KEY - Chem 4410 Exam I - 09_21_2023 (2)

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CHEM 4410 Exam I September 21, 2023 Name ID# R = 8.314 J/mol-K 1.( 8 points) You have isolated 400 milligrams of a previously unknown protein. Explain how you would determine the amino acid composition of this protein. Explain any possible limitations of your methods. All the steps below were discussed in detail at various times in lecture: • Acid hydrolysis: 6N HCl, 110 ° C, sealed tube, 24 hours • React with Edman’s reagent to form PTH derivatives • Ion-exchange chromatography as in Figure 4.13. • Measure fractions for UV absorbance (to detect the PTH-amino acids) Limitations: Trp completely destroyed – must be determined separately. Ser, Thr slowly destroyed, must extrapolate back to start of hydrolysis. Some hydrophobics (Val, Ile) slowly hydrolyzed. Asn and Gln are hydrolyzed to Asp and Glu and the total number must be determined from amount of ammonium released. 2. (8 points) Give the one-letter codes for the amino acids that have: One chiral center A, C, D, E, F, H, K, L, M, N, P, Q, R, S, V, W, Y No chiral centers G Two chiral centers I, T Three chiral centers None

3. 1. (20 points) Amino acid analysis of an octapeptide gave the following results: D I C E A F NH 4 + The following facts were observed: Edman analysis of the octapeptide yielded: Treatment of the octapeptide with chymotrypsin yielded a free amino acid, a tetrapeptide, and a tripeptide. The tripeptide was found to contain Ala. Treatment of the octapeptide with an endoprotease that cleaves on the C-side of glutamate residues yielded a tripeptide and a pentapeptide. The tripeptide contained no Phe. Treatment of the mixture of the tri- and penta-peptides with LiAlH 4 (which reduces C-terminal amino acids to amino alcohols) yielded only a single amino alcohol. Edman analysis of the tripeptide obtained from this endoprotease treatment yielded: Give the sequence of this octapeptide using three letter abbreviations . You must explain your logic for full credit. Solution: A reasonable logic would be the following: • The first clue shows that the N-terminus is Cys. • The second clue says there are two Phe, and the free amino acid is Phe. • Finding a single amino alcohol with LiAlH 4 treatment tells you that there are two Glu residues and that Glu is C-terminal in the octapeptide.

• Edman analysis of the tripeptide from endoprotease treatment gives Ile and tells you that this tripeptide is C-terminal in the original octapeptide. This means that Glu is also at position 5 in the octapeptide. • This means that the two Phe residues must be at positions 3 and 4 in the octapeptide and that Ala is residue 2 in the octapeptide. • The finding of NH 4 + in the amino acid analysis tells you that the “D” observed is really “N” and the only possible location for Asn is at position 7 in the octapeptide. The correct sequence is Cys-Ala-Phe-Phe-Glu-Ile-Asn-Glu 4.(18 points) The fatty-acyl-CoA synthetase reaction is: a) D G ° ¢ = -0.8 kJ/mol. Use data found elsewhere in this exam to determine the standard state free energy of hydrolysis of fatty-acyl CoA. Summing the D G values for the first two reactions (0.8 – 32.3) yields -31.5 kJ/mol for the overall reaction. So D G o ’ = -31.5 kJ/mol for the free energy of hydrolysis of fatty-acyl CoA. b) The concentrations of various metabolites in the human red blood cell are: [CoASH] = 0.12 mM, [fatty acid] = 0.001 mM, [AMP] = 0.14 mM, [ATP] = 1.85 mM, PP i = [0.5 mM], [fatty-acyl-CoA] = 0.1 micromolar. What is the equilibrium constant for the fatty-acyl CoA synthetase reaction? What is the cellular free energy change for the fatty-acyl CoA synthetase reaction at 37ºC? Fatty − acyl − CoA + AMP + PP i → Fatty − acid + CoASH ATP + H 2 O → AMP + PP i Fatty − acyl − CoA + H 2 O → Fatty − acid + CoASH − 800 = − (8.314)(298)ln K eq ln K eq = 800 (8.314)(298) K eq = 1.38

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5. (15 points) Consider a solution of 0.15 M phosphate buffer at pH 6.9. How many mL of 1M HCl would you need to add to 1 L of this solution to lower the pH to 2.75? The pK a values for phosphoric acid are 2.15, 7.2 and 12.4. Δ G = 800 J / mol + (8.314)(298)ln (1 × 10 − 7 )(0.14 × 10 − 3 )(0.5 × 10 − 3 ) (1 × 10 − 6 )(1.85 × 10 − 3 ) Δ G = − 31.38 kJ mol

6. (8 points) Draw an appropriate helical wheel for the peptide below and determine whether the peptide forms an amphiphilic α -helix. EDRVRSVIAELRSILDRV This helix is indeed amphiphilic, being nonpolar on the left side and polar on the right. 7. (8 points) Consider the following peptide sequences: a) CNMKHGDSQCDDERTYP b) YTREQSDGHIPKMNCDS c) QQNWGGLVVILTLVWFLM d) EANQIDEMLYNVQCSLTTLEDTVPW e) LGVHLDITVPLSWTWTLYVKL f) AGPFGPDGPTIGPK Which of the preceding sequences would be likely to be found in each of the following? (Add the correct letter to the blank following each description below.) A parallel b -sheet c A tropocollagen molecule f An antiparallel b -sheet e The helical portions of a protein found in a bluebird’s beak d

8. (15 points) Consider a protein structure consisting of a βαβ loop, with β -bends linking the ends of the α -helix to the β -strands. If the length of the α -helix and the lengths of each β -strand are all the same (4.2 nm): a. How many amino acid residues does the α -helical segment contain? 28 residues b. How many turns does this α -helix have? About 8 c. How many amino acid residues does each β -strand contain? Assuming parallel β -strands, about 13. d. What is the total number of amino acid residues in the complete βαβ structure? 2 x 13 ( β -strands) + 28 ( α -helix) + 2 x 4 (each β -bend) = approximately 62. e. Approximately how many hydrogen bonds do you think this βαβ structure has? 2 for the β -bends, n-4 = 24 for the α -helix, and about 13 for the pair of β -strands equals approximately 39.

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