The standard free energy of binding of Fab derived from an antiviral IgG is -7 k

Question

The standard free energy of binding of Fab derived from an antiviral IgG is -7 kcal mol-1 at 25 degrees Celsius.

A. Calculate the dissociation constant (Kd) of this interaction.

B.When we assume that both combining sites of the antibody can interact with viral epitopes (bivalent binding) and that the free energy cost of assuming a favorable hinge angle is +3 kcal mol-1, the resulting free energy is -11 kcal mol-1. Predict the dissociation constant (Kd) of this intact IgG.

C. Compare the dissociation constants for bivalent binding (part b) to univalent binding (part a). Which binding is stronger and what is the magnitude difference in binding strengths?

Explanation / Answer

delta G =2.303 RT ln K where K= dissociation constant. R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1.

7 *4.2 = 2.303 * 0.008314 * 298 log K

log K = 5.15

K= 1.41 *105

B. log K = 11*4.2 / 2.303 *0.008314*298

log K = 8.09

K = 1.23 *108

Binding in A is stronger than in B. Because B dissociates faster than A.

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