Full file at

Transcription

1 Chapter a. DG=DH-TDS Δ G = 80 kj ( 98 K) kj = 44.6 kj K b. ΔG = T m. Unfolding will be favorable at temperatures above the T m. Δ G =Δ H TΔ S 0 kj kj 80 ( xk) K 0 Δ G = = K = x x = K or 81.3 C.. a. Reversing the direction of the reaction as written requires a reversal of the sign on ΔH. Also, when summing two or more reactions, the species that appear on both sides cancel; thus: H O H ½ O ΔH = 4 kj 1 H H ΔH = 436 kj 1 ½[O O ΔH = 495 kj 1 ] H O H O ΔH = ½ (495) ΔH = 96 kj 1 b. The reaction has two O-H bonds broken; therefore, the energy of a single O H bond is 96/ = 463 kj a. ΔG = ΔH TΔS ΔS = (ΔH ΔG )/T = [(9.6 kj 1 ) ( 30.5 kj 1 )]/98 K = 470 J K 1 1 b. The decomposition products are both gasses, and therefore have significantly more translational and rotational freedom compared to the initial solid. Thus, entropy increases. 11

2 4. a. ΔG = ΔH TΔS = ( 816 kj 1 ) (3 K)(0.181 kj K 1 1 ) = 87.1 kj 1 b. From Table 3.5: ADP P i H ATP H O ΔG = 3. kj 1 (3. kj 1 ) (3 ATP) = 30.4 kj 1 C 6 H 1 O 6 6O 3ADP 3P i 3H 6CO 38H O 3ATP ΔG = ( 87.1 kj 1 ) (30.4 kj 1 ) = kj 1 c. %efficiency = ΔG ATP synthesis / ΔG total available 0% = (30.4/87.1) 0 = 35.9% 5. a. glucose P i G6P H O ΔG ' = 13.8 kj 1 / K eq = ([G6P][H O]/[glucose] [P i ]) = e Δ G RT Note: In the biochemical standard state, the activity of H O is assigned a value of 1. ( 13.8)/( ) (3) ([G6P](1))/((0.005) (0.005)) = e [G6P] = e 5.36 [G6P] = 1. 7 M b. ATP H O ADP P i H ΔG = 3. kj 1 Glucose P i G6P H O ΔG = 13.8 kj 1 ATP glucose ADP G6P H ΔG = 18.4 kj 1 / c. K eq = ([G6P] [ADP] [H])/([ATP] [glucose]) = e Δ G RT Note: The activity of H is referenced to a biochemical standard state concentration of 1 7 M. K eq = ([G6P] ( 18.4)/( ) (3) )/( ) = e [G6P] = ( e 7.14 )/( ) = (0.0189)/( ) = 47.5 M This G6P concentration is never reached because G6P is continuously consumed by other reactions, and so the reaction never reaches true thermodynamic equilibrium. / 6. a. K eq = ([GAP]/[DHAP]) = e ΔG RT 7.5/( ) (3) K eq = ([GAP]/[DHAP]) = e K eq = [GAP]/[DHAP] = 5.4 1

3 b. ΔG = ΔG RTln([GAP]/[DHAP]) = (7.5 kj 1 ) ( kj 1 K 1 )(3 K)ln(0.01) = 4.37 kj 1 7. a. ΔS must be positive because the increase in isoenergetic conformations in the denatured state increases the entropy of the denatured state relative to the folded state. b. Since ΔG = ΔH TΔS, a positive value for ΔS results in a negative contribution to ΔG (T is always positive in the Kelvin scale). Thus, for proteins to be stably folded (which requires ΔG to be positive for the process described in the question), it would appear that ΔH must be large and positive. This is generally true; however, this simple analysis does not consider the hydrophobic effect, which results in a more negative overall ΔS (see Problems 8 and 9, and further elaboration of this topic in Chapter 6). Thus, the requirement that ΔH is large and positive is not absolute because the hydrophobic effect reduces the magnitude of ΔS. 8. This process reduces the entropy of the solvent water, which becomes more ordered in the clathrate structures. This is the hydrophobic effect. 9. a. We expect ΔS for denaturation to be positive due to the increase in conformational entropy. If ΔH is also positive (energy is required to break the noncovalent interactions in the folded state), the, situation described in Table 3.3 pertains. Under these conditions, the denaturation process goes from being unfavorable at lower temperature to being favorable at higher temperature. b. In cases of cold denaturation, the sign on ΔS is dominated by the hydrophobic effect and is therefore negative. In these cases, ΔH must also be negative (see Table 3.3) for the process to become favorable as temperature decreases. Clathrate cage formation would account for the sign of ΔH being negative (favorable) for cold denaturation.. ΔG = ΔH TΔS and ΔG = RTlnK, thus ΔH TΔS = RTlnK (ΔH /RT) ΔS /R = lnk Divide both sides by RT. This is the van't Hoff equation. lnk = ( ΔH /R) (1/T) ΔS /R If ΔH and ΔS are independent of temperature, a graph of lnk versus 1/T should be a straight line with slope ΔH /R. These values can also be determined from direct fits to the K versus T data using nonlinear curve-fitting software. 11. a. See Problem. Plot lnk w versus 1/T and fit a line to the points. The slope will correspond to ΔH /R. 13

4 1/T ln K w lnk w = ( K)(1/T) 8.46 slope = K = ΔH /R K = ΔH /( kj 1 K 1 ) ΔH = 59.0 kj 1 b. lnk = ΔH /RT ΔS /R ΔS = [lnk (ΔH /RT)] R = [RlnK (ΔH /T)] ΔS = (( kj 1 K 1 ) (ln( 14 )) (59.0 kj 1 /( 98 K)) ΔS = ( 0.68 kj 1 K 1 ) (0.198 kj 1 K 1 ) ΔS = kj 1 K 1 or 70 J 1 K 1 1. a. G1P H O glucose P i H ΔG ' = 0.9 kj/ glucose P i H G6P H O ΔG ' = 13.8 kj/ NET: G1P G6P ΔG ' = 7.1 kj/ Δ G ' 7.1 RT K = e = e K = 17.6 b. The favored direction for the reaction can be determined by comparing K to Q. From part (a) we know that K = Q = 1/(0.001) = 00. Thus, Q > K, and therefore the reverse reaction is favored. This conclusion is also borne out by the (much more time-consuming) calculation of ΔG: [ ] [ ] G6P kj kj 1 kj Δ G=Δ G ' RTln = ( 98K) ln = 5.93 G1P K kj kj kj Δ G = =.0 Since ΔG > 0 the reverse reaction, formation of G1P, is favored. 13. To be favorable, the reaction must have ΔG < 0. 14

5 [ ][ ] [ malate] NAD oxaloacetate NADH H 0 >Δ G =Δ G ' RTln [ oxaloacetate][ ][ 1] [ ][ 0.00] kj kj 0 > ( 3K) ln K [ oxaloacetate][ ][ 1] [ ][ 0.00] 11.5 > ln = ln 37.5 e = 9.89 > ( [ oxaloacetate] ) [ oxaloacetate] [ ] 7.63 > oxaloacetate Thus, the reaction is unfavorable under these conditions when [oxaloacetate} exceeds.63 7 M kj/hour 4 hours/day = 8400 kj/day Palmitate combustion = kj kj/( kj 1 ) = 0.84 s palmitate Palmitate formula: C 16 H 3 O Molar mass = 16(1.011) 3(1.0079) (16) = 56.4 g/ 56.4 g/ 0.84 s = 16 g palmitate 15. a. For a e of protein ecule, ΔS = R ln W R ln 1, where W is the number of conformations available to each and R is the gas constant, J/K. Because there are 99 bonds between 0 residues, W = a. ΔS = 9.04 J/K. 15. b. folded ½ denaturation, [folded] = [unfolded] Therefore, K eq = 1. This defines the melting temperature of the protein. Since, ΔH TΔS = RTlnK eq = 0 15

6 Then, ΔH = TΔS = (33 K) 904 J 1 K 1 = 9. kj c. Since both ΔH and ΔS are positive, the fraction denatured will increase with increasing temperature. 16. a. [ glucose ] in kj kj kj Δ G=Δ G ' RTln = ( 3K) ln = 5.93 [ glucoseout ] K For transport of 6 glucose: kj Δ = 5.93 ( ) = G kj b. [ glucose ] in kj kj kj Δ G=Δ G ' RTln = ( 3K) ln = 5.93 [ glucoseout ] K 0.0 For transport of 6 glucose: kj Δ = ( ) = 6 6 G kj c. 6 energy required for transport 5.93 kj = = 1.94 energy available from ATP hydrolysis kj 30.5 ATP 7 ATP Thus, > ATP are required to drive the reaction in part (a) forward (i.e., to make ΔG < 0). No ATP hydrolysis is required for the reaction in part (b) as ΔG is already < a. Δ G =Δ G ' RTln ethanol NAD CO [ ][ ] pyruvate NADH H 16

7 M 15 torr ( ethanol) kj kj 1 M 750 torr 38.3 = 64.4 RTln M 15 M M M 1 M M kj kj kj ( ) 6 ( ethanol) ( 7.0 ) 38.3 = K ln Thus, [ethanol] = 0.57 M. kj kj 6.1 =.577 ln 4.75 kj 6.1 kj.577 ( ethanol) ( ) 4 ( ethanol) ( ) e = 4.75 =.50 ( ethanol) 4.50 = = b. [ ] [ pyruvate] ATP] H O] NADH H Δ G=Δ G ' RTln [ glucose] ADP] P i ] NAD Evaluation of ΔG ' requires combination of ΔG ' values for ATP hydrolysis (given) and glucose oxidation (which can be calculated from the E ' info provided): glucose pyruvate 6H 4e This is the e donor with E ' = V. NAD H 4e NADH This is the e acceptor with E ' = V. NET: glucose NAD pyruvate NADH 4H ΔE ' = ( V) ( V) = 0.75 V Thus, ΔG ' = nfδe ' = (4)(96.5 kj/ V)(0.75V) = 6. kj/. 17

8 glucose NAD pyruvate NADH 4H ΔG ' = 6. kj/ ADP P i H ATP H O ΔG ' = 61.0 kj/ NET: glucose NAD ADP P i pyruvate ATP H O NADH H ΔG ' = 45. kj/ kj (6 ) (0.0031) (1) (15 ) ( ) Δ G = 45. RTln 4 4 ( )(. ) (0.0059) (3.5 ) kj kj 3 kj Δ G = ( 3K) ln( 1.5 ) =