Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Chapter 13, Problem 9P
Interpretation Introduction
Interpretation:
The reason for
Concept introduction:
Diffusion is defined as the movement of particles such as atoms, molecules or ions from a higher concentration region to a lower concentration region. This process is continued till the concentration of the particles is same on both sides of the membranes or boundary.
In the body, the diffusion of ions/molecules/atoms takes place inside or outside of the cell through the cell membrane.
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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)
←
<
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 13 Solutions
Biochemistry (Looseleaf)
Ch. 13 - Prob. 1PCh. 13 - Prob. 2PCh. 13 - Prob. 3PCh. 13 - Prob. 4PCh. 13 - Prob. 5PCh. 13 - Prob. 6PCh. 13 - Prob. 7PCh. 13 - Prob. 8PCh. 13 - Prob. 9PCh. 13 - Prob. 10P
Ch. 13 - Prob. 11PCh. 13 - Prob. 12PCh. 13 - Prob. 13PCh. 13 - Prob. 14PCh. 13 - Prob. 15PCh. 13 - Prob. 16PCh. 13 - Prob. 17PCh. 13 - Prob. 18PCh. 13 - Prob. 19PCh. 13 - Prob. 20PCh. 13 - Prob. 21PCh. 13 - Prob. 22PCh. 13 - Prob. 23PCh. 13 - Prob. 24PCh. 13 - Prob. 25PCh. 13 - Prob. 26PCh. 13 - Prob. 27PCh. 13 - Prob. 28PCh. 13 - Prob. 29PCh. 13 - Prob. 30PCh. 13 - Prob. 31PCh. 13 - Prob. 32P
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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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