E1 - Ch 24a Winter Term 2009‐10 Introduction to...

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Unformatted text preview: Ch 24a Winter Term 2009‐10 Introduction to Biophysical Chemistry Midterm Examination Distributed: Friday 12 February 2010 Due: Wednesday 17 February 2010, 1 p.m. Instructions: • Open this exam only when you are ready to take it. • This is a three‐hour exam. You are expected to take the exam in one continuous stretch of 3 hours. The 3 hours include a toilet break of up to 10 minutes. • This exam is OPEN book. You may consult your class notes (Ch 24a Introduction to Biophysical Chemistry), the supplementary notes provided by the UTAs, problem sets and solutions, and your text (James P. Allen). No other reference material is allowed. You may use a computer for graph plotting ONLY. You are, however, NOT required to print out your graphs; simply writing down the equations obtained from your graphs will be sufficient. • This exam contains 16 pages. Answers to the various questions should be written in the blank space provided following each question. Use back page, if necessary. • Please use a ball‐point or an ink pen in writing your answers. • Please make sure that the completed midterm exam is stapled before you turn it in class. • As soon as you open the exam, please write your name and student ID number at the places indicated on pp. 2 and 3. 1 NAME: __________________________________ Student ID Number: ______________________ SCORE 1 _____ /50 2 _____ /30 3 _____ /65 4 _____ /40 Total Score: ____ /300 5 ____ /40 6 /30 7 ____ /45 2 Name ____________________________ Student ID Number _________________ 1. (50 points) According to the laws of physics, the electric field of the scattered light produced by an atom is z E Figure 1 φ r Es Es = e a sin φ / r c2 , where −e is the electronic charge, a its acceleration during interaction of the electron with the electric field of the incident light E polarized along the z‐ direction, φ is the angle of observation measured from the z‐axis (see Figure 1), r is the distance from the electron or atom, and c is the speed of light. (a) Show that the acceleration of the electron depends on the frequency of the light ν, and on the polarizability αo, namely, the in‐phase polarizability. Note that if ν is far away from any absorption bands, α = αo and α90 = 0, so that the induced dipole moment (μin) is given by μin = − e x = αo E. (15 points) acceleration = d2x/d2t = −(1/e) αo d2E/d2t Now E = Eo sin 2πvt, so that dE/dt = Eo (2πv) cos 2πvt and d2E/d2t = − Eo (2πv)2 sin 2πvt Therefore acceleration = (1/e) αo Eo (2πv)2 sin 2πvt 3 (b) Use the result obtained in (a) to show that the electric field of the scattered light is Es = (1/rc2) 4π2 ν2 αo sin φ Eo sin 2πvt. (15 points) Es = e (1/e) αo Eo (2πv)2 sin 2πvt sin φ (1/ r c2) = (1/rc2) 4π2 ν2 αo sin φ Eo sin 2πvt. (c) Is the E field of the scattered light (Es) in‐phase or out‐of‐phase vis a vis the E field of the incident light (E)? (5 points) The E field of the scattered light is in‐phase vis a vis the E field of the incident light. (d) From the result obtained in (b), obtain an expression for the scattered light intensity in terms of the intensity of incident light intensity. (15 points) Iscattered/Iincident = Es2/E2 = (1/r2c4) 16π4 ν4 αo2 sin2 φ = (1/r2) 16π4 λ−4 αo2 sin2 φ 2. (30 points) (a) The mean lifetime of an electronically excited molecule is 1.0 × 10‐8 sec. If the emission of the radiation occurs at 610 nm, what are the uncertainties in frequency (Δν) and wavelength (Δλ)? (20 points) 4 From Heisenberg uncertainty principle : ΔE Δt ≥ ћ/2, or Δν Δt ≥ 1/4π So Δν ≥ 1/ 4πΔt = 1/ (12.6 × 10‐8) sec−1 = 8 × 106 sec−1 Since ν λ = c, λ = c / ν and Δλ =− c/v2 Δν = − (c2/v2) (Δν/c) |Δλ| = λ2 (Δν/c) = 610 nm [(8 × 106)/3 × 1010)] (610 × 10‐9 × 102) = 1 x 10‐5 nm, which is significantly smaller than the spectral widths observed. (b) Typically, what factors give rise to the resonance width in the UV‐visible spectrum of biological samples? (10 points) Typically, spectral widths in the UV‐visible region arise from spectral (overlapping transitions) and environmental heterogeneity (the same transition in different environments). 3. (65 points) The allyl system (Figure 2) has played a major role in the development of Hückel Molecular Orbital Theory. Figure 2: The allyl system: cation, radical, and anion. 5 The system is planar, so the bonding could be divided into a planar σ framework containing all the carbon and hydrogen atoms, and a π system. (a) Briefly describe the chemical bonding of the σ framework, highlighting the degree of hybridization of each of the carbon atoms, and the valence bond angles. (15 points) The 2s and 2p carbon atomic orbitals are sp2 hydridized for all three of the carbon atoms. Each of the terminal carbons forms one C−C valence bond with the central carbon using one of the sp2 orbitals from each carbon; and two C‐H valence bonds, each formed by overlapping one of the remaining two sp2 hybridized orbitals with an 1s atomic orbital of a hydrogen atom, yielding a H−C−H valence angle of 120°. The central carbon, in addition to forming two C−C valence bonds with the terminal carbons, also forms a C−H valence bond using the remaining sp2 carbon atomic orbital and an 1s hydrogen atomic orbital. The two C‐C‐H valence angles are 120°. The entire σ framework is planar. 6 (b) The π system involves the 2pz orbitals centered at each of the three carbon atoms, and the three Hückel molecular orbitals are given in Figure 3, in the order of increasing stability or decreasing energy; that is, the energy of ψMO1 is the lowest, followed by that of ψMO2, and then ψMO3. Figure 3 ψMO3 = (1/6)½ [ 2pzA − 2 2pzB + 2pzC] ψMO2 = (1/2)½ [2pzA − 2pzC] ψMO1 = (1/3)½ [2pzA + 2pzB + 2pzC] These molecular orbitals have been classified as bonding, non‐bonding, and anti‐ bonding. Indicate which molecular orbital can be classified as bonding, non‐ 7 bonding, and anti‐bonding, and provide one or two sentences to justify the assignment. (9 points) ψMO1 is a bonding molecular orbital. Adjacent 2pz orbitals overlap to yield enhanced electron density between the two carbons giving a C−C double‐bond. ψMO2 is a non‐bonding molecular orbital. There is no 2pz orbital from the central carbon to overlap with the 2pz orbitals from either of the terminal carbons, either constructively or destructively. ψMO3 is an anti‐bonding molecular orbital. Adjacent 2pz orbitals overlap destructively to yield reduced electron density between the two carbons at each end of the molecule. (c) The above molecular orbitals were obtained by neglecting the overlap integral between adjacent carbon 2pz atomic orbitals, that is, ∫ 2pzA* 2pzB dτ ≈ 0, etc., and assuming that ∫ 2pzA* 2pzC dτ is negligible. Show that ψMO3 is normalized, and ψMO3 is orthogonal to ψMO1. (6 points) ∫ ψMO3 * ψMO3 dτ = (1/6)½ (1/6)½ ∫ [ 2pzA − 2 2pzB + 2pzC]* [2pzA − 2 2pzB + 2pzC] dτ = (1/6) [∫ 2pzA* 2pzA dτ + 4 ∫ 2pzB* 2pzB dτ + ∫ 2pzC* 2pzC dτ + overlap integrals which we will ignore ≈ (1/6) [ 1 + 4 + 1] = 1 8 ∫ ψMO3 * ψMO1 dτ = (1/6)½ (1/3)½ ∫ [ 2pzA − 2 2pzB + 2pzC]* [2pzA + 2pzB + 2pzC] dτ = (1/6)(2)½ [∫ 2pzA* 2pzA dτ − 2 ∫ 2pzB* 2pzB dτ + ∫ 2pzC* 2pzC dτ + overlap integrals which we will ignore ≈ (1/6)(2)½ [ 1 – 2 + 1] = 0 (d) Note that the system has symmetry. Discuss the behavior of the ψMO1, ψMO2, and ψMO3 with respect to σxy (reflection about the plane of the molecule) and σyz (reflection about the plane bisecting the molecule through the central carbon). (10 points) Note that in addition to σxy and σyz, the system has one more symmetry operation C2, two‐fold rotation about the y‐axis. I shall include the effects of C2 as well, for sake of completeness. Molecular orbital σxy σyz C2 ψMO3 − + − ψMO2 − − + ψMO1 − + − (e) From the symmetry properties of the molecular orbitals, work out which electronic transitions among the three molecular orbitals are symmetry allowed, and which are forbidden by symmetry. (9 points) 9 Transition moment μx μy μz σxy σyz C2 + + − − + + − + − Integrand σxy σyz C2 ψMO2* ψMO1 + − − ψMO3* ψMO1 + + + ψMO3* ψMO2 + − − _______________________________________________________________ ψMO2* μx ψMO1 + + + ψMO3* μx ψMO1 + − _ ψMO3* μx ψMO2 + + + _______________________________________________________ ψMO2* μy ψMO1 + − − 10 ψMO3* μy ψMO1 + + + ψMO3* μy ψMO2 + − − ________________________________________________________________ ψMO2* μz ψMO1 − − + ψMO3* μz ψMO1 − + − ψMO3* μz ψMO2 − − + So the symmetry‐allowed transitions are ψMO2 ← ψMO1 (x‐polarized) ψMO3 ← ψMO2 (x‐polarized) ψMO3 ← ψMO1 (y‐polarized) None of the transitions are symmetry forbidden. (f) Indicate the distribution of electrons among the three molecular orbitals for the allyl radical, cation and anion in the ground electron states of these allyl species. (6 points) ______________ _____________ _______________ ψMO3 ______________ ______↑________ _______↑↓_____ ψMO2 ___________ ↑↓_______ _________↑↓________ _________↑↓________ ψMO1 11 cation radical anion (g) What is the electronic wavefunction, including spin, for the π electrons in the ground and first‐excited states of the allyl cation. (20 points) Wavefunction in the ground state: ψMO1(1) ψMO1(2) (1/2)½ [α(1) β(2) – β(1) α(2)] Wavefunctions in the first‐excited states: Singlet: (1/2) [ψMO1(1) ψMO1(2) + ψMO1(2) ψMO1(1)] [α(1) β(2) – β(1) α(2)] Triplet: (1/2)½ [ψMO1(1) ψMO1(2) − ψMO1(2) ψMO1(1)] α(1) α(2) (1/2)½ [ψMO1(1) ψMO1(2) + ψMO1(2) ψMO1(1)] β(1) β(2) (1/2) [ψMO1(1) ψMO1(2) − ψMO1(2) ψMO1(1)] [α(1) β(2) + β(1) α(2)] 4. (40 points) Plasocyanin† is a blue copper protein with a strong absorption at 597 nm with an extinction coefficient of ~ 6,000 cm‐1 M‐1 when the copper center is oxidized (Cu2+). It is colorless when the copper center is reduced to the Cu1+ state. 12 Design two experiments, one based on UV‐visible spectroscopy and the other on Raman spectroscopy, to follow the reduction of the blue copper center. Be succinct. The answers should not exceed one paragraph of 100 words in each case. †Plastocyanin is an important copper‐containing protein involved in electron‐ transfer. The protein is monomeric, with a molecular weight around 10,500 Da, and 99 amino acids in most vascular plants. In photosynthesis, plastocyanin functions as an electron transfer agent between cytochrome f of the cytochrome b6f complex from photosystem II and P700+ from photosystem I. Cytochrome b6f complex and P700+ are both membrane‐bound proteins with exposed residues on the lumen‐side of the thylakoid membrane of chloroplasts. Cytochrome f acts as an electron donor while P700+ accepts electrons from reduced plastocyanin. UV‐visible spectroscopy Since the plastocyanin is colorless when the copper ion is reduced, we could simply use the absorption at 597 nm as a measure of the oxidized blue copper center. First, we need to determine the extinction coefficient at 597 nm accurately so that we determine the concentration of the oxidized blue copper accurately via the absorbance A of a known concentration of the oxidized protein at this wavelength in a cuvette of known path length using the Beer‐ Lambert Law. Then we titrate the blue copper solution with a suitable reductant and follow the decrease in the A597nm. 13 Raman spectroscopy Since only the vibrations associated with the blue copper center in the oxidized protein will be resonance enhanced at 597 nm in a Resonance Raman experiment, we can use the intensity of one of the Raman bands as a measure of the oxidized protein in this experiment. The reduced protein does not absorb at this wavelength, so the vibrations associated with the copper center will not be resonance enhanced when the copper center is reduced. Again, we will start with a solution of the oxidized protein, titrate it with a suitable reductant, and follow the intensity of one of the Raman bands. 5. (40 points) (a) A molecule XY2 is known to be linear, but it is not clear whether it is Y−X−Y or X−Y−Y. How would you use IR spectroscopy to determine its structure? (20 points) For a linear triatomic molecule, there are 4 vibrations, the symmetric stretch, the asymmetric stretch, and the two degenerate bending vibrations. For the symmetric Y−X−Y structure, the symmetric stretching vibration is not IR active. So we would observe only the asymmetric stretching and the bending vibrations (degenerate) in the IR in this case. For the X−Y−Y structure, all four vibrations are in principle IR active. 14 (b) Could the structure be decided by Raman spectroscopy? Explain. (20 points) Yes, you can decide by Raman spectroscopy as well. For a symmetric linear triatomic molecule, only the symmetric stretching vibration is Raman active; the asymmetric stretch and the bending vibrations are not Raman active. For the X−Y−Y structure, all four vibrational modes are Raman active. 6. (30 points) Clegg et al. (PNAS 90, 2994-2998 (1993) used a series of DNAs ranging in length from 8 base-pairs to 20 base-pairs to study the conformation of double-stranded DNA. They covalently attached the fluorescein dye to one 5’-end of each double-stranded DNA and the rhodamine dye to the other 5’-end, and use FRET to measure the end-to-end distance for each member of the series. The following data (Figure 4) was obtained for the FRET transfer efficiency (E) versus the number of base pairs. Figure 4 Offer an explanation for why E does not decrease monotonically with the number of base pairs, but instead levels off as the end of a turn of the α-helix (~ 10 base-pairs) is approached. 15 The FRET efficiency should change most steeply as base‐pairs are added to form the DNA double‐helix, when the donor‐acceptor chromophore pair is not only increasing their vertical distance along the DNA helix but they are also moving away from each other across the DNA strand. This situation obtains when the two chromophores are on the opposite side of the DNA strand. This increase in the cross‐strand separation augments the incremental rise in the helix. At some point during the formation of the alpha helical turn, however, the donor‐acceptor pair will approach each other across the strand when additional base‐pairs are added to complete a turn of the alpha‐helix. The change in the separation of the donor‐acceptor pair then becomes progressively more gradual, as the vertical rise in the helix is no longer augmented by a cross‐ strand separation to the same extent. With the receding cross‐ strand separation between the donor‐acceptor pair, the two chromophores would at some point end up being on the same side of the DNA strand. The FRET data summarized in Figure 4 reflect the changing geometric disposition between the donor and acceptor attached to the 3’‐ and 5’‐ ends of an alpha helix of increasing length. 16 7. (45 points). Consider a pair of interacting chlorophyll molecules, like the “special pair” in the photosystem II of green plants (Figure 5). Assume that the two chromophores are coplanar and form a stacked dimer, but their transition dipoles are rotated 45º with respect to one another. The chlorophyll molecules absorb in the visible near 780 nm. (a) Calculate the excition splitting in nm if the transition dipoles μ1 and μ2 are 2 a.u. and the inter‐chromophore separation R is 8 a.u. (Dipole moment: 1 a.u. = 2.542 debyes; distance: 1 a.u. = 0.05292 nm; energy: ½ a.u. = 13.65 eV, and 1 eV = 8000 cm‐1). (30 points) Electric dipole‐dipole interaction: In this problem, only the first term contributes to the energy. Both μ1 and μ2 are perpendicular to R12, so the second term vanishes. Exciton interaction = μ1· μ2 / R3 17 The angle between the two transition dipoles is 45°. So the exciton interaction V12 = |μ1 | |μ2| cos 45°/ R3 = (2 a.u.)(2 a.u.) (0707) / 83 a.u. = 5.531 × 10‐3 a.u. = 5.531 ×10‐3 a.u. × 27.30 eV/a.u. × 8000 cm−1/eV = 1208 cm‐1 , and Exciton splitting = 2 V12 = 2416 cm‐1 Now |Δλ| = λ2 (Δν/c) so the exciton splitting in nm = 780 nm (2416) cm‐1 (780 nm) × 10−7 cm/nm = 780 nm × 1.884 × 10−1 = 147 nm (b) Determine the relative intensity of the two exciton transitions. (15 points) Since V12 > 0, intensity ratio of the higher frequency transition to the lower frequency transition = (1+ cos 45°)/(1−cos 45°) = 1.707/0.293 = 5.8 Figure 5 18 The end 19 ...
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This note was uploaded on 10/27/2010 for the course BI 110 taught by Professor Richards,j during the Winter '08 term at Caltech.

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