Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 10, Problem 8RQ
Summary Introduction
To review:
Definitions of the following terminologies:
(a) Ischemia
(b) Reperfusion injury
(c) MPTP
(d) Alpha-tocopherol
(e) Glutathione
Introduction:
During oxidative stress, the antioxidant mechanisms are overactive; reactive oxygen species increase, which may lead to the damage of cells. Damage results mainly from enzyme inactivation and the polysaccharide depolymerization as well as due to membrane destruction and DNA (deoxyribonucleic acid) breakage.
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Draw out the following peptide H-R-K-E-D at physiological pH (~7.4). Make sure toreference table 3.1 for pKa values.
The 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+
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 oxidized
Chapter 10 Solutions
Biochemistry: The Molecular Basis of Life
Ch. 10 - Prob. 1QCh. 10 - Prob. 2QCh. 10 - Prob. 3QCh. 10 - Prob. 4QCh. 10 - Prob. 5QCh. 10 - Prob. 6QCh. 10 - Prob. 7QCh. 10 - Prob. 8QCh. 10 - Prob. 1RQCh. 10 - Prob. 2RQ
Ch. 10 - Prob. 3RQCh. 10 - Prob. 4RQCh. 10 - Prob. 5RQCh. 10 - Prob. 6RQCh. 10 - Prob. 7RQCh. 10 - Prob. 8RQCh. 10 - Prob. 9RQCh. 10 - Prob. 10RQCh. 10 - Prob. 11RQCh. 10 - Prob. 12RQCh. 10 - Prob. 13RQCh. 10 - Prob. 14RQCh. 10 - Prob. 15RQCh. 10 - Prob. 16RQCh. 10 - Prob. 17RQCh. 10 - Prob. 18RQCh. 10 - Prob. 19RQCh. 10 - Prob. 20RQCh. 10 - Prob. 21RQCh. 10 - Prob. 22RQCh. 10 - Prob. 23RQCh. 10 - Prob. 24RQCh. 10 - Prob. 25RQCh. 10 - Prob. 26RQCh. 10 - Prob. 27RQCh. 10 - Prob. 28RQCh. 10 - Prob. 29RQCh. 10 - Prob. 30RQCh. 10 - Prob. 31RQCh. 10 - Prob. 32RQCh. 10 - Prob. 33RQCh. 10 - Prob. 34RQCh. 10 - Prob. 35FBCh. 10 - Prob. 36FBCh. 10 - Prob. 37FBCh. 10 - Prob. 38FBCh. 10 - Prob. 39FBCh. 10 - Prob. 40FBCh. 10 - Prob. 41FBCh. 10 - Prob. 42FBCh. 10 - Prob. 43FBCh. 10 - Prob. 44FBCh. 10 - Prob. 45SACh. 10 - Prob. 46SACh. 10 - Prob. 47SACh. 10 - Prob. 48SACh. 10 - Prob. 49SACh. 10 - Prob. 50TQCh. 10 - Prob. 51TQCh. 10 - Prob. 52TQCh. 10 - Prob. 53TQCh. 10 - Prob. 54TQCh. 10 - Prob. 55TQCh. 10 - Prob. 56TQCh. 10 - Prob. 57TQCh. 10 - Prob. 60TQCh. 10 - Prob. 61TQCh. 10 - Prob. 62TQCh. 10 - Prob. 63TQCh. 10 - Prob. 64TQCh. 10 - Prob. 65TQCh. 10 - Prob. 66TQ
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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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