The Hill Equation: Cooperative Binding and the Hill Plot
Some proteins bind their ligand in a simple way: each molecule binds once, independently. Others, such as the oxygen carrier haemoglobin, bind cooperatively, so the first ligand makes the next one bind more easily. The Hill equation is the simplest model that describes this behaviour. It gives a sigmoidal (S-shaped) binding curve instead of the hyperbola of simple binding. The topic appears in biophysical chemistry and enzyme kinetics, and it uses only logarithms and ratios, which makes it a good exam topic for IIT-JAM, GATE, CSIR-NET and CUET-PG candidates.
The equation
Hill plot: log [θ / (1 − θ)] = n log [L] − n log K0.5
θ is the fraction of binding sites occupied (0 to 1). [L] is the free ligand concentration. K0.5 is the ligand concentration at which half of the sites are occupied (θ = 0.5); it has concentration units. n is the Hill coefficient, a dimensionless number. For n = 1 the equation reduces to the simple hyperbolic binding θ = [L] / (Kd + [L]).
The Hill plot gives a straight line when log [θ / (1 − θ)] is plotted against log [L]. The slope is n and the line crosses zero on the vertical axis at [L] = K0.5.
What the Hill coefficient tells you
- n = 1: no cooperativity; sites bind independently.
- n > 1: positive cooperativity; binding at one site raises the affinity of the others.
- n < 1: negative cooperativity (or sites of different affinity).
- n is not the number of binding sites. It is an apparent value and is generally less than or equal to the number of interacting sites. It measures the steepness of the curve, not a count.
Worked example 1: simple and cooperative binding compared
A protein has K0.5 = 20 µM. Compare θ at [L] = 10, 20 and 40 µM for n = 1 and n = 3. (The numbers are for practice, not from a specific protein.)
n = 1: θ = 10/30 = 0.333; 20/40 = 0.500; 40/60 = 0.667.
n = 3: at 10 µM: 103 / (203 + 103) = 1000 / 9000 = 0.111. At 20 µM: 8000 / 16 000 = 0.500. At 40 µM: 64 000 / 72 000 = 0.889.
Both curves cross 0.5 at 20 µM, but the cooperative curve is far steeper: from 10 to 40 µM, θ goes from 0.11 to 0.89, against 0.33 to 0.67 for simple binding.
Worked example 2: how sharp is the switch?
By what factor must [L] rise to go from 10% to 90% saturation?
At θ = 0.9, θ / (1 − θ) = 9, and at θ = 0.1 it is 1/9. Their ratio is 81, so ([L]0.9 / [L]0.1)n = 81 and [L]0.9 / [L]0.1 = 811/n.
For n = 1: 81-fold. For n = 3: 811/3 = 4.33-fold. A cooperative protein responds to a much narrower range of ligand concentration, which is useful for taking up ligand in one place and releasing it in another.
Worked example 3: finding n and K0.5 from two data points
For a protein, θ = 0.20 at [L] = 10 µM and θ = 0.80 at [L] = 40 µM. Find n and K0.5.
At 10 µM: θ/(1 − θ) = 0.20/0.80 = 0.25, log = −0.602. At 40 µM: 0.80/0.20 = 4, log = +0.602.
Slope = (0.602 − (−0.602)) / (log 40 − log 10) = 1.204 / 0.602 = 2.0. So n = 2.
From the first point: 0.25 = 102 / K0.52, giving K0.52 = 100 / 0.25 = 400 and K0.5 = 20 µM.
Check at 40 µM: θ = 1600 / (400 + 1600) = 0.80. It matches the data.
Worked example 4: using experimental data
With real measurements you have many points. Calculate log [L] and log [θ/(1 − θ)] for each, then fit a straight line by least squares. The slope is n, and the intercept gives −n log K0.5. Using the model of example 3 (n = 2, K0.5 = 20 µM) the points are: [L] = 5 µM gives θ = 25/425 = 0.0588; 10 µM gives 0.200; 20 µM gives 0.500; 40 µM gives 0.800; 80 µM gives 6400/6800 = 0.941. All five points fall exactly on the line y = 2x − 2.602 (here x = log [L], y = log [θ/(1 − θ)]), because the data were generated from the equation. Real data scatter around the line and deviate at the very low and very high ends.
Limits of the Hill equation
- It is an empirical description. It does not tell you the mechanism, such as the MWC or the sequential model.
- It describes the middle of the curve best (roughly θ from 0.1 to 0.9). At very low or very high θ the real slope approaches 1.
- For ligands present at low total amounts, you must use free [L], not the total added.
Common mistakes
- Taking n as the number of binding sites. It is a measure of cooperativity.
- Plotting θ instead of θ/(1 − θ) on the Hill plot.
- Using natural log for one axis and log10 for the other. The slope stays the same only if both use the same base.
- Reading K0.5 as a dissociation constant for n ≠ 1. K0.5 is a concentration, whereas the quantity K0.5n carries units of concentration to the power n.
- Using total ligand instead of free ligand.
Exam relevance
| Question type | Approach |
|---|---|
| θ at a given [L] | Substitute in θ = [L]n/(K0.5n + [L]n) |
| Find n from data | Slope of the Hill plot |
| Compare two proteins | Larger n means a steeper, more switch-like curve |
| Range from 10% to 90% | 811/n-fold change in [L] |
The least-squares step of the Hill plot is a straight-line fit y = mx + c, so you can check your slope and intercept with a linear regression tool. For the exam syllabus and pattern in a given year, check the current official notification.
The Hill plot is a straight-line fit, so the linear regression tool can check your slope and intercept.
Open the Linear Regression Calculator →Step-by-step maths tools for algebra, calculus, matrices and statistics.
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