mol mol b) The movement of positive ions into the mito produces a free energy change of 7 in 6 out

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1 Chapter 5 Problems Page 1 of 5 11/1/ Chemiosmotic theory says that a mitochondrial proton gradient powers ATP synthesis. The mitochondrial membrane potential is -14 mv and ΔpH 1.5. Inside the mitochondrion ph 7., [ATP] 1 mm, [P i ] 2.5 mm, [ADP] 1 mm and T 298. (a How much chemical potential is needed to synthesize ATP the mitochondrion?; (b How much free energy is made available by movg 1 of protons from the side to the side of the mito? Is this enough to drive ATP synthesis? (c How many protons must be translocated per ATP synthesized? a ADP(aq P i (aq ATP(aq H 2 O(l We need to know ΔG for the synthesis of ATP. I found ΔG for ATP hydrolysis Voet & Voet s Biochemistry. It is ΔG -3.5 k/. The reverse of hydrolysis is synthesis, so ΔG 3.5 k/ for the above reaction. The ΔG for different (nonstandard-state concentrations of reactants and products for this reaction is [ ATP] k (1 1 ΔG' ΔG ' RT ln 3.5 (8.314 (298 ln [ ADP][ P ] i (1 1 (2.5 1 k k 2 k k k 3.5 (2.48 ln( , or it takes 45.4 k to produce 1. b The movement of positive ions to the mito produces a free energy change of 7 [ H ] (1 1 Δμ' RT ln ZFΦ (8.314 (298 ln (1(96,5 (.14V 6 [ H ] ( V k k k k k ( 2.48 ln( Thus the movement of 1 of protons gives 22.1 k sce ΔG n Δμ. This is not sufficient to power ATP synthesis (need 45.4 k/ from part a. c Two protons per ATP isn t quite enough (two protons would give 44.1 k/ of ATP. It will take 3 protons to make one ATP.

2 Chapter 5 Problems Page 2 of 5 11/1/ Tetrahydrofolate reductase has four identical, dependent ATP bdg sites. The osmotic pressure of a solution of THFR was measured to be atm at 2 ºC. This solution was put a dialysis bag and ATP bdg measured by equilibrium dialysis. In one run, [ATP] side was 1-4 M and the total [ATP] the bag was M. (a Calculate, the equilibrium constant for bdg ATP to THFR; (b Two Scatchard plots at different Ts were obtaed (as shown the question. Calculate ΔH and ΔS for the bdg of ATP to one site on THFR. (Assume ΔH and ΔS are dependent of T. ATP(aq THFR(aq THFR(ATP 4 (aq a The concentration of enzyme can be gotten from the first measurement of osmotic pressure: Π atm 5 [ THFR] RT L atm L (.825 (293 From the bdg equation we have N[ ATP] 4(1 [ ligand bnd] [ total lig] [ free lig] ν and we know that ν 1 [ ATP] 5 1 (1 [ prot total] (1 So ν 2 or 2 (2 1 (4 1 which says that ( b van t Hoff tells us that Δ(ln ( ln 1 Δ H R 1 ( 1 R R. The Δ ( 1 1 ( 1 1 T T2 T1 T2 T1 vertical axis tercept on a Scatchard plot equals m. At the two temperatures, it is clear that this tercept is twice as large for T 2 ºC compared to T 37 ºC, thus the ratio of 2 / 1 2 and van t Hoff gives us ΔH R ( ( ( 1 1 ( 1 1 T ( T1.693 k ( ( and 4 k k Δ G RT ln ( (293ln(1 (2.44 ( k k 3.3 ( 22.5 H G which leaves us with Δ Δ ΔS T 293

3 Chapter 5 Problems Page 3 of 5 11/1/ (a Assumg that Raoult s law holds, what it the boilg pot of a 2 M solution of urea? (b Urea actually forms complexes of two or more ecules. How will this effect the boilg pot? a Raoult s law says that P soln X solvent P pure solvent. So the vapor pressure of a 2 M urea n solution is P solvent 55 Ppure (1atm.965atm (where we note that the nsolute nsolvent 2 55 ar concentration of pure water is 55 M and when pure water boils at 1 ºC, its vapor pressure is 1 atm. So we must crease T to get the solution to have a vapor pressure equal to 1 atm. Eq the text shows that the temperature change of boilg pot is proportional to a constant, boil, times the al concentration of the solution. Also, the boil for water is stated to be.52 m -1. The al (es of solute per kg of solvent concentration of urea water is very close numerically to the ar concentration, so [urea] 2 m. Thus ΔT boil [urea] (.52 m -1 (2 m 1.4, thus the new boilg pot is 11.4 ºC. b If the urea is makg complexes, the net effective concentration of urea plus that of urea 2 plus urea 3 etc. will be lower than assumed a, so the boilg pot will crease less than we calculated.

4 Chapter 5 Problems Page 4 of 5 11/1/ The solubilities water and ethanol of Gly and Val are given. (a Calculate the standard free energy of transfer of 1 of Gly from solid to aqueous solution at 25 ºC. Consider the solution to be ideal. (b Calculate the standard free energy of transfer of 1 of Gly from ethanol to aqueous solution at 25 ºC. (c Assume that effects of the backbone and sidecha are additive, and calculate the standard free energy of transfer of 1 of the (CH 3 (CH 2 - side cha of Val from water to ethanol at 25 ºC. (d Will the mutation of a Gly to a Val the terior of a prote favor prote stability if solvent teraction is the domant effect? (e If the folded-unfolded transition of a prote is two-state, calculate the change the folded unfolded equilibrium constant due to the Gly to Val mutation. a This is an equilibrium situation. Thus ΔG and ΔG nδμ. The standard free energy of a Gly, solution 3.9 transfer is ΔG nδμ RT ln (8.314 / (298 ln( 2.8k agly, solid 1 where the activity of the solid is 1 and the equilibrium concentration of Gly is given the table. Note that because n 1, Δμ has the same value as ΔG. From here on let s just remember that n 1 and write down the chemical potential directly. b Apply the same relation, but the second phase is different: agly, water 3.9 Δμ RT ln (8.314 / (298ln( 22.2k. agly, EtOH Note that this works because both the water and EtOH solutions are equilibrium with the same thg, solid Gly. c We need to calculate Val from water to EtOH (as b, but the phases are reversed sce this is water to EtOH and then subtract the non-side group part of the ecule (which we can estimate from Gly that we got b notg that the phases are reversed, i.e the answer b was for EtOH to water, so Δμ Gly 22.2 k/ for water EtOH: aval EtOH , ΔμVal RT ln (8.314 / (298ln( 15.2k. So the aval, water.6 difference between this and Gly is the part that is due to the side cha, or Δμ sidecha Δμ Val - Δμ Gly 15.2 k/ 22.5 k/ -7.3 k/. d Yes, the more hydrophobic Val will have a lower chemical potential the EtOH-like prote terior compared to Gly accordg to our answer c. It will be 7.3 k/ lower energy. e The ratio of the equilibrium constants is equal to the ratio of the free energy of transfer (exponentiated of each of the species, the transfer energy difference which we just figured c to be equal to -7.3 k/. The ratio of the exponentials leaves us with a difference one exponential, with this difference beg the difference the transfer energies. Thus the ratio of the equilibria is Val e Gly the terior. Δμ 7.3k RT (8.314 / ( e e 19.. This says that Val is favored over Gly

5 Chapter 5 Problems Page 5 of 5 11/1/ A prote solution of.6 g prote per 1 ml has an osmotic pressure of 22 mm of water at 25 ºC and ph at which the prote has no net charge. What is the ecular weight of the prote? We know from the osmotic pressure equation that M wrt where w is the Π concentration g/l. We need to convert the pressure to better units first: Π (22 mm of water(1 mm Hg/13.65 mm water(1 atm/76 mm Hg atm. So wrt (6g / L(.825L atm (298 M 69,2g Π atm 5.26 A cell membrane at 31 is found to be permeable to Ca but not to anions. The side of the cell has [Ca ] 1 mm and the side 1 mm. (a What potential difference volts must exist for the calcium ion concentration gradient to be at equilibrium? (b If the measured side potential is 1 mv with respect to the side, what is the mimum reversible work required to transfer 1 of calcium ion from side to side? a This is a condition for a Donnan potential, so [ Ca ] [ Ca ] RT ln zf( Φ Φ or RT ln zf( Φ 2 Φ. This gives us [ Ca ] [ Ca ] 1 ( RT [ Ca ] (8.314 / (31 1 Φ Φ ln ln (.134V ( V, zf [ Ca ] (2(96,5 V 1 3 that is, the side is negative with respect to the side. b The Donnan condition says that [ Ca ].1 Δ μ RT ln zf( Φ (2.58k ln (2(96.5k V (.1V 2 Φ [ Ca ].1.1 ( 2.58k ln (2(96.5k V (.1V 11.9k 19.3k 31.2k.1 This will be the mimum required work to move calcium agast its concentration and potential gradients (if no PV work is done and sce ΔG n Δμ.