Chapter 16

1. Glucose is formed under prebiotic conditions. It is the most stable hexose sugar and consequently, has a low tendency, relative to other monosaccharides, to nonenzymatically react with proteins.

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2. Complete the interactive matching exercise to see answers.

3. In both cases, the electron donor is glyceraldehyde 3-phosphate. In lactic acid fermentation, the electron acceptor is pyruvate, converting it into lactate. In alcoholic fermentation, acetaldehyde is the electron acceptor, forming ethanol.

4. (a) 3 ATP; (b) 2 ATP; (c) 2 ATP; (d) 2 ATP; (e) 4 ATP

5. Glucokinase enables the liver to remove glucose from the blood when hexokinase is saturated, ensuring that glucose is captured for later use.

6. Glucokinase has a higher K_M value, which allows this enzyme to become more active at high glucose concentrations, conditions that saturate hexokinase.

7. Glucose cannot be cleaved into two three-carbon fragments, whereas fructose can, and three-carbon molecules are metabolized in the second stage of glycolysis. The conversion of fructose 6-phosphate into fructose 1,6-bisphosphate prevents the glucose isomer from being re-formed.

8. The GAP formed is immediately removed by subsequent reactions, resulting in the conversion of DHAP into GAP by the enzyme.

9. A thioester couples the oxidation of glyceraldehyde 3-phosphate to 3-phosphoglycerate with the formation of 1,3-bisphosphoglycerate. 1,3-Bisphosphoglycerate can subsequently power the formation of ATP.

10. Glycolysis is a component of alcoholic fermentation, the pathway that produces alcohol for beer and wine. The belief was that understanding the biochemical basis of alcohol production might lead to a more efficient means of producing beer.

11. The conversion of glyceraldehyde 3-phosphate into 1,3-bisphosphoglycerate would be impaired. Glycolysis would be less effective.

12. Glucose 6-phosphate must have other fates. Indeed, it can be converted into glycogen or be processed to yield reducing power for biosynthesis.

13. The energy needs of a muscle cell vary widely, from rest to intense exercise. Consequently, the regulation of phosphofructokinase by energy charge is vital. In other tissues, such as the liver, ATP concentration is less likely to fluctuate and will not be a key regulator of phosphofructokinase.

14. The ΔG°′ for the reverse of glycolysis is +90 kJ mol⁻¹ (+22 kcal mol⁻¹), far too endergonic to take place.

15. Pyruvate can be metabolized to ethanol in alcoholic fermentation, to lactate in lactic acid fermentation, or be completely oxidized to CO₂ and H₂O in cellular respiration.

16. The conversion of glucose into glucose 6-phosphate by hexokinase; the conversion of fructose 6-phosphate into fructose 1,6-bisphosphate by phosphofructokinase; the formation of pyruvate from phosphoenolpyruvate by pyruvate kinase

17. Lactic acid is a strong acid (problem 2.15). If it remained in the cell, the pH of the cell would fall, which could lead to the denaturation of muscle protein and result in muscle damage.

18. In liver, fructokinase converts fructose into fructose 1-phosphate. Fructose 1-phosphate is cleaved by a specific aldolase to yield glyceraldehyde and dihydroxyacetone phosphate, which is a component of the glycolytic pathway. Glyceraldehyde is converted into the glycolytic intermediate glyceraldehyde 3-phosphate by triose kinase. In other tissues, fructose is converted into fructose 6-phosphate by hexokinase.

19. Without triose phosphate isomerase, only one of the two three-carbon molecules generated by aldolase could be used to generate ATP. Only two molecules of ATP would result from the metabolism of each molecule of glucose. But two molecules of ATP would still be required to form fructose 1,6-bisphosphate, the substrate for aldolase. The net yield of ATP would be zero, a yield incompatible with life.

20. 3.06 × 10⁻⁵

21. The equilibrium concentrations of fructose 1,6-bisphosphate, dihydroxyacetone phosphate, and glyceraldehyde 3-phosphate are 7.8 × 10⁻⁴ M, 2.2 × 10⁻⁴ M, and 2.2 × 10⁻⁴ M, respectively.

22. 1. The fructose 1-phosphate pathway forms glyceraldehyde 3-phosphate, bypassing the control by phosphofructokinase.

    2. Since phosphofructokinase is bypassed, glycolysis proceeds in an unregulated manner. Lactic acidosis may result, and fatty liver may develop (Chapter 28).

23. The net reaction in the presence of arsenate is

    

    Glycolysis proceeds in the presence of arsenate, but the ATP normally formed in the conversion of 1,3-bisphosphoglycerate into 3-phosphoglycerate is lost. Thus, arsenate uncouples oxidation and phosphorylation by forming a highly labile acyl arsenate.

24. This example illustrates the difference between the stoichiometric and the catalytic use of a molecule. If cells used NAD⁺ stoichiometrically, a new molecule of NAD⁺ would be required each time a molecule of lactate was produced. As we will see, the synthesis of NAD⁺ requires ATP. On the other hand, if the NAD⁺ that is converted into NADH could be recycled and reused, a small amount of the molecule could regenerate a vast amount of lactate, which is the case in the cell. NAD⁺ is regenerated by the oxidation of NADH and reused. NAD⁺ is thus used catalytically.

25. Recall from our discussion of enzyme kinetics in Chapter 7 that substrates are usually present in much higher concentration than their enzymes. Consequently, converting a small amount of substrate into a potent activator of PFK will lead to a rapid increase in the rate of ATP synthesis.

26. Complete the interactive matching exercise to see answers.

27. Fructose 2,6-bisphosphate stabilizes the R state of the enzyme.

28. Galactose is a component of glycoproteins. Possibly, the absence of galactose leads to the improper formation or function of glycoproteins required in the central nervous system. More generally, the fact that the symptoms arise in the absence of galactose suggests that galactose is required in some fashion.

29. A potassium channel, a ligand-gated channel, is inhibited by binding ATP. This inhibition alters the voltage across the plasma membrane, which activates a calcium channel, a voltage-gated channel, allowing an influx of calcium ions. The calcium ions stimulate the fusion of insulin-containing granules with the plasma membrane, resulting in the secretion of insulin.

30. Using the Michaelis–Menten equation to solve for [S] when K_M = 50 µM and v_o = 0.9V_max shows that a substrate concentration of 0.45 mM yields 90% of V_max. Under normal conditions, the enzyme is essentially working at V_max.

31. 1. Curiously, the enzyme uses ADP as the phosphoryl donor rather than ATP.

    2. Both AMP and ATP behave as competitive inhibitors of ADP, the phosphoryl donor. Apparently, the P. furiosus enzyme is not allosterically inhibited by ATP.

32. ATP initially stimulates PFK activity, as would be expected for a substrate. Higher concentrations of ATP inhibit the enzyme. Although this effect seems counterintuitive for a substrate, recall that the function of glycolysis in muscle is to generate ATP. Consequently, high concentrations of ATP signal that the ATP needs are met and glycolysis should stop. In addition to being a substrate, ATP is an allosteric inhibitor of PFK.

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33. 1. Carbon dioxide is a good indicator of the rate of alcoholic fermentation because it is, along with ethanol, a product of alcoholic fermentation.

    2. Phosphate is a required substrate for the reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase. The phosphate is incorporated into 1,3-bisphosphoglycerate.

    3. It indicates that the amount of free phosphate must have been limiting.

    4. The ratio would be expected to be 1. A phosphate would be consumed for every pyruvate decarboxylated.

    5. Lack of phosphate would inhibit glyceraldehyde 3-phosphate dehydrogenase. This inhibition would “back-up” glycolysis, resulting in the accumulation of fructose 1,6-bisphosphate.

34. Glucose is reactive because its open-chain form contains an aldehyde group. In other words, glucose is a reducing sugar.

35. 1. The label is in the methyl carbon atom of pyruvate.

    2. 5 mCi mM⁻¹. The specific activity is halved because the number of moles of product (pyruvate) is twice that of the labeled substrate (glucose).

36. 1. Glucose + 2 P_i + 2 ADP → 2 lactate + 2 ATP

    2. ΔG°′ = −123 kJ mol⁻¹ (−29 kcal mol⁻¹)

    3. ΔG = −114 kJ mol⁻¹ (−27.2 kcal mol⁻¹)

37. GLUT 2 transports glucose only when the blood concentration of glucose is high, which is precisely the condition in which the β cells of the pancreas secrete insulin.

38. Fructose + ATP → fructose 1-phosphate + ADP: Fructokinase

    Fructose 1-phosphate → dihydroxyacetone phosphate + glyceraldehyde: Fructose 1-phosphate aldolase

    Glyceraldehyde + ATP → glyceraldehyde 3-phosphate + ADP: Triose kinase

    The primary controlling step of glycolysis catalyzed by phosphofructokinase is bypassed by the preceding reactions. Glycolysis will proceed in an unregulated fashion.

39. The ATPase activity of hexokinase is low in the absence of a sugar because it is in a catalytically inactive conformation. The addition of xylose closes the cleft between the two lobes of the enzyme. However, xylose lacks a hydroxymethyl group, and so it cannot be phosphorylated. Instead, a water molecule at the site normally occupied by the C-6 hydroxymethyl group acts as the acceptor of the phosphoryl group from ATP.

40. Consider the equilibrium equation of adenylate kinase:

    

    or

    

    Recall that [ATP] > [ADP] > [AMP] in the cell. As ATP is utilized, a small decrease in its concentration will result in a larger percentage increase in [ADP] because ATP’s concentration is greater than that of ADP. This larger percentage increase in [ADP] will result in an even greater percentage increase in [AMP] because the concentration of AMP is related to the square of [ADP]. In essence, equation 2 shows that monitoring the energy status with AMP magnifies small changes in [ATP], leading to tighter control.