Ch27 - Chapter 26: Lipids Solutions 7 46 OH C HAPTER 27 CH...

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747 Chapter 27: Amino Acids and Proteins Solutions CHAPTER 27 Solutions to the Problems . Problem 27.1 Of the 20 protein-derived amino acids shown in Table 27.1, which contain (a) no chiral cemer, (b) two chiral centers. The only amino acid with no chiral centers is glycine (Gly, G). Both isoleucine- (lie, I) and threonine (Thr, T) have two chiral centers as shown with asterisks in the structures below. OH H 3 C I I * _ CH3CH2~HrHCOO CH 3 CHCHCOO * I + NH~ NH 3 Glycine (Gly, G) Isoleucine (lie, I) Threonine (Thr, T) Problem.27.2 The isoelectric point of histidine is 7.64. Toward which electrode does histidine migrate on paper electrophoresis at pH 7.0? An amino acid will have at least a partial positive charge at any pH that is below its isoelectric point. A pH of 7.0 is below the isoelectric point of histidine (7.64), so it will have a partial positive charge. Therefore, at this pH histidine migrates toward the negative electrode. . Problem 27.3 Describe the behavior of a mixture of glutamic acid, arginine, and valine on paper electrophoresis at pH 6.0. The pI's for glutamic acid, arginine, and valine are 3.08,10.76, and 6.00, respectively. Therefore, at pH 6.0 glutamic acid is negatively charged, arginine is positively charged, and valine is neutral. Thus, on paper electrophoresis, glutamic acid will migrate toward the positive electrode, arginine will migrate toward the negative electrode, and valine will not move. '. Problem 27.4 Draw a structural formula for Lys-Phe-Ala. Label the N-terminal amino acid and the C-terminal amino acid. What is the net charge on this tripeptide at pH 6.0? ~N+ "Oi I 2 . H:!C" . ?H:! H H 0 ~C H:!C,.!i:' I II , .,~H + ....C" .... N" ....C" .... r". ~O- ~N' '"C. ... ....<t '"~' ""'C. ... H~:::- N-terminal 0 ,CH 2 H 0 amino acid C-terminal b H 6 5 amino acid Due to the presence of the basic lysine residue, this tripeptide will have a net positive charge at pH 6.0 Problem 27.5 Which of these tripeptides are hydrolyzed by trypsin? By chymotrypsin? (a) Tyr-Gln-Val (b) Thr-Phe-Ser (c) Thr-Ser-Phe, Based on the substrate specificities listed in Table 27.3, trypsin will not cleave any of these tripeptides because' there are no arginine or lysine residues. On the other hand, chymotrypsin will cleave peptides (a) and (b) between the Tyr-Gln and Phe-Ser residues, respectively, but not (c). .
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749 748 Solutions Chapter 27: Amino Acids and Proteins PrQblem 27.6 Deduce the aminQ acid sequence Qf an undecapeptide (11 aminQ acids) frQm the experimental results shQwn in the accQmpanying table'. Experimental Procedure Undecapeptide Edman degradatiQn AminQ Acid CQmpQsitiQn Trypsin-Catalyzed Hydrolysis Fragment E Fragment F Fragment G Fragment H Ala,GI~I~Arg Thr,Phe,Lys Lys IyIet,Ser,Trp,Val Chymotrypsin-Catalyzed Hydrolysis Fragment I Fmgment J Ala,Arg,Glu,Phe,Thr Lys2,Met,Set,Trp,Val Reaction with CyanJ>gen Bromide Fragment K Fragment L Ala,Arg,G lu,Lys2,Met,P,he,Thr,Val Trp,Ser Based on the Edman degradation result, alanine (Ala) is the N-terminal residue of the peptide.
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This note was uploaded on 07/09/2011 for the course CHE 301 taught by Professor Diver during the Spring '08 term at SUNY Buffalo.

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Ch27 - Chapter 26: Lipids Solutions 7 46 OH C HAPTER 27 CH...

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