Question

A scientist is trying to explain the free energy of binding (ΔG_{bind) between a ligand and protein receptor. Experimental measurements indicate a binding enthalpy of} ΔHbind = -59.7 kcal/mol. After analyzing the crystal structure of the receptor-ligand complex, the scientist identifies three interface charge-charge contacts and reasons that all other energetic contributions can be ignored. The three charge-charge contacts have the following properties,

Contact 1 involves a pair of positively (+1.0) and negatively (-1.0) charged atoms separated by 4.0 Å

Contact 2 involves a pair of positively (+1.0) charged atoms separated by 4.0 Å.

Contact 3 involves a pair of positively (+1.0) and negatively (-1.0) charged atoms separated by 3.8 Å.

　

The scientist further assumes that ΔHbind ≈ ΔEbind. By using this data and Coulomb's law, the scientist estimates the total change in electrostatic receptor-ligand interaction (ΔEelec) energy to be,

　

　

Using the above information answer the following questions, show your work

　

1a. The common dielectric constant (D) the scientist assumed in calculating ΔEelec is roughly,

A. 10.46

B. -10.46

C. 1.46

D. 3.46

E. 1.64

F. None of the above

1b. It turns out that the experimental binding affinity (ΔGbind) is measured to be -10.7 kcal/mol. Using the above information, estimate the entropy change (ΔSbind) for the protein-ligand binding reaction (assume T = 298 K).

A. -164.4 cal/K∙mol

B. -0.164 kJ/mol

C. +0.164 kcal/mol

D. +164.4 cal/K∙mol

E. None of the above

Answer 1

A scientist is trying to explain the free energy of binding (dG(bind) between a ligand and a proetin receptor.

the dH(bind) was found to be -59.7 kcal/mol

With the data given above the answer to the following questions would be,

1a. The common dielectric constant the scientist assumed in calculating dEelec is roughly,

D. 3.46

1b. Experimental binding affinity

dG(bind) = -10.7 kcal/mol

From the data given above,

dH(bind) = -59.7 kcal/mol

Using the relation,

dG = dH - TdS

with,

T = 298 K

To calculate entropy change dS(bind) for the protein-ligand binding reaction,

-10.7 = -59.7 - 298 x dS(bind)

So,

dS(bind) = (-59.7 + 10.7)/298 = -0.164 kcal/mol

Answer,

E. None of the above