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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
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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. 2.4The 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 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 2.4Cells manage to synthesize a variety of seemingly thermodynamically unfavored molecules every day. Phosphorylated glucose is an example of one of these molecules. The direct phosphorylation reaction is depicted below. As indicated, the forward reaction is unfavorable with a positive standard free energy change: glucose + phosphate⇄ glucose-6- phosphate + water ΔG°’ = +14 kJ/mol Describe specifically a thermodynamically favorable route by which the cell can produce glucose-6- phosphate from glucose. Please note cells cannot change either the temperature or the pressure at which the reaction is occurring. Please help me understand by using steps. Thank you!The first reaction in glycolysis is the phosphorylation of glucose to form glucose 6-phosphate: P₁+ glucose glucose 6-phosphate + H₂O This is a thermodynamically unfavorable reaction, with AG° = +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 is the equilibrium concentration of glucose 6- phosphate, according to the above reaction? b) This very low concentration of glucose 6-phosphate is unfavorable for glycolysis. In fact, the reaction is coupled to the hydrolysis of ATP: ATP + H₂O2 ADP + P₁ +H* AG°¹ = -32.2 kJ/mol Write the expression for the overall reaction and calculate AG°¹. c) In addition to the glucose and phosphate concentrations listed in part (a) above, assume that the liver cell concentrations of ATP and ADP are 3 mM and 1 mM, respectively. Calculate the theoretical equilibrium concentration of glucose 6-phosphate at pH = 7.4 and 37 °C. C ZOOM +
- Which of the following statements is most directly described by the first law of thermodynamics? A B с D The synthesis of highly-ordered biomolecules is coupled with the production of heat energy. All energy-transferring processes involve the loss of some of that energy into an unus- able form. Enzymes reduce the activation energy of chemical energy by stabilizing substrates at their transition states. The energy for the synthesis of glucose is provided by the absorption of light energy in chlorophyll pigments.In biological systems, enzymes are used to accelerate the rates of certain bio- logical reactions. Glucoamylase is an enzyme that aids in the conversion of starch to glucose (a sugar that cells use for energy). Experiments show that 1 pg mol of glucoamylase in a 4% starch solution results in a production rate of glucose of 0.6 ug mol/(mL)(min). Determine the production rate of glucose for this system in the units of Ib mol/(ft³)(day). Ibmol Answer: 0.0639 ft day True FalseWhat 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)
- A certain metabolic pathway can be diagrammed as: X Y Z A B C D C²D where A, B, C, and D are the metabolic intermediates and X, Y, and Z are the enzymes responsible for each conversion. The physiological free energies for each enzyme catalyzed reaction are: X AG = -0.2 kJ/mol Y AG¹ = -1.3 kJ/mol ZAG' = -12.3 kJ/mol a. Which reaction likely represents a major regulatory point for this pathway? b. If your answer for part a. was, in fact, the case, if an inhibitor of Enzyme Z was present, would the concentrations of metabolites A, B, C, and D be increased, decreased, or not affected? c. Identify which of the enzymes X, Y, and Z, if any, would likely need to be bypassed for the metabolite D to be converted to A.The reaction below occurs in nearly all human cells. It is catalyzed by the enzyme enolase. CO- CO- enolase -OPO3 OPO3 + H20 ČH,OH AGO'=+1.7 kJ/mol CH2 2PGA PEP Under cellular conditions, the AG is -3.3 kJ/mol. Which of the following could account for these changes in cellular AG compared to the standard biological free energy? The concentration of 2PGA is 7 times higher than PEP. The concentration of PEP is 7 times higher than 2PGA Enolase lowers the activation energy of the transition state. The release of water from 2PGA is a form of energy coupling.What is the standard free-energy change, ∆G°, under physiological conditions(E. coli grows in the human gut, at 37 °C) for the following reaction?Glucose + ATP → glucose 6-phosphate + ADP