Question

Affinities of protein inhibitors. A receptor protein \(P\) binds a metabolite molecule \(M\) with affinity \(K_{\text {assoc }}=\) \(10^{6} \mathrm{M}^{-1}\). (a) What ligand concentration in solution is required so that \(75 \%\) of the receptor sites are occupied? (b) A competitive inhibitor drug binds to the same protein with nanomolar affinity, i.e., \(R_{\operatorname{assoc}}=10^{9} \mathrm{M}^{-1}\). If the inhibitor has concentration \(y=10^{-7} \mathrm{M}\) in solution, in the situation described above, what is the fractional occupancy of \(P\) now?

Short answer

a) \([L] = 3 \times 10^{-6} \text{M}\), b) \(\theta' \approx 0.0288\)

Step-by-step solution

- Understand the binding affinity

The binding affinity for a receptor protein and a metabolite molecule is given by the association constant. For protein P and metabolite M, the affinity is \(K_{\text{assoc}} = 10^6 \text{M}^{-1}\). This means the higher the affinity constant, the tighter the protein binds to the metabolite.

- Calculate the ligand concentration for 75% occupancy

To find the ligand concentration for 75% occupancy, use the formula for fractional occupancy: \[ \theta = \frac{[L]}{[L] + K_d} \] where \(\theta\) is the fractional occupancy and \(K_d\) is the dissociation constant (\(K_d = \frac{1}{K_{\text{assoc}}} = \frac{1}{10^6} \text{M}^{-1}\)). For 75% occupancy, \(\theta = 0.75\), set up the equation: \[ 0.75 = \frac{[L]}{[L] + 10^{-6}} \] Solve for \([L]\): \[ 0.75([L] + 10^{-6}) = [L] \] \[ 0.75[L] + 0.75 \times 10^{-6} = [L] \] \[ 0.75 \times 10^{-6} = 0.25[L] \] \[ [L] = 3 \times 10^{-6} \text{M} \]

- Understand competitive inhibition

A competitive inhibitor binds to the same site as the metabolite, competing for the receptor protein. The affinity for the inhibitor is given as \(R_{\text{assoc}} = 10^9 \text{M}^{-1}\), so the dissociation constant for the inhibitor is \(K_d = 10^{-9} \text{M}\). The concentration of the inhibitor is given as \([I] = 10^{-7} \text{M}\).

- Calculate the fractional occupancy with inhibitor

With the presence of a competitive inhibitor, the modified equation for fractional occupancy \(\theta'\) becomes: \[ \theta' = \frac{[L]/K_d}{[L]/K_d + 1 + [I]/K_d} \] Using \([L] = 3 \times 10^{-6} \text{M}\), \(K_d = 10^{-6} \text{M}\), \([I] = 10^{-7} \text{M}\), and \(K_{d, \text{inhibitor}} = 10^{-9} \text{M}\): \[ \theta' = \frac{(3 \times 10^{-6})/(10^{-6})}{(3 \times 10^{-6})/(10^{-6}) + 1 + (10^{-7})/(10^{-9})} \] Simplify: \[ \theta' = \frac{3}{3 + 1 + 100} = \frac{3}{104} \] \[ \theta' \approx 0.0288 \]

Key concepts

Dissociation Constant

The dissociation constant, often denoted as \(K_d\), reflects the binding affinity between a receptor and a ligand. It is the inverse of the association constant \(K_{\text{assoc}}\). Mathematically, the dissociation constant is calculated as: \[ K_d = \frac{1}{K_{\text{assoc}}} \] For example, if a protein and a metabolite have a high association constant of \(10^6 \text{M}^{-1}\), the dissociation constant will be: \[ K_d = \frac{1}{10^6} \text{M} = 10^{-6} \text{M} \] A lower \(K_d\) value indicates a higher binding affinity because it means that the ligand can bind to the receptor more tightly and is less likely to dissociate.

Fractional Occupancy

Fractional occupancy \(\theta\) represents the fraction of receptors that are occupied by the ligand at a specific ligand concentration. It is calculated using the formula: \[ \theta = \frac{[L]}{[L] + K_d} \] where \([L]\) is the ligand concentration. For instance, if we want to find the ligand concentration to achieve 75% occupancy (\(\theta = 0.75\)), we can set up the equation: \[ 0.75 = \frac{[L]}{[L] + 10^{-6}} \] By solving for \([L]\), we find: \[ 0.75 ([L] + 10^{-6}) = [L] \] which simplifies to: \[ [L] = 3 \times 10^{-6} \text{M} \] This concentration ensures that 75% of the receptor sites are occupied by the ligand.

Competitive Inhibition

Competitive inhibition occurs when a molecule, often a drug, competes with the ligand for the same binding site on the receptor. In this scenario, the presence of the inhibitor affects the fractional occupancy of the ligand.
The association constant for the inhibitor and its dissociation constant \(K_{d, \text{inhibitor}}\) are crucial. Given an inhibitor concentration \( [I] \), and the dissociation constant for the inhibitor, the modified equation for fractional occupancy \(\theta'\) is: \[ \theta' = \frac{[L]/K_d}{[L]/K_d + 1 + [I]/K_{d, \text{inhibitor}}} \] For instance, with \( [L] = 3 \times 10^{-6} \text{M}\), \( K_d = 10^{-6} \text{M}\), \( [I] = 10^{-7} \text{M}\), and \( K_{d, \text{inhibitor}} = 10^{-9} \text{M}\), the calculation becomes: \[ \theta' = \frac{(3 \times 10^{-6})/(10^{-6})}{(3 \times 10^{-6})/(10^{-6}) + 1 + (10^{-7})/(10^{-9})} \] Simplifying this, we get: \[ \theta' = \frac{3}{3 + 1 + 100} = \frac{3}{104} \rightarrow \theta' \thickapprox 0.0288 \] This shows that the ligand's occupancy significantly decreases due to the competitive inhibitor.

Binding Affinity

Binding affinity is a measure of the strength of the interaction between a receptor and its ligand. It is often expressed through the association constant \(K_{\text{assoc}}\) or the dissociation constant \(K_d\). Higher binding affinity means that the receptor and ligand bind more tightly.
The higher the association constant \(K_{\text{assoc}}\), the higher the binding affinity. For instance, if \(K_{\text{assoc}}\) is \(10^6 \text{M}^{-1}\), it implies a strong interaction between the receptor and the ligand. Conversely, a lower dissociation constant \(K_d\) also indicates higher binding affinity, as it implies the complex formed is stable and less likely to dissociate.
Understanding binding affinity is crucial in many biological and pharmaceutical contexts, such as drug design, where higher binding affinity often translates to more effective therapies because the drug remains bound to its target longer, producing a more sustained biological response.