Keep the concentration of the FAM-oligo constant, at as low a concentration as you can manage with your equipment to get a good signal. Titrate the enzyme concentration [E]. Fit to the binding equation by nonlinear regression.

Y = Yo + MAX*[E]/(Kd + [E])

where Y is the polarization at [E], Yo is the polarization when [E] = 0, and Yo + MAX is the polarization at saturation.

This equation applies to the case where the Kd is well above the oligo concentration. If not, you have to use the quadratic Morrison equation instead.

To verify that the binding is to the oligo and not the fluorophore, demonstrate that you can fully compete the binding with unlabeled oligo having the same sequence.

@Adam B Shapiro How would I decide which concentration and what is the "good signal" for this instrument ? Second, is this the same equation used for "one site specific binding" in graph pad prism

(Y= Bmax*X/[Kd+X] that is used in the non linear regression to determine the Kd ? First I don't know the Kd of this substrate, so how would I confirm that obtained Kd is the right one and then what would make me decide to use other equation in order to determine the Kd ? It's not clear to me. Please explain.

I consider a good signal for fluorescence polarization to be achieved when the concentration of the fluorophore results in a total fluorescence intensity that is about 20-fold higher than the background. However, this is arbitrary. You can use less fluorophore if you have to. It will result in a higher background polarization because the background fluorescence is highly polarized. Since you are only interested in the change in polarization, rather than the absolute polarization, you can get away with a relatively high background polarization due to a low fluorophore concentration.

The equation I showed is the same as the one you showed, except for the Yo. If you subtract the baseline value of polarization from your titration, you can use the equation you showed.

The point is to measure the Kd, since you don't know what it is. You can use other methods to measure it besides fluorescence polarization and see if the results match.

If the equation gives a poor fit to the data, then you must consider whether the binding model is correct. One reason for a poor fit would be that you are working in the tight-binding regime, in which the Kd is similar to or lower than the concentration of the fluorophore. Another reason could be the existence of multiple interacting binding sites. Both of these cases would require using a different equation. The one we are discussing is the simplest case of a single binding site.

Adam B Shapiro Thank you for your explanation, I got your points now. Yesterday I ran the experiment as per your suggestion titrated the enzyme at keeping the substrate (probe) constant. Now, as I used the substrate concentration 25 nM which is lower than the previously used. This substrate has been shown to be lesser rate of enzyme degradation over the another. Please see the attached figures of the plot and data obtained (need to be repeated). My concern now is that the other substrate which is expectedly with higher rate of enzyme degradation or with higher affinity, what concentration should I take in that case ? Should I proceed with similar experiment design for other tight binding substrate ? As you said tighter binding would have a different way of determining, so which non-linear equation I will be using in that case ? Is that quadratic equation available in Graph pad (There are many equation present in the graph pad but not sure which one would be used in that case).

Please see this web page for tight binding nonlinear regression in GraphPad:

https://www.graphpad.com/guides/prism/latest/curve-fitting/reg_morrison.htm

The one shown is for enzyme inhibition, but essentially the same one can be used for binding, just omit the 1- at the beginning.

If the enzyme is degrading the substrate during the binding experiment, you are likely to get inaccurate Kd measurements, so I do not recommend making such a measurement under those conditions. Instead, you could measure the Km for the degradation reaction.

Adam B Shapiro Thanks again for this great explanation. Well, Mg2+ is an essential cofactor for the enzyme for degradation. However, here I am using Ca2+ instead to halt the reaction. But still I will test this on gel for full confirmation.

Thank you

Hi Adam B Shapiro,

Thank you very much for your sharing on FP. Regarding the signal-to-noise ratio of total fluorescence intensity for a reliable FP assay in general, could you please bring some relevant documents to me?

Thank you very much!

I don't have any documentation, although you can probably find some on-line resources. The following is based on my personal experience, measuring fluorescence polarization with plate readers.

If you prepare a set of dilutions of your fluorescent molecule, including buffer alone, and measure the fluorescence polarization (or anisotropy) with a suitably equipped plate reader, you will see that the polarization decreases as the concentration of the fluorophore increases from zero. This is due to the fact that the background noise has a very high polarization but the fluorophore has a much lower one (assuming it is a relatively small molecule). The reason the background noise has a high polarization is that it consists, in part, of stray light in the instrument. This light is highly polarized either because it has passed through the excitation polarizer filter or because it has bounced off of surfaces within the instrument (reflected light is highly polarized).

Since the amount of stray light is very small, it makes a substantial contribution to the overall fluorescence only when the fluorophore concentration is very low. As the fluorophore concentration increases, the fluorescence polarization approaches that of the fluorophore. In my experience, once the fluorophore fluorescence is about 20 times the background fluorescence, the polarization measured is almost as low as that of the fluorophore. This is just a rule-of-thumb. What you see will depend on the instrument you are using.

With a bright (high quantum yield) fluorophore, a high-quality instrument and optimized filters, it should be feasible to make fluorescence polarization measurements with as little as 1 or 2 nM fluorophore, if necessary, even if the fluorescence is not 20 times the background fluorescence at that concentration.

Another issue is the noise (as distinct from the background) in the polarization measurement. By this I mean the variation between individual measurements of the same sample. With plate readers, the measurement consists of the average of multiple lamp flashes. By increasing the number of flashes used per measurement, you can increase the signal-to-noise. This comes at the cost of increasing the amount of time it takes to make each measurement, and it may also cause bleaching of the fluorophore if you are using a fluorophore that is prone to rapid bleaching, such as fluorescein. I recommend using the highest number of flashes that are consistent with the time available for the measurement and with avoiding substantial bleaching.

I'm doing this now with an endonuclease. Adam's suggestion to titrate the enzyme against the fluorescent ligand is the easiest way to go. I believe the signal also depends on the path-length of the sample, so increasing volume in a plate well would probably allow you to reduce the concentration of the fluorophore without losing signal.

Personally, I want to use small sample volumes to conserve reagents, so I use 12.5 nM FAM labelled oligo. I could maybe go down a bit, but it doesn't matter as long as you fit the data with the tight binding equation.

I believe you can also titrate the fluorescent species against the non-fluorescent binding partner, or maybe that only works with fluorescence intensity measurements...

Dear Adam B Shapiro ,

Thank you very much, I appreciate your detailed comments! I learn and recognize a lot from them.

I am doing FP with FITC-oligos and see a decrease of FPs over an increase of FITC-oligo concentrations. Then, FP values keeps stable upon a high concentration. For choosing working concentration of oligo, do you have any rule-of-thumb? Will you choose the concentration at which FPs keep stable or at a concentration with the ratio of fluorophore fluorescence intensity to background is ~20?

With many thanks!

Measurements of fluorescence polarization have higher signal-to-noise when the fluorophore concentration is chosen so that the fluorescence intensity is high compared to the background. For FITC, this might be in the range 10-100 nM, but it depends on the instrument and other factors. You might want to conserve your supply of FITC-oligo by using a concentration on the lower end of this range.

To measure binding affinity of a macromolecule to a fluorescent ligand, it is only necessary to measure the change in polarization, not the actual polarization. Therefore, it is not necessary to use a fluorophore concentration that is high enough to reach the minimum stable polarization. You can use a lower concentration if it suits your needs.

It is sometimes desirable to use the lowest practical concentration of the fluorophore. This is when the dissociation constant of the interaction is close to the fluorophore concentration, which is known as the tight-binding regime. In this regime, the simple binding isotherm is inaccurate* and a more complicated calculation of Kd is required. You may be able to avoid this situation by reducing the fluorophore concentration.

*This is because the simple binding isotherm assumes that the free ligand concentration is equal to the total ligand concentration, which is not the case in the tight-binding regime. The correct analysis would be to use the so-called Morrison equation, which uses the total ligand concentration.

Adam B Shapiro Dr. Shapiro, Here is the result from Fluorescence Polarization experiment of two different ligands. In this, I used 20nM fixed concentration of substrate (FAM labelled) and varied the Enzyme concentration. I plotted in Graph pad using non linear regression (One binding site-specific) after normalizing the data. The curve obtained has a fit with R square value ~80%. I am worried if this is a good fit. If not, then how would I be sure which model is a good fit for this data ? I have attached the results for your reference. Please suggest.

It's not a great fit, but it is important to distinguish between a poor fit caused by scatter in the data and a poor fit caused by an incorrect binding model. More elaborate binding models will always fit data better because there are more adjustable parameters. You should be careful not to assume that the more complex model is correct just because it fits the data better.

Having said that, to my eye it looks like these data would be well fit by a model involving a high affinity binding mode and a low affinity mode. I suggest you repeat the experiment a few more times to see if you get consistent results.

I'm not sure what you mean by normalizing the data. Was it just subtracting the polarization at zero protein? It should not be necessary to do that in order to do the nonlinear least squares analysis, as long as you include a constant term for the polarization at zero protein.

• Adam B Shapiro Hello Dr. Shapiro ! Sorry for delayed response. As per your suggestion, I repeated the experiment about seven times and each time the polarization values came out very similar and consistent. The Kd values for substarte1 and substrate2 has been calculated to 254 (+ - 50) nM and 260 (+ - 50) nM respectively (with varying enzyme conc. and fixed substrates conc. at 20nM), which is almost identical (data shown in the attached plot). I have fit the data in non-linear regression with one specific binding site as done before. The data looks consistent. However, the kinetics of the enzyme are apparently differ for both the substrates (qualitatively observed). Is it possible that Kd could be almost identical but the kinetics be different ? Both the substrates are identical but with two nucleotide recessed from 3' end in one called "processed". So, how should I rationalize this observation of result, please suggest.

Yes, it is possible for two substrates to have the same Kd but different kinetics.

For a single-substrate enzyme reaction, the definition of the Kd is k-1/k+1.

The definition of the Michaelis constant for steady-state kinetics Km is

(k-1 + k+2)/k+1 and kcat is k+2.

E + S ES (forward rate constant is k+1, reverse rate constant is k-1)

ES => E + P (forward rate constant is k+2=kcat)

In the special case where kcat is very slow compared to the rate of equilibration of the substrate with the enzyme (rapid equilibrium), then Kd = Km.

If kcat differs for the two substrates and the reaction is not a case of rapid equilibrium, then the kinetics will be different for the two substrates.

Dr. Shapiro, Thanks a lot for your great help. It is ectremely helpful to every one here. Thank you again.