Q.44 The dissociation constant for a receptor-ligand pair is 0.25 × 10-7𝑀 . The ligand
was added to a solution of the receptor such that the receptor was 50% saturated
at equilibrium. Assume that the receptor has one ligand binding site. The
concentration of the free ligand at equilibrium in 𝑛𝑀 , correct to the nearest
integer, should be _____.
The free ligand concentration at equilibrium is 2.5 nM.
Receptor-Ligand Binding Basics
Receptor saturation follows the equation θ = [L] / (Kd + [L]), where θ is the fractional saturation, [L] is free ligand concentration, and Kd is the dissociation constant.
At 50% saturation (θ = 0.5), this simplifies to 0.5 = [L] / (Kd + [L]), so [L] = Kd.
The problem gives Kd = 0.25 × 10-7 M = 2.5 nM, and assumes one binding site with no ligand depletion specified.
Step-by-Step Calculation
Convert Kd to nM: 0.25 × 10-7 M = 0.25 × 10-7 × 109 nM/M = 2.5 nM.
At 50% saturation, free [L] equals Kd regardless of total receptor concentration, as the equation depends only on free [L].
Rounded to the nearest integer, the answer is 3 (though exactly 2.5 nM).
Ligand Depletion Consideration
If receptor concentration [Rt] ≈ Kd, depletion occurs, and total ligand [Lt] = Kd + [Rt]/2 for 50% saturation.
However, the query asks for free ligand at equilibrium, which remains Kd = 2.5 nM.
No [Rt] is provided, confirming the direct use of Kd.
Core Binding Equation
The fractional saturation θ = [RL]/[Rt] = [L]/(Kd + [L]), where [L] is free ligand.
For θ = 0.5, [L] = Kd by direct substitution.
Here, Kd = 0.25 × 10-7 M = 2.5 × 10-9 M = 2.5 nM.
Detailed Solution Walkthrough
- Identify θ = 0.5 (50% saturation).
- Set 0.5 = [L] / (2.5 nM + [L]); solve: [L] = 2.5 nM.
- No total receptor [Rt] given, so ignore depletion (valid when [Lt] >> [Rt]).
- Convert and round: 2.5 nM → 3 (nearest integer).
Common Pitfalls Explained
- Option-like errors: Mistaking total ligand for free (would need [Rt]).
- Unit confusion: 0.25 × 10-7 M = 2.5 nM, not 0.25 nM.
- Depletion trap: Only if [Rt]/2 significant; here unspecified.
This matches CSIR NET expectations for precise equilibrium calculations.
💡 CSIR NET Quick Tip
Key Memory Point: At 50% receptor saturation, free [L] = Kd always (independent of total receptor concentration).
