exam_2_key - Answer. \Lg Name CHM 3 2 l 8 Spring 2008...

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Unformatted text preview: Answer. \Lg Name CHM 3 2 l 8 Spring 2008 Examination #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 two hours to complete this exam. All books, notes and other aids are prohibited, but molecular models and calculators are allowed. Be sure to budget your time and answer questions briefly but completely. To receive partial credit for incorrect answers, show your work, particularly in problems involving calculations. Write your name on each page. 1. A single substrate, single-product enzyme—catalyzed reaction can be described by the following complete kinetic scheme: k k k 1 BS 2 EoP 3 K1 K2 K3 Using the process described in lecture, this can be simplified to the more familiar Michaelis- Menten form: E+S E+P k k E+S+ E-s ~fl> E+P -1 Use your knowledge of enzyme kinetics to answer the following questions. (Total 18 points). a. Note that the Michaelis-Menten treatment ignores binding of the product to the enzyme. Which of the three classes of enzyme inhibitors discussed during lecture would describe this type of inhibition? (4 points). CO‘W?L¥\\;‘V2‘ SVGLK w‘w fok QV —\-o W (Lu. (M1ij b. Why can the binding of product to the enzyme be neglected in the Michaelis-Menten treatment? (4 points). Vs‘m“*‘\w mavens We» "WM “WM seesaw hm tel: o. Au'swEQ ‘49 Name c. For the enzyme described above whose molecular weight is 32,000, the velocities were measured for [S]0 = 0.30 mM and 1.5 mM and found to be 19 uM-s'1 and 20 uM-s'l, respectively (enzyme amount was the same in both cases, 1.5 ng). If the total reaction volume was 4.5 mL, what is the numerical value for km for this enzyme? (6 points). Nok Wow a Cxuu-go\§ WC.ng \‘o [310 Cf“; C’\m°$\' no woweg‘ '“ V 60 \ V-mcm ‘cm - v Leta : x ( g l M ""\06.qmo\ ‘000 y(( 5 o i W x (._._.____. = " ‘4- A lxxoqyg 32,00025 \flr L ‘0’“) H") warts“ keel : I ,5 LO * ‘0 “If 2 2‘0 x 5“ (1. Initial velocity conditions require that no more than 5% of the initial substrate concentration be converted to product. For the enzyme described in part c and an initial substrate concentration of 0.30 mM, how long will it take the enzyme to consume 5% of the substrate? (4 points). v}: \q we 5% oQ 0.30 M3 5 0.0\5 mg 3 \‘SMhfi ISM»; 5" ‘- Ans») ea \Lej Name Bacteria such as Escherichia col i develop a proton gradient across their cytoplasmic membranes ([H+out] > [H+,-n]) and this gradient is used for a variety of metabolic purposes including ATP synthesis. (Total 32 points). Lactose permease catalyzes the import of the disaccharide lactose into E. coli cells. To which class of membrane proteins (not class of transport proteins) do you expect lactose permease to belong? Briefly explain your answer. (4 points). Ho OH o Hogwo 0“ OH HO O H Lactose HO O fixno- A, he.) Arc movt RNA-0R (30035 G Mambvovxu. Mn pvt/mam. \5 W305\- \\\u,\ 8-9 ‘06. cm WAC, VOA MLMbVGWL veereiws ‘1 3 P b. How many anomers for lactose exist? Briefly explain your answer (4 points). C. (we won“! MA “(Shhhatqg Q,thng my) 5039” v.) m eqUfi‘OQva le/m We Que ORBEMNIBQ“ 'r’wc M’va Qwomb/\L po-SGr‘mm ‘5 33°” ‘R CW3 attack so 96 SXQNtooMe’w/WVbW“, \5 (2‘)“) Lactose import involves proton symport. Given a pH gradient across the cytoplasmic membrane of 1 pH unit, what is the maximal concentration gradient of lactose that can be overcome by lactose permease if one proton is imported along with each lactose molecule? (4 points). A °m l)“ “mu EASE/"WW;- mLCmS a \O-COQ, Bight/«mu. M bill \C {Hq Moog-3 bbw‘fi a \DnCmb cowvaAve\\0vi Svo'b\fi.~'\*\ \ecficu some meat (o\ mos-l) Up a ‘0 -Co\b QCQUMU\Q\\OW (SIGMMX. AmswEK (631 Name (1. Lysozyme catalyzes the cleavage of glycosidic bonds analogous to that found in the disaccharide lactose. The active site of lysozyme contains two amino acid residues essential for catalysis: Glu 35 and Asp 52. The pKa values for the carboxy side chains of these residues are 5.9 and 4.5, respectively. Draw these two amino acid side-chains in their predominant protonation states at pH 5.2, the pH optimum of lysozyme. (4 points). 0 2km“ Gm 35 pm“ = '53 (51 a \oc.\ow ?KQ.5O evolenafiou COMOVQB3 ’. ~ 9 W0 . \c 8: ‘nM n :35? 5'2 ‘9qu H5 (5.2 o a‘oovc \9 <1. 50 5°?” °"’°‘ o Coven/o). e. How can the ionization states of Glu 35 and Asp 52 (see part (1) explain the pH—activity profile of lysozyme shown below? (4 points). 100 \l 0'! ActIVIty (percent of maximal value) M 01 01 O 2 4 6 8 10 pH ’f—M \oe\-\ow\ \‘Vu' ‘5 WC“ W Ccfioklhc, mechanwm VLC‘UWt/S GVU- lf'B’H and 0M gvouv,‘ cmb GM,‘ 35 M33 A5? 52 Q“ m“ vowfi Bow; pH 5-2‘ W gawCQmV0*‘0W 0% A3? '32 \v1 W3 bivvoitovscka ‘CUVW‘ \OUJOWUD +00 \cw‘, o‘oOvL ’PH 5.7,, W concevikct‘no‘o 0Q GMA 35 \n i¥> pvo¥0mo¥® QDVVV‘ Cc'\\5 0W. “ck Wag PM 5,2 ‘5 b‘xwu’“ % V‘C‘k “MW”. so *‘qn maxwmtcb «Wt CoWCUn’tVQMmflS cg ‘oolm ‘ocswcb Womakflw Shh/H. AuwaR Kev Name f. Once imported into the cytoplasm, lactose is hydrolyzed to its two constituent sugars by the enzyme B-galactosidase. Like lysozyme, B-galactosidase uses two amino acid carboxylate side chains in its chemical mechanism whose pKa values are essentially idential to those of lysozyme. Use curved arrows to show a chemical mechanism for [3-ga1actosidase—catalyzed hydrolysis of lactose. (8 points). \—\ . “0‘ O '6‘? n 0Q e ISA-O \ (MA fiw 9 \ on é‘wv e v) HO HO h HQ E OWOW‘E’V \w \\ A W0 5 0L by L , 7:73— .....‘> (a, 0\A A) 1“", O \ \/ he aka-«o ~ 0 0 O 6\'c.\’\ L ‘L e; Wk O’\"\ 5\&‘5\\‘®\‘°“ W O K) H0 0‘4 on Eg +“°’\~Lz \A'O \ \ \‘\O \ “O o —\ HQ (9 \r O O O Wko—H g. One of the sugars liberated by the hydrolysis of lactose is galactose. What is the name of the other sugar? Draw the other (non—galactose) sugar in its Fischer projection form and determine whether it is a D- or L-sugar. (4 points). ’rm 0% oust}! \5 SWLOK 0H (“email ‘ 040 < f‘ 0 0‘4 4 HO —— a R K!” >‘ ‘ ___ H 0‘ \-+0 3 no _ no H HO OH '5 3“ H OH H OH '\ CH SH t L CowagwehtM \\ \5 c. D« bug?" A «.3 3w 6 <1 \Ce‘I Name 3. The major glycoprotein (protein with carbohydrate attached) of the human red blood cell membrane is called glycophorin, which has the following properties. (Total 16 points). 0 Its molecular weight is 50,000 g / mole (60% protein, 40% carbohydrate) o It is an integral membrane protein 0 Five fragments are produced by CNBr treatment. Three of these fragments have carbohydrates attached (see diagram below). 0 Lactoperoxidase is an enzyme that labels exposed tyrosine residues with iodine. When intact red blood cells are treated with lactoperoxidase, then with CNBr, fragments CNBr- l, CNBr-2 and CNBr-3 are labeled with iodine. When red blood cell ghosts (ruptured, empty cells with holes in their membranes) are treated with lactoperoxidase, then with CNBr, fragments CNBr-l, CNBr—2, CNBr-3 and CNBr-5 are labeled with iodine. Ii 11 [l 8 fl = CNBr cleavage site :va MO MO MO C02 _ covalently-attached ' carbohydrate chain Om Ow Ow Ow <————><—————><—-><———><——> \— N (‘0 V In L'— .L i L L'— on CD to m m 2 Z Z Z Z O O O O O a. Why is fragment CNBr-5 labeled with iodine by lactoperoxidase treatment of red blood cell ghosts only? (4 points). Ghost) have Web.» We} Q\\0u_) \ab‘re?c“°>‘\.5013t “\43 ain‘t-v W unmet 0i Wax“ CM'fiv-S mus): be, CW) \va\’vc.o..\\u\ocv VUBHmA oQ (3\\1co?\oovim b. Why does fragment CNBr-4 fail to be labeled with iodine by lactoperoxidase treatment of intact red blood cells or ghosts (this portion of the protein does contain Tyr residues). (4 points). 2* \s \ncuabsx‘au wwov on covaWtcwfi RSQ’ 90 "\’\m& vt‘iwavw 0005‘» be— w‘fi’hm W mom'bvou c. What would you predict about the amino acid composition of fragment CNBr-4? Briefly explain your answer. (4 points). <5W5U. A \5 “)1me Ha» W7mean Han attire“ hwmfib \nouL o\w\o°o‘r echu‘swa 'hyblophobxc. «.335st ___&wC-_<L\é§1— Name d. Using the diagram below, sketch the way in which glycophorin interacts With the membrane. Be sure to indicate locations of CNBr fragments in your diagram. (4 points). CNS; .2 g“ 3 CM?” 00$ 4 outside eggs _ H .,, } cell membrane cell interior 04?): "5 J C026 AnswE 1 KEj Name 4. All parts of this problem deals with a newly-isolated NDP kinase and its mechanism. This new enzyme may or may not follow the same mechanism as the NDP kinase discussed during lecture. The enzyme catalyzes the reaction shown below. (Total 26 points). GDP + ATP ‘——- GTP + ADP a. Would you expect the reaction shown above to favor reactants strongly, favor products strongly, or favor neither reactants nor products strongly? Briefly explain your answer. (4 points). ,3 W‘S‘fl—Q‘OQJV%V] bovbb on M!»- 3 hng- emeNofi bombs cw wo>\n\* Do “’3‘ Unow%l‘ be «u:th 5xbc \‘s aha/MAN gm)on b. Draw the complete structure of GDP in its predominant ionization form at pH 8.0. (4 points). 0 “‘9? 0 0 (N N/XNH /© 2 ovP-ou?~o’\ 7 ‘ \ 0 o ’ \ He) on 0. Tracking the chirality of phosphate groups is a useful method for uncovering enzyme mechanisms. In the reaction shown above, indicate which phosphate group in the reactants should be made chiral to delineate the mechanism of the newly-discovered NDP kinase. Briefly explain your choice. (4 points). «‘5 i)\’\ @ 3-?MQK ‘5 . E? O 0 A QOFV~O-?~o~‘P-o/\/OE) ‘ \ \ 0 e 60 C90 HO ou we, Wen/W 5-D MGM W ‘WOMfiQQNu/LA PWOSQWOAL Q)\’3\VG~\ \-o ‘02}me m wvmbm’ 0Q “VUVOVV‘V‘Q o\-\o.n.\L» 5mg» octuv m. 3H!“ > A m awe R \Cev Name (1. When the appropriate reactant phosphate group was made chiral using oxygen isotopes, its configuration was inverted when found in the product. Based on this information, use curved arrows to propose a chemical mechanism for the NDP kinase-catalyzed reaction between GDP and ATP that is consistent with these data. (8 points). \nVUVSK'J‘n mccms a g‘chbk vsucAcopMKVL MAM/Nd, so. no ?W03?\00¢\1\Gx\\b comm \s mooweka V7ch 0 C Q a N \ n O A agav‘oe 0—?«0»§n0«?ro/\/ I G3 G) (90 ’ \ O HO 0‘; \ 0:3?"C3 0Q ~to I x O HO OH 0 9 1| Q (a 0‘ O A 0—?“O~?~09 mime—9-0% o 09 e ‘ G) 60 HO 0L3 ‘ \ / m HO OH ADP \h Answ‘ez \Ce‘1 Name e. The diagrams shown below use Cleland notation to depict four possible kinetic mechanisms for the newly-isolated NDP kinase. Circle all of the kinetic mechanisms that would be consistent with the information in given in part d. Briefly explain your answer, indicating why each mechanism was or was not selected. (6 points). GDP ATP GTP ADP l i i i T E E-GDP-ATP = E-GTP-ADP E 1 E-ATP T l E-GTP i ATP GDP ADP GTP GDP ATP GTP ADP l l i i 2' E E-GDP E-GDP-ATP~—-—‘ E‘GTP-ADP E-ADP E ATP GDP ADP GTP l l i i 3' E E-ATP E~GDP~ATP= E-GTP-ADP E-GTP E ATP ADP GDP GTP l l l l E E-ATP ~——‘ F-ADP F F-GDP = E-GTP E ‘BuuB 0V) ow NWQWDM WC conno\ Btubc thA-Mv W I 00%qu wow?) Show vomeow (9+1) 0v OTBWQ-B (PIP-3) \Qflc‘f‘fia WILQMOV‘WW“), \OVV a“ QC AM.“ \nvokm WW7 a SW303U1 «WC/\LODWVQ Q‘AOL‘P on PBOPVV’OVW was womb wwcfiwc ‘9‘ COWSDST‘m‘r w\\‘\n Th s-\~Ucoc\qewx\c.o\ ham. “Cu. Qwuw (ng- pox/€30 mnhenum (Lem b; «ukb eu‘r 3mm A swam)» vL’remkson a? Skvcocwcwxn‘w) ox pvmsafinovus. 10 “132% Name 5. Energetics of membrane transport. Note that R (universal gas constant) = 1.987 cal / mole-K and 9"(Faraday’s constant) = 23.1 kcal / mole-V. (Total 8 points). a. For a typical vertebrate cells with a transmembrane potential of +0.070 V (inside negative), what is the free energy change for transporting 1 mole of Na+ out of the cell and into the blood at 310 K? Assume that the concentration of Na+ inside the cell is 12 mM and in blood plasma, the Na+ concentration is 145 mM. (4 points). [N&‘)uu\. [UA‘lm * AC;me : \‘fi (“‘5 ‘ Wm m\)mo\-\C)(3\D a xv, (‘2 Mm * (musnoow\)mo\-v)(+0~ovov) .1. 3I\Sl \Q,&\)mo\ :- keelmob. b. Suppose that you determined experimentally that a cellular transport system for glucose, driven by symport of Na+, could accumulate glucose to concentrations 25—fold greater than in the external medium, while the external [Na+] was only 10-fold greater than intracellular [Na+]. Would this violate the laws of thermodynamics? If not, how could you explain this observation? (4 points). Ma, \S: two N03 mun, cov’rv‘cmspovteb wfiM QQCM (SKACOSL. MM"- woo\b \OL bvmxum’r one/VB” 3H) OVWCOW W 15-Co\’é wwLwVeMo» %V05\EM\'. ll ...
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This note was uploaded on 05/21/2008 for the course BCH 3218 taught by Professor Johnsteward during the Spring '08 term at University of Florida.

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exam_2_key - Answer. \Lg Name CHM 3 2 l 8 Spring 2008...

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