30 - Friday November 5 2010 Lecture 30 Announcements 1 Quiz...

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Friday, November 5, 2010 Lecture 30 Announcements: 1. Quiz 7 results: A 24.4 B 24.9 C 25.2 2. Today is THE LAST DAY to submit your midterm exam for regrading. Any exams submitted after today will simply be returned. 3. Today at 2:55PM: The HCEC process-- how to write the personal statement; how to get the best set of 3 recommendation letters. B108 Comstock Hall. 4. No class on Wed 11/24. 5. Public health panel discussion: Tue 11/9 at 4:30PM in Malott 253. Wednesday's Lecture: . max rates of ATP production . pentose phosphate pathway to synthesize NADPH and pentoses . pyruvate enters the mitochondrion and encounters the pyruvate dehydrogenase complex . p. 210 The first reaction in the PDH Complex involves TPP , thiamine pyrophosphate. This coenzyme is derived from vitamin B1, thiamine. This reaction (upper left on the page) occurs in the active site of the pyruvate dehydrogenase subunit, E 1, of the PDH complex. Thiamine is crucial to our metabolism because every case of decarboxylation of an alpha-keto acid (like pyruvate) involves TPP. Why? Because this decarboxylation involves transient negative charge on a carboxyl carbon, yet we know that this carbon is stable with a partial positive charge. TPP functions as the "electron sink" (a jargon term) to accept the extra electron density. . The carbons of this 2-carbon intermediate are at the same oxidation level as in pyruvate. . The acetyl group (pyruvate minus one carbon) is still attached to the TPP, and now reacts with the disulfide of the lipoamide in the active site of E 1 . Note that lipoamide has a long, flexible chain. This chain, about 14 Angstroms long, moves the sulfur-containing ring from the active site of one enzyme to the active site of another enzyme (“channeling”). . The reactive part of the lipoamide molecule is the disulfide. The acetyl group, the remnant of the pyruvate that has lost CO2, is transferred to one S of the disulfide, and a H is transferred to the other S. Oxidation of one carbon has now occurred, as well as reduction of both sulfur atoms. Part of the free energy is now "trapped" in the high energy thioester attached to lipoamide.
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Now the reason for the name of this E 2 enzyme, trans acetyl ase, becomes clear: water-soluble CoA binds to its active site in E 2 , and the acetyl is transferred to CoA, forming the "high-energy compound", acetyl -CoA. Acetyl- CoA is now ready to enter the CAC, and it diffuses out of E 2 . But the PDH Complex still has more rxns to catalyze, since the reducing power in the reduced lipoamide must be made available in a different form. So, the long, flexible lipoamide then swings into the active site of the next enzyme, E 3 : . Within the active site of dihydrolipoyl dehydrogenase or E 3 , the sulfhydryl groups react with tightly-bound FAD forming FADH 2 , still tightly-bound to the enzyme. A disulfide bond reforms in the ring of lipoamide (oxidation of the two sulfhydryls). . NAD
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30 - Friday November 5 2010 Lecture 30 Announcements 1 Quiz...

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