UCSB_MCDB110_FinalExam_Winter2009-1497285260178.pdf

Business end 4 points given the reactions atp h 2 o

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Business end
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Name ____________________ -15- 52. (4 points). Given the reactions: ATP + H 2 O ° ADP + Pi G ° = -31 kJ/mol X ° Y + Pi G ° = -22 kJ/mol Calculate the G ° for the phosphorylation of Y. To receive full credit, you must show how you arrived at this value. ATP + H 2 O ° ADP + Pi G ° = -31 kJ/mol Y + Pi ° X G ° = +22 kJ/mol ATP + Y + H 2 O ° ADP + X G ° = -9 kJ/mol 53. (4 points). If glucose were labeled on the last (number 6) carbon and fermented by yeast to ethanol, where would the label end up? Answer the same question if it were labeled on the number 3 carbon. #6 *CH 3 – CH 2 - OH #3 CO 2 54. (8 points). We discussed two different chains of electron donors and acceptors (i.e., electron transport systems). They are similar in several ways and different in several others. Name the two systems in which a series of electron donors and acceptors sequentially pass electrons. Where do these electron donors and acceptors reside in cells (be specific)? What are the initial electron donors in each of these systems? What are the ultimate electron acceptors in each system? a) First system: System cell location e - donor e - acceptor mitochondrial inner membrane NADH, O 2 e- transport chain succinate b) Second system: System cell location e - donor e - acceptor light reactions thylakoid H 2 O NADP+ photophos- membrane phorylation
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