Hill Dose–Response: EC50 from a Logit Fit
Unknown agonist. Measure effect E at several concentrations and recover EC50 from a fit of logit(E/Emax) versus log₁₀ C. Emax is known.
University / research· 24 min·Related simulator: Biophysics, Fluids & GeoscienceDose-Response & Hill Curves
Goal
Determine EC50 from E = Emax C^{n_H} / (EC50^{n_H} + C^{n_H}). A plot of logit versus log₁₀ C has intercept/slope that give log₁₀ EC50.
Equipment
- Unknown agonist
- Known Emax
- Concentration series
- Effect assay
Experiment
Theory
With E₀ = 0, logit(E/Emax) = ln(E/(Emax−E)) = n_H ln(10) (log₁₀ C − log₁₀ EC50). The slope is n_H ln 10 and EC50 = 10^{−intercept/slope}. The bench hides EC50, n_H and live E. Emax is known. This is a dose–response lab, not Michaelis–Menten (enzyme rate).
Procedure
- Emax is fixed and known (100). You only change the agonist concentration C. Avoid the far tails where E ≈ 0 or E ≈ Emax (logit blows up).
- Record. An assay logs E with small noise. There is no live E, EC50 or n_H.
- The notebook computes log₁₀ C and logit. Repeat for at least 6 concentrations spanning about 0.3–30 around the unknown midpoint.
- Fit logit versus log₁₀ C; EC50 = 10^{−intercept/slope}. Compare with the reference.
Conclusion
The fitted EC50 agrees with the hidden agonist. Main uncertainties: assay noise and staying off the logit tails. n_H is not graded.