Homework_12_02-06-09_C&A

Homework_12_02-06-09_C&A - Biology 5A Homework 12...

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Biology 5A Homework 12 February 6-9, 2009 1. In the overall process of cellular respiration, in which the potential energy in the chemical bonds of glucose is saved in the form of ATP, the final fate of the six carbon atoms of glucose is that: A. they become part of the ATP molecule B. they become part of two pyruvate molecules C. they become part of the NADH and FADH 2 molecules D. they are released as CO 2 in the glycolysis-Krebs cycle junction step and in the Krebs cycle E. all four answers are correct (Ans. D) 2. In the reaction shown in the accompanying figure from the Krebs cycle, the identities of the molecules X and Y are (Instead of looking at your lecture notes or the textbook, try to figure out the answer by balancing the chemical equation.): A. X is H 2 O and Y is NADH B. X is CO 2 and Y is NADH C. X is NADH and Y is FADH 2 D. X is pyruvate and Y is FADH 2 E. X is HCOOH and Y is NADH (Ans. B) 3. What is the efficiency of energy savings in the cellular respiration of glucose? The largest savings of energy in the form of ATP in the cellular respiration of glucose occurs in oxidative phosphorylation, the final portion of the pathway. Oxidative phosphorylation is the process of generating ATP from the high energy electrons carried by NADH and FADH 2 . These reduced forms of the coenzymes are oxidized (hence the name oxidative phosphorylation) as they start the electron transport chain in the mitochondrial inner membrane. As the electron
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transport chain operates, it moves protons from the matrix side of the membrane to the intermembrane space. Chemiosmosis is the use of the energy stored in the resulting proton gradient to phosphorylate (hence the name oxidative phosphorylation ) ADP to generate ATP as protons are allowed to move down
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Homework_12_02-06-09_C&A - Biology 5A Homework 12...

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