Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
8th Edition
ISBN: 9780134015187
Author: John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
Publisher: PEARSON
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Chapter 15, Problem 15.58CP
Complete the following equations (refer to “Summary of Reactions” in Chapters 13 and 14 if necessary):
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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 Answer
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
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The 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 Answer
Chapter 15 Solutions
Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
Ch. 15.1 - Prob. 15.1PCh. 15.1 - Prob. 15.2PCh. 15.2 - Draw structures corresponding to the following...Ch. 15.2 - Prob. 15.1CIAPCh. 15.2 - Prob. 15.2CIAPCh. 15.2 - Prob. 15.4PCh. 15.2 - Draw the line structures and provide common names...Ch. 15.2 - Prob. 15.6KCPCh. 15.3 - For each compound shown next (ad), indicate...Ch. 15.3 - Prob. 15.8KCP
Ch. 15.4 - Prob. 15.9KCPCh. 15.5 - Indicate whether the following compounds will give...Ch. 15.6 - Draw line structures of the following compounds...Ch. 15.6 - Prob. 15.12PCh. 15.7 - Prob. 15.13PCh. 15.7 - Prob. 15.14PCh. 15.7 - Prob. 15.15PCh. 15.7 - Draw the structure of each acetal or ketal final...Ch. 15.7 - Prob. 15.17PCh. 15.7 - Prob. 15.18PCh. 15.7 - Prob. 15.19PCh. 15.7 - Prob. 15.1MRPCh. 15.7 - Prob. 15.2MRPCh. 15.7 - Prob. 15.3MRPCh. 15.7 - Prob. 15.3CIAPCh. 15.7 - (a) What are mirotubules? (b) Why would drugs that...Ch. 15.7 - Tetrodotoxin, found in the puffer fish, has been...Ch. 15 - Prob. 15.20UKCCh. 15 - Prob. 15.21UKCCh. 15 - Prob. 15.22UKCCh. 15 - Prob. 15.23UKCCh. 15 - Prob. 15.24UKCCh. 15 - Prob. 15.25UKCCh. 15 - ALDEHYDESAND KETONES (SECTIONS 15.1 AND 15.2)...Ch. 15 - Draw a structure for a compound that meets each of...Ch. 15 - Prob. 15.28APCh. 15 - Prob. 15.29APCh. 15 - Draw structures corresponding to the following...Ch. 15 - Draw structures corresponding to the following...Ch. 15 - Prob. 15.32APCh. 15 - Prob. 15.33APCh. 15 - The following names are incorrect. What is wrong...Ch. 15 - Prob. 15.35APCh. 15 - Prob. 15.36APCh. 15 - Prob. 15.37APCh. 15 - Prob. 15.38APCh. 15 - Prob. 15.39APCh. 15 - Prob. 15.40APCh. 15 - Draw the structures of the aldehydes that might be...Ch. 15 - Write the structures of the hemiacetal or...Ch. 15 - Write the structures of the hemiacetal or...Ch. 15 - Prob. 15.44APCh. 15 - Prob. 15.45APCh. 15 - Prob. 15.46APCh. 15 - Prob. 15.47APCh. 15 - Prob. 15.48APCh. 15 - Prob. 15.49APCh. 15 - Prob. 15.50CPCh. 15 - Prob. 15.51CPCh. 15 - Prob. 15.52CPCh. 15 - Prob. 15.53CPCh. 15 - Prob. 15.54CPCh. 15 - Name the following compounds: (a) (b) (c) (be sure...Ch. 15 - Draw the structural formulas of the following...Ch. 15 - Prob. 15.57CPCh. 15 - Complete the following equations (refer to Summary...Ch. 15 - Prob. 15.59CPCh. 15 - How could you differentiate between 3-hexanol and...Ch. 15 - The liquids 1-butanol and butanal have similar...Ch. 15 - Prob. 15.62CPCh. 15 - Prob. 15.63GPCh. 15 - Prob. 15.64GPCh. 15 - Using the ketone structural form of fructose...Ch. 15 - Prob. 15.66GP
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- 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
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