2- The conversion of 1 mol of fructose 1,6-bisphosphate to 2 mol of pyruvate by the glycolytic pathway results in a net formation of:

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1 Section 8 Key 1- During strenuous exercise, the NADH formed in the glyceraldehyde 3-phosphate dehydrogenase reaction in skeletal muscle must be reoxidized to NAD + if glycolysis is to continue. The most important reaction involved in the reoxidation of NADH is: A) dihydroxyacetone phosphate glycerol 3-phosphate B) glucose 6-phosphate fructose 6-phosphate C) isocitrate α-ketoglutarate D) oxaloacetate malate E) pyruvate lactate 2- The conversion of 1 mol of fructose 1,6-bisphosphate to 2 mol of pyruvate by the glycolytic pathway results in a net formation of: A) 1 mol of NAD+ and 2 mol of ATP. B) 1 mol of NADH and 1 mol of ATP. C) 2 mol of NAD+ and 4 mol of ATP. D) E) 2 mol of NADH and 2 mol of ATP. F) 2 mol of NADH and 4 mol of ATP. 3- In glycolysis, fructose 1,6-bisphosphate is converted to two products with a standard freeenergy change ( G' ) of 23.8 kj/mol. Under what conditions (encountered in a normal cell) will the free-energy change ( G) be negative, enabling the reaction to proceed to the right? A) If the concentrations of the two products are high relative to that of fructose 1,6- bisphosphate. B) The reaction will not go to the right spontaneously under any conditions because the G' is positive. C) Under standard conditions, enough energy is released to drive the reaction to the right. D) When there is a high concentration of fructose 1,6-bisphosphate relative to the concentration of products. E) When there is a high concentration of products relative to the concentration of fructose 1,6-bisphosphate.

2 4- Inorganic fluoride inhibits enolase. In an anaerobic system that is metabolizing glucose as a substrate, which of the following compounds would you expect to increase in concentration following the addition of fluoride? A) 2-phosphoglycerate B) Glucose C) Glyoxylate D) Phosphoenolpyruvate E) Pyruvate 5- Which one of the following statements about gluconeogenesis is false? A) For starting materials, it can use carbon skeletons derived from certain amino acids. B) It consists entirely of the reactions of glycolysis, operating in the reverse direction. C) It employs the enzyme glucose 6-phosphatase. D) It is one of the ways that mammals maintain normal blood glucose levels between meals. E) It requires metabolic energy (ATP or GTP). 6- All of the following enzymes involved in the flow of carbon from glucose to lactate (glycolysis) are also involved in the reversal of this flow (gluconeogenesis) except: A) 3-phosphoglycerate kinase. B) aldolase. C) enolase. D) phosphofructokinase-1. E) phosphoglucoisomerase. 7- The main function of the pentose phosphate pathway is to: A) give the cell an alternative pathway should glycolysis fail. B) provide a mechanism for the utilization of the carbon skeletons of excess amino acids. C) supply energy. D) supply NADH. E) supply pentoses and NADPH.

3 8- Cellular isozymes of pyruvate kinase are allosterically inhibited by: A) high concentrations of AMP. B) high concentrations of ATP. C) high concentrations of citrate. D) low concentrations of acetyl-coa. E) low concentrations of ATP. 9- Glycogen phosphorylase a can be inhibited at an allosteric site by: A) AMP. B) calcium. C) GDP. D) glucagon. E) glucose. 10- There are two reactions in glycolysis in which an aldose is isomerized to a ketose. For one of these reactions draw the structures of the aldose and the ketose. For both reactions the G' is positive. Briefly explain how the reactions are able to proceed without the input of additional energy. Ans: The two reactions are those catalyzed by phosphohexose isomerase and triose phosphate isomerase: glucose 6-phosphate fructose 6-phosphate (aldose) (ketose) dihydroxyacetone phosphate glyceraldehyde 3-phosphate (ketose) (aldose) Although both of these reactions have standard free-energy changes ( G' ) that are positive, they can occur within cells because the products are immediately removed by the next step in the pathway. The result is a very low steady-state concentration of the products, making the actual free-energy changes ( G) negative: G = G' + RT ln ([products]/[substrates])

4 11- Briefly describe the possible metabolic fates of pyruvate produced by glycolysis in humans, and explain the circumstances that favor each. Ans: Under aerobic conditions, pyruvate is oxidized to acetyl-coa and passes through the citric acid cycle. Under anaerobic conditions, pyruvate is reduced to lactate to recycle NADH to NAD +, allowing the continuation of glycolysis. 12- The conversion of glucose into glucose 6-phosphate, which must occur in the breakdown of glucose, is thermodynamically unfavorable (endergonic). How do cells overcome this problem? Ans: Cells often drive a thermodynamically unfavorable reaction in the forward direction by coupling it to a highly exergonic reaction through a common intermediate. In this example, to make glucose 6-phosphate formation thermodynamically favorable, cells transfer phosphoryl groups from ATP to glucose. ATP hydrolysis is highly exergonic, making the overall reaction exergonic. (Numerical solution below not required.) Glucose + Pi glucose 6-phosphate + H2O G' = kj/mol ATP + H 2 O ADP + Pi G' = 30.5 kj/mol Sum: ATP + glucose ADP + glucose 6-phosphate G' = 16.7 kj/mol See p. 498; see also Chapter Explain why the phosphorolysis of glycogen is more efficient than the hydrolysis of glycogen in mobilizing glucose for the glycolytic pathway. Ans: Phosphorolysis yields glucose 1-phosphate, which can be converted into glucose 6- phosphate without the investment of energy from ATP. Hydrolysis of glycogen yields free glucose, which must be converted into glucose 6-phosphate (at the expense of ATP) before it can enter glycolysis.

5 14- What are the regulatory implications for the cell with regard to ATP and AMP, given that the former are generally high, and the latter are low? Ans: Normally, [ATP] is 5-10 mm, while [AMP] is < 0.1 mm, thus AMP is a much more sensitive indicator of a cell s energetic state. Small changes in ATP concentration are amplified into large changes in AMP concentration (see Table 15-1), hence many regulatory processes hinge on changes in the concentration of AMP. 15- In the glycolytic path from glucose to phosphoenolpyruvate, two steps are practically irreversible. What are these steps, and how is each bypassed in gluconeogenesis? What advantages does an organism gain from having separate pathways for anabolic and catabolic metabolism? What are the disadvantages? Ans: The two irreversible steps in glycolysis are conversion of glucose to glucose 6- phosphate, catalyzed by hexokinase, and conversion of fructose 6-phosphate to fructose 1,6- bisphosphate, catalyzed by phosphofructokinase-1 (Table 15-2, p. 573). The first reaction is bypassed during gluconeogenesis by the reaction catalyzed by glucose 6-phosphatase, an enzyme unique to the liver. The second is bypassed by fructose 1,6-bisphosphatase-1 (FBPase-1). By having separate pathways that employ different enzymes, an organism is able to control anabolic and catabolic processes separately, thus avoiding futile cycles. A potential disadvantage is the need to produce separate sets of enzymes for catabolism and anabolism. 16-Explain the distinction between metabolic regulation and metabolic control in a multienzyme pathway. Ans: Regulation refers to rebalancing the levels of metabolites along a pathway in response to a change in flux through the pathway, while control is what determines the total flux through the pathway. 17- What are the regulatory implications for the cell with regard to ATP and AMP, given that the former are generally high, and the latter are low? Ans: Normally, [ATP] is 5-10 mm, while [AMP] is < 0.1 mm, thus AMP is a much more sensitive indicator of a cell s energetic state. Small changes in ATP concentration are amplified into large changes in AMP concentration (see Table 15-1), hence many regulatory processes hinge on changes in the concentration of AMP.

6 18- In the glycolytic path from glucose to phosphoenolpyruvate, two steps are practically irreversible. What are these steps, and how is each bypassed in gluconeogenesis? What advantages does an organism gain from having separate pathways for anabolic and catabolic metabolism? What are the disadvantages? Ans: The two irreversible steps in glycolysis are conversion of glucose to glucose 6- phosphate, catalyzed by hexokinase, and conversion of fructose 6-phosphate to fructose 1,6- bisphosphate, catalyzed by phosphofructokinase-1 (Table 15-2, p. 573). The first reaction is bypassed during gluconeogenesis by the reaction catalyzed by glucose 6-phosphatase, an enzyme unique to the liver. The second is bypassed by fructose 1,6-bisphosphatase-1 (FBPase-1). By having separate pathways that employ different enzymes, an organism is able to control anabolic and catabolic processes separately, thus avoiding futile cycles. A potential disadvantage is the need to produce separate sets of enzymes for catabolism and anabolism. 19- What is a futile cycle? Give an example of a potential futile cycle in carbohydrate metabolism, and describe methods used by cells or organisms to avoid the operation of the futile cycle. Ans: A futile cycle is a pair of reactions or pathways in one of which A is converted into B, and in the other, B into A. For example, conversion of fructose 6-phosphate into fructose 1,6-bisphosphate (catalyzed by phosphofructokinase-1) is effectively reversed by the reaction catalyzed by fructose 1,6-bisphosphatase. The sum of the two reactions is the hydrolysis of ATP and the dissipation of energy as heat, a wasteful process except when the organism needs to generate heat to maintain body temperature. Fructose 6-phosphate + ATP fructose 1,6-bisphosphate + ADP Fructose 1,6-bisphosphate fructose 6-phosphate + Pi Sum: ATP ADP + Pi (+ heat) Cells use reciprocal regulation of the pathways such that when a reaction in the pathway in one direction is stimulated, the reaction in the reverse pathway is inhibited, and vice versa.