Fundamentals of Biochemistry: Life at the Molecular Level
5th Edition
ISBN: 9781118918401
Author: Donald Voet, Judith G. Voet, Charlotte W. Pratt
Publisher: WILEY
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Question
Chapter 9, Problem 14E
Summary Introduction
To explain: The reason why a bilayer containing only gangliosides would be unstable.
Concept introduction: Gangliosides are molecules that consist of a glycosphingolipid with one or more sialic acids that linked on the sugar chain. These molecules were first isolated from the brain tissue by Ernst Klenk. Excluding some echinoderms, they are characteristic of vertebrate nervous tissues.
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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
0
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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 9 Solutions
Fundamentals of Biochemistry: Life at the Molecular Level
Ch. 9 - Prob. 1ECh. 9 - 2. Which triacylglycerol yields more energy on...Ch. 9 - 3. Draw the structure of a glycerophospholipid...Ch. 9 - 4. Name this lipid.
Ch. 9 - Prob. 5ECh. 9 - 6. What products are obtained when...Ch. 9 - 7. Which of the glycerophospholipid head groups...Ch. 9 - Prob. 8ECh. 9 - 9. Identify the hormone molecule shown here. What...Ch. 9 - 10. Draw the structure of the fatty acid amide...
Ch. 9 - Prob. 11ECh. 9 - Prob. 12ECh. 9 - Prob. 13ECh. 9 - Prob. 14ECh. 9 - Prob. 15ECh. 9 - Prob. 16ECh. 9 - 17. (a) How many turns of an ? helix are required...Ch. 9 - 18. The distance between successive C? atoms in a...Ch. 9 - 19. Are the following lipid samples likely to...Ch. 9 - 20. In order to consume and dispose of dying...Ch. 9 - 21. Sphingosine-1-phosphate can serve as a cell...Ch. 9 - 22. Lipids known as sulfatides occur in cells of...Ch. 9 - 23. In some autoimmune diseases, an individual...Ch. 9 - 24. Most hormones, such as peptide hormones, exert...Ch. 9 - 25. E. coli outer membranes include a component...Ch. 9 - 26. Archaebacteria and Eubacteria produce...Ch. 9 - 27. Shigella bacteria cause severe diarrhea by...Ch. 9 - 28. The inner membranes of mitochondria are rich...Ch. 9 - 29. Describe the labeling pattern of glycophorin....Ch. 9 - 30. Identify a 20-residue segment that could form...Ch. 9 - 31. Macropinocytosis is a variation of endocytosis...Ch. 9 - 32. Explain why the action of phospholipase A2 on...Ch. 9 - 33. Predict the effect of a mutation in signal...Ch. 9 - 34. Explain why a drug that interferes with the...Ch. 9 - 35. The symmetries of oligomeric integral proteins...Ch. 9 - Prob. 36CQCh. 9 - MORE TO EXPLORE Some of the more unusual membrane...
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- for each pair of biomolecules, identify the type of reaction (oxidation-reduction, hydrolysis, isomerization, group transfer, or nternal rearrangement) required to convert the first molecule to the second. In each case, indicate the general type of enzyme and cofactor(s) c reactants required, and any other products that would result. R-CH-CH-CH-C-S-COA A(n) A(n) A(n) A(n) Palmitoyl-CoA R-CH-CH=CH-C-S-CoA ° trans-A-Enoyl-CoA reaction converts palmitoyl-CoA to trans-A2-enoyl-CoA. This reaction requires and also produces Coo HN-C-H CH₂ CH₂ CH CH CH, CH, L-Leucine CH, CH, D-Leucine 8/6881 COO HÌNH: reaction converts L-leucine to D-leucine. This reaction is catalyzed by a(n) H-C-OH H-C-OH C=0 HO-C-H HO-C-H H-C-OH H-C-OH H-C-OH CH,OH Glucose H-C-OH CH,OH Fructose OH OH OH CH-C-CH₂ reaction converts glucose to fructose. This reaction is catalyzed by a(n) OH OH OPO I CH-C-CH H Glycerol Glycerol 3-phosphate H reaction converts glycerol to glycerol 3-phosphate. This reaction requires H,N- H,N H…arrow_forwardAfter adding a small amount of ATP labeled with radioactive phosphorus in the terminal position, [7-32P]ATP, to a yeast extract, a researcher finds about half of the 32P activity in P; within a few minutes, but the concentration of ATP remains unchanged. She then carries out the same experiment using ATP labeled with 32P in the central position, [ẞ-³2P]ATP, but the 32P does not appear in P; within such a short time. Which statements explain these results? Yeast cells reincorporate P; released from [ß-³2P]ATP into ATP more quickly than P¡ released from [y-³2P]ATP. Only the terminal (y) phosphorous atom acts as an electrophilic target for nucleophilic attack. The terminal (y) phosphoryl group undergoes a more rapid turnover than the central (B) phosphate group. Yeast cells maintain ATP levels by regulating the synthesis and breakdown of ATP. Correct Answerarrow_forwardCompare the structure of the nucleoside triphosphate CTP with the structure of ATP. NH₂ 0- 0- 0- ·P—O—P—O—P—O—CH₂ H H H H OH OH Cytidine triphosphate (CTP) Consider the reaction: ATP + CDP ADP + CTP NH 0- 0- 0- ¯0— P—O— P—O—P-O-CH₂ H Η о H H OH OH Adenosine triphosphate (ATP) NH₂ Now predict the approximate K'eq for this reaction. Now predict the approximate AG for this reaction. Narrow_forward
- 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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