Suppose I am using a standard inhibition model to find biochemical parameters that fit my data. The equation is:

$y = \frac{A}{{1 + \exp \left( {\ln \left[ S \right] - \ln IC_{50}} \right)}} $

where $\left[ S \right]$ is the concentration of my inhibitor and $IC_{50}$ is the concentration of my inhibitor at which the measurement (with a maximum $A$) is reduced by half. Which of these approaches should I take?

  1. enter the equation into the NLR procedure as-is (FYI: I am using SPSS) and let it fit the values of $A$ and $IC_{50}$ in the same manner, i.e. with all of the assumptions of OLS regression parameters. Or...

  2. enter the equation as

    $y = \frac{A}{{1 + \exp \left( {\ln \left[ S \right]- \ IC_{50}^* } \right)}} $

    where $ IC_{50}^* = \ln IC_{50}$. This of course would require a transform on the output parameter and confidence limits of $ IC_{50}^*$ giving me asymmetric error bars.

Which of these strategies is most rigorous? My instincts suggest the 2nd option, as $ IC_{50}$ is actually bound by 0 and is thus more likely to be log-normal rather than normal. Any help (direct answers, references, etc) is appreciated.


I assume that the log(S) vs. Y curve is symmetrical, so the uncertainty of the log(IC50) is symmetrical and the uncertainty of the IC50 is not. Accordingly, you are much better off using your second model. You'll get the same the best-fit curve, the same best-fit value for the IC50 (after transforming), and the same sum-of-squares and R2. But the two methods will give you different confidence intervals of the IC50 (even after transforming) and the second method will give you a more useful (more accurate) confidence interval.

  • $\begingroup$ By symmetrical, you mean the variance of $y$ is symmetrical around the mean of $y$, correct? In the generalized case where this is true, is the uncertainty of each parameter also symmetrical around the prediction of that parameter? If that's true, then what you say makes perfect sense. $\endgroup$
    – DocBuckets
    May 14 '13 at 17:27
  • $\begingroup$ No, I mean the log(dose) v. response graph is symmetrical -- that the steepness from low concentrations to the IC50 is about the same as the steepness from the IC50 to high concentrations. If this is symmetrical when X is log(dose), it won't be symmetrical when X is concentration. $\endgroup$ May 15 '13 at 14:42
  • $\begingroup$ Oh, ok. So this answer is applicable in this case but not necessarily in the generalized case. Got it. On reading up more on the lognormal distribution, it seems that even method 2 isn't the best way to do it, but it's a lot better than method 1. That's answer enough for me- thanks for your help. $\endgroup$
    – DocBuckets
    May 15 '13 at 17:37

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