EBK BIOCHEMISTRY
6th Edition
ISBN: 8220102959286
Author: GRISHAM
Publisher: CENGAGE L
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Chapter 28, Problem 16P
Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137
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Chapter 28 Solutions
EBK BIOCHEMISTRY
Ch. 28 - Semiconservative or Conservative DNA Replication...Ch. 28 - The Enzymatic Activities of DNA Polymerase I (a)...Ch. 28 - Multiple Replication Forks in E. coli I Assuming...Ch. 28 - Multiple Replication Forks in E. coli II On the...Ch. 28 - Molecules of DNA Polymerase III per Cell vs....Ch. 28 - Number of Okazaki Fragments in E. coli and Human...Ch. 28 - The Roles of Helicases and Gyrases How do DNA...Ch. 28 - Human Genome Replication Rate Assume DNA...Ch. 28 - Heteroduplex DNA Formation in Recombination From...Ch. 28 - Homologous Recombination, Heteroduplex DNA, and...
Ch. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Chemical Mutagenesis of DNA Bases Show the...Ch. 28 - Prob. 14PCh. 28 - Recombination in Immunoglobulin Genes If...Ch. 28 - Helicase Unwinding of the E. coli Chromosome...Ch. 28 - Prob. 17PCh. 28 - Functional Consequences of Y-Family DNA Polymerase...Ch. 28 - Figure 28.11 depicts the eukaryotic cell cycle....Ch. 28 - Figure 28.41 gives some examples of recombination...Ch. 28 - Prob. 21PCh. 28 - Prob. 22P
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- Fura-2 Fluorescence (Arbitrary Unit) 4500 4000 3500 3000 2500 2000 1500 1000 500 [Ca2+]=2970nM, 25°C [Ca2+] 2970nM, 4°C [Ca2+]=0.9nM, 25°C [Ca2+] = 0.9nM, 4°C 0 260 280 300 340 360 380 400 420 440 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. <arrow_forwarddraw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been included. Draw the structure for glycine, alanine, valine, isoleucine, methionine, proline, phenylalanine, tryptophan, serine, threonine, asparagine, glutamine, lysine, arginine, aspartic acid, glutamic acid, histidine, tyrosine, cysteinearrow_forwarddraw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been includedarrow_forward
- draw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been includedarrow_forwardDraw out the following peptide H-R-K-E-D at physiological pH (~7.4). Make sure toreference table 3.1 for pKa values.arrow_forwardThe table provides the standard reduction potential, E', for relevant half-cell reactions. Half-reaction E'° (V) Oxaloacetate² + 2H+ + 2e malate²- -0.166 Pyruvate + 2H+ + 2e → lactate -0.185 Acetaldehyde + 2H+ + 2e¯ →→→ ethanol -0.197 NAD+ + H+ + 2e--> NADH -0.320 NADP+ + H+ + 2e →→ NADPH Acetoacetate + 2H+ + 2e¯ - -0.324 B-hydroxybutyrate -0.346 Which of the reactions listed would proceed in the direction shown, under standard conditions, in the presence of the appropriate enzymes? Malate + NAD+ oxaloacetate + NADH + H+ Malate + pyruvate oxaloacetate + lactate Pyruvate + NADH + H+ lactate + NAD+ Pyruvate + p-hydroxybutyrate lactate + acetoacetate Acetaldehyde + succinate ethanol + fumerate Acetoacetate + NADH + H+ → B-hydroxybutyrate + NAD+arrow_forward
- Arrange the four structures in order from most reduced to most oxidized. Most reduced R-CH2-CH3 R-CH2-CH₂-OH R-CH,-CHO R-CH₂-COO Most oxidizedarrow_forwardfor 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_forward
- Compare 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_forwardThe 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_forward
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