Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Chapter 26, Problem 43P
Interpretation Introduction
(a)
Interpretation:
The effect of cholesterol feeding on the amount of HMG-CoA reductase mRNA should be determined.
Concept introduction:
Cholesterol is an organic lipid molecule. It is a steroid which is synthesized in all animal cells and is an important component of cell membrane. Cholesterol also functions to serve as a precursor for steroid hormones, vitamin D, and bile acid. In vertebrates, cholesterol is produced by the hepatic cells.
Interpretation Introduction
(b)
Interpretation:
The purpose isolating actin mRNA must be determined.
Concept introduction:
Cholesterol is an organic lipid molecule. It is a steroid which is synthesized in all animal cells and is an important component of cell membrane. Cholesterol also functions to serve as a precursor for steroid hormones, vitamin D, and bile acid. In vertebrates, cholesterol is produced by the hepatic cells.
Interpretation Introduction
(c)
Interpretation:
The effect of cholesterol diet on the HMG CoA reductase protein should be determined.
Concept introduction:
Cholesterol is an organic lipid molecule. It is a steroid which is synthesized in all animal cells and is an important component of cell membrane. Cholesterol also functions to serve as a precursor for steroid hormones, vitamin D, and bile acid. In vertebrates, cholesterol is produced by the hepatic cells.
Interpretation Introduction
(d)
Interpretation:
The reason for surprising results in graph B.
Concept introduction:
Cholesterol is an organic lipid molecule. It is a steroid which is synthesized in all animal cells and is an important component of cell membrane. Cholesterol also functions to serve as a precursor for steroid hormones, vitamin D, and bile acid. In vertebrates, cholesterol is produced by the hepatic cells.
Interpretation Introduction
(e)
Interpretation:
The explanation for the graph B must be determined.
Concept introduction:
Cholesterol is an organic lipid molecule. It is a steroid which is synthesized in all animal cells and is an important component of cell membrane. Cholesterol also functions to serve as a precursor for steroid hormones, vitamin D, and bile acid. In vertebrates, cholesterol is produced by the hepatic cells.
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Fura-2 Fluorescence (Arbitrary Unit)
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[Ca2+] = 0.9nM, 4°C
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draw 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, cysteine
draw in the structure of each amino acid (as L-amino acids) using the Fischer projection style. an example has been included
Chapter 26 Solutions
Biochemistry (Looseleaf)
Ch. 26 - Prob. 1PCh. 26 - Prob. 2PCh. 26 - Prob. 3PCh. 26 - Prob. 4PCh. 26 - Prob. 5PCh. 26 - Prob. 6PCh. 26 - Prob. 7PCh. 26 - Prob. 8PCh. 26 - Prob. 9PCh. 26 - Prob. 10P
Ch. 26 - Prob. 11PCh. 26 - Prob. 12PCh. 26 - Prob. 13PCh. 26 - Prob. 14PCh. 26 - Prob. 15PCh. 26 - Prob. 16PCh. 26 - Prob. 17PCh. 26 - Prob. 18PCh. 26 - Prob. 19PCh. 26 - Prob. 20PCh. 26 - Prob. 21PCh. 26 - Prob. 22PCh. 26 - Prob. 23PCh. 26 - Prob. 24PCh. 26 - Prob. 25PCh. 26 - Prob. 26PCh. 26 - Prob. 27PCh. 26 - Prob. 28PCh. 26 - Prob. 29PCh. 26 - Prob. 30PCh. 26 - Prob. 31PCh. 26 - Prob. 32PCh. 26 - Prob. 33PCh. 26 - Prob. 34PCh. 26 - Prob. 35PCh. 26 - Prob. 36PCh. 26 - Prob. 37PCh. 26 - Prob. 38PCh. 26 - Prob. 39PCh. 26 - Prob. 40PCh. 26 - Prob. 41PCh. 26 - Prob. 42PCh. 26 - Prob. 43PCh. 26 - Prob. 44PCh. 26 - Prob. 45PCh. 26 - Prob. 46PCh. 26 - Prob. 47P
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- 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
- From 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_forwardMacmillan Learning The phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by P, is described by the equation Glucose + P ← glucose 6-phosphate + H₂O AG = 13.8 kJ/mol Coupling ATP hydrolysis to glucose phosphorylation makes thermodynamic sense, but consider how the coupling might take place. Given that coupling requires a common intermediate, one conceivable mechanism is to use ATP hydrolysis to raise the intracellular concentration of Pi. The increase in P; concentration would drive the unfavorable phosphorylation of glucose by Pi- Is increasing the P; concentration a reasonable way to couple ATP hydrolysis and glucose phosphorylation? No. The phosphate salts of divalent cations would be present in excess and precipitate out. Yes. Increasing the concentration of P; would decrease K'eq and shift equilibrium to the right. Yes. The extra ATP hydrolysis would provide enough free energy to drive the…arrow_forwardThe phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by P, is described by the equation Glucose + P → glucose 6-phosphate + H₂O AG' = 13.8 kJ/mol In principle, at least, one way to increase the concentration of glucose 6-phosphate (G6P) is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and Pj. The maximum solubility of glucose is less than 1 M, and the normal physiological concentration of G6P is 250 μM. Assume a fixed concentration of P, at 4.8 mM. The calculated value of K'cq is 4.74 × 10-³ M-¹. Calculate the intracellular concentration of glucose when the equilibrium concentration of glucose 6-phosphate is 250 μM, the normal physiological concentration. [glucose] = 10.99 Correct Answer Would increasing the concentration of glucose be a physiologically reasonable way to increase the concentration of G6P? No. Because the concentration of P,…arrow_forward
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