Enzymes Calculate the AG between glucose and ATP catalyzed by hexokinase: głucose + ATP glucose-6-phosphate + ADP Is the reaction spontaneous?| Given data: Glucose-6-phosphate + H,O= glucose + P. AG' =-3138 cal/mole ATP+H,0 =ADP+ P. AG'=-7700 cal/mole
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The spontaneous process occurs when free energy is lower in value than 0 that is G< 0.
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- Enzymes Calculate the AG between glucose and ATP catalyzed by hexokinase: glucose + ATP glucose-6-phosphate + ADP Is the reaction spontaneous?| Given data: Glucose-6-phosphate + H2O glucose + P; AG' = -3138 cal/mole ATP+H,O =ADP + P. AG' = -7700 cal/moleWhat 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)In the third step of glycolysis, the given reactions are coupled. reaction 1: fructose-6-phosphate + Pi ⟶ fructose-1,6-bisphosphate + H2O (Δ? = −28 kJ/mol) reaction 2: ATP + H2O ⟶ ADP + Pi (Δ? = +13.8 kJ/mol) Calculate the overall ΔG (kJ/mol) for the coupled reaction.
- 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. 2.4The 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. 1120The enzyme aldolase catalyzes the reaction shown in the glycolytic pathway: Fructose 1,6-bisphosphate dihydroxyacetone phosphate + glyceraldehyde 3-phosphate The AG" for the reaction is +23.8 kJ mol¯¹ (+5.7 kcal mol−¹), whereas the AG in the cell is −1.3 kJ mol¯¹ (−0.3 kcal mol¯¹). Calculate the ratio of products to reactants under standard (equilibrium) conditions at 37°C. [products] [reactants] 7 x10-5 [products] [reactants] Incorrect Aldolase ===== Calculate the ratio of products to reactants under intracellular conditions at 37°C. 4 ×10-5 Incorrect Complete the statement using your results. under standard conditions under intracellular conditions A reaction that is endergonic under standard conditions can be converted into an exergonic reaction by maintaining the ratio of products to reactants below the equilibrium value.
- The table contains the standard free energies of hydrolysis of phosphorylated compounds. Use the values in the table to answer questions about phosphorylation reactions in glycolysis. Compound kJ mol-1 kcal mol-1 Phosphoenolpyruvate -61.9 -14.8 1,3-Bisphosphoglycerate -49.4 -11.8 Creatine phosphate -43.1 -10.3 ATP (to ADP) -30.5 -7.3 Glucose 1-phosphate -20.9 -5.0 Pyrophosphate -19.3 -4.6 Glucose 6-phosphate -13.8 -3.3 Glycerol 3-phosphate -9.2 -2.2Calculate all the glucose data into cmol. What is the number of cmol glucose at timepont 10.42 (19.65 g/L)?If the DGo for ATP hydrolysis into ADP + inorganic phosphate is -7.3 kcal/mole, and the DGo for sucrose synthesis from glucose + fructose is +5.5 kcal/mole, calculate standard free energy change for the combined reaction of ATP + glucose + fructose g ADP + sucrose + inorganic phosphate. DGo = -12.8 kcal/mole DGo = -1.8 kcal/mole DGo = 0 kcal/mole DGo = +1.8 kcal/mole DGo = +12.8 kcal/mole
- Coupled reactions occur where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. This approach is common in biological settings. Determine if ATP could be generated by this biochemical reaction. You have calculated that cell potential is +0.637V. An example of a coupled reaction is the first step of glycolysis, the phosphorylation of glucose to form glucose-6-phosphate shown below. kJ/mol The net AG° for this reaction is 1 2 3 н он H. H- H- H H H H 4 6. H. glucose phosphate anion glucose-6-phosphate AG = +14.0 kJ/mol 7 8 9. АТР ADP phosphate anion AG = -30.5 kJ/mol +/- LOThe 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 2.4The conversion of lactate to glucose in the liver requires the input of 6 mol of ATP for every mol of glucose produced. Investigators can monitor the extent of this process in a rat liver preparation by administering [¹4C]lactate and measuring the amount of [¹4C]glucose produced. Because the stoichiometry of O, consumption and ATP production is known (about 5 ATP per O₂), investigators can predict the extra O₂ consumption above the normal rate after administering a given amount of lactate. However, when they actually measure the extra O₂ used in the synthesis of glucose from lactate, it is always higher than what the stoichiometric relationships predict. What could explain this observation? Unproductive cycling of ADP and ATP occurs, with extra O, consumption. The initial production of lactate requires extra O₂. The conversion of [C]lactate to [C]glucose requires less ATP and less O₂. The production of [C]glucose uses more O, than the production of unlabeled glucose.