Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 7, Problem 19P
Interpretation Introduction
Interpretation:
The rate of the cellulose synthase reaction that adds glucose units to the growing cellulose molecules is to be calculated.
Concept introduction:
Carbohydrates are made up of elements, carbon, hydrogen and oxygen. They are one of the essential macronutrient in our diet. It is the major source of energy and is important for balanced diet. The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Polysaccharides are those types of sugars which contain more than ten units of monosaccharides. Cellulose is a polysaccharide that consists of large number of glucose monomers.
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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.
Glycine 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 1.0
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.]
Chapter 7 Solutions
Biochemistry
Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Prob. 10P
Ch. 7 - Prob. 11PCh. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Prob. 14PCh. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Prob. 18PCh. 7 - Prob. 19PCh. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...Ch. 7 - Answers to all problems are at the end of this...
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- 1) 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_forwardOne 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_forward
- The 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_forwardGive 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_forward
- The 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_forwardSuppose you wanted to make a buffer of exactly pH 7.00 using KH2PO4 and Na2HPO4. If the final solution was 0.18 M in KH2PO4, you would need 0.25 M Na2HPO4. Now assume you wish to make a buffer at the same pH, using the same substances, but want the total phosphate molarity ([HPO42−]+[H2PO−4]) to equal 0.20 M. What concentration of the Na2HPO4 would be required?arrow_forward
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