15597.eh26.454-46810/20/0515:39Page45d EQA 26 Amino Acids,Peptides,Proteins,and Nucleic Acids: Nitrogen-Containing Polymers in Nature chemistry in particula mea molec of eviee the organic chemipoint of vThe nert s i drectn biochemistry. Outline of the Chapter 风 2Sucture nd Properieof Amino Acids 26-2,26-3 Preparation of Amino Acids Fancy footwork involving two very different functional groups. 26-4 Peptides and Proteins:Amino Acid Oligomers and Polymers The whole is greater than the sum of its parts. 265eamnienotPgibmrysmctre:5oqpencing 26,2678-DoeesAce29Cee0alknge 26-8 Polypeptides in Nature:The Proteins Myoglobin and Hemoglobin -1f Nucleic Acids w Nature appears to do 26-11 DNA Sequencing and Synthesis:Gene Technology The future is now! Keys to the Chapter 26-1.Structure and Properties of Amino Acids The th 454
26 Amino Acids, Peptides, Proteins, and Nucleic Acids: Nitrogen-Containing Polymers in Nature Here it is: the last chapter! It is, however, as much a beginning as an end. Chemistry in general, and organic chemistry in particular, are not isolated fields. Organic chemistry is the basic stuff of biology, and this chapter bridges the two. The basic principles that govern the behavior of organic molecules in general are shown here to be directly applicable to molecules of greater and greater complexity. You see here some of the most fundamental molecules of life viewed from the organic chemist’s point of view. The next step in this direction is biochemistry. Outline of the Chapter 26-1 Structure and Properties of Amino Acids Nature’s most versatile building blocks. 26-2, 26-3 Preparation of Amino Acids Fancy footwork involving two very different functional groups. 26-4 Peptides and Proteins: Amino Acid Oligomers and Polymers The whole is greater than the sum of its parts. 26-5 Determination of Polypeptide Primary Structure: Sequencing Exercises in logic, plus a lot of hard work. 26-6, 26-7 Synthesis of Polypeptides: A Protecting Group Challenge Even fancier footwork; how does Nature do it so easily??!! 26-8 Polypeptides in Nature: The Proteins Myoglobin and Hemoglobin 26-9, 26-10 Biosynthesis of Proteins: Nucleic Acids How Nature appears to do it so easily. 26-11 DNA Sequencing and Synthesis: Gene Technology The future is now! Keys to the Chapter 26-1. Structure and Properties of Amino Acids Read the introduction to the text chapter and this text section, and then come back here. All done? O.K., here’s the big picture. The chemistry of life is complicated. Structures have to be built to hold things together, and a lot of chemical reactions have to be going on to perform the various functions that maintain life, such as 454 1559T_ch26_454-468 10/20/05 15:39 Page 454
1559T_ch26_454-46810/20/0515:39Pa9e455 EQA Keys to the Chapter·455 R onstrained set of conditions:Water is the y very narrow ranges of temperature and pH are acceptable.Other The answer begn only in the polar groups.both small and large. capable of various egrees of There are uncharged bu polar hic will be eof this varicty. ofbe depoontod and negaivly chargodp lists pk valu s for all rele oups.Notice.for one thing.that the way the amino acids have been drawn up to no wrong.In so amino acids in fact never exi to pH.one or both of theseg oups is always charged.with the pH 7structure COOBe. gteeiectsofpHon ation of Amine acids 一2i子and子6r3eP时thrntinowr combinations aimed at solving the problem of introducing basic and acidic groups into the same molecule. 26-4 and 26-5.Peptides and Proteins:Amino Acid Oligomers and Polymers determining peptide and pro extremely well of fo 26-6an 26 Synth esis of Polypeptides:A Protecting Group Challenge ptide chains from condensations(Sections 18-6 and 23-1) e trick y nu y el at linkine two mol les of a.a.I to ther.two mo oh a.a to each er,or a.a. o a.a. in the wrong se tional gr will let you try them for yourself. 26-8 through 26-11.Proteins in Operation;Biosynthesis; DNA and Gene Technology units gives rise to structures of such highly elaborate function.This stuff is really neat.You might like to read more about it some time
energy storage and utilization. These all have to occur under a very constrained set of conditions: Water is the only available solvent, and in general only very narrow ranges of temperature and pH are acceptable. Otherwise everything falls apart. So how is it done? The answer begins with the amino acids. Look first at the structures of the 20 most common examples (Table 26-1). They differ only in the group attached to the -carbon. The variety in these groups establishes the versatility of amino acids. There are nonpolar groups, both small and large, capable of various degrees of steric interaction. There are uncharged but polar groups, capable of hydrogen bonding. There are nitrogen-containing groups of various base strengths, some of which will be protonated and positively charged at pH 7. There are oxygen and sulfur groups of various acid strengths, some of which will be deprotonated and negatively charged at pH 7. Because of this variety, Nature can choose from among these 20 compounds just the right one to fill any of a number of chemical needs. One feature that is emphasized in this section is the acid-base behavior of the amino acids. Table 26-1 lists pKa values for all relevant groups. Notice, for one thing, that the way the amino acids have been drawn up to now, H2NOCHROCOOH, is wrong. In solution, amino acids in fact never exist to any significant extent in this form, with neutral amino and carboxylic acid groups present at the same time. Depending on the pH, one or both of these groups is always charged, with the pH 7 structure being H3NOCHROCOO. Because of this feature, amino acids are very good buffers at a variety of different pH’s, depending on R. Obviously the effects of pH on amino acid structure are important, and Problem 28 will give you several examples to work on so you can get the feel for it. 26-2 and 26-3. Preparation of Amino Acids Although none of the individual reactions in these sections are new, the sequences present them in powerful combinations aimed at solving the problem of introducing basic and acidic groups into the same molecule. 26-4 and 26-5. Peptides and Proteins: Amino Acid Oligomers and Polymers Techniques for determining peptide and protein structure are extremely well worked out (and Problems 42–47 will give you plenty of chances to try them for yourself). The results, especially in the subtleties of folding of these polymeric chains, reveal the extent to which the characteristics of the different amino acids are combined and used in nature to generate large molecular assemblies perfectly suited for very specific biological roles. Problems 38–41 are intended to give you something to think about in this regard. 26-6 and 26-7. Synthesis of Polypeptides: A Protecting Group Challenge Don’t ever lose sight of the fact that the linkage between amino acids is nothing more than a simple amide bond: OHNOCOO. Nonetheless, the construction of peptide chains from simple amino acids is a major challenge for the same reasons that, say, mixed aldol or Claisen condensations (Sections 18-6 and 23-1) are tricky things to do: Each amino acid involved has both a potentially nucleophilic atom (the N) and a potentially electrophilic one (the carboxy C). Thus, an attempt at linking, say, the amine of amino acid 1 with the carboxy group of amino acid 2 is going to be complicated by the need to prevent the simultaneous linkage of either two molecules of a.a.1 to each other, two molecules of a.a.2 to each other, or a.a.1 to a.a.2 in the wrong sense: carboxy of 1 to amine of 2. The solution to the problem lies in, again, a very well worked out array of functional group protection–deprotection procedures, the simplest of which are presented here. Problems 48 and 49 will let you try them for yourself. 26-8 through 26-11. Proteins in Operation; Biosynthesis; DNA and Gene Technology Obviously only a tiny taste of what’s involved with these topics can be presented in the space of four chapter sections. Nonetheless, you should be able to sense the remarkable way in which the linkage of relatively small units gives rise to structures of such highly elaborate function. This stuff is really neat. You might like to read more about it some time. Keys to the Chapter • 455 1559T_ch26_454-468 10/20/05 15:39 Page 455
1559r.ch26_454-46810/20/0515:39Pag0456 456.chapter 26 AMINO ACIDS,PEPTIDES,PROTEINS,AND NUCLEK ACIDS:NITROGEN-CONTAINING POLYMERS IN NATURE Solutions to Problems 26. COOH COOH HN- H.cH HN- CH2CHs CHs L-Isoleucine t-Threonine Systematic name for Lthreonine:()-2-amino-3-hydroxybutanoic acid. 27. COOH HN-H H-CHS Systematic name for allo-L-isoleucine:(2S.3R)-2-amino-3-methylpentanoic acid 28.Structures are presented in order of increasing pH(in parentheses). COOH C00- C00- aH衣H④HHm,N H(12) CH CH CH COOH C00 (b)HaN--H (1)HN-H (7) H2N--H (12) CH-OH CH2OH CH-OH COOH C00 C00 C00 (e)HN-H (1) HaN- H -H(9.5) N +H(12) (CH24-NH; (CH2)4-NHs (CH2)4-NH3 (CH2)4-NH2 COOH C00- C00 900 @HNH是a)HNH() HN十H() H.N- -H思(12 H COOH C00- C00- H (7)H-H (9)HN- H(12) CH-SH COOH C00- C00 C00 (f)HN-H (1)HN-H (3)HsN-H (7) HN-H(12) CH-COOH CH-COOH CH-COO CHCOO
Solutions to Problems 26. Systematic name for L-threonine: (2S, 3R)-2-amino-3-hydroxybutanoic acid. 27. Systematic name for allo-L-isoleucine: (2S, 3R)-2-amino-3-methylpentanoic acid. 28. Structures are presented in order of increasing pH (in parentheses). (a) (b) (c) (d) (e) (f) H3N (1) H COOH CH2COOH CH2COOH CH2COO CH2COO H3N (3) H COO H3N (7) H COO H2N H (12) COO H3N (1) H COOH CH2SH H3N (7) H COO CH2SH H3N (9) H COO CH2S H2N H (12) COO CH2S H3N (1) (5) (7) H H N NH COO COO COO CH2 H3N H H N NH COOH CH2 H2N H H N N CH2 H3N H H N N CH2 (12) H3N (1) NH3 H COOH (CH2)4 COO H2N H (9.5) COO H3N (7) NH3 H (CH2)4 NH3 (CH2)4 H2N H (12) COO (CH2)4 NH2 H3N (1) H COOH CH2OH CH2OH CH2OH H3N (7) H COO H2N H (12) COO H3N (1) H COOH CH3 H3N (7) H COO CH3 H2N H (12) COO CH3 H H2N S H COOH CH2CH3 CH3 Allo-L-isoleucine R H3C H2N S H H COOH CH2CH3 L-Isoleucine S H H2N S OH H COOH CH3 L-Threonine R 456 • Chapter 26 AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: NITROGEN-CONTAINING POLYMERS IN NATURE 1559T_ch26_454-468 10/20/05 15:39 Page 456
1559T_ch26_454-46810/20/0515:53Pa9e457 EQA Solutionsto Problems457 COOH 900 (g)HN-H NH2 (1)HN-H NH2 (7) (CH)NHCNH> (CH2)3NHCNHz C00 C00 HaN--H NH2 (12) H2N-H NH (14) (CH2)NHCNHz 900 H (1)H. H (7)H2 H(9.5)H, H(12) 12 CH OH OH 0- 29.(a)Arg.Lys:(b)Ala.Ser.Tyr.His,Cys:(e)Asp single and a single roup.Their average is the pl. ()(9.0+10.5/2=9.7 (d(9.2+6.1)/2=7.6 (e(8.2+2.0/2=5.1 (0(6.7+1.92=2.8 (g(12.5+9.0M2=10.8 h)(9.1+2.2)/2=5.7 31.(a)Because the R group is secondary.alkylation routes should be avoided.Use the Strecker synthesis. NH3 CH(CO,CH,CH)
(g) (h) 29. (a) Arg, Lys; (b) Ala, Ser, Tyr, His, Cys; (c) Asp 30. First identify the net charge-neutral structure, which is always the one with a single and a single charge. Choose the two pKa values that bracket that structure. They are, repectively, the pKa for deprotonation of the most acidic group in that structure and the pKa for protonation of the most basic group. Their average is the pI. (c) (9.0 10.5)/2 9.7 (d) (9.2 6.1)/2 7.6 (e) (8.2 2.0)/2 5.1 (f) (3.7 1.9)/2 2.8 (g) (12.5 9.0)/2 10.8 (h) (9.1 2.2)/2 5.7 31. (a) Because the R group is secondary, alkylation routes should be avoided. Use the Strecker synthesis. (b) The R group is primary; now you have a choice. Either the Strecker synthesis starting with (CH3)2CHCH2CHO or a Gabriel-based method will do. 1. NaOCH2CH3, CH3CH2OH 2. BrCH2CH(CH3)2 3. H, H2O, NH3 N O O CH(CO2CH2CH3)2 (CH3)2CHCH2CHCOO H, H2O 1. NH3 2. HCN (CH3)2CHCHO (CH3)2CHCHCN NH2 (CH3)2CHCHCOO NH3 (1) COO COO COO O H3N H COOH CH2 OH H3N (7) H CH2 OH H2N H (9.5) CH2 OH H2N H (12) CH2 H3N NH2 (1) H COOH (CH2)3NHCNH2 H3N (7) H COO H2N H (12) COO H2N H (14) COO NH2 (CH2)3NHCNH2 NH2 (CH2)3NHCNH2 NH (CH2)3NHCNH2 Solutions to Problems • 457 1559T_ch26_454-468 10/20/05 15:53 Page 457
1559r.eh26.454-46810/20/0515:39Page458 458 chapter 26 AMINO ACIDS,PEPTIDES,PROTEINS,AND NUCLEK ACIDS:NITROGEN-CONTAINING POLYMERS IN NATURE Even the Hell-Volhardt-Zelinskyamination sequence works just fine here Br CH)CHCHCH-COOH(CH)CHCH-CHCOOH NHs (CH-CHCH CHCOO- (c)Several ways to go,but you have to first recognize the need for a three-carbon building block with leaving groups at 1.NaOCH.CH.CH.CH.OH -CH(COCH-CH COC.OCCH. NH3 Cone.HBr NH; BrCH,CH2CH-CHCOO- aving group raaawr"y 一N NCHCO.CH.CH HO NHs C(CO,CH,CH).H. CH:CHCHCOO- HOCHCH (e)The extra amine group must be present in protected form.irrespective of the method used.Here'sa Gabriel-type sequenc
Even the Hell-Volhardt-Zelinsky–amination sequence works just fine here. (c) Several ways to go, but you have to first recognize the need for a three-carbon building block with leaving groups at each end to allow linkage to both the -carbon and (later on) the amine nitrogen to form the ring. Start with a Gabriel-type sequence. (d) Use a Gabriel-based method. Instead of forming the necessary carbon–carbon bond by SN2 reaction with a haloalkane, use an aldol-type condensation with CH3CHO. (e) The extra amine group must be present in protected form, irrespective of the method used. Here’s a Gabriel-type sequence. H, H2O, NH3 HO CH3CHCHCOO N O O C(CO2CH2CH3)2 HOCHCH3 1. NaOCH2CH3 2. CH3CHO O O NCH(CO2CH2CH3)2 NH2 BrCH2CH2CH2CHCOO COO Br H2N H Conc. HBr (Converts OH to Br, a good leaving group) NH3 BrCH2CH2CH2CHCOO 1. NaOCH2CH3, CH3CH2OH 2. BrCH2CH2CH2OCCH3 O 3. H, H2O, N O O CH(CO2CH2CH3)2 NH3 HOCH2CH2CH2CHCOO NH3, H2O (CH3)2CHCH2CHCOO NH3 Br2, PBr3 (CH3)2CHCH2CH2COOH Br (CH3)2CHCH2CHCOOH 458 • Chapter 26 AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: NITROGEN-CONTAINING POLYMERS IN NATURE 1559T_ch26_454-468 10/20/05 15:39 Page 458
1559T_ch26_454-46810/20/0515:39Pa9e459 EQA Soutions o Problems59 人oa兰e NH. (b)The use of an opically active amine (is shown)means that the addition product s actually a mixture ond stereocenter is generated,which can be e don't necess reatly predominate and,after hydrolysis and removal of the phenylmethyl group with H2,mainly Samino acid is obtained. 33.Allicin is structurally related to cysteine,which should be readily available if you can first devise an approach to serine. NCH(CO-CHCH3)- NaOH.CH:-0 C(CO.CH2CHa2 SH C(CO-CH2CHa2 HOCH2 HSCH2
32. (a) (b) The use of an optically active amine (an S enantiomer is shown) means that the addition product is actually a mixture, because a second stereocenter is generated, which can be either R or S. This is, therefore, a mixture of R,S and S,S products. Because these are diastereomers of each other, they don’t necessarily form in identical yields. In fact, the S,S product (illustrated) greatly predominates, and, after hydrolysis and removal of the phenylmethyl group with H2, mainly S amino acid is obtained. 33. Allicin is structurally related to cysteine, which should be readily available if you can first devise an approach to serine. 1. PBr3 2. Na SH N O O C(CO2CH2CH3)2 HOCH2 Hydrolysis at this stage would make serine N O O C(CO2CH2CH3)2 HSCH2 O O NCH(CO2CH2CH3)2 NaOH, CH2 O Aldol cond. [compare Problem 31(d)] CH2CHO 3. H, H2O 1. NH3 2. HCN NH3 CH2CHCOO Stereocenter Chiral, but racemic (i.e., not optically active) O O O N O N(CH2)4CH(CO2CH2CH3)2 H, H2O, NH3 H3N(CH2)4CHCOO NaOCH2CH3 O O N(CH2)4Cl O O NCH(CO2CH2CH3)2, Br(CH2)4Cl Introduces the extra amine in protected form. N K O O Solutions to Problems • 459 1559T_ch26_454-468 10/20/05 15:39 Page 459
1559r.ah26.454-46910/20/0515:39Page460 460 chapter 26 AMINO ACIDS,PEPTIDES,PROTEINS,AND NUCLEI ACIDS:NITROGEN-CONTAINING POLYMERS IN NATURE Treament of this product with hotaqueous acid gives cysteine irectly.Otherwise. O +NH 上2odCH=CHCH SCH.CHCO0 C(CO.CH.CH) CH,=CHCH-SCH, 34.The alloisoleucines are diastereomers of the isoleucines.so simple recrystallization will separate them Crystals (+)-and(-)-isoleucine Mixture Continue with each mixture of enantiomers separately,making use of brucine as a resolving agent. NH3 (eg-R=) salt of (+)-acid +M salt of (-)-acid Finally,releases each pure amino acid enantiomer in tur. 35.The本er For example in tripeptide (a): (CH3)2CH O CH3 O HSCH2 HN-CH-C.NH-CH-C.NH-CH-COO
Treatment of this product with hot aqueous acid gives cysteine directly. Otherwise, 34. The alloisoleucines are diastereomers of the isoleucines, so simple recrystallization will separate them. Continue with each mixture of enantiomers separately, making use of brucine as a resolving agent. Finally, H, H2O treatment releases each pure amino acid enantiomer in turn. 35. (a) tripeptide; (b) dipeptide; (c) tetrapeptide; (d) pentapeptide. The peptide bonds are simply the amide linkages, For example, in tripeptide (a): (CH3)2CH CH3 HSCH2 CH C NH CH COO CH C NH O O H3N C NH O 1. Brucine CH3OH, 0C 2. Separate (crystallization) salt of ()-acid salt of ()-acid CH3CH2CH(CH3)CHCOOH RCNH O ()/() NH3 CH3CH2CH(CH3)CHCOO ()/() Mixture RCOCR, O O (e.g., R C6H5) 1. Dissolve in hot 80% ethanol 2. Cool to 0 Solutions Crystals Mixture ()- and ()-isoleucine ()- and ()-alloisoleucine O O N CH2 CHCH2SCH2 C(CO2CH2CH3)2 1. NaOH Alkylate on sulfur 2. CH2 CHCH2Br 2. H, H2O, 1. H2O2 O NH3 CHCH2SCH2CHCOO CH2 460 • Chapter 26 AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: NITROGEN-CONTAINING POLYMERS IN NATURE 1559T_ch26_454-468 10/20/05 15:39 Page 460
1559T_ch26_454-46810/20/0515:39Pa9e46 ⊕ EQA Solutions to Problems.461 (a)Val-Ala-Cys:(b)Ser-Asp:(c)His-Thr-Pro-Lys:(d)Tyr-Gly-Gly-Phe-Leu R物e以e Amino acids (Problem 28):(a).(b).(d).(e)N.(f)A.(c).(g)C Peptides (Problem 35):(a)N.(b)A.(c)C.(d)N 38. -CH3,or- sheets.where only small groups will fit casily.The nonpolar nature of five of thesix groups isaso vers of compatible with their location,a relatively nonpolar region with few hydrogen-bonding groups in the vicinity. a-Helix Amino acid numbers a-Helix Amino acid numbers 3-18 58-77 20-35 8694 3642 G 100-118 51-57 H 125-148 istics of the five-memberedn 40.Fxce ssociated with the heme-hound iron atom all the r side chains are well positioned for hydrogen bonding with solvent molecules(water).In contrast.all the nonpolar side chains adopt interior positions,avoiding contact with polar solvent molecules. 41.(a)The shee side chains are exposed to the solvent.solubilizing the entire molcu (Problem 0).Similar effects
36. By convention, the short notation format always begins with the end of the peptide chain with the amino group (the “N-terminal” or “amino-terminal” end). (a) Val-Ala-Cys; (b) Ser-Asp; (c) His-Thr-Pro-Lys; (d) Tyr-Gly-Gly-Phe-Leu 37. Determine the net charge on the amino acid or peptide at pH 7 and then recall that negative species migrate to the anode (A), positive to the cathode (C), and neutrals do not migrate at all (N). Amino acids (Problem 28): (a), (b), (d), (e) N, (f) A, (c), (g) C Peptides (Problem 35): (a) N, (b) A, (c) C, (d) N 38. The side chains are all small (OH, OCH3, or OCH2OH), and mostly nonpolar. In the illustrations, especially Figure 26-3, note that the sheet structure packs the R groups into small channels between layers of sheets, where only small groups will fit easily. The nonpolar nature of five of the six groups is also compatible with their location, a relatively nonpolar region with few hydrogen-bonding groups in the vicinity. 39. The -helical stretches are fairly noticeable by their spiral shape (compare Figure 26-4). Myoglobin in fact contains eight significant -helical stretches, which are labeled by the letters A–H: -Helix Amino acid numbers -Helix Amino acid numbers A 3–18 E 58–77 B 20–35 F 86–94 C 36–42 G 100–118 D 51–57 H 125–148 In the figure, all but -helix D (which is viewed on-end from this perspective) are fairly easy to pick out. The four prolines are located at or near the ends of -helices and coincide with “kinks” in the overall tertiary structure of the molecule, a result of the conformational characteristics of the five-membered ring. 40. Except for the two histidines associated with the heme-bound iron atom, all the polar side chains are well positioned for hydrogen bonding with solvent molecules (water). In contrast, all the nonpolar side chains adopt interior positions, avoiding contact with polar solvent molecules. 41. (a) The sheet structure is favored by amino acids with small, nonpolar side chains and has very little ability to hydrogen bond to a polar solvent like water (Problem 38). (b) In globular proteins the polar side chains are exposed to the solvent, solubilizing the entire molecule (Problem 40). Similar effects are seen in micelles formed by soap molecules, in which polar groups are located on the surface, facilitating water solubility, whereas nonpolar groups are buried in the inside (Chemical Highlight 19-1). (c) If the tertiary structure of a globular protein is disrupted, its nonpolar amino acid side chains become exposed to the polar solvent, greatly reducing the overall solubility of the protein molecule. Solutions to Problems • 461 1559T_ch26_454-468 10/20/05 15:39 Page 461
1559Tch26454-46910/20/0515:39Page462 462 chapter 26 AMINO ACIDS,PEPTIDES,PROTEINS,AND NUCLEI ACIDS:NITROGEN-CONTAINING POLYMERS IN NATURE determine the sequence of amino acids.Because ony nine are present.the entire chain can be sequenced in his way. 3 (a) -CHCH(CH)+Al-Cys NH H NH +Thr-Pro-Ly 44.Because the peptide is cyclic.the Edman process will not give a normal result.It will simply form thiourea derivatives at the two"extra"Om amino groups: Because there is no a-amino group available to react.mild acid treatment will cleave no bonds in the product,and the cyclic polypeptide structure will remain intact. 45.Complete hydrolysis indicates that a total of nine amino acid units are present.Examine the four fragments of incomplete hydrolysis.You know that the peptide begins with Arg (iv)so the fragmen Arg-PTO-PIO be overlapping all the Ag品5678 Gly-Phe-Ser Ser-Pro-Phe
42. (1) Purify and cleave the disulfide bridge (Section 9-10). (2) On a portion of the sample, degrade the entire chain by amide hydrolysis (6 N HCl, 110°C, 24 h) to determine amino acid composition by using an amino acid analyzer. (3) On another portion of this material, apply repetitive Edman degradation to determine the sequence of amino acids. Because only nine are present, the entire chain can be sequenced in this way. 43. (a) (b) (c) (d) 44. Because the peptide is cyclic, the Edman process will not give a normal result. It will simply form thiourea derivatives at the two “extra” Orn amino groups: Because there is no -amino group available to react, mild acid treatment will cleave no bonds in the product, and the cyclic polypeptide structure will remain intact. 45. Complete hydrolysis indicates that a total of nine amino acid units are present. Examine the four fragments of incomplete hydrolysis. You know that the peptide begins with Arg (given), so the fragment Arg-Pro-Pro-Gly must be first. There is only one Gly present, so the last Gly of this tetrapeptide must be the same one that is at the start of the tripeptide fragment Gly-Phe-Ser. You can use the same logic to start overlapping all the pieces to generate the whole solution. Thus, because only one Ser is present, the last Ser in the above tripeptide must be the same as the first one in Ser-Pro-Phe. So far, you have 123 4 5 67 8 Arg-Pro-Pro-Gly Gly-Phe-Ser Ser-Pro-Phe NHCNH(CH2)3 S CHCH2 S O N OH NH Gly-Gly-Phe-Leu CHCH2 S N O NH Thr-Pro-Lys N N H CHCH2OH S N O NH Asp CHCH(CH3)2 S N O NH Ala-Cys 462 • Chapter 26 AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: NITROGEN-CONTAINING POLYMERS IN NATURE 1559T_ch26_454-468 10/20/05 15:39 Page 462
1559T_ch26_454-46810/20/0515:39Pa9e463 ⊕ EQA Solutions to Problems.463 The final fragp 46.Products of degradation,in order of appearance CH-CHSCH The last product to appear from leu-enkephalin is CH-CH(CH 47.Look first for a piece that ends with an amino acid that should not be a site of cleavage by one of the chymotrypsin fragments end in Pne end of the intact hormone.Now itsa matter of matching up all the pieces.Start with this end picce from trypsin hydrolysis and overlap it with chymotrypsin fragments. (Trypsin piece) Val-Tyr-Pro-Asp-Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe (Chymotrypsin pieces) Pro-Asp-Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro Leu-Glu-Phe Ser-Tyr-Ser-Met-Glu-His-Phe-ArgTrp-Gly-Lys Pro-Val-Gly-Lys Pro-Val-Lys Val-Tyr Ser-Tyr Ser-Met-Glu-His-Phe Arg-Trp Gly-Lys-Pro-Val-Gly-Lys-Lys-Arg-Arg-Pro-Val-Lys-Val-Tyr .d the o h SerTy-SeMet Hi he Arg Arg-Arg-Pro-Val-Lys-Val-Tyr-Pro-Asp-Ala-Gly-Glu-Asp-GIn-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe 48.Start at the carboxy-terminal end. 00 1.Phe +(CH3)COCOCOC(CH3)3-Boc-Phe (N-protected Phe)
The final fragment, Phe-Arg, clearly is at the end, overlapping the Phe in position 8. So the answer is Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg. 46. Products of degradation, in order of appearance The last product to appear from leu-enkephalin is 47. Look first for a piece that ends with an amino acid that should not be a site of cleavage by one of the enzymes. All the chymotrypsin fragments end in Phe, Trp, or Tyr, so that’s no help. The trypsin results are more useful. It only cleaves after Arg or Lys, so the 18-amino acid fragment ending in Phe must be at the end of the intact hormone. Now it’s a matter of matching up all the pieces. Start with this end piece from trypsin hydrolysis and overlap it with chymotrypsin fragments. (Trypsin piece) Val-Tyr-Pro-Asp-Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe (Chymotrypsin pieces) Pro-Asp-Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro Leu-Glu-Phe Now identify a chymotrypsin piece to overlap with the Val-Tyr- front end of the trypsin piece and then continue the process, all the way to the N-terminal end (the “beginning”) of the entire hormone. Ser-Tyr-Ser-Met-Glu-His-Phe-Arg Trp-Gly-Lys Pro-Val-Gly-Lys Pro-Val-Lys Val-TyrSer-Tyr Ser-Met-Glu-His-Phe Arg-Trp Gly-Lys-Pro-Val-Gly-Lys-Lys-Arg-Arg-Pro-Val-Lys-Val-Tyr The complete answer is read directly, starting at Ser-Tyr- (just above), overlapping Val-Tyr with the large trypsin piece, and then on to the end: Ser-Tyr-Ser-Met-Gln-His-Phe-Arg-Trp-Gly-Lys-Pro-Val-Gly-Lys-LysArg-Arg-Pro-Val-Lys-Val-Tyr-Pro-Asp-Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe 48. Start at the carboxy-terminal end. 1. Phe (CH3)3COCOCOC(CH3)3 O O Boc-Phe (N-protected Phe) S O N NH CH2CH(CH3)2 S O N NH CH2 S O N NH CH2CH2SCH3 S O N OH NH S O 2 N NH CH2 Solutions to Problems • 463 1559T_ch26_454-468 10/20/05 15:39 Page 463