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DOC091209-012 - Chemistry 3785 Third Hour Exam...

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Unformatted text preview: Chemistry 3785 Third Hour Exam Name "Shred 203:“ Summer 2009 100 points Fri. June 19, 2009 Points Score Points Score 1 48 2/g IV 12 [O 11 14 V 14 III 12 Total 1 00 i i f/eaw W A table of reduction potentials for selected biochemical half-reactions is on page 8. 4W 1. (48) Multiple Choice. Circle the letter corresponding to the best answer. Three points each. 1. Adrian Brown proposed that the overall reaction for B—fructofuranosidase occurred as two elementary reactions that could be written as t, k; E+S—“—"ES _.—>E+P ‘_.— k4 Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression: 3- k1([E]r -[ESD- b. k1([E]r -[ESD[S]- c. ”k; [ES]. gt, [ES] + k2 [ESM 1"! [ES] 2. In the derivation of the Michealis-Menten equation, the steady state assumption a. assumes that the rate of formation of the ES complex is equal to its rate of dissociation. b. assumes that product is formed as rapidly as the ES complex. @ assumes that the concentration of the ES complex is unchanged during the time course of the experiment.>f‘ d. assumes that the k; = k -1 = k; e. assumes that the substrate is depleted at a steady rate with time. 3. The following data were obtained for an enzyme in a study of an enzyme known to follow Michaelis- Menten kinetics. [31 v0 mol/min [S]( M) \fo : \[mtx 'l::;-l._§l 49 1 \ 96 2 349 8 621 50 km: AV...“ 676 100 698 l000 699 5000 .\ What is the KM for the reaction? G) uM b. 700 umol/min 0. 5,000 uM @ 350 umol/min e. 100 am it 4. a. @ c. d. (9 5. d. © a. G) 10. a. An enzyme is said to have reached catalytic perfection when Vmax is on the order of 108 to 109 lVLs‘l. I kcm= 108 to 109 M-s". / oi" wt KM is on the order of 10'8 to 10'9 M. “d“ ‘i l“ ‘0 [Coat/KM: 1 none of the above are truest Which statement is true conceming the structure and function of aspartate transcarbamoylase (ATCase)? ATCase is allosterically activated by glutamine. CTP is a feedback inhibitor of the enzyme. The regulatory subunits bind carbamoyl phosphate and aspartate. The catalytic subunits are related by a two-fold rotation axis. both b and d are true. What is true concerning the regulation of ATCase by ATP? ATP is an allosteric activator, increasing the enzyme’s affinity for its substrate} The KM for aspartate increases in the presence of ATP. ATP preferentially binds to the T state of ATCase. all of the above are true. none of the above are true. / For the reaction A —-—> B at 25°C, the change in enthalpy is —7 kJ-mol'] and the change in entropy is -25 J -mol'1. Based on this information we can say AH , ,7 \“5/Mo' the reaction is spontaneous at all temperatures. ’ { 37m o\ the reaction is non-spontaneous at all temperatures. D3 3 "9‘ 9 , if?” E the reaction can be made spontaneous by increasing the temperature. D“ fl 03‘; ) the reaction can be made spontaneous by decreasing the temperature. Y DH 5" gall” ‘ there is not enough information to answer the questiOn. / /\ ”’3‘ u \ a® $06 The AG“ for the isomerization of glucose-6-phosphate to fructose—6-phosphate is 2.2 kJ-mol'l at pH 7.0 and 25°C. The equilibrium constant for the reaction is R = 8.314 J-K'lomol'l. 0.4114. s Kg 3.95 x104 . ,. ewe/4 Sign ?..53><10_5 / gt/ .‘v‘ 1 hi 4 6/ ’ W is 0 ' q i“ W“ I" Anabolism is characterized by reactions that result in the release of free energy. the synthesis of biological molecules from a few precursors». reactions involving the reductiori of substrate. 3? all of the above. only b and cat The coenzyme accepts 2 electrons in 2, one electron transfers. mm+ b. NADP+ @ FAD . d. TPP 6. both a and b flared \ng LYN ll. All the following contribute to the large, negative, free—energy change upon hydrolysis of “high- energy” compounds except: the low free energy of ATp as compared to its products of hydrolysis}? . electrostatic repulsion in the reactant. b c. stabilization of the products by solvation. d. stabilization of products by extra resonance forms. e both a and b 12. Which underlined substance has the greatest reducing power? a. Lactate' ——> pyruvate' + 2 Hr + 2 e' __H_;Q—%V202+2H++26‘ Q- 0. [i-hydroxybutyrate —-—> acetoacetate + 2 H+ + 2 675- g» . Dihydrolipoic acid ——> lipoic acid + 2 1-1+ + 2 e' e. Succinate' ——> fumarate' + 2 H“ + 2 e' \ 13. The steps of glycolysis between glucose-6—phosphate and 1,3-bisphosphoglycerate involve all the following except? a. consumption of one ATP @ reduction of NAD+ to NADH c. an aldol cleavage @ a substrate level phosphorylation it e. isomerization of substrate 14. Certain hereditary metabolic defects arise due to the loss of single enzymes of catabolism. What is the most likely consequence of the lack of triose phosphate isomerase? . Impaired ability to use glucose as an energy source. . Inability to use fructose as an energy source. C“? Q Inability to use galactose as an energy source. . lethal: prevents the use of carbohydrates in carbohydrate catabolism. both b and c 15. ATP is produced in which of the following transformations? a. Glyceraldehyde-3-phosphate ——> 1,3-bisphosphoglycerate ' (I) Phosphoenolpyruvate —-—> pyruvate c Glucose—6-phosphate ———> fructose-6-phosphate d. 3-Phosphoglycerate ——> 2-phosphoglycerate e both a and b. 16. All the following statements about the regulation of phosphofructokinase (PFK) are true except: . Fructose-é—phosphate preferentially binds to the R state of PFK. . The most potent activator of PFK is fructose-2,6-bisphosphate. PFK is an allosteric enzyme that shows sigmoidal kinetics in the presence of substrate. a 115 g. 1 ,df Citrate is an inhibitor of PFK. @ Elevated levels of ATP increase the enzymes’s affinity for fructose-6-phosphate. Scared 1263;!" II. (14) The AG“ for the conversion of fructose—1,6—bisphosphate to dihydroxyacetone phosphate and glcyeraldehyde—3—ph05phate is 22.8 kJJnol'l. Answer the following two questions based on this information. a. The cytosolic concentrations of the metabolites are as follows: fructose-1,6-bisphosphate, 8.0 x 10‘5 M; dihydroxyacetone phosphate, 1.6 x 10“1 M, and glyceraldehyde-3-phosphate, 8.0 x 10'5 M. Calculate the free energy change, AG, for the reaction at 37°C. R = 8.314 J-K'l-mol"]. Please show your work. AG; fi'} {LT 31% : ‘3\0.\S \c . [aflosnmwdq @310.IS\(8-3H) w TT ( {55: 0.413 5‘/.M\ Cacb Aé : .416“; gram /%._ /(€ 9/ b. Is the reaction spontaneous? If not, what can be done to make it spontaneous? 9/ 12+ 1, “Us, Watched-as. we so sewers Mi the rents» l) awe J“: Equilibrium Jrku; five. equflibfihm wk“ ‘9‘! OK We trod-aw Fl 5- ‘ u “A“ ‘ " , id‘s?“ 5r; 3— S\d€¢ .er order 55..., ,mn'V—Ea. Mn.” rec-(LHUH Swoinhnewvé ‘H‘e- qddi H0“ 9‘: d0 IL. HAVE would Lrww—J- ‘ pkrwyfifiom we.» lope! bl:- car/K fush W‘— «new: HM DE“ ‘30 " (j Cmmmnk') 35¢. {‘nguifld" +0 5 LT (oi-14:5” l'fovz MDMP] («‘7 $667422] ,g/ C3 ja (“Fol Q05 r; k“ III. (12) You and your laboratory partner are studying oxamate which is said to inhibit glycolysis. Oxamate and glucose were added to a cell free muscle extract under anaerobic conditions. After 30 minutes, samples are assayed for levels of various glycolytic interemediates. Compared with a control experiment performed in the absence of oxamate, you note an accumulation of glyceraldehyde—3— phosphate, dihydroxyacetone phosphate, and fructose-1,6-bisphosphate. Levels of phosphenol- pyruvate and pyruvate are relatively low, and there is virtually no lactate. Your partner concludes that glyceraldehyde-3—phosphate dehydrogenase is inhibited by oxamate. However, you suspect that another enzyme is affected. You repeat the experiment, but this time you include NAD+. After 30 minutes and compared to the control there is no accumulation of any hexose or triose phosphates, and only pyruvate appears to accumulate in the extract. Your partner immediately agrees that another enzyme is inhibited by oxamate. Which enzyme is it? Briefly explain your answer. y, r; . o , . _ a , , . gar, H3NLii—coo' We” MW Cy ”pf/M mate / M rt w W Betrj that: was 09 V E Q and P‘l’” «fee NE \W a “1%,“ng Rpm; HA9. Carr“:- '4’“ 0; ““15?— lruqo W‘ €¢U\€) c “a. Eileen} 90’“ M cob-1" Hm o£ TL' wufit} mnver’l’ Jena DHIK? il'o GA? as GAP i'i (Unverx‘e! Tim‘s “we. ecc'ide‘)‘ enzyme ”W5“ l: 6 G-N’D H. 15" G N) D H were Motl- gQCQfJ'CL-J level: .7“ G A? uuuld \93/7 lower a5 hell .5 w, W“ A C; aMV/dmdz‘ £232? IV.(12) Under typical cellular conditions the AG of hydrolysis for ATP is -50.2 kJ/mol. Given thiséfl’f’fl: information is it energetically possible for bacteria to survive using hydrogen sulfide, H28, as a sole hydrogen source and nitrate, NOg' as the sole environmental electron acceptor? In your answer include the following: (a) the relevant half-reactions, (b) the overall reaction, and (c) calculations of A2“ and AG‘". Faraday’s constant is 96.485 kJ-V'l-mol'l. Briefly explain your answer in a sentence or two. Please show your work. by: so; +244 1.1/5 :1 neg +HLO 0.423. +0.2. up :2 s + Met! 9 0.25 1/ no; in; :2" N07. + b + H10 Alf, .. “fig/3i”. )é/ “;Ll a [nu/3’ m ' 97360185, be - q 99,485)(U.5,s) N.” 0 95' ricrflkdh .ts &r(‘i€€j 6"“A (Set wr' ed 1205.2. 1" V. (14) The protease papain cleaves the substrate analog p-nitrophenylhippurate. This reaction is inhibited by N—benzoylarnino acetaldehyde. 1.2 x 10'7 moles of papain were used to generate kinetic data in the presence and absence of 3.3 x 10'5 M N—benzoylamino acetaldehyde. The resulting kinetic data were plotted using a Lineweaver-Burk plot. The plot is on page 7. Answer the following questions based on these data. a. Calculate the values of KM and Vmax in the presence and absence of inhibitor. Include the proper units in your answer. Please show your work. do : Vmax ‘ ‘0“th uh\n\r§\\°\\e¢3‘ “MAJ-Aral. \ . \ - [5'3 ‘1 F F‘- . H l -‘ me 5 / s MM run — Vmbv m '3' ‘\ VWMJ: 4.5mM/m;n \f x q S / Km I .— lt 9‘74 .5 3 "5 a '1 an». “Menage "‘ ‘ DAM he, 9 [st/MM iii—‘62 X. ink ' h it. 1. _ E ”—3. ESF‘L*? *1 it “" ET. b. Calculate the turnover number as number of molecules converted to product per second. Include C. the proper units in your answer. LLEMM lanai“ . _n . p / “rid“ 1;; if; 1 7-3“ l0 5 Wigs/sen (7 "\ g: . ’ 75 M0“ Me @913 'MUB‘W‘WUM “$3 :7 E l 6 Me— '— ,_. (o 'q '\ \ / ( “Ev-W. LLB I? X m9 2w ‘25 5%, bin 9/ What is the dissociation constant for this inhibitor? Please show your work. / j 3 l + «7” 6’2 I as" '1 (ha: K: M as a an m (1“an (-6021 m€.\ ...
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