1. An article (J. Agric. Food Chem. 1998, 46, 2671-2677) measured the binding co
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1. An article (J. Agric. Food Chem. 1998, 46, 2671-2677) measured the binding constants for ANS binding to BSA and ovalbumin (OVA). In each case the reaction is ANS + protein protein-ANS complex. K(BSA) = 6 X 106 and K(OVA) 2 X 10, ANS only fluoresces strongly when bound to protein. Set up equilibrium (K) calculations to calculate the following answer the one sig fig: You have two solutions with equal molar concentrations of BSA and OVA. You add some ANS to the BSA solution and observe a certain amount of fluorescence. How many times more grams of ANS would you have to add to the OVA solution to observe the same amount of fluorescence? (You do not need any more information to solve this problem if you assume the ratio of bound protein to free protein is similar for the two tubes when fluorescence is the same!) Compare this answer to the results you observed in lab comparing ANS fluorescence for egg white samples and BSA samples.Explanation / Answer
ANS (1-anilinonapthalene-8-sulfonate) which is hardly fluoroscent in aqueuos enviroment but becomes highly effective in apolar, organic solvents and adsorption on solid phases. BSA and OVA have molecular weight of 68700 and 45500 Da approx respectively.
In this experiment basically there is binding between the ligand and the receptor or the available binding sites. Assumption that all the binding sites are occupied. The interaction of any ligand to receptor is characterized by several binding parameters: no of binding sites, stiochiometry and binding constant of each binding mode. The binding constant (K) of ligand to receptor is determined by the ration of ligand - receptor complex concentration i.e, concentration of bound ligand (Cb) to the product of concentration of free binding sites of receptor (nCp - Cb) and concentration of free ligand (Cf)
K = Cb / (nCp - Cb) Cf
Cp = concentration of receptor
n= number of ligand binding sites
In our case we are taking ratio of bound protein to free protein is equal that is why our k become equal to
K = [Cb / (nCp - Cb) Cf]BSA /[Cb / (nCp - Cb) Cf]OVA
K = 6 X 106 / 2 X 103
K = 3000
We need 3 times more ANS for binding with OVA as compared to BSA. Binding sites available in BSA are 5.
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