Cooperative Oxygen Binding

Introduction

Haemoglobin's sigmoidal oxygen-binding curve arises from cooperativity: binding the first O₂ molecules makes it easier for the rest to bind, and losing some makes it easier to lose the others. This behaviour turns a simple equilibrium into a switch. The explanation links coordination chemistry at a single iron atom to the movement of whole protein subunits, and it provides the classic model of allostery used throughout biochemistry.

Core explanation

The trigger at the heme. In deoxy-Hb, each iron is five-coordinate high-spin Fe(II), too large for the porphyrin cavity and displaced about 0.4 Å out of the plane towards the proximal histidine, with the porphyrin slightly domed. When O₂ binds, iron becomes low spin, shrinks and moves into the plane. The proximal histidine, attached to helix F, is pulled along, shifting the F helix by a small distance.

Transmission to the interface. The helix movement is transmitted to the α₁β₂ interface, where subunit contacts must rearrange. Deoxy-Hb is held in the tense (T) state by salt bridges between and within subunits, including interactions at the chain C-termini. These constraints resist the movement of iron into the plane, so T-state hemes have low O₂ affinity. Binding O₂ breaks some of these interactions. When enough are broken, the tetramer switches to the relaxed (R) quaternary structure, in which one αβ dimer rotates by about 15° relative to the other. R-state hemes bind O₂ with much higher affinity, close to that of isolated chains.

The concerted (MWC) model. Monod, Wyman and Changeux proposed that the tetramer exists in two states, T and R, in equilibrium. All subunits in a molecule share one state. Without O₂, T is strongly favoured. Each O₂ bound shifts the equilibrium towards R because R binds O₂ more tightly. After about two or three O₂ bind, R predominates, and the remaining sites fill easily. This model explains why the curve steepens abruptly.

The sequential (KNF) model. Koshland, Némethy and Filmer proposed that each subunit changes conformation individually when it binds a ligand, and these changes alter neighbours' affinity. Real haemoglobin shows features of both models: the quaternary switch is largely concerted, but tertiary changes within T-state subunits also affect affinity.

The Hill equation. Empirically, Y / (1 − Y) = (pO₂ / P₅₀)^n H. A plot of log[Y / (1 − Y)] against log pO₂ gives a line of slope n H in the middle region. For myoglobin, n H = 1. For haemoglobin, n H ≈ 2.8-3.0. If binding were perfectly concerted with all four O₂ binding simultaneously, n H would equal 4; the real value shows strong but incomplete cooperativity. At very low and very high saturation, the Hill plot for haemoglobin has slope close to 1, reflecting binding to pure T and pure R states respectively.

Physiological consequence. Positive cooperativity makes saturation fall steeply over the tissue pO₂ range, enabling large O₂ release for small pressure changes.

Formulae

Hill equation: Y = pO₂^n / (P₅₀^n + pO₂^n). Linear form: log[Y / (1 − Y)] = n H log pO₂ − n H log P₅₀. Non-cooperative binding has n H = 1.

Step-by-step reasoning

1. Start with deoxy-Hb in the T state with high-spin iron out of the plane. 2. O₂ binds to one heme, making iron low spin and pulling it into the plane. 3. The proximal histidine and F helix move, straining subunit contacts. 4. Salt bridges break as more O₂ binds, destabilising T. 5. The tetramer switches to R, raising the affinity of remaining empty sites. 6. The overall curve becomes sigmoidal with n H near 3.

Visual explanation

Draw four squares (T state) linked by thick bars representing salt bridges. Show one O₂ binding and one bar cracking. After two or three O₂, redraw the tetramer as circles (R state) with no bars, and the empty sites filling easily. Alongside, plot a Hill graph with a slope-1 line at each end joined by a steeper slope-3 middle section.

Real-world analogy

Four people pushing open a heavy double door each find it hard at first; once two have pushed it partly open, it swings the rest of the way easily for the others. The door's state is shared by all four pushers, like the quaternary state of the tetramer. Unlike people, the subunits "sense" each other only through structural changes, not communication.

Real-world example

Some haemoglobin mutants that disrupt the α₁β₂ interface, such as Hb Kempsey, stabilise the R state. They show high O₂ affinity, reduced cooperativity and poor O₂ delivery, leading to increased red blood cell production (polycythaemia). Such variants confirm that the interface is central to cooperativity.

Why?

Why is the Hill coefficient less than 4 for a four-site protein? n H = 4 would require no intermediate species — every molecule either fully empty or fully loaded. In reality, partially liganded tetramers exist, so cooperativity is strong but not all-or-nothing.

Common misconception

"The Hill coefficient equals the number of binding sites." It is an empirical measure of cooperativity and is always less than or equal to the number of sites. Another misconception is that oxygen physically travels between subunits; information passes through protein conformational change, not by moving O₂.

Worked example

Question: Haemoglobin has P₅₀ = 3.5 kPa and n H = 2.8. Calculate Y at 13 kPa and at 5.3 kPa.

Reasoning: At 13 kPa, (13/3.5)^2.8 = 3.71^2.8 ≈ 39.4, so Y = 39.4 / 40.4 ≈ 0.98. At 5.3 kPa, (5.3/3.5)^2.8 = 1.51^2.8 ≈ 3.2, so Y = 3.2 / 4.2 ≈ 0.76.

Answer: About 98% in the lungs and 76% in resting tissue, matching the physiological curve.

Quick check

1. What change in the iron atom starts the T-to-R transition in haemoglobin? Answer: Iron becomes low spin on binding O₂ and moves into the porphyrin plane, pulling the proximal histidine.

Exam focus

Explain the mechanism from spin state change to quaternary switch. Compare MWC and KNF models. Use the Hill equation to calculate saturation and interpret n H values for myoglobin (1) and haemoglobin (about 3).

Advanced insight

Modern studies combine crystal structures of fixed T and R states with solution kinetics, showing that ligand affinity in T-state subunits differs between α and β chains and that tertiary structure changes within T can modulate affinity. Extensions of the MWC model incorporate these tertiary effects to account for the effects of allosteric effectors on both states.

Summary

Cooperative O₂ binding in haemoglobin begins when iron moves into the porphyrin plane on binding O₂. Movement of the proximal histidine and F helix strains subunit contacts, breaking salt bridges and switching the tetramer from low-affinity T to high-affinity R. The MWC and KNF models describe this, and the Hill coefficient of about 3 quantifies the strong cooperativity that shapes the sigmoidal curve.

Practice questions

1. What are the T and R states of haemoglobin? Answer: T is the tense, low-affinity quaternary state favoured by deoxy-Hb; R is the relaxed, high-affinity state favoured when O₂ is bound. 2. What Hill coefficient would you expect for myoglobin, and why? Answer: n H = 1, because it has a single binding site and no cooperativity. 3. Distinguish the concerted and sequential models of cooperativity. Answer: The concerted model has all subunits switching state together; the sequential model has each subunit changing individually on binding and influencing its neighbours. 4. A mutation removes a salt bridge that stabilises the T state. Predict the effect on O₂ affinity and cooperativity. Answer: The R state becomes more favoured, raising O₂ affinity and lowering cooperativity, so less O₂ is released in tissues.