Solutions to Problem Set #4

Solutions to Problem Set #4 - BIOCHEMISTRY lOOB SECOND...

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Unformatted text preview: BIOCHEMISTRY lOOB SECOND MIDTERM EXAM February l9, 2004 Winter 2004 page i Bogomolni Name: Student number: (last four digits) Score: 1 (40pts) 2 (25pts) 3 (20pts) 4 (20pts) 5 (IOpts) 6 (iOpts) 7 (25 pts) T(150) l-(40 pts). indicate whether the following statements are true or false. if true briefly justify your answer. If false explain why and write a true statement that applies to the case. a) The three essential amino acid residues in the active site of chymotrypsin: histidine, serine and aspartate, fulfill the following roles during catalysis: i.-‘ The histidine facilitates the reaction by acting as an acid-base catalyst. legl . 1+ antes as a bass mags: when fibet'rach'mj ‘memg "Fr-om ear—DH or Hagar-t also acts as an dud (disc/33+ with“ 3“ d“”“*"* a robe. +0 Hm ammo lg - and 1-star r Tana/arm. P “wgglnig’ilifig 23 3 ii- The serine'residue acts as an electr the reaction. Fri/5e sen m re swam n Cris 9 5 a n lJfi/fid‘f/LJ/‘e ., iii- The aspartate residue initiates the diac lation step by a nucleophilic attack on the carbonyl carbon of the acyl intermediate. Fa 1'91, 7% a paw/2‘ ad a 23.15971? firm, Asp ’v‘egefiu 50/ “"3 geaerwe a showy Mata, UM 5U win-u. Rea “#5 51‘ m Cdrrbprgfll/r eygffimoffl iW%J'W_iI'd;€5 b) Both low and hi pH are expected to decreas e otripsin catalyzed peptide bond hydrolysis TM 5123,1485 are (41‘?! flat! we‘ve only a; cu+aflm PH mkjg' H [ow/lugk pH, 06“ mg my“ Aewafuréj, 1‘) Sammie can he momma/wrote”? 4"" wmmw - made/e 5) m 621m of (Naif;th ammo 4am 'feydtaps Hm dqmge max c) Competitive inhibitors do dot alter the ofan enmeeatalyzed reaction. ’39 0 W1 Fame—f parh‘cfpah'nj in acid—big? (45%} . 77w AHe..r' Km d) The most important factor in the lysozyme mechanism is the stabilization of the substrate conformation that approaches the conformation of the the transition state. 1 [mil ti. ‘Siziei/imtipn a; 42-19- incl/{*f‘llatr'f (Dwarjmi-goh fiftymwé whammy \j ’ fiflfi i‘;;5{‘}-a{!f__.5?;r\ (‘2‘! A2,“: with the (“fifbpmflpm H‘a‘f’w’ar: (CW-ram trim”??? a! it; w- f—L *3 a waviffiuem’a, MM? gr'DM-ps titty/'1‘? fr’cismr -€‘H4~E"t..1" PVOWI'Dfi'l/ifl- {inf-4933 BIOCHEMISTRY 1003 SECOND MIDTERM EXAM February I9, 2004 Winter 2004 page 2 Name: Student number: (last four digits) 2- (25 pts) a) lOpts. 'On the basis of the following NAG (abbreviated as (3) binding energies for lysozyme's sugar bmdlflg Sites Athrough F, bmhyfl Eng rm: cl AF 9) t (J Eng i: 49, tun. J 3‘ 4‘ SITE Binding Energy (KcalJ'Mole) D *L‘ W ‘ N/i Q Can bind; A ".10 _W«..__H_ee_m_.-fi k; 7‘0 am. caf WW: [3} -3.5 I . tutu a n w I V . C -5,0 I d r A At SIX elks’ ) wk, I? +4.0 e- D 9H? plum, wot flueuwolcd—E (may in 1+5 WL‘A-J‘ 0”“ 7““ -' v be (a 0191 it, NE)” _2‘5 I Jr “busy-w: .qu 03:65 bark-V A“ kw“ Ova—H, 3Wu'ere ' (0th m 5 D p. F‘) -3.0 (5"‘195‘3 i:de 9F sfiuncnfi M D'Ylh3 {mm H: I J Pvt-{Erad- oluv-tr writer mwn WHML W’- WWW predict which positions will be occupied in the most prevalent r wni-Drmad’i on « complex with each of the following oligosaccharid. Assume that sites that can accommodate NAM (abbreviated as M) would have the same affinity for NAG and NAM). .55 + (—5.0) = —§.S’ teas/Mot 1.66 -——.> saw, EC. : —2.o +1433): or *Z.S+ (—3.0) =- —$.5 Lad/moi - \ “ 'r— é -. i ‘ n Agni) , CDE‘F ~1.o+i—%.§)+ (“5103+ LLD 5 5 C /W 5- GGMM-‘Wfiw AECD) CDEF or ‘~s.o+n.o+[—7.s)+(—a.o)= -6-S tau/m, 0" ll} bihd%,+ke_,h-yo DEF W “AS WM hanging 0&5. VA» ,IV,.»---~.vx "* 0H ‘ ,i A __ irral .21“ j i O l ‘ . “2| WM Mm We“ \ ‘ b-lO pts. An analog oftetra-NAG containing -H in place of -CH20H at tut; , LLL: v : e k‘ ’D Hist— C-5 of residue D binds to lysozyme much more strongly than does tetra-NAG. Propose a structural basis for this difference in binding affinity. *' 0H NI _ I , f 0 Jr Vg ' D , “i “W + 0H “Hm NHAL Wil’kiOfifl Amlflo amt oft/UK At’L? Qd-DPJV 0" “1L 'sz‘A D’i Mk6” MAKES; un‘hflUDrable ‘50 tee-dues oi— web-ewe Dkibln 90mg, we aim ‘ ' * bi din LEE-9 'melovnhle half-“CNUV’ Lowgormm wifia ion-def V‘ 9 . . u __ mu Manet). it: 444on .e (€?\am_d'wl+k -H J m tom. can now 3 - ‘ - Cowiorwnimx ‘10 W D'fiHfi ih H”? ‘me emf-{Malaise low EMQJfl\fi dwqu BIOCHEMISTRY lOOB SECOND MIDTERM EXAM February 19, 2004 Winter 2004 page 3 Name: Student number: (last four digits) 3— 20 pts. Many irreversible inhibitors react quantitatively and stoichiometrically with specific groups of the enzyme they inactivate. This offers a convenient experimental method to deten'nine if an enzyme has more than one catalytic site (or catalytic subunit) per molecule. Such an experiment follows: An enzyme of molecular weight 100,000hD was completely inactivated when its only cystein residue(s) known to be located at the catalytic site reacted with iodoacetamide (MW 185 #D). It was found that 3.7 x 10'2 mg of iodoacetamide were required to completely inactivate 5 ml of a 1.0mg/ml solution of the enzyme. Calculate the number of active sites (or catalytic subunits) present in each enzyme molecule. MwE : Walow p =LUMD§3/Mol S=iw3d0ac,L+-am‘tdf . EVEHHP _ .L was = Vesta-$125 3mm "‘6’ 37"“) mg ME = 5m(l'2113>- 5.01:“5 ml ‘lmoiee of S '." -1 ' -- 33x10 WW) (T273; .mol >ZD‘IDX\O}le Wj H386 home a oi. E ". _ l ' —2 5.0 J45. ( "‘0- = S.Dx.lo mot m6 ( [one was LDK'Iogs . . . is, o£ €.ln'—?:L,W~-Q m U?€-M‘mem shows; W 401’ each MD J ‘i'kug (Lire alour mmecg Di: Subgrka mokmlee midi/id .' .a The sat/raft.ng Jrkog. anon thetjwe mole one have. «COLL? NHWL 9H9“; (0V ammo)er aubuut’fi) . BIOCHEMISTRY IOOB SECOND'MIDTERM EXAM February I9. 2004 Winter 2004 page 4 flame: Student number: (last four digits) 4 (20 pts). Complete the following diagram to indicate the first step in the catalysis of a peptide j/ substrate by a serine type protease. Be sure to indicate all relevant nucleophilic attacks, charge states and stereochemistry. 5- 10 pts. How would the Eadie-Hofstee and the Hanes-Woolf linearized plots look for the following cases: 1- Competitive inhibition 2- Pure non—competitive inhibition gee M4? @133. W I. ComEeth‘va Inhibi’fion'. BIOCHEMISTRY IOOB ‘ SECOND MlDTERM EXAM February l9, 2004 Winter 2004 page 5 Bogomolni Name: Student number: (last four digits) 6-10 pts The KM ofa typical enzyme is 1 uM a~ Calculate the concentration of substrate necessary to fill 30% of the active sites. b- A competitive inhibitor for this reaction has a K] = 10 uM. Calculate the fi-action of active sites occupied by substrate when this inhibitor is present at a concentration of 20 uM. What would be under those conditions the fi'action of enzyme occupied by inhibitor? What is the fraction of the enzyme? km = not io‘e'u t} = 9307., = 0.3 A) B= 4:11... Km+ [a] swans}? [31 Giant: 951: [53 [53 -9ESJ= 9%“ [s] (ti—B) = 3 Km f5] = i Km l'a Dug '6 — ~1- = Mono N3» LLB “to M [5] (may ~5- b) Kc=\.ouo'su “FAN”: H [s] , Lay“ artifice lwl‘ibiw‘ '- B 2 Pi Km(l 1' 1- [5a] Liamo‘q‘HM .__ 0.x?) = 175% Zinc-JOSH 4 1.0: o" ( m) + “.35le H l M 1+ homo";le a: 'F'V'mL-k-ioy‘ oi Hiram. outset" ‘05 mummy 1 1592—4370 -; WA FTCRLHOW o-C 9m: thatqu ‘. \OQZ~3o°/Q = ‘l’O'A ‘ BIOCHEMISTRY lOOB SECOND MIDTERM EXAM February l9, 2004 Winter 2004 page 6 Name: Student number: (last four digits) 7-25pts. An enzyme that exhibits Michaelis—Menten kinetics was studied and yielded the following data relating initial reaction rates with the given substrate concentrations. (Note: l.00E-03 is equivalent scientific notation for 1.00 x 103): [S] (M) v (M sec") “[8] (M") W (M" sec) vi (M sec-54 Wi (MI see) Loos-03 LSSE-O'I memos 663* We “'5 A ‘0 l. 0 MD + 2505-03 3.61E-07 4.00:101’ 2.}?x109 3-3 ‘ ‘04 H- 1i " ‘0 6 5005-03 03113-07 100 “53,406 ' H. 2* 10'? 1A a 10‘ 10013-02 [.01E-06 [0.0 945210110? 7L. i '1. :04 l. H MD" 20015-02 L44E-06 50 QIQQMDE M x W" 0H x to" 5005-02 1.94E-06 3.0 5.5110? |.3Lxlo“-’ 4.0 x 10$ 1.00E—01 22013-05 no H , 55 mos 1'0 x \0-5 5; 0 K \o 5 i .’ L l i t ’ ‘ s l v i > , ...,,m.., .1. ._,. I a E r l i l l I __ Fr 1.. ".2: , i . 2 v t r _. . -,. p“. ,., l i l t E . i 2.0qu- uml 0.0.11" lane‘— thifi Itsw’ M 7 uni"— uni" uni" unla- umt‘ , {a} M l _ , l u Nut-“WWI “i E i ' ° ‘ E l E Km” \ - DMD'J'H a) Using the graph (above right), plot the raw data b) b) Fill in the two empty columns in the table above for the "double reciprocal" form of the data, and plot using the above graph on the left . ‘ L .. i...— = _ I -I _ : Lo ‘ '- M c) What is the value 0me for this enzyme? Km = i 0" ‘0; H '9 Kt“ ‘ D . . ' _ -‘ . -.l d) What isthevalueofvmforthisenzyme? {EL-f _ xlor H‘s-u. -% VM‘M :25 xtD HS“ M e If the enzyme concentration is 5 .05 x mm what is the value of kw? /@ If a competitive inhibitior having a enzyme-inhibitor dissociation constant, K = 10'2 M, is added at a see ‘ concentration of 1.5 x 10'2 M calculate the velocities in presence of inhibitor, Vi, for the substrate My“, concentrations given in the table, and enter those values and their inverse in the last two columns. Vega, g) l:lhot in the same graph the data calculated for the reaction in the presence of this competitive ' ibitor. e). '7 EE] = 5.05: 10‘“ H Vmax : kw; {at ‘-' VMK 2.5. K ‘65 H - gee" felt 9' UQ A \o" M = H95 K '02— sea" [I] = 151.161“ ('3' r51= LODXIOJE‘H km : 1-OX\D-LH VW = 2,-5uo" H .5“; V: MLL _ (mm-5420.owo’3~3 *3 L ‘ = “Lama “.54 kthr Ell. *1 LSMUl M M35] (Leno M§<‘ + '04” M) + "DONE-5M Thies‘l’ a“: M \Mues cue In fig Jam; ...
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This note was uploaded on 04/07/2010 for the course BIOC 100B taught by Professor Sethrubin during the Winter '10 term at UCSC.

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Solutions to Problem Set #4 - BIOCHEMISTRY lOOB SECOND...

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