Biochemistry, The Molecular Basis of Life, 6th Edition
6th Edition
ISBN: 9780190259204
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 9, Problem 12RQ
Summary Introduction
To review:
The balanced reactions of the citric acid cycle.
Introduction:
The citric acid cycle is a biochemical pathway in which chemical energy present in the acetyl-CoA (coenzyme A) is released and hence, energy is produced. This pathway has eight reactions that occur in two phases. The precursor of acetyl CoA is pyruvate, which is a three carbon containing molecule derived from glucose by the process of glycolysis.
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The standard free energy, AGO, of hydrolysis of inorganic polyphosphate, polyP, is about −20 kJ/mol for each P; released. In a
cell, it takes about 50 kJ/mol of energy to synthesize ATP from ADP and Pi.
○
P
O
Inorganic polyphosphate (polyP)
Is it feasible for a cell to use polyP to synthesize ATP from ADP? Why or why not?
No. The reaction is unidirectional and always proceeds in the direction of polyP synthesis from ATP.
Yes. If [ADP] and [polyP] are kept high, and [ATP] is kept low, the actual free-energy change would be negative.
No. The synthesis of ATP from ADP and P; has a large positive G'o compared to polyP hydrolysis.
Yes. The hydrolysis of polyP has a sufficiently negative AG to overcome the positive AGO of ATP synthesis.
Correct Answer
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.
H
H
|
H-C-OH
H-C-OH
C=0
HO-C-H
HO-C-H
phosphohexose
isomerase
H-C-OH
H-C-OH
H-C-OH
H-C-OH
CH₂OPO
CH₂OPO
Glucose 6-phosphate
Fructose 6-phosphate
What does the isomerization step accomplish from a chemical perspective?
Isomerization alters the molecular formula of the compound, allowing for subsequent phosphorylation.
Isomerization moves the carbonyl group, setting up a cleavage between the central carbons.
Isomerization causes the gain of electrons, allowing for the eventual release of NADH.
Isomerization reactions cause the direct production of energy in the form of ATP.
From data in the table, calculate the AG value for the reactions.
Reaction
AG' (kJ/mol)
Phosphocreatine + H₂O →>> creatine + P
-43.0
ADP + Pi → ATP + H₂O
+30.5
Fructose +P → fructose 6-phosphate + H₂O
+15.9
Phosphocreatine + ADP creatine + ATP
AG'O
ATP + fructose → ADP + fructose 6-phosphate
AG'°
kJ/mol
kJ/mol
Chapter 9 Solutions
Biochemistry, The Molecular Basis of Life, 6th Edition
Ch. 9 - Prob. 1QCh. 9 - Prob. 2QCh. 9 - Prob. 3QCh. 9 - Prob. 4QCh. 9 - Prob. 5QCh. 9 - Prob. 6QCh. 9 - Prob. 7QCh. 9 - Prob. 1RQCh. 9 - Prob. 2RQCh. 9 - Prob. 3RQ
Ch. 9 - Prob. 4RQCh. 9 - Prob. 5RQCh. 9 - Prob. 6RQCh. 9 - Prob. 7RQCh. 9 - Prob. 8RQCh. 9 - Prob. 9RQCh. 9 - Prob. 10RQCh. 9 - Prob. 11RQCh. 9 - Prob. 12RQCh. 9 - Prob. 13RQCh. 9 - Prob. 14RQCh. 9 - Prob. 15RQCh. 9 - Prob. 16RQCh. 9 - Prob. 17RQCh. 9 - Prob. 18RQCh. 9 - Prob. 19RQCh. 9 - Prob. 20RQCh. 9 - Prob. 21RQCh. 9 - Prob. 22RQCh. 9 - Prob. 23RQCh. 9 - Prob. 24RQCh. 9 - Prob. 25RQCh. 9 - Prob. 26RQCh. 9 - Prob. 27RQCh. 9 - Prob. 28FBCh. 9 - Prob. 29FBCh. 9 - Prob. 30FBCh. 9 - Prob. 31FBCh. 9 - Prob. 32FBCh. 9 - Prob. 33FBCh. 9 - Prob. 34FBCh. 9 - Prob. 35FBCh. 9 - Prob. 36FBCh. 9 - Prob. 37FBCh. 9 - Prob. 38SACh. 9 - Prob. 39SACh. 9 - Prob. 40SACh. 9 - Prob. 41SACh. 9 - Prob. 42SACh. 9 - Prob. 43TQCh. 9 - Prob. 44TQCh. 9 - Prob. 45TQCh. 9 - Prob. 46TQCh. 9 - Prob. 47TQCh. 9 - Prob. 48TQCh. 9 - Prob. 49TQCh. 9 - Prob. 50TQCh. 9 - Prob. 51TQCh. 9 - Prob. 52TQCh. 9 - Prob. 53TQCh. 9 - Prob. 54TQCh. 9 - Prob. 55TQCh. 9 - Prob. 56TQCh. 9 - Prob. 57TQCh. 9 - Prob. 58TQCh. 9 - Prob. 59TQ
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- Macmillan Learning The phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by P, is described by the equation Glucose + P ← glucose 6-phosphate + H₂O AG = 13.8 kJ/mol Coupling ATP hydrolysis to glucose phosphorylation makes thermodynamic sense, but consider how the coupling might take place. Given that coupling requires a common intermediate, one conceivable mechanism is to use ATP hydrolysis to raise the intracellular concentration of Pi. The increase in P; concentration would drive the unfavorable phosphorylation of glucose by Pi- Is increasing the P; concentration a reasonable way to couple ATP hydrolysis and glucose phosphorylation? No. The phosphate salts of divalent cations would be present in excess and precipitate out. Yes. Increasing the concentration of P; would decrease K'eq and shift equilibrium to the right. Yes. The extra ATP hydrolysis would provide enough free energy to drive the…arrow_forwardThe phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by P, is described by the equation Glucose + P → glucose 6-phosphate + H₂O AG' = 13.8 kJ/mol In principle, at least, one way to increase the concentration of glucose 6-phosphate (G6P) is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and Pj. The maximum solubility of glucose is less than 1 M, and the normal physiological concentration of G6P is 250 μM. Assume a fixed concentration of P, at 4.8 mM. The calculated value of K'cq is 4.74 × 10-³ M-¹. Calculate the intracellular concentration of glucose when the equilibrium concentration of glucose 6-phosphate is 250 μM, the normal physiological concentration. [glucose] = 10.99 Correct Answer Would increasing the concentration of glucose be a physiologically reasonable way to increase the concentration of G6P? No. Because the concentration of P,…arrow_forwardCalculate the equilibrium constant for the phosphorylation of glucose to glucose 6-phosphate at 37.0 °C. K'eq = M-' In the rat hepatocyte, the physiological concentrations of glucose and P, are maintained at approximately 4.8 mM. What is the equilibrium concentration of glucose 6-phosphate (G6P) obtained by the direct phosphorylation of glucose by P.? [G6P] = Does this reaction represent a reasonable metabolic step for the catabolism of glucose? Why or why not? Yes, because the value of AG" is positive. No, because the K'eq is too large for the reaction to proceed in the forward direction. Yes, because AG is negative at the calculated value of K'eq No, because [G6P] is likely to be higher than the calculated value. Marrow_forward
- The pKa values for glutamic acid are 2.19, 9.67, 4.25. Sketch out the titration curve for this amino acid and include all of the pKa values and the pl.arrow_forwardCalculate the isoelectronic point, pl, from the pKa values for histidine, arginine and asparagine.arrow_forwardThe free energy released by the hydrolysis of ATP under standard conditions is -30.5 kJ/mol. If ATP is hydrolyzed under standard conditions except at pH 5.0, is more or less free energy released? Why? More free energy is released because the increased [H+] stabilizes the negative charge on the ADP molecule. Less free energy is released because an acidic environment depletes cellular ATP levels. Less free energy is released because the reaction favors ATP production over hydrolysis due to the higher [H+] in solution. More free energy is released because the total cellular concentrations of ATP, ADP, and P; are greater at the lower pH. Correct Answerarrow_forward
- 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. How does the entropy change in both the system (developing chick) and suroundings (the egg environment) drive the irreversible process of chick development? ☐ The release of glucose from sucrose, which produces energy needed for chick development, decreases entropy in the surroundings. Chick development increases entropy in the system, which causes a concominant decrease in entropy in the surroundings. Carbohydrates, proteins, and lipids within the egg break down into CO2 and H2O, which increases entropy in the surroundings. Chick development decreases entropy in the system, but this is smaller than the concominant increase in entropy in the surroundings.arrow_forwardThe amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in the protonated form -NH or as the free base -NH2, because of the reversible equilibrium R-NH =R-NH₂ + H+ In what pH range can glycine be used as an effective buffer due to its amino group? pH 8.6 to pH 10.6 In a 0.1 M solution of glycine at pH 9.0, what fraction of glycine has its amino group in the -NH form? Correct Answer Correct Answer 45 How much 5 M KOH must be added to 1.0 L of 0.1 M glycine at pH 9.0 to bring its pH to 10.0? 10 mL When 99% of the glycine is in its -NH form, what is the numerical relation between the pH of the solution and the pKa of the amino group? pH = pKa - 2 Correct Answer Correct Answerarrow_forwardThe glycolytic enzyme Phosphofructokinase (PFK) catalyzes the following reaction: Fructose-6-phosphate (F6P) + ATP → Fructose-1,6-bisphosphate (F1,6BP) + ADP AG"=-14.2 kJ/mol This is considered the enzymatic step that commits a sugar substrate to glycolysis. a) Calculate the standard free energy of hydrolysis of fructose-1,6-bisphosphate. b) What is the equilibrium constant for this coupled reaction? c) ATP is a known inhibitor of PFK. If the cellular concentrations of ATP and ADP are 5 mM and 1.0mM respectively, and the concentrations of F6P and F1,6BP are 2mM, what is the free energy change of the system?arrow_forward
- 2) Consider the following reaction: A + 2B 3C + D At equilibrium the concentration of the reactants and products are: [A] = 20.0 mM [C] = 3.0 mM [B] = 4.0 mM [D] = 50.0 mM Calculate (a) the equilibrium constant and (b) AG". Comment on which side of this reaction is more likely to occur.arrow_forwardGlycine is a diprotic acid, which can potentially undergo two dissociation reactions, one for the a-amino group (NH), and the other for the carboxyl (-COOH) group. Therefore, it has two pK₁ values. The carboxyl group has a pK₁ of 2.34 and the α-amino group has a pK2 of 9.60. Glycine can exist in fully deprotonated (NH2-CH2-COO¯), fully protonated (NH3-CH2-COOH), or zwitterionic form (NH3-CH2-COO¯). Match the pH values with the corresponding form of glycine that would be present in the highest concentration in a solution of that pH. fully deprotonated form NH2-CH2-COO- fully protonated form NH–CH,–COOH zwitterionic form NH–CH,−COO Answer Bank pH 7.0 pH 11.9 pH 6.0 pH 8.0 pH 1.0arrow_forwardThe AG of hydrolysis of a sugar phosphate (S-O-P) to the free sugar (S-OH) is -26.6 kJ/mol in a hypothetical cell in which the steady-state concentrations of sugar phosphate, free sugar, and inorganic phosphate are 1.0 mM, 0.20 mM, and 50.0 mM, respectively. S-O-P + H2O S-OH + Pi (a) What is the AG°' for this reaction? (b) In the cell, S-O-P is formed by the transfer of a phosphate group from ATP. What would the AG be for the transfer of the g-phosphate from ATP to this sugar (S-OH)? [AG for ATP hydrolysis is -31 kJ/mol.]arrow_forward
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