Chapter Review in Chapter 13 Introduction to Metabolism

Chapter Review

KEY TERMS

Terms in bold are defined in the glossary.

autotroph
heterotroph
metabolite
intermediary metabolism
catabolism
anabolism
energy transduction
free energy, G
exergonic
endergonic
enthalpy, H
exothermic
endothermic
entropy, S
standard transformed constants
mass-action ratio, Q
homolytic cleavage
radical
heterolytic cleavage
nucleophile
electrophile
carbanion
carbocation
aldol condensation
Claisen condensation
kinases
phosphorylation potential $(Δ G_{p})$ $bold left-parenthesis bold upper Delta bold-italic upper G Subscript bold p Baseline bold right-parenthesis$
thioester
adenylylation
inorganic pyrophosphatase
nucleoside diphosphate kinase
adenylate kinase
creatine kinase
phosphagens
electromotive force (emf)
conjugate redox pair
dehydrogenases
reducing equivalent
standard reduction potential $(E °)$ $bold left-parenthesis bold-italic upper E bold degree bold right-parenthesis$
pyridine nucleotide
oxidoreductase
flavoprotein
flavin nucleotides
glucose 6-phosphate
homeostasis
transcription factor
response element
turnover
transcriptome
proteome
metabolome
metabolic regulation
metabolic control
adenylate kinase
AMP-activated protein kinase (AMPK)

Problems

1. Entropy Changes during Egg Development Consider a system consisting of an egg in an incubator. The white and yolk of the egg contain proteins, carbohydrates, and lipids. If fertilized, the egg transforms from a single cell to a complex organism. Discuss this irreversible process in terms of the entropy changes in the system and surroundings. Be sure that you first clearly define the system and surroundings.
2. Calculation of $Δ G^{'} °$ $bold upper Delta bold-italic upper G bold prime degree$ from an Equilibrium Constant Calculate the standard free-energy change for each of the three metabolically important enzyme-catalyzed reactions, using the equilibrium constants given for the reactions at $25 ° C$ $25 degree upper C$ and pH 7.0.
3. Calculation of the Equilibrium Constant from $Δ G^{'} °$ $bold upper Delta bold-italic upper G bold prime degree$ Calculate the equilibrium constant $K_{eq}^{'}$ $upper K prime Subscript eq$ for each of the three reactions at pH 7.0 and $25 ° C$ $25 degree upper C$ , using the $Δ G^{'} °$ $upper Delta upper G prime degree$ values in Table 13-4.
4. Experimental Determination of $K_{eq}^{'}$ $bold-italic upper K bold prime Subscript bold eq$ and $Δ G^{'} °$ $bold upper Delta bold-italic upper G bold prime degree$ Incubating a 0.1 m solution of glucose 1-phosphate at $25 ° C$ $25 degree upper C$ with a catalytic amount of phosphoglucomutase transforms some of the glucose 1-phosphate to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are

$\begin{matrix} Glucose 1 -phosphate ⇌ glucose 6 -phosphate \\ \begin{matrix} 4.5 \times 10^{- 3} M & 9.6 \times 10^{- 2} M \end{matrix} \end{matrix}$ $StartLayout 1st Row Glucose 1 hyphen phosphate right harpoon over left harpoon glucose 6 hyphen phosphate 2nd Row StartLayout 1st Row 1st Column 4.5 times 10 Superscript negative 3 Baseline upper M 2nd Column 3rd Column 9.6 times 10 Superscript negative 2 Baseline upper M EndLayout EndLayout$

Calculate $K_{eq}^{'}$ $upper K prime Subscript eq$ and $Δ G^{'} °$ $upper Delta upper G prime degree$ for this reaction.
5. Experimental Determination of $Δ G^{'} °$ $upper Delta bold-italic upper G prime degree$ for ATP Hydrolysis A direct measurement of the standard free-energy change associated with the hydrolysis of ATP is technically demanding because the minute amount of ATP remaining at equilibrium is difficult to measure accurately. The value of $Δ G^{'} °$ $upper Delta upper G prime degree$ can be calculated indirectly, however, from the equilibrium constants of two other enzymatic reactions having less favorable equilibrium constants:

$\begin{matrix} Glucose 6 -phosphate + H_{2} O \overset{}{\to} glucose + P_{i} K_{eq}^{'} = 270 \\ ATP + glucose \overset{}{\to} ADP + glucose 6 -phosphate K_{eq}^{'} = 890 \end{matrix}$ $StartLayout 1st Row Glucose 6 hyphen phosphate plus upper H Subscript 2 Baseline upper O right-arrow Overscript Endscripts glucose plus upper P Subscript i Baseline upper K prime Subscript eq Baseline equals 270 2nd Row ATP plus glucose right-arrow Overscript Endscripts ADP plus glucose 6 hyphen phosphate upper K prime Subscript eq Baseline equals 890 EndLayout$

Using this information for equilibrium constants determined at $25 ° C$ $25 degree upper C$ , calculate the standard free energy of hydrolysis of ATP.
6. Difference between $Δ G^{'} °$ $upper Delta bold-italic upper G prime degree$ and ΔG Consider the interconversion shown, which occurs in glycolysis (Chapter 14):

$\begin{array}{l} Fructose 6 -phosphate ⇌ glucose 6 -phosphate & K_{eq}^{'} = 1.97 \end{array}$ $StartLayout 1st Row 1st Column Fructose 6 hyphen phosphate right harpoon over left harpoon glucose 6 hyphen phosphate 2nd Column upper K prime Subscript eq Baseline equals 1.97 EndLayout$
1. What is $Δ G^{'} °$ $upper Delta upper G prime degree$ for the reaction ( $K_{eq}^{'}$ $upper K prime Subscript eq$ measured at $25 ° C$ $25 degree upper C$ )?
2. If the concentration of fructose 6-phosphate is adjusted to 1.5 m and that of glucose 6-phosphate is adjusted to 0.50 m, what is ΔG?
3. Why are $Δ G^{'} °$ $upper Delta upper G prime degree$ and ΔG different?
7. Free Energy of Hydrolysis of CTP Compare the structure of the nucleoside triphosphate CTP with the structure of ATP.

Cytidine triphosphate (C T P) has a five-membered ring with O substituted for C at position 6 at the top vertex, C 1 prime bonded to N of the lower right vertex of a six-membered ring above and to H below, C 2 prime and C 3 prime bonded to H above and O H below, C 4 prime bonded to C 5 prime above and to H below, and C 5 prime bonded to 2 H and to O bonded to P double bonded to O above, bonded to O minus below, and bonded to P to the left double bonded to O above, bonded to O minus below, and bonded to P on the left double bonded to O above, bonded to O minus below, and bonded to O minus to the left. The six-membered ring shares its left side with a six-membered ring. It has N substituted for C at its upper right vertex at position 1, a double bond at its left side between C 5 and C 6; N substituted for C at its upper right vertex at position 3 and further bonded to H, its lower right vertex at position 2 double bonded to O, and its top vertex at position 4 bonded to N H 2. Adenosine triphosphate (A T P) has a five-membered ring with O substituted for C at position 6 at the top vertex, C 1 prime bonded to N of the lower right vertex of a five-membered ring above and to H below, C 2 prime and C 3 prime bonded to H above and O H below, C 4 prime bonded to C 5 prime above and to H below, and C 5 prime bonded to 2 H and to O bonded to P double bonded to O above, bonded to O minus below, and bonded to P to the left double bonded to O above, bonded to O minus below, and bonded to P on the left double bonded to O above, bonded to O minus below, and bonded to O minus to the left. The five-membered ring shares its right side with a six-membered ring. It has N substituted for C at its upper left vertex at position 7, a double bond at its upper left side between positions 7 and 8, a double bond at its right side between positions 4 and 5, and N substituted for C at its lower right vertex at position 9 and further bonded to C 1 prime of the ring below. The six-membered ring has double bond at its left side between positions 4 and 5 shared with the five-membered ring, a double bond at its upper right side between positions 1 and 6, a double bond at its lower right side between positions 2 and 3, N substituted for C at its upper right vertex at position 1, N substituted for C at its bottom vertex at position 3, and its top vertex, C 6, bonded to N H 2.

Now predict the $K_{eq}^{'}$ $upper K prime Subscript eq$ and $Δ G^{'} °$ $upper Delta upper G prime degree$ for the reaction:

$ATP + CDP \to ADP + CTP$ $ATP plus CDP right-arrow ADP plus CTP$
8. Dependence of ΔG on pH The free energy released by the hydrolysis of ATP under standard conditions is $- 30.5 kJ/mol$ $negative 30.5 kJ slash mol$ . If ATP is hydrolyzed under standard conditions except at pH 5.0, is more or less free energy released? Explain.
9. The $Δ G^{'} °$ $upper Delta bold-italic upper G prime degree$ for Coupled Reactions Glucose 1-phosphate is converted into fructose 6-phosphate in two successive reactions:

$\begin{matrix} Glucose 1 -phosphate \overset{}{\to} glucose 6 -phosphate \\ Glucose 6 -phosphate \overset{}{\to} fructose 6 -phosphate \end{matrix}$ $StartLayout 1st Row Glucose 1 hyphen phosphate right-arrow Overscript Endscripts glucose 6 hyphen phosphate 2nd Row Glucose 6 hyphen phosphate right-arrow Overscript Endscripts fructose 6 hyphen phosphate EndLayout$

Using the $Δ G^{'} °$ $upper Delta upper G prime degree$ values in Table 13-4, calculate the equilibrium constant, $K_{eq}^{'}$ $upper K prime Subscript eq$ , for the sum of the two reactions:

$Glucose 1 -phosphate \overset{}{\to} fructose 6 -phosphate$ $Glucose 1 hyphen phosphate right-arrow Overscript Endscripts fructose 6 hyphen phosphate$
10. Effect of [ATP]/[ADP] Ratio on Free Energy of Hydrolysis of ATP Using Equation 13-4, plot ΔG against ln Q (mass-action ratio) at $25 ° C$ $25 degree upper C$ for the concentrations of ATP, ADP, and $P_{i}$ $upper P Subscript i$ in the table shown. $Δ G^{'} °$ $upper Delta upper G prime degree$ for the reaction is $- 30.5 kJ/mol$ $negative 30.5 kJ slash mol$ . Use the resulting plot to explain why metabolism is regulated to keep the ratio [ATP]/[ADP] high.

Concentration (mm)

ATP 5 3 1 0.2 5

ADP 0.2 2.2 4.2 5.0 25

$P_{i}$ $upper P Subscript i$ 10 12.1 14.1 14.9 10
11. Strategy for Overcoming an Unfavorable Reaction: ATP-Dependent Chemical Coupling The phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by $P_{i}$ $upper P Subscript i$ is described by the equation

$Glucose + P_{i} \overset{}{\to} glucose 6 -phosphate + H_{2} O Δ G^{'} ° = 13.8 kJ/mol$ $Glucose plus upper P Subscript i Baseline right-arrow Overscript Endscripts glucose 6 hyphen phosphate plus upper H Subscript 2 Baseline upper O upper Delta upper G Superscript prime Baseline degree equals 13.8 kJ slash mol$
1. Calculate the equilibrium constant for this reaction at $37 ° C$ $37 degree upper C$ . In the rat hepatocyte, the physiological concentrations of glucose and $P_{i}$ $upper P Subscript i$ are maintained at approximately 4.8 mm. What is the equilibrium concentration of glucose 6-phosphate obtained by the direct phosphorylation of glucose by $P_{i}$ $upper P Subscript i$ ? Does this reaction represent a reasonable metabolic step for the catabolism of glucose? Explain.
2. In principle at least, one way to increase the concentration of glucose 6-phosphate is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and $P_{i}$ $upper P Subscript i$ . Assuming a fixed concentration of $P_{i}$ $upper P Subscript i$ at 4.8 mm, how high would the intracellular concentration of glucose have to be to give an equilibrium concentration of glucose 6-phosphate of 250 μm (the normal physiological concentration)? Would this route be physiologically reasonable, given that the maximum solubility of glucose is less than 1 m?
3. The phosphorylation of glucose in the cell is coupled to the hydrolysis of ATP; that is, part of the free energy of ATP hydrolysis is used to phosphorylate glucose:
  $\begin{matrix} (1) Glucose + P_{i} \overset{}{\to} glucose 6 -phosphate + H_{2} O \\ Δ G^{'} ° = 13.8 kJ/mol \\ \underline{(2) ATP + H_{2} O \overset{}{\to} ADP + P_{i} Δ G^{'} ° = - 30.5 kJ/mol} \\ S u m : Glucose + ATP \overset{}{\to} glucose 6 -phosphate + ADP \end{matrix}$ $StartLayout 1st Row left-parenthesis 1 right-parenthesis Glucose plus upper P Subscript i Baseline right-arrow Overscript Endscripts glucose 6 hyphen phosphate plus upper H Subscript 2 Baseline upper O 2nd Row upper Delta upper G prime degree equals 13.8 kJ slash mol 3rd Row ModifyingBelow left-parenthesis 2 right-parenthesis ATP plus upper H Subscript 2 Baseline upper O right-arrow Overscript Endscripts ADP plus upper P Subscript i Baseline upper Delta upper G prime degree equals negative 30.5 kJ slash mol With bar 4th Row upper S u m colon Glucose plus ATP right-arrow Overscript Endscripts glucose 6 hyphen phosphate plus ADP EndLayout$
  
  Calculate $K_{eq}^{'}$ $upper K prime Subscript eq$ at $37 ° C$ $37 degree upper C$ for the overall reaction. For the ATP-dependent phosphorylation of glucose, what concentration of glucose is needed to achieve a 250 μm intracellular concentration of glucose 6-phosphate when the concentrations of ATP and ADP are 3.38 mm and 1.32 mm, respectively? Does this coupling process provide a feasible route, at least in principle, for the phosphorylation of glucose in the cell? Explain.
4. Although coupling ATP hydrolysis to glucose phosphorylation makes thermodynamic sense, we have not yet specified how this coupling is to take place. Given that coupling requires a common intermediate, one conceivable route is to use ATP hydrolysis to raise the intracellular concentration of $P_{i}$ $upper P Subscript i$ and thus drive the unfavorable phosphorylation of glucose by $P_{i}$ $upper P Subscript i$ . Is this a reasonable route? (Think about the solubility product, $K_{sp}$ $upper K Subscript sp$ , of metabolic intermediates.)
5. The ATP-coupled phosphorylation of glucose is catalyzed in hepatocytes by the enzyme glucokinase. This enzyme binds ATP and glucose to form a glucose-ATP-enzyme complex, and the phosphoryl group is transferred directly from ATP to glucose. Explain the advantages of this route.
12. Calculations of $Δ G^{'} °$ $upper Delta bold-italic upper G prime degree$ for ATP-Coupled Reactions From data in Table 13-6, calculate the $Δ G^{'} °$ $upper Delta upper G prime degree$ value for each reaction:
1. $Phosphocreatine + ADP \overset{}{\to} creatine + ATP$ $Phosphocreatine plus ADP right-arrow Overscript Endscripts creatine plus ATP$
2. $ATP + fructose \overset{}{\to} ADP + fructose 6 -phosphate$ $ATP plus fructose right-arrow Overscript Endscripts ADP plus fructose 6 hyphen phosphate$
13. Coupling ATP Cleavage to an Unfavorable Reaction To explore the consequences of coupling ATP hydrolysis under physiological conditions to a thermodynamically unfavorable biochemical reaction, consider the hypothetical transformation $X \to Y$ $upper X right-arrow upper Y$ , for which $Δ G^{'} ° = 20.0 kJ/mol$ $upper Delta upper G Superscript prime Baseline degree equals 20.0 kJ slash mol$ .
1. What is the ratio [Y]/[X] at equilibrium?
2. Suppose X and Y participate in a sequence of reactions during which ATP is hydrolyzed to ADP and $P_{i}$ $upper P Subscript i$ . The overall reaction is
  $X + ATP + H_{2} O \overset{}{\to} Y + ADP + P_{i}$ $upper X plus ATP plus upper H Subscript 2 Baseline upper O right-arrow Overscript Endscripts upper Y plus ADP plus upper P Subscript i Baseline$
  
  Calculate [Y]/[X] for this reaction at equilibrium. Assume that the temperature is $25.0 ° C$ $25.0 degree upper C$ and the equilibrium concentrations of ATP, ADP, and $P_{i}$ $upper P Subscript i$ are 1 m.
3. We know that [ATP], [ADP], and ${[P}_{i}]$ $left-bracket upper P Subscript i Baseline right-bracket$ are not 1 m under physiological conditions. Calculate [Y]/[X] for the ATP-coupled reaction when the values of [ATP], [ADP], and ${[P}_{i}]$ $left-bracket upper P Subscript i Baseline right-bracket$ are those found in rat myocytes (Table 13-5).
14. Calculations of ΔG at Physiological Concentrations Calculate the actual, physiological ΔG for the reaction

$Phosphocreatine + ADP \overset{}{\to} creatine + ATP$ $Phosphocreatine plus ADP right-arrow Overscript Endscripts creatine plus ATP$

at $37 ° C$ $37 degree upper C$ , as it occurs in the cytosol of neurons, with phosphocreatine at 4.7 mm, creatine at 1.0 mm, ADP at 0.73 mm, and ATP at 2.6 mm.
15. Free Energy Required for ATP Synthesis under Physiological Conditions In the cytosol of rat hepatocytes, the temperature is $37 ° C$ $37 degree upper C$ and the mass-action ratio, Q, is

$\frac{[ATP]}{[ADP] [P_{i}]} = 5.33 \times 10^{2} M^{- 1}$ $StartFraction left-bracket ATP right-bracket Over left-bracket ADP right-bracket left-bracket upper P Subscript i Baseline right-bracket EndFraction equals 5.33 times 10 squared upper M Superscript negative 1$

Calculate the free energy required to synthesize ATP in a rat hepatocyte.
16. Chemical Logic In the glycolytic pathway, a six-carbon sugar (fructose 1,6-bisphosphate) is cleaved to form two three-carbon sugars, which undergo further metabolism. In this pathway, an isomerization of glucose 6-phosphate to fructose 6-phosphate (as shown in the diagram) occurs two steps before the cleavage reaction. The intervening step is phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate (p. 516).

Glucose 6-phosphate is a six-carbon chain with C 1 bonded to H to the upper left and double bonded to O to its upper right; C 2, C 4, and C 5 bonded to O H on the right and H on the left; C 3 bonded to H on the right and O H on the left; and C 6 bonded to 2 H and O P O subscript 3 superscript 2 minus. Right- and left-pointing arrows are shown above text reading, phosphohexose isomerase. This yields fructose 6-phosphate, a six-carbon chain with C 1 bonded to H above and to the left and to O H to the right; C 2 double bonded to O; C 3 bonded to H on the right and O H on the left; C 4 and C 5 bonded to O H on the right and H on the left; and C 6 bonded to 2 H and O P O subscript 3 superscript 2 minus.

What does the isomerization step accomplish from a chemical perspective? (Hint: Consider what might happen if the $C — C$ $upper C em-dash upper C$ bond cleavage were to proceed without the preceding isomerization.)
17. Enzymatic Reaction Mechanisms I Lactate dehydrogenase is one of the many enzymes that require NADH as coenzyme. It catalyzes the conversion of pyruvate to lactate:

Pyruvate is shown as a three-carbon chain with the top two carbons enclosed in a light red box. C 1 is double bonded to O to the upper left and bonded to O minus to the upper right within the box, C 2 is double bonded to O within the box, and C 3 is bonded to 3 H outside of the box. A right-pointing arrow is accompanied by a curved arrow showing that N A D H plus H plus are added and N A D plus leaves. A shorter arrow points to the left and text below reads, lactate dehydrogenase. The reaction produces L-lactate, which has a similar structure to pyruvate except that C 2 is bonded to H to the right and O H to the left within the light red box containing C 1 and C 2.

Draw the mechanism of this reaction (show electron-pushing arrows). (Hint: This is a common reaction throughout metabolism; the mechanism is similar to that catalyzed by other dehydrogenases that use NADH, such as alcohol dehydrogenase.)
18. Enzymatic Reaction Mechanisms II Biochemical reactions often look more complex than they really are. In the pentose phosphate pathway (Chapter 14), sedoheptulose 7-phosphate and glyceraldehyde 3-phosphate react to form erythrose 4-phosphate and fructose 6-phosphate in a reaction catalyzed by transaldolase.

Sedoheptulose has a vertical seven-carbon chain with the top three carbons and their substituents highlighted in red. The structure is red highlighted C 1 bonded to 2 H and O H, red highlighted C 2 double bonded to O; red highlighted C 3 bonded to H on the right and O H on the left; C 4, C 5, and C 6 bonded to O H on the right and H on the left; and C 7 bonded to 2 H and O P O subscript 3 superscript 2 minus. There is a red bond between C 3 and C 4. Glyceraldehyde 3-phosphate has a vertical three-carbon chain with C 1 double bonded to O to the upper left and bonded to H to the upper right; C 2 bonded to O H to the right and H to the left; and C 3 bonded to 2 H and to O P O subscript 3 superscript 2 minus. Arrows point to the right and left above the word transaldolase. This reaction yields erythrose 4- phosphate and fructose 6- phosphate. Erythrose 4- phosphate has a vertical four-carbon chain with C 1 double bonded to O to the upper left and bonded to H to the upper right; C 2 and C 3 bonded to O H to the right and H to the left; and C 4 bonded to 2 H and to O P O subscript 3 superscript 2 minus. Fructose 6-phosphate has a vertical six-carbon vertical chain with the top three carbons and their substituents highlighted in red. Red highlighted C 1 is bonded to 2 H and to O H; red highlighted C 2 is double bonded to O to the right; red highlighted C 3 is bonded to H to the right and to O H to the left; C 4 and C 5 are bonded to O H to the right and to H to the left; and C 6 is bonded to 2 H and to O P O subscript 3 superscript 2 minus. There is a red bond between C 3 and C 4.

Draw a mechanism for this reaction (show electron-pushing arrows). (Hint: Take another look at aldol condensations, then consider the name of this enzyme.)
19. Recognizing Reaction Types For each pair of biomolecules, identify the type of reaction (oxidation-reduction, hydrolysis, isomerization, group transfer, or internal rearrangement) required to convert the first molecule to the second. In each case, indicate the general type of enzyme and cofactor(s) or reactants that would be required, and any other products that would result.
1. Palmitoyl- Co A has R bonded to C H 2 bonded to C H 2 bonded to C H 2 bonded to C double bonded to O below and bonded to S bonded to Co A. Italicized trans end italics Greek letter delta superscript 2- enoyl- Co A has R bonded to C H 2 bonded to C H double bonded to C H bonded to C double bonded to O below and bonded to S bonded to Co A.
2. L- leucine has a vertical four-carbon chain with C 1 forming C O O minus; C 2 bonded to H to the right and to N H 3 to the left with a positive charge on N; C 3 bonded to 2 H; and C 4 bonded to H and to C H 3 to the lower left and right. D- leucine is similar except that C 2 is bonded to N H 3 to the right with a positive charge on N and to H on the left.
3. Glucose is a six-carbon chain with C 1 bonded to H to the upper left and double bonded to O to its upper right; C 2, C 4, and C 5 bonded to O H on the right and H on the left; C 3 bonded to H on the right and O H on the left; and C 6 bonded to 2 H and O H. Fructose is a six-carbon chain with C 1 bonded to H above and to the left and to O H to the right; C 2 double bonded to O; C 3 bonded to H on the right and O H on the left; C 4 and C 5 bonded to O H on the right and H on the left; and C 6 bonded to 2 H and O H.
4. Glycerol has a horizontal three-carbon chain with C 1 bonded to 2 H and to O H above, C 2 bonded to O H above and H below, and C 3 bonded to 2 H and to O H above. Glycerol 3- phosphate has a horizontal three-carbon chain with C 1 bonded to 2 H and to O H above, C 2 bonded to O H above and H below, and C 3 bonded to 2 H and to O P O subscript 3 superscript 2 minus above.
5. Glycylalanine has N H 3 with a positive charge on N bonded to C bonded to H above and below and to C double bonded to O above further bonded to N bonded to H below and bonded to C to the right bonded to H above, to C H 3 below, and to C to the right double bonded to O above and bonded to O minus to the right. Glycine has N H 3 with a positive charge on N bonded to C bonded to H above and below and to C double bonded to O above and to O minus to the right. Alanine has N H 3 with a positive charge on N bonded to C bonded to H above, C H 3 below, and C to the right double bonded to O above and bonded to O minus to the right.
20. Effect of Structure on Group Transfer Potential Some invertebrates contain phosphoarginine. Is the standard free energy of hydrolysis of this molecule more similar to that of glucose 6-phosphate or of ATP? Explain your answer.
21. Polyphosphate as a Possible Energy Source The standard free energy of hydrolysis of inorganic polyphosphate (polyP) is about $- 20 kJ/mol$ $negative 20 kJ slash mol$ for each $P_{i}$ $upper P Subscript i$ released. We calculated in Worked Example 13-2 that, in a cell, it takes about 50 kJ/mol of energy to synthesize ATP from ADP and $P_{i}$ $upper P Subscript i$ .

From left to right, the structure is O minus bonded to P double bonded to O above, bonded to O minus below, and bonded to O to the right further bonded to P double bonded to O above, bonded to O minus below, and bonded to O to the right further bonded to P double bonded to O above, bonded to O minus below, and bonded to O to the right further bonded to P double bonded to O above, bonded to O minus below, and bonded to O with a dashed horizontal line extending to the right.

Is it feasible for a cell to use polyphosphate to synthesize ATP from ADP? Explain your answer.
22. Daily ATP Utilization by Human Adults
1. The synthesis of ATP from ADP and $P_{i}$ $upper P Subscript i$ requires a total of 30.5 kJ/mol of free energy when the reactants and products are at 1 m concentrations and the temperature is $25 ° C$ $25 degree upper C$ (standard state). However, the actual physiological concentrations of ATP, ADP, and $P_{i}$ $upper P Subscript i$ are not 1 m, and the physiological temperature is 37 °C. Thus, the free energy required to synthesize ATP under physiological conditions is different from $Δ G^{'} °$ $upper Delta upper G prime degree$ . Calculate the free energy required to synthesize ATP in the human hepatocyte when the physiological concentrations of ATP, ADP, and $P_{i}$ $upper P Subscript i$ are 3.5, 1.50, and 5.0 mm, respectively.
2. A 68 kg (150 lb) adult requires a caloric intake of 2,000 kcal (8,360 kJ) of food per day (24 hours). The body metabolizes the food and uses the free energy to synthesize ATP, which then provides energy for the body’s daily chemical and mechanical work. Assuming that the efficiency of converting food energy into ATP is 50%, calculate the weight of ATP used by a human adult in 24 hours. What percentage of the body weight does this represent?
3. Although adults synthesize large amounts of ATP daily, their body weight, structure, and composition do not change significantly during this period. Explain this apparent contradiction.
23. Rates of Turnover of γ and β Phosphates of ATP After adding a small amount of ATP labeled with radioactive phosphorus in the terminal position, $[γ -^{32} P]$ $left-bracket gamma hyphen Superscript 32 Baseline upper P right-bracket$ ATP, to a yeast extract, a researcher finds about half of the $^{32} P$ $Superscript 32 Baseline upper P$ activity in $P_{i}$ $upper P Subscript i$ within a few minutes, but the concentration of ATP remains unchanged. Explain. She then carries out the same experiment using ATP labeled with $^{32} P$ $Superscript 32 Baseline upper P$ in the central position, $[β -^{32} P]$ $left-bracket beta hyphen Superscript 32 Baseline upper P right-bracket$ ATP, but the $^{32} P$ $Superscript 32 Baseline upper P$ does not appear in $P_{i}$ $upper P Subscript i$ within such a short time. Why?
24. Cleavage of ATP to AMP and $P P_{i}$ $bold upper P bold upper P Subscript bold i Baseline$ during Metabolism Synthesis of the activated form of acetate (acetyl-CoA) is carried out in an ATP-dependent process:

$Acetate + CoA + ATP \overset{}{\to} acetyl-CoA + AMP + {PP}_{i}$ $Acetate plus CoA plus ATP right-arrow Overscript Endscripts acetyl hyphen CoA plus AMP plus PP Subscript i Baseline$
1. The $Δ G^{'} °$ $upper Delta upper G prime degree$ for hydrolysis of acetyl-CoA to acetate and CoA is $- 32.2 kJ/mol$ $negative 32.2 kJ slash mol$ . The $Δ G^{'} °$ $upper Delta upper G prime degree$ for hydrolysis of ATP to AMP and ${PP}_{i}$ $PP Subscript i$ is $- 30.5 kJ/mol$ $negative 30.5 kJ slash mol$ . Calculate $Δ G^{'} °$ $upper Delta upper G prime degree$ for the ATP-dependent synthesis of acetyl-CoA.
2. Almost all cells contain the enzyme inorganic pyrophosphatase, which catalyzes the hydrolysis of ${PP}_{i}$ $PP Subscript i$ to $P_{i}$ $upper P Subscript i$ . What effect does the presence of this enzyme have on the synthesis of acetyl-CoA? Explain.
25. Activation of a Fatty Acid by Reaction with Coenzyme A In the reaction sequence for fatty acid breakdown, coenzyme A (CoA), with its thiol $(— SH)$ $left-parenthesis em-dash SH right-parenthesis$ group, joins to the fatty acid as a thiol ester, as ATP is converted into AMP and ${PP}_{i}$ $PP Subscript i$ :

$\begin{matrix} R — {COO}^{-} + ATP + CoA — SH \underset{}{\to} \\ AMP + {PP}_{i} + R — CO — S — CoA \end{matrix}$ $StartLayout 1st Row upper R em-dash COO Superscript minus Baseline plus ATP plus CoA em-dash SH right-arrow Underscript Endscripts 2nd Row AMP plus PP Subscript i Baseline plus upper R em-dash CO em-dash upper S em-dash CoA EndLayout$

The oxidation of fatty acids as fuels requires two steps. The first step transfers an activating group from ATP to the carboxyl group of the fatty acid. In the second step, $CoA — SH$ $CoA em-dash SH$ displaces the activating group to form fatty acyl- $S — CoA$ $upper S em-dash CoA$ . Given the known products of the reaction, what is the activating group?
26. Energy for $H^{+}$ $upper H Superscript plus$ Pumping The parietal cells of the stomach lining contain membrane “pumps” that transport hydrogen ions from the cytosol (pH 7.0) into the stomach, contributing to the acidity of gastric juice (pH 1.0). Calculate the free energy required to transport 1 mol of hydrogen ions through these pumps. (Hint: See Chapter 11.) Assume a temperature of 37 °C.
27. Most-Reduced Carbon Compounds Arrange the four structures in order from most reduced to most oxidized.
1. $R — {CH}_{2} — {CH}_{2} — OH$ $upper R em-dash CH Subscript 2 Baseline em-dash CH Subscript 2 Baseline em-dash OH$
2. $R — {CH}_{2} — {COO}^{-}$ $upper R em-dash CH Subscript 2 Baseline em-dash COO Superscript minus$
3. $R — {CH}_{2} — CHO$ $upper R em-dash CH Subscript 2 Baseline em-dash CHO$
4. $R — {CH}_{2} — {CH}_{3}$ $upper R em-dash CH Subscript 2 Baseline em-dash CH Subscript 3 Baseline$
28. Standard Reduction Potentials The standard reduction potential, $E^{'} °$ $upper E prime degree$ , of any redox pair is defined for the half-cell reaction

$Oxidizing agent + n electrons \underset{}{\to} reducing agent$ $Oxidizing agent plus n electrons right-arrow Underscript Endscripts reducing agent$

The $E^{'} °$ $upper E prime degree$ values for the ${NAD}^{+} /NADH$ $NAD Superscript plus Baseline slash NADH$ and pyruvate/lactate conjugate redox pairs are $- 0.32 V$ $negative 0.32 upper V$ and $- 0.19 V,$ $negative 0.19 upper V comma$ respectively.
1. Which redox pair has the greater tendency to lose electrons? Explain.
2. Which pair is the stronger oxidizing agent? Explain.
3. Beginning with 1 m concentrations of each reactant and product at pH 7 and 25 °C, in which direction will the following reaction proceed?
  $Pyruvate + NADH + H^{+} ⇌ lactate + {NAD}^{+}$ $Pyruvate plus NADH plus upper H Superscript plus Baseline right harpoon over left harpoon lactate plus NAD Superscript plus$
4. What is the standard free-energy change $(Δ G^{'} °)$ $left-parenthesis upper Delta upper G prime degree right-parenthesis$ for the conversion of pyruvate to lactate?
5. What is the equilibrium constant ( $K_{eq}^{'}$ $upper K prime Subscript eq$ ) for this reaction?
29. Simple Biobattery Suppose you set up a simple battery using half-reactions as pictured in Figure 13-23. One electrode contains pyruvate and lactate at 1 mm, and the other electrode contains fumarate and succinate at 1 mm (see Table 13-7).
1. In which direction will electrons initially flow?
2. Calculate the standard reduction potential and standard free-energy change for your biological battery.
3. When a flashlight battery “runs out,” net electron movement has essentially ended. What is the equivalent situation for your biobattery?
30. Energy Span of the Respiratory Chain Electron transfer in the mitochondrial respiratory chain may be represented by the net reaction equation

$NADH + H^{+} + \frac{1}{2} O_{2} ⇌ H_{2} O + {NAD}^{+}$ $NADH plus upper H Superscript plus Baseline plus one-half upper O Subscript 2 Baseline right harpoon over left harpoon upper H Subscript 2 Baseline upper O plus NAD Superscript plus$
1. Calculate $Δ E^{'} °$ $upper Delta upper E prime degree$ for the net reaction of mitochondrial electron transfer. Use $E^{'} °$ $upper E prime degree$ values in Table 13-7.
2. Calculate $Δ G^{'} °$ $upper Delta upper G prime degree$ for this reaction.
3. How many ATP molecules can theoretically be generated by this reaction if the free energy of ATP synthesis under cellular conditions is 52 kJ/mol?
31. Dependence of Electromotive Force on Concentrations Suppose that you place an electrode into solutions of various concentrations of ${NAD}^{+}$ $NAD Superscript plus$ and NADH at pH 7.0 and 25 °C. Calculate the electromotive force (in volts) registered by the electrode when immersed in each solution, with reference to a half-cell of $E^{'} °$ $upper E prime degree$ 0.00 V.
1. 1.0 mm ${NAD}^{+}$ $NAD Superscript plus$ and 10 mm NADH
2. 1.0 mm ${NAD}^{+}$ $NAD Superscript plus$ and 1.0 mm NADH
3. 10 mm ${NAD}^{+}$ $NAD Superscript plus$ and 1.0 mm NADH
32. Electron Affinity of Compounds List the four compounds or reactions in order of increasing tendency to accept electrons:
1. $α -ketoglutarate + {CO}_{2}$ $alpha hyphen ketoglutarate plus CO Subscript 2 Baseline$ (yielding isocitrate)
2. oxaloacetate
3. $O_{2}$ $upper O Subscript 2$
4. ${NADP}^{+}$ $NADP Superscript plus$
33. Direction of Oxidation-Reduction Reactions Which of the reactions listed would you expect to proceed in the direction shown, under standard conditions, in the presence of the appropriate enzymes?
1. $Malate + {NAD}^{+} \overset{}{\to} oxaloacetate + NADH + H^{+}$ $Malate plus NAD Superscript plus Baseline right-arrow Overscript Endscripts oxaloacetate plus NADH plus upper H Superscript plus$
2. $Acetoacetate + NADH + H^{+} \overset{}{\to} β -hydroxybutyrate + {NAD}^{+}$ $Acetoacetate plus NADH plus upper H Superscript plus Baseline right-arrow Overscript Endscripts beta hyphen hydroxybutyrate plus NAD Superscript plus$
3. $Pyruvate + NADH + H^{+} \overset{}{\to} lactate + {NAD}^{+}$ $Pyruvate plus NADH plus upper H Superscript plus Baseline right-arrow Overscript Endscripts lactate plus NAD Superscript plus$
4. $Pyruvate + β -hydroxybutyrate \overset{}{\to} lactate + acetoacetate$ $Pyruvate plus beta hyphen hydroxybutyrate right-arrow Overscript Endscripts lactate plus acetoacetate$
5. $Malate + pyruvate \overset{}{\to} oxaloacetate + lactate$ $Malate plus pyruvate right-arrow Overscript Endscripts oxaloacetate plus lactate$
6. $Acetaldehyde + succinate \overset{}{\to} ethanol + fumarate$ $Acetaldehyde plus succinate right-arrow Overscript Endscripts ethanol plus fumarate$

Concentration (mm)
ATP	5	3	1	0.2	5
ADP	0.2	2.2	4.2	5.0	25
$P_{i}$ $upper P Subscript i$	10	12.1	14.1	14.9	10

34. Measurement of Intracellular Metabolite Concentrations Measuring the concentrations of metabolic intermediates in a living cell presents great experimental difficulties — usually, a cell must be destroyed before metabolite concentrations can be measured. Yet enzymes catalyze metabolic interconversions very rapidly, so a common problem associated with these types of measurements is that the findings reflect not the physiological concentrations of metabolites but the equilibrium concentrations. To prevent changes in metabolite concentrations during sample preparation, cells were quick-frozen in liquid nitrogen, then extracted under conditions that prevented enzymatic activity.

The table gives the intracellular concentrations of the substrates and products of the phosphofructokinase-1 reaction in isolated rat heart tissue.

Metabolite	Concentration (μM)^a
Fructose 6-phosphate	87.0
Fructose 1,6-bisphosphate	22.0
ATP	11,400
ADP	1,320
Data from J. R. Williamson, J. Biol. Chem. 240:2308, 1965. ^aCalculated as μmol/mL of intracellular water.

Calculate Q, [fructose 1,6-bisphosphate][ADP]/[fructose 6-phosphate][ATP], for the PFK-1 reaction under physiological conditions.
Given a $Δ G^{'} °$ $upper Delta upper G prime degree$ for the PFK-1 reaction of $- 14.2 kJ/mol$ $negative 14.2 kJ slash mol$ , calculate the equilibrium constant for this reaction.
Compare the values of Q and $K_{eq}^{'}$ $upper K prime Subscript eq$ . Is the physiological reaction near or far from equilibrium? Explain. What does this experiment suggest about the role of PFK-1 as a regulatory enzyme?

35. Are All Metabolic Reactions at Equilibrium?
1. Phosphoenolpyruvate (PEP) is one of the two phosphoryl group donors in the synthesis of ATP during glycolysis. In human erythrocytes, the steady-state concentration of ATP is 2.24 mm, that of ADP is 0.25 mm, and that of pyruvate is 0.051 mm. Calculate the concentration of PEP at 25 °C, assuming that the pyruvate kinase reaction (see Fig. 13-13) is at equilibrium in the cell.
2. The physiological concentration of PEP in human erythrocytes is 0.023 mm. Compare this with the value obtained in (a). Explain the significance of this difference.
36. Michaelis Constant $K_{m}$ $upper K Subscript bold m$ Compared with Substrate Concentration Malate synthase in E. coli catalyzes the reaction

$Acetyl-CoA + glyoxylate + H_{2} O \underset{}{\to} malate + CoA-SH + H^{+}$ $Acetyl hyphen CoA plus glyoxylate plus upper H Subscript 2 Baseline upper O right-arrow Underscript Endscripts malate plus CoA hyphen SH plus upper H Superscript plus$

The experimentally measured $K_{m}$ $upper K Subscript m$ for acetyl-CoA is $9 \times 10^{6} M$ $9 times 10 Superscript 6 Baseline upper M$ . In a growing culture of E. coli, the measured concentration of acetyl-CoA is $6 \times 10^{- 4} M$ $6 times 10 Superscript negative 4 Baseline upper M$ . Is malate synthase operating at its $V_{\max}$ $upper V Subscript max$ under these conditions?

DATA ANALYSIS PROBLEM

37. Thermodynamics Can Be Tricky Thermodynamics is a challenging area of study and one with many opportunities for confusion. An interesting example is found in an article by Robinson, Hampson, Munro, and Vaney, published in Science in 1993. Robinson and colleagues studied the movement of small molecules between neighboring cells of the nervous system through cell-to-cell channels (gap junctions). They found that the dyes Lucifer yellow (a small, negatively charged molecule) and biocytin (a small zwitterionic molecule) moved in only one direction between two particular types of glia (nonneuronal cells of the nervous system). Dye injected into astrocytes would rapidly pass into adjacent astrocytes, oligodendrocytes, or Müller cells, but dye injected into oligodendrocytes or Müller cells passed slowly if at all into astrocytes. All of these cell types are connected by gap junctions.

Although it was not a central point of their article, the authors presented a molecular model for how this unidirectional transport might occur, as shown in their Figure 3:

A left-hand circle labeled astrocyte is next to a right-hand circle labeled oligodendrocyte. A structure connecting them is wider on the left, within the astrocyte, and narrower on the right, within the oligodendrocyte. In part a, two spheres are shown in the astrocyte, one of which has an arrow pointing to the right. Three spheres are in the structure connecting them, two of which have arrows pointing to the right. In part b, all of the spheres are in the oligodendrocyte against the right-hand side of the structure connecting the astrocyte to the oligodendrocyte.

The figure legend reads: “Model of the unidirectional diffusion of dye between coupled oligodendrocytes and astrocytes, based on differences in connection pore diameter. Like a fish in a fish trap, dye molecules (black circles) can pass from an astrocyte to an oligodendrocyte (A) but not back in the other direction (B).”

Although this article clearly passed review at a well-respected journal, several letters to the editor (1994) followed, showing that Robinson and coauthors’ model violated the second law of thermodynamics.
1. Explain how the model violates the second law. Hint: Consider what would happen to the entropy of the system if one started with equal concentrations of dye in the astrocyte and oligodendrocyte connected by the “fish trap” type of gap junctions.
2. Explain why this model cannot work for small molecules, although it may allow one to catch fish.
3. Explain why a fish trap does work for fish.
4. Provide two plausible mechanisms for the unidirectional transport of dye molecules between the cells that do not violate the second law of thermodynamics.

References

Letters to the editor. 1994. Science 265:1017–1019.
Robinson, S.R., E.C.G.M. Hampson, M.N. Munro, and D.I. Vaney. 1993. Unidirectional coupling of gap junctions between neuroglia. Science 262:1072–1074.