Re: How to fit a Hill model using braidrm package?

Hi,

If your main goal is Kd estimation, I would try drc :: drm approach

Best
SV


> Le 24 sept. 2025 à 11:52, Luigi Marongiu <[email protected]> a écrit :
> 
> Hello,
> I have a set of data coming from a dissociation experiment
> (protein/ligand). Since the data is required to calculate the constant
> of dissociation (Kd) of this pair, I am looking for a way of fitting a
> Hill function to the data.
> I have seen that the package braidrm
> (https://cran.r-project.org/web/packages/braidrm/index.html) provides
> this function, but when I launch the function `evalHillEqn`,
> `findBestHill` and so forth I get the error of function not found.
> Yet, the package is given as properly installed by the system.
> How can I run this package?
> Is there an alternative way to fit a Hill function to these data?
> Thank you
> 
> ```
> df = data.frame(Response =    c(890.72,    895.46,    895.63,
> 894.72,    895.49,    893.59,
>                892.53,    895.06,    897.21,    889.27,    876.05,
> 857.96,    862.02,    858.36,
>                890.94,    890.8,    887.22,    888.91,    890.83,
> 889.92,    891.76,    890.32,
>                886.35,    878.11,    866.57,    859.04,    863.64,
> 880.16,    884.15,    879.57,
>                878.89,    882.27,    881.59,    880.98,    881.45,
> 876.19,    868.32,    859.16,
>                850.53,    853.21,    859.34,    859.73,    861.19),
>                Dose =    c(0.0000000015,    0.000000003,
> 0.000000006,    0.000000012,
>                       0.000000024,    0.000000048,    0.000000095,
> 0.00000018,
>                       0.00000038,    0.00000078,    0.0000015,
> 0.000013,    0.000025,
>                       0.00005,    0.0000000015,    0.000000003,
> 0.000000006,
>                       0.000000012,    0.000000024,    0.000000048,
> 0.000000095,
>                       0.00000018,    0.00000038,    0.00000078,
> 0.0000015,    0.000025,
>                       0.00005,    0.0000000015,    0.000000003,
> 0.000000006,
>                       0.000000012,    0.000000024,    0.000000048,
> 0.000000095,
>                       0.00000018,    0.00000038,    0.00000078,
> 0.0000015,    0.000003,
>                       0.000006,    0.000013,    0.000025,    0.00005)
> )
> plot(Response~log10(Dose), df)
> library(braidrm)
> evalHillEqn(Response~Dose, df)
> findBestHill(Response~Dose, df)
> ```
> 
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