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Consider an original ideal model with sample volume $V_s$, bath volume $V_b$, and initial sample solute concentration $c_0$. Assume each volume is fixed, a freely permeant, nonbinding solute, no chemical reactions, complete mixing on each side, and equal concentrations at equilibrium. Each fresh bath initially contains none of this solute. The protein is perfectly retained in this model.

The first equilibration redistributes the initial amount of substance $c_0{V}_s$ through both compartments. Conservation givesThis denominator includes the sample itself. For the chosen $V_s=1\mathrm{mL}$ and $V_b=9\mathrm{mL}$, $c_1/c_0=1/10$, rather than $1/9$. Nine-tenths of the initial solute is in the bath and one-tenth remains in the sample; nothing has vanished during equilibration.

Discarding the equilibrated bath removes its solute. A second fresh bath starts with only $c_1{V}_s$ of solute in the entire system, so the same argument applies again. Induction gives the geometric progressionHere $n$ counts completed equilibrations with fresh baths: $n=1$ is the first bath, $n=2$ follows one replacement, and $n=3$ follows two replacements. The normalized sample concentrations are respectively $0.1$, $0.01$ and $0.001$. Leaving the first bath in place for longer cannot lower the ideal equilibrium value below $0.1$.

The same conservation argument handles a fresh bath already containing a permeant solute at fixed concentration $b$:Thus the sample approaches $b$, not necessarily zero. For example, if a hypothetical new buffer component starts at zero in the sample and at $b$ in every bath, its sample concentration reaches $0.9b$, $0.99b$ and $0.999b$ in the three equilibrations. Outward removal and inward replacement follow the same mathematics.

Equilibrium fractions do not specify elapsed time. An illustrative well-mixed linear ordinary differential equation model within one bath isHere $K$ is an effective transfer coefficient with units of volume per time. Adding the two equations conserves $V_s{c}_s+V_b{c}_b$; subtracting their normalized forms givesThe difference undergoes exponential decay. A bath replacement resets $c_b$, renewing the driving difference without making $c_s$ jump instantly. The plotted points are equilibrium endpoints, not a measured time course or a timetable for an arbitrary protein.

Real sample volume can change through osmosis; proteins can be lost through leakage or adsorption to the semipermeable membrane. Consequently, a measured decrease in protein concentration alone cannot distinguish dilution from loss of protein. Measuring recovered volume as well as protein concentration permits an estimate of recovered protein mass.^[ref]

An experimental comparison joined a two-ended diol chain, HOPCLOH, with a bifunctional isocyanate, at reported isocyanate-to-diol ratio $1.14:1$. The researchers used oven-dried $25\mathrm{mL}$ round-bottom flasks and an IKA RET magnetic hotplate with a metal carrier. They varied solvent, concentration, reaction temperature ($50$ or $61{}^{\circ }\mathrm{C}$), and duration ($1$ or $8\mathrm{h}$).^[ref]

A Bruker $500\mathrm{MHz}$ NMR spectroscopy instrument measured residual hydroxy group-adjacent signal relative to an ester signal. Independently, size-exclusion chromatography used ResiPore columns at $70{}^{\circ }\mathrm{C}$, a refractive-index detector, and polystyrene calibration. Figure 8 reports sharply increasing apparent number-average molar mass beyond the reported conversion $0.95$. A deliberately water-exposed isocyanate comparison reproduced a $1625\mathrm{c}{\mathrm{m}}^{-1}$ infrared spectroscopy band assigned to urea-containing products. The authors warn that tiny residual NMR spectroscopy signals reduce integration precision.^[ref]

The inference is increasing chain size alongside depletion of reactive groups, consistent with step-growth polymerization. It is not an exact measurement of $X_n$ in the toy model. The feed contains already extended diol chains: counting their constituent repeat units differs from counting each feed molecule as one unit. Their varying molar masses, calibration against a different polymer, residual starting material, and competing chemical reactions further obstruct a direct conversion-to-$X_n$ substitution. Report which functional groups the conversion readout measures before comparing it with stoichiometric imbalance in step growth.

Useful proposed controls, distinct from the reported water comparison, are pre-reaction spectra and matched mixtures omitting each joining partner in turn. Keep solvent, heating, and signal normalization matched. Disappearance of an isocyanate signal in the omitted-diol control would reveal competing consumption rather than establish chain joining. Agreement between residual-group and independently calibrated chain-size readouts is stronger evidence than calculating both quantities from the same Carothers equation.

The respiratory control ratio divides the oxygen consumption rate during ADP-stimulated respiration by that after supplied ADP has been used. It compares reaction rates, not ATP formed per oxygen.^[ref]

A large ratio means strong stimulation under the comparison's conditions; it is not a universal measure of all mitochondrial functions.

For a constructed example, choose constant suspension volume $V=2\mathrm{mL}$ and oxygen concentration declines of $0.030$ and $0.006\mathrm{mmol}{\mathrm{L}}^{-1}\mathrm{mi}{\mathrm{n}}^{-1}$ during the two comparison windows. Assume no gas exchange or dilution and a correctly calibrated readout. A derivative converts each concentration slope into an amount of substance consumption rate:The common volume cancels in the ratio, so dividing the positive slope magnitudes gives the same answer. These are chosen inputs, not the 1955 observations. If an uncorrected influx of $3\mathrm{nmol}\mathrm{mi}{\mathrm{n}}^{-1}$ of oxygen occurred in both windows, the apparent rates would instead be $57$ and $9$, giving $19/3\approx 6.33$. The same absolute error can inflate a ratio because it removes a larger fraction of its smaller denominator.

A reversal potential $E_{\mathrm{rev}}$ is a membrane potential at which a specified open ion channel, or specified combination of ion channels, carries zero net electric current, with the surrounding conditions fixed; the electric current changes sign across that voltage. Specifying which current matters: an individual ion contribution and a sum of several contributions can reverse at different voltages. Zero net electric current does not mean that all ions have stopped moving. See Purves et al., Neuroscience, 2nd edition (2001): reversal of a mixed-ion channel current.

Throughout these examples, membrane potential is $V=V_{\mathrm{inside}}-V_{\mathrm{outside}}$, and conventional current is inward-positive. Positive electric current corresponds to net positive electric charge entering or net negative electric charge leaving. Negative electric current corresponds to the reverse charge transfer. A convenient Ohmic approximation iswhere $g$ is the available electrical conductance. For fixed $g>0$, $V<E_{\mathrm{rev}}$ gives positive electric current, $V=E_{\mathrm{rev}}$ gives zero, and $V>E_{\mathrm{rev}}$ gives negative electric current; the line has slope $-g$. If $g=0$, the modeled pathway carries zero electric current at every voltage: closing it does not locate its reversal potential. 3.1 Synapses (§3.1: conductance-based synaptic current) uses outward-positive $g(V-E)$; the formula here reverses that sign to match this wiki's inward-positive convention.

For the stipulated single-pathway example from ion channel, $E_{\mathrm{rev}}=-80\mathrm{mV}$ and $g=10\mathrm{nS}$ giveThe open ion channel carries zero net electric current at $-80\mathrm{mV}$ even though $g$ remains positive. Near that point a change of membrane potential changes the electric current, which is why zero current at one voltage does not make an open pathway electrically irrelevant; see shunting inhibition.

For an ion channel passing only one ion species by passive transport, its reversal potential can equal that species' Nernst equilibrium potential, where the chemical and electrical contributions to electrochemical potential balance. At that thermodynamic equilibrium, passage in both directions can continue with no net transport. For an ion channel admitting several ion species, opposing contributions can cancel while each species continues to have net transport. Such a reversal potential is not generally any one species' Nernst equilibrium potential. See the Purves reference above.

To work out that cancellation explicitly, put two Ohmic ion channel populations across the same cell membrane. Stipulate constant electrical conductances and fixed reversal potentials:The potassium and sodium contributions add because both pathways experience the same membrane potential:Thus $I_{\mathrm{total}}=12\mathrm{nS}(E_{\mathrm{eff}}-V)$. This weighted mean follows simply by expanding and regrouping the two current terms; it assumes additive Ohmic contributions and a positive total electrical conductance. At $V=E_{\mathrm{eff}}$, the inward-positive individual electric currents areThe potassium contribution is outward and the sodium contribution inward. Equal and opposite electric currents give $I_{\mathrm{total}}=0$; neither pathway has stopped transporting its ions. The calculation treats two distinct pathways. Applying it to one mixed-species ion channel would additionally require a justified decomposition into independent Ohmic contributions; a general mixed-channel reversal potential cannot be obtained by assuming those hypotheses silently.

This is an instantaneous balance of the specified passive electric currents. It would be the stationary voltage of a passive capacitance model containing only these constant pathways and no applied electric current, but it is not a complete prediction of a living cell's resting membrane potential. Other electric currents, voltage-dependent gating, and pumps involved in active transport can matter. Pumps that transfer net electric charge are electrogenic and contribute their own electric current. See Purves et al.: functional properties of the sodium/potassium pump.

The toy model holds ion concentrations and therefore the stipulated reversal potentials fixed. Continuing inward sodium and outward potassium transport would change finite reservoirs unless active transport or external control maintained them. Transport can be a small fraction of the reservoir's ions over a short interval, making fixed concentrations a useful approximation; it is not a claim that the reservoirs never change. See Gerstner et al., Neuronal Dynamics (2014), §2.1: equilibrium potentials, channel currents and maintained gradients.

A reversal potential makes the effect of opening an ion channel predictable from the present membrane potential. The same open pathway can carry inward or outward electric current as $V$ crosses its reversal potential. The two-pathway example reveals a particularly useful distinction: zero net charge transfer can coexist with continuing exchange of different substances. That distinction connects membrane transport to electric current measurements without mistaking electrical balance for thermodynamic equilibrium.

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.

Construct a minimal model with one substrate and no appreciable accumulated reaction product. Only the enzyme-substrate complex $ES$ produces product, with catalytic rate constant $k_{\mathrm{cat}}$; $EI$ and $ESI$ are inactive. Write $s=[S]$ and $I=[\mathrm{I}]$ for free molar concentrations, and $E_T$ for total enzyme molar concentration. Assume a suitable quasi-steady-state approximation, rapid inhibitor binding and release relative to turnover, and negligible depletion of free inhibitor. In this construction the population relations areHere $K_M$ is the uninhibited Michaelis constant, and $K_i,K_i^{\prime }$ are the inhibition constants for the two indicated binding branches. The symbol $I$ denotes a free concentration, not an amount added to the tube. Define $V_{\max }=k_{\mathrm{cat}}{E}_T$. Conservation of enzyme gives an original short derivation:Thus fitting a Michaelis-Menten equation at a fixed $I$ would return the apparent parameters“Apparent” means conditional on that inhibitor concentration. The uninhibited Michaelis constant has not been redefined. Dimensional analysis checks the denominator: both $\alpha K_M$ and ${\alpha }^{\prime }s$ have units of molar concentration; the resulting $v$ has units of reaction rate.

The shared numerical choices make the mechanisms comparable. At $I=0$ and $s=20\mu \mathrm{M}$, all three models give $v=50\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. At $I=10\mu \mathrm{M}$, the respective reaction rates are $100/3$, $25$, and $30\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$; at $I=30\mu \mathrm{M}$ they are $20$, $12.5$, and $50/3\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. These are independently chosen teaching models, not measured inhibition constants for a named enzyme.

Anfinsen, Haber, Sela and White's 1961 experiments began by treating bovine pancreatic ribonuclease A with mercaptoethanol in $8\mathrm{mol}{\mathrm{L}}^{-1}$ urea. Urea promotes unfolding; mercaptoethanol breaks protein disulfide bonds. The peptide bonds along the chain remain intact. After separating the chain from those reagents, the authors allowed its reactive sulfur groups to form protein disulfide bonds in the presence of molecular oxygen, at pH $8.2$ and temperature $23$–$24{}^{\circ }\mathrm{C}$. They examined protein concentrations of $0.1$ and $2.0\mathrm{mg}\mathrm{m}{\mathrm{L}}^{-1}$. Measurements showed recovery of native properties under these conditions; the isolated major component had normal activity toward both RNA and a synthetic substrate. See Anfinsen, Haber, Sela and White, The Kinetics of Formation of Native Ribonuclease during Oxidation of the Reduced Polypeptide Chain, PNAS 47, 1309–1314 (1961), pp. 1309–1311.

The observation supports the sufficiency of that intact sequence to recover native protein structure in the tested surroundings without a folded template. It does not establish that every protein refolds, that the original living biological cell needed no assistance, or that activity always returns immediately. Protein folding, forming protein disulfide bonds and recovering an active site are related processes rather than interchangeable measurements.

Anfinsen's lecture also discusses forming protein disulfide bonds while the chain remains in concentrated urea. This gave mostly inactive material; after removing urea, a small amount of mercaptoethanol permitted rearrangement toward the native form. As an exact combinatorial check, eight labeled reactive sulfur groups have $7\times 5\times 3\times 1=105$ ways to be paired into four protein disulfide bonds: choose the partner of one group, then repeat with the unpaired groups. Only one pairing is the native one. Equal counting does not mean that all pairings are equally likely experimentally, nor that actual protein folding searches them uniformly. See Anfinsen's Nobel lecture, 1973 publication: scrambled ribonuclease and the 105 pairings, p. 223.

A distinct experiment in An Automated, Highly Selective Reaction Monitoring Instrument Using Molecular Rotational Resonance Spectroscopy by Ann Adele Byars and others (Figure 1 and Figure 3) tracked exchange of hydrogen for deuterium in ethyl acetoacetate mixed with deuterated methanol. A pump circulated chemical solution through a valve; a calibrated loop injected aliquots into a temperature-controlled inlet, held at room temperature for this exchange experiment. Neon carried the vaporized sample to a cavity spectrometer. Separately recorded broadband spectra and fitted rotational Hamiltonians identified transitions for the four isotopic forms before targeted monitoring.^[ref]

The readout was each assigned line's signal over reaction time, normalized by the sum of the four signals. Transitions with the same quantum numbers reduced response-factor differences. A solvent wash addressed observed sample carryover. The signals distinguished the unsubstituted, two singly substituted, and doubly substituted forms. Only one gas-phase structural form was detected; the authors explicitly caution that their readout does not recover the structural chemical equilibrium in chemical solution.^[ref]

This example uses nonlinear molecules, so neither the single-constant linear rigid rotor nor the synthetic comb above is a fit to its spectrum. Its connection to the model is the sensitivity of moments of inertia to isotope placement. Line position helps assign a species; signal size additionally depends on sample transfer, initial quantum state populations and instrument response. Matching quantum numbers is a useful control, not a proof of identical response factors. These distinctions explain why resolving species and determining their concentrations are separate inference steps.