Biochemistry
8th Edition
ISBN: 9781464126109
Author: Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr., Lubert Stryer
Publisher: W. H. Freeman
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Chapter 11, Problem 34P
Interpretation Introduction
Interpretation:
The specificity of sucrose is to be explained on the basis of potential substrates.
Concept introduction:
Sucrose is a disaccharide that is formed by the combination of the substrates glucose and fructose. These substrates combine with each other in the presence of certain enzymes.
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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
the surroundings.
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 11 Solutions
Biochemistry
Ch. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - Prob. 3PCh. 11 - Prob. 4PCh. 11 - Prob. 5PCh. 11 - Prob. 6PCh. 11 - Prob. 7PCh. 11 - Prob. 8PCh. 11 - Prob. 9PCh. 11 - Prob. 10P
Ch. 11 - Prob. 11PCh. 11 - Prob. 12PCh. 11 - Prob. 13PCh. 11 - Prob. 14PCh. 11 - Prob. 15PCh. 11 - Prob. 16PCh. 11 - Prob. 17PCh. 11 - Prob. 18PCh. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - Prob. 21PCh. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - Prob. 26PCh. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Prob. 30PCh. 11 - Prob. 31PCh. 11 - Prob. 32PCh. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36P
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Similar questions
- 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
- One prominent theory on life origins was that RNA enzymes came into existence early inthe prebiotic history of Earth and were able to do basic chemical catalyses. Eventually,this “RNA-world” was overtaken by the stability of DNA as an information carrier and thediversity of catalytic functions capable of being performed by polypeptides. Is the RNA world hypothesis is a well-founded model?arrow_forwardThe AG" of hydrolysis (ATP + H2O --> ADP + Pi) is -31.0 kJ/mol. Answer the following questions assuming that the steady-state concentrations in the cell are as indicated below. (Note: Steady-state refers to a non-equilibrium situation that exists due to a balance between reactions that supply and remove these substances.) [ADP] = 0.40 mM, [ATP] = 4.0 mM, and [Pi] = 40.0 mM a) Calculate the equilibrium constant for this reaction. b) What would the AG' for ATP hydrolysis be in the cell? c) Is this reaction at equilibrium in the cell? Briefly explain your answer.arrow_forward5) Theoretically, ATP did not have to become our bodies' main energy currency. Two alternative carriers, acetyl phosphate and S-adenosylmethionine could have been utilized, rather than ATP. AG" for acetyl phosphate hydrolysis is -43.3 kJ/mol and AG" for S- adenosylmethionine hydrolysis is -25.6 kJ/mol. (a) Calculate the weight of each alternative energy carrier that would need to be consumed by humans on a 2000 calorie per day diet if our bodies could not recycle it. Assume a 50% absorption of energy from our diet. (b) If our bodies contain 25g of each alternative energy carrier and they CAN be recycled, how many times would each molecule of each energy carrier need to be recycled? (c) Comment on the special properties of ATP and why it is unlikely that these alternative carriers would be utilized biologically.arrow_forward
- Give three reasons why evolution may have selected for phosphates compared to othersimilar leaving groups such as conjugated carboxylic acids or amides. Explain whatbenefit each of your reasons has granted to the living organism.arrow_forwardThe preferred substrate is T because the enzyme half-saturates at 7.00 mM for T, but requires 28.0 mM for U, and 112 mM for S. b Question Content Area The rate constant k 2 with substrate S is 9.60×107 sec-1, with substrate T, k 2 = 6.00×104 sec-1, and with substrate U, k 2 = 2.40×106 sec-1. Calculate the catalytic efficiency with S, T, and U. Catalytic efficiency with S = Catalytic efficiency with T = Catalytic efficiency with U = Does enzyme A use substrate S, substrate T, or substrate U with greater catalytic efficiency?arrow_forwardFumerase catalyzes the conversion of fumerate to malate. fumerate + H2O ⇋ malate The turnover number, kcat, for fumerase is 8.00×102 sec-1. The Km of this enzyme for fumerate is 5.00×10-3 μmol mL-1. a In an experiment using 2.00×10-3 μmol·mL-1, what is Vmax?arrow_forward
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