Cooperative Binding (Hill Equation) Calculator

Find the fractional occupancy of a cooperative receptor or protein from ligand concentration, half-saturation constant, and the Hill coefficient.

🩸 Cooperative Binding (Hill Equation) Calculator
Ligand concentration [L]40.00
mmHg or µM
0200
Half-saturation constant Kd (P50)
same units as [L]
Hill coefficient (n)2.80
dimensionless
0.26
Percent saturation
—
Fractional occupancy (θ)
—
Hill plot y = log[θ/(1−θ)]
—
Cooperativity
—
Step-by-step working

🩸 What is Cooperative Binding and the Hill Equation?

Cooperative binding describes a protein with several binding sites where occupying one site changes how readily the remaining sites bind the same ligand. The Hill equation, published by Archibald Hill in 1910 while he was analysing oxygen uptake by haemoglobin, captures that behaviour in a single compact expression: theta = [L]^n / (Kd^n + [L]^n). Here theta is the fraction of sites occupied, [L] is the free ligand concentration, Kd is the concentration that fills exactly half the sites, and n is the Hill coefficient, an empirical steepness index. When n exceeds 1 the binding curve stops being a gentle hyperbola and becomes a sigmoid, an S shape with a shallow foot, a steep middle, and a flat plateau.

The equation shows up across biology and medicine. Respiratory physiologists use it to model the oxygen-haemoglobin dissociation curve, where Kd is called P50 and equals roughly 26 mmHg in healthy adult blood, so arterial blood at 100 mmHg is about 98% saturated while mixed venous blood at 40 mmHg is about 77% saturated, and that difference is exactly the oxygen delivered to tissue. Pharmacologists reuse the same algebra as the sigmoid Emax model, replacing Kd with EC50 and n with the Hill slope, to fit dose-response curves. Molecular biologists apply it to transcription factors binding cooperatively to promoters, and biophysicists apply it to ligand-gated ion channels, where high Hill coefficients produce near switch-like gating.

Two misconceptions are worth clearing up. First, the Hill coefficient is not a count of binding sites. Haemoglobin has four oxygen sites yet fits n = 2.8, because a Hill coefficient equal to the site count would require perfectly all-or-nothing binding that no real protein achieves. Treat n as a lower bound on the number of interacting sites. Second, Kd in this formulation is a real concentration with the same units as [L], not a dissociation constant for a single site. Some textbooks write theta = [L]^n/(K + [L]^n) with K carrying units of concentration to the power n. The two forms are related by K = Kd^n and give identical curves.

This calculator is useful because it turns three inputs into an immediate, interpretable answer: the fractional occupancy, the percent saturation, the position of your point on the Hill plot, and the transition steepness 81^(1/n) that tells you how sharply the system switches. The chart plots your full saturation curve against a non-cooperative n = 1 reference, so you can see the sigmoid emerge, watch the half-saturation point stay pinned at Kd, and judge at a glance whether your operating concentration sits on the steep working part of the curve or out on the saturated plateau.

📐 Formula

θ  =  [L]n ÷ (Kdn + [L]n)
θ = fractional occupancy, from 0 (no sites bound) to 1 (all sites bound); percent saturation = 100 × θ
[L] = free ligand concentration (mmHg for oxygen partial pressure, µM or nM for solution-phase ligands)
Kd = half-saturation constant, the ligand concentration giving θ = 0.5; called P50 for haemoglobin and oxygen
n = Hill coefficient (dimensionless); n > 1 positive cooperativity, n = 1 independent sites, n < 1 negative cooperativity
Hill plot (linearised form): log10[θ ÷ (1 − θ)] = n × log10[L] − n × log10Kd, a straight line of slope n
Transition steepness: [L]90 ÷ [L]10 = 811/n, the ligand ratio spanning 10% to 90% occupancy
Example: With [L] = 26, Kd = 26 and n = 2.8, the numerator and the Kd term are equal, so θ = 0.5000 exactly, the defining property of the half-saturation constant. Values follow the convention of the oxygen-haemoglobin dissociation curve as described by Wikipedia and standard respiratory physiology texts.

📖 How to Use This Calculator

Steps

1
Enter the ligand concentration. Type the free ligand concentration [L], for example the oxygen partial pressure in mmHg or a drug concentration in micromolar.
2
Enter the half-saturation constant. Type Kd, the ligand concentration that occupies exactly half the binding sites, in the same units as [L]. For haemoglobin this is P50, about 26 mmHg.
3
Set the Hill coefficient. Drag the Hill coefficient n to match your system, using 2.8 for haemoglobin, 1 for non-cooperative binding such as myoglobin, and higher values for switch-like ion channels.
4
Read the occupancy and curve. Read the fractional occupancy, percent saturation, and Hill plot coordinate, then check where your point sits on the sigmoidal saturation curve against the non-cooperative reference.

💡 Example Calculations

Example 1 — Haemoglobin in Mixed Venous Blood

pO2 = 40 mmHg, P50 = 26 mmHg, n = 2.8

1
[L]n = 402.8 = 30,603.3 and Kdn = 262.8 = 9160.6
2
θ = 30,603.3 ÷ (9160.6 + 30,603.3) = 30,603.3 ÷ 39,763.9 = 0.7696
3
Percent saturation = 100 × 0.7696 = 76.96%, the classic mixed venous saturation figure. Hill plot y = log10(0.7696 ÷ 0.2304) = 0.5238.
θ = 0.7696 (76.96% saturated, positive cooperative)
Try this example →

Example 2 — Haemoglobin in Arterial Blood

pO2 = 100 mmHg, P50 = 26 mmHg, n = 2.8

1
[L]n = 1002.8 = 398,107.2 and Kdn = 262.8 = 9160.6
2
θ = 398,107.2 ÷ (9160.6 + 398,107.2) = 398,107.2 ÷ 407,267.8 = 0.9775
3
Percent saturation = 97.75%. Comparing with Example 1, moving from arterial 100 mmHg to venous 40 mmHg releases 97.75% − 76.96% = 20.79% of haemoglobin's oxygen load to the tissue.
θ = 0.9775 (97.75% saturated, on the plateau)
Try this example →

Example 3 — Myoglobin, Non-Cooperative Binding

pO2 = 5 mmHg, P50 = 3.5 mmHg, n = 1 (a single binding site)

1
With n = 1 the Hill equation collapses to the simple hyperbola θ = [L] ÷ (Kd + [L]) = 5 ÷ (3.5 + 5)
2
θ = 5 ÷ 8.5 = 0.5882, so myoglobin is 58.82% saturated even at the low oxygen tension found inside working muscle
3
Because n = 1, moving from 10% to 90% occupancy requires an 81 fold ligand increase, and the Hill plot y value is log10(0.5882 ÷ 0.4118) = 0.1549
θ = 0.5882 (58.82% saturated, non-cooperative)
Try this example →

Example 4 — A Switch-Like Ligand-Gated Ion Channel

[L] = 20 µM agonist, Kd = 12 µM, n = 4 (strong positive cooperativity)

1
[L]n = 204 = 160,000 and Kdn = 124 = 20,736
2
θ = 160,000 ÷ (20,736 + 160,000) = 160,000 ÷ 180,736 = 0.8853, or 88.53% of channels open
3
With n = 4 the 10% to 90% transition needs only 811/4 = 3.00 fold more agonist, so only 1.67 times the Kd already drives the channel population close to fully open. That is switch-like behaviour.
θ = 0.8853 (88.53% occupied, positive cooperative)
Try this example →

❓ Frequently Asked Questions

What is the Hill equation for cooperative binding?+
The Hill equation is theta = [L]^n / (Kd^n + [L]^n), where theta is the fractional occupancy (0 to 1), [L] is the free ligand concentration, Kd is the half-saturation constant (the ligand concentration that occupies half the sites), and n is the Hill coefficient describing cooperativity.
What does the Hill coefficient n tell you?+
n measures cooperativity. n greater than 1 is positive cooperativity, where binding one ligand makes further binding easier and the curve becomes sigmoidal. n equal to 1 is non-cooperative binding, which collapses to a simple hyperbola identical to Michaelis-Menten saturation. n less than 1 is negative cooperativity, giving a flatter, more gradual curve.
Why is the Hill coefficient of haemoglobin 2.8 and not 4?+
Haemoglobin has four oxygen binding sites, but the measured Hill coefficient of about 2.8 is an empirical steepness index, not a site count. A Hill coefficient equal to the number of sites would require perfectly all-or-nothing binding. Real haemoglobin binds with strong but incomplete cooperativity, so n falls between 1 and 4.
How do you calculate percent saturation from the Hill equation?+
Percent saturation is 100 times theta. With P50 = 26 mmHg, n = 2.8, and pO2 = 40 mmHg (typical mixed venous blood), theta = 40^2.8 / (26^2.8 + 40^2.8) = 0.7696, so haemoglobin is 76.96% saturated. At arterial pO2 = 100 mmHg the same equation gives 97.75%.
What is P50 and how does it relate to Kd?+
P50 is the partial pressure of oxygen at which haemoglobin is exactly 50% saturated, roughly 26 mmHg in healthy adult arterial blood. It is the haemoglobin-specific name for the half-saturation constant Kd used in this calculator. A higher P50 shifts the curve right and means lower oxygen affinity.
What is a Hill plot and what is its slope?+
A Hill plot graphs log10[theta / (1 - theta)] against log10[L]. Rearranging the Hill equation gives log10[theta/(1-theta)] = n log10[L] - n log10(Kd), so the plot is a straight line whose slope is the Hill coefficient n and whose x-intercept is log10(Kd). This calculator reports the y coordinate of your current point on that plot.
How much ligand does it take to go from 10% to 90% occupancy?+
The required ratio is 81^(1/n). For non-cooperative binding (n = 1) you need an 81 fold increase in ligand. For n = 2.8 you need only about a 4.8 fold increase, and for n = 4 exactly a 3 fold increase. This is the quantitative reason cooperative systems behave like biological switches.
Is the Hill equation the same as the Michaelis-Menten equation?+
They are the same only when n = 1. Setting n = 1 turns theta = [L]^n/(Kd^n + [L]^n) into theta = [L]/(Kd + [L]), which is identical in form to v = Vmax[S]/(Km + [S]). The Hill equation generalises that hyperbola by allowing the exponent n to bend the curve into a sigmoid.
What are the units of Kd in the Hill equation?+
In the form used here, Kd is a genuine concentration and carries the same units as [L], for example mmHg for oxygen partial pressure or micromolar for a solution-phase ligand. Some textbooks write theta = [L]^n/(K + [L]^n), in which case K has units of concentration to the power n, and K = Kd^n.
What causes the oxygen dissociation curve to shift right?+
A rightward shift means a higher P50 and lower oxygen affinity. It is caused by increased carbon dioxide, lower pH (the Bohr effect), higher temperature, and higher 2,3-bisphosphoglycerate. All four occur in metabolically active tissue, so the shift automatically releases more oxygen exactly where it is needed.
Can the Hill equation model drug dose-response curves?+
Yes. The identical algebra appears in pharmacology as E/Emax = [D]^h / (EC50^h + [D]^h), where h is the Hill slope and EC50 replaces Kd. Enter your EC50 as the half-saturation constant and the Hill slope as n, and the fractional occupancy this calculator returns is the fractional effect.
What does negative cooperativity look like on the curve?+
With n less than 1 the curve is shallower than a simple hyperbola: it rises quickly at very low ligand then flattens out over a very wide concentration range. With n = 0.6 you need a 1516 fold ligand increase to move from 10% to 90% occupancy, compared with 81 fold for independent sites.
Does the Hill equation assume the ligand concentration is free or total?+
It assumes free (unbound) ligand concentration. In dilute experiments where the protein concentration is far below Kd, total and free ligand are nearly identical and the distinction does not matter. When protein concentration approaches or exceeds Kd, ligand depletion becomes significant and fitting total ligand to the Hill equation inflates the apparent Kd.
Tags: Hill Equation Calculator Cooperative Binding Calculator Hill Coefficient Fractional Occupancy Oxygen Saturation Curve Calculator Formula Free Online

🔗 Related Calculators

🧫
Enzyme Kinetics (Michaelis-Menten) Calculator
Science - Biophysics
⚡
Nernst Potential Calculator
Science - Biophysics
🧠
Goldman-Hodgkin-Katz Equation Calculator
Science - Biophysics
🧬
Protein Folding Free Energy Estimator
Science - Biophysics
🔬
pH Calculator
Science - Chemistry
🧪
Molarity Calculator
Science - Chemistry

What is the Hill equation for cooperative binding?

The Hill equation is theta = [L]^n / (Kd^n + [L]^n), where theta is the fractional occupancy (0 to 1), [L] is the free ligand concentration, Kd is the half-saturation constant (the ligand concentration that occupies half the sites), and n is the Hill coefficient describing cooperativity.

What does the Hill coefficient n tell you?

n measures cooperativity. n greater than 1 is positive cooperativity, where binding one ligand makes further binding easier and the curve becomes sigmoidal. n equal to 1 is non-cooperative binding, which collapses to a simple hyperbola identical to Michaelis-Menten saturation. n less than 1 is negative cooperativity, giving a flatter, more gradual curve.

Why is the Hill coefficient of haemoglobin 2.8 and not 4?

Haemoglobin has four oxygen binding sites, but the measured Hill coefficient of about 2.8 is an empirical steepness index, not a site count. A Hill coefficient equal to the number of sites would require perfectly all-or-nothing binding. Real haemoglobin binds with strong but incomplete cooperativity, so n falls between 1 and 4.

How do you calculate percent saturation from the Hill equation?

Percent saturation is 100 times theta. With P50 = 26 mmHg, n = 2.8, and pO2 = 40 mmHg (typical mixed venous blood), theta = 40^2.8 / (26^2.8 + 40^2.8) = 0.7696, so haemoglobin is 76.96% saturated. At arterial pO2 = 100 mmHg the same equation gives 97.75%.

What is P50 and how does it relate to Kd?

P50 is the partial pressure of oxygen at which haemoglobin is exactly 50% saturated, roughly 26 mmHg in healthy adult arterial blood. It is the haemoglobin-specific name for the half-saturation constant Kd used in this calculator. A higher P50 shifts the curve right and means lower oxygen affinity.

What is a Hill plot and what is its slope?

A Hill plot graphs log10[theta / (1 - theta)] against log10[L]. Rearranging the Hill equation gives log10[theta/(1-theta)] = n log10[L] - n log10(Kd), so the plot is a straight line whose slope is the Hill coefficient n and whose x-intercept is log10(Kd). This calculator reports the y coordinate of your current point on that plot.

How much ligand does it take to go from 10% to 90% occupancy?

The required ratio is 81^(1/n). For non-cooperative binding (n = 1) you need an 81 fold increase in ligand. For n = 2.8 you need only about a 4.8 fold increase, and for n = 4 exactly a 3 fold increase. This is the quantitative reason cooperative systems behave like biological switches.

Is the Hill equation the same as the Michaelis-Menten equation?

They are the same only when n = 1. Setting n = 1 turns theta = [L]^n/(Kd^n + [L]^n) into theta = [L]/(Kd + [L]), which is identical in form to v = Vmax[S]/(Km + [S]). The Hill equation generalises that hyperbola by allowing the exponent n to bend the curve into a sigmoid.

What are the units of Kd in the Hill equation?

In the form used here, Kd is a genuine concentration and carries the same units as [L], for example mmHg for oxygen partial pressure or micromolar for a solution-phase ligand. Some textbooks write theta = [L]^n/(K + [L]^n), in which case K has units of concentration to the power n, and K = Kd^n.

What causes the oxygen dissociation curve to shift right?

A rightward shift means a higher P50 and lower oxygen affinity. It is caused by increased carbon dioxide, lower pH (the Bohr effect), higher temperature, and higher 2,3-bisphosphoglycerate. All four occur in metabolically active tissue, so the shift automatically releases more oxygen exactly where it is needed.

Can the Hill equation model drug dose-response curves?

Yes. The identical algebra appears in pharmacology as E/Emax = [D]^h / (EC50^h + [D]^h), where h is the Hill slope and EC50 replaces Kd. Enter your EC50 as the half-saturation constant and the Hill slope as n, and the fractional occupancy this calculator returns is the fractional effect.

What does negative cooperativity look like on the curve?

With n less than 1 the curve is shallower than a simple hyperbola: it rises quickly at very low ligand then flattens out over a very wide concentration range. With n = 0.6 you need a 1516 fold ligand increase to move from 10% to 90% occupancy, compared with 81 fold for independent sites.

📌 Quick Tips

💡The Hill coefficient n is the single most informative number here. n greater than 1 means positive cooperativity (binding one ligand makes the next bind more easily), n equal to 1 means independent non-cooperative binding that reduces to a simple hyperbola, and n less than 1 means negative cooperativity.
💡The Hill coefficient is an empirical measure, not a literal count of binding sites. Haemoglobin has four oxygen sites but its measured Hill coefficient is about 2.8, so n is best read as a lower bound on the number of interacting sites, never as an exact site count.
💡Kd in this calculator is the half-saturation constant, the ligand concentration that fills exactly half the sites. For haemoglobin and oxygen this quantity is called P50, roughly 26 mmHg in healthy adult blood at pH 7.4 and 37 degrees Celsius.
💡The ratio of ligand needed to move from 10% to 90% occupancy is 81 raised to the power 1/n. For n=1 that ratio is 81 fold, for n=2.8 it is about 4.8 fold, and for n=4 it is exactly 3 fold, which is why cooperative systems act like near switches.
💡A rightward shift of the curve (larger P50) means lower affinity and easier unloading. In blood this is the Bohr effect: rising carbon dioxide, falling pH, higher temperature, and higher 2,3-BPG all raise P50 and push oxygen off haemoglobin at working tissue.