Chem/Biochem 471 Exam 2 11/18/09 Page 1 of 7 Name:

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1 Chem/Biochem 471 Exm 2 11/18/09 Pge 1 of 7 Plese leve the exm pges stpled together. The formuls re on seprte sheet. This exm hs 5 questions. You must nswer t lest 4 of the questions. You my nswer more questions if you wish. Answerg 5 questions cn be n dvntge if you re unsure of some of your nswers (this will distribute the risk. Answerg 4 questions is dvntgeous if you re very sure of your nswers. A good strtegy is to completely nswer 4 questions nd then work on the fifth if you hve time. Ech pge is worth 20 pots. The totl exm grde will be normlized so tht the mximum vlue for the exm is 100 pots. For exmple, gettg 80 pots on 4 questions will normlize to 100 pots. Gettg 100 pots on 5 questions will normlize to 100 pots. Gettg 60 pots on 4 questions will normlize to 75 pots. Gettg 60 pots on 5 questions will normlize to 60 pots. If you leve pge bk, it will not be cluded the grdg. If you work on pge nd then decide tht you do not wnt it to be grded, be sure to mrk the DO NOT GRADE THIS PAGE box t the bottom of the pge. If you work on the pge nd fil to mrk the box, the pge will be grded. Work t lest 4 problems (of your choosg or more, s you prefer. Answers with exptions (where dicted re not complete.

2 Chem/Biochem 471 Exm 2 11/18/09 Pge 2 of 7 1. Oxlocette (OA cn be reduced to mlte (M by iron. The two redox hlf cells tht re coupled for this rection re oxlocette + 2 H e - mlte, ε = V nd 5 pts ech Fe e - Fe, ε = V. Wht is the bced net rection for the reduction of OA to M by iron t T = 298? The top eqution is written the correct direction. The bottom rection needs to be reversed. To bce electrons, the top eqution needs to be multiplied by 3 nd the bottom by 2 which will leve 3 oxlocette + 6 H Fe 3 mlte + 2 Fe 3+ b Wht is the stndrd stte electrochemicl potentil for the rection prt? The top rection is written the correct direction but the bottom is reversed, so the net stndrd stte electrochemicl potentil is ε rxn = ε top - ε bottom = V (-0.036V = V. c Wht is the equilibrium constnt for the rection prt? eq ' = e o nefε ' RT = e (6(96,500Jmol V Note tht this is 6-electron rection. ( 0.13V (8.3145J mol (298 = e 30.4 = d Wht is the Gibbs free energy chnge the rection of prt if ll species re t stndrd stte? When ll species re t stndrd stte, ΔG = ΔG = -n e Fε = -(6(96,500 J mol -1 V -1 (-0.13 V = 75,300 J/mol = 75.3 kj/mol. The nswer cn lso be gotten from ΔG = -RT eq.

3 4 pts ech Chem/Biochem 471 Exm 2 11/18/09 Pge 3 of 7 2. Suppose tht you dd sgle-strnded DNA to solution A. Given the mss of DNA tht you hve dded, you compute tht the totl concentrtion of DNA the solution is 5 mm. The temperture is 30 o C. You my ssume tht the solution behves idelly. The concentrtion of DNA is well with the "wek solution" rnge. By how much will the osmotic pressure of the solution crese when DNA is dded to solution A? (You my ssume tht the DNA rems s sgle-strnded DNA. If the DNA is sgle-strnded, its concentrtion is 5 mm. The osmotic pressure for this much mteril solution cn be clculted from Π = crt; this mount of osmotic pressure will represent the crese from the strtg osmotic pressure of the solution before the DNA ws dded. Π = crt = ( mol L -1 ( L tm -1 mol -1 (303 = tm. b You relize tht sgle strnded DNA will dimerize to mke double-strnded DNA ccordg to 2 ssdna dsdna. If you ssume tht ll of the DNA solution is double-strnded, wht would be the crese of osmotic pressure when DNA is dded to solution A? If ll the DNA is double-strnded, its concentrtion would be 2.5 mm. The osmotic pressure for this much mteril will be hlf tht clculted prt : Π = crt = ( mol L -1 ( L tm -1 mol -1 (303 = tm. c You mesure the ctul osmotic pressure the solution to be hlfwy between the nswers to prts nd b. Therefore, only portion of the ssdna hs dimerized. Wht is the sum of the concentrtions of ssdna nd dsdna solution? From the nswers to nd b, the ctul osmotic pressure is (0.124 tm tm/2 = tm. Rewritg the osmotic pressure formul gives c = Π/RT = (0.093 tm/( L tm -1 mol -1 (303 = mol L -1 = 3.74 mm. This is the concentrtion of ll the extr DN both ss nd ds, the solution. d Wht is the concentrtion of ssdna the solution? Cll S the concentrtion of ssdna nd D the concentrtion of dsdna. From c we know S + D = 3.74 mm. From the totl mount of DNA dded, we know tht S + 2D = 5 mm. Subtrctg the first eqution from the second leves D = 1.26 mm. Usg this plus either eqution gives S = 2.48 mm. e Wht is the ssocition constnt A for this ssdna solution? The ss nd dsdna re t equilibrium, nd the eqution prt b tells us tht A = ds /( ss 2 = ( /( = 205. (If you re ctg like biochemist, the units of A re 1/M.

4 Chem/Biochem 471 Exm 2 11/18/09 Pge 4 of 7 3. Frog muscle cells nd the solution bthg the cells cont ions t different concentrtions. The concentrtions of N +, C 2+, + nd Cl - re shown the tble. The temperture is 25 C. ion N + C 2+ + Cl - trcellulr concentrtion (mm extrcellulr concentrtion (mm pts ech The restg frog muscle membrne potentil is determed by the equilibrium of + cross the cell membrne. Wht is the voltge side the frog cell if the side solution is t zero volts? The formul on the formul sheet is RT. Here let s cll ΔΦ eq = Φ Φ zf extrcellulr nd trcellulr. Evlutg the eqution gives 3 (8.3145J mol (298 (10 10 = (0.0257V ( = V = 70.4mV 3 ( + 1(96,500Jmol V ( Φ Φ RT zf. We re given tht Φ = 0 mv, so tht mens tht Φ = mv. b Wht is the chemicl potentil difference μ extrcell - μ trcell for Cl - cross the restg frog muscle membrne? We hve tht Δμ A = RT + zfδφ. Ag, we note tht this is μ - μ, so here Cl, ex _ cell μ μ = RT + zf ( Φ = (8.3145J mol (298 + ( (96,500JV mol { 0V ( V } ex _ cell _ cell ex _ cell Φ _ cell Cl, _ cell = 6791Jmol 6794Jmol 3 J/mol. This is so smll it is likely tht the difference is ctully zero, i.e. the Cl - is lso t equilibrium. c Which of the four ions is t equilibrium cross the frog cell membrne? Expl. 3 3 N + C 2+ + Cl - Clcultg the chemicl potentil difference for ech of the ions (s we did prt b gives Δμ N = 12.5 kj/mol, Δμ C = 39.1 kj/mol, Δμ = 0 kj/mol, nd Δμ Cl = kj/mol. Clerly N + nd C 2+ re fr from equilibrium while + is exctly t equilibrium. One cn rgue tht Cl - is with roundoff error of 0, so it too is t equilibrium. If not, it is very close. d Suppose mutnt frog hd no + chnnels (i.e. + could not pss through the cell membrne. The voltge side the frog muscle cell is mesured to be 59 mv more positive thn the side solution. Which ion is controllg the membrne potentil for the mutnt frog muscle cell? (You my ssume tht only one ion is control nd tht it is t equilibrium cross the membrne. For ny ion t equilibrium, the Donnn potentil must hold, or ΔΦ eq = Φ Φ RT zf. We know tht Cl - won t stisfy this eqution from our results prt b. Checkg the vlues for N + nd C 2+ we hve tht (8.3145J mol ( ΔΦ mv nd N 59 ( + 1(96,500JV (8.3145J mol ( ΔΦ C = 03mV. So N + is t equilibrium becuse Φ - Φ = 7 ( + 2(96,500JV mol mv. mol 0.012

5 5 pts ech Chem/Biochem 471 Exm 2 11/18/09 Pge 5 of 7 4. The enzyme formyltetrhydrofolte synthetse employs ATP hydrolysis to form crbonnitrogen bond between tetrhydrofolte nd formte. The enzyme hs four bdg sites for ATP. In n equilibrium dilysis experiment, 1 10 M enzyme (E is plced dilysis bg. The bg is put solution contg ATP. When the bdg comes to equilibrium, there is 1 10 M ATP side the bg nd 3 10 M ATP side the bg. Wht is the concentrtion of ATP tht is bound to E the bg? The totl mount of ATP the bg is the sum of free plus bound. We know tht [ATP] = [ATP] becuse ATP cn freely move cross the dilysis membrne. Thus [ATP] bound + [ATP] free = 3 10 M nd [ATP] free = 1 10 M, so [ATP] bound = 2 10 M. b If the four ATP bdg sites re identicl nd do not terct with ech other, wht is the bdg ssocition constnt for ATP bdg to sgle site on E? [ bnd N The expression for the verge # of bound ATPs is =ν = which, when known [ P] tot 2 10 M 4 A vlues re serted gives = 2 =. We cn rerrnge this to get A = 1 10 M ( 1 10 M ( 1 10 M A 1/( M = M -1. (Note: if you re ctg like chemist rther thn biochemist, the concentrtions bove re relly ctivities, so A would hve no units. c If the four ATP bdg sites re identicl nd do not terct with ech other, t wht concentrtion of ATP will hlf of the enzyme sites be occupied by ATP? We know tht 4 N 4( 1 10 M [ ν =. When ν = 2, this eqution becomes 2 4 [ ( 1 10 M [ A =. It s firly esy to see tht this is true when [ = 1/ A = 1/( M -1 = M = 100 μm. d If the four ATP bdg sites re positively coopertive with Hill coefficient of 2.3 nd the bdg constnt is the sme s prt b, t wht concentrtion of ATP will hlf of the enzyme sites be occupied by ATP? n Here we hve f =. When f = 0.5, then n A [ n = 1. Tht rerrnges to 1 1 [ L ] = =. If your clcultor cn t do this sort of root, you cn solve this by n 2.3 ( A tkg the log of both sides: [ L ] = ( 1 10 ( , nd then exponentitg to get the fl nswer [ = We hve hd to work like chemists with unitless A nd ctivity here becuse it mkes no sense to tke the 2.3 root of molr, so the units of the fl nswer when it is converted to concentrtion will be molr.

6 Chem/Biochem 471 Exm 2 11/18/09 Pge 6 of 7 5. The three isotopes of hydrogen, H (hydrogen, D (deuterium nd T (tritium, ll hve collisionl dimeters σ equl to 2.5 Å, or m. The moleculr weight of H 2 is 2 g/mol, of D 2 is 4 g/mol nd of T 2 is 6 g/mol. You my ssume tht smples of ll three gses behve idelly nd tht they re t 0 C nd 1 tm pressure. 5 pts ech Wht is the most probble velocity of H 2 molecule under these conditions? 2kBT ump =. The mss of sgle hydrogen molecule is m = M/N A = (2 g/mol/( m 1/mol = g = kg. Pluggg to the formul gives -23 2( J / (273 = = 1510 m/sec kg u mp 27 b Which of the three gses, H 2, D 2 or T 2, hs the smllest vlue for the root men squre speed of molecule? Expl. 3k BT Becuse urms = depends versely on the squre root of molecule mss, the gs with the m lrgest mss will hve the smllest vlue of u rms. Tht would be T 2. c Which of the three gses, H 2, D 2 or T 2, hs the smllest vlue for the verge ketic energy of molecule? Expl. 3 Averge.E. is. E. = RT. All three gses re t the sme temperture, so ll three hve the 2 sme vlue for verge.e. of molecule. You cn get this nother wy by computg the verge of.e. = ½ mu 2 : <.E.> = ½ m<u 2 >. Pluggg the vlue for the rms speed cuses m to cncel nd you get the sme result. d For which of the three gses, H 2, D 2 or T 2, will gs molecule fly frthest on verge before collidg with nother gs molecule? Expl. We know tht the men free pth, the verge distnce flown by gs molecule, is 1 l =. The three gses hve the sme collisionl dimeter σ nd sme density N/V N 2 2π σ V (this from the idel gs lw nd the fct tht they re t the sme T nd P. Thus, ll fly, on verge, the sme distnce between collisions.

7 Chem/Biochem 471 Exm 2 11/18/09 Pge 7 of 7 Scrtch pd: For grdg purposes: question Tot score