BIS102 Final S97 Key Hjelmeland

BIS102 Final S97 Key Hjelmeland - Name Student ID. No. S 8...

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Unformatted text preview: Name Student ID. No. S 8 7 Last First W / FINAL ' :3 ’ SPRING QUARTER: JUNE 9, 1997 BIS-102: Leonard M. Hjelmeland INSTRUCTIONS log 1.0 = 0.00 log 1.5 = 0.18 log 2.0 = 0.30 log 2.5 = 0.40 log 3.0 = 0.48 log 3.5 = 0.54 Alanine 2.4 9.7 - log 4‘0 " 0‘60 Arginine 2.2 9.0 12.5 103 4-5 = 0'65 Asparagine 2.0 3.3 - 108 5-0 = 0-70 Aspartic acid 2.1 9.3 3.9 103 5-5 = 0-74 Cysteine 1.7 10.3 3.3 log 6.0 = 0.78 Glutamic acid 2.2 9.7 . log 6.5 = 0.81 Glutamine 2.2 9.1 log 7.0 = 0.85 Glycine 2.3 9.6 log 75 = 033 Histidine 1.8 9.2 log 31) = 090 Isoleucine 2.4 9.7 _ Leucine 2.4 9.6 :03 3'3 _ 8'3: Lysine 2.2 9.0 0g ' ' ' Methionine 2.3 9.2 log 9'5 = 0'93 Phenylalanine 1.3 9.1 ‘03 10°: 1-00 Proline 2.0 10.6 Serine 2.2 9.2 'I'hreonine 2.6 10.4 Tryptophan 2.4 9.4 Tyrosine 2.2 9.1 LKa VALUES 0F AMINO ACIDS There are 10 pages to this examination. Please count them before you start to make sure they are all present. Amino Acid pKa values or-COOH cx-NH3+ Side Group Group Chain Check the Appropriate Lines Below : Undergraduate Graduate Incomplete Concurrent Audit Write your name at the top of each page. SCORING Place your answer in the space provided beneath the question. 10.1 Valine 2.3 9.6 - Question Value 25 Score Do not use any other paper; if you N terminus of a peptidelpmtein has a need more room, use the back of these sheets, but clearly state that you are continuing your answer on the back. Partial credit will be given at times if you show your reasoning in the space provided, and if your calculations are clearly labeled. /. STUDENT AUTHORIZATION: 1, exam (e.g., handed out in class or [ch in a bin for me to pick up). Signature pKa 0f 8.0 C terminus of a peptidelprotein has a 9K; of 3.1 Date 25 25 25 25 25 25 200 , authorize the University to distribute publicly this graded Useful Constants and Equations 273 °K = 0 °C R = 8.314 J/mol°K Boltzmann’s constant K=1.381 x 10 ‘23J/°K Kw = 1 )(lO'M pH = pKa + 10g([conjugate basel/[aci d]) For the reaction, aA \2 bB AG = AH - TAS AG = AG" + RTln([Blb/[A]a) AG° = -RT1nKeq ch = e(-AG°/'RT) AS = R1n(Vrma1/le) Vo = (Vmax[s])/(Km+[S]) l/Vo = (KWVmaxXI/[SD + 1Ivmax Type of Inhibition mm + [II/K1) we + um KW(1+[I]/KL) VIM/(1 + [11/19) 9 = PW(Poz+P50) _ Hill Equation: loge/(1-9)] = logPo2 - logPso ea *- v ' €353} {mm/CT? .}m\3/;:uu4p 3 1. The fallowing equation is a restatement ofth Henderson-Hasselbalch equation in terms of pOH and prv 3_ DD}; = pr + 10g(conjugate acid/base) JA Calculate the pOH ofa solution of 100 ml. of 0.01 M tn'sodium citrate to which has been added 80 mL of0.02 M HCI. The pK. values ofCitric acid are 3.06, 4.74, 5.40. Show an work for fiJll credit. fl moleg A?" -: (loomL) (0 .DM PK“. + r -: l mmol fig—3r. £ 1 1 moleg H“ = (ROmL) (0.01m : [.9 Mo’gmol ‘ a Ill0_}u?‘. *: LC: 0 \ 2 efiut'uqlents cer H+ K2830 5~Pb - K:‘?,7_(, , K: a ~—__#——4 HA: L. HZA Mi: HBA 3". ’r b 2— 7 R H HA 1 name! H” (l equiva/mf H+) tri- lflfhol l-"fhffl0| O 49d 3-— z... {139 "Inmol “Imam HWUOJ P .M q“ A +b HA ) ‘—'—-—“—~———-————————______‘ lemma 0.5 mmol H", 1 o 0.5mmol Immat HAQ‘ + H* :3 Hzn' n l "1010* 0.9 name I O (VIE 'O.'01Hfflvl ’Q6flimo‘ 1-0.6 mmul 0.“! "Mia! O O, 6 man; I 30H: PKb + £09 cue ’r goamzt CHM Cowman/mom.) 2. The following are molecular weights and relative mobility values (R0 for a set of standard proteins run on an SDS gel. Graph; loSiMVV w§_ An unknown protein is run in a separate lane and the relative mobility is experimentally determined to be Rf: 0.53. Using the graph paper provided below, plot a standard curve . and determine the molecular weight of the unknown protein X. Be sure to label all axes on your graph and indicate the correct units. log MW? L«(4'7 MW: .O”‘L‘_7 N 2"“)!r BOO Dalto'ws :4 ' —: ._—_i—J.—++v- _ l Ill-IE _ 4‘ _ ‘ ‘ 3 Wfign; lilacs... i é ! l i ' ;r t a ' : 'H-llll i ' ‘ I --:=.. k I... III-“II I I .- Il-I-I-IIIE Ill-IEIIEII- .- I EE====EEEEIIII§EIIEIIIII i ' 5: "ar- a III-IIIIIIIIIIIIIE III-II IQEIIflIIIIIIIIIIIII III-III- IIIIIIEIIIIIIIIIIIIIuSiI'IHH E IIQIIEIIIIIIIIIIIIII!III-I Ila-IIIII-IIIIIIIIIIIEIIII III-II- Iii-IIIIIIIIIIIIIIIIIIEIIIIII-I IIIIIIIIIIIIIIIIIIIIIIIEIIIIIII I IIIIIIIIIII-IIIIIIIIIIIEIIIIIIII-II II!!!III-IIIIIIIIIIIIIIIEIIIII III“HIIII-IIIIIIIIIIIIIIIIIIIIII-I- IIII-IIIIIIIIIIIIIIIIIIIIIII-III..- EEEEIIIIEEEIEEEIEEEII-IEEIIIIE IE.- IIIEIIIIIIIIIIIIIIIIIIIIIIII-IIII IIIi'IIIIIIlla-IEEIIIHEIIIIII ' EIIIEEEIIIIIIII .- III-III... 4-- II- III- “l 0"“ FTG‘FJ A . "—J D as l 3 . DOW? [£5341 WU? Consider the following folding reactiOn for a small protein Q. Qunfolded # Qfolded You are given the following information: The folded stmcture of Q has 60 hydrogen bonds and 10 salt bridges which are not found in the unfolded structure. The enthalpies of the hydrogen bond and a typical salt bn'dge are -5 and -10 kJOules/mole respectively. At 25°C the ratio offolded to unfolded forms on is 10:1. What is the enthalpy of folding (AH) for the folding reaction? What is the value ofAS? What is the AG° for the folding reaction at 25°C? Suppose the ratio of products to reactants is instantaneously adjusted to 2:1. What is AG at 25°C? A ,4 .1 (at)? S lat%0,) + 10 (— I tied/1,13,) A ;' #3965 ‘l‘ ‘lC‘Z’ [5 1+ > ’ 4C?) l/r-T/mg f 'N-jk,\1 Ear/.lLlVFl/‘l I :1 O be} ; EH ' O : f40‘og— (251-5 KM; mm 4 at (2:29 XK) 13$ (55: a /7’C7Cll13/ry_.ui u _ l 974' he), 4%” K J (Lin/0‘ Awe ’ “RT/V? K12 Alia—33H)” r C ’ "n l-' ( /m<.l K “5(4) liq i—fi—igfi'wL W “6 Abyt‘ .1 CH/ ‘— 5 :P- L5 Mo RCA ‘ o AL] + Rim [Bf ca — Am, AL.) ‘SQLS i BDl‘irj (2cjgv) CH 1 HlUli 4. The following table gives values offractiOnal binding (9) vs PC.z for the binding of oxygen to hemocyanin in shrimp. ' \ Hill plot“ @9385. X-nms Q." R ?5b" P50: l28.3 mm H3 Extend pioi‘ For his“ qPfl'an S‘l‘crl-e- 3W Q‘i’ Q03 P50: LEG: Psalm," = 7231mm H3 D. Mber otcsvtbwiifi is a'i lea” “M, So Edenci piu‘i 59f [OW ftblmblsf H is duvet/1A Militant, stable- Intent: Li or getaways «1+ 303 P50: 2&8 Chet “it‘s” Use this data to construct a Hill plot on the graph paper provided below and answer the following questions. A. What is the P50 for this oxygen binding protein? B . Estimate the P50 for the high affinity state and the low affinity state of this oxygen binding protein. C. Estimate a value for the Hill coefificient for this protein. D. Use the Hill coefficient to make a guess about the number of oxygen binding subunits which this protein ma have. ---N---- IIIII'IIIIIIIIEIII I‘IEII .E' IIIIIIIIIIIII III I I II , fig. IIIIIIIIIIIII IIIIIIIIEII- ' AI — IIIIIIIIIIIII IIIIIIII lama-lg?- ' :==""=.-======-==:==='“3Lnunu ' In. In 1 x 3 IIIIIIIIIIIIIIIIIIIIIIIIIfiEE’IIEI IIIIIIIIIIIIIIIIIIIIIII!Z EIEI II IIIIIIIIIIIIIIIIIIIIIEIflflIII I—III IIIIIIIIIIIIIIIIIIIIIIAEMIIIIIIIIII III-flIIIEIIIIIIIIIIZ‘IIIIIIIIIII IIIIIEII IIIIIIIIEEIIIIIIIIIII II 'E'E" III-IIIEEEIIEEEIIEIIE I 5;! .‘l IIIIIIIII-IIEIIIIIIIIIIIIIIIIIIII L III IE lgp'l=llll: III. I ,, Ll 4f: .. .1 II _j i Lyn, ___ ismhk. _1 .. A __ l i 4—71—- k: 5. A. Draw the structure of di—trans, trans-12:2 (38,1l-phosphetidylinositol. B. Snake venom typically contains phospholipase A2, an enzyme which cleaves , glycerophospholipids at the interior (A2) ester position to produce a lysophospholipid having only one fatty acid ester which is in the A1 position. Why do snake venoms disrupt the integrity of cell membranes? C. Why does phospholipase A: not alter the structure of ceramldes'? » 9 P“ CHL—Q—QWVW \ 0 II "‘ CH-Q-C/WVW/ I)" Q l H Ice \ ITPLAL - H 0- P - O (— L A I, P‘) 0H1 .— tne \naopm¢pnc.iipltl with-54 (yr-la one £51k) QL;L1 1a.]! grander §crm o. -b1‘lk\d4.r flmmbf-KN-L. It; :Irucmn mom (\3gmhlc4 m? “9 CL fi‘crgm" anJt 1T vol” Qatar-haoflt. Mow; in‘u. Qo\'.\dr'.ut or wuda,‘ Sin-\PLJ SIT!“ ugl’uffi) KCtl‘LLi m;UJt\M _ pna;‘:nal.‘p«.;t_ [\k Lmvufi The 6.3Ttr bond 91 TR hL PN\.\}QA LM Q'ru‘") - {urgmwus d3 rml' nU‘Vk- an LLIU bend or ‘h-mc {‘1 Pentium nA‘L my LA-\_ Um;(l\_)- boa} which i; noT LLLxAvui “'3 Phos?k(lépqu A}. 6. Given below are values ofVo as a function of substrate concentration [S] in the absence A3h/ P‘CH/ LAM-l LAN! Mlvi and presence of an inhibitor ([1] = 500 pM). Use the graph paper below to estimate the values: A. Km and V,” for the uninhibited reaction? B. Km and Vm for the inhibited reaction? C. Identify the type of inhibitiou (competitive,lfiintiiinpetitive, lot uncompetitive). D. Calculate the value of K1 IEIII . I _l IIIIIIII III. ‘ ' ‘ - III III‘KIIIIIIIIIIII IE..- IIIIIIIIIIIIIIBQEIIIIIIIIIIII I II IIIEEEIIIIIII _‘ .a.» II I IIIIIIIIIIIIIIIIIIIIIIII” I I g. IIIIIIIIIIIIIIIIIIII IHEIIIIII : IIIIIIIIIIIIIIIIIE' IIIIII IEIIIIIIIIIIIE‘EEIIIIIIII = III III-[Iii 'EIIIIIIII III. III IIIIIIIII III ‘ IIII IIIIIIIIIIIIII III III! . via-IEEIEIIRHI E 0 u g n i _ "E i l' 7. The following is the structure ofTPCK (Tosyl-L—phenylalaninechloromethyl ketone), an irreversible inhibitor of chymotrypsin. ' O :‘M‘l‘l'y L‘; / || w/fi CH2—Cl3H—Cw—CH2—CLU TH ' O=ISI CH3 0 A. Suggest (you do not need to draw) a related stmcture which would be an appropriate inhibitor for trypsin. (Hint: change the amino acid side chain in the structure of TPCK). B. Given that TPCK does not form an ester bond to serine 195 in the same fashion as diisopropyl fluorophosphate, identify the most liker active site residue to which TPCK is b0und. th 4t VPSW \IF‘VV/E Oh I( C_ fiche w-al JL (2 E L \/ ti :1 6' "1 m 0‘ A l’ Kl mi 1-: (a f ‘K’ r- \-/ 1:) 5 h \A/ :I . L r ( v I’ ‘- o_, _\ ' C ‘ _. ‘4 ./ .“D ‘— F‘ \‘/ '1’ m n y ‘1 ( _‘, " “ " v’! ‘ ‘ \71 .‘x I“ (a 6 CL C — w—- L C "\ r" ) R C L— grctl. 113‘ L Cm/K +2 'l-JcaJ KO+4 g '7 u S. mm ~ H ‘lJf‘ r (a c- 3c :25: ET PQ ‘(l D} ’U'F— S +L“-( "“Ifir‘gg'd— fijt‘ [CIL'Pfl """r x ‘5"\— it‘ll Q Phobkr'fi Sl““l "HR / h __ l l “JHFF dJV/f J \fcf i—xl) [ Roul— ijva-«U\ ’T D F l" J , st 1 ' m ' ‘ ‘l 0 (~- . _J \/ \ C' (“x T \ ‘7 ‘3 (/I "N g; 1' FJ? \ ( ‘C 1 ( / ijr; — {\ r‘;(\/C \ Y L)’\/ ‘ Ah l-\£- [7/ F L‘-L\}Cll Lt“ — KC {’L] I“ (L(3V{ “C i \ C: "l 5 . \‘m A! ( \\ i\ [3 - "t C "x ,7 \h/ ref -k 1 \z ‘ ‘1 \/ e) r— 10 8. One microgram ofa pure enzyme (MW = 92,000) dissolved in 1 mL ofbufi‘er catalyzed a A. ‘B. C. reaction with a me of8.0 x104 W5 and a K... of7.3 x10‘3 M. Calculate the turnover number for this enzyme. Calculate the catalytic efficiency for this enzyme. Is this catalytic efficiency within one order of magnitude ofthe theoretical limit for a diffilsion controlled enzymatic reaction? / » Lang MIC-(3% 1.1%! \itvtzm i” “Wei {,sz 7;; 6210275 3AM Kid, ; Vettoox ‘ £01] .1 801/0 5% ; 7% >4cm [,1 MO’?5 Wk '1 ...
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This note was uploaded on 10/22/2010 for the course BIS 102 taught by Professor Hilt during the Winter '08 term at UC Davis.

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BIS102 Final S97 Key Hjelmeland - Name Student ID. No. S 8...

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