LEHNINGER PRINCIPLES OF BIOCHEMISTRY - VOLUME 1. THE FOUNDATIONS OF BIOCHEMISTRY: STRUCTURE AND CATALYSIS - 2011

PART I. STRUCTURE AND CATALYSIS

7. CARBOHYDRATES AND GLYCOBIOLOGY

Questions and Problems

1. Sugar Alcohols.

In sugar alcohols (Monosaccharide Derivatives), the carbonyl oxygen is reduced to form a hydroxyl group. For example, D-glyceraldehyde can be reduced to glycerol. However, the latter is not designated with D or L prefixes. Explain why.

2. Identification of Epimers.

Using Fig. 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. Melting Point of Osazone Monosaccharide Derivatives.

Many CARBOHYDRATES react with phenylhydrazine (С₆Н₅МНNН₂) to yield bright yellow crystalline derivatives known as osazones:

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The melting points of these compounds are easily determined and used to characterize each of them. This property was widely employed to identify Monosaccharides before the advent of High-Performance Liquid Chromatography (HPLC) and Gas-Liquid Chromatography. The melting points (m.p.) of several aldose osazone derivatives are listed below.

As seen from the table, certain pairs of derivatives share identical melting points, even though the parent monosaccharides differ in this parameter. Why do glucose and mannose, as well as galactose and talose, form osazones with the same melting points?

4. Interconversion of D-Glucose Forms.

A monosaccharide stereoisomer whose solution rotates the plane of polarized light to the left (counterclockwise) is termed levorotatory and designated by (-); the second stereoisomer, which rotates the plane of polarized light to the right (clockwise), is termed dextrorotatory and designated by (+). Equimolar mixtures of both forms do not rotate the plane of polarized light passing through them.

The optical activity of a stereoisomer is quantitatively expressed in terms of optical rotation, i.e., defined by the number of degrees by which the plane of polarized light is rotated after passing through a solution of a given concentration and path length. The specific rotation ([а]25D°C) of an optically active substance is given by the formula:

In this case, the solution Temperature and the wavelength of the incident light (typically the Na D-line, i.e., 589 nm) must be specified.

A freshly prepared solution of α-D-glucose has an optical rotation of +112°. Over time, the optical rotation gradually decreases to an equilibrium value of +52.5°. Conversely, a freshly prepared solution of β-D-glucose exhibits an optical rotation of +19°. The optical rotation of this solution increases over time until it reaches the same equilibrium value.

(a) Draw the Haworth projection formulas for the α and β forms of D-glucose. What is the fundamental difference between them?

(b) Explain why the optical rotation of a stored solution of the α form decreases. Why do solutions of the α and β forms of D-glucose at equal concentrations eventually reach the same optical rotation value?

(c) Calculate the percentage of both D-glucose forms present at equilibrium.

5. Configuration and Conformation.

Which bonds in the α-D-glucose molecule must be broken for the molecule to adopt the configuration of β-D-glucose? Which bonds must be broken to convert D-glucose into D-mannose? Which bonds must be broken for D-glucose to transition from one chair conformation to another?

6. Deoxysugars.

Does D-2-deoxygalactose share the same chemical properties as D-2-deoxyglucose? Explain your answer.

7. Structure of sugars.

List the shared structural features and differences for each pair of substances: (a) Cellulose and Glycogen; (b) D-glucose and D-fructose; (c) maltose and sucrose.

8. Reducing sugars.

Draw the structural formula of α-D-glucosyl-(1 —> 6)-D-mannosamine and identify the part of the structure that makes this substance a reducing sugar.

9. Hemiacetals and the glycosidic bond.

Explain the difference between a hemiacetal and a glycoside.

10. The taste of honey.

The fructose present in honey exists predominantly as β-D-pyranose. It is one of the sweetest known carbohydrates, with a sweetness twice that of glucose. Fructose in the β-D-furanose form is considerably less sweet. Upon heating honey, its sweetness gradually decreases. Furthermore, high-fructose corn syrup (a commercial product in which a large portion of glucose is converted to fructose) is used to sweeten cold rather than hot beverages. What chemical property of fructose underlies these observations?

11. Reducing Disaccharides.

A disaccharide, which is either maltose or sucrose, is treated with Fehling's solution, yielding a red color. Which disaccharide was present in the solution? Explain your answer.

12. Use of glucose oxidase for Blood glucose determination.

The enzyme glucose oxidase from the mold Penicillium notatum catalyzes The oxidation of β-D-glucose to D-glucono-δ-lactone. This enzyme specifically catalyzes the reaction of the β-anomer only and has no effect on the α-anomer. Despite this Specificity, the glucose oxidase reaction is routinely used in clinical practice to determine total blood glucose, which is a mixture of α- and β-glucose. Why is this possible? What advantages other than The ability to detect low glucose concentrations does the glucose oxidase method have over Fehling's method?

13. Invertase "inverts" sucrose.

The Hydrolysis of sucrose (specific optical rotation +66.5°) yields an equimolar mixture of D-glucose (specific rotation +52.5°) and D-fructose (specific rotation -92°). (For the determination of optical rotation, see Problem 4.)

a) Propose a METHOD FOR DETERMINING The rate of sucrose hydrolysis using an enzyme obtained from the Cells lining the Small Intestine.

b) Explain why an equimolar mixture of D-glucose and D-fructose is referred to in the food industry as invert sugar.

c) A 10% sucrose solution is treated with the enzyme invertase (now frequently called sucrase) until complete hydrolysis is achieved. What will be the optical Rotation of the solution in a 10 cm path-length Cell? (The minor contribution of the enzyme can be neglected.)

14. Production of liquid-filled chocolates.

The manufacture of chocolates with liquid centers can be viewed as an interesting Application of Enzymes in the food industry. The flavored liquid filling is primarily an aqueous sugar solution enriched with fructose for sweetness. A technical challenge arises: to form the chocolate shell, the solid (or nearly solid) center must be surrounded by hot molten chocolate, yet the finished product must contain a liquid interior beneath a solid shell. Propose a way to solve this problem. (Hint: sucrose is much less soluble than a mixture of glucose and fructose.)

15. Anomers of sucrose.

Lactose exists in two anomeric forms, yet no anomeric forms of sucrose have been detected. Why?

16. Gentiobiose.

Gentiobiose (D-Glc (β —> 6) D-Glc) is a disaccharide found in certain plant Glycosides. Draw The structure of this compound. Is this substance a reducing sugar? Does it undergo mutarotation?

17. Identification of reducing sugars.

Is N-acetyl-β-D-glucosamine (Fig. 7-9) a reducing sugar? What about D-gluconate? Is the disaccharide GlcNAc (α1 ⇄ α1) Glc a reducing sugar?

18. Cellulose degradation.

Cellulose could potentially serve as a widely available and cheap source of glucose, yet humans are unable to digest it. If you were offered a specific Procedure that would grant you this ability, would you accept it? Explain your answer.

19. Physical Properties of cellulose and glycogen.

Virtually pure cellulose, obtained from the fibers surrounding the seeds of plants of the genus Gossypium (cotton), is a tough, fibrous substance that is completely insoluble in Water. Unlike cellulose, glycogen isolated from Muscle or Liver readily disperses in hot water to form a turbid solution. Despite the difference in their physical properties, both substances are constructed from D-glucose residues joined by 1 -> 4 linkages and have approximately the same molecular weights. What Structural Features of these substances account for the differences in their physical properties? What is the Biological Significance of these characteristics in the two Polysaccharides?

20. Dimensions of polysaccharides.

Compare the geometric dimensions of cellulose and amylose molecules, each having a Molecular Weight of 200,000.

21. Growth rate of bamboo.

Under optimal conditions, stems of the tropical grass bamboo can grow phenomenally fast—up to 0.3 m/day. Assuming the stems consist almost entirely of cellulose with fibers oriented in the direction of growth, how many sugar residues per second are added to the growing cellulose chain during enzymatic reaction at this growth rate? The length of each D-glucose residue in a cellulose molecule is ~0.5 nm.

22. Glycogen as an energy store: how long can game birds fly?

Since ancient times, it has been known that game birds such as partridges, quails, and pheasants tire very quickly. The Greek historian Xenophon wrote: "The bustard can easily be caught if suddenly flushed, because, like the partridge, it tires quickly and cannot fly far; its meat is exquisitely delicious." The flight Muscles of game birds are supplied with energy (in the form of ATP) almost entirely through The breakdown of glucose-1-phosphate (see Chapter 14), which is formed during the degradation of muscle glycogen catalyzed by the enzyme Glycogen phosphorylase. The rate of ATP synthesis is limited by the rate of glycogen breakdown. In a state of panic, the rate of glycogen breakdown in birds is very high, approximately 120 µmol/min of glucose-1-phosphate per gram of fresh tissue. Assuming that flight muscles contain about 35% glycogen by weight, calculate how long a game bird can fly. (Assume the average molecular weight of a glucose residue in glycogen is 160 g/mol.)

23. Relative stability of two conformers.

Explain why the two structures shown in Fig. 7-19 differ so significantly in energy level (stability). Hint: see Fig. 1-21.

24. Volume occupied by chondroitin sulfate in solution.

One of the important Functions of chondroitin sulfate is joint lubrication, as this substance creates a gel-like medium that protects joints from friction and Shock. This function is directly related to a well-known property of chondroitin sulfate: the volume occupied by its molecule in solution is much greater than in the solid state. Why does the molecular volume of chondroitin sulfate increase so drastically in solution?

25. Reactions of heparin.

Heparin, a heavily negatively charged glycosaminoglycan, is used clinically as an anticoagulant. The action of heparin involves binding to certain Plasma Proteins, including antithrombin III, an inhibitor of Blood Coagulation. The binding of heparin to antithrombin III in a 1:1 ratio induces Conformational Changes in the protein that significantly enhance its inhibitory capacity. Which amino acid residues of antithrombin III are likely to participate in the interaction with heparin?

26. Trisaccharides.

How can one determine the number of trisaccharides that can be formed from N-acetylglucosamine-4-sulfate (GlcNAc4S) and glucuronic acid (GlcA)? Illustrate 10 Examples.

27. Effect of sialic acid on the electrophoretic mobility of Glycoproteins.

Suppose you have four forms of a protein with the identical Amino Acid Sequence but differing in the number (zero, one, two, and three) of oligosaccharide chains, each terminating in a single sialic acid residue. Sketch the results of separating a mixture of these four glycoproteins by Polyacrylamide gel Electrophoresis in the presence of SDS (see Fig. 3-18). Indicate which glycoprotein corresponds to each band.

28. Information content of Oligosaccharides.

The carbohydrate moiety of certain glycoproteins can serve as a recognition site for The Cell. To fulfill this function, glycoprotein oligosaccharides must be capable of existing in A wide variety of forms. Which chain can exist in a greater number of variants: an oligopeptide consisting of five different amino acid residues, or an oligosaccharide built from five different monosaccharide units? Explain your answer.

29. Determining the degree of branching of amylopectin.

The degree of branching (the number of α1 —> 6 glycosidic bonds) in amylopectin can be determined using the following procedure. A sample of amylopectin is subjected to complete methylation, meaning it is treated with a methylating agent (methyl iodide) that converts all hydrogen atoms in the sugar hydroxyl groups into methyl groups (-ОН —> -ОСН₃). Subsequently, all glycosidic bonds are subjected to acid hydrolysis, and The amount of 2,3-di-O-methylglucose is quantified.

a) Explain the rationale behind this procedure for determining the number of α1 —> 6 bonds in the amylopectin molecule. What happens to the glucose residues in the unbranched regions of amylopectin during methylation and hydrolysis?

b) As a result of such Treatment, 12.4 mg of 2,3-di-O-methylglucose was obtained from 258 g of amylopectin. Determine the percentage of glucose residues in amylopectin involved in forming α1 —> 6 bonds. (Assume the average molecular weight of a glucose residue in amylopectin is 162 g/mol.)

30. Structural Analysis of a Polysaccharide.

A polysaccharide of unknown structure was isolated, subjected to complete methylation, and hydrolyzed. Product Analysis revealed the presence of three methylated sugars in a 20:1:1 ratio. These sugars were 2,3,4-tri-O-methyl-D-glucose, 2,4-di-O-methyl-D-glucose, and 2,3,4,6-tetra-O-methyl-D-glucose. Determine the STRUCTURE OF THE original polysaccharide.

Analysis of Experimental Data

31. Determining the Structure of ABO Blood Group Antigens.

The human ABO blood group system was first described in 1901, and in 1924 it was demonstrated that this trait is inherited as a single Gene locus with three alleles. In 1960, W. T. J. Morgan published a paper summarizing everything known at the time about the ABO antigens. At the time of publication, the complete structure of the A, B, and O antigens was not yet known; this article serves as a case study illustrating the accumulation of scientific knowledge.

When attempting to determine the structure of an unknown biological material, any scientist faces two major challenges. First, if you do not know what the substance is, how can you be sure it is pure? Second, if you do not know what the substance is, how can you be sure that the Isolation and Purification process will not alter its structure? To address the first question, Morgan employed several Methods. One method, described in his paper, relies on "constancy of analytical values upon fractional solubility testing." In this context, "analytical values" refer to measurements of chemical composition, melting point, etc.

a) Based on your knowledge of chemical analysis methods, what did Morgan mean by "fractional solubility testing"?

b) Why are the analytical values obtained through such testing constant for a pure substance, but not for an impure one?

To address the second question, Morgan measured the immunological activity of substances present in various samples.

c) Why was it important for Morgan's research—and particularly for answering the second question—that the method for determining activity be quantitative (measuring the level of activity) rather than qualitative (determining the presence or absence of activity)?

The Structure of Blood group antigens is shown in Figures 10–15. In his paper, Morgan listed several Properties of the A, B, and O antigens that were known at the time:

1. Antigen B contains more galactose than antigens A and O.

2. Antigen A contains more amino sugars than antigens B and O.

3. The glucosamine-to-galactosamine ratio is 1.2 for antigen A and 2.5 for antigen B.

d) Which of Morgan's Conclusions are consistent with current data on the structure of the A, B, and O antigens?

e) How would you explain the discrepancies between Morgan's assumptions and modern data?

In a later study, Morgan and his coworkers used an ingenious method to gain insight into the structure of blood group antigens. Enzymes capable of specifically degrading the antigens were discovered. However, these enzymes existed only as crude preparations, potentially contaminated with other enzymatic activities. The degradation of blood group antigens by these enzyme preparations could be prevented by adding specific sugars to the reaction mixture, specifically those sugars found within the antigen structure. One enzyme preparation, isolated from the protozoan Trichomonas foetus, degraded all three antigens, and its action was inhibited by The addition of specific sugars. The results of these experiments are summarized in the table below, which shows the amount of blood group antigens remaining intact after treatment with the T. foetus enzyme in the presence of various sugars.

In the case of the O antigen, a comparison between the control and the experiment with added L-fucose shows that L-fucose inhibited antigen degradation. This is an example of end-product inhibition, where an excess of a reaction product shifts the reaction equilibrium and prevents further substrate breakdown.

f) Although the O antigen contains galactose, N-acetylglucosamine, and N-acetylgalactosamine, none of these sugars inhibited O antigen degradation. Based on these data, determine whether the enzyme preparation from T. foetus was an endoglycosidase or an exoglycosidase? (Endoglycosidases cleave bonds between internal residues in a chain, whereas exoglycosidases remove single residues sequentially from the chain end.) Explain your reasoning.

g) Fucose is also present in the A and B antigens. Based on the structures of these antigens, explain why fucose did not prevent their Cleavage by the enzyme preparation.

3) Which of the results given in items (e) and (g) correspond to the structures shown in Fig. 10-15? Explain your reasoning.