exam_1_key - Aoswérz \(e/ Name CHM 3218 Spring 2008...

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Unformatted text preview: Aoswérz \(e/ Name CHM 3218 Spring 2008 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 two hours in which to complete this exam. All books, notes and other aids are prohibited, with the exception of molecular models and 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. Consider amino acid derivative 1 and its hydrolysis products for the questions in this section. (Total 20 points). (5) . N a. Assign correct (R) / (S) designations to each asymmetric center in structure 1. (4 points). b. Which of the compounds shown below in their Fischer projection forms (2 — 5) corresponds to the hydrolysis product of 1? Is the correct structure a D- or L-amino acid?. (4 points). L- owi‘no amb c. What is the name and one-letter code for the hydrolysis product of 1? (3 points). $«oiifl-k ‘ F? d. If 1 is added to water that is initially at neutral pH and the hydrolysis is allowed to proceed to completion in the absence of a buffer, will the final pH of the solution be acidic, neutral or basic? Show the relevant equilibrium (or equilibria) to support your answer. (4 points). Page 1 A mica? \Ctjj Name .4 N UCQLH :3 </-]C.C?z Consider the four graphs shown below (Curves A — D). Which of these corresponds to the titration behavior expected for the complete aqueous hydrolysis product of 1? Briefly explain your answer, indicating the places on your chosen curve from which pKa values can be determined (and the approximate values). (5 points). 12- 12- 10 10 0I | I I 0 I I I I I I 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Equivalents of NaOH S Equivalents of NaOH ?\CQ,1 0 I I I I I I 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 Equivalents of NaOH Equivalents of NaOH Page 2 NAME: 2 E91 Name CUVUQ- 1'; CWVCCA’. "/WO ¥‘)‘VO\O\.L PVORWS \aJ-Y‘n onJOxIvnc.\-L\\, cow/cc}: \>\Cq void» '\>\Lq,\ 3 Z ?\¢q.13 H 0-33 a? Cqu A make)» no W$L Snag. FA bbbhw (3.3 \oqu n CULva \3 shows M We; vo\vu Cuwt C, obs ohms MINU— 37K: Who; Page 3 ___Lw&\ég____ Name 2. This problem explores the effect of structure on pKa values using alanine and homopeptides of alanine (Table l) as examples. (Total 20 points). Table 1. Ionization behavior of alanine and alanine homopeptides. In Amino acid or peptide Ala—Ala—Ala-Ala-Ala-Ala a. Draw the structure of Ala—Ala in its predominant ionization state at pH 7.0. Identify the functional groups associated with pK1 and pKz. (4 points). b. Why does the value of pK1 increase With each addition of an Ala residue to the Ala oligopeptide? (4 points). As H,» Bxs’remvc \QQA'VU bun HA. Covbcmfll an ? 4 o *owo‘ka amine, $1 00? \ncvecvcs, H» comombic. slosmmhoq bourwa 17m: mus A mm E\\QQ\Q\>\“\‘ *0 vgmovt W emboxq\ chfi-ovy‘ Cu \an by G“) I\WCVCQ%b pkg; chuu. c. Why does the value of pKz decrease with each addition of an Ala residue to the Ala oligopeptide? (4 points). A: 3A. CwagV\C.\'C bw‘rcmu. \vowunbs QM w ?VO*OWG\Q ammo (Stoop, Have \3 W5 CW\W\0\L a-lcb‘xlxz—oA-‘oh avoidabu. "Tim, WWW W come] 5:9 WM Wm. pO‘s‘A-NL thorax. X», bapvwvowcxoom d. What is the concentration of the form of Ala With no net charge at pH 9.0 if the total concentration of Ala is 25 mM? (4 points). _. , G3 6) I Foam ‘wA’V‘ \,\ N [HA—KT a 2.5 mt; Va : CLO wag = ago. 3 Y 1 V30 M Unwa ? Page 4 by;sz (l \Cei Name [Ix] ?'\-\ ; ?\(e\ * \oc:> [MBA {ANX= "2055 MY} INK = WAT — WM (WAT ' hum M vfi ‘— V342 + \03 [HA1 (\HMY - [me [H AX : —- 'PKC‘ [HA1 .Y ' [HM Wu» gm {MM 2 \o ‘- (a) [HMT = tmwo‘“ ‘" * we : mom—mew \ -‘r \o e. What volume of 5 M NaOH must be added to 20 mL of the buffer solution described in part (1 to increase its pH to 10.00? (4 points). mow 0Q HA 0» gm <10: mu, (A AA c.» in; \0002 20.% MW“ a‘ou wscfi _ . I -L‘ $.11 mm“ 0.01L L = ‘77 x . ““‘°*‘0 "W ’1‘“ 7““ meow Lg»! Ho" mm x ‘9 , ., [HACK a p \o_0o AHA: qJQqafimugLG-Uflqu (15mg) 1 2521\O‘L‘MGM / . [HA] 1 (handyman <“‘ * ‘0 ) 2,52HO‘MMM L '——' z \ x 5 m0\ : $.27. ML“ Page 5 Aubwat [a] Name 3. Fill in the missing information in the table. You should draw the predominant species at pH 7, paying attention to stereochemistry. (2 points each; Total 16 points). One or Three-Letter Code and Side-chain ionizable over the Structure pH range 3 - 10?* (R) / (S) designation Page 6 Auswez \l-E‘1 Name 4. Delta sleep inducing peptide (DSIP) has been investigated for its clinical potential in treating insomnia, pain and withdrawal. Its linear sequence is Trp-Ala-Gly—Gly—Asp-Ala—Ser-Gly— Glu. (Total 10 points). a. Draw the complete structure of DSIP as it would exist as pH 8.0. You may neglect stereochemistry in this drawing. (6 points). ‘H (v) H 0 L09 COG « \r‘Q " i \ _/ ‘M Q ’l H b. What is the net charge of DSIP at pH 8.0? Briefly explain your answer. (4 points). ,\-\ “\ -\ '\ Z -2 about WVGVJ)\) Page 7 A M SNER \(Erj Name ’ 5. 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. (Total 10 points). Total acid hydrolysis: Arg, Asp, Ile, Met, Phe, Pro, Tyr, Val Trypsin cleaves the octapeptide to yield a dipeptide containing Asp and Arg, and a hexapeptide with all the remaining amino acids. CNBr cleavage of the octapeptide yields a dipeptide containing Phe and Pro and a hexapeptide with all the remaining amino acids. Chymotrypsin cleaves the octapeptide into two tetrapeptides. Complete hydrolysis and amino acid composition analysis yielded the following data: #1 (Arg, Asp, Tyr, Val); #2 (Ile, Met, Phe, Pro). Carboxypeptidase treatment of the octapeptide gives Phe. ’(flh’sm ~ Cr“‘l"‘° CNTbV I Cmec‘ich’fiBh l x Alaiifilkflkpvo 31V}: l v\ ,mu3\' \Qc cow N-kxm \Pl' \/0» emu. 5:2 me no 59th 'fix‘ Ps‘u/x . ASP' “L, MA} .Vm, (We ’11,. \Ic‘\ L (PM V”) («L—WM CNB’i (ur‘wv) AVS‘ As?‘ \\&, m¥|f\]1. Val (ihvw wnDl ‘09— C-l'vVM “0 C,sz pang‘“ womh Mad?- besznrs no Mal—w“) CV‘DGXY?L?\”IBQ)L (an/v5 ?V\L owwbfi— WV“ (j Most MB “a “Tip! +0 r m») (H accounl- Q“ u’v’waoVfi? 3“? CquekfipS‘m : bwtbxk"? Wm Vol oven \\0.. ‘ WM 9%. ’9‘0 (Gleam) (my Page 8 Luswet Z?) Name 6. To identify amino acids, they are first derivatized with phenyisothiocyanate (PITC) in the reaction shown below. (Total 14 points) H H H2N COZH N N COZH N ‘n’ 0 sC\ pH 9.0 s + \S ————-> H20 NH2 NH2 a. Use curved arrows to show a reasonable chemical mechanism for this reaction. Note that the ionization state of the functional groups in Lys may not be correct for pH 9.0. You must write the correct ionization state as part of your mechanism. (10 points). 94.5 H (1%,9 (av “10) W e ?“ 9v» 5' x“ w M \N'. ‘N s u‘ \ 9163 2.95 __, e . \(/@ 6 u \f’ 9 {M05 56.95 J. V“ H- NJ £01 :2. g - N \(m)1 =2 (“50% W?) H W 39242wa “H3 \‘ g, 6 \-\*b: NH we 1:353 y: \ G.) 3 N“: H 6) b. Lysine contains two amino groups, but reaction with PITC occurs almost exclusively on the (Jr-amino group under the reaction conditions shown above. Briefly explain Why the side— chain amino group does not react with PITC. (4 points). A» -\::\—\ qr), w skegno‘ws 9x056 QMMOSV uquwug m ‘H« wckomakb Cevm- LO‘NH no \om. qu oudwqbu, \\' cowva cc) 0.» 0 OUW—OPWU. Page 9 Auing Key Name 7. 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 cenditions until the reaction was complete. This reagent reacts specifically with free (it-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: H K) OZN CH3 CH3 If no other labeled amino acids were detected, calculate the number of polypeptide chains in the original protein. (6 points). mobs» 02 Be»! modva X\9¢3/% WW\ . x f. ‘ 39% LC‘B *lOns mob.» MOD» 0% pvokim QEQ v \ )r , Wm "3 5.o><\0‘c’v~m>km \ $2,000)? mob.» buonwfi 1 mew T’V‘okfin 5.0 X lO'C’ mobs) : a L} MUQL be c. \—e.\—vcom</v\c, wax—«Ln b. What can you deduce from the data in part a about the sequence of amino acids in the chain(s) found in the original protein? (4 points). E1an Os“ Coov memos 0/!- \Bm\weo\‘ av 0\\\ "couv m'Vi w Sewn). M-«kxm‘rnod awxno ac"), (VOA) Page 10 Arousal LEN Name Amino Acid pKa values Name pKa,1 pKa,2 pKa,3 Alanine 2.34 9.69 Arginine 2.17 9.04 12.48 Asparagine 2.02 8.80 Aspartate 1.88. 9.60 3 .65 Cysteine 1.96 10.28 8.18 Glutamate 2.19 9.67 4.25 Glutamine 2.17 9.13 Glycine 2.34 9.60 Histidine 1.82 9.17 6.00 Isoleucine 2.36 9.68 Leucine 2.36 9.60 Lysine 2.18 8.95 10.53 Methionine 2.28 9.21 Phenylalanine 1.83 9.13 Proline 1.99 10.96 Serine 2.21 9.15 Threonine 2.1 l 9.62 Tryptophan 2.38 9.39 Tyrosine 2.20 9.11 10.07 Valine 2.32 9.62 Reagents for peptide cleavage ArgC: Cleaves after Arg AspN: Cleaves before Asp Carboxypeptidase: Liberates the C—terminal amino acid as long as this residue is not Pro. If the C—terminal residue is Pro, there is no reaction Chymotrypsin: Cleaves after Leu, Phe, Trp, Tyr CNBr: Cleaves after Met LysC: Cleaves after Lys Staphylococcus aureus V8: Cleaves after Asp, Glu Thermolysin: Cleaves before most hydrophobic residues Trypsin: Cleaves after Lys, Arg Page 11 ...
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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_1_key - Aoswérz \(e/ Name CHM 3218 Spring 2008...

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