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CHEM 560 Dynamic systems: Page 1 UNIVERSITY OF VICTORIA

CHEMISTRY 560, MODULE ON THE CHARACTERIZATION OF DYNAMIC SYSTEMS

FINAL EXAM — AUGUST 17, 2012

NAME: STUDENT ID:______

INSTRUCTOR: FRASER HOF

TOTAL MARKS = 34

DURATION: 2 HOURS

THIS EXAMINATION PAPER HAS 5 PAGES, INCLUDING THIS COVER PAGE. COUNT THE NUMBER OF PAGES IN THIS EXAMINATION PAPER BEFORE YOU START TO WRITE, AND REPORT ANY DISCREPANCY IMMEDIATELY TO THE INVIGILATOR.

QUESTIONS ARE TO BE ANSWERED IN THE SPACE PROVIDED.

Formulae provided

The generalized 1:1 binding isotherm for X binding to Y to make a 1:1 complex XY is: f11 = K11[Y] / (1 + K11[Y]).

For a 1:1 complexation event, the fraction of compound bound (f11) relates to NMR chemical shift as: f11 = ∆dobs / ∆dmax CHEM 560 Dynamic systems: Page 2

1. The following encapsulation complex works well with CH4 as the guest in the solvent CDCl3.

An excerpt of the NMR spectrum of this solution taken at 300 MHz looks like this:

1a. [2 marks] What is the upper limit for the rate of exchange between free and bound states of CH4? k = π • ∆n/(root 2) ∆n = (0.2 ppm – –0.9 ppm) * 300 MHz = 330 Hz k = π • 330 s-1 / (root 2) = 733 s–1

1b. [4 marks] Let’s abbreviate this complex as AB2. Please provide two stepwise association constants and all mass balance equations for this equilibrating system.

A + 2 B AB2

K11 = [AB]/[A][B] AT = [A] + [AB] + [AB2] K12 = [AB2]/[AB][B] BT = [B] + [AB] + 2 [AB2]

1c. [1 mark] If you wanted to simplify these expressions by making the assumption that [A] ~ 0, what would you do experimentally?

Add an excess of B, to drive the equilibrium to the right so that no free A existed.

/7 CHEM 560 Dynamic systems: Page 3 2. [4 marks] For the following equilibrating system, please provide all stepwise association constants and mass balance equations for all species. Also, provide an expression for a, the fraction of monomeric A.

3 A A3

2 K2 = [A2]/[A] AT = [A] + 2 [A2] + 3 [A3] a = [A]/AT K3 = [A3]/[A2][A]

3. [4 marks] Now let’s say that two solutions of A and B, that form a 1:1 complex AB, have been quickly added to each other but have not yet reached equilbrium. Please give the three mass balance equations that relate actual concentrations at a given time to concentrations at equilibrium.

Progress of reaction at a certain time is “x” x = ∆ [A] = ∆ [B] = – ∆ [AB]

[A] = [A]eq + x [B] = [B]eq + x [AB] = [AB]eq – x

/8 CHEM 560 Dynamic systems: Page 4 4a. [5 marks] The assembly of D (a fluorescent dye) and P (a protein) to form 1:1 complex DP is observed by binding-induced changes in steady state fluorescence intensity, I. The fluorescence increases upon addition of protein. Given the data below, what is the value of K11?

Ifree = 100 rfu (relative fluorescence units)

Ibound = 1100 rfu Dt = 1 uM Pt = 5 uM

Iobs = 900 rfu

D + P  DP f11 = ∆ Iobs / ∆ Imax = 800/1000 0.8 = [DP]/Dt f11 to ∆ I 1 mark [DP] = 0.8 uM Dt = [D] + [DP] = 1 uM, [D] = 0.2 uM use of mass balance 1 mark Pt = [P] + [DP] = 5 uM substitution/values [D] and [P] 1 mark each [P] = 5 uM – 0.8 uM = 4.2 uM

5 –1 K11 = [DP]/[D][P] = 0.8 uM / 0.2 uM • 4.2 uM = 9.5 x 10 M K definition/value 1 mark

4b. [4 marks] Consider now that the dye does not change emission intensity upon binding, but instead changes in the anisotropy (r) described by:

r = (I‖ – I⊥ ) / ( I‖ + 2 I⊥)

Define the terms I⊥ and I‖ by making specific reference to the experimental processes that allow one to measure them. Does r go up or down upon binding of the protein? Why?

When the dye is excited with vertically polarized light, I⊥ and I‖ are the intensities of emitted photons observed through perpendicular and parallel polarizing filters, respectively (2 marks). Anisotropy (r) goes up upon binding to protein (1 mark), because the rate of fluorophore tumbling slows, and that causes more intensity at parallel detection and less at perpendicular detection than before protein binding (1 mark).

/9 CHEM 560 Dynamic systems: Page 5 5. [4 marks] Please give acceptable units for each of the four indicated k or K values in the expression below. E = enzyme; S = substrate, P = product, I = Inhibitor.

k+ k2 E + S ES E + P + k– I Ki

Ki = units M or uM or nM –1 –1 k+ units are M s –1 k– units are s –1 k2 units are s

6. [6 marks] Describe in detail two methods for determining the enthalpic and entropic contributions for a particular association event. Include a discussion of experimental aspects of each method (i.e. how you get the data), a description of what you do with the data to determine ∆ H and ∆ S, and a discussion of their errors and limitations.

Method 1: van’t Hoff plot. Determine Kassoc at various values of T.

∆G = –RTlnK = ∆ H – T ∆ S, lnK = –∆ H/RT + ∆ S/R 1 mark

So a plot of lnK vs. 1/T gives you a line with slope –1/∆ H/R and an intercept of ∆ S/R. This method assumes that ∆ H is constant with temperature 1 mark, which is only true over small changes in temperature of ~40 K. Large error in ∆ S is introduced by extrapolation to the intercept. 1 mark

Method 2: Isothermal titration calorimetry.

This method involves the addition of a titrant to a receiving solution in a calorimeter. The power required _to maintain constant temperature_ 1 mark is measured and integrated to give the enthalpy for that addition. The total ∆ H for each addition is measured and plotted to give a titration curve of ∆ H vs. molar equivalents 1 mark added that can be fitted to determine ∆ H, stoichiometry, and Kassoc for the reaction (giving ∆ S indirectly). 1 mark