SME
Cell & Molecular Biology
Signal transduction
Use this for “A receptor has Kd = 2 nM. What fraction is bound at 6 nM, and what ligand level gives 90%?”
Chart
θ(6) = 0.75
Half bound at L = K = 2: θ(K) = 0.5, half of 1
This is an example. Type your own values to start a new problem.
θ
=
[L] (nM)
Kd (nM)
+
[L] (nM)
Formulas
θ
=
[L]
÷
(Kd
+
[L])
0.75
=
6
2 + 6
Assumptions
•
Each receptor has one binding site, and ligand binds and leaves until it reaches equilibrium.
•
Ligand is in large excess: binding barely lowers the free ligand [L].
•
Kd is the ligand level that fills half the receptors: a smaller Kd means tighter binding.
•
θ climbs toward 1 but never reaches it: 90% needs 9 Kd, 99% needs 99 Kd.
Step-by-step
WE KNOW
Kd = 2 nM, [L] = 6 nM
FIND
fraction bound (θ)
Step 1 · Find fraction bound (θ)
Use θ = [L] ÷ (Kd + [L])
The bound share is the ligand level over Kd plus the ligand level.
θ
=
6
2 + 6
θ
=
6
8
θ
=
0.75
Check
Put the numbers back in. Both sides should match.
0.75
=
6
2 + 6
✓
Topic check
See what you know: 5 questions on this topic
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