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Metabolism basic concepts qa

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1. Energyflow. What is the direction of each of the following reactions when the
reactants are initially present in equimolar amounts? Use the data given in the
Table below.
2. A proper inference. What information do the ∆G° data given in Table 13.6
provide about the relative rates of hydrolysis of pyrophosphateand acetyl
phosphate?
See answer
3. A potent donor. Consider the following reaction:
(a) Calculate ∆G° and K eq at 25°C for this reaction, by using the data given in
Table 13.6.
(b) What is the equilibrium ratio of pyruvate to phosphoenolpyruvate if the ratio of
ATP to ADP is 10?
See answer
4. Isomeric equilibrium. Calculate ∆G° for the isomerization of glucose 6-
phosphateto glucose 1-phosphate. What is the equilibrium ratio of glucose 6-
phosphateto glucose 1-phosphate at 25°C?
5. Activated acetate. The formation of acetyl CoA from acetate is an ATP-driven
reaction:
(a) Calculate ∆G° for this reaction by using data given in this chapter.
(b) The PPi formed in the preceding reaction is rapidly hydrolyzed in vivo because
of the ubiquity of inorganic pyrophosphatase. The ∆G° for the hydrolysis of PPi is
-4.6 kcal mol-1. Calculate the ∆G° for the overall reaction.
What effect does the hydrolysis of PPi have on the formation of acetyl CoA?
6. Acid strength. The pK of an acid is a measure of its proton-group-transfer
potential.
(a) Derive a relation between ∆G° and pK.
(b) What is the ∆G° for the ionization of acetic acid, which has a pK of 4.8?
See answer
7. Raison d'être. The muscles of some invertebrates are rich in argininephosphate
(phosphoarginine). Proposea function for this amino acid derivative.
8. Recurring motif. What is the structural feature common to ATP, FAD, NAD+,
and CoA?
See answer
9. Ergogenichelp or hindrance? Creatine is a popular, but untested, dietary
supplement.
(a) What is the biochemical rationale for the use of creatine?
(b) What type of exercise would most benefit from creatine supplementation?
10. Standardconditions versus real life. The enzyme aldolase catalyzes the
following reaction in the glycolytic pathway:
G° for the reaction is +5.7 kcal mol- G in the cell is -0.3
kcal mol-1. Calculate the ratio of reactants to products under equilibrium and
intracellular conditions. Using your results, explain how the reaction can be
endergonic under standard conditions and exergonic under intracellular conditions.
See answer
11. Not all alike. The concentrations of ATP, ADP, and Pi differ with cell type.
Consequently, the release of free energy with the hydrolysis of ATP will vary with
G for the hydrolysis of ATP
in muscle, liver, and brain cells. In which cell type is the free energy of ATP
hydrolysis greatest?
12. Runningdownhill. Glycolysis is a series of 10 linked reactions that convert one
molecule of glucose into two molecules of pyruvate with the concomitant synthesis
of two mole G° for this set of reactions is -8.5
kcal mol-1 (-35.6 kJ mol- G is -18.3 kcal mol-1 (-76.6 kJ mol-
1). Explain why the free-energy release is so much greater under intracellular
conditions than under standard conditions.
See answer
Chapter Integration Problem
13. Activated sulfate. Fibrinogen contains tyrosine-O-sulfate. Proposean activated
form of sulfate that could react in vivo with the aromatic hydroxyl group of a
tyrosine residue in a protein to form tyrosine-O-sulfate.
See answer
Data Interpretation
14. Opposites attract G for the hydrolysis
of ATP varies as a function of the Mg2+ concentration (pMg = log 1/[Mg2+]).
G of hydrolysis for ATP?
(b) How can you explain this effect?
Answers
1. Reactions in parts a and c, to the left; reactions in parts b and d, to the right.
See question
2. None whatsoever.
See question
3. G° = +7.5 kcal mol-1 (+31.4 kJ mol-1) and K eq = 3.2 × 10-6. (b) 3.28 ×
104.
See question
4. G° = -1.7 kcal mol-1 (-7.1 kJ mol-1). The equilibrium ratio is 17.8.
5. G° = +7.5 kcal mol-1
(+31.4 kJ mol- G° = -10.9 kcal mol-1 (-45.6 kJ mol-1).
G° = -3.4 kcal mol-1 (-14.2 kJ mol-1).
G° = -8.0 kcal mol-1 (-33.4 kJ mol-1).
See question
6. (a) The pK is defined as pK = -log10 K G° is the standard free energy
change at pH 7. G° = -RT ln K = -2.303 log10 K = -2.303 (pK - 7) kcal
mol-1 since [H+] = 10-7 M.
G° = -2.303 (4.8 - 7) = -5.1 kcal mol-1 (-21.3 kJ mol-1).
See question
7. Arginine phosphatein invertebrate muscle, like creatine phosphatein vertebrate
muscle, serves as a reservoir of highpotential phosphorylgroups. Arginine
phosphatemaintains a high level of ATP in muscular exertion.
See question
8. An ADP unit.
9. (a) The rationale behind creatine supplementation is that it would be converted
into creatine phosphate and thus serves as a rapid means of replenishing ATP after
muscle contraction.
(b) If it is beneficial, it would affect activities that depend on short bursts of
activity; any sustained activity would require ATP generation by fuel metabolism,
which, as Figure 14.7 shows, requires more time.
See question
10. G° = -RT ln [product]/[reactants]. Substituting
+5.7 kcal mol- G° and solving for [products]/[reactants] yields 7 × 10-5.
In other words, the forward reaction does not take place to a significant extent.
G is 0.3 kcal mol-1. If one uses the equation
G G° + RT ln [product]/[reactants] and solves for [products]/[reactants],
the ratio is 3.7 × 10-5. Thus, a reaction that is endergonic under standard
conditions can be converted into an exergonic reaction by maintaining the
[products]/[reactants]ratio below the equilibrium value. This conversion is usually
attained by using the products in another coupled reaction as soonas they are
formed.
11. Liver: -10.8 kcal mol-1 (-45.2 kJ mol-1); muscle: -11.5 kcal mol-1 (-47.8 kJ
mol-1); brain: -11.6 kcal mol-1(-48.4 kJ mol-1).
See question
12. G G° + RT ln [products/reactants]. Altering the ratio of
G to vary. In glycolysis, the concentrations of
G greater than that of
G° .
See question
13. The activated form of sulfate in most organisms is 3 -phosphoadenosine5 -
phosphosulfate.
See question
14. G of hydrolysis rises. Note that
pMg is a logarithmic plot, and so each number on the x-axis represents a 10-fold
change in [Mg2+].
(b) Mg2+ would bind to the phosphates of ATP and help to mitigate charge
repulsion. As the [Mg2+] falls, charge stabilization of ATP would be less, leading
G on hydrolysis.

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Metabolism basic concepts qa

  • 1. 1. Energyflow. What is the direction of each of the following reactions when the reactants are initially present in equimolar amounts? Use the data given in the Table below. 2. A proper inference. What information do the ∆G° data given in Table 13.6 provide about the relative rates of hydrolysis of pyrophosphateand acetyl phosphate? See answer
  • 2. 3. A potent donor. Consider the following reaction: (a) Calculate ∆G° and K eq at 25°C for this reaction, by using the data given in Table 13.6. (b) What is the equilibrium ratio of pyruvate to phosphoenolpyruvate if the ratio of ATP to ADP is 10? See answer 4. Isomeric equilibrium. Calculate ∆G° for the isomerization of glucose 6- phosphateto glucose 1-phosphate. What is the equilibrium ratio of glucose 6- phosphateto glucose 1-phosphate at 25°C? 5. Activated acetate. The formation of acetyl CoA from acetate is an ATP-driven reaction: (a) Calculate ∆G° for this reaction by using data given in this chapter. (b) The PPi formed in the preceding reaction is rapidly hydrolyzed in vivo because of the ubiquity of inorganic pyrophosphatase. The ∆G° for the hydrolysis of PPi is -4.6 kcal mol-1. Calculate the ∆G° for the overall reaction. What effect does the hydrolysis of PPi have on the formation of acetyl CoA? 6. Acid strength. The pK of an acid is a measure of its proton-group-transfer potential. (a) Derive a relation between ∆G° and pK. (b) What is the ∆G° for the ionization of acetic acid, which has a pK of 4.8? See answer 7. Raison d'être. The muscles of some invertebrates are rich in argininephosphate (phosphoarginine). Proposea function for this amino acid derivative.
  • 3. 8. Recurring motif. What is the structural feature common to ATP, FAD, NAD+, and CoA? See answer 9. Ergogenichelp or hindrance? Creatine is a popular, but untested, dietary supplement. (a) What is the biochemical rationale for the use of creatine? (b) What type of exercise would most benefit from creatine supplementation? 10. Standardconditions versus real life. The enzyme aldolase catalyzes the following reaction in the glycolytic pathway: G° for the reaction is +5.7 kcal mol- G in the cell is -0.3 kcal mol-1. Calculate the ratio of reactants to products under equilibrium and intracellular conditions. Using your results, explain how the reaction can be endergonic under standard conditions and exergonic under intracellular conditions. See answer 11. Not all alike. The concentrations of ATP, ADP, and Pi differ with cell type. Consequently, the release of free energy with the hydrolysis of ATP will vary with G for the hydrolysis of ATP in muscle, liver, and brain cells. In which cell type is the free energy of ATP hydrolysis greatest?
  • 4. 12. Runningdownhill. Glycolysis is a series of 10 linked reactions that convert one molecule of glucose into two molecules of pyruvate with the concomitant synthesis of two mole G° for this set of reactions is -8.5 kcal mol-1 (-35.6 kJ mol- G is -18.3 kcal mol-1 (-76.6 kJ mol- 1). Explain why the free-energy release is so much greater under intracellular conditions than under standard conditions. See answer Chapter Integration Problem 13. Activated sulfate. Fibrinogen contains tyrosine-O-sulfate. Proposean activated form of sulfate that could react in vivo with the aromatic hydroxyl group of a tyrosine residue in a protein to form tyrosine-O-sulfate. See answer Data Interpretation 14. Opposites attract G for the hydrolysis of ATP varies as a function of the Mg2+ concentration (pMg = log 1/[Mg2+]). G of hydrolysis for ATP? (b) How can you explain this effect? Answers 1. Reactions in parts a and c, to the left; reactions in parts b and d, to the right. See question 2. None whatsoever. See question
  • 5. 3. G° = +7.5 kcal mol-1 (+31.4 kJ mol-1) and K eq = 3.2 × 10-6. (b) 3.28 × 104. See question 4. G° = -1.7 kcal mol-1 (-7.1 kJ mol-1). The equilibrium ratio is 17.8. 5. G° = +7.5 kcal mol-1 (+31.4 kJ mol- G° = -10.9 kcal mol-1 (-45.6 kJ mol-1). G° = -3.4 kcal mol-1 (-14.2 kJ mol-1). G° = -8.0 kcal mol-1 (-33.4 kJ mol-1). See question 6. (a) The pK is defined as pK = -log10 K G° is the standard free energy change at pH 7. G° = -RT ln K = -2.303 log10 K = -2.303 (pK - 7) kcal mol-1 since [H+] = 10-7 M. G° = -2.303 (4.8 - 7) = -5.1 kcal mol-1 (-21.3 kJ mol-1). See question 7. Arginine phosphatein invertebrate muscle, like creatine phosphatein vertebrate muscle, serves as a reservoir of highpotential phosphorylgroups. Arginine phosphatemaintains a high level of ATP in muscular exertion. See question 8. An ADP unit. 9. (a) The rationale behind creatine supplementation is that it would be converted into creatine phosphate and thus serves as a rapid means of replenishing ATP after muscle contraction. (b) If it is beneficial, it would affect activities that depend on short bursts of activity; any sustained activity would require ATP generation by fuel metabolism, which, as Figure 14.7 shows, requires more time. See question 10. G° = -RT ln [product]/[reactants]. Substituting +5.7 kcal mol- G° and solving for [products]/[reactants] yields 7 × 10-5. In other words, the forward reaction does not take place to a significant extent. G is 0.3 kcal mol-1. If one uses the equation G G° + RT ln [product]/[reactants] and solves for [products]/[reactants], the ratio is 3.7 × 10-5. Thus, a reaction that is endergonic under standard
  • 6. conditions can be converted into an exergonic reaction by maintaining the [products]/[reactants]ratio below the equilibrium value. This conversion is usually attained by using the products in another coupled reaction as soonas they are formed. 11. Liver: -10.8 kcal mol-1 (-45.2 kJ mol-1); muscle: -11.5 kcal mol-1 (-47.8 kJ mol-1); brain: -11.6 kcal mol-1(-48.4 kJ mol-1). See question 12. G G° + RT ln [products/reactants]. Altering the ratio of G to vary. In glycolysis, the concentrations of G greater than that of G° . See question 13. The activated form of sulfate in most organisms is 3 -phosphoadenosine5 - phosphosulfate. See question 14. G of hydrolysis rises. Note that pMg is a logarithmic plot, and so each number on the x-axis represents a 10-fold change in [Mg2+]. (b) Mg2+ would bind to the phosphates of ATP and help to mitigate charge repulsion. As the [Mg2+] falls, charge stabilization of ATP would be less, leading G on hydrolysis.