Chemistry 27 Professor Matthew Shair. Harvard University Spring Hour Exam 3 Friday, April 28, :07 AM 12:00 PM

Transcription

1 ame: Chemistry 27 pring 2006 Exam III Chemistry 27 Professor Matthew hair arvard University pring 2006 our Exam 3 Friday, April 28, :07 AM 12:00 PM Discussion ection (Day, Time): TF: Directions: 1. Do not write in red ink. othing in red ink will be graded. 2. Write your name on every page of the exam. 3. LUTI Provide your answers in the designated space. 4. The last page of the exam has a table of amino acids. You may use the back of this amino acid page for scratch work but nothing written on this page will be graded. 5. You have 9 total pages. We have provided plenty of space for each answer. 6. You EED to draw out proton transfers as distinct steps unless explicitly stated in the question not to. 7. You do not need to draw out full structures of co-factors. Just draw out the reactive portions when required. 8. Molecular model kits and calculators are permitted. 9. A significant fraction of your exams are photocopied before they are returned. Question core 1 /24 2 /20 3 /24 4 /32 Total /100 Page 1 of 9

2 ame: 1. [24 points] It is known that 7-methylation of guanine bases in DA (deoxyribonucleic acid) results in the eventual cleavage of the phosphodiester bond. 7 P 5-2 P P - P 2,3-dideoxygluconucleic acid (2,3-DDGA) 3-deoxygluconucleic acid (3-DGA) A glucose analog of DA was synthesized using 2,3-DDGA monomers to form a similar double helix. owever, the phosphodiester backbone of this non-natural double helix was found to be insensitive to 7-methylation of guanine. In addition, when another glucose analog of DA was synthesized using 3-DGA monomers, the resulting double helix was much more susceptible to phosphodiester backbone cleavage upon 7- methylation of guanine. (a) uggest a plausible mechanism for the phosphodiester backbone cleavage upon 7- methylation of guanine in 3-DGA. You may start with the 7-methylated guanine adduct. [20 points] P - Me 2 P - P - P - P - P P - Page 2 of 9

3 ame: 1. (a) P - P - - strand cleavge - P Total = 20 points 4 for hydrolysis of methylated guanine to give lactol 4 for ring opening of lactol to give aldehyde 4 for rearrangement of aldehyde to ketone (through ene-diol) 6 for β-elimination through enolate 2 for showing final strand cleavage products (b) In one sentence, indicate why phosphodiester backbone cleavage is not observed in 2,3-DDGA double helix. [4 points] There is no way to create a carbonyl group at the 2 position to effect β-elimination. Page 3 of 9

4 ame: 2. [20 points] Carbonyl sulfide (C) is a component of volcanic gas emissions on present-day Earth. It has been discovered that C is an effective reagent for the synthesis of pyrophosphates from inorganic phosphates (P i ). Please provide an arrow-pushing mechanism to account for this amino acid-catalyzed pyrophosphate formation reaction. You D T need to draw out tetrahedral intermediates. [20 points] C 2 P i C 2 - P P C C - P P C - 2 C - - P C 2 - P - 20 points total 4 for 2 attack of C, 4 for elimination of 2 through formation of isocyanate intermediate, 1 for tautomerization to form intermediate shown in question, 5 for using P i to attack ester carbon in ring, 4 for second P i attack, 2 for showing decarboxylation. - Page 4 of 9

5 ame: 3. [24 points] The following molecule is designed to first alkylate DA under reducing conditions and then undergo ergmann cyclization to generate an aryl diradical that will result in DA double strand cleavage. 2 2 e - DA alkylation, then DA- 2 Aryl diradical formation (a) Identify the feature in the molecule (by circling it) that prevents ergmann cyclization from occurring spontaneously, and explain why this feature does so, in no more than two sentences. [4 points] Upon ergmann cyclization, there would be a trans double bond in a seven-membered ring, which is very strained and thus unlikely to form. (b) Provide a mechanism for how the molecule alkylates DA and how it is then converted to the aryl diradical. [20 points] Page 5 of 9

6 ame: 3. (b) 2 - e - 2 e rotation about indicated bond 2 -DA formation of cis double bond in ring - - DA - - DA 20 points total 10 for showing quinone reduction, and kicking out LG. (8 only for not showing C 2 and 3, 5 only if quinone mechanism not shown explicitly) 10 for alkylation first, then ergmann cyclization (6 for reverse sequence, 5 for only ergmann) -1 if final adduct is a semiquinone only. (you were given two electrons) Page 6 of 9

7 ame: 4. [32 points] In non-ribosomal peptide synthesis (P), the C n (condensing) domain catalyzes amide coupling between the amino acid in the domain and the preceding amino acid in the -1 domain. 2 C n domain n-1 n-1-1 n n (a) In some P, the C n domain in selected modules could be replaced by the CY n (cyclase) domain, which catalyzes the heterocyclization of a cysteine residue in the domain with the preceding amino acid in the -1 domain, to form thiazolyl rings as shown. 2 CY n domain n-1 n-1-1 Given the following P gene sequence, please draw out the peptide synthesized by the P. The final TE domain catalyzes the marcocyclization of the peptide. You can abbreviate the side chains of amino acid residues using one-letter codes if those side chains are not important in your answer. [10 points] [A 1 (V)-PCP 1 ]-[C 2 -A 2 (A)-PCP 2-2 ]-[CY 3 -A 3 (C)-PCP 3 ]-[CY 4 -A 4 (C)-PCP 4 ]-[C 5 -A 5 (M)- PCP 5-5 -TE 5 (cyclization)] V M A 10 points total 1 for each correct amino acid and module (5 points) 1 for each correct stereochemistry by domain (2 points) 1 for each correct cyclization to thiazoline ring by CY domain (2 points) 1 for macrocyclization by TE domain(1 point) Page 7 of 9

8 ame: 4. (b) In a novel P, a new type of domain function was discovered that catalyzes the formation of disulfide bridges ( domain). domains are always found in modules that are selective for cysteines, and the n domain is found to catalyze disulfide bond formation between the cysteine residue on ACP n and the nearest preceding cysteine residue in the growing peptide. n domain ased on the given gene sequence, draw the resulting gene product. You can abbreviate the side chains of amino acid residues using one-letter codes if those side chains are not important in your answer. [10 points] [A 1 (M)-PCP 1-1 ]-[C 2 -A 2 (C)-PCP 2 ]-[C 3 -A 3 (C)-PCP 3-3 ]-[C 4 -A 4 (V)-PCP 4 ]-[C 5 -A 5 (C)- PCP 5-5 ]-[C 6 -A 6 (C)CP 6-6 -TE 6 (hydrolysis)] 2 M V 10 points total 1 for each amino acid and module (6 points) 1 for each correct stereochemistry by domain (2 points) 1 for each correct disulfide bond by domain (2 points) -1 for extra a.a., -1 for lack of one a.a, 6 only if C to. Page 8 of 9

9 ame: 4. (c) ased on the following peptide structure, write out the gene sequence of the P responsible for its formation. ote: The peptide is composed of all cysteine residues. [12 points] 2 C [A 1 (C)-PCP 1 ]-[C 2 -A 2 (C)-PCP 2-2 ]-[CY 3 -A 3 (C)-PCP 3 ]-[C 4 -A 4 (C)-PCP 4-4 ]-[CY 5 - A 5 (C)-PCP 5 ]-[C 6 -A 6 (C)-PCP 6-6 -TE 6 (hydrolysis)] 12 points total 1 for each correct module (6 points) 1 for each correct stereochemistry by domain (1 point) 1 for each correct CY domain (2 points) 1 for each correct domain (2 points) 1 for correct TE (hydrolysis) domain (1 point) -1 for each wrongly placed or extra domain -1 for extra modules Page 9 of 9