Second Midterm Exam Key

Second Midterm Exam Key - BIOCHEMISTRY IOOB SECOND MIDTERM...

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Unformatted text preview: BIOCHEMISTRY IOOB SECOND MIDTERM EXAM Winter 2007 Bogomolni Name: Score: 1 (40pts) 2 (20pts) 3 (20pts) 4 (20pts) 5 (lfipts) 6 (20pts) 7 (lSpts) T(1 50) February 15. 2007 page 1 Student number: (last four digits), 1-(40 pts). Indicate whether the following statements are true or false. If true briefly justify your answer. If false explaintwhy and write a true statement that applies to the case. a. The addition of a reversible com etitive inhibitor will increase the aEEarent Km for the substrate but not alter V?m FIKL} Wiftla‘m thZLI‘iW Linca’j 13m. 6— onL'l 5;) lie 3)) it] 'l/W can LL rigging“ can be overcome if you add a high b. The effects of a reversible non-com etitive inhibi enough substrate concentration - m fukki‘bf C49 11" m‘; {55] Mil/()6. 3: mafia c. A reversible noncompetive inhibitor will typically bind at the enzyme actiye site HGLYE’.’ n—comp DnLiL-J‘lr (5-5 Genny/WV aw; n4" aura/Lib Cw ‘43 Wlifi. 655} ‘5 C53. d. A reversible com etitive inhibitor will typicallly be structurally similar to the real substrate flat; in 94w '65 canted! why 6 {4. Myth-tir- Phil/kill: SM Mil/4%“th losteric heterotro ic effectors must bind to the enz 'me active sit ‘ {in/Jig Mm draft? a/iU‘lit/rc agaier Warsaw», ma afldsfié Al e. ‘(3 LWL‘JG Claim. J: 642430: .ré. la. BIOCHEMISTRY IOOB SECOND MIDTERM EXAM February 15, 2007 Winter 2007 page 2 Name: Student number: {last four digits) 2- (20 pts) The mechanism of the cysteine proteases is very similar to that of the serine proteases. The main differences are that a cysteine residue is used instead ofa serine residue in the initial attack to the peptide group. and that no evidence for a catalytic triad exists. There is, however. evidence that the cysteine residue is activated by general acid-base catalysis as with serine proteases. Consider the following peptide bond hydrolysis: ') ‘. L R1 x/Rz f/Jlxkc H\\ «*RZ N, | ———- | H i) Write a detailed mechanism for the enzymatic hydrolysis of the peptide bond presented above, accountin for all nucleo hilic attacks and _ hydrogen transfers H‘s ‘5 “"2 mtg 0W3 ii) In the reaction mechanism, indicate which steps can be as general 510' base show catalysis. Sale p5 \ , 2 ’ 1‘ . 5' iii) Indicate all the transition states that form during the reaction. Hi should be two of them. .. [16' or 5.31 Sivan YF‘AAUJS/ (an pmhcipate :There l“ “04’ L358? (Magus. 0‘ has mati'in'hj Mae mun a: fig 'gasabase OF 6'0, ii (rm be loo-HA in the reaction? Protoymwd and unpvomuflhd Additional data‘ GI” ‘ PK = 4-25 LYS ' K: 10-53 ‘m We awn hr am ~ ,. ASP - PK = 3-65 Arg - pK = 12-48 __ ‘___ guufitfijg PH range 0 o H \ (5-5? — 6 s) \ L,H__..__M_n,__fl s R/KN/Rl O Ill I \ l \H l l l C‘r'l ————> E g H Pt' “h if. Hg‘gE " arm in 3. E 8 cap %5_U\H —i E A I 9 l Zhd +rmnu iii!- qfi-gt'r '3, BIOCHEMISTRY 10013 SECOND MIDTERM EXAM Winter 2007 page 3 Name: Student number: [last four digits)_ February 15. 2007 3- 20 pts. The table below contains data collected for the rate of an enzymatic reaction in the I presence of a 4 mM concentration of inhibitor 1. The graph on the next page is a double-reciprocal plot, with the data for the same enzyme in the absence of inhibitor already plotted. for the same substrate concentrations. r. :mM iEIIITI. lil— 4:n\l 1mm I ll.” l lift-r. tan}! mml. Ill —ll 'HhUnJ\ finnedi Hill 737' ll.( '13 “."iil tillfifififi [H lhlll ll] lltllSS-r’l {illhhif’ Label the axes of the graph, including units, and give the symbolic values for the x-intercept, yintercept, and slop of the plot (for example, the y-intercept is<1Nmax): x-intercept : "‘ i M , y-intercept :lN max , slope : l : iii-«ML Double—Reciprocal Plot III-III- III-Illa gm. III- III. I HIE-E. II II illfllfl! k Plot the data for the inhibited reaction on the same graph, and determine Km and Vmax in the presence and absence of the inhibitor. Enter the numerical values here, with units: No inhibitor: Km 3.6 mfi’l . Vmax l riff gil 4mM inhibitor: Km ngm/fl ,Vmax OJa-u“ "en-l PROBLEM CONTINUES NEXT PAGE Comffllw‘l'Nfl, LiLo/t'. Calculations in the reverse 1+: BIOCHEMISTRY 1008 Winter 2007 Name: (problem 3 continued) SECOND MIDTERM EXAM February 15. 2007 page 4 Student number: (last four digits) What kind of inhibition is exhibited here? Explain how you know. Determine k2 .-‘Krn (uninhibited) for this enzyme OR specify what further information you would need in order tp do so (use the definition of Vmax): .So'iwv‘l'U CO "lib {fin-J.” if”, = litter), WM A =_E’:‘5_r€ /k;/ wwlai lac. mlmldlqll :10 *écrlcflc 4— 20pts The first step in the lysozyme mechanism is shown below. What is the role ot‘Asp 52 in the reaction? f‘ri Glu 35 H ' a (Jr-tEOH H/ dis“ ‘0’ i If MOE W CH 0 ———F H r J H i . H02 NHAL 1| Asp 526’ flefirosl'fiiz’ncfi: I Q P A; STADILI-EES 4m ‘90er afi bio Garbo-nim—.‘6ma5741‘ah3€afi 0n C, 0-? MAM ravfi in °4 ’Wrm- (omen. wish) Based on your answer to (a), which one of the following compounds could be a transition state analogue (and therefore a good inhibitor) of the enzyme? Explain your reasoning. A5005; Siva. Kama Vim an (Milan COM {10+- K'“ Mia/ma, OOth mikwm.&l BIOCHEMISTRY [00B SECOND MIDTERM EXAM February 15. 2007 Winter 2007 page 5 Name: ___ u_ Student number: {last four digits) 7 _ 5- (15 pts). The structure below is the acyl-enzyme intermediate in the hydrolysis of an ester by a serine protease. which has a mechanism similar to the amide hydrolysis we have studied. A- Draw the structures for the next two steps. paying attention to protonation states. The first step isjust the replacement of the alcohol leaving group with water. There is no need to draw the Asp side chain. m i - ‘ "'5 bar L} - a B- As pH is varied, the km! for chymotrypsin is found to be directly proportional to the amount of the Hi357 in the catalytic triad which is in the His as opposed to HisH+ form, while Kdt‘ss for substrate is constant. The pKa for HisS? is 7.2. Is the histidine acting as a general acrd or as a general base? How do you know? . MM [8] 44 [a] or“ [5144414m 1526*“; rem/t. =&nw~¢ I Since, Matt; l3 Gms'i’cmi' pix/6“ 14:35.? is new or lilLHQ) (4153 PrO‘i’DM‘l’IOA of FA; JOQSHL£ so %a«7£ 24‘: 5?’ raga/Iii'éi'és gag «fit/CIA 04"”! gm M34“ dam»?! glldwei'im. 9F kawwpfim; A ' 6g ’i at} M M 32mm. Wt my a5 0’4 “4’5 g ' a: maww Swiss”, WK as a 4 “f (‘0. ' 0 qu ml won « thme‘i-‘rv? Cow‘G‘WH-f . BIOCHEMISTRY 100B SECOND MIDTERM EXAM February 15. 2007 Winter 2007 page 6 Name: I ‘ Student number: (last four digits)_ 6- (20pts) a— We have mentioned that the pKa of an amino acid side chain can change substantially in different protein contexts. Ifa histidine side chain were found buried among isoleueines in the center of a proteindwhat effect would this have on the actual pKa of that particular side chain, and why?3What if there were a buried aspartate as well? (9 Hjfimpkobk, environmaniis Cat-4% iwmrfi' iiree Clméfi. ' u H“ i __s Hxx'“ ’9 l. 77; «Pg an isoluecinat GK; 0? His wealth ‘--— dgurie. ZZCM; depri'i-Ona‘i'ad Siva Q..- i’M'O ""— V? i it: His Ali; '2' A5536 could. ehni'vosiaitmii Sufiav’t tits, H-‘t Misha PKQO b-Why are serine proteases synthesized as inactive zymogens, wh ch must be proteOlyzed to become active? fl, zsécflb Cit/Lila” COMFJHANXL! fiowt PYQiLO 6L.C:"i"i\l6~ Oh i t c PEWLATIOMI c- Why is ATCase activated by ATP and inhibited by CTP? What kind of inhibition is this? is «pan/int am& CTP it a. l: rim'nfijnm. ATOM-g 5 h i “iv-idle ' H W i if? who, pmi ‘is cancer»th or 1:37?) ATP allfii‘i'MCA-iiiy ac “WA‘M Emmi; ) Amie-fa imam WNW“ “4" b‘~°$vn““°~"= “J‘hm‘flq‘lf p” MA “Plum” r M23 .1110 re ‘A bm'da reads—(3m, Asxéh aQ-flig 4% NM; 9?; VC‘t’P ATCOJIL-n gilep-gnfiffidu' ar- msmifg nuggyfiw). d- Serme prote 1 cata yze peptide hydroly51s (slo y w en er lS rep ace y i a. 1. What principle of enzymatic catalysis explains this observation? 2. Explain why replacement of the other two residues of the catalytic triad with alanine does not further reduce the reaction rate. 3. What class of artificial enzymes work according to the same principle? (I) Gpnh'nl Qeii‘bmp dad/fr} tiara/V151), (flirt-C049“ c2.) Emilia; Ritakfik’ MJ Q-CLGMfiFJ'L {4H {flak émliMii'Y) 3.) Ala—wine; WE’YME-I) e— List two ways in which the regulation of enzymatic activity is achieved. and give an example of an enzyme which is regulated by each mechanism. Discuss briefly the advantages and disadvantages of one of these mechanisms for the cell. a [50: ( fl? ' Cade 3 PMSFl‘U'rirIifl'L 2) 4//asrearc. MJm/asiiov: V - COM “Miranda; mew. (See. a W- M C3 fly a 14W 0%., “J”! ‘11.“ W: o eff (Z) frv‘WWIiA/e fia‘i’w lyiis : f } Jud/M Eyre/vth ... BIOCHEMISTRY 100B SECOND MIDTERM EXAM February 15. 2007 Winter 2007 page .7 Name: Student number: {last four digits) W __ 7-15pts. Di-isopropylfluorophosphate (DH-P) is a potent irrex'ersihle inhibitor ot‘serine proteuses. The phosphorus atom is extremel} electrophilie due to the presence ot‘the electrunegative flour-ine. which is also a good leaving group. Complete the mechanism below For the reaction between Ser 105 and DIFP. Explain why DIFP does not react with other serines on the protein. (You can answer the last part even if you were unable to draw the mechanism). ililCl'lHL‘JluiL’ . I U‘r’t\\¥2 04%“ get ‘me 3 Car 4+th E in AW: name at H: \c,‘ Pm t at a V‘thi‘v'fl ' if {1‘ 31 r \"mf—E‘LIJLY ‘r-L in u.) 'r'xt i 1A “ma 0—1, \p g Y but) we t~.-\_bl?,tr-’_, at UM; (Cl i‘ \O} A hmj‘nc \w rod 1 ASP 31% HHS? 2h . . . « y‘ame “"1 Wat-hwch of r-«Prfif bij W‘Aem‘r‘q ti ‘fém r-xurlrnrzixihc and I’m Itl ui-l am! VY‘OI€ ?VOV"€ +0 awnt ‘Hn-E DIFP ...
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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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Second Midterm Exam Key - BIOCHEMISTRY IOOB SECOND MIDTERM...

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