Download Thermodynamic Analysis of a Pharmaceutical Drug: Cureall - Prof. Stefan Franzen and more Assignments Physical Chemistry in PDF only on Docsity! North Carolina State University Homework #7 Name _____________________________ Section ______ Physical Chemistry 331 Due: November 7, 2003 1. A pharmaceutical company is investigating a new lead for a drug known as Cureall. The drug binds to the active site of enzyme Blahase, which tends to make sick people feel lousy if it is not inhibited. The binding equilibrium is known to be: Cureall + Blahase ßà Complex (1) The substrate for Blahase is the carbohydrate Yuckose. Cureall is a competitive inhibitor of Yuckose and prevents the formation Hyperyuckose in the reaction: Yuckose + Blahase ßà Hyperyuckose (2) A chemist reports that the association constant for Cureall is greater than that for Yuckose and that it has a high binding enthalpy. You are an analyst for the investment firm Smartmoney, Inc. and you are asked to examine the thermodynamic data. To determine whether Cureall will be a success in clinical trials you should do the following: a.) Determine the binding (association) enthalpy for Cureall and Yuckose, respectively, with Blahase. Solution: Use a van’t Hoff plot. 1/T (K-1) ln(Ka) for (1) ln(Ka) for (1) 0.00357 22.34 21.49 0.00345 21.16 20.74 0.00333 20.05 20.05 0.00322 19.01 19.41 0.00312 18.05 18.83 For (1) the slope is 9640 K. Thus, ∆Ho = -R(9640 K) = -8.31 J/mol-K(9640 K) = - 80.1 kJ/mol For (2) the slope is 5968 K. Thus, ∆Ho = -R(5968 K) = -8.31 J/mol-K(6958 K) = - 49.5 kJ/mol Notice that the slope of the van’t Hoff plot is positive because ? Ho < 0. ∆Ho (1) = _________________. ∆Ho (2) = _________________. b.) Determine the binding (association) entropy for Cureall and Yuckose, respectively, with Blahase. Solution: Calculate ∆Go at 300 K. Note that it is the same for both the substrate and the inhibitor at this temperature. ∆Go = -RT ln Ka = -(8.31 J/mol-K)(300 K) ln (0.51 x 109) = - 49.98 kJ/mol ~ - 50 kJ/mol ∆So (1) = (∆Ho (1) - ∆Go (1))/T ~ (- 80 kJ/mol + 50 kJ/mol)/300 K = -100 J/mol- K ∆So (2) = (∆Ho (2) - ∆Go (2))/T ~ (- 50 kJ/mol + 50 kJ/mol)/300 K = -0 J/mol-K ∆So (1) = _________________. ∆So (2) = _________________. c.) Cureall is a mimic for a carbohydrate, but it is a floppy molecule (i.e. it has many ether linkages and there any many possible conformations of the drug in solution). Your boss asks you to explain the sign of the entropy of binding of the drug. The entropy change upon binding is large and negative. The reason for this is that the drug loses conformational entropy when it interacts with the protein. There is more freedom for the drug molecule to assume different conformations (and hence greater conformational entropy) when the drug is in solution.
