In glycolysis, fructose 1,6-bisphosphate is converted to two products with a standard free-energy change (AG) of 23.8 kJ/mol. Under what cellular conditions will the free-energychange be negative, enabling the reaction to proceed spontaneously to form products?O In a cell where the concentrations of products and reactants yield a low value of the mass-action ratioQ○ The reaction will not go to the right spontaneously under any conditions because the AG is positive○ The reaction will proceed spontaneously to the right if there is a high concentration of productsrelative to the concentration of fructose 1,6-bisphosphate○ The reaction will proceed spontaneously to the right if there is a high concentration of enzyme tocatalyze the reactionO None of these conditions is sufficient

under what cellular conditions the free-energy change (ΔG) will be negative, we need to consider the…

Answered: In glycolysis, fructose…

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 standard free energy change for the reaction catalyzed by phosphoglucomutase is -7.1 kJ/mol. Calculate the equilibrium constant for the reaction. Calculate AG at 37°C when the concentration of glucose-1-phosphate is 1-mM and the concentration of glucose-6-phosphate is 25-mM. Is the reaction spontaneous under these conditions? HOCH, -20;POCH, H H H H H H ОН Н НО ОН Н H НО ОН H ОН H. ОН Glucose-l-phosphate Glucose-6-phosphate
The formation of glutamine from glutamate and ammonium ions requires 14.2 kJ mol-1 of energy input. It is driven by the hydrolysis of ATP to ADP mediated by the enzyme glutamine synthetase. {a) Given that the change in Gibbs energy for the hydrolysis of ATP corresponds to ΔG = -31 kJ mol-1, under the conditions prevailing in a typical cell , can the hydrolysis drive the formation of glutamine? (b) What amount {in moles) of ATP must be hydrolysed to form 1 mol glutamine? (c) Suppose that the radius of a typical cell is 10 μm and that inside it 106 ATP molecules are hydrolysed each second. What is the power density of the cell in watts per cubic metre (1W = 1 Js- 1)? (d) A computer battery delivers about 15 Wand has a volume of 100 cm3 Which has the greater power density, the biological cell or the battery?
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.
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)
Given that the AG" values for the hydrolysis of glucose 1-phosphate and glucose 6-phosphate are approximately -21 kJ/mol and –14 kJ/mol, respectively, which statement is true regarding the isomerization shown? glucose 1-phosphate → glucose 6-phosphate O The reaction can likely proceed in either direction and will therefore depend on the concentrations of glucose 1-phosphate and glucose 6-phosphate. O Only the anabolic direction will occur because the free energy change is positive. O The reaction will proceed in both directions simultaneously. This reaction can only proceed in one direction and must be the rate limiting reaction for this point in metabolism.
The standard free energy change for this reaction in the direction written is +23.8 kuimol. The tabie shows the concentrations of the three intermediates in the hepatocyte of a mammal. Intermediate Concentration (M) Fructose 1.0-bisphosphate 0.000028 Gyoeraldehyde 3phosphate 0.0000068 Ditydroxyacetone phosphate 0.000032 At body temperature (37 "C). what is the actual free energy change for the reaction (in kimol) ?
The glucose oxidase used in this experiment has a concentration of 27 U/mg. Calculate the mass (in g) of solid glucose oxidase required to react with all the glucose in 4.00 µL of10.0- mM glucose sample assuming the reaction proceeds for 30 minutes.
Most glucose in the human diet comes from plants. Unlike humans, plants transport sucrose instead of glucose. The conversion of sucrose to glucose and fructose is a reversible reaction as shown in Reaction 25.1: Reaction 25.1 Sucrose (aq) + H2O (l) ⇄ Glucose (aq) + Fructose (aq) Under conditions typical of human metabolism, the Keq for Reaction 25.1 is 140,000. At equilibrium, would you expect more of the reactant sucrose or more of the products glucose and fructose to be present? Explain reasoning.
ATP can be hydrolyzed into ADP + inorganic phosphate (P.) with a AG of -30500 J/mol. If, galactose-1-phosphate can be hydrolyzed into galactose + P, with a AG of -15.3 kJ/mol what is the Gibbs free energy change for the reaction of galactose + ATP = galactose-1-phospate + ADP? O a. -45.8 kJ/mol O b. 15.2 kJ/mol Oc. -30484.7 J/mol O d. -15.2 kJ/mol Oe. -1520 J/mol
For an enzyme catalyzed reaction of the form: S + E → P + E, the rate of product formation, [P], is given by: d[P]/dt = k2[E}total [S] /(Km + [S]) = For the enzymatically catalyzed hydrolysis of ATP at 25 °C and pH 7.0, the Michaelis Menten constant, Km was found to be equal to 16.8 μmol L 1 and the value of k2[E]total was found to be 0.220 μmol L-¹s¹. Find the initial rate at an initial ATP concentration of 30.0 μmol L-1
Acetyl CoA + 2H* + 2e = pyruvate + COASH E = -0.48 V Ubiquinone + 2H* + 2e = Ubiquinol E" = +0.04 V Consider the redox rxn wherein a pair of e passes from pyruvate to ubiquinone. Calculate the change in standard Gibbs free energy (kJ/mol). Report answer to two decimal places.
Many metabolites are maintained at steady-state concentrations that are far from equilibrium. A comparison of Kéq and Q, the mass-action ratio, can determine whether a metabolic reaction is far from equilibrium. The equation for this equilibrium is, fructose 6-phosphate + ATP = fructose 1,6-bisphosphate + ADP Calculate Kg for this reaction at T = 25.0 °C. Ka 1.8 x10-6 AG' = -14.2 kJ/mol Incorrect Calculate the mass-action ratio, Q, from the approximate physiological concentrations for rat heart tissue shown in the table. Metabolite Concentration (µM) Q = 3 fructose 6-phosphate 81.0 fructose 1,6-bisphosphate 15.0 Incorrect ATP 12,500 ADP 1,340

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