The first reaction in Glycolysis is the phosphorylation of Glucose: Pi + Glucose → Glucose 6-Phosphate + water This is a thermodynamically unfavorable process with a ∆G0’ = +13.8 kJ/mol. In a liver cell at 370C the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H2O] = 1.0 a. 0.00000012 b. 0.0000025 c. 0.00000025 d. 0.00474 e. 0.005 f. 1.0 g. 0.1 h. 0.25 i. 2.4

The first reaction in Glycolysis is the phosphorylation of Glucose: Pi + Glucose → Glucose 6-Phosphate + water This is a thermodynamically unfavorable process with a ∆G0’ = +13.8 kJ/mol. In a liver cell at 370C the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H2O] = 1.0 a. 0.00000012 b. 0.0000025 c. 0.00000025 d. 0.00474 e. 0.005 f. 1.0 g. 0.1 h. 0.25 i. 2.4

Basic Clinical Laboratory Techniques 6E

6th Edition

ISBN:9781133893943

Author:ESTRIDGE

Publisher:ESTRIDGE

Chapter6: Basic Clinical Chemistry

Section6.5: Blood Glucose And Hemoglobin A1c

Problem 8RQ

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Question

The first reaction in Glycolysis is the phosphorylation of Glucose:
Pi + Glucose → Glucose 6-Phosphate + water
This is a thermodynamically unfavorable process with a ∆G₀’ = +13.8 kJ/mol. In a liver cell at
37⁰C the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H₂O] = 1.0

a. 0.00000012

b. 0.0000025

c. 0.00000025

d. 0.00474

e. 0.005

f. 1.0

g. 0.1

h. 0.25

i. 2.4

Science that deals with the amount of energy transferred from one equilibrium state to another equilibrium state.

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The first reaction in Glycolysis is the phosphorylation of Glucose:Pi + Glucose → Glucose 6-Phosphate + waterThis is a thermodynamically unfavorable process with a ∆G0’ = +13.8 kJ/mol. In a liver cell at370C the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H2O] = 1.0 (this will help with the question below) When the reaction described in question number 3 is coupled to the hydrolysis of ATP the equilibrium concentration of Glucose 6-Phosphate goes to: a. 10.5 b. 350 c. 3000 d. 77 e. 4.5 f. 55 g. 550 h. 652 i. 1120
The first reaction in glycolysis is the phosphorylation of glucose: Pi + glucose → glucose-6-phosphate + H2O This is a thermodynamically unfavorable process, with ∆G°′ = +13.8 kJ/mol. (a) In a liver cell at 37 °C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What would be the equilibrium concentration of glucose-6-phosphate, according to the above data? (b) This very low concentration of the desired product would be unfavorable for glycolysis. In fact, the reaction is coupled to ATP hydrolysis to give the overall reaction ATP + glucose → glucose-6-phosphate + ADP + H+ What is ∆G°′ for the coupled reaction? (c) If, in addition to the constraints on glucose concentration listed previously, we have in the liver cell ATP concentration = 3 mM and ADP concentration = 1 mM, what is the theoretical concentration of glucose6-phosphate at equilibrium at pH = 7.4 and 37 °C? The answer…
The first reaction in Glycolysis is the phosphorylation of Glucose: Pi + Glucose - Glucose 6-Phosphate + water This is a thermodynamically unfavorable process with a AG,' = +13.8 kJ/mol. In a liver cell at 37°c the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H2O] = 1.0 Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a 0.00000012 0.0000025 0.00000025 0.00474 e 0.005 1.0 0.1 h 0.25 i 2.4
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8
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