CHEM2523 Chemistry of Biological Molecules – Week 5 Tutorial Suggested Responses This set of questions covers Lectures 7-11 on nucleic acids. Each question tests one of the learning objectives from each lecture. Lecture 7 1. The amino acid lysine has three pKa values: 2.18, 8.95, and 10.79 (side-chain). a) Draw lysine in its predominant charge state at I. pH 1 II. pH 7 III. pH 13 Answer: b) Calculate the pI of lysine. Answer: (8.95 + 10.79)/2 = 9.87 We choose to average these two pKa values because the pKa 8.95 refers to the transition between the net +1 and 0 charge state, and pKa 10.79 refers to the transition between the net 0 and -1 charge state. Lecture 8-9 2. Consider the following two protected amino acids. a) Draw the protected di-peptide product that forms between these amino acids. Answer: The best way to draw it would be like this. Note because the entire Asp residue is flipped downwards, the wedged bond needs to be changed into a dashed bond. If you were lazy and didn’t want to do mental gymnastics to flip the Asp residue to make the structure look neat, you could have drawn it as follows… however the bond angles around the amide bond don’t look right. b) Provide a full curly arrow mechanism for the peptide coupling using DCC/HOBt. (This question takes a long time to draw, best to leave it to the end of the tute…) Feel free to use “R1” and “R2” to denote the side chains to making drawing easier. Cy = cyclohexane 3. This question uses the same amino acids as Q2. Aspartic acid has a protected carboxylic acid on its side chain, while lysine has a protected amine on its side chain. If you juggled your protecting groups wisely, you could couple Asp and Lys via their side chain functional groups. This is called an isopeptide bond. To enable a selective reaction between the two side chains, you need to: I. Protect the free amine of H-Asp(OtBu)-OBn and carboxylic acid of Fmoc-Lys(Boc)-OH II. Deprotect the side chain protecting groups OtBu and Boc. III. Couple the side chains using DCC/HOBt. a) What protecting groups would you use to achieve step I? Answer: Fmoc for the amine of Asp. OBn for the carboxylic acid of Lys. b) Why would a Boc protecting group be a poor choice for protecting the amine of Asp? Answer: In the next step, you’re going to remove the side-chain OtBu of Asp using acid. This would also remove the Boc group you just installed… c) In Step II, what reagents would deprotect the OtBu and Boc groups from the two side chains? Answer: TFA (trifluoroacetic acid) Lecture 10 4. The structure below shows a peptide containing a chemical modification that promotes the formation of an alpha helix, designed by Paramjit Arora at New York University. Why does this modification favour alpha helix formation? Answer: The cyclic modification is known as a “hydrogen-bond surrogate”. Compare to the structure below. The alkene takes the place of a carbonyl C=O, and the CH2 linkage takes the place of the normal hydrogen bond that occurs between amino acid residues 4 positions apart in the sequence. By converting the hydrogen bond (weaker, can more easily be formed and broken) into a covalent bond (stronger, pretty irreversible), this rigidifies the peptide and forces it to adopt an alpha helical conformation. A 3D-like representation is also given (although you are not expected to be able to draw this from scratch!) Lecture 11 5. Consider the following SN2 reaction between a thiophosphate-bearing oligo and an alkyl iodide oligo, studied by Eric Kool at Stanford. Explain why the reaction is accelerated by using the oligo reagent shown in the box. Answer: Each of the two starting materials will hybridise to the oligo reagent, forming a DNA duplex that brings the thiophosphate S– anion and the C–I alkyl iodide bond close together. This template effect promotes a much more rapid reaction between the two groups, by positioning the reactive groups close together, thereby increasing their ‘effective concentration’. Extra challenge question (Lecture 8-9): 6. One of the challenges with Fmoc-SPPS is the repeated exposure of the growing peptide chain to piperidine (during Fmoc deprotection in every cycle). Peptides that contain aspartic acid can undergo rearrangement under prolonged exposure to base (even when side-chain protected as the tert-butyl ester, tBu). The scheme below shows this ‘base-promoted aspartimide rearrangement’ process. a) Can you draw out a full curly arrow mechanism for this process? b) Super-extra-bonus question: The aspartimide (5-membered ring) is also extremely prone to epimerisation by piperidine. Draw the mechanism for this process. (Super technical note: The scheme is an oversimplification, and not 100% correct as drawn. t-BuOH is a poor nucleophile and won’t attack to re-open the aspartimide ring. The aspartimide form is usually where the reaction stops, until final TFA cleavage of the peptide, when a water molecule opens the aspartimide.)
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