A3 - 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 Problem Set # 3 Distributed: 24 February 2010 Due: 8 March 2010, 1 p.m. 1. (15 points) When CD spectroscopy is performed on three solutions, each containing 1.0 × 10‐4 M of DNA, 1.0 × 10‐4 M of RNA, and a mixture of the two with unknown concentrations, the following data are obtained. All spectra are taken in a cell of path length 1 cm. λ (nm) DNA ‐4 300 0.10 ‐0.50 ‐0.25 RNA Mixture 280 3.00 1.60 1.16 260 0.00 6.00 3.12 240 ‐2.20 0.00 ‐0.24 AL‐AR (×10 ) (a) What are the concentrations of DNA and RNA in the mixture? (9 points) (b) If the mixture is hydrolyzed into its component nucleotides, will each of the following increase or decrease? Briefly explain why. (6 points) (i) Circular dichroism at 260 nm; (ii) Absorbance at 260 nm. 2. (15 points) The folding of a protein from the unstructured or random coil to the native state is often followed in real time by time‐resolved circular dichroism (CD) and tryptophan fluorescence spectroscopy. (a) Indicate what property of the protein structure is being monitored by the two spectroscopic methods. Briefly describe how the two methods provide information about the structure of the protein. (10 points) CD monitors the secondary structure of the polypeptide. Tryptophan fluorescence spectroscopy monitors formation of tertiary structure, specifically, the formation of hydrophobic domains (the so‐called molten globule). The tryptophan fluorescence is observed in a hydrophobic environment; it is quenched in an aqueous environment. (b) The two methods, unfortunately, are sensitive to different stages of the folding of the polypeptide. The CD method seems to be particularly sensitive to the early stages of the folding process, whereas tryptophan fluorescence is primarily sensitive to the late phases. What could be the explanation for this difference in behavior between the two methods? (5 points). Secondary structure is formed rapidly followed by sequestering of the hydrophobic domains to form tertiary structure at a slower pace. 3. (15 points) James P. Allen, Chapter 14, problem 14.22. For background, see Chapter 14 of text, pp. 312‐315. 4. (5 points) James P. Allen, Chapter 16, problem 16.19. 5. (25 points) (a) Recall that the nuclear spin of 15N is ½ and that of 14N is 1. The A‐values for the 15N‐ nitroxide radical could be obtained by scaling the A‐values of the 14N nitroxide radical according to the ratio of the corresponding nuclear magnetogyric ratios (γI) of the two nitrogen nuclei, since A is proportional to γI. If Axx = 5.90 Gauss, Ayy = 5.40 Gauss, and Azz = 32.9 Gauss for the 14N‐ TEMPO, what are the corresponding values for the 15N analogue? At 1 T, the NMR frequencies of 14N and 15N occur at 3.076 MHz and 4.315 MHz, respectively. Also the magnetogyric ratios of the two nitrogen isotopes have opposite signs, negative in the case of 15N. (15 points) (b) Based on the EPR spectrum recorded for a 10−4 M TEMPO in aqueous buffer at room temperature, predict the derivative spectrum that you would observe for 15N TEMPO under the same conditions at X‐band frequency. Sketch the spectrum as a function of magnetic field, and include the g‐value of the nitroxide radical and the nuclear hyperfine interaction or splitting in Gauss. (10 points) 6. (10 points) James P. Allen, Chapter 15, problem 15.12. 7. (15 points) James P. Allen, Chapter 15, problem 15.9. 8. (40 points) (a) An instrument manufacturer is presently developing a NMR spectrometer that is capable of observing 1H at 1 GHz. What is the magnetic field of the superconducting magnet that the manufacturer is attempting to attain? Express your answer in Tesla. (10 points) The magnetic field to achieve resonance is directly proportional to the resonance Frequency. Observation of 1H NMR at 500 MHz is accomplished at a magnetic field of 11.7 T. Therefore, to observe 1H NMR at 1 GHz or 1,000 MHz requires a magnetic field of 23.4 T. (b) What is the amplitude of the B1 field that would be required to obtain a π/2 pulse of 10 μs in a single‐pulse 1H FT NMR experiment with this 1 GHz spectrometer? Express your result in gauss, or in frequency units, either as γB1 or γB1/2π. (10 points) Since γB1 tπ/2 = π/2, to obtain a π/2 pulse of 10 μs, that is for tπ/2 10‐5 s γB1 = (π/2)/10‐5 radians/s (c) If the spectrometer under development is also capable of observing 13C and 15N, what would be the NMR frequency of observation for each of these nuclei? (10 points) At a magnetic field of 1 T, the frequencies of observing 13C and 15 N NMR are 10.705 MHz and 4.315 MHz, respectively. The corresponding frequencies at 23.4 T are therefore, 10.705 x 23.4 MHz = 250.497 MHz and 4.315 x 23.4 MHz = 100.971 MHz, respectively. (d) Estimate the minimum B1 field required to excite all the resonances in a 15N FT NMR experiment with this spectrometer if the 15N chemical shifts span a range of the order of 100 ppm? (10 points) We require γB1 >> 2π Δ, where is the spectral width in Hz For 15N NMR at 23.4 T, 100 ppm would correspond to a spectral width of 100.971 x 106 x 100 x 10‐6 Hz = 104 Hz. So γB1 /2π >> 104 Hz or B1 >> 2 π x 104 /γ For 15N NMR, γB0 /2π = 4.315 MHz when B0 = 10,000 Gauss Or γ /2π = 4.315 x 106 / 104 Gauss = 4.315 x 102 Gauss‐1 Therefore B1 >> 2 π x 104 /γ = 104 / (4.315 x 102) Gauss = 23.2 Gauss. 9. (15 points) James P. Allen, Chapter 16, problem 16.14. The two‐dimensional NMR referred to in the problem is the homonuclear 1H COSY experiment. _____________________________________________________________________________________ The end ...
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