A particular genomic caretaker protein, Protein X, has an affinity for both Liga

Question

A particular genomic caretaker protein, Protein X, has an affinity for both Ligand Y and Ligand Z. When you have 0.23 microM of Protein Xin a solution and mix it with 0.11 microM of Ligand Y, the resulting solution contains 0.20 microM of free Protein X, 0.09 microM of free Ligand Y, and 0.02 microM of the protein-ligand complex, after equilibrium has been reached. However, when you have 0.23 microM of Protein X in a solution and mix it with 0.11 microM of Ligand Z, the resulting solution contains 0.14 microM of free Protein X, 0.02 microM of free Ligand Z, and 0.09 microM of the protein-ligand complex, after equilibrium has been reached. Which ligand has a greater affinity for Protein X? Choose one: A. Ligand Z B. Ligand Y C. Ligands Y and Z have equal affinity for Protein X. D. It is not possible to determine with the given information.

Explanation / Answer

Given data,

Protein X concentration = 0.23microM

Ligand Y concentration = 0.11microM

Complex XY concnetration = 0.02microM

Protein X concentration = 0.23microM

Ligand Z concentration = 0.11microM

Complex XZ concnetration = 0.09microM

The equation at equilibrium will be

[X]eq + [Y]eq                      [X-Y]eq

[X]eq + [Z]eq                      [X-Z]eq

Ka =   [X-Y]eq /[X]eq [Y]eq    

Ka =   0.02 /0.23*0.11 = 0.79

Ka =   [X-Z]eq /[X]eq [Z]eq

Ka =   0.09/0.23*0.11 = 3.5   

From the calculation of binding affinity ligand Z has strong affinity to protein X.     

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