The first reaction in Glycolysis is the phosphorylation of Glucose:Pi + Glucose - Glucose 6-Phosphate + waterThis is a thermodynamically unfavorable process with a AG,' = +13.8 kJ/mol. In a liver cell at37°c the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be theequilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state [H2O] =1.0Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.a0.000000120.00000250.000000250.00474e0.0051.00.1h0.25i2.4

Glycolysis is a metabolic process that occurs in the cytoplasm irrespective of the presence or…

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Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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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 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 + 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 conversion of glucose-1-phosphate to glucose-6-phosphate by the enzyme phosphoglucomutase has a △G°' of -7.6 kJ/mol. Calculate the equilibrium constant for this reaction at 298 K and a pH of 7. (R = 8.315 J/K-mol) A. 0.003 B. 0.047 C. 1.00 D. 21
If the hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase has a ΔG′∘ of −11.386 kJ/mol at 25 °C, what percentage of substrate remains once the reaction reaches equilibrium assuming no product was initially present? (Round answer to the nearest whole number)
Assume that you have a solution of 0.1 M glucose 6-phosphate. To this solution you add the enzyme phosphoglucomutase, which catalyzes the reaction: glucose-6-P ⇔ glucose-1-P ∆G°′ = +1.7 kJ/mol (a) Does this reaction proceed at all as written at 25 °C, and if so, what are the final concentrations of glucose 6-P and glucose 1-P? (b) What effect would omitting the enzyme have on the reaction. Be specific. (c) Under what cellular conditions, if any, would this reaction continuously produce glucose 1-P at a high rate?
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…
Within biological systems, there are always reactions that seem to occur when thermodynamically, they should not. An example is in the process of glycolysis (the conversion of glucose to pyruvate) which has ΔG°' = 2183.6 kJ/mol. How is glycolysis possible with such a large, positive ΔG°', when cells are governed by the laws of thermodynamics?
What terms would best describe the above coupled reaction? (If the DGo for ATP hydrolysis into ADP + inorganic phosphate is -7.3 kcal/mole, and the DGo for maltose synthesis from glucose + glucose is +3.7 kcal/mole, calculate the standard free energy change for the combined reaction of ATP + glucose + glucose g ADP + maltose + inorganic phosphate.) it is non-spontaneous and endothermic (because the overall DGo is negative) it is spontaneous and exothermic (because the overall DGo is negative) it is non-spontaneous and endothermic (because the overall DGo is positive) it is spontaneous and exothermic (because the overall DGo is positive) it is non-spontaneous and exothermic (because the overall DGo is negative)
One of the examples that we have used to illustrate the concept of equilibrium is the isomerization of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P), which is the second step in g ycolysis. Draw a graph to show how the reaction Gibbs energy varies with the fraction fof F6P in solution.Label the regions of the graph that correspond to the formation of F6P and G6P being spontaneous, respectively.
The enzyme malate dehydrogenase catalyzes the conversion of malate to oxaloacetate in the final step that completes the Citric Acid Cycle. The ∆G of this reaction is +29.7 kJ/mol. Explain how this thermodynamically unfavorable reaction could possibly be accomplished by the cell even though the reaction is not coupled to a favorable reaction.
Consider the following equilibrium at 25ºC :Glucose-1-Phosphate Glucose-6-PhophateUsing the equilibrium concentrations of [Glucose-1-Phosphate] = 0.35 M and [Glucose-6-Phosphate] = 1.65 M, calculate BOTH K′eqand Gº′ for this reaction. Is this reaction exergonicor endergonic? R = 8.314 J/K·mol
A bacterial enzyme catalyzes the hydrolysis of maltose as shown in the reaction given below: Maltose + H2O -> 2 glucose If the reaction has a Km of 0.135 mM and a V max of 65 umol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM? (Please take note of the units)

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