Carbon Monoxide and Competitive Heme Binding

Introduction

Carbon monoxide is a small, colourless, odourless gas that binds to exactly the same site as oxygen: the ferrous iron of a haem group. Because it binds far more tightly, even a low concentration in air can occupy a significant fraction of haemoglobin. The chemistry of this competition shows how a protein pocket tunes metal–ligand selectivity, and why a poison that "only" blocks some sites can do disproportionate damage to oxygen delivery.

Core explanation

The same site, two ligands. In myoglobin and haemoglobin, each haem iron is Fe(II), bound to four porphyrin nitrogens and one proximal histidine. The sixth position faces a pocket where a diatomic ligand can bind. O₂, CO and NO all bind Fe(II) haem. The equilibrium Hb + CO ⇌ HbCO competes directly with Hb + O₂ ⇌ HbO₂.

Why CO binds so strongly. CO is a good σ-donor through the carbon lone pair and a good π-acceptor through its empty π orbitals. Low-spin Fe(II) has filled t₂g-type orbitals that back-donate into CO π , strengthening the Fe–C bond. The resulting Fe–C–O unit is close to linear. O₂ binding is different: the bound ligand adopts a bent Fe–O–O geometry (about 120°) and has considerable Fe(III)–superoxide character, so the terminal oxygen carries partial negative charge.

The pocket narrows the preference. For a simple haem in solution, CO is preferred over O₂ by a factor of roughly 10⁴. In myoglobin and haemoglobin the preference falls to a few hundred. The main reason is the distal histidine (His E7). Its N–H hydrogen-bonds to the polar, bent Fe–O–O unit and stabilises bound O₂. Bound CO is nearly non-polar and gains little from this hydrogen bond. Older textbooks emphasised steric bending of CO by the pocket; spectroscopy and crystal structures show that CO tilts only slightly, and that selective electrostatic stabilisation of O₂ is the larger effect. Mutating the distal histidine to leucine in myoglobin lowers O₂ affinity sharply while barely changing CO affinity, which confirms this interpretation.

The Haldane relationship. At equilibrium, the ratio of carboxyhaemoglobin to oxyhaemoglobin is proportional to the ratio of partial pressures:

[HbCO] / [HbO₂] = M × p(CO) / p(O₂)

For human haemoglobin, M is about 200–250. A partial pressure of CO about 1/240 of the O₂ partial pressure therefore produces equal amounts of HbCO and HbO₂ at equilibrium.

Two harmful effects, not one. First, every haem occupied by CO cannot carry O₂, so oxygen capacity falls. Second, CO bound to one subunit favours the high-affinity R state of the tetramer, so the remaining haems hold their O₂ more tightly. The oxygen dissociation curve shifts to the left and becomes less sigmoidal. Blood with 50% HbCO delivers far less oxygen to tissue than blood from an anaemic person with half the normal haemoglobin, because in the anaemic case the remaining haemoglobin still unloads normally.

Beyond haemoglobin. CO also binds myoglobin and the reduced haem a₃ site of cytochrome c oxidase, interfering with cellular respiration. Paradoxically, cells make small amounts of CO themselves when haem oxygenase breaks down haem; this endogenous CO acts as a signalling molecule.

Formulae

Haldane relationship: [HbCO]/[HbO₂] = M × p(CO)/p(O₂), with M ≈ 200–250 for human haemoglobin. Fraction COHb (ignoring deoxy sites) = r/(1 + r), where r = M × p(CO)/p(O₂).

Step-by-step reasoning

1. Identify the binding site: Fe(II) haem with a free sixth position. 2. Compare ligand bonding: CO is a strong σ-donor and π-acceptor, linear on iron; O₂ is bent and polar. 3. Add the protein: the distal histidine hydrogen-bonds to O₂, not effectively to CO, reducing CO selectivity. 4. Apply the Haldane relationship to estimate equilibrium occupancy. 5. Add allostery: CO binding shifts remaining subunits towards the R state, hindering O₂ release.

Visual explanation

Sketch a haem plane with iron at the centre. Below it, draw the proximal histidine. Above it, draw two alternatives: a straight Fe–C–O stick, and a bent Fe–O–O stick with a dashed hydrogen bond to the distal histidine N–H. Then draw two oxygen dissociation curves: the normal sigmoidal curve, and a lower, left-shifted, more hyperbolic curve for blood containing HbCO.

Real-world analogy

Imagine a car park where a few long-stay vehicles occupy spaces all day. Fewer spaces are available, and the remaining short-stay cars are also locked in by the barrier, so they leave more slowly. CO reduces capacity and also slows the unloading of the oxygen that is still bound.

Real-world example

Faulty or poorly ventilated combustion appliances can release CO indoors, which is why CO alarms are recommended in homes. In hospitals, standard two-wavelength pulse oximeters cannot distinguish HbCO from HbO₂, so a poisoned patient may show a misleadingly normal oxygen saturation; blood co-oximetry, which measures more wavelengths, is used instead. Breathing high-concentration oxygen shortens the half-life of HbCO from several hours to roughly an hour by mass action.

Why?

Why does the body not simply evolve haemoglobin that ignores CO? Any site that binds O₂ reversibly through Fe(II) will also bind CO, because both ligands use σ-donation and π-acceptance. The protein can only bias the competition. Natural CO levels were historically low, so a few-hundred-fold preference was an acceptable compromise that preserves efficient O₂ transport.

Common misconception

"CO poisoning is just a reduced number of oxygen carriers." That ignores cooperativity. CO-bound subunits push the tetramer towards the high-affinity state, so the oxygen that is still bound is released less readily in tissues. The left shift of the curve is as important as the lost capacity.

Worked example

Question: Alveolar p(O₂) is about 13.3 kPa. Air containing 100 ppm CO at 101 kPa gives p(CO) ≈ 0.0101 kPa. Taking M = 240, estimate the equilibrium percentage of HbCO relative to HbCO + HbO₂.

Reasoning: r = 240 × 0.0101 / 13.3 = 0.182. Fraction HbCO = 0.182 / 1.182 = 0.154.

Answer: About 15% at equilibrium. Uptake is slow, so reaching this value takes hours; the calculation gives the limiting value, not the level after a short exposure.

Quick check

1. Which pocket residue selectively stabilises bound O₂ over bound CO, and how? Answer: The distal histidine, which forms a hydrogen bond to the polar, bent Fe–O–O unit but interacts only weakly with the nearly non-polar Fe–C–O unit.

Exam focus

Be ready to explain CO selectivity in terms of bonding (σ-donation and π back-bonding) and pocket effects (distal histidine hydrogen bonding). State both consequences of HbCO: fewer available sites and a left-shifted, less sigmoidal dissociation curve. Use the Haldane relationship with consistent pressure units.

Advanced insight

Ligand discrimination has kinetic as well as thermodynamic components. Photolysis experiments break the Fe–CO bond with a laser pulse and follow rebinding on picosecond to millisecond timescales, revealing internal cavities where the ligand pauses before escaping. NO, which binds even more strongly than CO, reacts with oxyhaemoglobin to form methaemoglobin and nitrate, a chemically distinct fate. Comparing the three diatomics shows that affinity reflects the full electronic structure of the Fe–ligand unit, not size alone.

Summary

CO binds the same Fe(II) haem site as O₂ but much more strongly, thanks to σ-donation and π back-bonding in a linear Fe–C–O unit. The distal histidine stabilises polar bound O₂ and lowers the CO preference from about 10⁴ for free haem to a few hundred in haemoglobin. The Haldane relationship predicts equilibrium HbCO levels. CO reduces capacity and shifts the dissociation curve left, impairing oxygen release.

Practice questions

1. Why is the Fe–C–O unit nearly linear while Fe–O–O is bent? Answer: CO binds through carbon with strong π back-bonding and no extra electrons to occupy a bent geometry, whereas bound O₂ has superoxide-like character with an extra electron in a π orbital, favouring a bent arrangement. 2. With M = 240 and p(O₂) = 13.3 kPa, what p(CO) gives equal HbCO and HbO₂? Answer: p(CO) = 13.3 / 240 ≈ 0.055 kPa, about 550 ppm at atmospheric pressure. 3. Explain why 50% HbCO is more dangerous than losing half of one's haemoglobin. Answer: With HbCO, the remaining O₂-carrying haems are held in a higher-affinity state and unload poorly, whereas in anaemia the remaining haemoglobin still releases O₂ normally. 4. Why does breathing high-concentration oxygen speed CO elimination? Answer: Raising p(O₂) shifts the competitive equilibrium towards HbO₂, displacing CO by mass action and shortening the half-life of HbCO.