Final Answers - 2009SPRING

Final Answers - 2009SPRING - Name Final Exam MCB C100A/...

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Unformatted text preview: Name Final Exam MCB C100A/ Chem 0130 Spring 2009 i I ## pages total including cover 1a). Below is the structure of oseltamivir, an antiviral drug marketed as TamifluTM. Tamiflu is etfectiye against the recently emerged H1 N1 influenza strain. The compound is an orally bioavailable “prodrug” that is hydrolyzed in vivo to the carboxylate. ls TamitluTM a nucleoside analogue and if so, What base does it mimic? (Explain briefly) (5 pts.) *5 ; Not at— tS not' a flbkwwftdi comato‘jwcu HM Eli/mp: + {immersed rower Olin/H— MCIL+C/tf\ 9% ' -Mr€ (S no “aircrew m lch Hing I I I. ’ Hkg no? am.WLh"¢ GK '— dQQJ‘fi ’r V‘an at Sugar «irraton 1b). If a mutation weakens drug binding by 11.7 lemole at 37 C, calculate the ratio of dissociation constants of drug for the mutant and wild type enzymes at 37 C. (5 pts.) AGO ': ébfimu’sszn:=~u1ryhsiz—srmkmueeTmKw+ - " LC il‘ “H71 l‘T/Wl : @TUoKwtzlnL/mwA' iZTm Tim”? Karl- r KG. illeOii/lmoiw /él51q§ f/mcftysm ; O. 0101 \CMUJ’ \chi [if a5. mumfida& kid ‘5? 10). Using 1b), how much higher concentration of the drug in 00 would be required to inhibit the mutant enzyme to the same degree as the normal dose inhibits the wild—type protein? (5 pts.) 1 __ 1 I i t v ' ii i: 7 ‘ ’ [L3 lCJ Ira-wit] VEL‘E; — + {Mob - (Ll-Md (LILQ-i' i _+ l:de m if [civic [UM-t ~ cog—— KJMU} _ [demot- Kdmd : 1d). The enzyme to which Tamiflu binds has a molecular weight of 50 kD. What approach would you use to determine structural details of the binding site around the drug molecule? (5 pts.) Cngfiwsrr‘aka: r N‘Mifl (crux do ‘S‘Wpt‘l/V‘fj 9%» lb 301C Dq. 1f. The target of Tamiflu is a viral enzyme of ~450 acids. In different influenza isolates, the length of the protein can vary by over 20 amino acids. In What part of the structure do you think the insertions and detetions are most likely to occur? Briefly justify your answer. (5 pts.) iiifiit- Pix-r'i-m‘é.‘ ‘T. 7‘31,“ L C» «‘7‘! ,4, is.) ‘63?!) mi i€ >9 J Pom; {'vdi. I iuqvkci-t t\'\) q "t Alfie, termini 19. Suppose you are investigating a Tamiflu-resistant strain isolated from a student at Malcolm X Elementary School in Berkeley. Compared to the closest related drug—sensitive strain in your collection, you discover 26 changes in the nucleotide sequence of the gene but only 10 changes in the amino acid sequence of the target protein. Briefly explain why there could be more changes in the gene sequence than the protein sequence. (5 pts.) PX ch“ 3 C 6 many ti ta\ fig 9:; W 5 . 531 .« ww,‘ 1 c"- . it 7 ,. the“ L‘XC’L' ‘ V“ §i't\ Sid-r Ci" ‘~'\ (tan (Wit‘x (31; (R (K. 1h. Do you think the ten amino—acid substitutions in the Tamlflu—resistant target are likely to change the fold of the protein? Why or why not? (5 pts.) 1‘3 , ‘ -. . . . 5M 5“, I”; ('Ltt'xit" pl i’\!' i?,¥1két.ll" u- ‘t 13‘ a av‘J- “x We 9414?“; id aux-Latier d‘ 1i. A key question to answer in order to understand the mechanism of drug resistance is to define which sequence changes weaken drug binding. Propose a computational method you would use analyze the sequence of the mutant target protein to identify which of the 10 substitutions are most likely to weaken drug binding. Using this method, what characteristic (result) would be likely to identify the residues that confer resistance and why? Hint: There are many sequences of different flu isolates available. (5 pts.) M 55’} '1 K €13 r-' r~z\o‘~'if Va“) 1 e 1% rue-K re LEA“ 5 . . ‘ , .r - i .7 {testifate Li a}; ‘l’Lx as EMA; W3 (alt-11*i a Leo ’\ r). + - a . ‘ ‘ 'L (a. 1'.“ ‘ 7' r, _ ,- _. ‘ {fl whrck {Ma-(- ‘7’“ 5x747, 'i’m ifiv "‘ ' (at; x: gun-v ca“ *2“ “5‘ ‘i “' (Mk c cvxhk mg) i E; Lt M“ i.'\ <4; ‘1: ‘Y'irfi‘a‘i‘ “‘3 2. True/false. (2 pts. each) 2a. Loops in proteins are characterized by different mai ' dihedral angles than secondary structures. TRUE ALSE 2b. RNA viruses (like HIV) make DNA copies that are tr "Ted directly into proteins. TRUE " ALSE; // 2c. Membrane protein structures generally sequester e nonpolar groups away from the membrane. TRUE FALSE») 2d. Glycine causes alpha helices to bend. - TRUE FALSE 2e. A hit in a BLAST search using a protein query means that the query and subject have the same function. r 7-“ TRUE FALS 2f. More rotamers will be discovered as more protein structures are determined. TRUE FALSE) 3) Give a brief explanation: 3a. When nucleic acids make outer-sphere contacts with metal ions, what penalty must be paid that is not paid in diffuse interactions? (3 pts.) ,- . f K ‘ ‘ 106‘: cpl-6‘62""; c5: ‘ts’au'wélfii‘; for; Jr. R 3b. What further penalty is paid in inner—sphere interactions with metal ions (that is not paid in outer-sphere contacts)? (3 pts.) C>\'fi?l\tttt.\“’\€t\'Jl 11-3: NR r ” Q‘TW’JPE Pénfiwfi 3c. What interactions are likjlo lead to a specific magnesium binding site Within a folded RNA structure? (2 PtS-) , mm 1i, $1M? C'x“ {IV/V“? PC 5 ‘ up“? A l V\ t; K0 '3 c7. Ks?) ‘th 3 “M kw, lvx FAQ-4.;— W1 ac flay? ion. 4) A researcher wanted to send a sample to a collaborator, and needed to make sure it stayed frozen during shipment. He wanted to pack it in a styrofoam box filled with 1 kg dry ice (at —78 C). He knew dry ice sublimated (evaporated directly from solid to vapor), and found the AHsubnmauon = 571 Jlg at — ~78 C. He looked up equations for heat transfer, and found that the rate of heat transfer, dqldt, depends on the temperature difference across the heat barrier (the styrofoam box), with a coefficient that depends on the material and on the surface area: dqldt = k (Tenv — Tinside). To calibrate in a manner similarto calibrating a calorimeter, he carefully measured 100 g of water at O C and put it into the styrofoam container, which was in a room at 25 C. He measured the temperature of the water, and found it had increased by 1 C after 10 minutes. He knew Cp liq water = 4.18 JlglK. Calculate how long the styrofoam box would keep the sample at — 78 C in surroundings at 25 C (tie. how long before all the dry ice evaporated). Show your calculations. (20 pts.) “2;. K s was“ by 2.59 =1 k (Tc-av ' Tingl‘l...) 4% Ow) : K<1S-*0) [0 ml» H113 = K05) Tcnv fl L'I-d-L) k( T A11 :- Ejl—‘E—J : 13310 M3“ (LG? rm“ icSk.) 2 gash :2 2-3 ij5 2: lol‘izoo Sect. 51%) The enzyme T4 lysozyme has two charged residues that are near each other and relatiirely buried in the inside of the protein. a histidine and an aspartic acid. The graphfibelowgshowga measurement of one proton resonance of the side chain of the histidine for the wild type T4 lysozyme, and for a mutant in which the aspartic acid was replaced by an asparagine. To 4‘ . . . . . . . .. . . . . . . . . . . . . . Wwwmmf. yo . ' _ 2 I _.__ i «to j - _ 65 "1 a I I I ' I ' ibo —- r/d 3%” a: 53° A . 9 ':~ ,. 3‘" .2: ' x/ as, R ' E ‘ i t i ; w ‘ ‘3 $5 '1 m M} m: w 2 . if f; I g E 1 ago i... XX, h "In . a ‘ j a 2 t ' _ “‘ §‘ ; E ‘l ' — 45 -‘ if in ,3 3 . . ; z? - ? V’s: _. - x 4D - i “‘ O ’ i - E f. 7.5 «E I . m 35 A if") , - ' j; . 7°WWW“W"—T‘_TT~TTT‘ 3°““““WWW 4 5 a 9" a .9 to u 12 2 :3 A a 6 r a 9 la b) identify which curve you think‘corresponds to the wild type protein, and which to the mutant, explain your reasoning briefly. ’ 1 t1" c) The stability of the protein (as assessed by melting temperature on the vertical axis in the graph gelatin) was measured at different pH values for the wiid type protein (open squares) and the mutant "protein (solid triangles). Draw a structure of how the histidine and aspartate might interact at pH 5. . v 5 t .- ‘ : xi: ‘ 1, ‘: R - n d) EXpla‘in briefly Why the Wild type protein 10593 Stability at PH 10, and why it loses more stability at pH 2.5. .13"; 1;. 2 ‘t x ' ‘z/ff-tl. “fit...--. [if if 6) A student read about some bacteria which preferred to use lactic acid as a carbon and energy source rather than glucose. She thought since part ofthe tactic acid as more oxidized than the carbons in glucose it might be a poorer energy source, but then realized lactic acid also has a methyl group that is more reduced than in glucose - so she wasn’t sure which was better. She found the following data in a table of thermodynamic values: AHr° AGr° lactic acid (03H503) -694 kJ/mol —523 kJ/mol glucose (C6H1206) 4274 kJ/mol -910 kJ/mol carbon dioxide (C02) -393 lemol —394 kJ/mol water (H20) —286 kJ/mol -237 kJ/mol Assuming that both compounds are ‘burned’ completely by the cells to give 002 and water, decide whether the lactic acid or glucose has the potential to produce more useful energy for the cells w gram of fuel utilized (for example for ‘swimming’ in chemotaxis toward nutrients). Show your work. (20 pts) \ .;‘ ,. K ' ix (ii-hid l," «_ ‘ _ 1 .QGIW‘MA ml) by]; H13“? ~: ’46»; + 46c; ; 2 5(5L;> i (—Sfi‘lr23a . ._ ‘ Hit it“; _‘ a I i . fig-gig. .13 COL-Jr h7Q ; -—f_\(,-j. + ALE? __ filo ., (@1113?) ‘ ’ \ t: (K . \ C(‘Wui‘CQKC A G L a Jud“? .» .E‘e 3‘04 i 9° u \ *3 may 1w i 4 1-“ i. A (if cit. r695)?!“ if” 27‘ f5 ‘KKH .Ul I Ii’Q‘l "Iii/gal iv“? t1§/.:.L..i > “ill 14/”; . . {L .{ H, c. lgb‘LU‘T‘f LC We} 1 65+ (3‘4l' (“u \Lk 5 a, g Shiv-"‘5" (HAL If) I”) it? 6M); \abkt 7) Consider the stability curves drawn below for four different monomeric proteins, numbered 1 — 4. Temperature a) If the temperature of samples of these four proteins is increased slowly starting at 10 C, so that equilibrium is maintained, which protein Will unfold first? (5 pts) 45‘ ‘VCV‘GSBU AQCG “Vi-“5“? b) Which protein would be most susceptible to cold denaturatlon ? (5 pts) *3: l .-—-'V Nom‘i ARA/Lk-kwe 509on 0°C c) Which protein has the lowest value ofACp (the difference in heat capacity between the unfolded and folded states)? (5 pts) fig-Vlow :10” Lurvk+wfi d) Estimate the fraction of protein 4 which would be unfolded at body temperature of 37 C. (5 pts) Afiuqvlg ‘2’: 2-7 KT/Y‘nal (Alf 37°C- [M] ,110003— K21: 1:. e Wink) 2 2.8)‘19‘5 8) A student was looking at reactions for reducing disulfide bonds in proteins. She found that a natural reductant in cells is glutathione = GSH, made from three amino acids: She found another chemical, dithiothreitol = DTT, OWN - . = t H that also can form a disulfide as shown. GSH (imam/k ’3; ijgm}; 9H “(3% 52H : SH ” ,3 ‘2 DTT = “SW (C3 0H H0 3) What is unusual about the structure of GSH as a polypeptide ? (5 pts) 3 ‘ .. 64mm? etc/{A 13 shmlvtcl m m Witch bead 'Wujk (l1 Edit/Wm“. K b) She found standard half cell potentials (both with H2/H+ as reference) for the two reactions: GS—SG (disulfide) + 2e' ——> 2 GSH 5° = - 0.23 V DTT (disulfide) + 2e" ——> DTT (2 x SH) 5" = - 0.33 V (gnu, If she mixed 1 mM DTT(2xSH) and 1 lel GS-SG, what would dominant species be when the system reached equilibrium? (5 pts) Ovfiw mm: @‘SSCB at b’TT (QXSQ—fi b'i‘thugoAMb *— QGSH _,.—— "13‘ - 3 O‘ \C) V DOM\Q.¢;\‘J’\+ BWUHfS Ctr—C bl 0AA c) The student knew that a protein she was interested in had a folded structure that was stabilized by an intramolecular disulfide bond. She wanted to reduce the disulfide with DTT(2xSH) to study the stability of the reduced protein. Will it be easier to reduce the protein’s disulfide or the disulfide in GS— SG? (explain briefly). (5 pts) Cease mu w tastier i-o redone, (M ’lVUch ET ova 0W1 ache/\tficr +c ‘ ' ~\- ~‘ - . GSH ‘3‘ W lg not ctJ‘ SH AFFQWV oil M ['Y ‘\Z G356? \A’f'D “Ur-3° \ _v > ‘ y \ n W \‘ EEC ELCfiMtu - (h M moi-mm. N'l—Unouh , “AL 9mm 5 (145%th ('3 ll ' 1‘0 lat. lowrccl En 3M Infieler vmsicua w l (at; Solvm’f ach-y I,th d) The student read that the reaction of reduced DTT with an oxidized protein is much more effective at high pH, with the implication that it accelerates the rate of reaction. She remembered that sulfur often does nucleophilic displacement reactions, give a kinetic scheme that is consistent with a higher rate of reaction at high pH. (5 pts) bTT —Si—l + 0H — i.) fiber—l: 36 ., Web—SIS -—=:. “’t" "DTT-S'me +8” Pots-P \nffiw’ {all : men: OH’ _ MPmchtJ 'bW’T—ge (x R W nucfiwpl/trk 9) A postdoctoral researcher was studying an enzyme. He had a compound that was very closely related to the real substrate, but lacked a key functional group that allowed it to react. He studied the binding of this compound to the enzyme and found that both binding and dissociation were very fast, and that the dissociation constant of this compound from the enzyme was 10'3 M. He then measured the rates of reaction of the enzyme and got the following results: temperature 7 C: k2 = 100 5'1 KM = 1.0 x10‘4 M temperature 27 C: k2 = 200 3'1 KM = 1.5 x 10'4 M a) He estimated that in a cell the enzyme would be present at 1 x 1O"5 M concentration, and substrate would be present at 10 mM concentration. Calculate the rate of formation of product that he should expect at 27 C. (7 pts) [)0 r. VFW-xx k1 Clap 5-‘(1‘1‘lu—bM5 _ 2 MT 1 GABON”? Ms“ lat—KM H_ KM W Lg‘l‘h: M [5] C53 -1, Fr-=='— M = fl 4341. b) If he did an assay in solution, again with enzyme at 1 x 10'5 M and at 27 C, in which he had a 10 mM concentration ofthe substrate mimic compound as well as 10 mM concentration of the substrate, what would the rate of product formation be? (7 pts) . . ‘ ‘ . 2 2 l0- Coch-lci-l-i/c \H'Hi‘al‘l-K 1 ~‘t -3 was, KN}: arr mil); Wm (H. lJr— KW‘ 3 1 10—3 [33 3: I.L57‘lo" V0 1:. KL‘CE-Jo @DO Sec—bl)(1$1°* M) "‘ z 1 0.00 H in] H l(.S>u.'3 WM E53 lOflV} ’“ 11%; 0) Estimate the enthalpy and entropy of binding of substrate to this enzyme. i p SIUQLL bfiio‘F“: Fact ‘FGI-rb'k Kc! 1 . - '3 {or \mmvh “Grok Kg *‘P 27 1: J" -.-. L ‘-‘ yr] KLKnolifij KM [.S‘Ho’H (ch/(“D t , Li Klalk'vtllnj 7‘ 1:5 ‘- 1° QC) (,n'Hnu-«Sef'H pa“ =1 "KT‘LAKI =1 AHFT‘AS Sula. .15”. a? ELM-{£ibn3 .- AGML = “RTLlV‘KIL‘ AH-‘TLAS fl "fi'TflvxK, '* RTLLAKL a O “Tubs +1315; R(T‘L\V‘K1 "T:‘“K\) 7‘ AS(T1*T') AS: Tl\\'\kl) fl - 9L (TL “To K1“ Kuhn“? 17'" K2. 7- Bihiiwj = (8-3\"\)(3c>o x“ (9.7m?) « ZQOMACIOLU) f__dffifl__fl*_wf___m 36b “2.590 e. v; 29.6; 1 \p med-K «cw AH”- pcq -= ~R7M<= aH—T» AH; *KTlhK 4 T65. = ,(giglfi)(30b) \n(£.7~+to’) + (goox 2.4.4) : rnflfis'fhwl AH : — \3 .q “fr/moi 10) After learning about membranes in cells a student realized that proteins must cross the membranes to get to their final locations in organelles (membrane separated compartments within the cell). She had recently learned about membrane potentials, and also had read that there are protein channels through which the proteins must pass. a) Suppose a protein has a net charge of —4, and is transported across a membrane for which the membrane potential is 150 mV, moving from the more negative side to the more positive side. Assume the protein is initially folded, and must be unfolded to fit through the channel. What is the maximum stability of a protein that could be pulled through the pore by this electrical driving force? (7 pts) As a w PM, : s timeszvofilo. leis} b) She also heard that some channels use energy from ATP hydrolysis rather than a potential to accomplish the transport of proteins across the membrane. She remembered that AG" for ATP ’ hydrolysis is -40 lemole, and concentrations in the cell of [ATP] = 10 mM ; [ADP] = 100 [M ; [P04] = ‘10 lel. Assuming that a transported protein has a free energy of folding of —45 kJ/mole, and needs to be completely unfolded to be transported, what is the minimum number ofATP molecules would have to be hydrolyzed to accomplish transport at 25 C. (Show your work) (7 pts) \MJ/ ,» , a . : AMER] r (Loo tM}( 10mm) -6 At) ‘ 4' if) Q Q _ f V /’ : 0 7‘7 M £36 1 “’40 KT/W‘Ol +é73‘rls‘tl03rr moi-L]{ahl>h\ In lGOK/Otc 1 *’ (E752. 8" [CT/moi QWEL at PFW “fish-ler 1-inch." outbach madman/Hr qémjiff imai HTPX : oé—iifo m-oumlli‘rtfi‘ Wok gruff-M C007,? [(T/runl : Msum imam a h. Moi Pfi'i'C—‘UH c) The student had focused on stability as an important factor in transport, but when she looked up rates oftransport for a number of different proteins that all had very similar stability (in terms of free energy of folding at the temperature that transport occurred) and length in amino acids, she found that they varied widely. Explain briefly Why these results are not contradictory. (6 pts) W M a} whim our—L MFDiCLtci Mai. \irotAt'lpafl—{éi s: be merw—Q [3 a PM”: {Mum HOv'm- lixbrmajla‘xx CLO/13$ F g I 7 about} but grab,-[.i (Le “Mic/Vi odjmcrm r553 O; W Frwfim S. Wmubzs no insith ‘13 by ‘W‘\/ )CU/bUhZ [gel/lawful“. ...
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Final Answers - 2009SPRING - Name Final Exam MCB C100A/...

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