Chapter Review in Chapter 7 Carbohydrates and Glycobiology

Chapter Review

Problems

1. Sugar Alcohols In the monosaccharide derivatives known as sugar alcohols, the carbonyl oxygen is reduced to a hydroxyl group. For example, d-glyceraldehyde can be reduced to glycerol. Why can the sugar alcohol glycerol no longer be designated d or l?
2. Recognizing Epimers Using Figure 7-3, identify the epimers of (a) d-allose, (b) d-gulose, and (c) d-ribose at C-2, C-3, and C-4.
3. Configuration and Conformation Which bond(s) in α-d-glucose must be broken to change its configuration to β-d-glucose? Which bond(s) must be broken to convert d-glucose to d-mannose? Which bond(s) must be broken to convert one “chair” form of d-glucose to the other?
4. Sugar Structures Compare and contrast the structural features of each pair: (a) Cellulose and glycogen (b) d-Glucose and d-fructose (c) Maltose and sucrose
5. Haworth Structures Draw the Haworth perspective formulas for α-d-mannose and β-l-galactose.
6. Reducing Sugars Draw the structural formula for α-d-glucosyl- $(1 → 6)$ $left-parenthesis 1 right-arrow 6 right-parenthesis$ -d-mannosamine, and circle the part of this structure that makes the compound a reducing sugar.
7. Hemoglobin Glycation The measurement of glycated hemoglobin (the HbA1c level) to monitor the regulation of blood glucose (Box 7-2) is not reliable in individuals with certain conditions or diseases. Can you suggest an explanation for a lower-than-normal HbA1c level in a patient who definitely is not diabetic?
8. Hemiacetal and Glycosidic Linkages Explain the difference between a hemiacetal and a glycoside.
9. A Taste of Honey The sweetness of honey gradually decreases at a high temperature. Also, high-fructose corn syrup (a commercial product in which much of the glucose in corn syrup is converted to fructose) is used for sweetening cold drinks but not hot drinks. What chemical property of fructose could account for both of these observations?
10. Gluconolactone and Glucose Oxidation States The cyclic glucose derivative 6-phosphogluconolactone is an intermediate in the pentose phosphate pathway (discussed in Chapter 14). Compare the oxidation state of C-1 for the cyclic form of both gluconolactone and β-d-glucose.

6-phosphogluconolactone is a six-membered ring with O at the top right vertex between C 5 and C 1. C 1 is double bonded to O; C 2 and C 4 are bonded to H above the ring and O below the ring; C 3 is bonded to O H above the ring and H below the ring; C 5 is bonded to C 6 above the ring and H below the ring; and C 6 is bonded to 2 H and O P O subscript 3 end subscript superscript 2 minus. Greek letter beta D glucose is a six-membered ring with O at the top right vertex between C 5 and C 1. C 1 and C 3 are bonded to O H above the ring and H below the ring; C 2 and C 4 are bonded to H above the ring and O H below the ring; C 5 is bonded to C 6 above the ring and H below the ring; and C 6 is bonded to 2 H and O H.
11. Invertase “Inverts” Sucrose As sweet as sucrose is, an equimolar mixture of its constituent monosaccharides, d-glucose and d-fructose, is sweeter. Besides enhancing sweetness, fructose has hygroscopic properties that improve the texture of foods, reducing crystallization and increasing moisture.

In the food industry, hydrolyzed sucrose is called invert sugar, and the yeast enzyme that hydrolyzes it is called invertase. The hydrolysis reaction is generally monitored by measuring the specific rotation of the solution, which is positive $(+ 66.4 °)$ $left-parenthesis plus 66.4 degree right-parenthesis$ for sucrose but becomes negative (inverts) as more d-glucose $(specific rotation = + 52.7 °)$ $left-parenthesis specific rotation equals plus 52.7 degree right-parenthesis$ and d-fructose $(specific rotation = − 92 °)$ $left-parenthesis specific rotation equals negative 92 degree right-parenthesis$ form.

From what you know about the chemistry of the glycosidic bond, how would you hydrolyze sucrose to invert sugar nonenzymatically in a home kitchen?
12. Manufacture of Liquid-Filled Chocolates The manufacture of chocolates containing a liquid center is an interesting application of enzyme engineering. The flavored liquid center consists largely of an aqueous solution of sugars rich in fructose to provide sweetness. The technical dilemma 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 soluble than a mixture of glucose and fructose.)
13. Anomers of Sucrose? Lactose exists in two anomeric forms, but no anomeric forms of sucrose have been reported. Why?
14. Gentiobiose Gentiobiose ( $D -Glc (β 1 → 6) D -Glc$ $upper D hyphen Glc left-parenthesis beta Baseline 1 right-arrow 6 right-parenthesis upper D hyphen Glc$ ) is a disaccharide found in some plant glycosides. Draw the Haworth structure of gentiobiose based on its abbreviated name. Is it a reducing sugar? Does it undergo mutarotation?
15. Identifying Reducing Sugars Is N-acetyl-β-d-glucosamine (Fig. 7-9) a reducing sugar? What about d-gluconate? Is the disaccharide $GlcN (α 1 → 1 α) Glc$ $GlcN left-parenthesis alpha Baseline 1 right-arrow 1 alpha right-parenthesis Glc$ a reducing sugar?
16. Physical Properties of Cellulose and Glycogen The almost pure cellulose obtained from the seed threads of Gossypium (cotton) is tough, fibrous, and completely insoluble in water. In contrast, glycogen obtained from muscle or liver disperses readily in hot water to make a turbid solution. Despite their markedly different physical properties, both substances are $(1 → 4)$ $left-parenthesis 1 right-arrow 4 right-parenthesis$ -linked d-glucose polymers of comparable molecular weight. What structural features of these two polysaccharides underlie their different physical properties? Suggest possible biological advantages of their respective properties.
17. Dimensions of a Polysaccharide Compare the dimensions of a molecule of cellulose and a molecule of amylose, each of $M_{r} 200,000$ $upper M Subscript r Baseline 200,000$ .
18. Growth Rate of Bamboo The stems of bamboo, a tropical 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 ~0.5 nm to the length of a cellulose molecule.
19. Glycoproteins versus Proteoglycans Which characteristics describe glycoproteins and which describe proteoglycans?
1. Exclusively located at the cell surface and in the extracellular matrix
2. May contain N-linked glycosidic bonds
3. Found in Golgi complexes, secretory granules, and lysosomes
4. Include the heparan sulfate family
5. Sulfated glycosaminoglycan chains can only be covalently linked to a Ser residue
6. Form highly specific sites for recognition and high-affinity binding by lectins
20. Relative Stability of Two Conformers Explain why the two structures shown in Figure 7-16 are so different in energy (stability). Hint: See Figure 1-21.
21. 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 related 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?
22. Heparin Interactions Heparin, a highly negatively charged glycosaminoglycan, is used clinically as an anticoagulant. It acts by binding several plasma proteins, including antithrombin III, an inhibitor of blood clotting. The 1:1 binding of heparin to antithrombin III seems to cause a conformational change in the protein that greatly increases its ability to inhibit clotting. What amino acid residues of antithrombin III are likely to interact with heparin?
23. Permutations of a Trisaccharide Three different hexoses (A, B, and C) can be combined to form a large number of trisaccharides. What structural features of trisaccharides allow so many permutations and combinations?
24. 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 oligosaccharide chains, each ending in a single sialic 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. 3-18) and stained for protein. Identify any bands in your drawing.
25. Information Content of Oligosaccharides The carbohydrate portion of some glycoproteins may serve as a cellular recognition site. To perform this function, the oligosaccharide(s) must have the potential to exist in a large variety of forms. Which can produce a greater variety of structures: oligopeptides composed of five different amino acid residues, or oligosaccharides composed of five different monosaccharide residues? Explain.
26. Castor Bean Toxin The seed of the castor bean (Ricinus communis) contains large amounts of ricin, a poison that is deadly to animals, including humans. One of the two subunits of this toxin is a lectin that binds terminal N-acetylgalactosamine residues on glycoproteins on the surface of eukaryotic cells, allowing the other subunit to enter the cell and kill it by preventing proteins from being made. Suggest a possible antidote to prevent or reverse ricin-mediated entry of the toxin.
27. Determination of the Extent of Branching in Amylopectin A biochemist wants to determine the amount of branching in amylopectin, defined by the number of $(α 1 → 6)$ $left-parenthesis alpha Baseline 1 right-arrow 6 right-parenthesis$ glycosidic bonds present. First, she treats the sample with methyl iodide, a methylating agent that replaces the hydrogen of every sugar hydroxyl with a methyl group, converting —OH to $— {OCH}_{3}$ $em-dash OCH Subscript 3 Baseline$ . She then hydrolyzes all the glycosidic bonds in the treated sample in aqueous acid and measures the amount of 2,3-di-O-methylglucose formed.

The structure is a six-membered ring with O at the top right vertex between C 5 and C 1. C 1 is bonded to H above the ring by a wavy line to the right and to O H below the ring by a wavy line to the right; C 2 is bonded to H above the ring and O C H 3 below the ring; C 3 is bonded to O C H 3 above the ring and H below the ring; C 4 is bonded to H above the ring and O H below the ring; C 5 is bonded to C 6 above the ring and H below the ring; and C 6 is bonded to 2 H and O H.

In this representation of 2,3-di-O-methylglucose, the wavy bonds at C-1 indicate that the structure represents both anomers (α and β).
1. Explain the basis of this procedure for determining the number of $(α 1 → 6)$ $left-parenthesis alpha Baseline 1 right-arrow 6 right-parenthesis$ branch points in amylopectin. What happens to the unbranched glucose residues in amylopectin during the methylation and hydrolysis procedure?
2. A 258 mg sample of amylopectin that was treated by the methylation and hydrolysis procedure described yielded 12.4 mg of 2,3-di-O-methylglucose. Determine what percentage of the glucose residues in the amylopectin contained an (α1 → 6) branch. (Assume that the average molecular weight of a glucose residue in amylopectin is 162 g/mol and the molecular weight of 2,3-di-O-methylglucose is 208 g/mol.)

DATA ANALYSIS PROBLEM

28. Determining the Structure of ABO Blood Group Antigens The human ABO blood group system was discovered in 1901, and in 1924 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 molecules. When the paper was published, the complete structures of the A, B, and O antigens were not yet known; this paper is an example of what scientific knowledge looks like “in the making.”

In any attempt to determine the structure of an unknown biological compound, researchers must deal with two fundamental problems: (1) If you don’t know what it is, how do you know if it is pure? (2) If you don’t know what it is, how do you know that your extraction and purification conditions have not changed its structure? Morgan addressed problem 1 through several methods. One method is described in his paper as observing “constant analytical values after fractional solubility tests” (p. 312). In this case, “analytical values” are measurements of chemical composition, melting point, and so forth.

Based on your understanding of chemical techniques, what could Morgan mean by “fractional solubility tests”?
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 substance present in different samples.
Why was it important for Morgan’s studies, and especially for addressing problem 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)?
The structure of the blood group antigens is shown in Figure 10-13. In his paper, Morgan listed several properties of the three antigens, A, B, and O, that were known at that time (p. 314):
1. Type B antigen has a higher content of galactose than A or O.
2. Type A antigen contains more total amino sugars than B or O.
3. The glucosamine:galactosamine ratio for the A antigen is roughly 1:2; for B, it is roughly 2:5.
Which of these findings is (are) consistent with the known structures of the blood group antigens?

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 structural 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 could be inhibited by the addition of particular sugar molecules to the reaction. Only sugars found in the blood type antigens would cause this inhibition. One enzyme preparation, isolated from the protozoan Trichomonas foetus, would degrade all three antigens and was inhibited by the addition of particular sugars. The results of these studies are summarized in the table below, showing the percentage of substrate remaining unchanged when the T. foetus enzyme acted on the blood group antigens in the presence of sugars.

Sugar added	A antigen	B antigen	O antigen
	Unchanged substrate (%)
Control — no sugar	3	1	1
l-Fucose	3	1	100
d-Fucose	3	1	1
l-Galactose	3	1	3
d-Galactose	6	100	1
N-Acetylglucosamine	3	1	1
N-Acetylgalactosamine	100	6	1

For the O antigen, a comparison of the control and l-fucose results shows that l-fucose 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.

Although the O antigen contains galactose, N-acetyl-glucosamine, and N-acetylgalactosamine, none of these sugars inhibited the degradation of this antigen. Based on these data, is the enzyme preparation from T. foetus an endoglycosidase or an exoglycosidase? (Endoglycosidases cut bonds between interior residues; exoglycosidases remove one residue at a time from the end of a polymer.) Explain your reasoning.
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?
Which of the results in (f) and (g) are consistent with the structures shown in Figure 10-13? Explain your reasoning.

Reference

Morgan, W.T.J. 1960. The Croonian Lecture: a contribution to human biochemical genetics; the chemical basis of blood-group specificity. Proc. R. Soc. Lond. B Biol. Sci. 151:308–347.