Amino Acids Peptides and Proteins - Review Sheet | CHEM 501, Study notes of Chemistry

Download Amino Acids Peptides and Proteins - Review Sheet | CHEM 501 and more Study notes Chemistry in PDF only on Docsity! S-26 Amino Acids, Peptides, and Proteins chapter 3 1. Absolute Configuration of Citrulline The citrulline isolated from watermelons has the structure shown below. Is it a D- or L-amino acid? Explain. Answer Rotating the structural representation 180 in the plane of the page puts the most highly oxidized group—the carboxyl (OCOO) group—at the top, in the same position as the OCHO group of glyceraldehyde in Figure 3–4. In this orientation, the a-amino group is on the left, and thus the absolute configuration of the citrulline is L. 2. Relationship between the Titration Curve and the Acid-Base Properties of Glycine A 100 mL solution of 0.1 M glycine at pH 1.72 was titrated with 2 M NaOH solution. The pH was monitored and the results were plotted as shown in the following graph. The key points in the titration are designated I to V. For each of the statements (a) to (o), identify the appropriate key point in the titration and justify your choice. Note: before considering statements (a) through (o), refer to Figure 3–10. The three species involved in the titration of glycine can be considered in terms of a useful physical analogy. Each ionic species can be viewed as a different floor of a building, each with a different net charge: 12 2 4 6 8 0 11.30 0.5 OH (equivalents) pH 1.0 1.5 2.0 (V) 9.60 (IV) (III) 2.34 (I) (II) 5.97 10 C C O )H (CH NH2 NH22 2 P H C N  H3 COO 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 26 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins S-27 H3NOCH2OCOOH 1 H3NOCH2OCOO  0 (zwitterion) H2NOCH2OCOO  1 The floors are connected by steep stairways, and each stairway has a landing halfway between the floors. A titration curve traces the path one would follow between the different floors as the pH changes in response to added OH. Recall that the pKa of an acid (on a halfway landing) represents the pH at which half of the acid is deprotonated. The isoelectric point (pI) is the pH at which the aver- age net charge is zero. Now you are ready to consider statements (a) through (o). (a) Glycine is present predominantly as the species H3NOCH2OCOOH. (b) The average net charge of glycine is  . (c) Half of the amino groups are ionized. (d) The pH is equal to the pKa of the carboxyl group. (e) The pH is equal to the pKa of the protonated amino group. (f) Glycine has its maximum buffering capacity. (g) The average net charge of glycine is zero. (h) The carboxyl group has been completely titrated (first equivalence point). (i) Glycine is completely titrated (second equivalence point). (j) The predominant species is H3NOCH2OCOO . (k) The average net charge of glycine is 1. (l) Glycine is present predominantly as a 50:50 mixture of H3NOCH2OCOOH and H3NOCH2OCOO . (m) This is the isoelectric point. (n) This is the end of the titration. (o) These are the worst pH regions for buffering power. Answer (a) I; maximum protonation occurs at the lowest pH (the highest [H]). (b) II; at the first pKa, or pK1 (2.34), half of the protons are removed from the a-carboxyl group (i.e., it is half deprotonated), changing its charge from 0 to  . The average net charge of glycine is ( )  1  . (c) IV; the a-amino group is half-deprotonated at its pKa, or pK2 (9.60). (d) II; from the Henderson-Hasselbalch equation, pH = pKa + log ([A ]/[HA]). If [A]/[HA]  1, or [A]  [HA], then pH  pKa. (Recall that log 1  0.) (e) IV; see answers (c) and (d). (f) II and IV; in the pKa regions, acid donates protons to or base abstracts protons from glycine, with minimal pH changes. (g) III; this occurs at the isoelectric point; pI  (pK1  pK2)/2  (2.34  9.60)/2  5.97. (h) III; the pH at which 1.0 equivalent of OH has been added, pH 5.97 (3.6 pH units away from either pKa). (i) V; pH 11.3 (1.7 pH units above pK2). (j) III; at pI (5.97) the carboxyl group is fully negatively charged (deprotonated) and the amino group is fully positively charged (protonated). (k) V; both groups are fully deprotonated, with a neutral amino group and a negatively charged carboxyl group (net charge  1). (l) II; the carboxyl group is half ionized at pH  pK1. 1  2 1  2 1  2 1  2 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 27 ntt 102:WHQY028:Solutions Manual:Ch-03: 5. Separation of Amino Acids by Ion-Exchange Chromatography Mixtures of amino acids can be analyzed by first separating the mixture into its components through ion-exchange chromatography. Amino acids placed on a cation-exchange resin (see Fig. 3–17a) containing sulfonate (OSO3 ) groups flow down the column at different rates because of two factors that influence their movement: (1) ionic attraction between the sulfonate residues on the column and positively charged functional groups on the amino acids, and (2) hydrophobic interactions between amino acid side chains and the strongly hydrophobic backbone of the polystyrene resin. For each pair of amino acids listed, determine which will be eluted first from an ion-exchange column by a pH 7.0 buffer. (a) Asp and Lys (b) Arg and Met (c) Glu and Val (d) Gly and Leu (e) Ser and Ala Answer See Table 3–1 for pKa values for the amino acid side chains. At pH  pI, an amino acid has a net positive charge; at pH pI, it has a net negative charge. For any pair of amino acids, the more negatively charged one passes through the sulfonated resin faster. For two neutral amino acids, the less polar one passes through more slowly because of its stronger hydrophobic interactions with the polystyrene. 6. Naming the Stereoisomers of Isoleucine The structure of the amino acid isoleucine is S-30 Chapter 3 Amino Acids, Peptides, and Proteins pI values Net charge (pH 7) Elution order Basis for separation (a) Asp, Lys 2.77, 9.74 1, 1 Asp, Lys Charge (b) Arg, Met 10.76, 5.74 1, 0 Met, Arg Charge (c) Glu, Val 3.22, 5.97 1, 0 Glu, Val Charge (d) Gly, Leu 5.97, 5.98 0, 0 Gly, Leu Polarity (e) Ser, Ala 5.68, 6.01 0, 0 Ser, Ala Polarity H C H3N H C COO H CH2 CH3 CH3 (a) How many chiral centers does it have? (b) How many optical isomers? (c) Draw perspective formulas for all the optical isomers of isoleucine. Answer (a) Two; at C-2 and C-3 (the a and b carbons). (b) Four; the two chiral centers permit four possible diastereoisomers: (S,S), (S,R), (R,R), and (R,S). 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 30 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins S-31 (c) 7. Comparing the pKa Values of Alanine and Polyalanine The titration curve of alanine shows the ionization of two functional groups with pKa values of 2.34 and 9.69, corresponding to the ionization of the carboxyl and the protonated amino groups, respectively. The titration of di-, tri-, and larger oligopeptides of alanine also shows the ionization of only two functional groups, although the experi- mental pKa values are different. The trend in pKa values is summarized in the table. (a) Draw the structure of Ala–Ala–Ala. Identify the functional groups associated with pK1 and pK2. (b) Why does the value of pK1 increase with each additional Ala residue in the Ala oligopeptide? (c) Why does the value of pK2 decrease with each additional Ala residue in the Ala oligopeptide? Answer (a) The structure at pH 7 is: Note that only the amino- and carboxyl-terminal groups ionize. (b) As the length of poly(Ala) increases, the two terminal groups move farther apart, sepa- rated by an intervening sequence with an “insulating” nonpolar structure. The carboxyl group becomes a weaker acid, as reflected in its higher pKa, because the electrostatic repulsion between the carboxyl proton and the positive charge on the NH3  group diminishes as the groups become more distant. (c) The negative charge on the terminal carboxyl group has a stabilizing effect on the positively charged (protonated) terminal amino group. With increasing numbers of intervening Ala residues, this stabilizing effect is diminished and the NH3  group loses its COO  CH2 H3N H CH3 C CH CH3 COO  CH2 NH3H CH3 C CH CH3 COO  CH2 NH3H CH3 C C HH3CH3C COO  CH2 H3N H CH3 C C H Amino acid or peptide pK1 pK2 Ala 2.34 9.69 Ala–Ala 3.12 8.30 Ala–Ala–Ala 3.39 8.03 Ala–(Ala)n–Ala, n 4 3.42 7.94 CH3 pK2  8.03 pK1  3.39 CH3H  H3N CH CH3 C O N H CH C N CH O C O O 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 31 ntt 102:WHQY028:Solutions Manual:Ch-03: proton more easily. The lower pK2 indicates that the terminal amino group has become a weaker base (stronger acid). The intramolecular effects of the amide (peptide bond) linkages keep pKa values lower than they would be for an alkyl-substituted amine. 8. The Size of Proteins What is the approximate molecular weight of a protein with 682 amino acid residues in a single polypeptide chain? Answer Assuming that the average Mr per residue is 110 (corrected for loss of water in for- mation of the peptide bond), a protein containing 682 residues has an Mr of approximately 682  110  75,000. 9. The Number of Tryptophan Residues in Bovine Serum Albumin A quantitative amino acid analysis reveals that bovine serum albumin (BSA) contains 0.58% tryptophan (Mr 204) by weight. (a) Calculate the minimum molecular weight of BSA (i.e., assuming there is only one tryptophan residue per protein molecule). (b) Gel filtration of BSA gives a molecular weight estimate of 70,000. How many tryptophan residues are present in a molecule of serum albumin? Answer (a) The Mr of a Trp residue must be adjusted to account for the removal of water during peptide bond formation: Mr  204  18  186. The molecular weight of BSA can be cal- culated using the following proportionality, where n is the number of Trp residues in the protein:   A minimum molecular weight can be found by assuming only one Trp residue per BSA molecule (n  1).  32,000 (b) Given that the Mr of BSA is approximately 70,000, BSA has ~ 70,000/32,000  2.2, or 2 Trp residues per molecule. (The remainder from this division suggests that the estimate of Mr 70,000 for BSA is somewhat high.) 10. Subunit Composition of a Protein A protein has a molecular mass of 400 kDa when measured by gel filtration. When subjected to gel electrophoresis in the presence of sodium dodecyl sulfate (SDS), the pro- tein gives three bands with molecular masses of 180, 160, and 60 kDa. When electrophoresis is carried out in the presence of SDS and dithiothreitol, three bands are again formed, this time with molecular masses of 160, 90, and 60 kDa. Determine the subunit composition of the protein. Answer The protein has four subunits, with molecular masses of 160, 90, 90, and 60 kDa. The two 90 kDa subunits (possibly identical) are linked by one or more disulfide bonds. 11. Net Electric Charge of Peptides A peptide has the sequence Glu–His–Trp–Ser–Gly–Leu–Arg–Pro–Gly (a) What is the net charge of the molecule at pH 3, 8, and 11? (Use pKa values for side chains and terminal amino and carboxyl groups as given in Table 3–1.) (b) Estimate the pI for this peptide. (100 g)(186)(1)  0.58g n(186)  Mr BSA 0.58 g  100 g n(Mr Trp) Mr BSA wt Trp  wt BSA S-32 Chapter 3 Amino Acids, Peptides, and Proteins 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 32 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins S-35 Answer (a) From the percentage recovery of activity (units), we calculate percentage yield and specific activity (units/mg). (b) Step 4, ion-exchange chromatography; this gives the greatest increase in specific activity (an index of purity and degree of increase in purification). (c) Step 3, pH precipitation; two-thirds of the total activity from the previous step was lost here. (d) Yes. The specific activity did not increase further after step 5. SDS polyacrylamide gel electrophoresis is an excellent, standard way of checking homogeneity and purity. 16. Dialysis A purified protein is in a Hepes (N-(2-hydroxyethyl)piperazine-N -(2-ethanesulfonic acid)) buffer at pH 7 with 500 mM NaCl. A sample (1 mL) of the protein solution is placed in a tube made of dialysis mem- brane and dialyzed against 1 L of the same Hepes buffer with 0 mM NaCl. Small molecules and ions (such as Na, Cl, and Hepes) can diffuse across the dialysis membrane, but the protein cannot. (a) Once the dialysis has come to equilibrium, what is the concentration of NaCl in the protein sam- ple? Assume no volume changes occur in the sample during the dialysis. (b) If the original 1 mL sample were dialyzed twice, successively, against 100 mL of the same Hepes buffer with 0 mM NaCl, what would be the final NaCl concentration in the sample? Answer (a) [NaCl]  0.5 mM (b) [NaCl]  0.05 mM. 17. Peptide Purification At pH 7.0, in what order would the following three peptides be eluted from a col- umn filled with a cation-exchange polymer? Their amino acid compositions are: Protein A: Ala 10%, Glu 5%, Ser 5%, Leu 10%, Arg 10%, His 5%, Ile 10%, Phe 5%, Tyr 5%, Lys 10%, Gly 10%, Pro 5%, and Trp 10%. Protein B: Ala 5%, Val 5%, Gly 10%, Asp 5%, Leu 5%, Arg 5%, Ile 5%, Phe 5%, Tyr 5%, Lys 5%, Trp 5%, Ser 5%, Thr 5%, Glu 5%, Asn 5%, Pro 10%, Met 5%, and Cys 5%. Protein C: Ala 10%, Glu 10%, Gly 5%, Leu 5%, Asp 10%, Arg 5%, Met 5%, Cys 5%, Tyr 5%, Phe 5%, His 5%, Val 5%, Pro 5%, Thr 5%, Ser 5%, Asn 5%, and Gln 5%. Answer Protein C has a net negative charge because there are more Glu and Asp residues than Lys, Arg, and His residues. Protein A has a net positive charge. Protein B has no net charge at neutral pH. A cation-exchange column has a negatively charged polymer, so protein C interacts most weakly with the column and is eluted first, followed by B, then A. Procedure Protein (mg) Activity (units) % Yield Specific activity (units/mg) Purification factor (overall) 1 20,000 4,000,000 (100) 200 (1.0) 2 5,000 3,000,000 75 600  3.0 3 4,000 1,000,000 25 250  1.25 4 200 800,000 20 4,000  20 5 50 750,000 19 15,000  75 6 45 675,000 17 15,000  75 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 35 ntt 102:WHQY028:Solutions Manual:Ch-03: 18. Sequence Determination of the Brain Peptide Leucine Enkephalin A group of peptides that in- fluence nerve transmission in certain parts of the brain has been isolated from normal brain tissue. These peptides are known as opioids because they bind to specific receptors that also bind opiate drugs, such as morphine and naloxone. Opioids thus mimic some of the properties of opiates. Some re- searchers consider these peptides to be the brain’s own painkillers. Using the information below, deter- mine the amino acid sequence of the opioid leucine enkephalin. Explain how your structure is consis- tent with each piece of information. (a) Complete hydrolysis by 6 M HCl at 110 C followed by amino acid analysis indicated the presence of Gly, Leu, Phe, and Tyr in a 2:1:1:1 molar ratio. (b) Treatment of the peptide with 1-fluoro-2,4-dinitrobenzene followed by complete hydrolysis and chromatography indicated the presence of the 2,4-dinitrophenyl derivative of tyrosine. No free tyrosine could be found. (c) Complete digestion of the peptide with chymotrypsin followed by chromatography yielded free tyrosine and leucine, plus a tripeptide containing Phe and Gly in a 1:2 ratio. Answer (a) The empirical composition is (2 Gly, Leu, Phe, Tyr)n. (b) Tyr is the amino-terminal residue, and there are no other Tyr residues, so n  1 and the sequence is Tyr–(2 Gly, Leu, Phe). (c) As shown in Table 3–7, chymotrypsin cleaves on the carboxyl side of aromatic residues (Phe, Trp, and Tyr). The peptide has only two aromatic residues, Tyr at the amino terminus and a Phe. Because there are three cleavage products, the Phe residue cannot be at the carboxyl terminus. Rather, release of free leucine means that Leu must be at the carboxyl terminus and must be on the carboxyl side of Phe in the peptide. Thus the sequence must be Tyr–(2 Gly)–Phe–Leu  Tyr–Gly–Gly–Phe–Leu 19. Structure of a Peptide Antibiotic from Bacillus brevis Extracts from the bacterium Bacillus brevis contain a peptide with antibiotic properties. This peptide forms complexes with metal ions and seems to disrupt ion transport across the cell membranes of other bacterial species, killing them. The structure of the peptide has been determined from the following observations. (a) Complete acid hydrolysis of the peptide followed by amino acid analysis yielded equimolar amounts of Leu, Orn, Phe, Pro, and Val. Orn is ornithine, an amino acid not present in proteins but present in some peptides. It has the structure (b) The molecular weight of the peptide was estimated as about 1,200. (c) The peptide failed to undergo hydrolysis when treated with the enzyme carboxypeptidase. This en- zyme catalyzes the hydrolysis of the carboxyl-terminal residue of a polypeptide unless the residue is Pro or, for some reason, does not contain a free carboxyl group. (d) Treatment of the intact peptide with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis and chromatography, yielded only free amino acids and the following derivative: (Hint: note that the 2,4-dinitrophenyl derivative involves the amino group of a side chain rather than the a-amino group.) S-36 Chapter 3 Amino Acids, Peptides, and Proteins NO2 CH2 CH2 NH3 O2N COOCH2NH C H CH2 CH2 CH2 C COOH3N H NH3  2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 36 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins S-37 (e) Partial hydrolysis of the peptide followed by chromatographic separation and sequence analysis yielded the following di- and tripeptides (the amino-terminal amino acid is always at the left): Leu–Phe Phe–Pro Orn–Leu Val–Orn Val–Orn–Leu Phe–Pro–Val Pro–Val–Orn Given the above information, deduce the amino acid sequence of the peptide antibiotic. Show your reasoning. When you have arrived at a structure, demonstrate that it is consistent with each experimental observation. Answer The information obtained from each experiment is as follows. (a) The simplest empirical formula for the peptide is (Leu, Orn, Phe, Pro, Val)n. (b) Assuming an average residue Mr of 110, the minimum molecular weight for the peptide is 550. Because 1,200/550 ≈ 2, the empirical formula is (Leu, Orn, Phe, Pro, Val)2. (c) Failure of carboxypeptidase to cleave the peptide could result from Pro at the carboxyl terminus or the absence of a carboxyl-terminal residue—as in a cyclic peptide. (d) Failure of FDNB to derivatize an a-amino group indicates either the absence of a free amino-terminal group or that Pro (an imino acid) is at the amino-terminal position. (The derivative formed is 2,4 dinitrophenyl- -ornithine.) (e) The presence of Pro at an internal position in the peptide Phe–Pro–Val indicates that it is not at the amino or carboxyl terminus. The information from these experiments suggests that the peptide is cyclic. The alignment of overlapping sequences is Leu–Phe Phe–Pro Phe–Pro–Val Pro–Val–Orn Val–Orn Orn–(Leu) Thus, the peptide is a cyclic dimer of Leu–Phe–Pro–Val–Orn: where the arrows indicate the –CO → NH–, or C → N, direction of the peptide bonds. This structure is consistent with all the data. 20. Efficiency in Peptide Sequencing A peptide with the primary structure Lys–Arg–Pro–Leu– Ile–Asp–Gly–Ala is sequenced by the Edman procedure. If each Edman cycle is 96% efficient, what percentage of the amino acids liberated in the fourth cycle will be leucine? Do the calculation a second time, but assume a 99% efficiency for each cycle. Answer 88%, 97%. The formula for calculating the percentage of correct amino acid liber- ated after sequencing cycle n, given an efficiency x, is xn/x, or xn  1. If the efficiency is 0.96, the fraction of correct amino acid liberated in the fourth cycle is (0.96)3  0.88. If the effi- ciency is 0.99, the fraction is (0.99)3  0.97. Leu Orn Phe Pro Val Val Pro OrnPhe Leu 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 37 ntt 102:WHQY028:Solutions Manual:Ch-03: (b) What does differential centrifugation accomplish? Organelles differ in size and therefore sediment at different rates during centrifugation. Larger organelles and pieces of cell debris sediment first, and progressively smaller cellular components can be isolated in a series of centrifugation steps at increasing speed. The contents of each fraction can be determined microscopically or by enzyme assay. (c) What is the rationale for the two-step addition of ammonium sulfate? Proteins have characteristic solubilities at different salt concentrations, depending on the functional groups in the protein. In a concentration of ammonium sulfate just below the precipitation point of CS, some unwanted proteins can be precipitated (salted out). The ammonium sulfate concentration is then increased so that CS is salted out. It can then be recovered by centrifugation. (d) Why is a buffer solution without ammonium sulfate used for the dialysis step? Osmolarity (as well as pH and temperature) affects the conformation and stability of proteins. To solubilize and renature the protein, the ammonium sulfate must be removed. In dialysis against a buffered solution containing no ammonium sulfate, the ammonium sulfate in the sample moves into the buffer until its concentration is equal in both solutions. By dialyzing against large volumes of buffer that are changed frequently, the concentration of ammonium sulfate in the sample can be reduced to almost zero. This procedure usually takes a long time (typically overnight). The dialysate must be buffered to keep the pH (and ionic strength) of the sample in a range that promotes the native conformation of the protein. (e) What does the instruction to collect the first fraction tell you about the protein? The CS molecule is larger than the pore size of the chromatographic gel. Size-exclusion columns retard the flow of smaller molecules, which enter the pores of the column matrix material. Larger molecules flow around the matrix, taking a direct route through the column. Why is UV absorbance at 280 nm a good way to monitor for the presence of protein in the eluted fractions? The aromatic side chains of Tyr and Trp residues strongly absorb at 280 nm. (f) Explain the procedure on the cation-exchange chromatography column. CS has a positive charge (at the pH of the separation) and binds to the negatively charged beads of the cation-exchange column, while negatively charged and neutral proteins pass through. CS is displaced from the column by raising the pH of the mobile phase and thus altering the charge on the CS molecules. (g) Why were you unconvinced of the purity of the “single” protein band on your isoelectric focusing gel? Several different proteins, all with the same pI, could be focused in the “single” band. SDS polyacrylamide gel electrophoresis separates on the basis of mass and therefore would separate any polypeptides in the pI 5.6 band. Why is it important to do the SDS gel electrophoresis after the isoelectric focusing? SDS is a highly negatively charged detergent that binds tightly and uniformly along the length of a polypeptide. Removing SDS from a protein is difficult, and a protein with only traces of SDS no longer has its native acid-base properties, including its native pI. Data Analysis Problem 23. Determining the Amino Acid Sequence of Insulin Figure 3–24 shows the amino acid sequence of the hormone insulin. This structure was determined by Frederick Sanger and his coworkers. Most of this work is described in a series of articles published in the Biochemical Journal from 1945 to 1955. When Sanger and colleagues began their work in 1945, it was known that insulin was a small protein consisting of two or four polypeptide chains linked by disulfide bonds. Sanger and his coworkers had developed a few simple methods for studying protein sequences. S-40 Chapter 3 Amino Acids, Peptides, and Proteins 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 40 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins S-41 Treatment with FDNB. FDNB (1-fluoro-2,4-dinitrobenzene) reacted with free amino (but not amido or guanidino) groups in proteins to produce dinitrophenyl (DNP) derivatives of amino acids: Acid Hydrolysis. Boiling a protein with 10% HCl for several hours hydrolyzed all of its peptide and amide bonds. Short treatments produced short polypeptides; the longer the treatment, the more com- plete the breakdown of the protein into its amino acids. Oxidation of Cysteines. Treatment of a protein with performic acid cleaved all the disulfide bonds and converted all Cys residues to cysteic acid residues (Fig. 3–26). Paper Chromatography. This more primitive version of thin-layer chromatography (see Fig. 10–24) separated compounds based on their chemical properties, allowing identification of single amino acids and, in some cases, dipeptides. Thin-layer chromatography also separates larger peptides. As reported in his first paper (1945), Sanger reacted insulin with FDNB and hydrolyzed the result- ing protein. He found many free amino acids, but only three DNP–amino acids: -DNP-glycine (DNP group attached to the -amino group); -DNP-phenylalanine; and -DNP-lysine (DNP attached to the -amino group). Sanger interpreted these results as showing that insulin had two protein chains: one with Gly at its amino terminus and one with Phe at its amino terminus. One of the two chains also contained a Lys residue, not at the amino terminus. He named the chain beginning with a Gly residue “A” and the chain beginning with Phe “B.” (a) Explain how Sanger’s results support his conclusions. (b) Are the results consistent with the known structure of insulin (Fig. 3–24)? In a later paper (1949), Sanger described how he used these techniques to determine the first few amino acids (amino-terminal end) of each insulin chain. To analyze the B chain, for example, he carried out the following steps: 1. Oxidized insulin to separate the A and B chains. 2. Prepared a sample of pure B chain with paper chromatography. 3. Reacted the B chain with FDNB. 4. Gently acid-hydrolyzed the protein so that some small peptides would be produced. 5. Separated the DNP-peptides from the peptides that did not contain DNP groups. 6. Isolated four of the DNP-peptides, which were named B1 through B4. 7. Strongly hydrolyzed each DNP-peptide to give free amino acids. 8. Identified the amino acids in each peptide with paper chromatography. The results were as follows: B1: -DNP-phenylalanine only B2: -DNP-phenylalanine; valine B3: aspartic acid; -DNP-phenylalanine; valine B4: aspartic acid; glutamic acid; -DNP-phenylalanine; valine (c) Based on these data, what are the first four (amino-terminal) amino acids of the B chain? Explain your reasoning. (d) Does this result match the known sequence of insulin (Fig. 3–24)? Explain any discrepancies. Sanger and colleagues used these and related methods to determine the entire sequence of the A and B chains. Their sequence for the A chain was as follows (amino terminus on left):  HF O2N NO2N H RNH2R O2N NO2F Amine FDNB DNP-amine Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val– 1051 2015 Cys–Ser–Leu–Tyr–Glx–Leu–Glx–Asx–Tyr–Cys–Asx 2608T_ch03sm_S26-S43 2/1/08 1:38PM Page 41 ntt 102:WHQY028:Solutions Manual:Ch-03: Because acid hydrolysis had converted all Asn to Asp and all Gln to Glu, these residues had to be desig- nated Asx and Glx, respectively (exact identity in the peptide unknown). Sanger solved this problem by using protease enzymes that cleave peptide bonds, but not the amide bonds in Asn and Gln residues, to prepare short peptides. He then determined the number of amide groups present in each peptide by measuring the NH4  released when the peptide was acid-hydrolyzed. Some of the results for the A chain are shown below. The peptides may not have been completely pure, so the numbers were approximate— but good enough for Sanger’s purposes. (e) Based on these data, determine the amino acid sequence of the A chain. Explain how you reached your answer. Compare it with Figure 3–24. Answer (a) Any linear polypeptide chain has only two kinds of free amino groups: a single -amino group at the amino terminus, and an -amino group on each Lys residue present. These amino groups react with FDNB to form a DNP–amino acid derivative. Insulin gave two different -amino-DNP derivatives, suggesting that it has two amino termini and thus two polypeptide chains—one with an amino-terminal Gly and the other with an amino- terminal Phe. Because the DNP-lysine product is -DNP-lysine, the Lys is not at an amino terminus. (b) Yes. The A chain has amino-terminal Gly; the B chain has amino-terminal Phe; and (non- terminal) residue 29 in the B chain is Lys. (c) Phe–Val–Asp–Glu–. Peptide B1 shows that the amino-terminal residue is Phe. Peptide B2 also includes Val, but since no DNP-Val is formed, Val is not at the amino terminus; it must be on the carboxyl side of Phe. Thus the sequence of B2 is DNP-Phe–Val. Similarly, the sequence of B3 must be DNP-Phe–Val–Asp, and the sequence of the A chain must begin Phe–Val–Asp–Glu–. (d) No. The known amino-terminal sequence of the A chain is Phe–Val–Asn–Gln–. The Asn and Gln appear in Sanger’s analysis as Asp and Glu because the vigorous hydrolysis in step 7 hydrolyzed the amide bonds in Asn and Gln (as well as the peptide bonds), forming Asp and Glu. Sanger et al. could not distinguish Asp from Asn or Glu from Gln at this stage in their analysis. (e) The sequence exactly matches that in Figure 3–24. Each peptide in the table gives spe- cific information about which Asx residues are Asn or Asp and which Glx residues are Glu or Gln. Ac1: residues 20–21. This is the only Cys–Asx sequence in the A chain; there is ~1 amido group in this peptide, so it must be Cys–Asn: S-42 Chapter 3 Amino Acids, Peptides, and Proteins Peptide name Peptide sequence Number of amide groups in peptide Ac1 Cys–Asx 0.7 Ap15 Tyr–Glx–Leu 0.98 Ap14 Tyr–Glx–Leu–Glx 1.06 Ap3 Asx–Tyr–Cys–Asx 2.10 Ap1 Glx–Asx–Tyr–Cys–Asx 1.94 Ap5pa1 Gly–Ile–Val–Glx 0.15 Ap5 Gly–Ile–Val–Glx–Glx–Cys–Cys– Ala–Ser–Val–Cys–Ser–Leu 1.16 N–Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Glx–Leu–Glx–Asx–Tyr–Cys–Asn–C 1 5 10 15 20 2608T_ch03sm_S26-S43 2/1/08 11:45AM Page 42 ntt 102:WHQY028:Solutions Manual:Ch-03: