Protein binding: separate affinity, capacity and cooperativity

MCAT content library / Biology & biochemistry / Protein binding: separate affinity, capacity and cooperativity

Read the concentration axis, plateau and curve shape before deciding what changed.

About 15 minutes, including the worked examples and checks.

Useful foundations
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Measure occupied sites before naming a mechanism

A binding assay asks how much ligand is associated with a protein under specified conditions. That is different from measuring the speed of a catalyzed reaction. The same protein can bind a molecule, change conformation, and perform another function, but one measurement does not establish all three events. Binding and cooperative regulation are within the AAMC protein-function scope.1

Three quantities organize the interpretation: affinity, the equilibrium tendency to bind a particular ligand; capacity, the amount that can bind when available sites are saturated; and fractional occupancy, the fraction of those sites currently occupied. If an assay reports 8 nM bound ligand, that could represent nearly full occupancy of 10 nM sites or low occupancy of 100 nM sites. Recover the denominator before comparing proteins.

A simple binding rule, with its assumptions

For one class of independent, equivalent sites at equilibrium, write P + L ⇌ PL. Here [P] means free sites and [L] means free ligand. The dissociation constant is Kd = [P][L]/[PL]. If B is bound ligand and Bmax is the total concentration of functional sites, fractional occupancy is θ = B/Bmax = [L]/(Kd + [L]). The comparison assumes matched temperature and solution conditions and a measurement specific to binding.2

At [L] = Kd, the equation gives θ = 1/2. Lower Kd means half-occupancy is reached at less free ligand: stronger affinity under this model. Concentration units must agree. An apparent midpoint read from an incomplete curve is unreliable if its true plateau has not been established.3

The word free matters. Total ligand includes both unbound and bound molecules. If binding removes an appreciable fraction of the ligand, substituting total concentration into the equation overstates the ligand available to empty sites. A study can measure free ligand directly or justify an excess-ligand approximation. The examples below supply free concentrations, so no depletion correction is needed.

Worked comparison: affinity, shape and capacity

The following original model dataset compares three preparations with the same independently measured Bmax of 10 nM sites. Protein A follows θ = L/(2 + L); B follows θ = L/(8 + L), with L in μM. A cooperative preparation C is represented by the supplied empirical rule θ = L²/(2² + L²). This last equation describes the example’s curve; its exponent is not a count of subunits.

Equal capacity, different binding curvesFractional occupancy against free ligand from zero to ten micromolar on a linear axis. A and B are hyperbolas with dissociation constants two and eight micromolar. Cooperative model C has midpoint two micromolar and exponent two. All curves approach full occupancy; raw capacity is ten nanomolar sites.Fraction of sites occupied00.250.50.7510246810Free ligand (μM)A: midpoint 2 μMB: midpoint 8 μMC: cooperative, midpoint 2 μM
Original model curves on a linear concentration axis. The accompanying table gives selected values; all preparations have the same 10 nM binding capacity.
Free ligandA: Kd = 2 μMB: Kd = 8 μMC: supplied cooperative rule
0.50.2000.0590.059
20.5000.2000.500
80.8000.5000.941
Original model occupancies; free ligand in μM

At 2 μM, A binds 0.500 × 10 = 5.00 nM ligand, while B binds 0.200 × 10 = 2.00 nM. Their capacities are identical; their occupancies differ because their Kd values differ. Raising ligand sufficiently makes both approach the same 10 nM plateau. It does not make their affinities equal.

A and C both reach half-occupancy at 2 μM, yet their shapes differ. At 0.5 μM, C is less occupied than A; at 8 μM, it is more occupied. C therefore makes a sharper transition around the shared midpoint. A midpoint alone cannot distinguish a hyperbolic binding process from a cooperative one. On the plotted linear ligand axis, A and B are hyperbolic while C is sigmoidal. A logarithmic concentration axis changes visual shape, so “S-shaped” is not a sufficient diagnosis without reading the axis.

What cooperativity adds

In positive cooperativity, occupancy of one site favors additional binding elsewhere in the assembly. Coupled conformational changes can communicate between sites. An allosteric regulator binds at a separate site and shifts the protein’s conformational distribution; it can favor either a more strongly binding or a less strongly binding state. These ideas explain why an occupancy curve can change without changing the protein’s covalent sequence or its number of sites.4

A Hill-style expression uses a midpoint K0.5 and an empirical slope parameter. For the supplied cooperative expression, K0.5 is 2 μM, but it should not be treated as a microscopic dissociation constant for every binding step. Likewise, a fitted Hill coefficient does not directly enumerate subunits. A real mechanism must account for the protein’s architecture and the conditions under which the curve was measured.2

For a changed case, suppose a regulator shifts C’s K0.5 from 2 to 4 μM while the same exponent and capacity remain applicable. At free ligand 2 μM, occupancy changes from 4/(4 + 4) = 0.50 to 4/(16 + 4) = 0.20. The preparation now binds 2 nM instead of 5 nM. That is a rightward shift and reduced occupancy at this ligand concentration, without a loss of maximum binding capacity.

Normalized curves can hide lost sites

Imagine instead that half of A’s sites become unavailable while the remaining sites retain Kd = 2 μM. Bmax falls from 10 to 5 nM. At ligand 2 μM, raw binding falls from 5 to 2.5 nM, yet normalization to each preparation’s own plateau gives θ = 0.5 in both. The normalized curves overlap. Retain the raw plateau and the amount of protein assayed before concluding that a treatment left binding unchanged.

Conversely, doubling the concentration of active protein can double raw binding at every ligand concentration without changing occupancy or affinity. A capacity difference can reflect protein amount, available sites per protein, or inactive protein. The binding curve alone does not choose among those explanations. Matched protein concentrations and an independent functional-site estimate sharpen the inference.

Connect binding to function without replacing the measurement

Transport proteins use reversible association to load and release cargo; antibodies recognize particular molecular surfaces. Motor proteins coordinate binding and conformational changes in a working cycle. Their biological outputs are distinct. Stronger binding is not automatically better transport, recognition in every context, or faster movement. For example, a carrier must release its cargo as well as acquire it. A binding assay establishes an interaction; a functional assay tests what that interaction accomplishes.4

When approaching a new experiment, state the measured quantity and units, identify whether the horizontal axis gives free or total ligand, compare the plateau before normalizing, and then compare midpoint and curve shape. Only after those observations should you name an affinity change, a capacity change, or evidence consistent with cooperativity.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

Two proteins each obey θ = L/(Kd + L) with one class of independent sites. At free ligand L = 6 μM, P has θ = 0.75 and Q has θ = 0.40. Which conclusion follows under matched equilibrium conditions?

A. P has lower affinity because its Kd is 2 μM.

Review this choice. The value for P is correct, but lower Kd means higher affinity in the supplied model.

B. P has lower affinity because its Kd is 8 μM.

Review this choice. Solving 0.75 = 6/(Kd + 6) gives Kd = 2 μM, not 8 μM.

C. Q has lower affinity because its Kd is 4 μM.

Review this choice. Q has lower affinity, but its Kd is 6(1/0.40 − 1) = 9 μM. The unbound fraction alone is not Kd.

D. Q has lower affinity because its Kd is 9 μM.

Correct. Kd = L(1/θ − 1). P gives 2 μM and Q gives 9 μM; Q therefore has lower affinity.

The principle to carry forward

Compare dissociation constants using the supplied free-ligand model; a larger Kd means lower equilibrium affinity.

Check 2

For a protein with independent equivalent sites, treatment changes the measured saturation plateau from 30 to 60 nM bound ligand. Both before and after treatment, half of each preparation’s own plateau occurs at 4 μM free ligand. Which interpretation is supported?

A. Capacity decreases; Kd increases.

Review this choice. The raw plateau rises, so capacity increases. The unchanged half-occupancy concentration does not support an increase in Kd.

B. Capacity increases; Kd is unchanged.

Correct. The plateau measures Bmax and doubles. Under the given model, the half-occupancy concentration gives Kd = 4 μM in both conditions.

C. Capacity is unchanged; Kd decreases.

Review this choice. The plateau shows a capacity change; a lower Kd would shift half-occupancy to less than 4 μM.

D. Capacity is unchanged; Kd increases.

Review this choice. Capacity changes, while the given Kd indicator remains the same.

The principle to carry forward

Read capacity from raw binding and affinity from the appropriate occupancy relationship; normalization can conceal a capacity change.

Check 3

A cooperative protein follows θ = L²/(K0.5² + L²), with concentrations in μM. A regulator binds at a separate site. With regulator present, θ = 0.20 at free ligand L = 3 μM; without it, K0.5 = 3 μM. Capacity and the exponent remain unchanged. What is K0.5 with regulator present?

A. 1.5 μM

Review this choice. K0.5 = 1.5 would give θ = 9/(2.25 + 9) = 0.80, opposite to the observed decrease.

B. 3 μM

Review this choice. An unchanged midpoint would give θ = 0.50 at L = 3 μM.

C. 6 μM

Correct. 0.20 = 9/(K0.5² + 9), so K0.5² = 36 and K0.5 = 6 μM. The midpoint shifts to higher free ligand.

D. 12 μM

Review this choice. K0.5 = 12 would give θ = 9/153, about 0.059, not 0.20.

The principle to carry forward

A rightward midpoint shift can reduce occupancy at a fixed ligand concentration without reducing capacity.

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Sources

1. AAMC: B/B Content Category 1A

2. Carnegie Mellon University: Biochemistry I — Ligand Binding

3. Henry Jakubowski, College of Saint Benedict/Saint John’s University: Binding of a Single Ligand

4. Alberts et al.: Molecular Biology of the Cell — Protein Function