Q.44 A receptor binds to its ligand with a dissociation constant Kd = 10−8 M. The concentration of the ligand required to occupy 10% of the receptors would be 10−x M. The value of x is _____ .
The ligand concentration required to occupy 10% of receptors with Kd = 10⁻⁸ M is 10⁻⁹ M, so x = 9.
Core Concept
Receptor-ligand binding follows the law of mass action at equilibrium. The fractional occupancy θ equals [L] / ([L] + Kd), where [L] is free ligand concentration and Kd is the dissociation constant.
Step-by-Step Solution
For 10% occupancy, set θ = 0.1. Substitute into the equation: 0.1 = [L] / ([L] + 10⁻⁸).
Rearrange: 0.1[L] + 0.1 × 10⁻⁸ = [L].
0.1 × 10⁻⁸ = [L] – 0.1[L] = 0.9[L].
[L] = (0.1 / 0.9) × 10⁻⁸ = (1/9) × 10⁻⁸ = 0.111… × 10⁻⁸ = 1.11 × 10⁻⁹ M, or 10⁻⁹ M.
This assumes [L] ≈ total ligand (valid when receptor concentration << Kd).
Key Relationships
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θ = 0.5 when [L] = Kd (50% occupancy).
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θ = 0.1 when [L] = Kd/9 ≈ 0.11 Kd.
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θ = 0.9 when [L] = 9 Kd.
The receptor ligand dissociation constant Kd governs binding affinity in molecular biology. For a receptor with Kd = 10⁻⁸ M, determining the ligand concentration for 10% occupancy is crucial for CSIR NET exam success. This calculation uses the occupancy equation θ = [L] / ([L] + Kd), where θ = 0.1 yields [L] = 10⁻⁹ M.
Binding Equation Breakdown
The equation derives from equilibrium: RL ⇌ R + L, where Kd = [R][L]/[RL].
Fractional occupancy simplifies to θ = [L] / ([L] + Kd).
Solving θ = 0.1 confirms [L] = Kd/9 = 10⁻⁹ M exactly.
CSIR NET Exam Relevance
Competitive exams test this via numerical fill-ins. Common traps include confusing Kd with 50% occupancy or ignoring approximations.
Practice: For Kd = 10⁻⁷ M, 10% occupancy needs 10⁻⁸ M.
Practical Applications
In drug design, low [L] for partial occupancy predicts dosing. Biotechnology uses this for sensor optimization.
