Bi/Ch110 Problem Set 4 Problem 1: Overview of metabolism (33 points) a. (12 points) Metabolic pathways often reuse reaction motifs to perform the necessary transformations required. List six major reaction types and the generalized outcome achieved by each reaction/their reactive purpose. - Oxidation-reduction reactions: Derives energy from the oxidation of carbon compounds. - Ligation reactions: form bonds by using free energy from ATP cleavage; necessary for anabolism. - Isomerization reactions: rearrange atoms necessary for subsequent reactions such as oxidation-reduction. - Group transfer reactions: activate compounds for subsequent reactions or modify compounds for regulatory purposes (trap glucose in the cell for further catabolism) - Hydrolytic reactions: Hydrolysis for the degradation of large molecules for metabolism or to reuse some of the components for biosynthetic purposes - Carbon-carbon bond cleavage: release of CO2 or H2O during metabolic pathways that help drive reactions forward. b. (6 points) To ensure proper homeostasis, describe three principle ways that cells control metabolic processes. - Control the amounts of enzymes: Occurs at the level of transcription - Control catalytic activity: Allosteric control; feedback loops; covalent modifications; energy levels of the cell - Control the accessibility of substrates: Compartmentalization of substrates; substrate sequestration c. (5 points) Sometimes, the substrate (or sites on the substrate) for a metabolic pathway are structurally similar: A sample of deuterated reduced NAD was prepared by incubating CH3CD2OH and NAD+ with alcohol dehydrogenase. This reduced coenzyme was added to a solution of 1,3-DPG and glyceraldehyde 3-phosphate dehydrogenase. The NAD+ formed by this second reaction contained one atom of deuterium, whereas the glyceraldehyde 3-phosphate, the other product, contained none. What does this experiment reveal about the nature of the two enzymes?
The Alcohol Dehydrogenase and glyceraldehyde dehydrogenase have opposite selectivities as to which H is taken from/replaced to NADH. d. (5 points) Suppose you discovered a mutant yeast whose glycolytic pathway was shorter because of the presence of a new enzyme catalyzing the reaction glyceraldehyde-3-phosphate + NAD + 3-phosphoglycerate + NADH + H +. Would shortening the glycolytic pathway in this way benefit the cell? Explain. No. There would be no anaerobic productions of ATP; aerobic ATP production would be diminished only slightly. e. (5 points) By choosing glycolysis over the citric acid cycle, what tradeoffs do cancer cells choose? Cancer cells will produce less ATP but will produce many more important metabolites such as nucleic acids, amino acids, and fatty acids for their proliferation. Glycolysis also provides energy at a much quicker rate. Problem 2: The Citric Acid Cycle (40 points) a. (5 points) The output of glycolysis is pyruvate, but the citric acid cycle uses acetyl CoA. Briefly explain how glycolysis and the citric acid cycle are linked. The pyruvate dehydrogenase complex converts pyruvate into acetyl CoA by the reaction given below: Pyruvate + CoA + NAD + acetyl CoA + CO2 + NADH + H + b. (5 points) The citric acid cycle produces only one molecule of ATP per cycle, yet it is key to producing the majority of energy used by the cell. Explain how this is possible. While the citric acid cycle itself does not produce much ATP, it also produces NADH and FADH2 by reducing NAD + and FAD. It does so by transferring 8 high-energy electrons from acetyl CoA to three molecules of NAD + (6 e - ) and one molecule of FAD (2 e - ). When NADH and FADH2 are then oxidized during oxidative phosphorylation, those electrons then are used to produce nine ATP. c. (5 points) Explain how the citric acid cycle is a cycle. The 4-carbon molecule oxaloacetate is a reactant and then a product of the cycle such that there is no net gain or loss of oxaloacetate over one cycle. In the first step of the citric acid cycle, oxaloacetate and the acetyl group of
acetyl CoA are condensed to form a 6-carbon molecule. Over the citric acid cycle, that 6-carbon molecule loses two carbons in the form of CO2 such that the last reaction in the citric acid cycle, catalyzed by malate dehydrogenase, regenerates oxaloacetate. d. (15 points) The citric acid cycle is tightly regulated. Explain how each of the following molecules affect the activity of isocitrate dehydrogenase and why each of these outcomes are logical for the regulation of the citric acid cycle. i. ADP ADP is an allosteric activator of the enzyme. A high [ADP] in the cell indicates the cell is energy-poor; therefore the activation of the citric acid cycle to produce more energy is in the best interest of the cell. ii. ATP ATP is an allosteric inhibitor of the enzyme. A high [ATP] in the cell indicates that the cell is energy-rich; therefore, it is in the best interest of the cell to shut down the citric acid cycle since no more energy is required. iii. NADH NADH is a product/competitive inhibitor of the enzyme. A high [NADH] in the cell indicates that the cell is energy-rich; therefore, it is in the best interest of the cell to shut down the citric acid cycle since no more energy is required. e. (10 points) Pyruvate, labeled with 14 C as shown below in red, was added to cell extracts that contain functioning metabolic enzymes. Where does the 14 C label end up after one turn of the citric acid cycle? Show your work by tracing the atom through the relevant metabolic reactions.
After one turn of the citric acid cycle, the label ends up on the internal carbon atoms of oxaloacetate. Because succinate is a symmetric molecule, you cannot trace the label specifically after that step. Instead, 50% will be on one internal carbon (carbonyl) while the other 50% of the label will be on the other internal carbon (aliphatic). Problem 3: Gluconeogenesis (27 points)
a. (7 points) As you have learned, glycolysis has several irreversible steps. Yet, gluconeogenesis is essentially a reversal of the process of glycolysis. How is that possible, given that some of the reactions of glycolysis are so energetically favorable? While most of the reactions in gluconeogenesis are the reverse of those found in glycolysis, the three irreversible steps, which are very energetically favorable in the glycolysis direction, are replaced by more kinetically favorable reactions. Hexokinase, phosphofructokinase, and pyruvate kinase enzymes of glycolysis are replaced with glucose-6-phosphatase, fructose-1,6- bisphosphatase, and phosphoenolpyruvate carboxykinase. These enzymes are able to perform the same function as the original enzymes in a way that is favorable during gluconeogenesis. b. Consider the enzyme phosphoenolpyruvate carboxykinase (PEPCK), which is involved in gluconeogenesis, but not glycolysis. i. (10 points) What is its role, metabolically speaking? Which hormones regulate its expression and why? PEPCK is an essential enzyme in gluconeogenesis which catalyzes the conversion of oxaloacetate into phosphoenolpyruvate. It is one of the enzymes that circumvents the irreversible steps of glycolysis. Its gene expression is regulated by camp, insulin, cortisol and glucagon in order to maintain glucose homeostasis. ii. (5 points) What would happen if PEPCK was overexpressed? Name a disease that could occur as a result of this overexpression. Increasing gene levels of PEPCK would result in more conversion of oxaloacetate into phosphoenolpyruvate, which ultimately results in more production of glucose. The increased level of glucose could result in the development of Type II diabetes mellitus from the persistent high circulating blood glucose levels. iii. (5 points) What would happen if there were a deficiency in PEPCK? A lower level of PEPCK would result in less conversion of oxaloacetate into phosphoenolpyruvate, which means that less production of glucose would occur. This would result in hypoglycemia. Lactic acidemia can result from the increased acid in the bloodstream, the liver can become enlarged and impaired, and normal growth could be inhibited.