practice_final_exam_key_2

practice_final_exam_key_2 - Am'awé Q \C 6‘1 Name CHM...

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Unformatted text preview: Am'awé Q \C 6‘1 Name CHM 3218/ CHM 5305 Summer 201 1 Practice Final Examination — Part 2 University of Florida Honor Code Statement: "On my honor, I have neither given nor received unauthorized aid in doing this assignment. " Student signature Instructions: You would have 75 minutes to complete this section of the exam. All books, notes and other aids would be prohibited, but calculators and molecular models would be allowed (no sharing of calculators or molecular models, however). Be sure to budget your time and answer questions briefly but completely. To receive partial credit for incorrect answers, be sure to show your work, particularly in problems involving calculations. Write your name on each page. 1. A bacterial ribonucleotide reductase catalyzes the NADPH-dependent reduction of the 2'— hydroxyl of nucleotide triphosphates to the corresponding deoxyribonucleotide triphosphates for later incorporation into DNA. (Total 9 points). @O @O O B O B + NADPH + H+ + NADP+ + H20 ribonucleotide < H6 20H reductase H6 B = any base a. What is the value of AE" for this reaction? (4 points). cl ,6): At 1 (1.436 -k: i n .9 O UJ L < KN anoint \Lé‘i Name The currently—accepted mechanism for ribonucleotide reductase involves radical intermediates. Formation of a disulfide bond on the enzyme has been suggested to provide the reducing power for the ribonucleotide reduction (step 1), then this protein disulfide bond is supposed to be reduced by NADPH to regenerate the reduced form of the enzyme needed for subsequent turnovers (step 2). Is the proposed two-step sequence shown below thermodynamically feasible? Justify your answer with a calculation that clearly illustrates your logic. (5 points). ®®®O ®®®o O B o B Step 1: + Protein disulfldereduced =2 + Protein disulfideoxidized H5 3H H6 B = any base Step 2: Protein disulfideoxidized + NADPH + H+ Protein disulfidereduced + NADP+ AEB‘ $0» 5kg) 1 = boozy) - (~e.zsv) : VO_ZSV AC1"i = “WFAEDI = - (zXZSA \cm\/meu.v)( v» 0.25v) 3 ‘ \\.‘<‘> keel/Mal” 3%? ’1 x) bow-mm m Ema/"('81 -0! - AL (3,: sh» Z = (-ozsvflfiolszfl = ¥ <3.qu Ag“ : ~ L2)(;z3.i let-CA PM”X ” O’qu) : ‘ Li, Z \cccfl I'mok Z \3 \oo-thl“ M Q'oev%7 AM$-ER \LE‘i Name 2. A simplified scheme for oxidative phosphorylation is shown below. Use your knowledge of these reactions to answer the following questions. For all of these questions, you may assume that the substrates needed to run the citric acid cycle are always present in excess as are all of the intermediates of this cycle. (Total 12 points). lntermembrane space H+ ATP Synthase 1 (Complex V) Complex ll 4315‘" ;-;"1d“7-‘-, r‘r‘g‘fi": 2;":7- r- .. r " ~ :;~—»v"-s»~‘:».-" is; 2' r- ,g'"’.. ‘ Mitochondrial inner membrane Mitochondrial matrix a. Under steady-state conditions, active mitochondria generate AGtotal = +4 kcal/mole/H+. If the membrane potential (AT) is 0.11 V, what is the value for ApH? (3 points). Ac A? VT); -O.OSCiV-Ao\4 = A? ~A‘P l | v l . A? = M0 - ocsqV- Aw ) A?“ = AVA? _ , 0.05%! A : {kawmukl (*0‘HV)-(0,\‘IV) \ (Z?) \keoA/wwl'v) [ : i 0.05qv 2 Q. \'7 : _ Lea kwbcAr‘vs. \OLcetuxx 9%ch r3 lowiw 3646‘" Vum) b. What will happen to the rate of ATP synthesis by actively respiring mitochondria if an inhibitor specific for Complex I is added? Be specific and justify your answer. (3 points). "le vak w.“ Bro?‘ 501 '00" \0 W0. "Wt 0"?- SVWO 1353‘“) bx, ‘NV‘\W BXLVVO-n') CVA’V 0x~PV703>~ Z ——5 ill >5. “.4- Tl— —-‘I W a w vv 5”“ W 513"“2‘ l’c‘xfiwofil Cc”) Sill“ Cundfic‘“, W P” %Vob\w\' ml“ to be CWME but ON a \owv swéxlvslrak «>314. Com \r- \vsUL AA '> «Ext \éé‘i Name c. What will happen to the rate of ATP synthesis by actively respiring mitochondria if an inhibitor specific for Complex IV is added? Be specific and justify your answer. (3 points). fiw \L‘R“ £61050 $\0 v0 5‘1 VNWL§ \ 5 \x Q/VC/‘fi 3k 0 i7 3 ' 350% Pathway: Cw (mam-n Womch/v MUS? pens WVOUCAM 9”“?\in‘ \Q m PVO‘VQA %Vcb\hm¥ V30" WNQ\V)\*C~\V\&‘ 50 \V EGF> Lb E010, d. What will happen to the rate of 02 reduction if an inhibitor of ATP synthase (Complex V) is added to a suspension of actively respiring mitochondria? Be specific and justify your answer. (3 points). \n'\\’\o\\\l7, \‘LL va\—<, 0Q Ct VLbVQhUm ml“ 53mg W Saw, bu? Wm \X‘ w\\\ %v c): vow; Bvo p \0 am, As VALCMW ‘va39cv \' onb 9'0 50“ Pu m pxhcz con \”\ku1, 40 \—\® (aw obxcvxll CowX—wxubs \0 fig“) (‘5 '0’) UL "Nok’v’a \m “6* m ’ 9” H" W W‘CA’ *‘Vfi homo??? COMPULX . 5* 3 am ?O\V}\,‘ w chm,ng \xbova\r\b by MACH ovx‘o sutLWflciH OWBQA’WV‘ \5 \ VQC\U\VQb \“l‘ (30’35 6W0 V‘ O\ CA\V\)\’ Q COWW\'VQ\’\GV7 cbvc‘hiem*v Mr Hon» powh Ol conww‘phevi shag; usage \(eL Name 3. Vesicles are spherical lipid bilayers that have been used as models for cell membranes. In one famous study, vesicles were created from lipids that had been modified to incorporate a paramagnetic group (a nitroxyl functionality). The label was distributed on both leaflets of the vesicles. The presence of the paramagnetic probe did not alter the membrane structure in any way. The paramagentic nitroxyl group can be detected and quantitated by electron paramagnetic resonance (EPR), a technique closely related to NMR. (Total 12 points). a. Nitroxyl groups can be reduced to a non-paramagnetic derivative by ascorbic acid: ('3' OH CH3 N 0 CH3 :1; CH3 H0 H3 O \/ \< 7< + ‘,7 CH3 >=f ’ CH3 CH3 H0 06 Ascorbate CH3 Lipid bilayer Lipid bilayer g Paramagnetic; Non-paramagnetic; detectable by EPR undetectable by EPR In the experiment, paramagnetically-labelled vesicles were mixed with ascorbate and the intensity of the EPR signal was monitored over time. This yielded the data shown below: 100 400/ 50—! / 0 Percent EPR signal remaining 10—l 5i Time Using words and pictures, provide an explanation for these observations using your knowledge of membrane structures and properties. Be sure to account for the significance of the 40% level as well as the much slower loss of signal seen after this level has been reached (5 points). /“ Kw QVOL‘OOW 0Q wi‘kescx-p Keg?) vgéuué eunuch, (60% 90 AA» iota) Cavvesacmbs «\o W iaoa ow w saw anMr. Law: on W WW leech} are. vebucfifi MUG" W76“ EAm’JW 5W)“ QI)C&v-b C09. ‘3 oMCNW’fij: CV O W \ \ \3 \b b; \ (‘4\’ 1’ 5 \lkvv mused EYL [’61 Name b. Paramagnetically-labelled vesicles give a single peak in the EPR spectrum whose width reflects the motion of the lipid (narrow lines = fast motion). Vesicles containing the following probe were studied and the linewidth was measured as a function of temperature: CH O s\/\/\/\/\/\/\/\/\”/ L CngleCHs O 0 CH3 0’ 0_ CH3 ‘CH3 O The other lipids making up the vesicle were normal phosphatidylcholines (the five- membered ring in the structure above was replaced by a methyl group). Explain the shape of the curve using your knowledge of membrane structure. (4 points). S\VV\‘\\O*’ i0 W A; muVm tunic 6% Q So\\% midpoint Linewidth (H2) Temperature l/M “MN \meww53m Noam» cevvcspcv~k5 \—o H» “5035" me 0c M b‘i\o\\1u (ham); cv‘t’sfm\). A3 3“” *Q‘MFCWQWW “"Q’VE’Q‘MAI We \‘\?\}¢, (he (“*0 Ha Chi-‘6 (lifiVlb‘MkQ) Hoyt, \\?h’> \Ohgg'uvt qU‘\(;\k\\1 'm “A x GAB \/ EWQEANOVYD‘ which ngvti) narrow EPQ \thfim’y c. The experiment described in part b was then repeated using a different set of lipids in which the acyl chains were replaced with the following (the remaining parts were the same as in part b): CHSWMO CHBMA/ZWMO When these lipids were used, the same curve shape as in part b was observed; however, the midpoint occurred at a lower temperature. Provide a simple explanation for this observation. (3 points). ’71:» en boabbn. bases m‘rstcvt coma PGL\Lm%, 5'90 W bum; 9" muh o\~ a \cde/l Wowu‘rum. Awake“: \2 [‘94 Name 4. One biosynthetic route to isopentenyl pyrophosphate is shown below. (Total 17 points). H20 NADA O M O COASH O O CH3)J\SCOA HO CHSO “MVH 2 CH3/U\SCoA ' ‘ CHgMSCoA é? OszSCoA % 1 CoASH 2 CoASH NADPH NADP+ ATP ADP 0 ATP ADP _ CH3 _ CH3 OZCMH v—AA—n OZCWOH —;-4— 5 FAA; 3 4 NO'M CH HO CH3 3 O C 2 W0 ® ® % NO ® ® 6 coz, H20 7 a. Where a box appears above an arrow, predict the cofactor most likely to participate in the enzymatic reaction, if any. If no cofactor is predicted to be involved, write “None” in the box. Note that the names and structures of all cofactors discussed during lectures can be found on the final page of this exam. (3 points each). b. Draw the complete structure for intermediate 5. Would you expect the value of AG‘” for the reaction 4 + ATP 5 + ADP to be —4.8 kcal/mole, +0.1 kcal/mole or +6.3 kcal/mole? Briefly explain your answer. (3 points). Ho CH3 6 ‘ \/\/ i 9. (3 be .43 \chA web» v: m0“ \\\LC\\, ATP \ncn WC) H65} NO? “by 0"“ HEY) CW7} 5‘ \MN) WC: HEB» SW7“ W vac} Chowfifi‘ ‘5 \033 6Q 0M HEB! «Rx vwehdw abomb be WW1 *WW‘OBVV‘O‘M‘CQ‘N waabg. c. Use curved arrows to show how an enzyme would catalyze the conversion 6 7 + H20 + C02. You may use acid—base catalysis as needed as well as the cofactor indicated in your answer to part a. (8 points). o O ’C‘0 m e / 0 A1 ~—- *Lmeo CH)‘ ; 0®® ‘ ) OH who Amino acid pKa values Name Alanine Arginine Asparagine Aspartate Cysteine Glutamate Glutamine Glycine Histidine Isoleucine Leucine Lysine Methionine Phenylalanine Proline Serine Threonine Tryptophan Tyrosine Valine pKa,1 2.34 2.17 2.02 1.88 1.96 2.19 2.17 2.34 1.82 2.36 2.36 2.18 2.28 1.83 1.99 2.21 2.11 2.38 2.20 2.32 Useful physical constants R (Universal gas constant) = 1.987 cal/mole-K F (F araday’s constant) = 23.1 kcal/mole-V 3.65 8.18 4.25 6.00 10.53 10.07 A N'pwEZ N EL Name A933qu [2"] Name Reduction potentials Half-cell n E°’ (V) Succinate + C02 + 2H+/0l-ket0g1utarate + H20 2 -0.67 Acetate + 2H+/acetaldehyde + H20 2 -0.60 F erredoxinox/ Ferredoxinred 1 -0.43 CHsMs-ACP + 21r1+/CH::)O:/is-ACP 2 -0.35 NAD(P)+ + 2H+/NAD(P)H + H+ 2 —0.32 S + 2H+/H28 2 -0.23 Protein disulfideoxidized + 2H+/Protein disulfidereduced 2 -0.23 Acetaldehyde + 2H+/Ethan01 2 -O.20 FAD + 2H+/FADH2 2 -0.18 oc-Ketoglutarate + NH4+ + 2H+/g1utamate + H20 2 -O. 14 NTP + 2H+/dNTP + H20 2 0.02 F umarate + 2H+/Succinate 2 0.03 Q + 2H+/QH2 2 0.05 CuZ‘L/Cu1+ 1 0.15 Chlorophyll (P680'+)/Ch10rophy11 (P680) 1 0.40 NO3' + 2H+/N02' + H20 2 0.42 804'2 + 2H+/SO3'2 + H20 2 0.48 Fe3‘VFe2+ 1 0.77 1/2 02 + 2H+/H20 2 0.82 Chlorophyll (P700-+)/Ch10rophyll (P700) 1 0.90‘ Adswm Kg Name Names and Structures of Some Important Enzyme Cofactors NH2 NH2 Ho, H O O / N NH ® 9H3 </N I \N bl: I \> 9 9 W 2 HM NH HaN s N N4 \N N 0"?‘0‘?*0 N W005 \l/V® 0 w 0 O O CO - ‘ o , G G S 2 OH OH Biotin OH OH OH OR S-adenosylmethionine (SAM) R = H, Nicotinamide adenine dinucieotide (NADH) R = P03: Nicotinamide adenine dinucleotide phosphate (NADPH) J UCE ~/ , ~. \ O 0 CH3 N \NAO CH3 N \N/ko CH3)\\N L5 9 e H H H H H OH H OH Thiamine pyrophosphate (TPP) H OH H OH H OH H OH “Hz 9 o 9 /N \N 0:53—09 0:1:3—0—1'3—0 (N I NA 09 09 Q0 {-0:r Flavin mononucleotide (FMN) OH OR Fiavin adenine dinucleotide (FAD) O 002' NH2 /©)LfiJ\/\CO2. N H H OH _9. _9 </ 1 UN 0 H HN HS/VNWi/VNWO 5—0 ($3-0 0 N N/ HN N O OCH3 CH3 9 G \ I OH OR H2N N ” Coenzyme A (CoA) Tetrahydrofolate (TH F) H O _ G O O 002 O CH30 CH3 / I 043500 \ 0 CH3 N6) 9 CH30 \ H H 0 CH3 n \S/3 Pyridoxai phosphate n = 5.10 Lipoamide Ubiquinone (Q) 10 ...
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practice_final_exam_key_2 - Am'awé Q \C 6‘1 Name CHM...

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