Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 7, Problem 18RQ
Summary Introduction
To review:
The structure of a disaccharide uni, thich contains D-glucose linked to D-galactosaminethrough alpha 1,4in a polysaccharide.
Introduction:
Glucose and galactose are hexose (containing six carbons) monosaccharides and are epimers (difference in the arrangement of groups at fourthcarbon) of each other. They can join with each other through alpha 1,4 linkage to form a disaccharide. Galactosamine is a hexosamine derived from galactose by the replacement of hydrogen with the
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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
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…
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
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,…
Chapter 7 Solutions
Biochemistry: The Molecular Basis of Life
Ch. 7 - Prob. 1QCh. 7 - Prob. 2QCh. 7 - Prob. 3QCh. 7 - Prob. 4QCh. 7 - Prob. 5QCh. 7 - Prob. 6QCh. 7 - Prob. 7QCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQ
Ch. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 16RQCh. 7 - Prob. 17RQCh. 7 - Prob. 18RQCh. 7 - Prob. 19RQCh. 7 - Prob. 20RQCh. 7 - Prob. 21RQCh. 7 - Prob. 22RQCh. 7 - Prob. 23RQCh. 7 - Prob. 24RQCh. 7 - Prob. 25RQCh. 7 - Prob. 26RQCh. 7 - Prob. 27RQCh. 7 - Prob. 28FBCh. 7 - Prob. 29FBCh. 7 - Prob. 30FBCh. 7 - Prob. 31FBCh. 7 - Prob. 32FBCh. 7 - Prob. 33FBCh. 7 - Prob. 34FBCh. 7 - Prob. 35FBCh. 7 - Prob. 36FBCh. 7 - Prob. 37FBCh. 7 - Prob. 38SACh. 7 - Prob. 39SACh. 7 - Prob. 40SACh. 7 - Prob. 41SACh. 7 - Prob. 42SACh. 7 - Prob. 43TQCh. 7 - Prob. 44TQCh. 7 - Prob. 45TQCh. 7 - Prob. 46TQCh. 7 - Prob. 47TQCh. 7 - Prob. 48TQCh. 7 - Prob. 49TQCh. 7 - Prob. 50TQCh. 7 - Prob. 51TQCh. 7 - Prob. 52TQCh. 7 - Prob. 53TQCh. 7 - Prob. 54TQCh. 7 - Prob. 55TQCh. 7 - Prob. 56TQCh. 7 - Prob. 57TQCh. 7 - Prob. 58TQCh. 7 - Prob. 59TQCh. 7 - Prob. 60TQCh. 7 - Prob. 61TQCh. 7 - Prob. 62TQCh. 7 - Prob. 63TQCh. 7 - Prob. 64TQCh. 7 - Prob. 65TQCh. 7 - Prob. 66TQCh. 7 - Prob. 67TQCh. 7 - Prob. 68TQCh. 7 - Prob. 69TQCh. 7 - Prob. 70TQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- Calculate 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_forwardThe 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_forward
- The 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_forwardConsider 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_forward
- The 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_forward2) 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_forward
- The 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_forward1) Consider the reaction: A B + C (a) What is the Keg for this reaction? AG= -8.80 kJ/mol (b) The reverse reaction is initiated by creating a solution containing 20mM B, 1mM A and 150mM C. At the instant these are mixed, what is the free energy change associated with the reaction?arrow_forwardWhat is the chemical importance of the negative charge on the phosphate group? Be asspecific as possible. In what ways might this negative charge have beenthermodynamically useful during the evolution of ATP-binding proteins?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles Of Radiographic Imaging: An Art And A ...Health & NutritionISBN:9781337711067Author:Richard R. Carlton, Arlene M. Adler, Vesna BalacPublisher:Cengage Learning
Anatomy & PhysiologyBiologyISBN:9781938168130Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark WomblePublisher:OpenStax College
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage LearningEssentials of Pharmacology for Health ProfessionsNursingISBN:9781305441620Author:WOODROWPublisher:Cengage
Principles Of Radiographic Imaging: An Art And A ...
Health & Nutrition
ISBN:9781337711067
Author:Richard R. Carlton, Arlene M. Adler, Vesna Balac
Publisher:Cengage Learning
Anatomy & Physiology
Biology
ISBN:9781938168130
Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Publisher:OpenStax College
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
Essentials of Pharmacology for Health Professions
Nursing
ISBN:9781305441620
Author:WOODROW
Publisher:Cengage