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Be sure that you firstclearly define the system and surroundings.bated with a catalytic amount of phosphoglucomutase, the glucose-1-phosphate is transformed toglucose-6-phosphate until equilibrium is established. The equilibrium concentrations are2. Calculation ofAG°' from Equilibrium ConstantsCalculate the standard free-energy changes of thefollowing metabolically important enzyme-catalyzed reactions at 25 °C and pH 7.0 from the equilibrium constants given.5.
Experimental Determination of AG°' for ATPHydrolysis A direct measurement of the standardfree-energy change associated with the hydrolysisof ATP is technically demanding because the minute amount of ATP remaining at equilibrium is difficult to measure accurately. The value of AG°' canbe calculated indirectly, however, from the equilibrium constants of two other enzymatic reactionshaving less favorable equilibrium constants:aspartateaminotransferase(a) Glutamate + oxaloacetateaspartate + a-ketoglutarateK'eq = 6.8triose phosphateisomerase..._.._(b) Dmydroxyacetone phosphate vglyceraldehyde-3-phosphate, „,„,,_,K'eq = 0.0475Glucose-1-phosphate4.5 x 1 0 - 3 Mglucose-6-phosphate9.6 x 1 0 - 2 MCalculate K'eq and AG°' for this reaction at 25 °C.Glucose-6-phosphate + H2Oglucose +K'eq = 270ATP + glucoseADP + glucose-6-phosphateK'eq = 890phosphofructokinase(c) Fructose-6-phosphate + ATP ,fructose-l,6-bisphosphate + ADPK'eq = 2543.
Calculation of Equilibrium Constants from AG0'Calculate the equilibrium constants K'eq for each ofthe following reactions at pH 7.0 and 25 °C, usingthe AG°' values of Table 13-4:glucose6-phosphataseUsing this information, calculate the standard freeenergy of hydrolysis of ATP. Assume a temperatureof 25 °C.6. Difference between AGO/ and AG Consider thefollowing interconversion, which occurs in glycolysis (Chapter 14):Fructose-6-phosphateglucose-6-phosphateK'eq = 1.97(a) Glucose-6-phosphate + H2O ^=glucose + PjjS-galactosidase(b) Lactose + H2Oglucose + galactose(c) Malatefumarase— fumarate + H2O4. Experimental Determination ofK^q and AG°' Ifa 0.1 M solution of glucose-1-phosphate is incu-(a) What is AG°' for the reaction (assuming thatthe temperature is 25 °C)?(b) If the concentration of fructose-6-phosphateis adjusted to 1.5 M and that of glucoses-phosphate is adjusted to 0.5 M, what is AG?(c) Why are AG°' and AG different?7.
Dependence of AG on pH The free energy released by the hydrolysis of ATP under standardChapter 13 Principles of Bioenergeticsconditions at pH 7.0 is -30.5 kJ/mol. If ATP is hydrolyzed under standard conditions but at pH 5.0,is more or less free energy released? Why?8. The AG°' for Coupled ReactionsGlucose-1phosphate is converted into fructose-6-phosphatein two successive reactions:Glucose-1-phosphateGlucose-6-phosphateglucose-6-phosphatefructose-6-phosphateUsing the AG°' values in Table 13-4, calculate theequilibrium constant, K'eq, for the sum of the tworeactions at 25 °C:Glucose-1-phosphate> fructose-6-phosphate9.
Strategy for Overcoming an Unfavorable Reaction: ATP-Dependent Chemical CouplingThephosphorylation of glucose to glucose-6-phosphateis the initial step in the catabolism of glucose. Thedirect phosphorylation of glucose by Pj is describedby the equationGlucose + Pj> glucose-6-phosphate + H2OAG°' = 13.8 kJ/mol(a) Calculate the equilibrium constant for theabove reaction. In the rat hepatocyte the physiological concentrations of glucose and Pj are maintained at approximately 4.8 mM. What is the equilibrium concentration of glucose-6-phosphateobtained by the direct phosphorylation of glucoseby Pi? Does this route represent a reasonable metabolic route for the catabolism of glucose? Explain.(b) In principle, at least, one way to increase theconcentration of glucose-6-phosphate is to drivethe equilibrium reaction to the right by increasingthe intracellular concentrations of glucose and Pj.Assuming a fixed concentration of P, at 4.8 mM,how high would the intracellular concentration ofglucose have to be to have an equilibrium concentration of glucose-6-phosphate of 250 fiM (normalphysiological concentration)? Would this route be aphysiologically reasonable approach, given thatthe maximum solubility of glucose is less than 1 M?(c) The phosphorylation of glucose in the cell iscoupled to the hydrolysis of ATP; that is, part of thefree energy of ATP hydrolysis is utilized to effectthe endergonic phosphorylation of glucose:(1)Glucose +(2)ATP + H2OSum:Glucose + ATPglucose-6-phosphate + H2OAG°' = 13.8 kJ/molADP + P,AG°' = -30.5kJ/molglucose-6-phosphate + ADPCalculate K'eq for the overall reaction.
When theATP-dependent phosphorylation of glucose is carried out, what concentration of glucose is needed toachieve a 250 IJLM intracellular concentration ofglucose-6-phosphate when the concentrations ofATP and ADP are 3.38 and 1.32 mM, respectively?397Does this coupling process provide a feasible route,at least in principle, for the phosphorylation of glucose as it occurs in the cell? Explain.(d) Although coupling ATP hydrolysis to glucosephosphorylation makes thermodynamic sense,how this coupling is to take place has not beenspecified. Given that coupling requires a commonintermediate, one conceivable route is to use ATPhydrolysis to raise the intracellular concentrationof Pi and thus drive the unfavorable phosphorylation of glucose by Pi.
Is this a reasonable route?Explain.(e) The ATP-coupled phosphorylation of glucoseis catalyzed in the hepatocyte by the enzyme glucokinase. This enzyme binds ATP and glucose toform a glucose-ATP-enzyme complex, and thephosphate is transferred directly from ATP to glucose. Explain the advantages of this route.10. Calculations of AG°' for ATP-Coupled Reactions From data in Table 13-6 calculate the AG0'value for the reactions(a) Phosphocreatine + ADP(b) ATP + fructosecreatine + ATP>—>ADP + fructose-6-phosphate11. Coupling ATP Cleavage to an Unfavorable Reaction This problem explores the consequences ofcoupling ATP hydrolysis under physiological conditions to a thermodynamically unfavorable biochemical reaction.
Because we want to explorethese consequences in stages, we shall consider thehypothetical transformation, X> Y, a reactionfor which AG°' = 20 kJ/mol.(a) What is the ratio [Yj/[Xj at equilibrium?(b) Suppose X and Y participate in a sequence ofreactions during which ATP is hydrolyzed to ADPand Pj. The overall reaction isX + ATP + HoOY + ADP + ^Calculate [Y]/[X] for this reaction at equilibrium. Assume for the purposes of this calculationthat the concentrations of ATP, ADP, and Pj are all1 M when the reaction is at equilibrium.(c) We know that [ATP], [ADP], and [Pi] are not1 M under physiological conditions.
Calculate theratio [Y]/[X] for the ATP-coupled reaction when thevalues of [ATP], [ADP], and [PJ are those found inrat myocytes (Table 13-5).12. Calculations of AG at Physiological Concentrations Calculate the physiological AG (not AG°') forthe reactionPhosphocreatine + ADPcreatine + ATPat 25 °C as it occurs in the cytosol of neurons, inwhich phosphocreatine is present at 4.7 mM, creatine at 1.0 mM, ADP at 0.20 mM, and ATP at2.6 mM.398Part III Bioenergetics and Metabolism13. Free Energy Required for ATP Synthesis underPhysiological Conditions In the cytosol of rat hepatocytes, the mass-action ratio is[ATP][ADP][PJ = 5.33 xCalculate the free energy required to synthesizeATP in the rat hepatocyte.14.
Daily ATP Utilization by Human Adults(a) A total of 30.5 kJ/mol of free energy isneeded to synthesize ATP from ADP and Pj whenthe reactants and products are at 1 M concentration (standard state). Because the actual physiological concentrations of ATP, ADP, and Pi are not1 M, the free energy required to synthesize ATPunder physiological conditions is different fromAGO/. Calculate the free energy required to synthesize ATP in the human hepatocyte when the physiological concentrations of ATP, ADP, and Pi are 3.5,1.50, and 5.0 mM, respectively.(b) A normal 68 kg (150 lb) adult requires a caloric intake of 2,000 kcal (8,360 kJ) of food per day(24 h).
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