Practice Exam 1 Answers Spring 14

Practice Exam 1 Answers Spring 14 - 11M wEO\LEV Name CHM...

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Unformatted text preview: 11M.) wEO. \LEV Name CHM 3218 / 5305 Spring 2014 Examination #1 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 2 hours to complete this exam. All books, notes, cell phones, translators and other aids are prohibited, with the exception of molecular models and calculators. No sharing of models or calculators. 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. Amino acid properties. (Total 28 points). a. Write the one-letter code or codes for the amino acid or amino acids that would be produced by the complete acid—catalyzed hydrolysis of compound 1 (shown below). (6 points). HzNWN 3 (s) o fiKL Ls) “’7 <91 <9 b. Draw complete structures of all the carbon—containing products that would be obtained after completely hydrolyzing compound 1 in strong acid. Briefly explain your answer. No stereochemistry needs to be shown. (4 points). 6 H a” Fl H CO H x I ‘ 1| ‘ Owl” cola Page 1 [3»wa? may Name c. Use curved arrows to show a chemical mechanism for the complete hydrolysis of compound 1 in strong acid that yields all the carbon-containing products you wrote in part b. Note that no enzyme is involved in. this reaction. (8 points). Mel-L. «Hood- (warez? cam be, h7SVol7-tzb w: Q-mf c432» ‘5 OT” °~ "’ ° “/13 ”N H O “" N l 2""— u N “O :2 N ‘0 1+ g; f" 9 l4 \ H ‘ w u L n 7 ‘4 ‘N ,‘fi‘ Q ‘ . (D A I k.) k "4 H 0—H ‘4'" 0-“, .O. . H O 'LuH-‘t O‘H ' ‘H O '5 36’‘4 0 01 H ' H U ' \ @‘l-l H H FLO (a / “A p N? o O N N {N ‘0 T'Qf} Ll #0 5)" _._.;. Ofi N —-—-~'- :4 2:2 N 0.. q' -<- a V“ ‘_ ,n w—v u (an I -. 3H.o‘ '1‘ 0-H N o J” a ‘ HO H H H’ t .. \ 06?“ C) O'HVJO‘” HOC- _,O~ HOC. H “-0 LN,“ ' \ O 1— n ’4 I H -,° H H H \A- .- AX ‘ - Io. 1”ng \* “- ‘0 f \v\ 0 N o ' N/j Exfg “A REA) N {-0.14 Tfifijer—‘ln- *5 H ‘— '2: .- ‘ ’\ N Ii) 0‘14 f I \ O‘l-l a? 20/]: O 2 C02} HO C l 7, 7%))»: ‘40}; H ’l H “or?“ “01$ (a ‘t H~ H IMO-H . \A — _ .. 0%"?- ’H H- -o c7"A 0 SJ‘: r ('9 ‘.O"* Ho C / .———h H O _...A H 1' N O «- ~ Q #— ~ )9 ' I—l \ l f NH: 1 NI-l °° R Co “Oi-3 1 C‘02” H02; “332 fl 2H 1 H O G J. \k ,1 O . HOzQ W H + um “070‘ H ® rgHL H J“ H w ‘ 14;) K, “,9 :0 H # H-E-H 1H COLA HOLE WWIK'l NH Page2 Afiswtaz Kb] Name (1. Compound 1 was completely hydrolyzed in strong acid, then the carbon-containing products were separated and titration curves were obtained for each one individually. One set corresponds to the expected behavior for the hydrolysis products of 1. Which one? Briefly explain why, being specific as to why the curves have the shapes that they do. (6 points). OV‘) L1 \ one. pH x pH 0 1.0 0 1.0 2.0 3.0 Equiv. strong base Equiv. strong base PH pH 0 1 0 2.0 0 1.0 2.0 3.0 Equiv. strong base Equiv. strong base ?\CQI'L 0 1.0 2.0 0 1.0 2.0 3,0 Equiv. strong base Equiv. strong base Page 3 Amman Kai Name 30%, first») hoe @4093 pH I I 0 1.0 2,0 0 1.0 2.0 Equiv. strong base Equiv. strong base (I 0 °‘ 1 r“ 2‘9 “on rear 090114 \J \fi‘? COLHK. % "‘ z ?\<c.‘> mo>\- Shaw) '3 ?Kq‘l C ‘3 cow/CL)!“ m \Q;\- 5\.bc CovVD?°V)B> *0 VVO\\VU- can) “\ML vxt‘SA“ Ebb “\‘0 SDLMQK e. Label both chiral centers in the structure of 1 with the correct D / L designation. (4 points). Swan. hybwbh/bfi \0093n‘\’ Dagmar c)mvu\\§7, we. new) ‘05 3d; cmoklm) ova m QU- oummo aux}; (5) a c.0114 “(B :0th 4) Marc-01H q 20;“ (5) n e ~=:~ em .4 a «fee—w ‘4 H ”T” E \+ ‘4 ‘ L cola L Page 4 bfibuoEQ. \ée Name 2. This problem explores the effect of structure on pKa values and buffer behavior. (Total 24 points). a. pKa values for the same type of functional group in the three compounds shown below are 4.23, 3.63 and 2.34. Match the correct pKa value with the correct molecule. (2 points each). + + H + HaNACOZH H3N NCO? H3N/V\COZH glycine '3 B-alanine 3 y-aminobutyric acid\ 3e 2 3“! 3 L123 b. Why do the pKEl values for the same type of functional group differ between the three compounds shonw in part 21? Briefly explain, making it clear both why the values are different and why each compound has the particular value that it does. (6 points). TM BIQWQHW M 3\§\-om0g ‘09:qu \‘M, wa0xy\ nub m Posil-‘xock, 900.1%} ammo (Sou? on. vcspowshak 9,, W Bug-uwu \n waoxxflw CAME @543. UCAMM. ® © 9 H,N-W~COH :2 H3MMCOZ F; A Cov‘oMVH $ ahwo’tuao‘nw Beauty“. Lucwos-l-Qhe mixer)? Bqub 0..., \A‘L | gum”, (Koo?) mar) mow. em¢%.\-\c°.\lj ngvabk-L. H, \> “40,620“, cane/v *‘0 (8%“ a wage-3mm the? “0 8‘7““: ,0th mm. waobh. m p- clomw. “WE \Wfl' M Xrawumibvl'yvlt- 0943‘ Page 5 Antwan £21 Name Note that the remaining two parts of this question deal with a different pKa value in 1)— aminobutyric acid than the one discussed in parts a and b. c. At pH 10 0 if the concentration of the form of 1/— amino butyric acid with a net charge of— 1 is 6 8 111M and the concentration of the form of 7— amino butyric acid with a net charge of 0 IS 18. 2 mM what IS the pK value? (6 points). 5) “:23 \../~~./C-<Z>_L}‘|-—A x...— 1-1,?) w/CO? 1‘9- st) W90: 11 O em -1 “A A- (ti-1 l‘ifithfi 6‘8 ”"5 1H ‘ 9K1 + “3 [11m iA—l ) (3K: = e\-\ - \05 ([HAX d. If 0.50 mL of 2 M KOH is added to 150 mL of the solution described in part c, what is the final pH value? (6 points). HA1- oH-c—‘A ,1,sz gt” We)“, web. 0260\-\ c$§¢> \ mob- DQ HA 191.com A Mob»: 0?- HL‘A a1— ?\-\ 109-. \%.2mw/}K1 , mo\ ‘ 2 ,3 Lu P - O Klooeyzp ’Tv " 1000/4191 73"”) m E“! abhm‘fi 10x10 mob; eofl wL now View \-73"\0_3 "mu" 0? ”A eh»: 1.01410 new mom oQ A” mob.) a: A“ 0A 5’“ 10.03 C.3MM0\ 11f 150m. MM _.—— 3‘ 1.. F loos-m," I IC'OOMMDI ,_ '3 l .. Vat. W now have. H311 ID'Sw-peu: + \‘O ‘t-iO >1,wa '“ 2‘02 ’H0 m a“. A AF) 4‘ (2.02:\o‘>WJ NM TH 111 = 11141 +1179, ([11111A1 0° 113) *‘°€1((1.7Ms1mw) : ”’50 j “’1‘” we, UAW UM. Wink: 9mu,\)o\uwu_ 15 1.11 so“, C» HA QW‘BK ,. 1101x105 AuwaQ. \CEX Name 3. Fill in the missing information in the table. You should draw the predominant Species at pH 7, paying attention to stereochemistry. Show configurations of all chiral centers explicitly by drawing all four attached atoms. (2 points each; Total 14 points). One or Three-Letter Code and Structure Side-chain ionizable between pH 3 and 13? (R) / (S) designation G) a H (R)—Met > No S \ CH3 + _ H3N t z(:02 rt “ H No S — L2» 0 (s) , L CH3 Page 7 ANMAEAZ. \Lej Name 4. A peptide with the sequence C-A-G—P—E—A—V—K has been used in studies of sister chromatid separation (Stemmann et (11., Cell 2001, 107, 715-726). This problem explores the structure and properties of one example. (Total 16 points). a. Draw the complete structure of the peptide with the sequence C—A—G—P—E—A—V—K as it would exist as pH 7.0. You may neglect stereochemistry in this drawing. (8 points). GD /\/\ NH?’ 0 CH3 \% O o A “\ N /\( N ka J\ \VN \ H COG) ® 0 CH} 1“ OK ‘* O CH.)\ Z _ 3 CH- HsN \(K: \rNQ N to?) 3 SH 0 b. For mass spectrometry studies, the peptide in part a was carboxymethylatcd selectively on the side—chain of the first amino acid. This involves an 8N2 reaction of the peptide with iodoacetate (structure shown below). Draw the structure of the first two amino acids of the peptide in part a, showing the product of carboxymethylation. (4 points). IAco; iodoacetate c. What is numerical value for the pl of the peptide shown in part a? You may use pKa values of the free amino acids in your calculation. (4 points). N“"”"’"- to 2:6 +\ +\ *4 A4 - ‘ was less) CAP" c6.\°6 O O O ._.\ T : Q 3 G‘“: “1—7.8” 0 0 __‘ _‘ ?’ Z Laiv 1053 +\ +1 H H = 6.22 (1-sz 2-13 0 —\ -l .1 9H —’ 1-2. +\ JED—fl T k} k: v 2‘6 ““2 %.\‘6 hasxgee \CE\[ Name . Using the following information, deduce the sequence of the original octapeptide. Be sure to consult the list of peptide cleavage reagents on the last page of the exam. Note that amino acids are listed in alphabetical order in the data, and no sequence order is implied. (8 points). [Total acid hydrolysis: Arg, 7p, Lys, wt, §£4 966quivalents)7fi, Val AspN: 2 different products. Each was isolated and hydrolyzed with acid and their amino acid compositions were determined: #1 (Arg, Asp, Lys, Met, Val); #2 (Ser (2.0 equivalents), Tyr). Chymotrypsin: 2 different products. Each was isolated and hydrolyzed with acid and their amino acid compositions were determined: #1 (Arg, Asp, Lys, Met, Scr (2.0 equivalents), Val); #2 (Tyr). Cyanogen bromide: 2 different products. Each was isolated and hydrolyzed with acid and their amino acid compositions were determined: #1 (Asp, Met, Ser (2.0 equivalents), Tyr); #2 (Arg, Lys, Val). Trypsin: 2 different products. Each was isolated and hydrolyzed with acid and their amino acid compositions were determined: #1 (Asp, Lys, Met, Ser (2.0 equivalents), Tyr); #2 (Arg, Val). iSSDflLé—L—EA mu3I- ‘9': gov" H; C‘¥-MM|~\M can}, 50 D mual- \O‘- W 9904“" Out}, mi)“. ‘9?— k) 'k“ Page 9 AMbQER \LEV Name 6. This problem describes a method that can be used to determine the number of polypeptide chains in proteins suspected to possess quaternary structure. (Total 10 points). a. A sample (660 mg) of a protein with total molecular weight 132,000 g / mole was treated With an excess of 1-fluoro—2,4—dinitrobenzene under slightly alkaline conditions until the reaction was complete. This reagent reacts specifically with free u-amino groups in polypeptide chains. The peptide bonds were then completely hydrolyzed by heating in 6 M HCl. Amino acid analysis gave 5.5 mg of the following compound Whose molecular weight is 281 g / mole: H 6W1 02N CH3 CH3 If no other labeled amino acids were detected, calculate the number of polypeptide chains in the original protein. (6 points). "Grad mom 0?‘ vol-can: 3c ‘ K- : 5.0 x\O—G mob—a lOOG wax 62.0000? mom 0? N—l—ummql kwwol-WU 5‘5 ”‘9‘- 3 meb ~————0 3‘ 1 " \000 M3, 28\ 3, = \.95 into—5 wacky.) M°\-L> 0? N-‘l'stwoJ Bonoahue. — (quxlO—SW‘A») __'____’______,_4____’——- _.,_. ¥o\.fi\ MOD.» 0‘ PVC) \-L\n (S_O;< '0 -C mob.) 3 3.5“ a "l m on, bl chum: cynsml- “n W c¢‘%\nq\ ?~ro\—\:‘vx b. What can you deduce from the data in part a about the sequence of amino acids in the cliain(s) found in the original protein? (4 points). €t5rhw c.\\ row 00am: Wow. somhml Sgfiumm; ovl 5; My ERR—9., “ab! q“ bug‘n wilt“: Val FAWN; 'l’lma Wed may)” Ewtvchvg Couvg. Page 10 Amino Acid pKa values Name Alanine Arginine Asparagine Aspartate Cysteine Glutamate Glutaminc Glycine Histidine Isoleucine Leucine Lysine Methionine Phenylalanine Proline S erine 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 3.65 8.18 4.25 6.00 10.53 10.07 Reagents for peptide cleavage ArgC: Cleaves after Arg AspN: Cleaves before Asp Amswerz me: Name Carboxypeptidase: Liberates the C—terminal amino acid as long as this residue is not Pro. If the Chymotrypsin: Cleaves after Leu, Phe, Trp, Tyr CNBr: Cleaves after Met LysC: Cleaves after Lys Staphylococcus aureus V8: Cleaves after Asp, Glu C-terminal residue is Pro, there is no reaction Thermolysin: Cleaves before most hydrophobic residues Trypsin: Cleaves after Lys, Arg Page 12 ...
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