# Calculate IC50 from fluorescence kinetic

I have an assay in which a fluorescence signal is generated when the enzyme reaction progresses. The fluorescence signal increases until a maximum is reached. Than it keeps constant. The first 1 h appears to be linear.

My data therefore looks in principle like this

(taken from another publication, my exact data points look different of course)

Now I want to titrate an inhibitor into the solution which inhibits the enzyme. I expect, that the initial speed of the enzymatic reaction decreases and therefore the slope of the initial linear part. If I get the idea right, I compare the slope of the linear part of all inhibitor concentrations and I compare it further it to the slope of the control without inhibitor. Therefore I determine how much of activity is left in percent in comparison to the control in which no inhibitor is added. If the slope is 59 units is comparison to 100 units of the control the activity would would be 59 percent and the inhibition 41 percent if I am right.

When you check the literature, there are figures that look like this, where Inhibition is plotted against concentration of the inhibitor. A paper calls for the fitting of the curve by this equation:

$$I =\frac{I_{max}C}{IC_{50}+C}$$

where $$I$$ is the percentage inhibition $$I_{max}$$ is 100% inhibition, $$C$$ is the concentration of inhibitor, and $$IC_{50}$$ is the concentration of inhibitor for 50%.

I am wondering about multiple things here:

1. What is meant by $$I_{max}$$ ? Is it the absolute value of fluorescence at 100 % Inhibition? Or is this the max inhibition for a specific concentration? But then it would be the same as $$I$$?

2. Is there any guidance/tutorial to realize this plotting for a software like graphpad or a good resource you can recommend?

3. I still need to draw a conventional dose response curve (activity against lg c) or is the the plot inhibition against concentration sufficient?

• Hello!!! (1) It is mathematically impossible for your equation to have a value of $I_\mathrm{max}$, but it tends to that value when $C \rightarrow \infty$. For practical purposes, when $C$ is several times higher than $IC_{50}$, then essentially $I \approx I_\mathrm{max}$. (2) If your university has money use Origin or MATLAB, if not use Octave or code in Python. (3) Rearrange your equation, invert it, and check that a plot of $C/I$ vs $I$ gives a linear plot, in which the slope is $1/I_\mathrm{max}$ and the y-ordinate is $b = IC_{50} / I_\mathrm{max}$. Commented Jan 15 at 1:51
• The equation looks similar to.the Michaelis Menton equation. I am wondering, is $I_{max}$ always the same value in a titration row? For example always 98 percent inhibition, regardless of the inhibitor concentration? Commented Jan 15 at 2:23

1. You start with $$c=0$$ and increasing inhibitor concentration to $$c_{max}$$
2. Generate linear equations $$y=mx+b$$ for the first linear segment of the fluorescence signal, in my case I used the first 60 minutes from the fluorescence over time graphs. In this case the change of fluorescence is equivalent to the change of product concentration, and therefore the reaction speed over time:
$$\frac{dF}{dt}=\frac{dP}{dt}$$
4. The calculated Km value equals the $$IC_{50}$$