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 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, is lower than the concentration of glucose, P, is the limiting reactant. No. The glucose concentration would need to be higher than the maximum solubility of glucose. Yes. [Glucose] is lower than [G6P], so adding glucose would shift the equilibrium to the right. Yes. The glucose concentration would be higher than the concentration of G6P, so the reaction would proceed. Correct Answer M
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 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, is lower than the concentration of glucose, P, is the limiting reactant. No. The glucose concentration would need to be higher than the maximum solubility of glucose. Yes. [Glucose] is lower than [G6P], so adding glucose would shift the equilibrium to the right. Yes. The glucose concentration would be higher than the concentration of G6P, so the reaction would proceed. Correct Answer M
Biochemistry
6th Edition
ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
Publisher:Reginald H. Garrett, Charles M. Grisham
Chapter18: Glycolysis
Section: Chapter Questions
Problem 21P
Related questions
Question
![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
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, is lower than the concentration of glucose, P, is the limiting reactant.
No. The glucose concentration would need to be higher than the maximum solubility of glucose.
Yes. [Glucose] is lower than [G6P], so adding glucose would shift the equilibrium to the right.
Yes. The glucose concentration would be higher than the concentration of G6P, so the reaction would proceed.
Correct Answer
M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39f6380b-55e3-47ff-8091-2f92e2b304db%2F195ede73-e424-400a-b64e-21aef198310d%2F2gskt1d_processed.png&w=3840&q=75)
Transcribed Image Text: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
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, is lower than the concentration of glucose, P, is the limiting reactant.
No. The glucose concentration would need to be higher than the maximum solubility of glucose.
Yes. [Glucose] is lower than [G6P], so adding glucose would shift the equilibrium to the right.
Yes. The glucose concentration would be higher than the concentration of G6P, so the reaction would proceed.
Correct Answer
M
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