2007 first midterm key

2007 first midterm key - MCB 100B Spring 2007 Key First...

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MCB 100B First Midterm Exam Spring 2007 February 26, 2007 Key 1. (40 pts) When glucose is fermented to ethanol and CO 2 in a yeast extract in which there are no enzymes other than those involved in this pathway , six carbons, 12 hydrogens, 6 oxygens are put into the system and the same numbers of each of the atoms are found in the products. a. Why then does the pathway not proceed normally unless ADP is present at levels comparable to the levels of glucose? Explain fully. 10 pts. 2 ATP are produced for each glucose converted and thus 2 ADP and 2 P i must be added.(5 pts) ADP must be present to accept P i from 1,3 bis phosphoglycerate and allow the pathway to proceed to pyruvate. Thus conversion of glucose cannot be separated from the use of ADP to produce ATP. (5 pts) b. Your classmate says, “well then, we have to add NAD also”. Why do you think he thinks about NAD here. 7 pts. The pathway involves the oxidation of glyceraldehyde phosphate with the transfer of the two protons and electrons to NAD producing NADH + H + . c. Why then do we not have to add NAD. 8 pts. The NADH produced in the reaction that involves oxidation of glyceraldehyde phosphate is oxidized back to NAD when acetaldehyde is reduced to ethanol or pyruvate is reduced to lactate. Thus NAD is recycled and only a small amount is required. This small amount is present in the yeast extract.
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This note was uploaded on 01/12/2010 for the course MCB 100B taught by Professor Alper/calendar during the Spring '08 term at Berkeley.

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2007 first midterm key - MCB 100B Spring 2007 Key First...

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