The oxygen-haemoglobin dissociation curve

Concept

The oxygen-haemoglobin dissociation curve

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Statement

Let haemoglobin be a tetramer carrying four O₂-binding sites, in rapid reversible equilibrium with dissolved oxygen at fixed temperature, pH, carbon dioxide tension, chloride and 2,3-bisphosphoglycerate concentration, and let the partial pressure \(p\) of O₂ stand for its activity (Henry’s law). Then the fractional saturation \(Y\) is determined by the binding polynomial \(Z(p)=\sum_{i=0}^{4}\beta_i p^{i}\), whose coefficients are the overall Adair association constants \(\beta_i=K_1K_2\cdots K_i\) (\(\beta_0=1\)), through \[ Y=\frac{1}{4}\,\frac{d\ln Z}{d\ln p}=\frac{1}{4}\cdot\frac{\beta_1p+2\beta_2p^{2}+3\beta_3p^{3}+4\beta_4p^{4}}{1+\beta_1p+\beta_2p^{2}+\beta_3p^{3}+\beta_4p^{4}}, \] and the local Hill coefficient \(n_H=d\ln\!\left(Y/(1-Y)\right)/d\ln p\) obeys the exact identity \(n_H=4\operatorname{Var}(i)\big/\!\left(\langle i\rangle(4-\langle i\rangle)\right)\), where \(i\) is the number of occupied sites on a tetramer drawn at random. Consequently \(n_H\equiv1\) — a rectangular hyperbola — if and only if the four sites bind independently and identically; \(n_H\le 4\) always, with equality only if no partly liganded tetramer is populated; and the sigmoid measured for human adult haemoglobin, \(n_H\approx2.8\) with \(P_{50}\approx26\ \mathrm{mmHg}\) at pH 7.4 and 37 °C, is a statement about cooperativity between sites, not about their number.

Why it matters

Receptor-ligand binding derives the hyperbolic law that a single site obeys under mass action: half-saturation at the dissociation constant, and a response that needs an eighty-one-fold change in ligand to move from 10% to 90% occupancy. A transport protein obeying that law would be useless. Blood must load almost completely at an alveolar \(P_{\mathrm{O_2}}\) near \(100\ \mathrm{mmHg}\) and give up a large fraction of its cargo at a mixed-venous \(P_{\mathrm{O_2}}\) near \(40\ \mathrm{mmHg}\) — a range spanning a factor of only two and a half. The oxygen–haemoglobin dissociation curve is the quantitative demonstration that a protein can beat the hyperbola by making its sites talk to one another, and this page is where the curve stops being a shape to memorise and becomes an object with a derivation, an exact steepness bound, and a measurable free energy of interaction.

The curve is also the cleanest worked example in physiology of thermodynamic linkage, the idea that whatever changes a protein’s affinity for one ligand is itself bound differentially by that protein. Protons, CO₂, chloride and 2,3-BPG all bind more tightly to deoxyhaemoglobin than to oxyhaemoglobin, and the same single cross-derivative that says so also says that oxygen binding must expel them — the Bohr effect and the Haldane effect are one equation read in two directions, derived in Step 13 below. That machinery generalises directly: Allosteric regulation states the same relations for enzymes, and enzyme kinetics is the non-cooperative limit of the same binding polynomial. Clinically the curve is the reason a pulse oximeter reading of 90% is an emergency while 98% and 100% are nearly the same thing, and the reason anaemia, carbon monoxide poisoning and a stored unit of blood each damage oxygen transport in a different way, as gas exchange surfaces and the principle of mass transport set up and the worked examples below quantify.

Hypotheses

Binding is at equilibrium, and fast compared with the transit of a red cell through a capillary.The curve is a thermodynamic object: it says what saturation a cell would reach at a given \(P_{\mathrm{O_2}}\), given time. A pulmonary capillary transit lasts roughly \(0.75\ \mathrm{s}\) at rest and shortens with cardiac output, so in heavy exercise or in interstitial lung disease equilibration may be incomplete and the measured saturation falls below the value the curve predicts — a kinetic failure, not a shift of the curve.

The functional unit is one tetramer with four sites, and it does not dissociate.Haemoglobin dissociates reversibly into \(\alpha\beta\) dimers, and dimers bind oxygen with high affinity and no cooperativity at all. At the \(\approx5\ \mathrm{mM}\) tetramer concentration inside a red cell the dimer fraction is negligible; in a dilute solution in a cuvette it is not, and the apparent \(n_H\) of a diluted sample falls toward \(1\) for that reason alone.

All heterotropic conditions are held fixed: pH, \(P_{\mathrm{CO_2}}\), temperature, \([\mathrm{Cl^-}]\) and \([\text{2,3-BPG}]\).Each of these is a ligand in its own right and each shifts \(P_{50}\). A curve is therefore a curve at stated conditions; comparing two \(P_{50}\) values measured under different conditions conflates an intrinsic affinity difference with an environmental shift. Effector-stripped haemoglobin in dilute buffer has a far higher affinity than whole blood, which is why in-vivo \(P_{50}\) cannot be predicted from purified protein.

Free O₂ in solution is negligible next to bound O₂, and \(p\) is not depleted by binding.Physiological plasma dissolves about \(0.003\ \mathrm{mL}\) of O₂ per decilitre per mmHg, so at \(100\ \mathrm{mmHg}\) the dissolved pool is \(0.3\ \mathrm{mL\,dL^{-1}}\) against roughly \(20\ \mathrm{mL\,dL^{-1}}\) carried by haemoglobin. The assumption is what lets \(p\) be treated as a free variable rather than as a quantity to be solved for simultaneously; it fails in hyperbaric oxygen, where the dissolved term becomes large enough to supply resting metabolism on its own.

The haemoglobin population is chemically homogeneous.The identity \(n_H=4\operatorname{Var}(i)/(\langle i\rangle(4-\langle i\rangle))\) refers to the variance over ligation states of one species. Mixing species with different \(P_{50}\) — HbA with HbF, or normal tetramers with carboxy- and methaemoglobin — adds between-species variance to \(\langle i\rangle\) but flattens the observable curve, so a mixture cannot be read as a single Adair system, and its apparent Hill coefficient is not a measure of anyone’s cooperativity.

For the MWC form only: exactly two quaternary states, T and R, exist and all four subunits change together (the symmetry postulate).The Adair polynomial needs no structural assumption whatever — it is exact for any four-site protein. The two-state form of Step 12 is a mechanistic restriction that forces the four Adair constants to be functions of just two parameters \(L\) and \(c\), and it fails wherever tertiary changes within a quaternary state contribute to affinity, or where the sequential (KNF) pathway is realised, as it must be for negative cooperativity, which MWC cannot produce at all.

Proof

Write \(\mathrm{Hb}_i\) for a tetramer carrying \(i\) bound oxygens, \(p\) for the partial pressure of O₂, and let \(K_i\) be the macroscopic stepwise association constant for the \(i\)-th oxygen. Everything below is exact for a four-site protein; the two-state mechanism enters only at Step 12, and physiological numbers only in the worked examples.

\[ \mathrm{Hb}_{i-1}+\mathrm{O_2}\rightleftharpoons\mathrm{Hb}_i,\qquad K_i=\frac{[\mathrm{Hb}_i]}{[\mathrm{Hb}_{i-1}]\,p} \quad\Longrightarrow\quad [\mathrm{Hb}_i]=\beta_i p^{i}[\mathrm{Hb}_0],\quad \beta_i=\prod_{j=1}^{i}K_j \]

Apply the law of mass action to each of the four steps and telescope the product. Nothing is assumed about the mechanism: any equilibrium mixture of the five ligation states can be written this way, with \(\beta_i\) simply naming the equilibrium constant for adding \(i\) oxygens at once. A

\[ Z(p)\equiv\sum_{i=0}^{4}\beta_ip^{i},\qquad f_i=\frac{[\mathrm{Hb}_i]}{\sum_j[\mathrm{Hb}_j]}=\frac{\beta_ip^{i}}{Z(p)},\qquad \sum_{i=0}^{4}f_i=1 \]

Divide by total haemoglobin. \(Z\) is the binding polynomial — formally a grand partition function with \(p\) playing the role of an absolute activity — and \(f_i\) is the probability that a tetramer drawn at random carries \(i\) oxygens. Conservation of protein is the whole content of the normalisation. A

\[ \langle i\rangle=\sum_{i=0}^{4}i f_i=\frac{\sum_i i\beta_ip^{i}}{Z},\qquad \frac{d\ln Z}{d\ln p}=\frac{p}{Z}\frac{dZ}{dp}=\frac{\sum_i i\beta_ip^{i}}{Z}=\langle i\rangle,\qquad Y\equiv\frac{\langle i\rangle}{4} \]

Wyman’s linkage identity: the mean occupancy is the logarithmic derivative of the binding polynomial. Differentiating \(Z\) term by term multiplies each coefficient by \(i\), which is exactly the numerator of the mean. Fractional saturation is mean occupancy per site, so \(Y\) is now a derived quantity, not a definition to be fitted. B

\[ \frac{d\langle i\rangle}{d\ln p}=\frac{d^{2}\ln Z}{d(\ln p)^{2}}=\frac{\sum_i i^{2}\beta_ip^{i}}{Z}-\left(\frac{\sum_i i\beta_ip^{i}}{Z}\right)^{2}=\langle i^{2}\rangle-\langle i\rangle^{2}=\operatorname{Var}(i) \]

Differentiate Step 3 once more. The steepness of the binding curve on a logarithmic pressure axis is the variance of the ligation state: a curve is steep exactly when the population is spread over widely separated ligation states at the same pressure. This fluctuation–response relation is what makes the bound in Step 9 possible. B

\[ K_i=\frac{4-i+1}{i}\,k_i \quad\Longrightarrow\quad \beta_i=\binom{4}{i}\prod_{j=1}^{i}k_j,\qquad \text{sites independent and identical}\ (k_j\equiv k)\ \Longrightarrow\ Z=(1+kp)^{4} \]

Separate statistics from chemistry. There are \(4-i+1\) empty sites to fill on \(\mathrm{Hb}_{i-1}\) and \(i\) filled sites to vacate on \(\mathrm{Hb}_i\), so the macroscopic constant exceeds the intrinsic per-site constant \(k_i\) by that ratio of multiplicities. With all \(k_j\) equal the sum collapses by the binomial theorem — a factorised \(Z\) is the algebraic signature of independence. B

\[ Z=(1+kp)^{4}\ \Longrightarrow\ Y=\frac{1}{4}\frac{d\ln Z}{d\ln p}=\frac{1}{4}\cdot\frac{4kp}{1+kp}=\frac{kp}{1+kp}=\frac{p}{p+1/k} \]

The independent case reproduces the hyperbola of receptor-ligand binding, with \(P_{50}=1/k\) equal to the intrinsic dissociation constant. Myoglobin, a monomer, obeys this exactly. Any deviation from a hyperbola is therefore evidence that the sites are not independent — which is the whole logical basis for reading a sigmoid as cooperativity. A

\[ n_H\equiv\frac{d}{d\ln p}\ln\frac{Y}{1-Y}=\frac{1}{Y(1-Y)}\frac{dY}{d\ln p} =\frac{\operatorname{Var}(i)/4}{\frac{\langle i\rangle}{4}\left(1-\frac{\langle i\rangle}{4}\right)}=\frac{4\operatorname{Var}(i)}{\langle i\rangle\left(4-\langle i\rangle\right)} \]

Define the Hill coefficient as the local slope of the Hill plot, \(\ln[Y/(1-Y)]\) against \(\ln p\), then substitute \(dY/d\ln p=\operatorname{Var}(i)/4\) from Step 4 and \(Y=\langle i\rangle/4\) from Step 3. This identity is the centre of the page: it converts an empirical steepness index into a statement about the width of the ligation-state distribution, and it holds for every four-site system, cooperative or not. B

\[ \text{independent sites}\Rightarrow i\sim\mathrm{Binomial}(4,\theta),\ \theta=\frac{kp}{1+kp} \quad\Longrightarrow\quad n_H=\frac{4\cdot4\theta(1-\theta)}{4\theta\,(4-4\theta)}=1\ \ \text{for every }p \]

With a factorised \(Z\) each site is an independent Bernoulli trial, so \(\langle i\rangle=4\theta\) and \(\operatorname{Var}(i)=4\theta(1-\theta)\), and Step 7 returns \(1\) identically. Together with Step 6 this makes the converse precise: \(n_H\equiv1\) on an interval iff the Hill plot is a line of unit slope iff the curve is a hyperbola. A single measured \(n_H=1\) at one pressure does not prove independence; the identity must hold everywhere. B

\[ 0\le i\le4\ \Longrightarrow\ \mathbb{E}\big[(4-i)(i-0)\big]\ge0\ \Longrightarrow\ \operatorname{Var}(i)\le\langle i\rangle\big(4-\langle i\rangle\big) \quad\Longrightarrow\quad n_H\le4 \]

The Bhatia–Davis inequality. Expanding \(\mathbb{E}[(4-i)i]\ge0\) gives \(\langle i^{2}\rangle\le4\langle i\rangle\), hence \(\operatorname{Var}(i)\le4\langle i\rangle-\langle i\rangle^{2}=\langle i\rangle(4-\langle i\rangle)\); substituting in Step 7 bounds the Hill coefficient by the number of sites. Equality demands \((4-i)i=0\) with probability one, i.e. every tetramer is either empty or full: the all-or-none limit in which \(Z=1+\beta_4p^{4}\) and \(Y=\beta_4p^{4}/(1+\beta_4p^{4})\). A Hill coefficient is therefore a lower bound on the number of interacting sites and never an estimate of it. C

\[ p\to0:\ \frac{Y}{1-Y}\to\frac{\beta_1}{4}p=k_1p; \qquad p\to\infty:\ 1-Y\to\frac{\beta_3}{4\beta_4\,p}\ \Longrightarrow\ \frac{Y}{1-Y}\to4K_4p=k_4p \]

Keep the leading term at each end. At vanishing pressure only \(\mathrm{Hb}_0\) and \(\mathrm{Hb}_1\) are populated, so binding is that of the first site alone; at saturating pressure only \(\mathrm{Hb}_3\) and \(\mathrm{Hb}_4\) are, so it is that of the last site alone. Both asymptotes have unit slope on the Hill plot — cooperativity is invisible in the wings, because a lone binding event has no partner to cooperate with — and their intercepts are the intrinsic constants \(k_1\) and \(k_4\), using \(K_1=4k_1\) and \(K_4=k_4/4\) from Step 5. B

\[ \Delta G_{\mathrm I}=-RT\ln\frac{k_4}{k_1},\qquad \text{and near }Y=\tfrac12:\quad \ln\frac{Y}{1-Y}\approx n_H\ln\frac{p}{P_{50}} \ \Longleftrightarrow\ Y\approx\frac{p^{\,n_H}}{P_{50}^{\,n_H}+p^{\,n_H}} \]

The horizontal separation of the two asymptotes of Step 10 is \(\ln(k_4/k_1)\), and multiplying by \(-RT\) turns it into the free energy of interaction — the extra binding energy the fourth oxygen enjoys because three are already bound. The Hill equation itself is then just the statement that \(n_H\) is nearly constant over the central decade: integrate the definition in Step 7 with \(n_H\) treated as constant and fix the constant of integration by \(Y(P_{50})=\tfrac12\). It is an interpolation between the asymptotes, valid where it is fitted and wrong in both wings, where the true slope returns to \(1\). B

\[ Z_{\mathrm{MWC}}(\alpha)=(1+\alpha)^{4}+L(1+c\alpha)^{4},\qquad \alpha=\frac{p}{K_R},\ \ c=\frac{K_R}{K_T},\ \ L=\frac{[\mathrm T_0]}{[\mathrm R_0]} \quad\Longrightarrow\quad Y=\frac{\alpha(1+\alpha)^{3}+Lc\alpha(1+c\alpha)^{3}}{(1+\alpha)^{4}+L(1+c\alpha)^{4}} \]

A mechanistic realisation. Within each quaternary state the four sites are independent, so each contributes a factorised polynomial as in Step 5; the two states are alternatives, so their polynomials add, weighted by the unliganded equilibrium constant \(L\). Applying \(Y=\frac14 d\ln Z/d\ln\alpha\) gives the Monod–Wyman–Changeux saturation function with only two free parameters. Cooperativity requires both \(L\gg1\) (the ligand-free protein is mostly in the low-affinity state) and \(c\ll1\) (the states really differ); if \(c=1\) then \(Z=(1+\alpha)^{4}(1+L)\) and Step 6 returns a hyperbola for every \(L\). B

\[ \left(\frac{\partial\langle i\rangle}{\partial\ln x}\right)_{p}=\frac{\partial^{2}\ln Z}{\partial\ln p\,\partial\ln x}=\left(\frac{\partial\langle\nu\rangle}{\partial\ln p}\right)_{x} \quad\Longrightarrow\quad \left(\frac{\partial\log_{10}P_{50}}{\partial\mathrm{pH}}\right)=\frac{1}{n_H}\left(\frac{\partial\langle\nu\rangle}{\partial\ln p}\right)_{Y=1/2} \]

Heterotropic linkage. Let \(x=[\mathrm H^{+}]\) and let \(\langle\nu\rangle=\partial\ln Z/\partial\ln x\) be the mean number of protons bound; equality of mixed partial derivatives of \(\ln Z\) is Wyman’s reciprocity. Now define \(P_{50}\) implicitly by \(\langle i\rangle(\ln P_{50},\ln x)=2\) and differentiate: \(\operatorname{Var}(i)\,d\ln P_{50}+(\partial\langle i\rangle/\partial\ln x)\,d\ln x=0\), where \(\operatorname{Var}(i)=n_H\) at \(Y=\tfrac12\) by Step 7. Converting \(d\ln x=-\ln\!10\;d\mathrm{pH}\) gives the displayed Bohr coefficient. If proton release simply tracks oxygenation, \(\langle\nu\rangle=\nu_0-\Delta\nu_H\langle i\rangle\), it collapses to \(\partial\log_{10}P_{50}/\partial\mathrm{pH}=-\Delta\nu_H\): the Bohr coefficient is minus the number of protons expelled per oxygen bound. The same identity read the other way is the Haldane effect, deoxygenation increasing proton and carbamate binding. C

\[ \beta_4=K_1K_2K_3K_4=k_1k_2k_3k_4,\qquad P_m\equiv\beta_4^{-1/4},\qquad \Delta G^{\circ}_{\text{total}}=-RT\ln\beta_4=4RT\ln P_m \]

The statistical factors of Step 5 multiply to \(4\cdot\frac32\cdot\frac23\cdot\frac14=1\), so the overall constant is the plain product of the intrinsic constants and is free of combinatorial contamination. \(P_m\) is Wyman’s median ligand activity: the pressure that divides the area between the saturation curve and its asymptotes into two equal halves, and the only point on the curve that reports the total binding free energy independently of any model. When the Hill plot is symmetric about its midpoint (\(k_2k_3=k_1k_4\)) the median coincides with \(P_{50}\), which is why \(P_{50}\) is usable as a thermodynamic quantity for haemoglobin even though in general it is not one. C

Result

\[ Y=\frac{1}{4}\frac{d\ln Z}{d\ln p},\quad Z=\sum_{i=0}^{4}\beta_ip^{i} \qquad\Longrightarrow\qquad n_H=\frac{4\operatorname{Var}(i)}{\langle i\rangle\left(4-\langle i\rangle\right)}\le4, \qquad Y\approx\frac{p^{\,n_H}}{P_{50}^{\,n_H}+p^{\,n_H}} \]

Reading. Saturation is the logarithmic derivative of the binding polynomial, and the steepness of the curve is the variance of the ligation state. Independent sites make that variance binomial and force a hyperbola with \(n_H=1\); cooperativity broadens the distribution, raising \(n_H\) toward but never past the number of sites. The Hill equation on the right is not a mechanism but a two-parameter interpolation between the two unit-slope asymptotes, exact only in the all-or-none limit.

Scope. Equilibrium, one homogeneous tetrameric species, fixed pH, \(P_{\mathrm{CO_2}}\), temperature, chloride and 2,3-BPG. Standard reference values used throughout: human adult haemoglobin in whole blood at pH 7.4, \(P_{\mathrm{CO_2}}=40\ \mathrm{mmHg}\), 37 °C has \(P_{50}\approx26.6\ \mathrm{mmHg}=3.5\ \mathrm{kPa}\) and \(n_H\approx2.8\); myoglobin is hyperbolic with \(P_{50}\approx2.8\ \mathrm{mmHg}\); fetal haemoglobin in whole fetal blood has \(P_{50}\approx19\ \mathrm{mmHg}\); the Bohr coefficient of whole blood is \(\partial\log_{10}P_{50}/\partial\mathrm{pH}\approx-0.48\). Oxygen capacity is taken as \(1.34\ \mathrm{mL\,O_2}\) per gram of haemoglobin (Hüfner’s constant, in-vivo value) and plasma solubility as \(0.003\ \mathrm{mL\,dL^{-1}mmHg^{-1}}\). Conversion: \(1\ \mathrm{kPa}=7.50\ \mathrm{mmHg}\).

Corollaries & converses

Hyperbola theorem, both directions. \(Y\) is a rectangular hyperbola \(\iff\) \(Z\) factorises as \((1+kp)^4\) \(\iff\) the sites are independent and identical \(\iff\) \(n_H\equiv1\). Myoglobin realises this case; so does haemoglobin dissociated into \(\alpha\beta\) dimers.
The Hill coefficient is a lower bound, not a count. \(n_H\le4\) with equality only when intermediates are unpopulated (Step 9). Human \(n_H\approx2.8\) says the four sites interact and that partly liganded tetramers are genuinely present at equilibrium; a measured \(n_H\) of \(2.8\) is compatible with four sites and with any larger number, never with fewer than three.
Cooperative free energy. \(\Delta G_{\mathrm I}=-RT\ln(k_4/k_1)\) is the only model-free measure of cooperativity available from a binding curve, since it uses the two asymptotes rather than the fitted middle. It is independent of the standard state, being a ratio of constants with identical units.
Bohr and Haldane are one relation. Step 13 gives \(\partial\log_{10}P_{50}/\partial\mathrm{pH}=-\Delta\nu_H\): acid lowers oxygen affinity by exactly as much as oxygenation raises proton release. Hence CO₂-loaded, acidified capillary blood unloads more oxygen, and deoxygenated venous blood carries more CO₂ at the same \(P_{\mathrm{CO_2}}\), from a single cross-derivative.
Why the sigmoid is asymmetric in usefulness. Because the curve is flat above \(80\ \mathrm{mmHg}\) and steep between \(20\) and \(50\ \mathrm{mmHg}\), a right shift costs almost nothing at the lung and gains a great deal at the tissue (Example 2), which is what makes the Bohr effect adaptive rather than merely a curiosity.
2,3-BPG as a T-state ligand. One BPG molecule binds in the central cavity of the deoxy tetramer, present in the red cell at a concentration comparable with haemoglobin itself. Adding a ligand that binds only T raises \(L\) in Step 12, which lowers affinity and raises \(P_{50}\) — the mechanism of altitude adaptation and of the loss and restoration of affinity in stored blood.
Median ligand activity. \(P_m=\beta_4^{-1/4}\) reports the total oxygenation free energy exactly, whatever the mechanism (Step 14). For a symmetric Hill plot \(P_m=P_{50}\), so the everyday \(P_{50}\) doubles as a thermodynamic quantity for haemoglobin, though not in general.

Fails without

Equilibrium dropped (diffusion-limited transit): at rest a red cell spends about \(0.75\ \mathrm{s}\) in a pulmonary capillary and equilibrates within roughly a third of it, so the curve applies. In severe exercise the transit time falls with rising cardiac output, and where the alveolar–capillary barrier is thickened the flux is diffusion limited; end-capillary saturation is then below the curve’s prediction at the same \(P_{\mathrm{O_2}}\). The failure is diagnostic: it worsens with exercise and with hypoxic gas, whereas a genuine shift of the curve does not.
Independence assumed where it does not hold, or cooperativity assumed where it has been lost: the whole page is the statement that these two regimes give different physiology. Dilute haemoglobin dissociates into \(\alpha\beta\) dimers whose binding is hyperbolic and high-affinity, so a cell-free haemoglobin solution binds oxygen tightly and unloads it poorly — one of the reasons early haemoglobin-based oxygen carriers failed as transfusion substitutes. Example 1 quantifies the cost: with the same \(P_{50}\), a non-cooperative carrier unloads about \(13\%\) less oxygen per litre of blood and leaves the arterial blood only \(79\%\) saturated.
Fixed heterotropic conditions dropped: a \(P_{50}\) quoted without pH, temperature, \(P_{\mathrm{CO_2}}\) and BPG status is meaningless. Effector-stripped haemoglobin has a far higher oxygen affinity than the same protein inside a red cell, and blood stored in citrate–phosphate–dextrose loses 2,3-BPG over weeks, shifting the curve left; transfused, it holds its oxygen until BPG is resynthesised over hours to a day.
Chemical homogeneity dropped (CO, methaemoglobin): carbon monoxide binds haemoglobin roughly two hundred times more tightly than oxygen, so a small inspired fraction converts much of the pigment to carboxyhaemoglobin; worse, occupying some sites shifts the remaining sites toward the high-affinity state, so the residual curve is both lowered and shifted left. Anaemia halves the capacity while leaving \(Y(p)\) untouched; CO poisoning at the same reduction in capacity is far more dangerous, and standard pulse oximetry cannot distinguish the two.
Negligible dissolved oxygen dropped (hyperbaric oxygen): at \(3\) atmospheres of pure O₂ the dissolved term reaches roughly \(6\ \mathrm{mL\,dL^{-1}}\), enough to meet resting extraction with no contribution from haemoglobin at all. The saturation curve is then not the relevant transport law, which is the therapeutic basis of hyperbaric treatment for CO poisoning.

Common errors

“The Hill coefficient counts the binding sites.” It is the local slope of a log-odds plot and equals \(4\) only in the impossible all-or-none limit (Step 9). Haemoglobin has four sites and \(n_H\approx2.8\); the gap measures how much of the population sits in partly liganded states, and it is information, not error.
“\(P_{50}\) measures cooperativity.” \(P_{50}\) locates the curve; \(n_H\) shapes it. The Bohr effect, 2,3-BPG and temperature move \(P_{50}\) while leaving \(n_H\) nearly unchanged, whereas dissociation into dimers collapses \(n_H\) while raising affinity.
“Saturation is oxygen content.” Content is \(1.34\,[\mathrm{Hb}]\,Y+0.003\,p\) per decilitre. A patient with \([\mathrm{Hb}]=70\ \mathrm{g\,L^{-1}}\) and \(Y=0.98\) has a perfect oximeter reading and barely half the normal oxygen content. The pulse oximeter measures the curve, not the cargo.
“The Hill equation is derived from the reaction \(\mathrm{Hb}+n\,\mathrm{O_2}\rightleftharpoons\mathrm{Hb(O_2)}_n\).” That derivation requires \(n\) to be an integer stoichiometry and predicts a slope of \(n\) at every pressure; the measured slope is \(2.8\) in the middle and \(1\) in both wings. The equation is an interpolation (Step 11), and no molecular species with \(2.8\) sites exists.
“A right shift is harmful because it lowers affinity.” Delivery is a difference of saturations, not a saturation. Example 2 shows a right shift losing \(2\) percentage points at the lung and gaining \(15\) at the tissue — a factor of seven in favour, precisely because the curve is flat at one end and steep at the other.
“\(P_{50}\) is the dissociation constant.” Only when \(n_H=1\). For a cooperative protein \(P_{50}\) is a property of the whole polynomial; the constant that governs the first oxygen bound is \(k_1\), some two hundred times weaker than the \(k_4\) that governs the last (Problem 4).
“\(26\) and \(3.5\) are different values of \(P_{50}\).” They are the same number in mmHg and kPa. Mixing the two units is the commonest arithmetic failure in this material; \(1\ \mathrm{kPa}=7.50\ \mathrm{mmHg}\).

Discussion

The sigmoid shape was measured by Christian Bohr and colleagues at the end of the nineteenth century, and the pH dependence that bears Bohr’s name was reported in 1904; Archibald Hill wrote his interpolating equation in 1910 as a curve fit, explicitly aware that a fractional exponent could not be a stoichiometry. Gilbert Adair’s four-constant scheme of 1925 is the exact treatment, and it is what Steps 1 to 4 reconstruct. Structural explanation came much later: Max Perutz’s crystallographic work resolved the quaternary change between the deoxy (T) and oxy (R) states and identified the salt bridges, the proton-binding groups and the central cavity where 2,3-BPG sits. The order matters pedagogically — the thermodynamics was complete and quantitative decades before anyone knew what the protein looked like, because linkage relations of the kind in Step 13 constrain a mechanism without describing it.

The right way to hold the whole subject together is as one binding polynomial with several ligands in it. Oxygen, protons, CO₂ (as carbamate on the N-terminal amino groups), chloride and 2,3-BPG all appear in \(Z\); every physiological effect on the curve is then a statement that one of these species binds T and R differently, and every such statement automatically has a reciprocal partner. That is why the Bohr effect and the Haldane effect are not two facts but one, and why altitude acclimatisation (rising BPG), acidosis, fever and stored blood can all be described with the single parameter \(L\) of Step 12 without touching the intrinsic site chemistry at all.

The two-state model of Step 12 is nevertheless known to be an idealisation. Precise Adair analyses show that the four intrinsic constants are not reproduced by any single pair \((L,c)\), and that a substantial part of the cooperative free energy is associated with tertiary rearrangements within a quaternary state rather than with the T \(\to\) R switch itself; the modern account is a tertiary-two-state or ensemble picture in which the quaternary structures each contain low- and high-affinity tertiary conformations. None of this disturbs Steps 1 to 11 or 13–14, which assume no mechanism, and that is exactly the methodological point: the model-free layer of the theory — binding polynomial, variance identity, Hill bound, linkage reciprocity, median ligand — survives every revision of the structural story built on top of it. When a structural model is contradicted by data it is the model that goes; when the linkage relations appear to be contradicted, it is almost always the assumption of a single homogeneous species at equilibrium that has failed.

Common misconceptions. That cooperativity means the oxygens bind each other, or that they physically communicate through the haem irons: the coupling is conformational, mediated by movement at the \(\alpha_1\beta_2\) interface, and the haem irons are \(2.5\ \mathrm{nm}\) or more apart. That the curve explains why oxygen moves: it does not, diffusion down a partial-pressure gradient does, and the curve only says how much oxygen a given \(P_{\mathrm{O_2}}\) corresponds to. And that a left-shifted curve is always advantageous or always harmful — fetal haemoglobin is left-shifted because its job is to load from a low-\(P_{\mathrm{O_2}}\) placenta, whereas adult tissue delivery is helped by a right shift; whether a shift is good depends entirely on which end of the circulation is the bottleneck.

Worked examples

Example 1. A healthy adult at rest: \([\mathrm{Hb}]=150\ \mathrm{g\,L^{-1}}\), arterial \(P_{\mathrm{O_2}}=100\ \mathrm{mmHg}\), mixed venous \(P_{\mathrm{O_2}}=40\ \mathrm{mmHg}\), cardiac output \(5.0\ \mathrm{L\,min^{-1}}\), and a haemoglobin curve with \(P_{50}=26.6\ \mathrm{mmHg}\), \(n_H=2.8\). Find the arterial and venous saturations, the oxygen content of each, the delivery and consumption, and the extraction ratio — then repeat with a hypothetical non-cooperative haemoglobin of the same \(P_{50}\) to price the cooperativity.

\[ Y=\frac{(p/P_{50})^{n_H}}{1+(p/P_{50})^{n_H}},\qquad \left(\frac{100}{26.6}\right)^{2.8}=e^{2.8\ln 3.759}=e^{3.708}=40.8 \ \Longrightarrow\ Y_a=\frac{40.8}{41.8}=0.976 \]

The Hill form of Step 11, applied where it is valid — and note that at \(Y_a=0.976\) we are already in the flat wing, where the interpolation is least reliable; the conclusion drawn from it below (that arterial saturation is insensitive to \(P_{\mathrm{O_2}}\) here) is exactly the robust one. A

\[ \left(\frac{40}{26.6}\right)^{2.8}=e^{2.8\ln1.504}=e^{1.142}=3.134 \ \Longrightarrow\ Y_v=\frac{3.134}{4.134}=0.758,\qquad \Delta Y=0.976-0.758=0.218 \]

The same formula at the venous point. Everything the circulation achieves is in this difference of \(0.218\): only about a fifth of the cargo is unloaded at rest, which is the reserve that exercise and hypoxia draw on. A

\[ C_{\mathrm{cap}}=1.34\ \mathrm{mL\,g^{-1}}\times150\ \mathrm{g\,L^{-1}}=201\ \mathrm{mL\,L^{-1}},\qquad C=C_{\mathrm{cap}}Y+0.03\,p\ \ (p\ \text{in mmHg}) \]

Capacity first, symbolically, then content. The dissolved term is \(0.003\ \mathrm{mL\,dL^{-1}mmHg^{-1}}=0.03\ \mathrm{mL\,L^{-1}mmHg^{-1}}\); carrying it explicitly is what shows it to be small rather than assuming so. A

\[ \begin{aligned} C_a&=(201)(0.976)+(0.03)(100)=196.2+3.0=199.2\ \mathrm{mL\,L^{-1}}\\ C_v&=(201)(0.758)+(0.03)(40)=152.4+1.2=153.6\ \mathrm{mL\,L^{-1}} \end{aligned} \]

In the older clinical units these are \(19.9\) and \(15.4\ \mathrm{mL\,dL^{-1}}\). The dissolved contribution is \(1.5\%\) of arterial content, justifying the fourth hypothesis. A

\[ \dot D_{\mathrm{O_2}}=\dot QC_a=(5.0)(199.2)=996\ \mathrm{mL\,min^{-1}},\qquad \dot V_{\mathrm{O_2}}=\dot Q(C_a-C_v)=(5.0)(45.6)=228\ \mathrm{mL\,min^{-1}} \]

The Fick principle. Both numbers land on the textbook resting figures of roughly \(1\ \mathrm{L\,min^{-1}}\) delivered and \(250\ \mathrm{mL\,min^{-1}}\) consumed, which is the check that the parameters were physiological. A

\[ \mathrm{O_2ER}=\frac{C_a-C_v}{C_a}=\frac{45.6}{199.2}=0.229 \qquad\text{versus}\qquad n_H=1:\ Y_a=\frac{100}{126.6}=0.790,\ \ Y_v=\frac{40}{66.6}=0.601 \]

Extraction is \(23\%\), leaving a large reserve. Setting \(n_H=1\) at the same \(P_{50}\) — the hyperbola of Step 6 — isolates the effect of cooperativity alone, since affinity at the midpoint is unchanged. B

\[ n_H=1:\ \Delta Y=0.790-0.601=0.189,\quad C_a-C_v=(201)(0.189)+1.8=39.8\ \mathrm{mL\,L^{-1}},\quad \dot V_{\mathrm{O_2}}=(5.0)(39.8)=199\ \mathrm{mL\,min^{-1}}=0.87\times228 \]

Two separate penalties, with the dissolved term carried on both sides so that the two \(\dot V_{\mathrm{O_2}}\) values are like for like. The oxygen actually unloaded falls by \(13\%\) at the same cardiac output and haemoglobin concentration, and the arterial blood is only \(79\%\) saturated — a value that in a patient would read as respiratory failure, with no reserve left for any fall in alveolar \(P_{\mathrm{O_2}}\). B

\[ Y_a=0.976,\quad Y_v=0.758,\quad C_a=199\ \mathrm{mL\,L^{-1}},\quad \dot D_{\mathrm{O_2}}=996\ \mathrm{mL\,min^{-1}},\quad \dot V_{\mathrm{O_2}}=228\ \mathrm{mL\,min^{-1}},\quad \mathrm{O_2ER}=23\% \]

Reading. Cooperativity is worth about \(13\%\) of the oxygen unloaded at rest and nearly twenty points of arterial saturation, at identical \(P_{50}\) and identical haemoglobin mass. The sigmoid buys full loading on the flat top and generous unloading on the steep flank; a hyperbola of the same affinity can have one or the other, never both.

Scope. Steady state, equilibrium binding, one homogeneous haemoglobin species, and the same curve assumed in lung and tissue — an approximation that Example 2 removes.

Example 2. The Bohr shift in exercising muscle. Capillary blood in working muscle acidifies from pH 7.40 to \(7.20\); take the whole-blood Bohr coefficient \(\partial\log_{10}P_{50}/\partial\mathrm{pH}=-0.48\) and \(n_H=2.8\) unchanged. Find the shifted \(P_{50}\), the extra oxygen released at a tissue \(P_{\mathrm{O_2}}\) of \(25\ \mathrm{mmHg}\), and the cost at the lung at \(100\ \mathrm{mmHg}\); take \([\mathrm{Hb}]=150\ \mathrm{g\,L^{-1}}\).

\[ \Delta\log_{10}P_{50}=\left(\frac{\partial\log_{10}P_{50}}{\partial\mathrm{pH}}\right)\Delta\mathrm{pH}=(-0.48)(7.20-7.40)=+0.096 \]

Integrate the Bohr coefficient of Step 13 over a small pH change, treating it as constant — legitimate over \(0.2\) pH units, where the curve of \(\log P_{50}\) against pH is close to linear near \(7.4\). A

\[ P_{50}'=P_{50}\times10^{0.096}=(26.6)(1.248)=33.2\ \mathrm{mmHg} \]

A fifth of a pH unit moves the whole curve right by a quarter of its position. In words: \(0.48\) protons are taken up per oxygen released, which is precisely how the muscle’s own metabolic acid buys itself oxygen. A

\[ Y(25;\,26.6)=\frac{(0.940)^{2.8}}{1+(0.940)^{2.8}}=\frac{0.840}{1.840}=0.457,\qquad Y(25;\,33.2)=\frac{(0.753)^{2.8}}{1+(0.753)^{2.8}}=\frac{0.452}{1.452}=0.311 \]

Evaluate the same Hill form before and after the shift at the tissue pressure. Both points lie on the steep flank, where \(d Y/d\ln p\) is largest, so a modest horizontal displacement produces a large vertical one. A

\[ Y(100;\,33.2)=\frac{(3.012)^{2.8}}{1+(3.012)^{2.8}}=\frac{21.9}{22.9}=0.956,\qquad \Delta Y_{\text{lung}}=0.976-0.956=0.020 \]

The same shift evaluated on the flat top. Here \(dY/d\ln p\) is small — the variance of Step 4 has collapsed because nearly every tetramer is fully liganded — so the same displacement costs almost nothing. B

\[ \Delta Y_{\text{tissue}}=0.457-0.311=0.146,\qquad \frac{\Delta Y_{\text{tissue}}}{\Delta Y_{\text{lung}}}=\frac{0.146}{0.020}=7.3 \]

The asymmetry of the sigmoid, stated as a number: the acid shift gains seven times as much at the tissue as it loses at the lung. This is the quantitative content of “the Bohr effect is adaptive”. B

\[ \Delta C=1.34\times150\times0.146=29.3\ \mathrm{mL\,L^{-1}} \qquad\text{against}\qquad (201)(0.020)=4.0\ \mathrm{mL\,L^{-1}}\ \text{lost at the lung} \]

Convert saturation back to cargo with the capacity of Example 1. Roughly \(29\ \mathrm{mL}\) of extra oxygen per litre of blood is released in the acidified capillary — against a resting arteriovenous difference of \(46\ \mathrm{mL\,L^{-1}}\), an increase of nearly two-thirds from pH alone, before any change in flow, haemoglobin or \(P_{\mathrm{O_2}}\). A

\[ P_{50}:\ 26.6\to33.2\ \mathrm{mmHg},\qquad \Delta Y_{\text{tissue}}=+0.146,\qquad \Delta Y_{\text{lung}}=-0.020,\qquad \Delta C_{\text{released}}=+29\ \mathrm{mL\,L^{-1}} \]

Reading. Local chemistry retunes the carrier exactly where the retuning pays. The muscle that consumes the oxygen produces the acid that liberates it, and the lung, sitting on the flat part of the curve, barely notices; adding the temperature and CO₂ effects, which act in the same direction, makes the tissue gain larger still.

Scope. A \(0.2\)-unit pH change with the Bohr coefficient held constant, \(n_H\) unchanged by the shift, and equilibrium in both beds. Very large pH excursions violate the first, and the Bohr coefficient itself becomes less negative at extreme pH.

Problems

With \(P_{50}=26.6\ \mathrm{mmHg}\) and \(n_H=2.8\), find the \(P_{\mathrm{O_2}}\) needed for \(Y=0.75\), \(0.90\) and \(0.98\), in mmHg and in kPa. Use the answers to explain why a pulse oximeter reading of \(90\%\) is treated as serious while \(98\%\) and \(100\%\) are treated as equivalent.
Solution
Invert the Hill form symbolically first: from \(Y=(p/P_{50})^{n}/[1+(p/P_{50})^{n}]\) we get \((p/P_{50})^{n}=Y/(1-Y)\), so \[ p=P_{50}\left(\frac{Y}{1-Y}\right)^{1/n_H}. \] For \(Y=0.75\): \(Y/(1-Y)=3\), \(\ln3=1.0986\), divided by \(2.8\) gives \(0.3924\), and \(e^{0.3924}=1.480\), so \(p=(26.6)(1.480)=39.4\ \mathrm{mmHg}=5.25\ \mathrm{kPa}\). For \(Y=0.90\): \(Y/(1-Y)=9\), \(\ln9=2.1972\), \(/2.8=0.7847\), \(e^{0.7847}=2.192\), \(p=58.3\ \mathrm{mmHg}=7.77\ \mathrm{kPa}\). For \(Y=0.98\): \(Y/(1-Y)=49\), \(\ln49=3.8918\), \(/2.8=1.3899\), \(e^{1.3899}=4.015\), \(p=106.8\ \mathrm{mmHg}=14.2\ \mathrm{kPa}\).
The clinical reading follows from the spacing. Between \(98\%\) and \(90\%\) saturation lies a full \(48\ \mathrm{mmHg}\) of arterial \(P_{\mathrm{O_2}}\), whereas between \(90\%\) and \(75\%\) lies only \(19\ \mathrm{mmHg}\): the patient at \(90\%\) is at the shoulder of the curve, where the next small fall in alveolar oxygen produces a large fall in saturation, whereas the patient at \(98\%\) has tens of mmHg of reserve before saturation moves at all. The oximeter is a poor detector of early gas-exchange failure for the same reason — \(P_{\mathrm{O_2}}\) can fall from \(107\) to \(58\ \mathrm{mmHg}\) while the displayed number drops only from \(98\) to \(90\).
Myoglobin is a monomer, hyperbolic, with \(P_{50}=2.8\ \mathrm{mmHg}\). Compare its saturation with that of haemoglobin (\(P_{50}=26.6\ \mathrm{mmHg}\), \(n_H=2.8\)) at a muscle capillary \(P_{\mathrm{O_2}}\) of \(20\ \mathrm{mmHg}\), and at the \(3\ \mathrm{mmHg}\) reached in a maximally working fibre. What does the comparison say about the direction of oxygen transfer, and why must myoglobin be non-cooperative to do its job?
Solution
Myoglobin obeys Step 6, \(Y=p/(p+P_{50})\). At \(20\ \mathrm{mmHg}\): \(Y_{\mathrm{Mb}}=20/22.8=0.877\). Haemoglobin at the same pressure: \((20/26.6)^{2.8}=e^{2.8\ln0.7519}=e^{-0.798}=0.450\), so \(Y_{\mathrm{Hb}}=0.450/1.450=0.310\). At \(3\ \mathrm{mmHg}\): \(Y_{\mathrm{Mb}}=3/5.8=0.517\), while \((3/26.6)^{2.8}=e^{2.8\ln0.1128}=e^{-6.109}=0.00222\), giving \(Y_{\mathrm{Hb}}=0.0022\).
At every physiological pressure myoglobin is the more saturated of the two, so oxygen released by haemoglobin is taken up by myoglobin and not the reverse: the two proteins in series form a downhill staircase, haemoglobin unloading in the capillary and myoglobin holding the oxygen until the mitochondrial \(P_{\mathrm{O_2}}\) falls to a few mmHg. That last step is the point of being non-cooperative. A cooperative myoglobin would have a flat foot to its curve and would release nothing until the pressure had already collapsed, whereas the hyperbola is at its steepest at the very lowest pressures — \(dY/dp=P_{50}/(p+P_{50})^{2}\) is maximal at \(p=0\) — which is exactly where a tissue store must give up its contents. Cooperativity is an advantage for a shuttle working between two fixed pressures and a disadvantage for a buffer working at one very low pressure.
Fetal haemoglobin in whole fetal blood has \(P_{50}\approx19\ \mathrm{mmHg}\) against the maternal \(26.6\ \mathrm{mmHg}\), with \(n_H\approx2.8\) for both. At a placental \(P_{\mathrm{O_2}}\) of \(30\ \mathrm{mmHg}\), compute both saturations and the transfer advantage. Express the affinity difference as a free energy per oxygen at 37 °C, and identify the molecular cause.
Solution
Maternal: \((30/26.6)^{2.8}=e^{2.8\ln1.1278}=e^{0.3368}=1.400\), so \(Y_A=1.400/2.400=0.583\). Fetal: \((30/19)^{2.8}=e^{2.8\ln1.5789}=e^{1.279}=3.593\), so \(Y_F=3.593/4.593=0.782\). The fetal blood leaves the placenta \(19.9\) percentage points more saturated than the maternal blood it equilibrated with, at the same \(P_{\mathrm{O_2}}\); with a fetal haemoglobin concentration around \(170\ \mathrm{g\,L^{-1}}\) that is a large advantage in content as well.
The energy is startlingly small. Treating \(P_{50}\) as an apparent dissociation constant, \[ \Delta\Delta G=RT\ln\frac{P_{50,A}}{P_{50,F}}=(8.314)(310.15)\ln\frac{26.6}{19}\ \mathrm{J\,mol^{-1}}=(2578)(0.3364)=0.87\ \mathrm{kJ\,mol^{-1}}, \] about \(0.34\,RT\) per oxygen — a third of thermal energy, and far less than a hydrogen bond. It is worth twenty points of saturation only because the placental pressure sits on the steep flank of the curve, where \(dY/d\ln p=n_HY(1-Y)\) is near its maximum.
The cause is not a difference in intrinsic haem chemistry: stripped of effectors, adult and fetal haemoglobin have very similar oxygen affinities. It is 2,3-BPG. The \(\gamma\) chain of HbF carries serine where the \(\beta\) chain of HbA carries histidine at position \(143\), one of the residues lining the central cavity that binds BPG, so HbF binds BPG much more weakly. In the language of Step 12, less BPG bound means a smaller effective \(L\), so less of the population sits in the low-affinity state and \(P_{50}\) falls. The whole placental transfer therefore rests on a single lost salt bridge to a small anion — and on the sigmoid shape that amplifies \(0.87\ \mathrm{kJ\,mol^{-1}}\) into a physiological mechanism.
A Hill plot of a haemoglobin preparation is measured over the full range and its two asymptotes are extracted, with intercepts \(k_1=0.0026\ \mathrm{mmHg^{-1}}\) at the low-pressure end and \(k_4=0.55\ \mathrm{mmHg^{-1}}\) at the high-pressure end. (a) Compute the free energy of cooperativity at 37 °C. (b) Assuming the plot is symmetric about its midpoint (\(k_2k_3=k_1k_4\)), compute the median ligand activity \(P_m\) and check it against the measured \(P_{50}=26.6\ \mathrm{mmHg}\). (c) Compute the total free energy of binding four oxygens, and say why its numerical value is less informative than the answer to (a).
Solution
(a) From Step 11, \(\Delta G_{\mathrm I}=-RT\ln(k_4/k_1)\). The ratio is \(0.55/0.0026=212\), and \(RT=(8.314)(310.15)=2578\ \mathrm{J\,mol^{-1}}=2.58\ \mathrm{kJ\,mol^{-1}}\), so \[ \Delta G_{\mathrm I}=-(2.58)\ln(212)=-(2.58)(5.36)=-13.8\ \mathrm{kJ\,mol^{-1}}, \] about \(-3.3\ \mathrm{kcal\,mol^{-1}}\) — the standard magnitude quoted for haemoglobin’s cooperative free energy. The last oxygen binds two hundred times more tightly than the first, and that factor of \(212\), not the sigmoid’s appearance, is the physical content of cooperativity.
(b) By Step 14, \(\beta_4=k_1k_2k_3k_4\); with the symmetry assumption \(k_2k_3=k_1k_4\) this is \((k_1k_4)^{2}\), so \[ P_m=\beta_4^{-1/4}=(k_1k_4)^{-1/2}=\left[(0.0026)(0.55)\right]^{-1/2}=(1.43\times10^{-3})^{-1/2}=26.4\ \mathrm{mmHg}, \] agreeing with the measured \(P_{50}=26.6\ \mathrm{mmHg}\) to within the precision of the asymptote intercepts. The agreement is the check that the assumed symmetry is reasonable; had \(P_m\) and \(P_{50}\) differed appreciably, \(P_{50}\) could not have been used as a thermodynamic quantity.
(c) \(\Delta G^{\circ}_{\text{total}}=4RT\ln P_m=4(2.58)\ln(26.4)=4(2.58)(3.27)=+33.8\ \mathrm{kJ\,mol^{-1}}\). The positive sign and the size are artefacts of the standard state: \(P_m\) was expressed in mmHg, so this is the free energy relative to a hypothetical \(1\ \mathrm{mmHg}\) reference, a pressure far below \(P_{50}\), and choosing atmospheres instead would change the number by \(4RT\ln760\). The answer to (a) has no such defect, because \(k_4/k_1\) is a ratio of two constants with identical units: the standard state cancels. This is the general reason to quote cooperativity as an interaction free energy rather than as a binding free energy.
Take the MWC model of Step 12 with \(L=10^{5}\) and \(c=0.01\). (a) Prove that \(c=1\) gives a hyperbola for every \(L\). (b) Compute \(Y\) at \(\alpha=0.1,\ 1,\ 10,\ 20\) and locate the half-saturation point. (c) Estimate \(n_H\) there from a finite difference and explain, using Step 9, why it falls short of \(4\).
Solution
(a) With \(c=1\) the two polynomials are proportional: \(Z=(1+\alpha)^{4}+L(1+\alpha)^{4}=(1+L)(1+\alpha)^{4}\). A constant factor does not survive the logarithmic derivative, so \(Y=\frac14 d\ln Z/d\ln\alpha=\frac14\cdot\frac{4\alpha}{1+\alpha}=\alpha/(1+\alpha)\), the hyperbola of Step 6, for every \(L\). Two conformations that bind identically are, as far as binding is concerned, one conformation — a state change with no affinity change carries no cooperativity.
(b) Use \(Y=\left[\alpha(1+\alpha)^{3}+Lc\alpha(1+c\alpha)^{3}\right]/\left[(1+\alpha)^{4}+L(1+c\alpha)^{4}\right]\) with \(Lc=10^{3}\). At \(\alpha=0.1\): numerator \(=0.1[(1.1)^{3}+10^{3}(1.001)^{3}]=0.1[1.331+1003.0]=100.4\); denominator \(=(1.1)^{4}+10^{5}(1.001)^{4}=1.464+1.0040\times10^{5}=1.0040\times10^{5}\); \(Y=1.00\times10^{-3}\). At \(\alpha=1\): numerator \(=8+1030.3=1038.3\); denominator \(=16+1.0406\times10^{5}=1.0408\times10^{5}\); \(Y=9.98\times10^{-3}\). At \(\alpha=10\): numerator \(=10[1331+1331]=2.662\times10^{4}\); denominator \(=14641+1.4641\times10^{5}=1.6105\times10^{5}\); \(Y=0.165\). At \(\alpha=20\): numerator \(=20[9261+1728]=2.198\times10^{5}\); denominator \(=194481+2.0736\times10^{5}=4.018\times10^{5}\); \(Y=0.547\). Half-saturation therefore lies just below \(\alpha=20\); evaluating at \(\alpha=18\) gives \(Y=0.472\) and at \(\alpha=18.7\) gives \(Y=0.499\), so \(\alpha_{50}\approx18.7\), i.e. \(P_{50}\approx18.7\,K_R\). Note that the protein is far from its intrinsic R-state affinity: \(L\) has pushed half-saturation almost twenty-fold above \(K_R\).
(c) With \(Y(18)=0.472\) and \(Y(20)=0.547\), \[ n_H\approx\frac{\ln\!\frac{0.547}{0.453}-\ln\!\frac{0.472}{0.528}}{\ln(20/18)}=\frac{0.188-(-0.112)}{0.105}=2.85, \] very close to the \(2.8\) measured for human haemoglobin, which is why this parameter pair is a reasonable caricature of the real protein. It falls short of \(4\) because, by Step 9, \(n_H=4\) would require \(\operatorname{Var}(i)\) to attain its Bhatia–Davis maximum, i.e. every tetramer to be either empty or full. Here \(L=10^{5}\) is large but finite and \(c=0.01\) is small but non-zero, so at \(\alpha_{50}\) the population contains appreciable amounts of singly, doubly and triply liganded tetramers — in both quaternary states — and that spread of the distribution over intermediate ligation states is precisely the deficit \(4-n_H\). Driving \(L\to\infty\) and \(c\to0\) fast enough that \(Lc^{4}\to0\) (equivalently \(c\ll L^{-1/4}\), since half-saturation sits at \(\alpha_{50}\approx L^{1/4}\)) empties the intermediates and pushes \(n_H\to4\), at the price of a protein that would need an unattainably large pressure change to switch. Holding \(Lc^{4}\) fixed instead is not enough: at \(Lc^{4}=10^{-3}\) the limit \(L\to\infty\) stalls at \(n_H\approx3.03\), because the T state still binds appreciably at \(\alpha_{50}\) and keeps the intermediates populated.