Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 15, Problem 52FB
Summary Introduction
To review:
The given blank space in the statement“The bases of purine nucleotides are degraded to ______________.”
Introduction:
Purinebases are present in the nucleotides of RNA (ribonucleic acid) as well as DNA(deoxyribonucleicacid). They include nitrogenous bases adenine (A) and guanine (G). Purinesplay many central roles in the energy
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Fura-2 Fluorescence (Arbitrary Unit)
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[Ca2+]=2970nM, 25°C
[Ca2+] 2970nM, 4°C
[Ca2+]=0.9nM, 25°C
[Ca2+] = 0.9nM, 4°C
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Wavelength (nm)
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<
The figure on the LHS shows the excitation spectra of Fura-2 (Em
= 510 nm) in 2 solutions with two different Ca2+ ion concentration
as indicated. Except for temperature, the setting for excitation &
signal acquisition was identical.<
ப
a) The unit in Y-axis is arbitrary (unspecified). Why? <
<
b) Compare & contrast the excitation wavelength of the Isosbestic
Point of Fura-2 at 25 °C & 4 °C. Give a possible reason for the
discrepancy. <
c) The fluorescence intensity at 25 °C & 4 °C are different. Explain
why with the concept of electronic configuration. <
Chapter 15 Solutions
Biochemistry: The Molecular Basis of Life
Ch. 15 - Prob. 1QCh. 15 - Prob. 2QCh. 15 - Prob. 5QCh. 15 - Prob. 6QCh. 15 - Prob. 7QCh. 15 - Prob. 1RQCh. 15 - Prob. 2RQCh. 15 - Prob. 3RQCh. 15 - Prob. 4RQCh. 15 - Prob. 5RQ
Ch. 15 - Prob. 6RQCh. 15 - Prob. 7RQCh. 15 - Prob. 8RQCh. 15 - Prob. 9RQCh. 15 - Prob. 10RQCh. 15 - Prob. 11RQCh. 15 - Prob. 12RQCh. 15 - Prob. 13RQCh. 15 - Prob. 14RQCh. 15 - Prob. 15RQCh. 15 - Prob. 16RQCh. 15 - Prob. 17RQCh. 15 - Prob. 18RQCh. 15 - Prob. 19RQCh. 15 - Prob. 20RQCh. 15 - Prob. 21RQCh. 15 - Prob. 22RQCh. 15 - Prob. 23RQCh. 15 - Prob. 24RQCh. 15 - Prob. 25RQCh. 15 - Prob. 26RQCh. 15 - Prob. 27RQCh. 15 - Prob. 28RQCh. 15 - Prob. 29RQCh. 15 - Prob. 30RQCh. 15 - Prob. 31RQCh. 15 - Prob. 32RQCh. 15 - Prob. 33RQCh. 15 - Prob. 34RQCh. 15 - Prob. 35RQCh. 15 - Prob. 36RQCh. 15 - Prob. 37RQCh. 15 - Prob. 38RQCh. 15 - Prob. 39RQCh. 15 - Prob. 40RQCh. 15 - Prob. 41RQCh. 15 - Prob. 42RQCh. 15 - Prob. 43RQCh. 15 - Prob. 44RQCh. 15 - Prob. 45FBCh. 15 - Prob. 46FBCh. 15 - Prob. 47FBCh. 15 - Prob. 48FBCh. 15 - Prob. 49FBCh. 15 - Prob. 50FBCh. 15 - Prob. 51FBCh. 15 - Prob. 52FBCh. 15 - Prob. 53FBCh. 15 - Prob. 54FBCh. 15 - Prob. 55SACh. 15 - Prob. 56SACh. 15 - Prob. 57SACh. 15 - Prob. 58SACh. 15 - Prob. 59SACh. 15 - Prob. 60TQCh. 15 - Prob. 61TQCh. 15 - Prob. 62TQCh. 15 - Prob. 63TQCh. 15 - Prob. 64TQCh. 15 - Prob. 65TQCh. 15 - Prob. 66TQCh. 15 - Prob. 67TQCh. 15 - Prob. 68TQCh. 15 - Prob. 69TQCh. 15 - Prob. 71TQCh. 15 - Prob. 72TQCh. 15 - Prob. 73TQCh. 15 - Prob. 74TQ
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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 Answerarrow_forwardIn 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.arrow_forwardFrom 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/molarrow_forward
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