有机化学结构与功能5th答案sg_ch26

1、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 biolo

2、gy, 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 fun- damental molecules of life viewed from the organic chemists

3、point of view. The next step in this direction is biochemistry. Outline of the Chapter 26-1 Structure and Properties of Amino Acids Natures 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:

4、 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, 26-7Synthesis of Polypeptides: A Protecting Group Challenge Even fancier footwork; how does Nature do

5、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 Ami

6、no Acids Read the introduction to the text chapter and this text section, and then come back here. All done? O.K., heres 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 vari

7、ous functions that maintain life, such as 454 1559T_ch26_454-468 10/20/05 15:39 Page 454 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 acceptabl

8、e. Other- wise 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 at- tached to the ?-carbon. The variety in these groups establishes the versatility of amino acids

9、. There are non- polar 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. T

10、here 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 thi

11、s section is the acid-base behavior of the amino acids. Table 26-1 lists pKavalues 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 ex- tent in this form, wi

12、th 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?. Be- cause of this feature, amino acids are very good buffers at a variety of different pHs, depending on R. Obvi- ous

13、ly 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 co

14、mbinations 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 4247 will give you plenty

15、 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

16、 biological roles. Problems 3841 are intended to give you something to think about in this regard. 26-6 and 26-7.Synthesis of Polypeptides: A Protecting Group Challenge Dont ever lose sight of the fact that the linkage between amino acids is nothing more than a simple amide bond: OHNOCOO. Nonetheles

17、s, the construction of peptide chains from simple amino acids is a major chal- lenge 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 elec

18、- trophilic 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.

19、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 func- tional group protectiondeprotection procedures, the simplest of which are presented here. Problems 48 and 49 will let you try them for yourself. 26-8 through 2

20、6-11.Proteins in Operation; Biosynthesis; DNA and Gene Technology Obviously only a tiny taste of whats 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 r

21、ise 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 Solutions to Problems 26. Systematic name for L-threonine: (2S, 3R)-2-amino-3-hydroxybutanoic acid. 27. S

22、ystematic 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)(1)H3N ? H COOH CH2COOHCH2COOHCH2COO?CH2COO? (3)H3N ? H COO? (7)H3N ? H COO? (12)H2NH COO? (1)H3N ? H COOH CH2SH (7)H3N ? H C

23、OO? CH2SH (9)H3N ? H COO? CH2S? (12)H2NH COO? CH2S? H3N ? ? H(1)(5)(7) H N NH COO?COO?COO? CH2 H3N ? ? H H N NH COOH CH2 H2NH H N N CH2 H3N ? H H N N CH2 (12) (1)H3N ? NH3 ? H COOH (CH2)4 COO? (9.5)H2NH COO? (7)H3N ? NH3 ? H (CH2)4NH3 ? (CH2)4 (12)H2NH COO? NH2(CH2)4 (1)H3N ? H COOH CH2OHCH2OHCH2OH

24、(7)H3N ? H COO? (12)H2NH COO? (1)H3N ? H COOH CH3 (7)H3N ? H COO? CH3 (12)H2NH 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 PO

25、LYMERS IN NATURE 1559T_ch26_454-468 10/20/05 15:39 Page 456 (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 pKavalues that bracket that structure. They are,

26、 repectively, the pKafor deprotona- tion of the most acidic group in that structure and the pKafor 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)/

27、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

28、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 (7)H3N ? H CH2 OH (9.5)H2NH CH2 OH (12)H2NH CH2 (1)NH2H3N ? H COOH (CH2)3NHCNH2 (7)H3N ? H COO? (12)H2NH COO? (14)H2NH COO? NH2 ? (CH

29、2)3NHCNH2 NH2 ? (CH2)3NHCNH2 NH (CH2)3NHCNH2 Solutions to Problems 457 1559T_ch26_454-468 10/20/05 15:53 Page 457 Even the Hell-Volhardt-Zelinskyamination 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 gr

30、oups 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 carboncarbon bond by SN2 reaction with a haloalkane, use an aldol-type condensation with CH3

31、CHO. (e) The extra amine group must be present in protected form, irrespective of the method used. Heres 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

32、?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

33、: NITROGEN-CONTAINING POLYMERS IN NATURE 1559T_ch26_454-468 10/20/05 15:39 Page 458 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, ther

34、efore, a mixture of R,S and S,S products. Because these are diastereomers of each other, they dont 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.

35、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, CH2O Aldol cond. compare Probl

36、em 31(d) CH2CHO 3. H?, H2O 1. NH3 2. HCN ?NH3 CH2CHCOO? Stereocenter Chiral, but racemic (i.e., not optically active) O OO 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 Solut

37、ions to Problems 459 1559T_ch26_454-468 10/20/05 15:39 Page 459 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

38、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)2CHCH3HSCH

39、2 CHCCHCNHCHCOO?NH OO H3N ? NHC O 1. Brucine CH3OH, 0?C 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 (?

40、)- and (?)-isoleucine (?)- and (?)-alloisoleucine O O N CHCH2SCH2CH2 C(CO2CH2CH3)2 1. NaOH Alkylate on sulfur 2. CH2CHCH2Br2. H?, H2O, ? 1. H2O2 ?NH3 O CHCH2SCH2CHCOO?CH2 460 Chapter 26 AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: NITROGEN-CONTAINING POLYMERS IN NATURE 1559T_ch26_454-468 10/2

41、0/05 15:39 Page 460 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

42、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 smal

43、l (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 th

44、eir 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 AH: ?-HelixAm

45、ino acid numbers?-HelixAmino acid numbers A318E5877 B2035F8694 C3642G100118 D5157H125148 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 overal

46、l 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,

47、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 protei

48、ns 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 (Chemic

49、al 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 42. (1)

50、 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, 110C, 24 h) to determine amino acid composition by using an amino acid analyzer. (3) On another portion of this material, apply repetitive Edman degradation

51、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

52、: 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 hydr

53、olysis. 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 overlapp

54、ing 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 12345678 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

55、 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 The final fragment, Phe-Arg, clearly is at the end, overlapping the Ph

56、e 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 ch

57、ymotrypsin fragments end in Phe, Trp, or Tyr, so thats 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 its a matter of matching up all the pieces. Start with this end piece from

58、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- fr

59、ont 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-Tyr- Ser-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-Lys-