ࡱ> bdaq` *bjbjqPqP ..::v  $!%!'!'!'!'!'!'!$"h %LK!]K!!FFF%!F%!FF:5 ,  @ s"a %!!0!k RX%X% X% hFK!K!< ! < < Energy flow. What is the direction of each of the following reactions when the reactants are initially present in equimolar amounts? Use the data given in Table 14.1.  Answer: a. left 7.3-10.3=-3 b. right 7.3-2.2=5.1 c. left 7.3-14.8=-7.5 d. right 7.3-3.3=4.0 ATP standard free energy change is -7.3 kcal/mole Table 14.1. Standard free energies of hydrolysis of some phosphorylated compounds Compound kcal mol-1 kJ mol-1 Phosphoenolpyruvate -14.8 -61.9 1,3-Bisphosphoglycerate -11.8 -49.4 Creatine phosphate -10.3 -43.1 ATP (to ADP) - 7.3 -30.5 Glucose 1-phosphate - 5.0 -20.9 Pyrophosphate - 4.6 -19.3 Glucose 6-phosphate - 3.3 -13.8 Glycerol 3-phosphate - 2.2 - 9.2 A proper inference. What information do the G2 data given in Table 14.1 provide about the relative rates of hydrolysis of pyrophosphate and acetyl phosphate? Answer: None whatsoever. 3. A potent donor. Consider the following reaction:  (a) Calculate G2 and K2 eq at 25C for this reaction, by using the data given in Table 14.1. (b) What is the equilibrium ratio of pyruvate to phosphoenolpyruvate if the ratio of ATP to ADP is 10? Answer: G2 = +7.5 kcal mol-1 (+31.4 kJ mol-1) and K2 eq = 3.2 10-6. (b) 3.28 104. 4. Isomeric equilibrium. Calculate G2 for the isomerization of glucose 6-phosphate to glucose 1-phosphate. What is the equilibrium ratio of glucose 6-phosphate to glucose 1-phosphate at 25C? Answer: G2 = -1.7 kcal mol-1 (-7.1 kJ mol-1). The equilibrium ratio is 17.8. 5. Activated acetate. The formation of acetyl CoA from acetate is an ATP-driven reaction:  (a) Calculate G2 for this reaction by using data given in this chapter. The PPi formed in the preceding reaction is rapidly hydrolyzed in vivo because of the ubiquity of inorganic pyrophosphatase. The G2 for the hydrolysis of PPi is -4.6 kcal mol-1. Calculate the G2 for the overall reaction. What effect does the hydrolysis of PPi have on the formation of acetyl CoA? Answer: (a) Acetate + CoA + H+ goes to acetyl CoA + H2O, G2 = +7.5 kc 2 < C d Z \ LRz||~r !!|$~$****{p{{j6h1h'!U h1h'!U h1hEjj-h1hlXU h1hlX h1h8 h1hyv h1hpTj$h1h@U h1h@ h1h(2 h1h| h1hg h1hK~ h1hs h1hUkjh1hUkU h1hBw+ C 6 j b d  & Fgd@h^hgdsh^hgdBw & FgdBw* X Z ^ |"gd8  & Fgdyv h`hgd@"26z|r !"!!!!!!!z$$ RgdEj & F RgdEj RNR^N`RgdlX RgdlX$a$gdBwh^hgd8 al mol-1 (+31.4 kJ mol-1). ATP hydrolysis, G2 = -10.9 kcal mol-1 (-45.6 kJ mol-1). Overall reaction, G2 = -3.4 kcal mol-1 (-14.2 kJ mol-1). (b) With pyrophosphate hydrolysis, G2 = -8.0 kcal mol-1 (-33.4 kJ mol-1). 12. Not all alike. The concentrations of ATP, ADP, and Pi differ with cell type. Consequently, the release of free energy with the hydrolysis of ATP will vary with cell type. Using the following table, calculate the G for the hydrolysis of ATP in muscle, liver, and brain cells. In which cell type is the free energy of ATP hydrolysis greatest?  Answer: Liver: -10.8 kcal mol-1 (-45.2 kJ mol-1); muscle: -11.5 kcal mol-1 (-47.8 kJ mol-1); brain: -11.6 kcal mol-1(-48.4 kJ mol-1). 13. Running downhill. Glycolysis is a series of 10 linked reactions that convert one molecule of glucose into two molecules of pyruvate with the concomitant synthesis of two molecules of ATP (Chapter 16). The G2 for this set of reactions is -8.5 kcal mol-1 (-35.6 kJ mol-1), whereas the G is -18.3 kcal mol-1 (-76.6 kJ mol-1). Explain why the free-energy release is so much greater under intracellular conditions than under standard conditions. Answer: G = G2 + RT ln [products/reactants]. Altering the ratio of products to reactants will cause G to vary. 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