Oxygen–Haemoglobin Dissociation Curve

Set the pressures at the lung and at the tissue, then read what the blood picks up and what it puts down. The pH, temperature and CO₂ sliders move the tissue curve; the Hill coefficient turns cooperativity off.

Worked example — Oxygen–Haemoglobin Dissociation Curve

P₅₀ at the tissue (mmHg)	26.6
Arterial saturation	97.3%
Saturation leaving the tissue	75.1%
Released here (points of saturation)	22.2%
Arterial O₂ content (mL/dL)	19.85
O₂ delivered (mL/min)	232
Curve steepest at (mmHg)	19.9

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

The plateau protects you from thin air and does nothing at all for anaemia 🖖

Pull arterial PO₂ from 100 down to 60 mmHg — a 40% fall, which is pneumonia, or about 3,500 m of altitude — and saturation goes only from 97.3% to 90.0%. Arterial content falls from 19.85 to 18.27 mL/dL, a loss of 8.0%. Now put the pressure back and cut haemoglobin by the same 40%, 15 g/dL to 9: content falls to 12.03 mL/dL, a loss of 39.4%. The flat top of the curve is a reserve against a failing lung, and haemoglobin loss walks straight past it, because content is capacity times saturation and only one of those two has a plateau. This is also the limit of the finger probe: it reports the percentage, never the amount, so it reads 97% on the anaemic patient and tells you nothing.

Myoglobin is a store, and the difference from a courier is 18.4 points of saturation 🖖

Set n to 1 and P₅₀ to 2.8 mmHg and the tool is now modelling myoglobin, the pigment inside a muscle cell. It arrives 97.3% saturated, just as haemoglobin does. At a tissue PO₂ of 40 mmHg it is still 93.5% saturated: it has released 3.8 points. Haemoglobin between the same two pressures releases 22.2, which is 5.8 times as much, and it does that because four cooperating subunits make its curve sigmoid instead of hyperbolic. Myoglobin does not give up half its load until the pressure has fallen to 2.8 mmHg, a pressure blood never sees: drag the tissue slider down to its floor of 3 mmHg and myoglobin is still 51.7% saturated. That is what you want from an emergency reserve sitting inside the cell, and what you do not want from blood.

The curve is steepest at 19.9 mmHg, well below its own half-saturation point 🖖

The last card reads 19.9 mmHg while P₅₀ reads 26.6. Setting the second derivative of the Hill equation to zero gives P = P₅₀((n−1)/(n+1))^(1/n), so the inflection sits below P₅₀ for every finite n and creeps up towards it as cooperativity rises: push n to 4 and the steepest point moves to 23.4 mmHg. Drop n to 1 and the inflection disappears, because a hyperbola is steepest at zero pressure. So the region where a small further fall in tissue PO₂ buys a large extra release sits nearer 20 mmHg than 27 — down among the pressures a working muscle actually reaches, rather than at the number the curve is labelled by.

Problems solved in full

A patient with haemoglobin of 9 g/dL and arterial PO₂ 100 mmHg 5 steps
A patient with haemoglobin of 9 g/dL, arterial PO₂ 100 mmHg, sitting still. The heart pumps 5 L/min; the body burns 250 mL of oxygen a minute. Find the PO₂ of the blood coming back from the tissues, and say what that number means for them.
1. Start from Fick's principle. Whatever the tissues consume is the flow multiplied by the difference between what arrives and what leaves.
  
  $\dot{V}_{O_{2}} = \dot{Q} (C_{a} O_{2} - C_{v} O_{2})$
2. Rearrange for the content difference, and put it in the units the blood is measured in: millilitres of oxygen per decilitre.
  
  $C_{a} O_{2} - C_{v} O_{2} = \frac{250 mL/min}{5 L/min} = 50 mL/L = 5.00 mL/dL$
3. Now the arterial content. Each gram of fully saturated haemoglobin carries 1.34 mL of oxygen, and a further 0.003 mL per decilitre stays physically dissolved for every mmHg of PO₂.
  
  $C_{a} O_{2} = 1.34 \times 9 \times 0.9728 + 0.003 \times 100 = 11.73 + 0.30 = 12.03 mL/dL$
4. Subtract the 5.00, then turn content back into saturation. The dissolved part depends on the pressure the next step is solving for, so start without it: 7.03 ÷ 12.06 = 0.5829 carries through to 30.1 mmHg, and 0.003 × 30.1 = 0.09 mL/dL put back gives 6.94 ÷ 12.06 = 0.5755. One pass settles it, because the dissolved term is 1.3% of the content.
  
  $C_{v} O_{2} = 12.03 - 5.00 = 7.03; S_{v} = \frac{7.03 - 0.09}{1.34 \times 9} = \frac{6.94}{12.06} = 0.5755$
5. Run the Hill equation backwards. The tool maps pressure to saturation and has no card for the other direction, so this step is done by hand.
  
  $P_{v} = P_{50} (\frac{S _{v}}{1 - S _{v}})^{1/ n} = 26.6 (\frac{0.5755}{0.4245})^{1/2.7} = 29.8 mmHg$
Answer

29.8 mmHg. The tool prints the two figures this started from: 97.3% arterial saturation and 12.03 mL/dL of arterial content. Now do the same arithmetic twice more. At Hb 15 and the same resting workload the venous PO₂ is 38.7 mmHg. At Hb 15 with a workload of 1,000 mL a minute — a brisk walk up a hill — it is 30.2 mmHg. This patient, sitting still, has the mixed venous oxygen of a healthy person climbing. Their arterial blood is fine, their saturation probe reads 97%, and the reserve that would have covered the next flight of stairs is already spent, so the heart has to find it by beating faster. The venous side is where anaemia is visible, and nobody measures it.
Fourteen times the oxygen, and what the Bohr effect earns of it 6 steps
Load Hard exercise: tissue PO₂ down to 20 mmHg, pH 7.2, 39 °C, PCO₂ 60, cardiac output 20 L/min. The delivery card reads 3,301 mL/min against 232 at rest. Split that factor between the heart and the blood, then decide whether the Bohr shift or the falling tissue pressure does more of the blood’s half.
1. Delivery is flow multiplied by the content each litre gives up, so the factor separates at the first line: four from the heart, and whatever is left from the blood.
  
  $\dot{V}_{O_{2}} = \dot{Q} Δ C : 20 \times 165.0 = 3, 301 mL mi n^{- 1} against 5 \times 46.47 = 232 mL mi n^{- 1}$
2. The heart supplies a factor of four. The blood supplies 3.55, and the two multiply to the whole 14.2.
  
  $\frac{3301}{232} = 14.21 = 4.00 (flow) \frac{20}{5} \times 3.552 (blood) \frac{165.0}{46.47}$
3. All three tissue conditions push the curve the same way, and Severinghaus keeps them separate so the sizes are visible: the acid contributes 0.080 of the exponent, the two degrees 0.048, and the carbon dioxide 0.011.
  
  $Δ lo g_{10} P_{50} = 0.40 (7.40 - 7.20) + 0.024 (39 - 37) + 0.06 lo g_{10} \frac{60}{40} = 0.1386 ⟹ P_{50} = 26.6 \to 36.6 mmHg$
4. That right shift together with the lower tissue pressure takes saturation on the way out from three quarters to a sixth. The arterial end does not move at all, because the blood equilibrated in the lung, where none of these conditions apply.
  
  $S_{tissue} : 75.1% (40 mmHg, P_{50} 26.6) \to 16.4% (20 mmHg, P_{50} 36.6); S_{art} = 97.3% either way$
5. To split the blood’s 3.55, put one factor back at a time and read the delivery card again.
  
  $From 3, 301, put one factor back: P_{tissue} 20 \to 40 \Rightarrow 1, 696; Bohr off \Rightarrow 2, 686$
6. The two interact, so their sizes depend on the order you restore them in. Do it both ways.
  
  $232 \to 929 \to 2686 \to 3301 (4.00, 2.891, 1.229) 232 \to 929 \to 1696 \to 3301 (4.00, 1.825, 1.946)$
Answer

Four from the heart, 3.55 from the blood — and inside the blood the falling tissue pressure beats the Bohr shift in either ordering: 2.891 against 1.229, or 1.946 against 1.825.

The Bohr effect is worth roughly a fifth of the delivery here, which is a great deal for a mechanism that costs the body nothing to run. Most of the blood’s share comes from somewhere plainer: a working muscle burns oxygen fast enough to pull its own PO₂ down to 20 mmHg. The last card says why that number matters — under these conditions the curve is steepest at 27.4 mmHg, so the tissue has already come down through the steepest part of it. Put the tissue slider back to 40 and delivery falls from 3,301 to 1,696 with the acid and the heat still in place.

$2.891 \times 1.229 = 1.825 \times 1.946 = 3.552; 4.00 \times 3.552 = 14.21$

By the 42Math team · Updated 2026-08-26

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Textbook resting blood - PaO₂ 100 mmHg and a tissue PO₂ of 40 give the pair every physiology course opens with: 97.3% saturated arriving, 75.1% leaving, 22.2 points handed over. At 5 L/min that is 232 mL of oxygen a minute, which is roughly what a resting adult burns.
Thin air: PaO₂ 60 - Arterial PO₂ drops by 40%, to 60 mmHg, and the blood still leaves the lung 90.0% saturated carrying 18.27 mL/dL. This is the flat top of the curve doing its job.
Anaemia: Hb 9 g/dL - Same lungs, same 97.3% saturation, four-tenths less haemoglobin. Arterial content is 12.03 mL/dL and delivery falls to 143 mL/min: a saturation probe on this patient reads normal.
Myoglobin instead - Myoglobin, with n = 1 and P₅₀ 2.8 mmHg, arrives 97.3% saturated and is still 93.5% saturated at a tissue PO₂ of 40. It lets go of 3.8 points, which is not transport.
Hard exercise - Tissue PO₂ 20 mmHg, pH 7.20, 39 °C, PCO₂ 60: P₅₀ moves out to 36.6 mmHg and the blood leaves the muscle 16.4% saturated. At 20 L/min the delivery is 3,301 mL a minute.