Calculate 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. M

Biochemistry
6th Edition
ISBN:9781305577206
Author:Reginald H. Garrett, Charles M. Grisham
Publisher:Reginald H. Garrett, Charles M. Grisham
Chapter26: Synthesis And Degradation Of Nucleotides
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Calculate 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.
M
Transcribed Image Text:Calculate 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. M
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