Final Exam Spring 2007 - University ofHouston BCHS 3394/ J...

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Unformatted text preview: University ofHouston BCHS 3394/ J ’,' ‘ Final Exam Instructor Dr. Briggs May 10, 2007 Name l l ’1 4 1" I ' \‘ Answer all questions. LAST, FIRST Multiple choice 2 pts each (100 pts total) — Choose the most correct answer and fill it in on the Scantron sheet. Also circle your answer on the exam and show ALL work. l). The conversion of pyruvate to acetyl-CoA requires which of the following coenzymes? a. Biotin I b. Vitamin C i @Thiamine pyrophosphate ., d. NADH e. ATP 2). Incubation of gram-negative bacteria with lysozyme in an isotonic medium causes rod—shaped bacteria to assume a spherical shape. The cause of this phenomenon is: a.’ absorption of water AWSUbF P-}I®destruction of the cell wall . c. destruction of the cytoskeleton ,d. damage to the plasma membrane I e. change in gene expression 3). In an intact cell, the free energy change (AG’) associated with an enzyme-catalyzed reaction is frequently different from the standard free energy change (AGO’) of the same reaction because in the intact cell the x 7 ,a. activation energy is different A 7 '7 b'frreaction is always near equilibrium ,_ , ‘9/ enzyme may be regulated allosterically AMLM Al~-->l@eactants are not at 1M concentrations fie. reaction may be catalyzed by more than one enzyme 4). In animals, an enzyme unique to gluconeogenesis is a. enolase b. phosphoglycerate mutase c. glyceraldehyde 3-phosphate dehydrogenase d. aldolase A‘\SLC£L‘\ “L2,; @mctose bisphosphatase 5). DNA polymerase contains a lysine residue that is important for binding to DNA. Mutations were found that converted this lysine to either glutamate, glycine, valine, or arginine. Which mutations would be predicted to be the most and least harmful to the ability of the enzyme to bind DNA? / ) Most Least C‘.\‘;:x.,,‘~ \"w’n = r / r/agValine Aspartate I V L v .i OL/Glyarle ' /c.~Arg1n1ne Glycme ’ ‘ A ”‘ r -- Aifilutamate Valine r “>93 \\ @lutamate Arginine, of ~ I \p Jr“ \ 6). A purified peptide derived from a flu viral protein important in invasion of cells is subjected to treatment with cyanogen bromide and chymotrypsin. CNBr treatment of the peptide results in the fragments: - Arg-Thr / Phe—Ser-Met / Arg—Thr-Phe-Arg—Thr-Met I~ Chymotrypsin treatment of the peptide results in the fragments; U‘Ser—Met—Arg-Thr if?» t a; «_ V/Arg-Thr—Phe ‘ " Arg—Thr—Met—Phe Circle the letter corresponding to the correct amino acid sequence of the peptide. ,, a; Arg—Thr—Phe—Ser-Met-Arg—Thr-Phe-Arg-Thr—Met /’3W$Wef {@Arg-Thr—Phe—Arg—Thr-Met-Phe-Ser-Met-Arg-Thr \ c. Arg-Thr—Arg—Thr—Met-Phe-Ser-Met—Phe-Arg-Thr d. Phe-Arg-Thr—Met—Arg—Thr—Phe-Ser—Met—Arg—Thr e. Phe-Ser-Met-Arg-Thr—Phe—Arg-Thr-Met-Arg-Met 7). Which of the following statements are true of water and/or hydrogen-bonds? \ / ;:¥\The oxygen atom in water has a partial positive charge "’(ZTThe hydrogen atoms of water each bear a partial positive charge } N Eli-bonds can form only between water molecules l' ‘x I "(QB-bonds can form between different parts of polypeptide chains / jfffifiH-bonds are relatively weak compared to covalent bonds C(QZH-bonds in water contribute to its cohesiveness '”(‘7)TH~—bonds play no role in the solubility of proteins in water p—iw u (69> v Artsmc O. \- ® Wmv-bw-b moat/1.5m O\\]O\O\\) u M. pa 8). Using the following molecular weights and isoelectric points: Protein Molecular Weight pI Pepsin 78,000 1 Hexokinase 150,000 5.7 Myoglobin 38,000 6.8 ' ’j Lysozyme 33,000 11.0 What is the order of elution from a DEAE-cellulose ion exchange column at pH 7.0 of the proteins in the mixture? Answer “Km? @VUysozyme, 2nd-myoglobin, 3rd-hexokinase, 4‘h-pepsin I ’ b. lS‘-pepsin, 2"d-hexokinase, 3rd—myoglobin, 4‘h-lysozyme c. lS‘-hexokinase, 2nd—pepsin, 3’d-myoglobin, 4‘h-lysozyme ’ d._ lS'-lysozyme, 2nd—myoglobin, 3rd—pepsin, 4m—hexokinase fljfef’lst-myoglobin, 2nd-hexokinase, 3rd—lysozyme, filth—pepsin 9). If AGO’ for an isomerization reaction of [A] (—9 [B] is 15.0 kJ/mol, what is the equilibrium ratio of [Al/[B] at room temperature? 5; \r /,l’,“\ ‘7‘, .2 , A. 1 i z“ 1 \‘ _. r '. b. 0.994. 45.. N ' ~ V‘ ‘ ‘ ‘ ‘\ ‘ C. \ x, N W , 5 l d. 0.067 \ .u.»'. ‘r IT‘: if ‘21.; l 4. " 9 E‘n: in 4 «- _‘ _ . _ LA '1‘ .v‘ s ‘ ' I E l. K" ‘ kixs we»? @426 10). Calculate AG°’ for the reaction A + B <—> C + D at 25°C when the equilibrium conditions are [A] =10 uMw B] = 15 uM, [C] = 10 uM, [D] = 10 uM. AWL”? Mg» 52171004 J/mol A _ MR? a a, , *b. 10 kJ/mo] u v \J c.1J/mol K n C k I ‘ Time . d. insufficient data to determine answer ’V' < - ' r ' - > e. none ofthe above =9 =c- "‘ T 100 . 1 11). Th ,mosphere of early earth probably contained the following molecules: Answer . H20, coz, N2 , CH4, and NH3. / -. zb H2O, CO2, CH4, C5H606, and NH3. 0. H20, C02, CH4, C6H606, and COOHCHZNH3. V H20, CO2, CH4, and 6. none of the above 12). Amphiphilic molecules: a. have both oxidizing and reducing groups. b. are micelles. c. have chromophores in two different wavelength regions. d. have both acidic and basic groups. Any-sec ave both hydrophilic and hydrophobic groups. 13). If you add a drop (about 0.05 mL) of 1.0 M HCl to one liter of pure water (assume pH 7.0), the pH would be r I . v __ , 1W), " ...,itr>‘\ v s \ t [/02 \ K . 2.7 Answer ":9 l ' c. 9.7 d. 7.0 e. 5.0 14). If you add 1.0 mL of 1.0 M acetic acid (pK = 4.76, K: 1.74 x 10—5) to one liter of pure water, the pH would be \ x r; I I y if: '\\§\ _. ,1 {A . i k V. 10.1 x ti, . b.53.0 r. ~ “0’. 1.0 Answer s d‘ ,9 L 2 \ \pr H e. 1.32 k “x” , a W: k .. \_ i . ‘l . > : ‘n—A \J J) 5,, Lw‘ . a. ‘- . We i: “1.33 15). Hydrophobic interactions between nonpolar molecules or groups: ,aflesult from the tendency to maximize water’s contacts with nonpolar molecules. b. are the result of strong attractions between nonpolar regions. prime the result of strong repulsion between water and nonpolar regions. / d.'depend on the strong permanent dipoles of C—H bonds in nonpolar molecules. @equire the presence of surrounding water molecules. 16). To make a phosphate buffer at pH 6.82 starting with one liter of 10 mM phosphoric acid (sz of 2.15, and 12.38), you could add: ‘ _ , .\ - ' Ia. You can’t make a buffer by adding HCl or KOH. * ' xb’f'S millimoles ofHCl. - c. 20 millimoles ofK+. )d‘325 millimoles ofHCl. Al’lSédfl/imim‘9 millimoles of KOH. I" 17). At pH 1, lysine would be charged as follows: I , ~ -- a. 0 OL-carboxylate, —1 a-amino, —l e-amino, —2 net charge ” Lib/:1 OL-carboxylate, +1 oc-amino, +l s—amino, +1 net charge ‘ l k _ 0. +1 a-carboxylate, +2 OL-amino, +2 s-amino, +5 net charge 1 , » _ ,_ v AV‘R'X’F oc-carboxylate, +1 OL-amino, +1 e—amino, +2 net charge “‘ sf { e. +2 d-carboxylate, +1 OL-amino, +1 8-amino, +4 net charge ‘ \ 18). The, “salting in” of proteins can be explained by: NWJZ? @3211: counter-ions reducing electrostatic attractions between protein molecules. I / ' _. b. proteins attracting primarily salt cations. ‘ c. proteins attracting primarily salt anions. )df‘r'eleasing hydrophobic proteins from nonpolar tissue environments. ritef‘hydration of the salt ions reducing solubility of proteins. 19). Enzymes that hydrolyze the internal peptide bonds (not the peptide bonds of the terminal amino acids) ofa protein are: a. oxidoreductases. lyases. Mvdfif @)endopeptidases. \ / ' nucleases. e._\exopeptidases. \ 20). Trypsin degradation of a peptide yields the fragments H, MNK, IMR, and LMR. Cyanogen bromide treatment yields RM, NKIM and LM. The sequence of the intact peptide is: / / a. HMNKIMRLMR g, .b‘.'MNKHLMRIMR c: RI—IRMNKIMLM , /,.,d./\NKIMLMRHRM Am“? , 21). In a protein, the most conformationally restricted amino acid is ; the least conformationally restricted is a. T rp, Gly fldfne, Ala ‘ .‘ ,2 ref Ala, Pro -' 7 ‘ A 22).Whig;_h\of the following is not a ligand to the porphyrin ring Fe2+ ion in oxymyoglobin? a XQQflVcfl £115 E7 . ~ : His F8 ,.c;’Nitrogen atoms in the porphyrin ring idiOxygen "”e.\all are ligands 23). Which gas does not bind to the porphyrin ring Fe2+ ion in myoglobin? a. NO 1 5 - 2 6. H28 24). When the partial pressure of oxygen in venous blood is 30 torr, the Yog value for myoglobin is “ a. 0.55, 0.91 Answer/{x247 0.55 c. 2.8, 26 d. 0.91, 0.97 e. none of the above while the YO; value for hemoglobin is up ‘n ‘,-‘\i :r. 1. I '7“ | 1.. 1 3}!) 25). Consider a hypothetical hemoglobin with a Hill coefficient of lfland the same p50 value as normal hemoglobin. Choose the statement below that best describes the two proteins. [gt/There is a cooperative interaction between oxygen-binding sites in both the hypothetical / and normal hemoglobins. , The hypothetical hemoglobin has a greater oxygen affinity than normal hemoglobin. Aflgwef‘ fl c. ig‘he oxygen binding curve for the hypothetical hemoglobin is hyperbolic, and the curve for ormal hemoglobin is sigmoidal. [The two hemoglobins would be able to deliver about the same amount of oxygen to the / tissues. e.‘JAt p02 less than p50, normal hemoglobin has a greater 1’02 value. 26). The Hill plot shows that the fourth oxygen binds to hemoglobin with than the first. a. 2 b. 5 c. 10 . I g .20 Amour}? «'——~;a 00 27). How many stereoisomers are possible for an aldopentose such as ribose? I \ .‘ a ' a. 2 ‘i " ' . b. 4 ALi'iS‘oflF .-—7 ’ l ‘d. 16 ” g, e. 32 » 9' 28). Which of the sugars shown in the figure are L sugars? («af'A and B “ireNone of the above CHO OH HO . H H oH H OH GHZOH A CHO H OH HO H H OH HO H CHZOH —fold greater affinity CHO H OH H OH H OH H OH CHZOH/ ) /B CHO HO H H OH HO H Ho H CH20H D 29). Which sugar in the figure on the prior page is the enantiomer of sugar A? I e. none of the above 1_ / 30). Which two sugars shown in the figure on the prior page are epimers? A‘WSW amt; and B “i / . B and C \_ c.\C and D ("dh‘A and D “e. None of the above 31). A lead compound would be most promising if it had: a. K1: 4.7 X10S M. . a - b.K1=1.5x108M. /J-VY\S’W‘“'/f‘ M37 “’3‘ I: 1.5 x 10‘8 M. ' - .K1=4.7x10‘5M. 'etyKM = 4.7 x 105 M. 32). What is the velocity of a first-order reaction when the reactant concentration is 6 x 10—2 M and the rate constant is 8 x 103 sec—1? a. 1.33 x 105 M‘1-see‘1 b. 1.33 x 105 Mosee - ’ c.‘7.5 x 10‘2 Mosec Mme? ‘ _—«———;‘y @113 x 102 M-see‘1 not enough data are given to make this calculation KM is: ‘ La? a measure of the catalytic efficiency of the enzyme. b. the rate at which the enzyme dissociates from the substrate. A 19 ‘ a A game rate constant for the reaction ES e) E + P. Q We“ Lil/:gthe [S] that half—saturates the enzyme. e. the rate at which the enzyme binds the substrate. 34). In order for an enzymatic reaction obeying the Michaelis—Menten equation to reach 3/4 of its maximum velocity, 7 a. [S] would need to be 2KM " b. not enough information is given to make this calculation ” 1 V ‘. . . [S] would need to be 50% greater than KM Z,‘ A/MW‘Q’V ere—4? @[S] would need to be 3KM ‘ e. [S] would need to be 3/4KM \fl 35). An enzyme is considered to have evolved to its most efficient form if . V K a. kcat is a large number AHSwe/P [Cam/KM is near the diffusion—controlled limit " CE'KM is a large number de'kcat/KM is a very small number ,ef KM is a small number , 36). Find the initial velocity for an enzymatic reaction when Vmax = 6.5 x 10“5 molsec“, [S] = 3.0 x 10‘3 M, and KM = 4.5 x 10‘3 M. a. not enough information is given to make this calculation mxfirp b.52.6 x 10-5 mol'sec_l * I t, . 1.4 x 10’2 mol'sec'1 V V :2 , " d. 8.7 x 10”3 mol'sec'1 .5 e. 3.9 x 10'5 mol-sec‘l 37). Which of the following is at a higher level of oxidation than CH3CHO? a. CH3CH20H (if: i’ \» . i L » l . V ‘ v V .pH2=CH2 " ‘ Myers? was; ad.” chozH e1 none of the above 38). Consider the following metabolic reaction: Succinyl-CoA + Acetoacetate —> Acetoacetyl—CoA + Succinate AG°' = ~l.25 kJ/mol The A0" for the hydrolysis of Succinyl-CoA is —33.9 kJ/mol. What is the AG°' for the hydrolysis of Acetoacetyl-COA: Acetoacetyl-CoA —> Acetoacetate + CoA Ans f fag—35.2 kJ/mol ‘ by “if g ‘- B,\—32.7 kJ/mol _ - ' +32.7 kJ/mol /’ ‘1) d. +352 kJ/mol ‘ __I ,1 ‘_ e. none of the above ' ” ‘- "' l) J,» . . 39). The Keq is 0.503 at 25°C for the following reaction. What is the AG°' for this reaction? D-Glucose-6-phosphate —> D-Fructose-6—phosphate /a; £1,700 J/mol 5 _ 3 ._ .. \ _ ’ .ij—2,870 J/mol L .. ‘f /' A" 40). Consider the following metabolic reaction: 3-Phosphoglycerate —> 2—Phosphoglycerate AG°' = +4.40 kJ/mol What is the AG for this reaction when the concentration of 2-phosphoglycerate is 0.290 mM and the concentration of 3-phosphoglycerate is 2.90 mM at 37°C? . a. _+lO.3 kJ/mol ‘ ARSLV- “2F @3—1 .54 kJ/mol "x 5.73 ‘ * g . _ c. —l.30 kJ/mol ' _ \ f \ d. —5.93 kJ/mol . t g . _ j t ' . . e. —4.40 kJ/mol ‘ ‘ " ' " 41). Which is the'net equation of glycolysis as it occurs in aerobic cells such as brain cells? azj‘gilucose + 2 ATP —> 2 lactate + 2 ADP + 2 P,- i lucose + 2 ADP + 2 P,- + 2 NAD+ —> 2 pyruvate + 2 ATP + 2 NADH + 4 H+ flGlucose + 2 ADP + 2 P,- —> 2 lactate + 2 ATP + 2 PF ' / idiAGlucose + 2 ADP + 2 P,- —> 2 CH3CH2OH + 2 C02 + 2 ATP ' e."’Glucose + 2 ADP + 2 P,- + 2 NAD+—> 2 lactate + 2 ATP + 2 NADH + 4 F \ \ Amnerm 42). The reaction below is catalyzed by yeast alcoholdehydrogenase. Which of the following corresponds to X and Y? '- " ‘ ~ // Cr 9H " CH3CH + X‘k CH3CH2 T Y ,. 'a-XZNADH+ Y=NAD+HJr r , - = NAD+ Y = NADH + H)r ~ ‘ '. ' ‘ I. C firm “BF ' = NADH + H+ Y = NAD+ wiX=NAD+H+ Y=NADPP f e.‘~.x = NADPH + H Y = NADP+ 43). In skeletal muscle cells, the NADH that is produced by glycolysis under anaerobic conditions (vigorous exercise) is regenerated to NAD+ by the conversion of: a. acetaldehyde —> ethanol. bf lactate —> pyruvate. {a ‘ ‘ ., “ c. phosphoenolpyruvate —> pyruvate. AV\SLU€/S\ mm”? d.“ yruvate —~> lactate. \’ e. glyceraldehyde—3—phosphate 4) 1,3—bisphosphoglycerate. 44). Which of the following metabolic conversions is considered to be the major control point of glycolysis? ' k. .p s . p a. Fructose—1,6-bisphosphate' —> dihydroxyacetone phosphate + glyceraldehyde—3—phosphate . I/b'TGlucose ——> glucose-6—phosphate ‘, ' ’ / .2 a i A 2—phosphoglyerate —> phosphoenolpyruvate » 7 ~ /J\(VL5Wt3/l\ wructose-6-ph05phate —> fructose-1,6-bisphosphate g”'pyruvate —> lactate 45). What are the coenzyme forms that correspond to X and Y in the following gluconeogenic reaction catalyzed by glyceraldehyde—3-pho‘sphate dehydrogenasez 1; O . - 0 ll ‘ _ || '203POCHZcHCOPO3'2 + x —> 203POCH2C|3HCH + Pi + Y OH OH 1,3-Bisphosphoglycerate Z . I Glyceraldehydef3-Phosphate ,. . . , ‘ F L”) \ R. i «E ~. _ ,_ 6/,.\ g ‘ \ ‘. A‘N‘vbl = NADH + H, Y = NAD+ ' ‘ ' Vb’l’ X = NADPH + H, Y = NADP+ \_ " ‘. i c./X = NADZ Y = NADH + W " Edi X = FAD, _ Y = FADH2 / q/x = NADH + H*, Y = NADP+ 46). Biosynthesis requires the expenditure of free energy. Which of the following metabolic conversions involved in glucose synthesis requires the direct expenditure of ATP free energy (ATP is also a reactant)? va. 'Glyceraldehyde-3-phosphate + dihydrdxyacetone phosphate —> fructose-l,6—bisphosphate I " ".::.Jb*.’fFructose-l,6.-bisphosphate —> fructose-6-phosphate ” " i c. §LPhosphoglycerate —> 1,3-bisphosphoglycerate ‘1 ' \ A \ V 7‘1,3—Bisphosphoglycerate —> glyceraldehyde-3—phosphate ~ e. None of the above. 47). Which of these compounds is oxidized by a multienzyme complex that requires five different coenzymes? / It -02C—Cl::C—COZ- O n 'OgC-CHZ—C—COZ' C. '02C—CH2—CH—C02' t? 'OZC4CH2—CH2—C—Coz' (PI CH3—C—SCoA 10 48). One reason why arsenic is poisonous is that arsenite ion (A5033) reacts with reduced lipoamide I e), and C I _ 0_ to form a stable complex as shown to the right. This derivative of lipoamide cannot be reoxidized. Which of the following metabolic conversions cannot occur in the presence of arsenite? ’ O ' II a. Pyruvate + C02 —> oxaloacetate. Lys b. Pyruvate —\> acetyl—CoA + C02. I Wklwa c. a-Ketoglutarate"a succinyl-COA+ C02. 3\ s \‘Phosphoenolpyruvate + CO; ——> oxaloacetate As 49). Which of the following causes pyruvate dehydrogenase kinase to catalyze the phosphorylation 50). If acetyl-CoA labeled with MC, as shown in the figure to the right, were Haw/Pun? and inactivation of E1 in the pyruvate dehydrogenase complex? a. \’ levated concentrations of NADH and ATP . elevated concentrations ofNAD+ and ADP ix’CTC‘Sa2+ %’insulin e. elevated concentrations of acetyl-CoA used as the substrate for the citric acid cycle, which of the following intermediates would be produced during the first round of the cycle? 0 . n * CoAS—C—CH3 air/ we ‘oJ‘C—p—(lj—CHz—Co; f3\ H C02— 91“? ‘ozc—c—ci‘CHz—Co; H (:0; 19.1” ' 9H1? ‘ozc—p—c—CH2—*co‘2 H co; d. 9H? ‘02C—*c—(|:—CH2—Cog H co; e. - OHH _ | | _ OzC—(II—ik$—CH2- C02 H co; ll Standard free energies (AG0) of some hydrolysis and phosphorolysis reactions. (R = 8.3145x10'3 kJ/K- mol) Reactants :. Products AG0’(kJ/mol) Pi + Glucose 1: Glucose—6-P + H20 13.8 ATP + H20 2. ADP + Pi -30.5 Glucose-l-P + H2O <:. Glucose + Pi —20.9 Glucose + Pi r. Glucose-3-P + H2O 9.2 1,3-bisphosphoglycerate + H20 2. 3—bisphosphoglycerate + P. 49.4 Acetate + Pi <:. Acetylphosphate + H20 43.1 Phosphocreatine + H20 : Creatine + P. -43.1 H20 + PPi r, 2Pl- -33.5 Fructose-6-P + H20 :: Fructose + P. -13.8 Acetyl—CoA + H20 : Acetate + CoA —31.5 12 180 3242;s§i4z§iz45:},5413 ....:fli§i _ ‘ .,.1.!.i iflESEJ. ,1.-i-.z.:.3 . , . it? t. y]... Sil!&|.>ilxmxlzl ..?tll.1]ll.ll.!inlllil I!!!‘II.i111':virillvlruéplflerli.l1!!tl:l.s. O -90 .IL.-L.,-ILII,xiii: -180 9 ~18O (P (deg) ens. Inc. All dams reaewed » 3 and arm Wfieh/ > Tight $95953 -_ «2m E. 13 ...
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This note was uploaded on 01/05/2012 for the course CHEM 3321 taught by Professor Bean during the Spring '11 term at American Jewish University.

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Final Exam Spring 2007 - University ofHouston BCHS 3394/ J...

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