Just to get this clear for me: you measure trp fluorescence somewhere ex 280/295 and em 340 nm. You add substrate 1 (small molecule? protein?) to your protein solution and can observe a change in trp fluorescence. You add substrate 2 and also see some change in tryptophane fluorescence.

You just observe snapshots before and after addition of your substrate? I would try time course measurements at first. When fitting the data of S1 and S2 with monoexponential functions maybe you already see a difference in half life values/ tau? When adding both S1 and S2 and measuring time course and there is competitiveness I would expect either changes of tau or maybe even two populations that you can observe e.g. by a biexponential fit that has matches your data better than a monoexponential fit.

You can also calculate the Vmax and Michaelis menten constant of your reaction. There are distinct rules here: if you have a competetive substrate the menten constant increases by a factor and Vmax remains unaltered. This is different for allostery: Km remains the same and Vmax changes.

Another thing is introducing a second label - I recently tried labeling my substrate with a MANT group which gives an excellent FRET signal with the Trp Emission at 340 nm up to about 430 nm.

If I got anything wrong or you have any questions regarding my suggestions do not hesitate to ask

Greetings from Bochum, GER and good luck with your experiments

Daniel

I would perform a fluorescence titration with substrate S1in the presence of a fixed concentration of substrate S2 and in the absence of substrate S2. You will get two dissociation constants for S1, Kd1_S2 and Kd1, corresponding to the two previous situations. in addition, you must perform a titration with substrate S2, from which you obtain the dissociation constant Kd2. If Kd1_S2 and Kd1 are equal, the binding of the two substrates is independent; if Kd1_S2 is lower than Kd1, S2 promotes the binding of S1 (positive heteretropy); if Kd1_S2 is larger than K1d, S2 interferes with binding of S1 (negative heteretropy). In the last case it is posible to discriminate between negative cooperativivity and competitive ligands (maximal negative cooperativity):

if, within the experimental errors, the three dissociation constants fulfil this relationship, Kd1_S2 = Kd1 x (1 + Kd2 x [S2]), the two substrates are competitive; otherwise, they are not competitive.

Sorry, there is a mistake in the previous post. If the two substrate bind competitively the three dissociation constants fulfil this relationship:

Kd1_S2 = Kd1 x (1 + [S2]/Kd2)

Thanks very much Adrian and Daniel. I really appreciate your insightful answers.

Adrian, do you have a reference where I can read more about the methodology you have explained?

Ahmed

For a protein capable of binding two different ligands, S1 and S2, the following relationship holds:

Kd1_S2 = Kd1 x (1 + [S2]/Kd2) / (1 + alpha x [S2]/Kd2)

where Kd1 and Kd2 are the dissociation constants for S1 and S2 interacting with the protein, Kd1_S2 is the apparent dissociation constant for S1 in the presence of S2 at a concentration [S2], and alpha is the binding cooperativity interaction constant. You must perform a titration with protein and ligand S1 to determine Kd1 and another titration with protein and ligand S2 to determine Kd2. Then, you perform a titration with protein and ligand S1 in the presence of ligand S2 at a fixed concentration [S2], to determine Kd1_S2. The previous relationship will provide the information about the potential heterotropic interactions between ligands S1 and S2. If alpha is equal to 1, the binding of both ligands is independent; if alpha is larger than 1, the binding of both ligands shows positive cooperativity; if alpha is smaller than 1, the binding of both ligands shows negative cooperativity; if alpha equal to zero, the binding of both ligands is competitive (maximal negative cooperativity or excluding binding).

You may look up these two references:

http://www.ncbi.nlm.nih.gov/pubmed/16766617

http://www.ncbi.nlm.nih.gov/pubmed/20028974

which are focused on isothermal titration calorimetry, but the basic equations are equally valid for any technique for studying binding interactions. But a very comprehensive, must-read text on biological interactions is the book: Binding and Linkage. Functional Chemistry of Biological Macromolecules, by Jeffries Wyman and Stanley J. Gill.

Great comprehensive answer. I'm really grateful.

Thanks a lot.

Hi Ahmed and Adrian

I followed the same protocol and method you described in this post. I had same kind of study where I have a molecule A which binds with a protein. The protein also binds with B and C which are its natural substrate for normal enzymatic function. I wanted to access the binding site of A w.r.t. to B and C and figure out whether A is competitive to B or C.

I did fluorescence for A, B and C individually and then again titrated with A in presence of fix concentration of B and C separately. And as per above equation I tried to fit the data and in both case I got alpha = 0.480 and 0.469 respectively. Now what does this value indicates? Does it mean that the binding of both ligands shows negative cooperativity or the binding of both ligands is competitive?

I used the fixed concentration of B and C as twice the Kd obtained from titration of B and C.

An alpha value lower than 1 indicates negative cooperativity. For competitive binding (maximal negative cooperativity) alpha must be zero, although in practice a value sufficiently small is enough (according to my previous answer, when alpha