conversion, release and substrate sensitivity separately.

§Exercises

Four that can be done on paper, in the order the lecture built them. The first two are from the 2025 problem set.¹² The first extension mini-essay may be on any of them.

The blueprint and the machine. Which is bigger, an mRNA or the protein it encodes, in mass and in volume? Use 110 Da per amino-acid residue and 330 Da per RNA nucleotide. First compare material masses and volumes; then explain why the spatial extent of a folded or extended RNA needs an additional conformational assumption. (2025 hw1, 2.1.)
Head space in a culture tube. Estimate the oxygen a saturated 1 mL E. coli culture needs, at 0.2% glucose by mass and a biomass yield of 0.5, and check the five-fold head-space rule. Does dissolved oxygen matter? (2025 hw1, 2.3.)
Write one down both ways. Pick a mechanism you know: a two-component system, a kinase cascade, a repressed promoter. Write it as overall reactions, then as reaction steps. Count the species and conservation laws you gained, and name the arrow you could not open.
Check a proposed reduction. Use $K_M=100\,\mu\mathrm{M}$ , initial free substrate $S_0=0.18\,\mathrm{mM}$ and total enzyme $q_E=1\,\mu\mathrm{M}$ in a closed assay starting with free enzyme. Compute $\varepsilon=q_E/(K_M+S_0)$ . Which approximation does this support? Use A4's two microscopic mechanisms to show why it does not establish rapid equilibrium. Then identify which assumptions would need reconsidering for binding to a single DNA operator.

†References & note sources

Every number on this page traces to one of these or to lectures/research/lecture02/analysis/cell_ledger.py. Where a badge says read, the full text is in the course's local literature archive and was read for this page. The archive is not redistributed.

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B. D. Bennett, E. H. Kimball, M. Gao, R. Osterhout, S. J. Van Dien, J. D. Rabinowitz. Absolute metabolite concentrations and implied enzyme active site occupancy in Escherichia coli. Nature Chemical Biology 5:593–599, 2009. PDF Metabolite concentrations and substrate-to-K_M comparisons, with lower glycolysis as a named exception.
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