practice final 2 key

practice final 2 key - Chem 153A, Winter 2009 Practice...

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Chem 153A, Winter 2009 1/3 Practice Final 2 Answer Key 2. a, c, d, g, h 3. e 4. acetyl-CoA 5. The A→B reaction is highly spontaneous and would lead to a buildup of B in the cell. This buildup would be enough to make B→C favorable. 6. Glucose. The phosphoryl group of ATP is a better electrophile than C6 of glucose. 7. c 8. a, b, c, d 9. a. I, M, N, O, Q, R, V b. K, S c. E, M, N, Q 10. C, E, A, B, D 11. a. Lipoic acid is covalently bound to one or more lysines in E 2 , so if there are no lysines, there is no lipoic acid, and no catalytic activity. b. PDH is not used in anaerobic metabolism, so growth would be unaffected. c. PDH is necessary for aerobic metabolism, so growth would be severely reduced (to only what you’d get under anaerobic conditions). d. You mutated α-ketoglutarate DH E 2 instead of PDH E 2 , which would likewise eliminate aerobic metabolism. e. α-ketoglutarate 12. Before round 1: C3 & 4 leave (pyruvate → acetyl-CoA) 1/3 of C’s TCA round 1: no additional carbons released KG→succinyl-CoA) 1/3 of C’s tot. = 1/3 + 1/3 = 2/3 TCA round 3: ½ of C1 & 6 leave (same steps as round 2) ½ of 1/3 = 1/6 of C’s tot.= 2/3+1/6 = 5/6 13. The pyruvate carboxylase reaction uses ATP, so the cell shouldn’t run the reaction unless oxaloacetate is absolutely needed. (The presence of acetyl-CoA indicates that the energy
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This note was uploaded on 04/25/2010 for the course BIOCHEM 142637201 taught by Professor Dr.nelson during the Spring '09 term at UCLA.

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practice final 2 key - Chem 153A, Winter 2009 Practice...

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