problem_set_3_solutions_fa09 - MCB 450 Problem Set 3...

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MCB 450 Problem Set 3 – Answers 1. This gives pyruvate labeled in its methyl group. Newly-synthesized glycogen would have label in carbons 1 and 6 of its glucosyl groups. 2. Glucose labeled in carbon 2 would give pyruvate labeled in its middle carbon atom, and acetyl CoA labeled in its carboxyl carbon. Tripalmitin would have labels at each odd-numbered carbon of the palmitoyl groups, and the middle carbon of the glycerol backbone would be labeled. In some cases, the glucosyl residues might also be labeled at positions 2 and 5 well, if OAA equilibrates with the symmetrical TXA intermediates fumarate and succinate. 3. This would occur via an aminotransferase mechanism: Phe + enz-pyridoxal-phosphate = phenylpyruvate + enz-pyridoxamine-phosphate alphaKG + pyridoxamine-phosphate= Glu + pyridoxal phosphate. 4. This involves the glycolytic pathway and the actions of enzymes in the non- oxidative arm of the pentose phosphate pathway, notable transketolase and transaldolase. 1 F6P + ATP (glycolysis)
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This note was uploaded on 10/20/2010 for the course MCB 01 taught by Professor Mintel during the Summer '09 term at University of Illinois, Urbana Champaign.

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problem_set_3_solutions_fa09 - MCB 450 Problem Set 3...

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