hmk8ANS_07 - NAME T.A.name sect.no Page 1 Biochemistry 462a...

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NAME_____________________________ T.A.name & sect.no._________________________ Page 1 p. 1 Biochemistry 462a -- Problem Set #8 ANSWERS October 19, 2007 (due Fri., October 26, 2007) 15 points READING: Lehninger Principles of Biochemistry , 4th ed., Chapter 6 1. (4 pts) Suppose that a protein is a dimer with identical binding sites for a specific oligosaccharide (carbohydrate) found on the surface of a viral particle. When the dimer interacts with the oligo- saccharides on the surface of the virus, binding the viral carbohydrates at both sites requires a conformational change at the subunit interface of the dimer to arrive at the appropriate orientation to bind two carbohydrate groups simultaneously. Legend: [V] : viral particle C : carbohydrate on viral surface PP: dimeric protein P-P : conformation that binds only 1 carbohydrate at a time P ^ P : conformation oriented to bind 2 carbohydrates A. (1 pt) Suppose you measure the equilibrium dissociation constant (K c ) for the interaction of the viral particle with a monomeric mutant of the protein and find it to be 6.0 μM at 25 °C. P:VCC P + VCC K c = 6.0 μM Calculate the standard free energy change for the dissociation reaction, and then convert it to a free energy of binding. Show both numbers, making it clear which is which. Is binding or dissociation favored? (As always, show your work . Also, remember that if a mass action ratio has units, they must be in M or M –1 , etc., not in mM or μM, in a free energy calculation.) Δ G o ' = – RTlnK c = – 2.478 ln(6.0 x 10 –6 M) = –2.478(–12.02) = +29.8 kJ/mol for dissoc. Δ G o ' for binding = – 29.8 kJ/mol ( Δ G o ' < 0, so BINDING is obviously favored.) B. (1 pt) Suppose that you next measure the dissociation constant for viral particles bound to the normal dimeric protein (2 binding sites for the viral particle) and find that K dissoc to be 9.7 nM. PP:VCC PP + VCC K dissoc = 9.7 nM (25 °C) Calculate the standard free energy of dissociation and then the free energy of binding of the viral particle to the normal dimeric protein. Show both numbers, making it clear which is which. Δ G o ' for dissociation = – RTlnK dissoc Δ G o ' for dissoc. = – 2.478 kJ/mol ln(9.7x10 –9 M) = – 2.478 kJ/mol (–18.45) = +45.7 kJ/mol Δ G o ' for binding = – 45.7 kJ/mol C. (2 pts) From the free energy of binding of the viral carbohydrate to the monomeric mutant
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