Biochemistry, The Molecular Basis of Life, 6th Edition
6th Edition
ISBN: 9780190259204
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 1, Problem 12RQ
Summary Introduction
To review:
Definition of the following biological terminologies:
a. Elimination reaction
b. Hydrolysis
c. Addition reaction
d. Dehydration reaction
e. Hydration reaction
Introduction:
There are a number of reactions taking place within a cell. In order to understand the complexity of these reactions, theycan be categorized into different types. All the reactions taking place in the body follow the simple mechanism of organic reactions.
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Chapter 1 Solutions
Biochemistry, The Molecular Basis of Life, 6th Edition
Ch. 1 - Prob. 1RQCh. 1 - Prob. 2RQCh. 1 - Prob. 3RQCh. 1 - Prob. 4RQCh. 1 - Prob. 5RQCh. 1 - Prob. 6RQCh. 1 - Prob. 7RQCh. 1 - Prob. 8RQCh. 1 - Prob. 9RQCh. 1 - Prob. 10RQ
Ch. 1 - Prob. 11RQCh. 1 - Prob. 12RQCh. 1 - Prob. 13RQCh. 1 - Prob. 14RQCh. 1 - Prob. 15RQCh. 1 - Prob. 16RQCh. 1 - Prob. 17RQCh. 1 - Prob. 18RQCh. 1 - Prob. 19RQCh. 1 - Prob. 20RQCh. 1 - Prob. 21RQCh. 1 - Prob. 22RQCh. 1 - Prob. 23RQCh. 1 - Prob. 24RQCh. 1 - Prob. 25RQCh. 1 - Prob. 26RQCh. 1 - Prob. 27RQCh. 1 - Prob. 28RQCh. 1 - Prob. 29RQCh. 1 - Prob. 30RQCh. 1 - Prob. 31RQCh. 1 - Prob. 32RQCh. 1 - Prob. 33RQCh. 1 - Prob. 34RQCh. 1 - Prob. 35RQCh. 1 - Prob. 36RQCh. 1 - Prob. 37RQCh. 1 - Prob. 38RQCh. 1 - Prob. 39RQCh. 1 - Prob. 40RQCh. 1 - Prob. 41RQCh. 1 - Prob. 42RQCh. 1 - Prob. 43RQCh. 1 - Prob. 44RQCh. 1 - Prob. 45RQCh. 1 - Prob. 46RQCh. 1 - Prob. 47RQCh. 1 - Prob. 48FBCh. 1 - Prob. 49FBCh. 1 - Prob. 50FBCh. 1 - Prob. 51FBCh. 1 - Prob. 52FBCh. 1 - Prob. 53FBCh. 1 - Prob. 54FBCh. 1 - Prob. 55FBCh. 1 - Prob. 56FBCh. 1 - Prob. 57FBCh. 1 - Prob. 58SACh. 1 - Prob. 59SACh. 1 - Prob. 60SACh. 1 - Prob. 61SACh. 1 - Prob. 62SACh. 1 - Prob. 63TQCh. 1 - Prob. 64TQCh. 1 - Prob. 65TQCh. 1 - Prob. 66TQCh. 1 - Prob. 67TQCh. 1 - Prob. 68TQCh. 1 - Prob. 69TQCh. 1 - Prob. 70TQCh. 1 - Prob. 71TQCh. 1 - Prob. 72TQCh. 1 - Prob. 73TQ
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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
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- Macmillan 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_forwardCalculate the equilibrium constant for the phosphorylation of glucose to glucose 6-phosphate at 37.0 °C. K'eq = M-' In the rat hepatocyte, the physiological concentrations of glucose and P, are maintained at approximately 4.8 mM. What is the equilibrium concentration of glucose 6-phosphate (G6P) obtained by the direct phosphorylation of glucose by P.? [G6P] = Does this reaction represent a reasonable metabolic step for the catabolism of glucose? Why or why not? Yes, because the value of AG" is positive. No, because the K'eq is too large for the reaction to proceed in the forward direction. Yes, because AG is negative at the calculated value of K'eq No, because [G6P] is likely to be higher than the calculated value. Marrow_forward
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