The oxygen dissociation curve is sigmoid rather than hyperbolic, and my notes say this is because binding is cooperative. That is the name of the observation rather than an explanation of it.
What physically changes in the molecule after the first oxygen binds, and why does that make the next one easier?
Accepted Answer
Haemoglobin has four subunits and two quaternary states, called T for tense and R for relaxed.
In the T state, salt bridges between subunits hold the structure in a conformation with low oxygen affinity. When an oxygen binds, the iron moves into the plane of the haem ring, which drags the histidine attached to it, which pulls the helix, which breaks some of those salt bridges.
Now the whole tetramer is closer to the R state, and every remaining site has higher affinity. That is the cooperativity: the subunits are physically linked, so one binding event changes the shape of the others.
Myoglobin has one subunit, nothing to cooperate with, and a hyperbolic curve. The comparison is the cleanest evidence that the sigmoid shape comes from the quaternary structure.
Worth being careful with the phrase "binding changes the affinity of the others", because it is often read as the first oxygen sending a signal.
Nothing is signalled. The tetramer has two stable shapes, and binding shifts the equilibrium between them. Every site in the R state has high affinity whether or not it is occupied.
This matters for the exam because it is why 2,3-BPG, CO₂ and low pH shift the curve. They stabilise the T state directly, without any oxygen being involved.
The physiological payoff is worth stating: the sigmoid shape means haemoglobin can be nearly saturated in the lungs and still unload a large fraction in tissue, across a fairly small pressure difference. A hyperbolic curve could not do both.
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