ch13 - 1. 8-142 Bioenergetics and Biochemical Reaction...

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Unformatted text preview: 1. 8-142 Bioenergetics and Biochemical Reaction Types 1 Entropy Changes during Egg Development Consider a system consisting of an egg in an incubw tor. The White and yoik of the egg contain proteins, carbohydrates, and lipids. If fertilized, the egg is transformed from a single cell to a compiex organism. Discuss this irreversible process in terms of the entropy changes in the system, surroundings, and universe. Be sure that you first clearly define the system and surroundings. Answer Consider the developing chick as the system. The nutrients, egg shell, and outside world are the surroundings. Transformation of the single cell into a chick drastically reduces the entropy of the system (increases the order). initially, the parts of the egg outside the one bryo (within the surroundings) contain complex fuel molecules (a low-entropy condition). During incubation, some of these compiex molecules are converted to large numbers of 002 and H20 molecules (high entropy). This increase in entropy of the surroundings is larger than the decrease in entropy of the chick (the system). Thus, the entropy of the universe (the system + surroundings) increases. . Calculation of AG” from an Equilibrium Constant Calculate the standard freeenergy change for each of the foliow‘mg metabolically important enzyme~catalyzed reactions, using the equilibrium constants given for the reactions at 25 °C and pH 7.0. aspartate anénotransferase (a) Glutamate + oxaloacetate aspartate ~l~ a—ketoglutarate Kgq = 6.8 triose phosphate isomerase (b) Dihydroxyacetone phosphate glyceraldehyde 8-phosphate Kgq x 0.0475 phosphofructokinase (c) Fructose 6~phosphate + ATP fructose 1,6—bisphosphate we ADP Kgq m 254 Answer AG z AG” + ET in [productslffreactants] and [productsvireactants] is the massection ratio, Q. At equilibrium, AG 2 O and Q = K’eq, so AG” m “RTln Kgq where R m 8.815 J/mol - K and T = 25 0C = 298 K. Using the value RT 2 2.48 kJ/mol, we can calculate the AG” values from the K 1m for each reaction. (a) as” = ""(248 kJ/mol) in as = are IKJ/rnol (:1) AG” 2 —(2.48 kJ/mol) in 0.0475 = 7.56 lei/moi (c) AG’° m —(2.48 icJ/mol) in 254 = ~1s.r kJ/rnol Chapter 13 Bioenergetics and Biochemical Reaction Types 5-143 3. Calculation of the Equfiibrium Constant from AG” Calcrflate the equilibrium constant Kng for each of the following reactions at pH 7.0 and 25 °C, using the AG” values in rfable 13—4. glucose B—gihosphatase (a) Glucose 6-phosphate + H20 glucose +~ Pi ,B-galactosidase (b) Lactose + H30 glucose + galactose fumerase (c) Malaise furnarate + H20 Answer As noted in Problem 2, AG 2 AG” ~1- RT 111 Q, and at equilibrimn, Q m Kgq, AG m 0, and AG” m ~RT In mm 80, at equilibrium, meg“ m —AG'°/RT: or Kgq = e“(AG'°’Mj; at 25 °C,RT = 2.48 EsJ/mol. From these relationships, we can calculate Kgq for each reaction using the vaiues of AGm in Table 13—4. (3) For glucose 6-phosphatase: AG” = WEBB kJ/mol 1n K’eq = —(~13.8 kJ/mol)/{2.48 kJ/mol) x 557' Egg = e557 = 262 (b) For B—gaiactosidase: AG’0 2 MES}? kJ/mol in K’EQ = —(—-15.9 kJ/mol)/(2.48 kJ/mol) m 6.41 Kgq = 96-41 = 608 (c) For filmarase: AG’o = 3.1 kJ/moi in Kim = ~(3.l kJ/mol)/(2.48 kJ/mol) = —1.2 Kgq m 8-1'2 = 0.80 4. Experimental Determination of Kim anti AG” If a 0.1 M solution of glucose 1«phosphate at 25 °C is incubated with a catalytic amount of phOSphoglucomutase, the glucose 1-phosphete is trans- formed to glucose 6wphosphate. At equilibrium, the concentrations of the reaction components are Glucose i~phosphate :2 glucose 6-phosphate 4.5 x 10—3 M 9.6 x 10"“2 M Calculate Km and AG"0 for this reaction. Answer mm m [G6P]/§GlP§ m {9.6 X 10"? M)/(4.5 X 104’ M) m 21 so” m 4351’ in mm m — (2.48 k5/moi)(ln 21) 2 —7.6 kJ/moi 8-144 Chapter 13 Bioenergetics and Biochemical Reaction Types 5. Experimental Determination of AG” for ATP Hydrolysis A direct measurement of the standard free-energy change associated with the hydrolysis of ATP is technically demanding because the minute amount of ATP remaining at equilibrium is difficult to measure accurately. The vaiue of AG” can be calculated indirectly, however, from the equilibrium constants of two other enzymatic reactions having less favorable equilibrium constants: Glucose 6—phosphate + H20 ~—> giucose + Pi Kgq m 270 ATP + glucose —-~> ADP ~I~ glucose 6-phosphate mm = 890 Using this information for equilibrium constants determined at 25 °C, calculate the standard free energy of hydroiysis of ATP. Answer The reactions, if coupied together, constitute a “futile cycle” that results inthe net hydrolysis of ATP: (1) (l6:3 -i~ H20 We glucose .+ P; (2) ATP + glucose —> AD? + G6P Sum: ATP + H20 —-> ADP + P]- Calculating from A0” = “RT 111 K’eq: A01” = (—2.48 kJ/mol)(1n 270) m “14 kJ/mol AGZ” = (-2.48 kJ/molen 890) == -1’7 kJ/mol AGgfim = AGf" + AG? 3 —31 kJ/mol 6. Difference between AG’0 and AG Consider the following interconversion, Which occurs in glycoiy- sis (Chapter 14): Fructose 6-phosphate m giucose 6—phosphate Kgq = 1.97 (a) What is AG'° for the reaction (Kg... measured at 25 °C)? (b) If the concentration of fructose 6—phosphate is adjusted to 1.5 M and that of glucose 6-phosphate is adg'usted to 0.50 M, What is AG? (c) Why are AG'0 and AG different? Answer (a) At equilibrium, AG’° w —RT In Kim M “(2.48 kJ/mol) in 1.97 ":2 ~I.68 kJ/mol (b) AG = AG” + RTln Q Q = [SSH/{F613} = 0.5 ill/1.5 M = 0.88 AG = -l.68 icJ/tnol + (2.48 kJ/mol) in 0.83 = -4.4 kJ/mol (c) AG” for any reaction is a fixed parameter because it is defined for standard conditions of temperature (25 °C = 298 K) and concentration (both FSP and 66.? = 1 M). in com trast, AG is a variable and can be calculated for any set of product and reactant concen- trations. AG is defined as AG’° (standard conditions) plus Whatever difference occurs in AG on moving to nonstandard conditions. Chapter 13 Bioenergetics and Biochemicafi Reaction Types 3-}.45 7. Free Energy of Hydroiysis of CT}J Compare the structure of the nucleoside triphosphate CTP with the structure of ATP. NH2 (\NH or 0“ e“ E l E x i” "O—IEPE—O—Iu’~0— E—~O——CH2 O N O 0 O O H H H H OH OH Cytidine triphosphate {CTP) NH2 N \ N o~ 0‘ 0‘ </ | J m i | I N / Owlgwowln’wOMEWOWCHg O O O O H H H H OH OH Adenosine triphosphate {ATP) N ow predict the K’m and AG” for the following reaction: ATP + GDP —> ADP + CTP Answer AG” near 0; K2“, near 1. The high AG” of ATP is related to structurai features not of the base or the sugar, but primarily of the anhydride linkages between phosphate groups. in this structural feature, CTP is equivalent to ATP, and thus it most likely has about the same AG” as ATP. If this is the case, the reaction ATP + GDP --->~ ADP + GTP has a 1163’" very close to zero, and a K20, close to i (see Table 13m3). 8. Dependence of AG on pH The free energy reieased by the hydrolysis of ATP under standard condi- tions at pH 7.0 is w30.5 kJ/moi. If ATP is hydrolyzed under standard conditions except at pH 5.0, is more or less free energy released? Explain. Use the Living Graph to explore this reiationship. Answer Less; the Overall equation for ATP hydroiysis can be approximated as ATP“ + 320 m ADP“ + HPO4‘ + an (This is only an approximation, because the ionized species shown here are the maior, but not the only, forms present.) Under standard conditions (to, [ATP] : [ADP] m [13,} e i M), the concentration of water is 55 M and does not change during the reaction. Because H” ions are produced in the reaction, the lower the pH at which the reaction proceeds—that is, the higher the [H+]__the more the equilibrium shifts toward reactants. As a result, at lower pH the reac- tion does not proceed as far toward products, and less free energy is released. 9. The AG” for Coupled Reactions Glucose 1-phosphate is converted into fructose fimphosphate in two successive reactions: Glucose l—phosphate Mme glucose 6-phosphate Giucose 6-phosphate mm:- fructose Guphosphate 8-146 Chapter 13 Biaenergetics and Biochemicai Reaction Types Using the AG” values in Table 134, calculate the equiiibriuni constant, Egg, for the sum of the two reactions: Glucose I-phosphate —> fructose fiphosphate Answer i (E) 611? --> G6? A61” = ~—7.3 kJ/mol (2) GSP MW) F6P AG2m = 1.7‘ kJ/moi Sum: G1? —~w-~> FSP AGgfim = ~56 Eel/moi mrrgq m —AG’°/RT m —(-—5.6 kJ/mol)/(2.48 kJ/moi) x 2.3 mm m 10 10. Effect of {ATE/[ADP] Ratio 011 Free Energy of Hydrolysis of ATP Using Equation 13—4, plot AG against in Q (mess-action ratio) at 25 °C for the concentrations of ATP, ADP, and Pi in the tabie beiow. AG” for the reaction is “30.5 kJ/mol. Use the resulting plot to explain Why metabolism is reguw lated to keep the ratio {ATE/[ADP] high. :zzr.wwu;_a«aw=zaz.—u~ ‘nW—u‘*1":’yr_.m¥fisfi_wn wmmmmchrmmmmmmw- wm- -~-- E Concentration (mm) g i Answer The reaction is ATP —> ADP + Pi. From Equation 134, with Q (the mass action ratio) = [ADPHPfl/EATP], expressed as molar concentrations, the free-energy change for this reaction is: AG :2 AG” + RTln {{ADPHPfl/{ATPD Calculate In Q for each of the five cases: in Q; m In [(2 X 10“4)(1.0 >< i0“2}/(5 x 10—3)] = was in Q2 m In [(2.2 x 10“3)(1.21 x 10"2)/(3 x 10%} 2 “4.7 In cs 2 in [(4.2 x 10“3)(1.4i x 10‘2)/(1 x 10%} m —2.8 In Q4 =1n[{5.0 >< 10”3){i.49 >< uric/(2 x 10%)] m —1.0 In 625 = In [(2.5 >< 10”?)(10 >< urn/(5 x 10—3)} m —3,0 Substitute each of these values for in Q, —30.5 kJ/mol for AG”, and 2&8 kJ/mol for RT in Equation 13—4: AG]L m —30.5 kJ/mol + (2.48 kJ/moEX-TS) = “‘50 kJ/mol AGZ w —30.5 kJ/mol + (2.48 kJ/mol)(-4.7) = “42 kJ/mol AGg == —80.5 kJ/Inol + (2.48 kJ/mol)(-—2.8) = ~38 kJ/mol AG4 w ~80.5 RJ/mol ~i~ (2.48 Eel/Incl) (“1.0) = #33 kJ/moi AG5 a —30.5 kJ/mol + (2.48 kJ/mol) (M30) = w38 kJ/mol Chapter 13 Bioenergetics and Biochemical Reaction Types 3-147 Now plot AG versus in Q for each case: 0 AG {kJ/mol) l w c “40 -50 ~6 99 -8 ""7 m6 "m5 m4 —3 —2 -—1 0 ' 11162 The AG for ATP hydrolysis is smaller when [ATM/{ADP} is low («1) than when [ATE/[ADE is high. The energy available to a cell from a given amount of ATP is smaller when [ATE/{ADP} falls and greater when this ratio rises. 11. Strategy for Overcoming an Unfavorable Reaction: ATP-Dependent Chemical Coupling The phosphorylation of glucose to glucose 6-phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by Pi is described by the equation Glucose + P, ---~> glucose 6—phosphate + H20 AG” m 18.8 kJ/rnol (a) Calculate the equilibrium constant for the above reaction at 37 °C. In the rat hepatocyte the physiological concentrations of glucose and Pi are maintained at approximately 4.8 mM. What is the equilibrium concentration of glucose 6-phosphate obtained by the direct phosphorylation of glucose by P1? Does this reaction represent a reasonable metabolic step for the cataboiism of glucose? Explain. (ii) in principle, at least, one way to increase the concentration of glucose 6-phosphate is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and P1. Assuming a fixed concentration of Pi at 4.8 mM, how high would the intracellular concentra- tion of glucose have to be to give an equilibrium concentration of glucose 6-phosphate of 250 “M (the normal physiological concentration)? Would this route be physiologically reasonabie, given that the maximum solubility of glucose is iess than i M? (c) The phosphoryiation of glucose in the coil is coupled to the hydrolysis of ATP; that is, part of the free energy of ATP hydrolysis is used to phosphorylate glucose: (1) Glucose + P, W“? glucose 6-phosphate + H20 AG” r—r 13.8 kJ/moi (2) ATP + H20 --9 ADP «r P; AG” m —30.5 kJ/mol Sum: Glucose + ATP ——> glucose 6mphosphate + ADP Calculate Kg, at 37‘ °C for the overall reaction. For the ATP-dependent phosphorylation of giu- cose, what concentration of glucose is needed to achieve a 250 ,uM intracellular concentration of glucose 6-phosphate when the concentrations of ATP and ADP are 3.38 HM and 1.32 mM, respec— tively? Does this coupling process provide a feasible route, at least in principle, for the phospho— rylation of glucose in the celi‘? Explain. (d) Although coupling ATP hydrolysis to glucose phosphoryiation makes thermodynamic sense, we have not yet specified how this coupling is to take piace. Given that coupling requires a conunon intermediate, one conceivable route is to use ATP hydrolysis to raise the intracelluiar concentra- tion of P1 and thus drive the unfavorabie phosphoryiation of glucose by P1. Is this a reasonable route? (Think about the solubility products of metabolic intermediates.) (e) The ATP-coupled phosphorylation of glucose is catalyzed in hepatocytes by the enzyme glucokiw nase. This enzyme binds ATP and glucose to form a glucose-ATP-enzyme complex, and the phos— phoryl group is transferred directly from ATP to glucose. Explain the advantages of this route. 8-148 Chapter 13 Bioenergetics and Biochemicai Reaction Types Answer (3) AG” = “RTEII Kim langq = "AGm/RT = —(i3.8 laymen/(2.48 lei/moi) Kim = 6’556 = 3.85 X 10‘3 M—1 (Note: this value has units M"1 because the expression for Kim from the chemical equilibrium includes H20; see below.) , h [esp] Ki“ ‘ [Greta] [GGP] e Kim [GchPii m (3.85 x 10”“3 whats >< 10‘3M)(4.8 x 10"“3 M) a 8.9 X 10—8 M This would not be a reasonable route for giucose catabolism because the cellular {GBPE is likely to be much higher than 8.9 X 10""8 M, and the reaction wouid be unfavorabie. , _ {GSPI (1)) Because Keg — EGICHPI] _ {GBPE then {G10} — Kin—gm] 250 >< is“3 M (8.85 x 10"“3 M“1)(4.8 >< 10‘3M) This would not be a reasonabie route because the maximmn soiubility of giueose is less than 1 M. (e) (l) Glc + Pi —~——> G6P + HZO AG? m 13.8 irJ/Inol (2) ATP + H20 “we ADP + Pi AGQ’” 2 —30.5 kJ/moi Sum: G16 4“ ATP We G613 + ADP AGgfim = --16.7 kJ/mol in Egg = —AG’°/RT = "(—16.7 kJ/moi)/(2.48 RJ/rnol) E14M = Kgq = 837 , 2% Because Keq [G:C]{ATP] _ W then [Gic] “ Kgq [ATP] (250 x 10‘5M)(1.32 >< i0”3M) (837)(3.88 >< 10‘3M) m 1.2 x 10‘7M This route is feasible because the glucose concentration is reasonable. ((1) No; this is not reasonable. When glucose is at its physiological levei, the required Pi con- centration wouid be so high that phosphate saits of divalent cations would precipitate out. 03) Direct transfer of the phosphoryl group from ATP to giucose takes advantage of the high phosphoryl group transfer potential of ATP and does not demand that the concentration of intermediates be very high, unlike the mechanism proposed in (d). In addition, the usuai benefits of enzymatic cataiysis appiy, including binding interactions between the enzyme and its substrates; induced fit leading to the exclusion of water from the active site, so that only giucose is phosphorylated; and stabilization of the transition state. Chapter 13 Bioenergetics and Biochemical Reaction Types 8-149 I2. Calculations of A0” for ATP»Couplcd Reactions From data in Tobie 13w6 caiculate the AG” value for the reactions; (:1) Phosphocreatine + ADP ———> creatine + ATP (b) ATP we fructose —"~> ADP + fructose 6-phosphate Answer (a) The A6"0 value for the overali reaction is calculated from the sum of the AG” values for the coupled reactions. (1) Phosphocreatine ~i~ H20 ——> creatine + P, AG;o m —43.0 kJ/rnol (2) ADP + Pi W ATP + H20 A85o = 30.5 kit/moi Sum: Phosphocreatine + ADP We creatine é— ATP AGS’Em = “42.5 kJ/mol (b) (1) ATP + H20 —-> ADP + Pi AG;0 = “80.5 lei/moi (2) Fructose + P; —> F6]? + H20 AGéo m E59 kJ/mol Sum: ATP + fructose "—9 ADP + F613 saga,“ 2 use kJ/mol 13. Coupling ATP Cleavage to an Unfavorable Reaction To explore the consequences of coupling ATP hyrirolysis under physiological conditions to a thermodynamically unfavorable biochemicai reac- tion, consider the hypothetical transformation X -—> Y, for which AG” 2 20 kJ/mol. (a) What is the ratio [YE/{X} at equilibrium? (b) Suppose X and Y participate in a sequence of reactions during which ATP is hydrolyzed to ADP and P,. The overall reaction is X+ATP+H20WY+ADP+R Calculate [YJ/[X} for this reaction at equilibrium. Assume that the temperature is 25 “C and the equilibrium concentrations of ATP, ADP, and P, are all 1 M. (c) We know that [ATP], [ADP], and [Pi] are not 1 M under physiological conditions. Caicu— late [33/ [X] for the ATP-coupied reaction when the values of [ATP}, [ADI-“’3 and [P1] are those found in rat myocytes (Table 13—m5). Answer (a) The ratio [lam/[X]eq is equal to the equilibrium constant, Kgq. hiKgq = ~AG’°/RT & —(20 kJ/mol)/{2.48 kJ/mol) = MS Kg, m as m 3 X 1W m {Yleq/[Xleq This is a very small veins of Kgq; consequently, AG’0 is large and positive, making the re— action energeticain unfavorable as written. (b) First, we need to caicuiate AG” for the overall reaction. (1) X -> Y nor = 20 kJ/mol {2) ATP + H20 —-d—> ADP + P, AG; = M305 kJ/rnol Sum: X + ATP + H20 WW) ADP + P, + Y A053“, m “10.5 iii/mo} P- ADP Keg 2 [3,164 11" I 1e“ ; note: water is omitted {XieqlATPJeq 3-150 Chapter 13 Bloenergetics and Biochemical Reaction Types Because [ADPL {ATP}, and {Pi} are l M, this simplifies to Kg, = [SH/{X} in units of M. in Kg, : ~AG’°/RT : -~(—1o.5 kJ/mol)/(2.48 kJ/mol) a 4.23 Kg, = e433 a 68.7‘ m [vi/{X1 AG” is fairly large and negative; the coupled reaction is favorable as written. (c) Here we are dealing with the nonstandard conditions of the cell. Under physiological con- ditions, a favorable reaction (under standard conditions) becomes even more favorable. i K: m EY]Bq[Pl]6QEADP]eq er {XleqlATPleq If we hold the values of {Pi}, EADP], and [ATP] at the values knOWn to exist in the cell, we can calculate the values of [X] and [Y] that meet the equilibrium expression above, giving the equilibrium constant we calculated in (b). K; {ATP} [Yr/[X1 m m _ (68.7 were >< 10—3 M) — (ace x 10—3 recess x 10-3 m) m 7.4 X £04 So by coupling the conversion X ~—> Y to ATP hydrolysis, and by holding [ATP], [ADP], and [Pi] far from their equilibrium levels, the cell can greatly increase the ratio [proda new/{reactant}; the reaction goes essentially to completion. 14. Calculations of AG at Physiological Concentrations Calculate the actual, physiological AG for the reaction Phosphocreatine + ADP —w--> creatine + ATP at 37 °C, as it occurs in the cytosol of neurons, with phosphocreatine at 4.? RIM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6 mM. Answer Using AG” values from Table 13—6: (1) Phosphocreatine + H20 M—a creatine + Pi AGi" = ~43.0 kJ/mol (2) ADP + P, ——> ATP + H20 AGé" = 305 kJ/mol m Sum: Phosphocreatine = ADP ——-—> creatine + ATP AGng : -12.5 kJ/moi [products] W [creatinellATP] {reactants} “ [phosphocreatineHADP] (1 X10“3 M)(2.6 >< iO‘BM) (4.7 >< 10"3M)(7_s >< 10’4M) m 0.75 AG = A0” +122"an = —12.5 kJ/Inol + (8.315 J/mol - K)(310 K) In 0.75 m W13 kJ/mol Mass-action ratio, Q m Chapter 13 Bioenergetics and Biochemical Reaction Types $151 15. Free Energy Required for ATP Synthesis under Physiological} Conditions in the cytosoi of rat hepatocytes, the temperature is 37 °C and the mass~action ratio, Q, is m = 5.33 X 102M1 Calculate the free energy required to synthesize ATP in a rat hepatocyte. Answer The reaction for the synthesis of ATP is ADP + Pi —-—-> ATP + H20 AG” = 30.5 kJ/moi The mass—action ratio is {products} _ {ATP} {reactants} ""“ [amen Because AG 3 AG” ~i~ RTin [productsE/[reactantsh AG 2 30.5 kJ/Inol + (3.315 J/mol - K)(810 K) in 5.33 x 103 1W = 46.7 kJ/mol m 5.33 x 10‘2 M“ 16. Chemical Logic In the glycoiytic pathway, a six-carbon sugar (fructose 1,6—bisphosphate) is cleaved to form two three~carbon sugars, Wluch undergo further metabolism (see Fig. 145). in this pathway, an isomerization of glucose 6—phosphate to fructose 6‘phosphate (shown below) occurs two steps before the cleavage reaction (the intervening step is phosphorylation of fructose 6-phosphate to fructose 1,6" bisphosphate (p. 532)). H H O / | \o/ He“ch H— (,3 —OH CII=O HO“ O " H phosphohexose HO "‘ C " H i isomerase I Hm (I) MOH H-£EWOH H— C!) ——OH H~$-—OH CHQOP03 " (SI-1201303 ' Glucose Euphosphate Fructose 6—phosphate What does the isomerization step accomplish from a chemical perspective? (Hint: Consider What might happen if the CWC bond cleavage were to proceed without the preceding isomerizationl) Answer C—C bond cleavage is faciéitated by the presence of a carbonyl group one carbon re- moved from the bond being cleaved. Isomerization moves the carbonyl group from (3-1 to C~2, set— ting up a carbon-carbon bond cieavage between 0—3 and (3-4. Without isomerization, bond cleav- age would occur between 0-2 and 0-3, generating one have-carbon and one four-carbon compound. 17. Enzymatic Reaction Mechanisms I Lactate dehydrogenase is one of the many enzymes that require NADH as coenzyme. It catalyzes the conversion of pyruvate to lactate: ADH + H" NAD T'— ‘- lactate , V éeh {ire enase CH3 3’ g 033 L—Laetate 8—152 Chapter 13 Bioenergetics and Biochemical Reaction Types Draw the mechanism of this reaction (Show electronspushing arrows}. (Hint: This is a common reaction throughout metabolism; the mechanism is similar to that cataiyzed by other dehydrogenases that use NADH, such as alcohol dehydrogenase.) Answer The mechanism is the same as that of the alcohol dehyorogenase reaction (Fig. lit—n13, p. 547). 0.. I Pyruvate {ICWO CH3 H. H 0 v 1/ I 9 “x j?! NH2 NADH i R H+\i H 10 0\\C/0 / C\ l i NH2 + CH3—0~OH \+ NAB+ If R L-Lactate 18. Enzymatic Reaction Mechanisms II Biochemical reactions often look more complex than they really are. In the pentose phosphate pathway (Chapter 14), sedoheptulose 7-phosphate and glycer— i aldehyde 3~phosphate react to form erythrose 4-phosphate and fructose 6wphosphate in a reaction catalyzed by transalciolase. canon HO "MED—H (K /H (gm 0 HM ([3 "- OH 0 H (£1 HO M ? “NH \ H-(I3--OH \CII/ H—(lz-won H-(|}WOH + + HWCEIWOH ngwflH transaldolase H—-(IJMOH ngwOH CHgOPOS“ CH20P03“ CHZOPOB“ CHQOPOg“ Setioheptulose Glyceraldehyde Erythrose Fructose 7-phosphate 3—phosphate 4-phosphate 6-phosphate Draw a mechanism for this reaction {show electronwpnshing arrows). (Hint: Take another look at aldol condensations, then consider the name of this enzyme.) Chapter 13 Bioenergetics and Biochemical Reaction Types S~153 Answer The first step is the reverse of an aldoi condensation (see the aidolase mechanism, Fig. 14—5, p. 534); the second step is an aldoi condensation (see Fig. 13—4, p. 497). O H \ / ‘33 H~ {I} - OH CH‘ZOH ougopog“ CHECK (linger: Glyceraldehyde I C m 0 ‘13: 0 3—phosphate (€320 H+ E HOWE-Cl,__H 5 “e ) HO—(ilr-H H—(|3-®—H«;B_ o\ /H \<,3~H H—-(l3—OH HmC___OH ([3 H—(Il—OH H—(IanI-I H“$_OH H—(II—OH CHEOPOE“ cazoroi‘ I _ ___ Fructose CHZOPOE“ H (E OH 6-phosphate Sedoheptulose CH20P03“ 7-phosphate Erythrose 4—phosphate 19. Daily ATP Utilization by Human Atiuits (a) A total of 30.5 kJ/mol of free energy is needed to synthesize ATP from ADP and P, when the re» actants and products are at 1 M concentrations and the temperature is 25 °C (standard state). Because the actual physiciogical concentrations of ATP, ADP, and P, are not 1 M, and the temper— ature is 37' °G, the free energy required to synthesize ATP under physiological conditions is differ- ent from AG”. Calculate the free energy required to synthesize ATP in the human hepatocyte when the physiological concentrations of ATP, ADP, and P, are 3.5, 1.50, and 5.0 mM, respectively. (b) A 08 kg (150 lb) adult requires a caloric intake of 2,000 kcai (8,800 id) of food per day (24 hours). The food is metabolized and the free energy is used to synthesize ATP, which then provides en- ergy for the body’s daily chemical and mechanical work. Assuming that the efficiency of convert— ing food energy into ATP is 50%, calculate the weight of ATP used by a human adult in 24 hours. What percentage of the body weight does this represent? (c) Although adults synthesize large amounts of ATP daily, their body weight, structure, and compo- sition do not change significantly during this period. Explain this apparent contradiction. Answer (a) ADP ~i~ P, —--> ATP + H20 AG’o = 30.5 kJ/mol . . _ are] Mw_ 2e MESS actlon {£1me "‘““ "" X 10'3 M1 X 10-31%] — 4.7 X M AG=AG’°+RTinQ . = 30.5 icJ/mol + (2.58 lei/moi) in (4.7 x 202 M_1) = 46 kJ/mol (b) The energy going into ATP synthesis in 24 in is 8,360 M X 50% = 4,180 RE. Using the vaiue of AG from (a), the amount of ATP synthesized is - (4,180 ALB/(46 icJ/mol) m 91 tool The molecular weight of ATP is 503 (calculated by surruning atomic weights). Thus, the weight of ATP synthesized is ‘ (91 mol ATP) (503 g/moi) = 46 kg As a percentage of body weight: 100% (46 kg ATE/(68 fag body weight) = 68% W 3-154 Chapter 13 Bioenergetics and Biochemical Reaction Types as needed. 20. Rates of Turnover of 7 and B Phosphates of ATP If a small amount of ATP labeled with radioac- tive phosphorus in the terminal position, {7-32PJATP, is added to a yeast extract, about half of the 3213 activity is found in Pi Within a few minutes, but the concentration of ATP remains unchanged Explain. If the same experiment is carried out using ATP labeled with 32F in the central position, fB—3QPJATR the 3213 does not appear in Pi within such a short time. Why? Answer We can represent ATP as A—P—P—P (the P farthest from A is the T P) and a radiola— baled phosphate group as *P. One possible reaction for 7«labeled ATP would be phosphoryla— tion of glucose: A~P—P-*P + Glc —w~—> A—P—P + G6*P --> —> —> *Pi or, more generally: A-P—P—*P + H20 -——>A«P-P 4— *Pi 21. Cleavage of ATP to AIM? and PPi during Metabolism Synthesis of the activated form of acetate (acetyl—CoA) is carried out in an ATP-dependent process: Acetate + CoA + ATP ——~«> acetyl—CoA + AMP + PP;- (a) The AG” for the hydrolysis of acetyl—CoA to acetate and 00A is “32.2 kJ/mol and that for hy- drolysis of ATP to AMP and PP; is —30.5 kJ/moI. Calculate AG” for the ATPmdependent synthesis Answer (a) The AG” can be determined for the coupled reactions: (1) Acetate + 00A —-—> acetyl—COA + H20 AG? = 32.2 kJ/mo} (2) ATP + H20 ——~«> AMP + PR AG;o = “430.5 kJ/mol Sum: Acetate + CoA + ATP -> acetyl—COA + AMl3 + PP; aoggm 2 1.7 kJ/mol (b) Hydrolysis of PR would drive the reaction forward, favoring the synthesis of acetyimCoA. Chapter 13 Bioenergetics and Biochemical Reaction Types 3-155 Answer The free energy required to transport 1 moi of if" from the interior of the cell, where [EVE is 10“? M, across the membrane to Where §H+] is i0“1 M is not = RT ln {cg/CI) = ET in (10—1/10‘7) 2 (8.315 J/niol - 19(310 K) in 1.06 t 86 kJ/mol 23. Standard Reduction Potentials The standard reduction potential, E”, of any redox pair is defined for the halfmceli reaction: Oxidizing agent + to electrons --—> reducing agent ri‘he E” values for the NADWNADH and pyruvateflactate conjugate redox pairs are wuss V and —-O.19 V, respectively. (a) Which redox pair has the greater tendency to lose eiectrons? Explain. (1)) Which pair is the stronger oxidizing agent? Explain. ((1) Beginning with i M concentrations of each reactant and product at pH 7 and 25 °C, in which direction will the following reaction proceed? Pyruvate + NADl-i + H4” Wiactate + NAB+ (d) What is the standard free-energy change (A6”) for the conversion of pyruvate to lactate? (e) What is the equilibrium constant (Kgq) for this reaction? Answer (a) The NADWNADH pair is more likely to lose electrons. The equations in Table 13—? are written in the direction of reduction (gain of electrons). E” is positive if the oxidized member of a conjugate pair has a tendency to accept electrons. E” is negative if the 0Xi~ diced member of a conjugate pair does not have a tendency to accept electrons. Both NADWNABH and pyrnvate/lactate have negative E” values. The E” of NADWNADH (—0.0820 V) is more negative than that for pyruvate/lactate 0—0185 V), so this pair has the greater tendency to accept electrons and is thus the stronger oxidizing system. (b) ri‘he pyruvate/lactate pair is the more likely to accept electrons and thus is the stronger oxidizing agent. For the same reason that NADH tends to donate electrons to pyruvate, pyruvate tends to accept electrons from NADi-l. Pyruvate is reduced to iactate; NADH is oxidized to new. Pymvate is the oxidizing agent; NADH is the reducing agent. (c) From the answers to (a) and (b), it is evident that the reaction will tend to go in the direction of lactate formation. (6) ri‘he first step is to caicuiate M” for the reaction, using the E'0 values in Table 13-121 Re— cail that, by convention, AE” m (E’° of electron acceptor) M (E'° of electron donor). For NADH + pyruvate —> NAD"* + lactate AB?” 3 CE” for pyrnvate/lactate) —- CE” for NAD‘VNADH) = —0.185V * (—0.320 V] = 0.185V AG'° = moi? AXE” _2(9s.5 kJ/V - moi) (0.135 V) "26.3 kJ/moi 5-156 Chapter 13 Bioenergetics and Biochemical Reaction Types (e) inn, = -AG’°/RT = — (—26.1 kJ/moI)/(2.48 kJ/mol) = -« 10.5 K5,, = e195 m 8.63 x 104 2s. Energy Span of the Respiratory Chain Electron transfer in the mitochondrial respiratory chain may be represented by the net reaction equation NADH ~+~ H+ + $02 2 H20 + NAD+ (a) Calculate AB?” for the net reaction of mitochondrial electron transfer. Use E”0 values from Table 1347. (1)) Calculate AG” for this reaction. (c) How many AT}? molecules can theoretically be generated by this reaction if the free energy of ATP synthesis under cellular Conditions is 52 kJ/mol? Answer (a) Using E” values from Table 13—7: For NADH + H+ + toy—mango + NAD+ AB” 2 (am for g Osage) — CE” for NADWNADH) = 0.816 V -— (—0.320 V) = 1.14 V (a) so” = —n3 At?” = w2(96.5 kaV' mol)(l.14 V) = M2320 kJ/rnol (c) For ATP synthesis, the reaction is ADP + P, -~> ATP The free energy required for this reaction in the cell is 52 kJ/mol. Thus, the number of ATP molecules that could, in theory, be generated is 220 kJ/rnol 52 kJ/mol _ 4‘2 ~ 4 25. Dependence of Electromotive Force on Concentrations Calculate the eiectromotive force (in volts) registered by an electrode immersed in a solution containing the following mixtures of NAB+ and NADH at pH 7.0 and 25 °C, with reference to a halfmcell of E” 0.00 V. (a) 1.0 mM NAD“ and 10 mM NADH (b) 1.0 mM NAIL?" and 1.0 mM NADH (c) 10 mM NAD+ and 3.0 HIM NADH Answer The relevant equation for calculating E for this system is w ,0 RT {NAN} E‘E + £3 “Wins: At 25 °C, the RT/nc? term simplifies to 0.026 We. (a) From Table 13w7, E”0 for the NADWNADH redox pair is "-0.320 V. Because two electrons are transferred, 1% = 2. Thus, E 2 (—0.320 V} + (0026 WE) 111G X 10“3)/(10 X 10—3) a "0.320 V + (— 0.03 V) = “0.35V Chapter 13 Bioenergetics and Biochemical Reaction Types 5-157 (b) The conditions specified here are “standard conditions,” so we expect that E’ = E”. As proof, we. know that in l m 0, so under standard conditions the term [(0.026 War) in 1] m 0, andE = E"0 a —0.320 V. (c) Here the concentration of Mil)+ (the electron acceptor) is 10 times that of NADH (the electron donor). This affects the value of E: E = (moses V) + (ooze/2 V) in (10 x 10'3)/(i >< 10*“33 = moses V + 0.03 V = ~0.29 V 26. Electron Affinity of Compounds List the following in order of increasing tendency to accept electrons: (a), wketoglutarate + 003 (yielding isocitrate); (b), oxaioacetate; (c), 02; (d), NADPJ’. Answer To solve this problem, first write the half—reactions as in Table 1347, and then find the value for E”0 for each. Pay attention to the sight 5mm:mmmeemwmwm3—m- “'menu.quawcram52.1%$21.32:tr.::.“.fl£12-——_““."'=V:;:~Lp—-.-‘“‘W'wa—fisww" W'MMJAHWNMWW'V“““mm” m—w-mwmuwv-"uww-w cw mam-amu-mmwrm g: E! O . 5 55 Halfareaction (a) wKetogluterate + C302 + 2H+ + 2e“ —-> isocitrate (b) Oxaloacetate + 2H“ + 26"" -—> maiate (c) 5.02 + 2a“ a 29— we H20 (I!) NADP+ -i- H““ + 28“ —-~> NADPH The more positive the E’°, the more likely the substance will accept electrons; thus, we can list the substances in order of increasing tendency to accept electrons: (a), (d), (b), (c). 27. Direction of Oxidation-Reduction Reactions Which of the following reactions would you expect to proceed in the direction shown, under standard conditions, assuming that the appropriate enzymes are present to catalyze them? (a) Malate + NAB+ —--> oxaioacetate + NADH ~i~ H+ (b) Acetoacetate + NADH + id” -—-> B-hydroxybutyrate + NAD+ (c) Pyruvate + NADH + H+ We lactate + NAD+ (d) Pyruvate + B-hydroxybutyrate ——> lactate + acetoacetate (e) Maiate + pyruvate —-> oxaioacetate + lactate (f) Acetaldehyde + succinate -—-> ethanol + furnarate Answer It is important to note that standard conditions do not exist in the celi. The value of AE”, as calculated in this problem, gives an indication of whether a reaction would or would not occur in a cell without additional energy being added (usually from ATP); but AE’o does not tell the entire story. The actual cellular concentrations of the electron donors and electron acceptors contribute significantly to the value of E” (e.g., see Problem 25). Under nonstam dard conditions, the potentiai can either add to an already favorable AE” or be such a large positive number as to “overwhelm” an unfavorable AE”, making Al? favorable. To solve this problem, calculate the ME,0 for each reaction. AE” 2 (EM of electron accep- tor in the reaction) — (E’° of electron donor in the reaction). Use 15”“ values in Table 13—7. (3.) Not favorable. AE” = (15”0 for oxaloacetate/malate) — (E’° for NADVNADH) = “41820 V — (—0.166 V) = —0.154 V 8-158 Chapter 33 Bioenergetics and Biochemical Reaction Types (b) Not favorable. AB” = {E’° for acetoacetate/B~hydroxybutyrate) - (Em 1°01" NAD‘P/NADH) = 90.345 V) m 90.320 V) = WOOZES V (c) Favorable. AE” = (E’° for pymvate/lactate) w (E'° for NADWNABH) = *0185 V t“ (“"0320 V) = 0.135 V (d) Favorable. AE” 2 CE”0 for pyruvate/lactate) m (E’0 for acetoacetate/fi-hydroxybutyrate) m —0.185 V —— 90.346 V) r 0.161 V (e) Not favorable. AB” m (1?” for pyruvate/lactate) — (E’O for oxaloacetate/maiate) = w0.185 V — (—0.166 V) = —0.019 V (1') Not favorable. AE” = (23” for acetaldehyde/ethanol) — (E’°for furnarate/succmate) = «"0197 V ~ {+0.03} V) = —0.228 V Data Analysis Problem 28. Thermodynamics Can Be Tricky Thermodynamics is a challenging area of study and one with many opportunities for confusion. An interesting example is found in an article by Robinson, Hampson, Munro and Vaney, published in Science in 1998. Robinson and colleagues studied the movement of small mole cules between neighboring cells of the nervous system through cell—to—cell channels (gap junctions). They found that the dyes Lucifer yellow (a small, negatively charged molecule) and biocytin (a email zwitterionic molecule) moved in only one direction between two particular types of glia (nonmeumnal cells of the nervous system). Dye injected into astrocytes would rapidly pass into adjacent astrocytes, oligodendrocytes, or Muller cells, but dye injected into oligodendrocytes or Muller cells passed slowly if at all into astrocytes. All of these cell types are connected by gap junctions. Although it was not a central point of their article, the authors presented a molecular model for how this urndirectional transport might occur, as shown in their Figure 3: 9 (A) Astrocyte Oligodendrocyte (B) Astrocyte Oligodendrocyte Chapter 13 Bloenergetics and Biochemical Reaction Types 55-159 The figure legend reads: “Model of the unidirectional diffusion of dye between coupled oligodendro» cytes and astrocytes, based on differences in connection pore diameter. Like a sh in a fish trap, dye molecules (black circles) can pass from an astrocyte to an oligodendrocyte (A) but not back in the other direction CB)” Although this article clearly passed review at a well-respected journal, several letters to the editor (2994) followed, showing that Robinson and coauthors’ model violated the second law of thermodynamics. (a) Explain how the model violates the second law. Hint: Consider what would happen to the en- tropy of the system if one started with equal concentrations of dye in the astrocyte and oligoden~ drocyte connected by the “fish trap” type of gap junctions. (1)) Explain why this model cannot work for small molecules, although it may allow one to catch fish. (c) Expiain why a fish trap does work for fish. ((1) Provide two plausibie mechanisms for the unidirectional transport of dye molecules between the cells that do not violate the second law of thermodynamics. Answer (a) The lowest—energy, highest-entropy state occurs when the dye concentration is the same in both cells. If a “fish trap" gap junction ailowed unidirectional transport, more of the dye would end up in the oligodendrocyte and tees in the astrocyte. This would be a higher-energy, tower-entropy state than the starting state, violating the second law or thermodynamics. Robinson et al.’s model requires an irnpossibie spontaneous decrease in entropy. In terms of energy, the model entails a spontaneous change from a lower-energy to a higher~energy state without an energy input—again, thermodynamically impossible. (b) Moiecules, unlike fish, do not exhibit directed behavior; they move randomly by Brown- ian motion. Diffusion results in not movement of molecules from a region of higher corn centration to a region of tower concentration simply because it is more iii<eiy that a mol- ecule on the rugh-concentration side will enter the connecting channel. Look at this as a pathway with a rate-limiting step: the narrow end of the channel. The narrower end lim- its the rate at which molecules pass through because random motion of the molecules is less likely to move them through the smaller cross section. The wide end of the channel does not act like a funnel for molecules, although it may for fish, because moiecales are not “crowded” by the sides of the narrowing funnel as fish wouid be. The narrow end iiinits the rate of movement equally in both directions. When the concentrations on both sides are equal, the rates of movement in both directions are equal and there will be no change in concentration. (c) Fish exhibit nonrandom behavior, adjusting their actions in response to the environment. Fish that enter the large opening of the channel tend to move forward because fish have behavior that tends to make them prefer forward movement, and they experience “crowding” as they move through the narrowing channel. It is easy for fish to enter the large opening, but they don’t move out of the trap as readily because they are less likely to enter the small opening. (d) There are many possible explanations, some of which were proposed by the letter—writers who criticized the article. Here are two. (i) The dye could bind to a molecule in the oltgodendrocyte. Binding effectively removes the dye from the bulk solvent, so it doesn’t “count” as a solute for thermodynamic considerations yet remains visible in the flue— rescence microscope. (2) The dye could be sequestered in a subcelltdar organelle of the ottgodendrocyte, either actively pumped at the expense of ATP or drawn in by its attraction to other molecules in that organelle. References Letters to the editor. (1994) Science 265, 1817—1019. Robinson, 8.1%., Hampson, E.C.G.M., Munro, M.N., 8: Vaney, DJ. (1993} Unidirectional coupling of gap junctions between neu- roglla. Science 262, 1072—1074. ...
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