ch4 - t The Three-Dimensional Structure of Proteins 5-44 1...

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Unformatted text preview: t The Three-Dimensional Structure of Proteins 5-44 1. Properties of the Peptide Bond In x—ray studies of crystalline peptides, Linus Pauiing and Robert Corey found that the G—N bond in the peptide link is hitermediate in length (1.82 A) between a typical C—~N single bond (1.49 A) and a czar double bond (1.27 A). They also found that the peptide bond is planar (all four atoms attached to the C—-—N group are located in the same plane) and that the two «ya-carbon atoms attached to the C—-—N are aiways trans to each other (on opposite sides of the peptide bond). (a) What does the iength of the C-N bond in the peptide linkage indicate about its strength and its bond order (Le, whether it is single, double, or triple)? (b) What do the observations of Pauling and Corey teli us about the ease of rotation about the C——N peptide bond? Answer (a) The higher the bond order (double or triple vs. single), the shorter and stronger are the bonds. Thus, bond length is an indication of bond order. For example, the (Jr—“N bond is shorter (1.27 A) and has a higher order (n = 2.0) than a typical C—«N bond (length a 1.49 A, n = 1.0). The length of the C—MN bond of the peptide linic (1.32 A) indicates that it is intermediate in strength and bond order between a singie and double bond. , (b) Rotation about a double bond is generaliy impossible at physiological temperatures, and the steric relationship of the groups attached to the two atoms involved in the doubie bond is spatially “fixed.” Since the peptide bond has considerable doublewbond character, there is essentially no rotation, and the _C=O and whim—ll groups are fixed in the trans configuration. 2. Structurai and Functional Relationships in Fibrous Proteins William Astbury discovered that the x—ray diffraction pattern of wool shows a repeating structural unit spaced about 5.2 A along the length of the wool fiber. When he steamed and stretched the wool, the x—ray pattern showed a new repeating structural unit at a spacing of 7.0 A. Steaming and stretching the wool and then letting it shrink gave an x—ray pattern consistent with the original spacing of about 5.2 A. Aithough these observations provided important clues to the molecular structure of wool, Astbury was unable to interpret them at the time. (3.) Given our current understanding of the structure of wool, interpret Astbury’s observations. (b) When wool sweaters or socks are washed in hot water or heated in a dryer, they shrink. Silk, on the other hand, does not shrink under the same conditions. Explain. 3. Chapter 4 The Three~0imensionai Structure of Proteins 3-45 Answer (a) rl‘he principal structural units in the wool fiber polypeptide, iii—keratin, are successive turns of the or helix, which are spaced at 5.4 A intervals; two or—keratin strands ‘ twisted into a coiled coil produce the 5.2 A spacing. The intrinsic stability of the helix (and thus the fiber) results from intro; chain hydrogen bonds {see Fig. 4—4a). Steaming and stretching the fiber yields an extended polypeptide chain with the i3 conformation, in which the distance between adjacent R groups is about 7.0 A. Upon resteaming, the polypeptide chains again assume the teas-extended a—helix conformation. Freshly sheared woo} is primarily in its oz—keratin {oz-helical coiled coil) form (see Fig. 4ml 0). Because raw wool is crimped or curly, it is combed and stretched to straighten it before being spun into fibers for ciothing. This processing converts the wool from its native oe—heiicai conformation to a more extended ,6 form. Moist heat triggers a conformational change back to the native oz-helical structure, which shrinks both the fiber and the clothing. Under conditions of mechanical tension and moist heat, wool can be stretched back to a fully extended form. In silk, by contrast, the polypeptide chains have a very stable ,G—pleated sheet structure, fully extended aiong the axis of the fiber (see Fig. 445), and have small, closely packed amino acid side chains (see Fig. 4"18}. These characteristics make silk resistant to stretching and Shrinking. (1)) Rate of Synthesis of Hair a-Keratin Hair grows at a rate of E5 to 20 cm/yr. All this growth is concentrated at the base of the hair fiber, where {pr-keratin filaments are synthesized inside living epidermal coils and assembled into ropelike structures {see Fig. 4—~iO). The fundamental structural element of car—keratin is the or helix, which has 3.6 amino acid residues per turn and a rise of 5.4 A per turn (see Fig. 4—4a). Assuming that the biosynthesis of rye—helical keratin chains is the rate~lirniting factor in the growth of hair, calcafiate the rate at which peptide bonds of wkeratin chains must be synthesized (peptide bonds per second) to account for the observed yearly growth of hair. Answer Because there are 3.6 amino acids (AAs) per turn and the rise is 5.4 A/turn, the length per AA of the or helix is 5.4 iii/turn o -10 WWW—m = 1. = . X 3'6 tum 5 NM 1 5 10 m/AA A growth rate of 20 sin/yr is equivalent to 20 Cm/yw «~— ea x 104ch 2 5.3 x 10"“9m/s (365 days/yr) (24 h/day) {60 min/h) (60 s/rnin) Thus, the rate at which anuno acids are added is 6.3 x 10'9m/s m = 42 AN3 2 42 peptide bonds per second Effect of pH on the Conformation of u-Helical Secondary Structures The unfolding of the o: helix of a polypeptide to a randomly coiled conformation is accompanied by a large decrease in a property called specific rotation, a measure of a solutions capacity to rotate plane-poiarized light. Polygiutamate, a polypeptide made up of only L—Glu residues, has the car-helical conformation at pH 3. 6-46 Chapter 4 The Three-Dimensionai Structure of Proteins When the pH is raised to ?, there is a large decrease in the specific rotation of the solution. Similariy, polyiysine (L—Lys residues) is an or helix at pH 10, but when the pH is lowered to 7 the specific rotation also decreases, as shown by the following graph. r’ Polthlu) ‘5'“ a Heiix Random conformation Specific rotation <— Random conformation L. E 1 i l E l O 2 4 6 8 10 12 14 pH What is the explanation for the effect of the pH changes on the conformations of poiy(Giu) and polyCLys)? Why does the transition occur over such a narrow range of pH? Answer At pH values above 6, deprotonation of the carboxylate side chains of poly£Glu) leads to repulsion between adjacent negatively charged groups, which destabilizes the or heiix and results in unfoiding. Simiiarly, at pH 7 protonation of the amino-group side chains of poly(Lys) causes repulsion between positively charged groups, which leads to unfolding. 5. Disulfide Bonds Determine the i’roperties of Many Proteins Some natural proteins are rich in dismfide bonds, and their mechanical properties (tensile strength, viscosity, hardness, etc.) are corre- lated with the degree of disulfide bonding. (a) Glutenin, a wheat protein rich in disulfide bonds, is responsible for the cohesive and eiastic char— acter of dough made from wheat flour. Simiiarly, the hard, tough nature of tortoise sheil is due to the extensive disulfide bonding in its a—iteratiri. What is the molecular basis for the correlation between disulfidewbond content and mechanical properties of the protein? (b) Most giobuiar proteins are denatured and iose their activity when briefly heated to 65 00. However, globular proteins that contain multiple disulfide bonds often must be heated ionger at higher temperatures to denature them. One such protein is bovine pancreatic trypsin inhibitor (BPTE), which has 58 amino acid residues in a single chain and contains three disulfide bonds. On cooling a solution of denatured BPTE, the activity of the protein is restored. What is the molecular basis for this property? Answer (a) Disuitide bonds are covalent bonds, which are much stronger than the noncovaient if interactions (hydrogen bonds, hydrophobic interactions, van der Waals interactions) that ' stabilize the three—dimensional structure of most proteins. Disulfide bonds serve to cross—link protein chains, increasing stiffness, hardness, and mechanicai strength. (b) As the temperature is raised, the increased therrnai motion of the poiypeptide chains and the vibrational motions of hydrogen bonds ultimater lead to thermal denaturation (unfoldnig) of a protein. Cystine residues (disulfide bonds) can, depending on their location in the protein structure, prevent or restrict the movement of folded protein domains, block access of solvent water to the interior of the protein, and prevent the compiete unfolding of the protein. Refolding to the native structure from a random conformation is seidom spontaneous, owing to the very large number of conformations i .. n.3,? Chapter 4 The Three~DimensionaE Structure of Proteins 8-47 possible. Disulfide bonds limit the number of conformations by allowing only a few minimally unfolded structures, and hence the protein returns to its native conformation more easily upon cooling. 6. Amino Acid Sequence and Protein Structure Our growing understanding of how proteins fold allows researchers to make predictions about protein structure based on primary amino acid sequence data. Consider the following amino acid sequence. Ile—Ala-His—Thr—'i‘yr—GlyuPromPhewGlumAla—Ala-Met~Cys_Lys-Trp“Gin—Ala—Gln~Pro-AspwGly—Mob—Giu—Cys—AiawPhewHismArg l2 3 4 5 6 7 8 9 20H1213141516371819202122232425262728 (a) Where might bends or 6 turns occur? (In) Where might intrachain disulfide cross—linkages be formed? (c) Assuming that this sequence is part of a larger globular protein, indicate the probable location (the external surface or interior of the protein) of the following amino acid residues: Asp, lie, Thr, Ala, Gin, Lys. Explain your reasoning. (Hint: See the hydropathy index in Table 84.) Answer (a) Bends or turns are most lilceiy to occur at residues 7 and 19 because Pro residues are often (but not always) found at bends in globular folded proteins. A bend may also occur at the Thr residue (residue 4) and, assuming that this is a portion of a larger polypep- tide, at the lie residue (residue 1). (b) lntrachain disulfide cross-linkages can form only between residues 13 and 24 (Cys residues). ((2) Amino acids with ionic (charged) or strongly polar neutral groups (cg, Asp, Gin, and Lys in this protein) are located on the external surface, where they interact optimally with solvent water. Residues with nonpolar side chains (such as Ala and lie) are situ- ated in the interior, where they escape the polar environment. Thr is of intermediate polarity and could be found either in the interior or on the exterior surface (see Table Sel). 7. BacteriorhodOpsin in Purple Membrane Proteins Under the proper environmental conditions, the salt—loving bacterium Halobacterium halobium synthesizes a membrane protein (Mr 26,000) known as bacteriorhodopsin, which is purple because it contains retinal (see Fig. 10—21). Molecules of this protein aggregate into “purple patches” in the cell membrane. Bacteriorhodopsin acts as a light- activated proton pump that provides energy for cell functions. Xuray analysis of this protein reveals that it consists of seven parallel oz—helical segments, each of which traverses the bacterial cell mem- brane (thickness 45 A). Calculate the nururnurn number of amino acids necessary for one segment of a helix to traverse the membrane completely. Estimate the fraction of the bacteriorbodopsin protein that is involved in membrane-spanning helices. (Use an average amino acid residue weight of 110.) Answer Us'mg the parameters from Problem 3 (3.6 AA/turn, 5.4 Arturn), we can calculate that there are 0.67 ANA along the axis of a helix. Thus, a helix of length 45 A (sufficient to span the membrane) requires a minimum of (45 A) (0.67 AA/A) = 30 amino acid residues. ri‘he membrane protein has an M, of 26,000 and average AA M, of 110. Thus the protein contains 26,000/110 : 240 AA. Of these, (30 All/helix) (7 helices) =1 210 AA are involved in membrane-spanning helices, which is 210/240 m 0.87, or 87%, of the protein. 5-48 Chapter 4 The Three-Dimensional Structure of Proteins 8. Protein Structure Terminology Is myoglobin a motif, a domain, or a complete three-dimensional structure? Answer Myoglobin is all three. The folded structure, the “globin fold,” is a motif found in all globins. The polypeptide folds into a single domain, which for this protein represents the entire threemdimensional structure. 9. Pathogenic Action of Bacteria That Cause Gas Gangrene The highly pathogenic anaerobic bac~ 10. terium Clostrt’d’ium pefimngens is responsible for gas gangrene, a condition in which animal tissue structure is destroyed. This bacterium secretes an enzyme that efficiently catalyzes the hydrolysis of the peptide bond indicated by an asterisk: 3k MXWGlymPro—Y~ figs ——xmc00“ + Han-oiy—mevm where X and Y are any of the 20 conunon amino acids. How does the secretion of this enzyme contribute to the invasiveness of this bacterium in human tissues? Why does this enzyme not affect the bacterium itself? Answer Collagen is distinctive in its amino acid composition, having a very high proportion of Gly (85%) and Pro residues. The enzyme secreted by the bacterium is a coliagenase, which breaks down collagen at the XwGly bonds and damages the connective—tissue barrier (skin, hide, etc.) of the host; this allows the bacterium to invade the host tissues. Bacteria do not contain collagen and thus are unaffected by collagenase. Number of Polypeptide Chains in a Multisnbunit Protein A sample (660 mg) of an oligomeric protein of Mr 132,000 was treated with an excess of l-fluoro-2A-dinitrobenzene (Sanger’s reagent) under slightly alkaline conditions until the chemical reaction was complete. The peptide bonds of the protein were then completely hydrolyzed by heating it with concentrated H01. The hydrolysate was found to contain 5.5 mg of the following compound: 2,4«Dinitrophenyl derivatives of the ins—amino groups of other amino acids could not be found. (3.) Explain how true information can be used to determine the number of polypeptide chains in an oligomeric protein. (1)) Calculate the number of polypeptide chains in this protein. (c) What other protein analysis technique could you employ to determine whether the polypeptide chains in this protein are similar or different? Answer (3.) Because only a single 2,4—dinitrophenyl (DNP) amino acid derivative is found, there is only one kind of amino acid at the amino terminus (to, all the polypeptide chains have the same amino-ternunal residue). Comparing the number of moles of this derivative to the number of moles of protein gives the number of polypeptide chains. Chapter 4 The Three-Dimensional Structure of Proteins 5-49 (h) The amount of protein 2 (0.66 g)/(132,000 g/rnol) = 5 X 10"“6 mol. Because Mr for DNP»Val (CHHBOSNS) -— 2.83, the amount of DNPNal m (0.0055 g)/ (283 g/mol) a 1.9 >< 10'5 mol. The ratio of moles of DNP-Val to moles of protein gives the number of amino- terrninai residues and thus the number of chains per ohgomer: 1.9 x 10-5 mol DNP-Vai =4 oi et'd h' 5X10“6rnolprotein p ypplecams An alternative approach to the probiem is through the proportionality (w, 2 number of polypeptide chains): n(283 g/rnol) : 5.5 mg 132,000 g/mol 660 mg _ WM _ W (660 mg)(283 g/moi) ‘ (c) Polyacrylarnide gel electrophoresis in the presence of a detergent (such as sodium dodecylsulfate [SDS]) and an agent that prevents the formation of disulfide bonds (such as fi—mercaptoethanol) would provide information on subunit structure of a protein. In the example here, an oligomeric protein of Mr 182,000 that had four identical subunits would produce a single band on the electrophoretic gel, with apparent Mr ~33,000 (132,000/33,000 = 4). If the protein were made up of different polypeptide subunits, they would likely appear as multiple discrete bands on the gel. it 3.9z4 11. Predicting Secondary Structure Which of the foliowing peptides is more likely to take up an cr—helical structure, and why? (a) LKAENDEAARAMSEA (b) CRAGGFPWDQPGTSN Answer By cursory inspection, peptide (a) has five Ala residues (most likeiy to take up an a—helical conformation), and peptide (b) has five Pro and Gly residues (least often found in an a helix). rl‘his suggests that (a) is more likely than (b) to form an a helix. Referring to Table 4~»1, (a) has 15 residues with a total AAG" of 18 kJ/reol, and (b) has 15 residues with a total AAG" of 41 kJ/nrol. Given that a lower AAG° indicates a greater tendency to take up an car-helical structure, this confirms that peptide (a) is much more likely to form an or helix. 12. Amykoid Fibers in Disease Several small aromatic molecules, such as phenol red (used as a non— toxic drug model), have been shown to inhibit the formation of amyloid in laboratory model systems. A goal of the research on these small aromatic compounds is to find a drug that would efficiently inhibit the formation of amyloid in the brain in people with incipient Alzheimer’s disease. (a) Suggest why molecules with aromatic substituents would disrupt the formation of amytoid. (1)) Some researchers have suggested that a drug used to treat Alzheimer’s disease may also be ef- fective in treating type 2 (adult onset) diabetes meliitus. Why might a single drug be effective in treating these two different conditions? Answer (a) Aromatic residues seem to play an important role in stabilizing amyloid fibrils. Thus, molecules with aromatic substituents may inhibit amyioid formation by interfering with the stacking or association of the aromatic side chains. (1)) Amyloid is formed in the pancreas in association with type 2 diabetes, as it is in the brain in Alzheimer’s disease. Although the amyloid fibrils in the two diseases involve different pro- teins, the fundamental structure of the amyloid is similar and similarly stabilized in both, and thus they are potential targets for similar drugs designed to disrupt this structure. 3—50 Chapter 4 ‘I‘he Three-Dimensional Structure of Proteins Biochemistry on the interest include Protein information Resource (PER; http://pir.georgetown.edu), Structural Classification of Proteins (SCOP; http://scop.mrc—lrnbcamacuk/scop), and Prosite (http://expasy.org/prosite). At your seiected database site, foliow links to the sequence comparison engine. Enter about 30 residues from the protein sequence in the appropriate search field and submit it for anaiysis. What does this analysis tell you about the identity of the protein? (1)) Try using different portions of the amino acid sequence. Do you always get the same result? (c) A variety of websites provide information about the three-dimensional structure of proteins. Find information about the protein’s secondary, tertiary, and quaternary structures using database sites, such as the Protein Data Bank (FDR; wwwrcsborg) or 800?. Answer (a) At the FIRM—International Protein Sequence Database (litth/pirgeorgetownedu), click on “Search/Analysis” and choose “BLAST search.” Paste the first 30 amino acid residues of other species match as well. Click on the “Help” button for explanations of the various options and table items. (b) As more proteins are sequenced, the number of hits returned from a SG-residue se- (c) At the PDB (resborg) search on “NF-kappawB 1365.” You will get more than a dozen hits. Adding “human” to the search limits the resuits further. Go back to the more genera} search on “NF-kappa—B p65" and scan through the returned items. NFKB has two sub- Chapter 4 The Three-Dimensionat Structure of Proteins 3-51 ((1) The various proteins that predominate in this search are eukaryotic transcription factors, which stirnuiate transcription of genes involved in development and some immune re« spouses. The proteins have two distinct domains, including an amino—terminai Rel ho— mology domain 300 amino acid residues long and a carboxyl—terminal domain invoived in gene activation. A search of the various links in the databases will reveai much additional information about the proteins’ structure and function. Bate Analysis Problem 14. Mirror-Image Proteins As noted in Chapter 3, “The amino acid residues in protein moiecules are exclusiveiy L stereoisomers." It is not clear whether this selectivity is necessary for proper protein function or is an accident of evolution. To explore this question, Miiton and coileagues (i992) pub— lished a study of an enzyme roads entireiy of D stereoisomers. The enzyme they chose was HIV pro- tease, a proteolytic enzyme made by HIV that converts inactive viral pre>proteins to their active forms. Previously, Wlodawer and coworkers (1989) had reported the compiete chemical synthesis of HIV protease from L-arnino acids (the L—enzyme), using the process shown in Figure $29. Normal HIV protease contains two Cys residues at positions 67 and 95. Because chemical synthesis of proteins containing Cys is technicaliy difficult, Wlodawer and colleagues substituted the synthetic amino acid L—a—aminomn-butyiic acid (Abe) for the two Cys residues in the protein. In the authors’ words, this was done so as to “reduce synthetic difficulties associated with Cys deprotection and ease product handhng.“ (a) The structure of Abe is shown below. Why was this a suitable substitution for a Cys residue? Under what circumstances would it not be suitabie? / l . ' n—c—cagmcaa +NH3 L-a-Amino-n—butyric acid ‘0 \O Wiodawer and coworkers denatured the newly synthesized protein by dissoiving it in 6 M guani» dine H01, and then aliowed it to fold slowiy by dialyzing away the guanidine against a neutral buffer (10% glycerol, 25 mM NaPO4, pH 7). (b) rI‘here are many reasons to predict that a protein synthesized, denatured, and folded in this man- ner would not be active. Give three such reasons. (c) interestingly, the resuiting L—protease was active. What does this finding teli you about the role of disulfide bonds in the native HIV protease molecule? In their new study1 Miiton and coworkers synthesized HIV protease from o—arnino acids, using the same protocol as the earlier study {Wlodawer et at). Formally, there are three possibilities for the folding of the prrotease: it would give (1) the same shape as the prrotease, (2) the mirror image of the Leprotease, or (3) something else, possibly inactive. (d) For each possibiiity, decide whether or not it is a liirely outcome and defend your position. in fact, the D—protease was active: it cleaved a particuiar synthetic substrate and was inhibited by specific inhibitors. To examine the structure of the 2:)- and L—enzymes, Milton and coworkers tested both forms for activity with D and L forms of a chira}. peptide substrate and for inhibition by D and a forms of a chiral peptide-analog inhibitor. Both forms were also tested for inhibition by the achiral in- hibitor Evans blue. The findings are given in the table. $-52 Chapter 4 The ThreenDimensionai Structure of Proteins renew-maime-s:mean-ms"enhancements“ -:.-u\-.,:4..~.:u.;w,,3:3.“ «Q' ~ - x . .."‘:vw..'33‘-“A'zm::..“i:2..:.m..¥qmrtm“raLiana:mlwmrfiiwwfimWr:lwfl-mm‘;fifl;fiww Substrate hydrolysis Evans i biue (achirai) i i HIV protease prrotease J o—protease M'SWL'A'EL‘L’M'" vxuu um um: amulzzhr'n‘dwawiad‘lnv'~"“=~L1434;:4'£.:-.L "—“*.'.1.4,:.¢a5':.".:::Lies-:S-uxuvtz'a. '72"W'EJHLMJ:KGfl'JwflJihfiflafl'wfififxfl'd—Vn «phi"Wm-qgwmsgwmwwfiérw (e) Which of the three models proposed above is supported by these data? Explain your reasoning. (f) Why does Evans blue inhibit both forms of the protease? (g) Would you expect chymotrypsin to digest the D-protease? Explain your reasoning. (h) Would you expect totai synthesis from Dmamino acids followed by renaturation to yield active enzyme for any enzyme? Explain your reasoning. Answer (a) Aha is a suitable replacement because Aloe and Cys have approximately the same sized side chain and are similarly hydrophobic. However, Aba cannot form disuifide bonds so it will not be a suitable replacement if these are required. (b) There are many important differences between the synthesized protein and HIV pro— . tease produced by a human cell, any of which could result in an inactive synthetic en» zyme: (1) Although Aba and Cys have simiiar size and hydrophobicity, Aba may not be similar enough for the protein to fold properly. {2) HIV protease may require disulfide bonds for proper functioning. (8) Many proteins synthesized by ribosomes told as they are produced; the protein in this study folded only after the chain was complete. (4) Proteins synthesized by ribosomes may interact with the ribosomes as they fold; this is not possible for the protein in the study. (5) Cytosol is a more complex solution than the buffer used in the study; some proteins may require specific, unknown proteins for study was synthesized as a single moiecuie. (c) Because the enzyme is functioned with Aba substituted for Cys, disulfide bonds do not play an important role in the structure of HIV protease. ((1) Model I : it would foid like the L-protease. Argument for: the covalent structure is the same (except for chirality), so it shouid foid like the L—protease. Argument against: chirath is not a trivial detail; three—dimensional shape is a key feature of biological molecules. The synthetic enzyme will not fold like the L—protease. Model 2: it would fold to the mirror image of the L—protease. For: because the individual components are mirror images of those in the bioiogical protein, it will foid in the mirror—image shape. Against: the interactions involved in protein folding are very compiex, so the synthetic nents are mirror images of those in the biological protein, it will fold in the mirror— irnage shape. (e) Model 1. The enzyme is active, but With the enantiomeric form of the biological sub strate, and it is inhibited by the enantiomeric form of the bioiogical inhibitor This is consistent with the D—protease being the mirror image of the t—protease. Chapter 4 The Three-Dimensional Structure of Proteins 5—53 (f) Evans blue is achiral; it binds to both forms of the enzyme. (g) No. Because proteases contain only namino acids and recognize only L—peptides, ehymotrypsm would not digest the o—protease. (1:) Not necessarily. Depending on the individual enzyme, any of the problems listed in (b) could result in an inactive enzyme. References Milton, R. 0., Milton, S. 0., 8: Kent, S. B. (E992) Total chemical synthesis of a ewenzyme: the enantiomers of HIV~1 protease Show demonstration of reciprocal chiral substrate specificity. Science 256, 14454448. WIodawer, A., Miller, M., Jaskélski, M., Sathyanarayana, B. K, Baldwin, E., Weber, I. it, Selk, L. M., Clawson, L., Schneider, 3., 8; Kent, S. B. (i989) Conserved folding in retroviral proteases: crystal structure of a. synthetic HIV~l protease. Science 245, 616M621. ...
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