Final Answers - 2008SPRING

Final Answers - 2008SPRING - Name E5 1 M08 C100'A Chem 0130...

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Unformatted text preview: Name E5 1 M08 C100'A / Chem 0130 Final Exam Spring 2008 9 pages total including cover 11 problems POSSIBLY USEFUL INFORMATION: R = 8.314 J/mol/K = 1.987 cal/moI/K = 0.082 L atm/K kB = 1.88 x 10‘23 J/K NA=6.02. X 1023 molec/mole F = 96,500 coulomb RT/F = 0.0257 1 V coulomb = 1 J - properties of water: ice liquid water water vapor CD 2.1 .J/K/g 4.2 J/K/g 1.8 J/K/g density of ice: 0.95 g/ml of water 1.00 g/ml of water vapor (at 1 atm) 8 x 10'4 g/ml - heat of melting of ice 333 J/g heat of vaporization of water 2216 J/g Problem 1. Truelfalse or multiple choice questions, circle the correct answer for each part (2 pts each). a) Disulfide bonds assist in correct protein folding mainly because they make the folded state more energetically favorable relative to the unfolded state. TRUE FALSE/ b) Alpha helices are the most common secondary structural motif in membrane proteins. C TRUE/i FALSE c) Loop regions in proteins are almost always unstructured and ree-floating in solution. - TRUE ALSE d) Which of the following amino acid best starts ,lpha helix? ' Y P G D e) Which of the following folds commonly binds metals? i) Helix—turn-helix ii oiled coil _ fill ossman fold Problem 2. (10 pts) Briefly explain Levinthai's paradox. What is one possible resolution of the paradox? ’h/w ci'rmsc “he with W [9%ch (me‘zjg. TIM-*1 i‘S Lia Frosting! 55mm cm “fiaan 5;, M455: WW Wit, lot/H01 Mwambm f5 IN. , Mala The poking Problem 3. a) Circle the amino acid that is more hydrophobic. Briefly rationalize your choice. (2 pts each) Tyr T}? L145 .a'. Fall}? '0“ apart?* Plat W m begafirgrgww} E’EM?‘ vai léy 5”“ W WWW! [W17 11-" I“?! a busy» All“); Limp. Lys (“if has a “S” W, wLu‘a‘; is not Fate/tr, W‘lg Lg; he «hit; 2 Ply)”, 0,3555% b) “Base~fiipping" enzymes that bind DNA with an unstacked base generally insert a hydrophobic residue into the space vacated by the flipped-out base. Explain how inserting a hydrophobic residue, rather than a polar one, might aid the functions of these enzymes. (Tpts) Wpiwkrk ink/«443% m DNA Hm ,‘Mfmkmf #6 DNA sbwfyuy who,“ a m; is Wit; Sw'ws {‘34 flit-fit WAJ‘a; 16‘! fix Q1251; 11M rfidcaf If Vam-frr- Stbb'i'lir‘g flue 523%va WW! Wm" “ hag-Av!)th Magus Shiite» flu» (aloud-bu . c) B-sheets in membrane proteins do not have the same variability as [ii-sheets in soluble proteins. Which class of B-sheet do you think is most enriched in membrane proteins and why? (7 pts) up — W» dowm i3 breakage. 7L4}; Ma «Ma 4; {46m Ferric-5’ Mr my W1; Wham, iv. . * K — mar-'5 rm. ” g rrk 11. m— . “Amway mattmdsm ? bf f0 w mém. [i IIIN 1.“. a”? fir r." if; :1 ——-— - Problem 4. (5 pts each part) a) Propose a possible explanation for why DNA contains thymidine and RNA contains uracil. lat .. 2? MW m w 14AM "*4 ewe?“ ’5’ “jg I“ DNA; TIM} W 5W?” Wflf‘i’a’m W? W 0/ l o T L” Puma Wch it" "3‘5 3 “*- I’W‘” 5“ W" b) What characteristic of folded RNAs aids in the prediction of RNA structure from the sequences of homotogs from different organisms? 77M Mtflcasr 33 RNA M flu A’m‘m Jaiml’egfas. c) What is an “A-minor” motif and how does it stabilize RNA structures? Eire/«Aw! RNA “190% Marlin! {ml-a flu WW» 3/1099'6 g a RM“, Imam? My??? mlmhm 1hr meat-a In“ shady”. d) Do you expect an A-minor motif to form in single stranded DNA? Why or why not? N0. 1415 a W W‘WW $1609? [nu/f Me-Wja'j’ alcpmagnh ’11,“), gent 5.3m: - Problem 5. a) The mycobacterial protein called RV3910 contains a domain that gives BLAST hits (in a search of the nonredundant database) with E-values >104D against a family of protein kinases found in all three kingdoms of life. For example, the ~280—amino-acid Rv3910 domain shows 521159 identities to the sequence of the top hit. What does an E-value of 10"" mean? (6 pts) ' 1‘: flat Fmbddfliig 7 fig}? flu Wm}, A W? law numbeejmé M 10'"), W 71w!“ TL: palm l:‘ W? g;m'l%- *0 WM” “W F M was dang; Midi-M. flu 5 iii-é”, 0’: “Expat-lnliwa ‘ b) What do these results imply about the evolution of Rv3910? (7 pts) PVSQID is M fa with. law“, aw»! [9959's]; W 1'1! W M W palm. M W avails-9., M {flaw-f Minn: WM‘ HS . c) List two kinds of characteristics you would look for in the sequence to judge. if the Rv3910 domain is a protein kinase. (7 pts) " ’4 ' Wm" “MI” W“- K «1 Woe-Mr er-vwfilg mg?) 1: A meme EMU-ta “gm/«u M“ ” if flu! Fwdxchj fill ,‘5 12¢ 3m M 0124/, Erma: mm 9M 01 Wm WM MW at I Efrw,m *’ am {Ir-9AM fast: £27579! Problem 6. Consider a system comprised of four normal coins, a penny. a nickel, a dime and a quarter. a) When the four coins are flipped, what is the probability that all are heads? (6 pts) (4.)” b) What is the probability that the nickel and dime are the same (both heads or both tails)? (7 pts), T (a) w~ c) What is the multiplicity of the result that two of the coins (any two) are heads? (7 pts) ' Lit ' Wife,” ’ [2 2.5. l \ Problem 7. - Consider a system of two flasks of 1 liter volume each that are connected by a tube with a valve. One of the flasks is Completely empty (a vacuum), while the other contains 0.05 mole of helium, an ideal gas. Both are at 25°C. The valve is opened connecting the two flasks, and the system is allowed to reach equilibrium. a) circle all of the statements below that are true: (8 pts total) described is isothermal. <5 P W/ wept-M" fleafians‘ionofthe gas did work against the surroundings. I+ slid “I”; é‘m’rm‘ ’n‘W W W 3“ t“ it. W“; m i. at, m The process described had a AH value < 0. bit a o _ Since this process was spontaneous the entropy of the surroundings increased. Summits? 21hr “fid‘ééfigfig - cg 11% 539% ,‘iflgwwg‘gg. b) Calculate the change in free energy of the system for the process described. (6 pts) W ’“m’ “Sm : 4&1“le : web/ii) , u- _ v II.» V: :(«35 at!" W c) At equilibrium calculate the approximate difference in the number of atoms of helium you might expect to see on each side of the flask. (6 pts) 7144"; is T!“ ilmeuu (“3”) 63’ flu New"! fix‘ifm‘imfivw lb 1M 3M x 2‘? “02? mm; 1' Ali/NU" P) " ffawe‘ftmfi)(i)[i) a 15x10” (3 “’3- 0- u 0” = 3 swat-9* 23713.7 X/Dgpj M This '5 “I? V] “0”? wife; We r I m £0 that mail. Problem 8. In the table of standard reduction potentials in the text two values are given for the reduction of flavin adenine dinucleotide (an important energy carrier in the cell). (3“ = biochemist's standard state, assume everything is 25°C) ' For free FAD: FAD + 2H+ + 2e' 9 FADHZ 8‘” = -0.219 V For protein bound FAD: (FAD)p + 2H+ + 2e‘ 9 (FADH2)p a“ -= -0.040 V (a 'tlavoprotein’) a) An electrochemcal cell is created, one half cell with an inert platinum electrode and 10 pM each of FAD and FADHZ and the other cell with the same electrode with 10 pM each of (FAD)p and (FADHQP both at pH=7. If these are connected with wires to a voltmeter, what would the cell potential be? specify which side wilt be the oxidation reaction and which will be the reduction. (6 pts) , — .0 0V + 0.; *7! - 136M “ye—qwim‘ I v FAD”; --" FAD 4~ 24,!" + 25 fixr‘abw'i‘w (FAD)‘,-s' 2g'+2+it _* (mom)? Mean II "’23" I: uh ti b) What would be the free energy of reaction under these conditions? (6 pts) Ag, 5 —o FM“ 1: —. (2)(?é§aaC3{9-l?‘7 V) = raasa? TIM " c) Now consider the process of FAD and FADHZ binding to the apoflavoprotein. Use the idea of a thermodynamic cycle to determine whether FAD or FADHZ binds more tightly to the flavoprotein, grit what the ratio of binding constants must be. (8 pts) (apoftavoprotein is just the flavin binding protein before any form of flavin binds) FA be; {FADE :2 be,” -— AG; :_ gets éT/mj DH.» “M I Pr 5- ..v n a w a r AGO j L M". 2r My. la, 64:23) at. a, 7, l6 _ FAD -"”’ (FAD)? I 16‘" i2:— : ’13.?“ AG? ‘ “EMF” EL 4 2 3 t? V K2: was“ :1: o 0 a 9 A656 I ‘5‘} A631: béznrbfitg +564 5 l ‘ F r; ' Ag" : FUFg IAGIQ ': ~2ffgg§9§€5(' Oil‘de +952 “L 1(7ér5‘905)(_’r (3.0?ov) ‘5‘ 5 3 p 3 “ Fm“ K‘ t 3 a, 5y? I .,. Act; A c a _ o W "J ‘ '= A6“ ' kuF’g‘ in”. A630 f / Problem 9. The Ca+2 ATPase pumps calcium ions from a low concentration to a high concentration. As shown in lecture, tflg Ca+2 ions are pumped at a time using the energy of hydrolysis of one molecule of ATP. a) For ATP hydrolysis to ADP + Pi AG° = -31 ktllfmole, assume everything is 25°C. Under cellular conditions [ATP] = 10 mM, [ADP] = 100 uM, [Pi] = 1 mM. Calculate the maximum concentration difference that this pump could achieve using the energy from ATP under these conditions (be careful about stoichiometry) (6 pts) A5: A60 + 127':in = M. + 3,1,3“ M] {ATP} 5 = "5 {lilo—Pl) egf L J. 33” I (yaoxm pt) I I m! WWW) ’3‘“ law-w +40 rut/Mr _ at? Mm Howsz mm): was => M— 2 W?» “W a g] 2—3 we I b) From structural studies of the CaATPase the two calcium binding sites were very close together, with one at the end of a short tube leading into the protein. The site in the back is completely occluded by the front site, and residues contacting the second calcium are brought into place by the binding of the first one. Draw a curve showing the fraction of calcium sites occupied as a function of the free calcium concentration. What would you eXpect the Hill coefficient to be? (7 pts) fagia‘ivc worwfi'n'f‘a / Sléumbfir, MUS .r_-__._______.‘_________ [a 71‘] c) The CaATPase cannot hydrolyze ATP unless it undergoes a conformational change induced by the binding of Ca+2 as described in b). Draw a curve showing rate of calcium pumping as a function of free calcium concentration when the ATP level is high. (7pts) (75%.: £542“) Problem 11. A new form of an enzyme was characterized, and it was found that it follows normal Michaelis—Menton behavior. At very high substrate concentration the rate of product formation was found to be 53 uM min‘, and by varying substrate concentration the rate dropped to 1/2 of this maximum value when [S] = 28 will. A compound that was known to inhibit a related enzyme was tested for activity with this new enzyme. The dissociation constant for the inhibitor to the enzyme was determined to be 17 uM. a) If the inhibitor competes with substrate for the active site of the enzyme. by what factor will the velocity (rate of product formation) be reduced reiative to uninhibited enzyme at the same substrate concentration if [S] = 0.37 leI and [i] = 0.48 lel ? (7 pts) 00 1‘53.) 9 up : Vm ‘1 lmjs‘ 5 M/ x 0 E M" . - ' m“ 7' ‘ 1 vi (ta/la swer Q ’33 ‘ 1’ M a” V3? 9 K» '— int/M TA” /"‘”‘ .j I K; s m “WWW _ i 63 Wm ' i do, M 2:. —\:-_'M_f_;_ 1.. 5" g7- M /Wh l fgj I 22?an 004W": g: 2.151;» = 5filmfl/m‘“ I a 53 “1’13- ; 14-5 M H far-:1 [‘4' IMO" 1/21) {*(13/M)[}+ 3.1 M is] {arr / 0.3? mm b) If the inhibitor instead binds to a site away from the active site, independently of substrate, by what factor will the velocity (rate of product formation) be reduced relative to uninhibited enzyme at the same substrate concentration if [S] = 0.37 mM and [I] = 0.48 mM ? (7 pts) IE 00; 15%: 1 Vmaxtl-fz) ’ M/m'a)(i‘0.‘?7’) V f5 1* 5'13 I+ 35L“. EW/M [:3 ii: I}? “t K; 0:75WM rf ' 0‘43 WM t 0.0!? “’M claw] C) Explain why the inhibitor is more effective for the situation in part b) than it is in a) given that it has the same dissociation constant in each case. (6 pts) Problem 10. Metals which are bound to proteins can have unpaired electrons in d—orbitals. The number of unpaired electrons is affected by the ligands around the metal (from the protein, or heme, or other cofactors). If _ electrons are unpaired then they will have a magnetic moment (they act like a tiny bar magnet), but if I they are paired there is no magnetic moment. These magnetic moments can be experimentally measured. Suppose there is an iron atom in a protein with electrons that can be either paired in a ground state with no magnetic moment, or unpaired in an excited state with a magnetic moment M, with no other states that can be accessed. a') If the energy difference between the ground and excited state is 1.5 Jlmoie, at what temperature will there be 1/3 of the molecules in the excited state and 2/3 in the ground state? (7 pts) _ ’ u N! A LS T M _‘ legfizq J— "; = a I”: puma” ' 2'- 1v Mfl é [.BE'Kloiwj/L'T e 3 2 e , , q a ,2.5Hg_:__3' % 2 a H6 :19.” Wig-r L ’ "9,13 ’6“ a ' as w—“”"‘“””'”“ T 1 ‘1...” s ‘0.” la] Lu W b) Draw a graph that shows the expected magnetic moment of a sample (proportional to the number of iron atoms in the excited state) as a function of temperature from very low to very high temperature. (7 pts) [I‘MJNt c) What will be the value of the heat capacity at high temperature for this system? (6 pts) am, am M11“ (Mu-4}; “We! {%b:[)l M W est—ix In absealucl 5-3 3%, gym. ...
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Final Answers - 2008SPRING - Name E5 1 M08 C100'A Chem 0130...

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