final_exam_key_part_2

final_exam_key_part_2 - Ammo 62 \(‘c :1 Name CHM 3218 /...

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Unformatted text preview: Ammo 62 \(‘c :1 Name CHM 3218 / CHM 5305 Summer 201 1 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 have 75 minutes to complete this section of the exam. All books, smart phones, translators, notes and other aids are prohibited, but calculators and molecular models are 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. Phenylalanine biosynthesis and properties. (Total 26 points). a. One biosynthetic route to phenylalanine is shown on the following page (note that no stereochemistry is shown). Where a box appears above an arrow, indicate the cofactor most likely to participate in the enzyme-catalyzed reaction. If no cofactor is required, write “None” in the box. A complete list of cofactors discussed during lecture can be found near the end of this exam. (3 points each). b. In D-erythrose-4-phosphate, C2 has the R—configuration. Draw the structure of D—erythrose- 4—ph0sphate in Fischer projection form. (3 points). Q40 A; (2) ' Conguau'vagovs \4 0H \' b W V‘AO3 r n on .4; Q ‘3 5 Ibt san. A“) 04206) a Dr sficv c. How does the enzyme-catalyzed conversion of chorismate to prephenate illustrate the concept of an entropy trap? (3 points). "Tim. ‘Q\BL~Q§V\CAW\ c {\qCVWfi‘F/fickk can QRUAAOYC‘V’ quwu“ some“ “WE , b. . $wa. cnl c, C‘Kockcflcfl Qflb (Abe VO\”C\\‘L CNOUVQ) M Q’0 OWE) \1 \ w \4, ewl‘jwu, ‘Cvuseu GOOVWW‘OK SoQC\q/\L COWCCVMCX\\CW \5 VQ°L¥\‘JQ'5 l Lo \’\'\\3 (xx/«cvsaQVALWV are CUE;\\LLTC‘~\-‘L QQ\,O\\1:>.‘) \v’ C 6 011:. K Answev. \(1 Name ‘ ‘ NW CO _ O OH H20 Pi H0 002 F,i H0 002 H2O %\ 2 + H M069 ¥ 0 “A “A 0 ® OH OH 0 OH phosphoenol- erythrose-4- ® 0 OH OH py ruvate phosphate DAHP 3-dehydroquinate NADA CO ' w co ’ co ' CO; 2 Nina? \A 2 ATP ADP 2 2‘06?) Pi éé \ 0 OH HO OH (90 OH OH OH OH 3-dehydro- shikimate shikimate-3- shikimate phosphate _ co; New. 002’ CO; F)i C02— 020 H20 @o o co; 0 00; OH OH OH 5-enolpyruvyl-shikimate-3- chorismate prephenate phosphate 0 co; 1 + O - NH Glu P l” P co; 3 (:02 5 f 002 phenyl- phenylalanine pyruvate d. Use curved arrows to show a mechanism for the enzyme—catalyzed conversion of 3- dehydroquinate to 3-dehydroshikimate + H20. Be sure to use the cofactor (if any) indicated in your answer to part b. You may use acid-base catalysis as needed and stereochemistry can be neglected in your structures. (8 points). \A a A ~ ~‘ (5 H 0 (BL HOG/Cb? 0 C691 u ~ _ :2 a :2 / :6 a on (2/ mi 0/; O” H OH H n on on $5; . ’75“ pkg; \-‘3~\\ d mamas \LE‘t 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 16 points). I ntermembrane space H+ H+ H+ H+ ATP Synthase x (Complex V) Mitochondrial inner membrane succinate H+ Mi tochon dr/ 8/ m atrix a. What is the value of AG‘" for the overall reaction catalyzed by complex I? (4 points). NAQH J, Q :i NADQ s. QHI -ev ,6“ ‘91 A ' L“ wk) " to; (roosv) — L‘ 031V) z t 0.17)“! V I AC4G = — “F’AEc 3 — (1)K212\XLLC\‘/-mo\ VX 0.3‘1v) : ._\—14\ \LCC-¥\ )Mc\ b. What is the value of AG’ for the overall reaction catalyzed by complex I at 298 K if the concentrations of NADH, NAD+, Q and QH2 are 10 uM, 130 uM, 1.2 mM and 660 uM, respectively? Note that at constant pH, the proton concentration can be neglected in your calculation. (4 points). I \NAO +—\\QH 2A Ag“ ‘ E‘TMQ Q: [NASH-HO} 1 1‘s, It My . 4', - mom EXGQQMShg) (\OMOJ’ «:5 )( LZ i 't'oflH = («WA \gcm/wu) » (\g‘mxig'3 \Lcex }m\ \c )tZQ%\<).\h : " )5 (K \cemlmuu Ans-v.2 Ext \L‘E‘l Name c. When the antibiotic valinomycin is added to actively respiring mitochondria, several things happen: the yield of ATP decreases, the rate of 02 consumption increases, heat is released and the pH gradient across the inner mitochondrial membrane increases. Does valinomycin act as an uncoupler or as an inhibitor of oxidative phosphorylation? Valinomycin transports K¢ ions across membranes. Use this information to explain all of the experimental observations listed above. (4 points). \ic\\me M70“ is aw xxx/Magpie”. ACATMM I AQQKQ’M L Agukml bawdy, \icx\\~/WW\10\V) :5 ($503; \X ’rvovnpcvb \Ce CLchb’: w mw‘bvcm \OKCGUK Wwvs. \5 a LMchK' \Mhmw‘u‘ (Agency #0) “3"va W “UMbV emu” KGB cw» H,» mst‘oe 09 \’\« vvi~\'\—OQ,\)OWB"\0V’\ M0¥LMV§ M4 Vivi/W5” 0? Q—XU‘S’ we M We, comm A: w mutocvcmbwo“. Maw = e. we Acioxm 1?; m? AC‘TeM‘A Gas" \‘0 LDv5¥CkW¥I 0‘ bwficxd A?“ ‘) 41vava anUM mtcw-Q Vacfl €\"‘-¥°V‘ ‘VWCV‘B?OJ\~ mug" (3)6 Shh” OZ Cembv’mv\5 DWI) _ d. The mitochondrial inner membrane contains an antiporter reponsible for transporting ADP into the matrix and simultaneously exporting ATP to the intermembrane space. Explain how this transport is coupled to the proton gradient. (4 points). intermembrane space ADPwt ATPOU‘ 1:: ADPin ATPin G} mitochondrial matrix i’aacuu 5’» [9“ Gay 3 \‘V‘km. R). oo‘tuxbg, has a -’\)o>\\\dk chem? “C,\C\3r\q‘L C? G) Le \"KA \v735\>3Q_ e o o A (“JO-rote taxi? («is ————-————> ()0 9Q ‘30 no on Q C? A Answem \("91 Name 3. The enzyme hexokinase catalyzes the first step in glucose metabolism in eukaryotes, following its transport into cells. (Total 10 points). Hexokinase Glucose + ATP —_—‘ Glucose-G—phosphate + ADP a. The rate of the hexokinase-catalyzed reaction is faster for the B-anomer of glucose compared to that of the u-anomer. On other other hand, even when pure (1- glucose is used as the starting material, the rate of phosphorylation is only slightly descreased compared to the reaction when pure B-glucose is used. Explain, using a curved arrow mechanism, why starting from Ot- glucose is not a problem, even for an enzyme like hexokinase that prefers the B-anomer. Note that you do not need to Show the phosphorylation step in your mechanism. (6 points). \3 HO g HQ OS: _ / W/C'> _ ‘40 H “:0:— \\/O\i 7 u—Of no 0,9 PD \ u: w ‘3 34,31 J \-‘i§~\y\ F , p341 H3“ Ar? 9 l "1'34. WkaL\—&\ l5 ‘V‘ CDV‘b CV35 C CD f) efivfiilbwum wily“ \Oel") W V ” \Mu. Gvn'bw‘ bl“ Qwé B -— (1w cvva ¢V> . - (E) ‘ NAN H30 ¥H¢OQ%M N g“ c\\cc\’x\1}<. b. The optical activity of a stereoisomer is expressed quantitatively by its optical rotation, the number of degrees by Which plane-polarized light is rotated on passage through a given path length of a solution of the compound at a given concentration. The specific rotation at 589 nm (the D line of sodium) [(1]D is defined thus: observed optical rotation (°) [0:] = D optical path length (dm) x concentration (9 / mL) A freshly prepared solution of a-D-glucose shows a specific rotation of +1 12°. Over time, the rotation of the solution gradually decreases and reaches an equilibrium value corresponding to [0L]D : +5250. In contrast, a freshly prepared solution of B-D-glucose has a specific rotation of +19°. The rotation of this solution increases over time to the same equilibrium value as that shown by the a anomer. Calculate the percentage of each of the two forms of D-glucose present at equilibrium. Space is provided on the following page for your answer. (4 points). Anzwm’l {6‘1 Name &r%\\)~Low ; + 13- (Nutov. “ * \C\° Qfiu‘dkb‘rmm ; “' 52-51? )g : mob. QweOnowq Ol—omom/v \' ; mob, Q’OKSHOW ' CWOVHQ/v x '~ \~—\; \[ = \~>c 1639. gown = X‘deuy * ‘l XG‘XD,T$ 1 x [ex—399 ~ {V‘QWGDJS ; x 1&0,“ + LAOS, — xKoJ Div, = x (\QQOA- KN‘KD'R’a‘) ‘* 161913 >‘ (\tflp'q ,‘~ [639») = [a] D|Qc|ul\ - {ck-lD' is Y : Mpg-loam , (~52-6°>~W) (+\\1a)—(*\q‘~') : 0C3; :7 ng; Qu’ov‘om‘u) \1: by = x- (0.59) z 5. Q“ :5 1'/t. L g-Iuwbmu) AmawEQ K27 Name Reduction potentials Half-cell n E°’ (V) Succinate + C02 + 2H+/0c-ketoglutarate + H20 2 -0.67 Acetate + 2H+/acetaldehyde + H20 2 -0.60 Ferredoxinox/Ferredoxinred 1 -0.43 o 0 OH 0 CH3MS-ACP + 2H+/CH3/\)J\S-ACP 2 _0.3 5 NAD(P)Jr + 2H+/NAD(P)H + H+ 2 -0.32 S + 2H+/H28 2 -0.23 Protein disulfideoxidgzed + 2H+/Protein disulfidereduced 2 -0.23 Acetaldehyde + 2H+/Ethanol 2 —0.20 FAD + 2H+/FADH2 2 -0.18 oc—Ketoglutarate + NH4+ + 2H+/ glutamate + H20 2 -O.14 NTP + 2H+/dNTP + H20 2 0.02 Fumarate + 2H+/Succinate 2 0.03 Q + 2H+/QH2 2 0.05 CuzJF/Cu1+ 1 0.15 Chlorophyll (P680-+)/Ch10rophyll (P680) 1 0.40 NO3' + 2H+/Noz- + 1—120 2 0.42 804'2 + 2H+/803‘2 + H20 2 0.48 Fe3+/Fe2+ 1 0.77 1/2 02 + 2H+/H20 2 0.82 Chlorophyll (P700°+)/Ch10r0phy11 (P700) 1 0.90‘ Useful physical constants R (Universal gas constant) = 1.987 cal/moleoK F (F araday’s constant) = 23.1 kcal/mole-V A gswEQ \Li‘1 Name Names and Structures of Some Important Enzyme Cofactors cHo o NH2 NH2 H JL CH3 N \N N/ I “\> (I) (H) WNW HN NH Hfiwé </N 1 N; KN N o—Ff—o—if—o N W002. (=9 0 o o o 602- ‘97 K 03’. e e b s OH OH OR OH OH OH Biotin S-adenosylmethionine (SAM) R = H, Nicotinamide adenine dinucleotide (NADH) R = POf, Nicotinamide adenine dinucleotide phosphate (NADPH) O O NH2 CH O o 3 N (D _ i_ _u_ G) | O O \ \ CH3 N N 0 CH3 N NAG CHg/RN L3 9 9 H H H H H OH H OH Thiamine pyrophosphate (TPP) H OH H OH H OH H OH NH2 0 O N \ N / o=F:'—o@ ozé—o—fi-o glib; O O 9 09 09 k j Flavin mononucleotide (FMN) OH OH Flavin adenine dinucleotide (FAD) \ =O3PO OH H2” N Coenzyme A (CoA) Tetrahydrofolate (THF) 9 H O 0 C02- 0 CH O CH O / I o—E—oe 3 3 CH3 \N® 09 CH30 \ H S H 0 CH3 n s’ Pyridoxal phosphate n = 5.10 Lipoamide Ubiquinone (Q) ...
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final_exam_key_part_2 - Ammo 62 \(‘c :1 Name CHM 3218 /...

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