please consider editing your question to include the model formula.

I copy Victor Tapia, it is also really important how your spectra look like: do you have enough points describing the transition (before / during / after)? Which is your fitting error? How much noise do you have to subtract?

Which value do you get for the affinity constant in either case? 

Could it be that the addition of EDTA weakens the interaction so much that you see only background peaks and no binding? I suggest you fix the concentration of your protein to the value which you have obtained without EDTA.

Besides that, what value of N do you get if you don't adjust the concentration of the polysaccharide? Is it possible that one polysaccharide binds more than one protein molecule?

Hi Arun,

I do not quite understand what do you mean with "EDTA TREATED" protein. Is there EDTA in the ITC cell together with the protein? Or the protein is dialyzed in EDTA and then EDTA is removed?

Cheers

Marina

@marina, Hi., the protein was incubated with EDTA for 2 hours followed by buffer exchange using Amicon ultra centrifugal filters (protein concentrator) to remove EDTA.

Dear Victor,unfortunately ,i looked for the formula but could not get it.  I use 'one set of sites' model of  the Origin software. i'l post as son as get it.

 Dear Luca, i think the spectra should be able to deliver the parameters, i have attached the results (along with the question) for your reference. By noise you mean the heat of dilution of ligand? Compared to the heat produced by protein-ligand interactions 'noise' were very low.

Hi Arun,

As I understand, you do EDTA treatment to remove ions. If you get different binding patterns before and after ion removal it is likely that conformation of your protein depends on the presence of ions. Consequently, the amount of polysaccharide binding sites depends on protein conformation.

I suggest to investigate EDTA treated and untreated protein with CD or another structural technique.

Marina

 Dear Raffaele, in the first experiment Ka was 1x103M-1  and in the second experiment  Ka was 1.9x103M-7 and in the second experiment .

Dear Gerrit , thank you for the suggestion , i will get back with the values soon.

Thanks Marina! The EDTA treatment was done to remove metal ions. I also have the same view that ions are affecting the protein structure and in turn the binding sites. I just would like to be able figure out the number of sites.

arun

Hi Arun, I have had a look at your experiments. As far as I can understand, with N = 1, the values of the affinity constant should be 1x10^3 e 1x10^7 for the first and the second experiment, respectively, which, together with the shape of your titration curves, suggests a different affinity of the polysaccharide for the EDTA-treated and the untreated protein. Once you have fixed the protein concentration, this does not necessarily mean that N is not 1 after EDTA treatment, but that the ligand concentration at which N = 1 cannot be the same, depending on the affinity constant (the higher the affinity constant, the lower the ligand concentration). I agree with Marina Kasimova that EDTA treatment could modify the protein conformation, possibly by changing the amount of protein-bound ions, but it could also happen that for some reason the total protein concentration is higher than the calorimetrically active protein amount. For optimal evaluation of the binding parameters, I suggest that you vary the protein concentration until you obtain a sigmoidal shape for your titration experiment (the optimal protein-binding constant product should be 10-100 with one binding site).

Dear Raffaele, i will surely try your suggested experiment. i also have second thoughts about the enthalpy change in second reaction ,it is unusually high around a 100kcal/mol. Any thoughts about that?

thanks

 Were you able to recover the protein after the titration?

The fit for the low affinity titration is quite bad. You can see that the continuous line does not represent well the experimental points. Very surely the fitting can be improved and the affinity will be higher. Accordingly, the binding enthalpy will be lower and the N value will be between 1 and 2 (around 1.5).

Regarding the high affinity titration, the binding enthalpy is enormous (-100 kcal/mol). I think that is due to the "normalization" you have imposed in order to get N = 1. Stoichiometry, enthalpy and ligand concentration are correlated. This is the problem when ligand concentration is not known (or the stoichiometry is not known). You should find a way to determine the ligand concentration (e.g., elemental analysis).

When performing a titration, it is very convenient that you know the concentration of at least one of the reactants. In the case of protein-ligand interaction, because the protein may be inactive in some percentage (misfolded, aggregated, oxidized, other contaminants...) and the ligand is usually less susceptible to degradation, a robust and reliable method for determining the ligand concentration is crucial. Remember that the word "titration" comes from Latin "titulus" (through French, "titre"), meaning "inscription or title", that is, the "title of a solution = concentration". Thus, with a solution of known concentration you may find out the unknown concentration of another solution. In your titrations, knowing the ligand concentration precisely will help in estimating the stoichiometry in both cases (before and after EDTA treatment) and/or the concentration of active protein. You have changed the ligand concentration in order to get N = 1 (judging from a bad fit in the case of the low affinity), which is not necesssary true (in any of the titrations).

On the other hand, it is posible, as Marina commented, that metal ions induce conformational changes that expose/bury ligand binding sites. Therefore, the stoichiometry is not necessary the same after EDTA treatment.

Dear Raffaele, yes we can recover the protein. You suggest to do some experiments with the recovered protein?

Dear Adrian, Thanks for the insights on the low affinity titration. I wold try it again. When you say 'fitting can be improved..' for the low affinity titration do you suggest to use the same data to change concentration and get a better fit or repeat the experiment.?

Yes, you do not have to repeat the experiment; you can reanalyze the titration, and there are several issues, being the main ones: 1) there is an inflection point, which indicates that the dissociation constant is smaller than the protein concentration; 2) the "background heat" or "dilution heat" is considerable, and this is something that the built-in functions in Origin cannot handle, you must subtract manually a quantity and shift the titration vertically through the y-axis.

Dear Adrian, you indicated that the N value would be around 1.5 when the low affinity titation is modeled correctly. Would you say that the protein has more than one binding sites, may be 2.?

Well, it depends on how accurate the concentration of ligand and protein are determined. If they are wrong (one of them or both), the resulting N may be incorrect.

Looking at your low affinity data I have two comments:

1. The standard ITC-fitting algorithm does not have an offset. This means that you usually have to substract a background value which may be explained by the heat of dilution. In your case it would be done my moving the data points up by about 1 kcal/mol. The function is called "smple math".

2. You should delete the first data point which is often smaller than it should be.

In the high affinity experiment you could also move the points further up up it matters less because the initial peaks are much larger.

In theory, one should do a control experiment by titrating the ligand into buffer and substract that measurement from the binding experiment. For me it didn't work very well and you have to use a large amount of ligand.

What I usually do is to move the data points up and down until I reach a minimum value for Chi^2. In most cases the vakue is very close to the size of the last injections especially if the affinity is high and the protein in the cell is saturated.