EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 25, Problem 25.37AP
Interpretation Introduction
Interpretation:
Calculate the value of
Concept Introduction:
The energy change for any reaction can be calculated as the sum of the energy changes for other reactions. Energy change for the hydrolysis of acetyl phosphate and the hydrolysis of acetyl-CoA are given in the question.
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When the following reaction reached equilibrium the concentration of glucose 1-phosphate is 34mM, and the concentration of glucose 6-phosphate is 190mM. At standard temperatures and pressure, calculate the Keq and the ΔGo'
glucose 6-phosphate <---(phosphoglucoisomerase)---> glucose 1-phosphate
Keq= ΔGo' =
The reaction for “activation” of a fatty acid (RCOO−), ATP + CoA + RCOO−⇌ RCOOCoA + AMP + PPi has ΔG°′ = +4.6 kJ · mol−1. What is the thermo dynamic driving force for this reaction?
The AG" of the dephosphorylation of phosphocreatine is -43.0 kJ/mol.
Phosphocreatine → creatine +Pi; -43.0kJ/mol
When coupled to the phosphorylation of ADP to ATP (+30.5kJ/mol)
ADP +Pi → ATP; +30.5 kJ/mol
calculate the actual, physiological AG for the following reaction in kJ/mol:
Phosphocreatine + ADP creatine + ATP
at 37°C, with concentrations as follows:
Phosphocreatine = 0.715 mM
creatine = 0.566 mM
ADP = 0.431 mM
ATP = 2.382 mM
Chapter 25 Solutions
EBK ORGANIC CHEMISTRY
Ch. 25 - Prob. 25.1PCh. 25 - Prob. 25.2PCh. 25 - Prob. 25.3PCh. 25 - Prob. 25.4PCh. 25 - Prob. 25.5PCh. 25 - Prob. 25.6PCh. 25 - Prob. 25.7PCh. 25 - Prob. 25.8PCh. 25 - Prob. 25.9PCh. 25 - Prob. 25.10P
Ch. 25 - Prob. 25.11PCh. 25 - Prob. 25.12PCh. 25 - Prob. 25.13PCh. 25 - Prob. 25.14PCh. 25 - Prob. 25.15PCh. 25 - Prob. 25.16PCh. 25 - Prob. 25.17PCh. 25 - Prob. 25.18PCh. 25 - Prob. 25.19PCh. 25 - Prob. 25.20PCh. 25 - Prob. 25.21APCh. 25 - Prob. 25.22APCh. 25 - Prob. 25.23APCh. 25 - Prob. 25.24APCh. 25 - Prob. 25.25APCh. 25 - Prob. 25.26APCh. 25 - Prob. 25.27APCh. 25 - Prob. 25.28APCh. 25 - Prob. 25.29APCh. 25 - Prob. 25.30APCh. 25 - Prob. 25.31APCh. 25 - Prob. 25.32APCh. 25 - Prob. 25.33APCh. 25 - Prob. 25.34APCh. 25 - Prob. 25.35APCh. 25 - Prob. 25.36APCh. 25 - Prob. 25.37APCh. 25 - Prob. 25.38APCh. 25 - Prob. 25.39APCh. 25 - Prob. 25.40AP
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- Determine the direction that each of the reactions will progress. Assume that the reactants and products are present in equimolar amounts. The standard free energy of hydrolysis of ATP is –30.5 kJ/mol. fructose+ATP ____fructose 6‑phosphate+ADP The standard free energy of hydrolysis for fructose 6‑phosphate is −15.9 kJ/mol. 3‑phosphoglycerate+ATP___1,3‑bisphosphoglycerate+ADP The standard free energy of hydrolysis for 1,3‑bisphosphoglycerate is −49.3 kJ/mol. creatine+ATP___creatine phosphate+ADP The standard free energy of hydrolysis for creatine phosphate is –43.0 kJ/mol.arrow_forwardAssume that in a certain cell, the ratio of products/reactants or Keq = 614.9 (Keq is dimensionless) for the reaction Glucose + ATP <> Glucose-6-P + ADP, at a particular instant, the concentrations of each compound were Glucose =4.3M, ATP =10.5M, ADP =17.3M and G-6-P =23.5M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five significant figures.arrow_forwardFor a given acid HA, it was determined that at pH 6.0 the concentration of the conjugate base [A] was 0.075 M and the acid [HA] was 0.025 M. What percent of this acid is ionized at pH 6.0? What is the pKa of this acid? What pH would this acid be 50% lonized?arrow_forward
- If 0.1 M glucose-1-phosphate is incubated with phosphoglucomutase, the glucose-1-phosphate is transformed to an isomer. At equilibrium, the concentration of glucose- 1-phosphate is 4.5 x 10'M and that of the isomer is 8.6 x 10 M. Calculate K' and AGo" for this reaction (i.e., in the direction of product formation). (R = 8.314 J/mol K)arrow_forwardConsider these three reactions and fill in the blanks below: Malate + NAD+ → Oxaloacetate + NADH + H+ AGO. = = +29.8 kJ/mol Oxaloacetate + acetyl-CoA –→ citrate + COASH AGO = -32.2 kJ/mol Citrate → isocitrate AGO. = +8.4 kJ/mol Under standard conditions the net formation of isocitrate from malate is thermodynamically with a standard free energy change (AGU) of kJ/mol.arrow_forwardAnaerobic glycolysis (i.e., lactic acid fermentation) produces pyruvate that is then converted to lactate through the activity of lactate dehydrogenase. The conversion of pyruvate to lactate would seem to be an unnecessary step, since this process does not result in any further release of energy. Explain the necessity for the production of lactate as the endpoint for anaerobic glycolysis.arrow_forward
- TPP is a coenzyme for transketolase, the enzyme that catalyzes the conversion of a ketopentose (xylulose5-phosphate) and an aldopentose (ribose-5-phosphate) to an aldotriose (glyceraldehyde-3-phosphate) and a ketoheptose (sedoheptulose-7-phosphate). Notice that the total number of carbons in the reactants and products is the same (5 + 5 = 3 + 7). Propose a mechanism for this reaction.arrow_forward1. A. phosphoenolpyruvate (PEP) is one of the two phosphoryl group donors in the synthesis of ATP during glycolysis. In human erythrocytes, the steady state concentration of ATP is 2.24 mM, that of ADP is 0.25 mM, and that of pyruvate is 0.051 mM. Calculate the concentration of PEP at 25° C, assuming that the pyruvate kinase reaction (see figure 13-13) is at equilibrium in the cell. B. The physiological concentration of PEP in human erythrocytes is 0.023 mM. Compare this with the value obtained in A. Explain the significance of this difference.arrow_forwardAnswer True or False The Na/K-ATPase pump is an example of a coupling element between a spontaneous process and a non spontaneous process.arrow_forward
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