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Unformatted text preview: it » Z¥M”fi_. “A, <4: Carbohydrates and Glycobiology w~w1r=vrrwr=q "a; 4 x" »- anemia A i g 1. Sugar Alcohols In the monosaccharide derivatives known as sugar aicohols, the carbonyl oxygen is reduced to a hydroxyi group. For exampie, Dwglyceraidehyde can be reduced to glycerol. However, this sugar alcohoi is no longer designated D or L. Why? Answer With reduction of the carbonyl oxygen to a hydroxyl group, the stereochemistry at 0—1 and 0-3 is the same; the glycerol molecuie is not chiral. 2. Recognizing Epimers Using Figure ?»«3, identify the epimers of (a) D-ailose, (b) D-guiese, and (c) D-ribose at 02, 0-3, and 0-4. Answer Epimers differ by the configuration about oniy one carbon. (a) D—aitroseCCw2),D-g1ucose (one), s-guiose (0-4) (h) cmidose (0-2), D-galactose (0-3), D—ailose (0-4) (c) ' D-arabinose ((3-2), D—xylose (O3) 3. Melting Points of Monosaccharide Osazcne Derivatives Many carbohydrates react with phenyiw hydrazine (05H5NHNI-i2) to form bright yeiiow eiystaliine defivatives known as osazones: H O \ / (i) H m ([3 =NNHCGH5 H~£l3~OH CGHSNHNfig CEJENNHCsfis onmcm—H 4—» OH—(Iz—H HWCIFM OH H-(“r— OH H W?— OH H — (IJ— OH CHZOH CHZOH Glucose Osazone defivative ‘ of giucose The melting temperatures of these derivatives are easily determined and are characteristic for each os- 5 _- atone. This information was used to help identify monosaccharides before the development of HPLC or ._ - gas—iiqxfid chromatography. Listed below are the reciting points (MP5) of some aldose-osazene derivatives: ‘tfifllerrz‘fifi23:73:41.:73'2?}:1W'qwmmmwm’i1m?231:.RLTJAEIFTyr-flfntmzfinTmflmwmammtmm’nT-“ETJZZE‘WK‘JX‘L? MP of asazone derivative (“Ci - 13mm mmwm, _; Glucose 205 Mannose 132 205 Gaiactose 165—168 201 Talose 328—130 201 j S-78 Chapter 7 Carbohydrates and Glycohiology 8-79 As the table shows, certain pairs of derivatives have the same melting points, although the underiva— tized monosaccharides do not. Why do glucose and mannese, and similarly galactose and talose, form osazone derivatives with the same melting points? Answer The configuration at (3-2 of an aldose is lost in its osazone derivative, so aldoses dif— fering only at the 0-2 configuration (Cl-2 epimers) give the same derivative, with the same melting point. Glucose and mannose are 0—2 epirners and thus form the same osazone; the same is true for galactose and talose (see Fig. 7—3). Interconversion of D—Glucose Forms A solution of one enantiomer of a given monosaccharide ro- tates plane-polarized light to the left (counterclockwise) and is called the levorotatory isomer, desig- nated (—); the other enantiomer rotates plane-polarized light to the same extent but to the right (clockwise) and is called the dextrorotatory isomer, designated (+). An equimolar mixture of the (+) and (M) forms does not rotate plane~polarized light. The optical activity of a stereoisomer is expressed quantitatively by its optical rotation, the number of degrees by which plane—polarized light is rotated on passage through a given path length of a solution of the compound at a given concentration. The specific rotation [05K of an optically active compound is defined thus: observed optical rotation {°) optical path length (din) K concentration (g/mL) The temperature (it) and the wavelength of the light (A) employed (usually, as here, the D line of sodium, 589 run) must be specified. A freshly prepared solution of a—n—glucose shows a specific rotation of +112? Over time, the rota— tion of the solution gradually decreases and reaches an equilibrium vaiue corresponding to [051%560 = +525”. in contrast, a freshly prepared solution of lGaimglucose has a specific rotation of +19°. The rota~ tion of this solution increases over time to the same eduilibriurn value as that shown by the a anomer. [all m (a) Draw the Haworth perspective formulas of the a and 8 forms of D-glucose. What feature distin- guishes the two forms? (b) Why does the specific rotation of a freshly prepared solution of the a form gradually decrease with time? Why do solutions of the or and [3 forms reach the same specific rotation at equilibrium? ((2) Calculate the percentage of each of the two forms of D—glucose present at equilibrium. Answer (a) CHZOH CHgOI-I O 0 E H H H H OH OH H OH H HO OH HO H H OH H OH a—D-Glucose ,8«D«Glucose The a and .8 forms of D—giucose differ only at the hemiacetal carbon (0—1; the anomeric carbon). A fresh solution of the or form of glucose undergoes mutarotation to an equilibrium mix- ture containing both the or and 6 forms. The same applies to a fresh solution of the ,6 form. ri‘he change in specific rotation of a solution in changing from 100% a rampart?" 112°) to 100% is form as]? 0 19°) is 93°. For an equilibrium mixture having [a112,5 0 525°, the fraction of D—glucose in the or form is 52.5°— 19° ~ 335° 112° — 19° ”“ (1)) (c) = 0.36 e 36% 93° 5-80 Chapter 7 Carbohydrates and Giycobioiogy Thus, ignoring the small portions of furanose forms {~0.5% each), the mixture contains about 36% awnmglucose and 64% .6~n«glucose. 5. Configuration and Conformation Which hondCs) in a—D—glucose must be broken to change its config- uration to [in-glucose? Which bond(s) to convert D-glucose to o—mannose? Which berths) to convert one “chair” form of D—glucose to the other? Answer To convert a—o~glucose to ,B—nglucose, the bond between 0-1 and the hydroxyl on 0-5 must be broken and reformed in the opposite configuration (as in Fig. 743). To convert D-glucose to D-mannose, either the -H or the «MOE on (3-2 must be broken and reformed in the opposite configuration. Conversion between chair conformations does not require bond breakage; this is the critical distinction between configuration and conformation. 6. Deoxysugars ls DwZ—deoxygalactose the same chemicai as n-Z-deoxyglucose? Explain. Answer No; glucose and galactose differ in their configuration at 6—4. 7. Sugar Structures Describe the common structural features and the differences for each pair: (a) cellulose and glycogen; (b) D~g1ucose and D—fructose; (c) maltose and sucrose. Answer (a) Both are polymers of D~g1ucose, but they differ in the glycosidic linkage: (files) for cellulose, Cal—>4) for glycogen. (b) Both are hexoses, but giucose is an aldohex- ose, fructose a ketohexose. (c) Both are disaccharides, but maltose has two (a1-;-4)—linked D—giucose units; sucrose has (alwzm-linked D-glucose and D—fructose. 8. Reducing Sugars Draw the structural formuia for wD-glucosylm{1w6)—n-mannosamine and circle the part of this structure that makes the compound a reducing sugar. Answer CHZOH H O H H HO OH H 0 L‘_—CH2 O H OH H H on OHH N reducing HO 2 H sugar H H 9. Hemiacetal and Glycosidic Linkages Explain the difference between a herniacetal and a glycoside. Answer A herniacetal is formed when an aidose or iretose condenses with an alcohol; a glyco- side is formed when a henuacetal condenses with an alcohol (see Fig. 7w5, p. 238). 10. A Taste of Honey ri‘he fructose in honey is mainly in the B«D~pyrarrose form. This is one of the sweetest carbohydrates known, about twice as sweet as glucose; the 5—D-furanose form of fructose is much less sweet. The sweetness of honey gradually decreases at a high temperature. Also, high fructose corn 33min (a commercial product in which much of the glucose in corn syrup is converted to fructose) is used for sweetening cold but not hot drinks. What chemical property of fructose could account for both these observations? Answer Straight-chain fructose can cyclize to yield either the pyranose or the furanose struc- ture. increasing the temperature shifts the equilibrium in the direction of the furanose form, reducing the sweetness of the solution. The higher the temperature, the less sweet is the fruc~ tose solution. " ' ‘.. .‘ 11. 13. Chapter 7 Carbohydrates and Glycobioiogy 5-81 Reducing Disaccharide A disaccharide, which you know to be either maltose or sucrose, is treated with Fehling’s solution, and a red color is fouued. Which sugar is it, and how do you know? Answer Maltese; sucrose has no reducing (oxidiza’ple) group, as the anomeric carbons of both monosaccharides are involved in the glycosidic bond. . Glucose Oxidase in Determination of Blood Glucose The enzyme glucose oxidase isolated from the mold Pantotlltum notation, catalyzes the oxidation of B-D—glucose to D—glucono—fi-lactone. This en zyme is highly specific for the i8 anomer of glucose and does not affect the or anomer. In spite of this specificity, the reaction catalyzed by glucose oxidase is commonly used in a clinical assay for total blood glucose—«that is, for soiutious consisting of a mixture of B— and awn-glucose. What are the cirw cumstances required to make this possible? Aside from allowing the detection of smaller quantities of glucose, What advantage does glucose oxidase offer over Fehling’s reagent for the determination of blood glucose? Answer The rate of mutarotation (interconversion of the or and {3 anomers) is sufficiently high that, as the enzyme consumes B-Dwgiucose, more a—D-glucose is converted to the ,8 form, and, eventually, ail the giucose is oxidized. Glucose oxidase is specific for glucose and does not dew tect other reducing sugars (such as galactose). Fehling’s reagent reacts with any reducuig sugar. Invertase “Inverts” Sucrose The hydroiysis of sucrose (specific rotation +66.5°) yields an equimoiar mixture of D-glucose (specific rotation +525“) and D-fructose (specific rotation ~92”). (See Problem 4 for details of specific rotation.) (at) Suggest a convenient way to determine the rate of hydrolysis of sucrose by an enzyme prepara- tion extracted from the iining of the small intestine. (3)) Explain Why, in the food industry, an equimoiar mixture of D-giucose and D-fructose formed by hydrolysis of sucrose is cailed invert sugar. (c) The enzyme invertase (now commonly called sucrase) is allowed to act on a 10% (iii g/mL) so- lution of sucrose until hydrolysis is complete. What will be the observed optical rotation of the solution in a 10 cm celi? (ignore a possible small contribution from the enzyme.) Answer (a) An equimolar mixture of D~glucose and D-fructose, such as that formed from sucrose hy— drolysis, has optical rotation = 525" + (“92.0% = —39.5°. Enzyme (sucrase) activity can be assayed by observing the change in opticai rotation of a solution of 100% sucrose (spe» cific rotation = +665") as it is converted to a 1:1 mixture of iii-glucose and D—fructose. The optical rotation of the hydrolysis mixture is negative (inverted) relative to that of the unhydrolyzed sucrose solution. The addition of 1 mol of water (Mr 18) in the hydrolysis of 1 mol of sucrose (Mr 842) gives the products an increase in weight of (18/842)(i00%) m 5.26% with respect to the starting sugar. Accordingly, a 10% sucrose solution yieids a [10 + (0.053 X 10}]% m 10.5% solution of invert sugar. Of this 10.5%, 5.25% (0.0525 g/rnL) is D-giucose and 5.25% is o~iructose By rearranging the equation in Problem 4, (b) (0) observed optical rotation (°) 25°C m MD optical path iength (aim) >< concentration (g/mL) we can determine the optical rotation of each sugar in the mixture in a 10 cm ceil: Opticai rotation of glucose = (525°) (1 dm) (0.0525 g/mL) = 2.76" Optical rotation of fructose = (92°)(1 din) (0.0525 g/mL) m M48" The observed optical rotation of the solution is 2.76“ + (W4.8°) = ~20" 3-82 Chapter 7 Carbohydrates and Giycobioiogy 14. Manufacture of Liqojd~Filled Chocolates The manufacture of chocolates containing a liquid center is an interesting application of enzyme engineering. The flavored liquid center consists largeiy of an aqueous solution of sugars rich in fructose to provide sweetness. The technical diiemma is the following: the chocolate coating must be prepared by pouring hot melted chocolate over a solid (or almost solid) core, yet the final product must have a liquid, fructose-rich center. Suggest a way to solve this problem. (Hint: Sucrose is much less solubie than a mixture of glucose and fructose.) Answer Prepare the core as a semisolid slurry of sucrose and water. Add a small amount of sucrose (inveitase), and quickly coat the semisolid mixture udth chocolate. After the chocolate coat has cooled and hardened, the sucrase hydrolyzes enough of the sucrose to form a more liquid center: a mixture of fructose, glucose, and sucrose. 15. Anomers of Sucrose? Lactose exists in two anoreeric forms, but no anomeiic forms of sucrose have been reported. Why? Answer Lactose (Gai(,81—>4)Glc) has a free anomeric carbon {on the glucose residue). in su~ crose (GlcCals—a2fifii‘ru), the anomeric carbons of both monosaccharide units are involved in the glycosidic bond, and the disaccharide has no free anoreeric carbon to undergo mutarotation. 16. Gentiobiose Gentiobiose (D~Glc(fll+6)D«Glc) is a disaccharide found in some piant glycosides. Draw the structure of gentobiose based on its abbreviated name. Is it a reducing sugar? Does it un- dergo mutarotation? Answer H omen o m H H 0 CH2 0 OH H no H H H OH OH H on H0 H H OK It is a reducing sugar; it undergoes mutarotation. 1?. Identifying Reducing Sugars is Nmacetyl-fi-D-glucosanune (Fig. 7—9) a reducing sugar? What about o-gluconate? is the disaccharide GlcN£al<->1a)Glc a reducing sugar? Answer NmAcetyi—B-D-giucosanune is a reducing sugar; its C—l can be oxidized (see Fig. mo, p. 241). b-Glucoriate is not a reducing sugar; its 0-1 is already at the oxidation state of a carboxylic acid. GleNCal<—>ia)Glc is not a reducing sugar; the anomeric carbons of both mono— saccharides are involved in the giycosidic bond. 18. Cellulose Digestion Ceiluiose could provide a wider available and cheap form of glucose, but humans cannot digest it. Why not? if you were offered a procedure that allowed you to acquire this ability, would you accept? Why or why not? Answer Humans cannot break down ceilulose to its monosaccharides because they lack cellu- lases, a family of enzymes, produced chiefly by fungi, bacteria, and protozoans, that catalyze the hydrolysis of cellulose to glucose. In rundnant ammals (such as cows and sheep), the ru- men (one of tour stomach compartments) acts as an anaerobic ferrnenter in which bacteria and protozoa degrade cellulose, making its glucose available as a nutrient to the annual. if celm lulase were present in. the human digestive tract, we could use foods rich in celiulose as outri- ents. This would greatly increase the forms of biomass that couid be used for human nutrition. This change might require some changes in the teeth that would allow cellulosic materials to be ground into small pieces to serve as cellulase substrates. Chapter? Carbohydrates and Giycobiology 5-83 19. Physical Properties of Cellulose and Glycogen The almost pure cellulose obtained from the seed threads of Gossyptum (cotton) is tough, fibrous, and completely insoluble in water. in contrast, glyco— gen obtained from muscle or liver disperses readily in not water to make a turbid solution. Despite their markedly different physical properties, both substances are (l->4}-linl<ed D—giucose polymers of comparable molecular weight. What structural features of these two polysaccharides underlie their different physical properties? Explain the biological advantages of their respective properties. Answer Native cellulose consists of glucose units linked by (filwzl) glycosidic bonds. The {3 linkages force the polymer chain into an extended conformation. Parallel series of these ex— tended chains can form intemleouldr hydrogen bonds, thus aggregating into long, tough, insoluble fibers. Glycogen consists of glucose units linked by (aimed) glycosidic bonds. The a 1in- ages cause bends in the chain, and glycogen forms helical structures with intramleculdr hydrogen bonding; it cannot form long fibers. in addition, glycogen is highly branched and, because many of its hydroxyl groups are exposed to water, is highly hydrated and therefore very water-soluble. It can be extracted as a dispersion in hot water. The physical properties of the two polymers are well suited to their biological roles. Cellu- lose serves as a structural material in plants, consistent with the side-by-side aggregation of long molecules into tough, insoluble fibers. Glycogen is a storage fuel in animals. The highly hydrated glycogen granules, with their abundance of free, nonreducing ends, can be rapidly hydrolyzed by glycogen phosphorylase to release glucose l-phosphate, available for oxidation and energy production. 20. Dimensions of a Polysaccharide Compare the dimensions of a molecule of cellulose and a molecule of amylose, each of M,r 200,000. Answer Cellulose is several times longer; it assumes an extended conformation, whereas amyiose has a helical structure. 21. Growth Rate of Bamboo The stems of bamboo, a tropicai grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. Given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth, calculate the number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate. Each D—glucose unit contributes ~05 nm to the length of a cellulose molecule. Answer First, calculate the growth per second: 0.3 m/day (24 h/day) (60 min/h} (60 s/rnin) Given that each glucose residue increases the length of the cellulose chain by 0.5 mm (5 X 10”“) m), the number of residues added per second is 3 X 10‘6 m/s 5 x 10—10 Ira/residue maxiO‘Bm/s m 6,000 residues/s 22. Glycogen as Energy Storage: How Long Can a Game Bird Fly? Since ancient times it has been observed that certain game birds, such as grouse, quail, and pheasants, are easily fatigued. The Greek historian Xenophon wrote, “The bustards . . . can be caught if one is quick in starting them up, for they will fly only a short distance, like partridges, and soon tire; and their flesh is delicious.” The flight muscles of game birds rely almost entirely on the use of glucose l~phosphate for energy, in the form of ATP (Chapter E4). The glucose luphosphate is formed by the breakdown of stored muscle glyco— gen, catalyzed by the enzyme glycogen phosphorylase. The rate of ATP production is limited by the rate at which glycogen can be broken down. During a “panic flight,” the game bird’s rate of glycogen breakdown is quite high, approadmately 3.20 cruel/min of glucose l—phosphate produced per gram of fresh tissue. Given that the flight muscles usually contain about 0.35% glycogen by weight, calculate how long a game bird can fly. (Assume the average molecular weight of a glucose residue in glycogen is 162 g/mol.) f 1 . 5-84 Chapter 7 Carbohydrates and Giycobiofogy Answer Given the average molecular weight of a glucose residue 2 162, the amount of us- able glucose (as glycogen) in l g of tissue is 3.5 X 10“3 g E62 g/rnol In 1 min, 120 pmol of glucose 1—phosphate is produced, so 120 amol of glucose is hydrolyzed. Thus, depletion of the glycogen would occur in (2.2 >< i0"5mol)(60 s/rrun) I} lZUXlOmBmoi/min "‘“ S 23. Relative Stabifity of Two Conformers Explain Why the two structures shown in Figure 7M1?) are so different in energy (stability). Hint: See Figure 1—21. 2 2.2 x 10"“5moi Answer The ball-and—stick model of the disaccharide in Figure 7—19 shows no steric interac- tions, but a space—filling model, showing atoms with their real relative sizes, would show several strong steric hindrances in the — 170°,— 170° conformer that are not present in the 30°,—4{}° conformer. 24. Volume of Chondroitin Sulfate in Solution One critical function of chondroitin sulfate is to act as a lubricant in skeletal joints by creating a gel-like medium that is resilient to friction and shock. This function seems to be reiated to a distinctive property of chondroitin sulfate: the volume occupied by the molecule is much greater in solution than in the dehydrated solid. Why is the volume so much larger in solution? Answer In soiution, the negative charges on chondroitin sulfate repel each other and force the molecule into an extended conformation. The polar molecule also attracts many water molecules {water of hydration), further increasing the molecular volume. In the dehydrated soiid, each negative charge is counterbalanced by a counterion, such as Na‘“, and the molecule collapses into its condensed form. . Heparin Interactions Heparin, a highly negatively charged glycosammoglycan, is used clinically as an anticoagulant. it acts by binding several plasma proteins, including antithrombin Ill, an inhibitor of blood clotting. The 1:1 binding of heparin to antithrombin ill seems to cause a conformational change in the protein that greatly increases its ability to inhibit clotting. What amino acid residues of anti» thrombin HI are likely to interact with heparin? Answer Positively charged amino acid residues would be the best candidates to bind to the highly negatively charged groups on heparin. In fact, Lys residues of antithrornbin ill interact with heparin. 26. Permutations of a Trisaecliaride Think about how one might estimate the number of possible trisac— charides composed of N—acetyiglucosamine 4-sulfate (GlcNAcaiS) and glucuronic acid (Gch), and draw 10 of them. Answer If GlcNActiS is represented as A, and Gch as B, the trimer could have any of these sequences: AAA, AAB, ABB, ABA, BBB, BBA, BAA, or BAR {8 possible sequences). The COMEC» tions between each pair of monosaccharides could be 1'66, l—wi, 1——>8, or l-el {4 possibilities for each of two bonds, or 4 X 4 m 16 possible linkages in all), and each linkage could involve either the a or the ,8 anomer of each sugar (2 possibilities for each of two bonds, so 2 X 2 = 4 stereochemical possibilities}. Therefore there are 8 X 16 X 4 m 512 possible pennutationsl' 27. Effect of Sialic Acid on SDS Polyacrylamide Gel Electrophoresis Suppose you have four forms of a protein, all with identical amino acid sequence but containing zero, one, two, or three oligosaccha— ride chains, each ending in a single siaiic acid residue. Draw the gel pattern you would expect when a mixture of these four glycoproteins is subjected to SDS polyacrylamide gel electrOphoresis (see Fig. 3&8) and stained for protein. identify any bands in your drawing. 28. 29. Chapter? Carbohydrates and Giycobioiogy 8-85 Answer I; Oligosacchsride 1_ chains: Electrophoresis The significant feature of siaiic acids is the negative charge of their carboxyl group. The four glycoproteins would have the same charge except for the additional 1, 2, or 3 negative charges of the siaiic acid residues. in SDS gel electrophoresis, the proteins are coated uniformly with a layer of sodium dodecyi sulfate (which is negativeiy charged, p. 89) and thus move toward the positive eiectrode. The giycoproteins with 1, 2, or 3 extra negative charges will move progres— sively faster than the form without sialic acid. Information Content of Oiigosaccharides The carbohydrate portion of some glycoproteins may serve as a cellular recognition site. In order to perform this function, the oligosaccharide moiety of glycoproteins must have the potential to exist in a large variety of forms. Which can produce a greater variety of structures: oiigopeptides composed of five different amino acid residues or oiigosaccharides composed of five different monosaccharide residues? Explain. Answer Oiigosaccharides; their monosacciiaride residues can be combined in more ways than the amino acid residues of oligopeptides. Each of the several hydroxyl groups of each mono- saccharide can participate in glycosidic bonds, and the configuration of each glycosidic bond can be either or or 5. Furthermore, the polymer can be iinear or branched. Oligopeptides are unbranched polymers, with all amino acid residues linked through identical peptide bonds. Determination of the Extent of Branching in Amyiopectin The amount of branching (number of (ml-+6) glycosidic bonds) in amylopectin can be determined by the following procedure. A sample of amylopectm is exhaustively methylated—treated with a methylating agent (methyl iodide) that replaces the hydrogen of every sugar hydroxyi with a methyi group, converting “OH to —OCH3. All the glycosidic bonds in the treated sample are then hydrolyzed in aqueous acid, and the amount of 2,8—di—O—methylglucose so formed is determined. CHgOH O H H H HO OCH3 H OH H OCR-5 2,3«DinO-methylglucose (a) Explain the basis of this procedure for deterrruning the number of (Oil—~96) branch points in amylopectin. What happens to the unbranched glucose residues in amylopectin during the methylation and hydrolysis procedure? (b) A 258 mg sampie of amylopectin treated as described above yielded 12.4 mg of 2,3-di~O—methyl~ glucose. Determine what percentage of the glucose residues in amyiopectin contain an (Oil—)6) branch. (Assume the average molecular weight of a giucose residue in amylopectin is 162 g/moi.) Answer (a) In glucose residues at branch points, the hydroxyi of 0-6 is protected from methylation because it is involved in a glycosidic finkage. During complete methylation and subse— quent hydrolysis, the branch-point residues yield 2,3—di—O-methyiglucose and the 1mm branched residues yield 2,3,6-tri—O-methylgiucose. 5-86 Chapter 7 Carbohydrates and Glycobiaiogy (b) Given the average moiecmar weight of a glucose residue = 162, then 258 mg of amy- lopectin contains 258 x 10*“3 g _ “3 W — L59 X 10 mol of glucose The i2.4 mg yield of 2,3~di~0~methylglucose (M, 208) is equivalent to 12.4 x 10"“3g _ #5 m «w 5.96 X 10 mol of glucose Thus, the percentage of glucose residues in amylopectin that yield 2,3-di—O-methylglucose is (5.96 x 10‘5mon(100%) a 1.59 x 10—3mo1 = 3-751) 30. Structured Analysis of a Polysaccharide A polysaccharide of unknown structure was isolated, sub— jected to exhaustive methylation, and hydrolyzed. Anaiysis of the products revealed three methylated sugars: 2,3,4»»tri»O«methyi—D-glucose, 2,4-di-O-methyl—D—glucose, and 2,3,4,6—tetra-O-methyl«Dwglucose, in the ratio 20:1:1. What is the structure of the poiysaccharide‘? Answer The polysaccharide is a branched glucose polymer. Because the predominant prod« not is 2,8,4—tri-O-methyl—D-glucose, the predominant glycosidic linkage must be (14-6). The formation of 2,4-di-O—methyl—D-glucose indicates that branch points occur through 0-3. The ratio of these two methylated sugars indicates that a branch ocours at an average frequency of once every 20 residues. The 2,3,4,6~tetra—O-methyl—oglucose is derived from nonreducing chain ends, which compose about "2%, or 5%, of the residues, consistent with a high degree of branching. Thus, the polysaccharide has chains of (imafiylinked Dwglucose residues with (1—>3)«1inl<ed branches, about one branch every 20 residues. - Data Analysis Problem 31. Determining the Structure of ABO Blood Group Antigens The human ABO blood group system was first discovered in 1901, and in i924 this trait was shown to be inherited at a single gene locus with three alleles. in 1960, W. T. J. Morgan published a paper summarizing what was known at that time about the structure of the ABO antigen moiecuies. When the paper was published, the complete structures of the A, B, and O antigens were not yet known; this paper is an exempts of what scientific knowiedge looks like “in the makmg." in any attempt to determine the structure of an unknown biologicai compound, researchers must‘deal with two fundamental problems: (I) if you don’t know what it is, how do you know if a is pure? (2) If you don’t know what it is, how do you know that your extraction and purification condi- tions have not changed its structure? Morgan addressed probiem 1 through several methods. One method is described in his paper as observing “constant analytical values after fractional solubility tests” (p. 812). in this case, “anaiyticai values” are measurements of chemical composition, melting point, and so forth. (a) Based on your understanding of chemical techniques, what could Morgan mean by "fractional solubility tests”? (1)) Why would the analytical values obtained from fractional solubility tests of a pure substance be constant, and those of an impure substance not be constant? Morgan addressed problem 2 by using an assay to measure the immunological activity of the sub- stance present in different samples. (c) Why was it important for Morgan’s studies, and especially for addressing probiem 2, that this activity assay be quantitative (measuring a level of activity) rather than simply qualitative ~ (measuring only the presence or absence of a substance)? Chapter 7 Carbohydrates and Glycobiology 3-37 The structure of the blood group antigens is shown in Figure 10—15. In his paper (p. 314), Morgan listed several properties of the three antigens, A, B, and O, that were known at that time: 1. Type B antigen has a higher content of galactose than A or O. 2. Type A antigen contains more totai amino sugars than B or O. 3. The glucosamine/galactosannne ratio for the A antigen is roughly 1.2; for B, it is roughly 2.5. ((1) Which of these findings is (are) consistent with the known structures of the blood group antigens? (e) How do you explain the discrepancies between Morgan’s data and the known structures? in later work, Morgan and his colleagues used a clever technique to obtain structurai information about the blood group antigens. Enzymes had been found that would specifically degrade the antigens. However, these were available only as crude enzyme preparations, perhaps containing more than one enzyme of unknown specificity. Degradation of the blood type antigens by these crude enzymes couid be inhibited by the addition of particular sugar moiecules to the reaction. Only sugars found in the blood type antigens wouid cause this inhibition. One enzyme preparation, isolated from the protozoan Mchomoncts foetus, would degrade ali three antigens and was inhibited by the addition of particular sugars. The results of these studies are summarized in the tabie beiow, showing the percentage of sub- strate remaining unchanged when the T foetus enzyme acted on the biooci group antigens in the pres— ence of sugars. kwnun—a-Mzw WM?”_«:¢;_®LP&LV3.:.Mt'tlfiemcs'mwmcuremmmr;acor-y:'W"Www&2u:l"rwfifi5'mfifi-m-1k‘5Ff-T—T—m‘“'"""‘V5335?M‘fiifi‘f-fim3-3‘Jfiwfikmfw“‘flwm”m"m“w:mXx'mlauY:L:::::m"“"“~:5::‘m"‘“#:3Wm—mflflm'i‘w . '2: Unchanged substrate (%) n h if Sugar added 0 antigen Controi—no sugar 1 L~Fucose 3,00 D-Fucose E L—Galactose 3 o—Galactose 1 N~Acetylgiocosamine 3 I N-Acetylgalactosamine For the 0 antigen, a comparison of the control and L—fucose results shows that Lufucose inhibits the degradation of the antigen. This is an example of product inhibition, in which an excess of reaction product shifts the equilibrium of the reaction, preventing further breakdown of substrate. (f) Although the 0 antigen contains gaiactose, N—acetylgiucosarnine, and N—acetylgalactosamine, none of these sugars inhibited the degradation of this antigen. Based on these data, is the en- zyme preparation from T. foetus an endo— or exogiycosidase‘? (Endoglycosidases cut bonds be tween interior residues; exoglycosidases remove one residue at a time from the enci of a poly— mer.) Explain your reasoning. - {g} Fucose is also present in the A and B antigens. Based on the structure of these antigens, why does fucose fail to prevent their degradation by the T foetus enzyme? What structure would be produced? (h) Which of the results in (f) and (g) are consistent with the structures shown in Figure 10—15? Explain your reasoning. ' Answer (a) The tests involve trying to dissoive only part of the sample in a variety of solvents, then an- alyzing both dissolved and undissolved materials to see whether their compositions differ. (b) For a pure substance, ali molecules are the same and any dissolved fraction will have the same composition as any undissolved fraction. An impure substance is a mixture of more than one compound. When treated with a particuiar solvent, more of one component may dissolve, leaving more of the other component(s) behind. As a result, the dissolved and undissoived fractions have different compositions. S~88 Chapter 7 Carbohydrates and Giycobiotogy (c) (d) (e) (f) (g) (10 Reference A quantitative assay allows researchers to be sure that none of the activity has been lost through degradation. When determining the structure of a molecule, it is important that the sample under analysis consist only of intact (undegraded) molecules. If the sample is contaminated with degraded material, this will give confusing and perhaps uninter— pretabie structural results. A qualitative assay would detect the presence of activity, even if it had become significantly degraded. Results 1 and 2. Result 1 is consistent with the known structure, because type B antigen has three molecules of galactose; types A and 0 each have only two. Result 2 is also con» sistent, because type A has two amino sugars (N-acetylgalactosamine and N—acetylglu- cosamine); types 8 and 0 have only one (Nmacetylglucosamine). Result 8 is not consistent with the known structure: for type A, the glucosaminegalactosamine ratio is 1:1; for type B, it is 1:0. The samples were probably impure and/or partly degraded. The first two results were correct possibly because the method was only roughly quantitative and thus not as sensi- tive to inaccuracies in measurement. The third result is more quantitative and thus more likely to differ from predicted values, because of impure or degraded samples. An exoglycosidase. if it were an endoglycosidase, one of the products of its action on 0 antigen would include galactose, N—acetylglucosamine, or N—acetylgalactosamine, and at least one of those sugars would be able to inhibit the degradation. Given that the enzyme is not inhibited by any of these sugars, it must be an exoglycosidase, removing only the terminal sugar from the chain. The terminal sugar of 0 antigen is fucose, so fucose is the only sugar that could inhibit the degradation of 0 antigen. The exoglycosidase removes N—acetylgalactosarnine from A antigen and galactose from B antigen. Because fucose is not a product of either reaction, it will not prevent removal of these sugars, and the resulting substances will no longer be active as A or B antigen. However, the products should be active as 0 antigen, because degradation stops at fucose. All the results are consistent with Figure i0—15. (i) 13-Fucose and L—galactose, which would protect against degradation, are not present in any of the antigens. (2) The termi« nal sugar of A antigen is N acetylgalactosamine, and this sugar alone protects this antigen from degradation. (3) The terminal sugar of B antigen is galactose, which is the only sugar capable of protecting this antigen. Morgan, WJI‘. (1960) The Croonian Lecture: a contribution to human biochemical genetics; the chemical basis of blood-group specifici- ty. Proc. R. Soc. Lond. BBz'ol. Sci. 151, 368—347. ...
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