S-78
Carbohydrates and
Glycobiology
chapter 7
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. However, this
sugar alcohol is no longer designated
D
or
L
. Why?
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.
Answer Epimers differ by the configuration about only one carbon.
3. Melting Points of Monosaccharide Osazone Derivatives Many carbohydrates react with phenyl-
hydrazine (C
6
H
5
NHNH
2
) to form bright yellow crystalline derivatives known as osazones:
OH
H
H
H
HO
OH
OH
OH
H
C
C
C
C
C
CH
2
OH
Glucose
OH
H
H
NNHC
6
H
5
OH
OH
H
CH
C
C
C
C
CH
2
OH
NNHC
6
H
5
Osazone derivative
of glucose
C
6
H
5
NHNH
2
The melting temperatures of these derivatives are easily determined and are characteristic for each os-
azone. This information was used to help identify monosaccharides before the development of HPLC or
gas-liquid chromatography. Listed below are the melting points (MPs) of some aldose-osazone derivatives:
Monosaccharide MP of anhydrous monosaccharide (ⴗC) MP of osazone derivative (ⴗC)
Glucose 146 205
Mannose 132 205
Galactose 165–168 201
Talose 128–130 201
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Chapter 7 Carbohydrates and Glycobiology S-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 mannose, and similarly galactose and talose, form
osazone derivatives with the same melting points?
4. Configuration and Conformation Which bond(s) in ␣–
D
-glucose must be broken to change its config-
uration to –
D
-glucose? Which bond(s) to convert
D
-glucose to
D
-mannose? Which bond(s) to convert one
“chair” form of
D
-glucose to the other?
5. Deoxysugars Is
D
-2-deoxygalactose the same chemical as
D
-2-deoxyglucose? Explain.
6. Sugar Structures Describe the common structural features and the differences for each pair:
(a) cellulose and glycogen; (b)
D
-glucose and
D
-fructose; (c) maltose and sucrose.
7. Reducing Sugars Draw the structural formula for ␣–
D
-glucosyl-(1n6)-
D
-mannosamine and circle the
part of this structure that makes the compound a reducing sugar.
Answer
H
H
8. Hemiacetal and Glycosidic Linkages Explain the difference between a hemiacetal and a glycoside.
9. A Taste of Honey The fructose in honey is mainly in the b–
D
-pyranose form. This is one of the
sweetest carbohydrates known, about twice as sweet as glucose; the b–
D
-furanose form of fructose is
much less sweet. 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 but not hot drinks. What chemical property of fructose could
account for both these observations?
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S-80 Chapter 7 Carbohydrates and Glycobiology
10. Glucose Oxidase in Determination of Blood Glucose The enzyme glucose oxidase isolated from
the mold Penicillium notatum catalyzes the oxidation of b–
D
-glucose to
D
-glucono-d-lactone. This en-
zyme is highly specific for the banomer of glucose and does not affect the aanomer. 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 solutions consisting of a mixture of b– and a–
D
-glucose. What are the circum-
stances required to make this possible? Aside from allowing the detection of smaller quantities of glu-
cose, what advantage does glucose oxidase offer over Fehling’s reagent for measuring blood glucose?
11. Invertase “Inverts” Sucrose The hydrolysis of sucrose (specific rotation ⫹66.5⬚) yields an
equimolar mixture of
D
-glucose (specific rotation ⫹52.5⬚) and
D
-fructose (specific rotation ⫺92⬚).
(See Problem 4 for details of specific rotation.)
(a) Suggest a convenient way to determine the rate of hydrolysis of sucrose by an enzyme prepara-
tion extracted from the lining of the small intestine.
(b) Explain why, in the food industry, an equimolar mixture of
D
-glucose and
D
-fructose formed by
hydrolysis of sucrose is called invert sugar.
(c) The enzyme invertase (now commonly called sucrase) is allowed to act on a 10% (0.1 g/mL) so-
lution of sucrose until hydrolysis is complete. What will be the observed optical rotation of the
solution in a 10 cm cell? (Ignore a possible small contribution from the enzyme.)
Answer
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.)
Chapter 7 Carbohydrates and Glycobiology S-81
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(1n6)
D
-Glc) is a disaccharide found in some plant glycosides.
Draw the structure of gentobiose based on its abbreviated name. Is it a reducing sugar? Does it un-
dergo mutarotation?
Answer
15. Identifying Reducing Sugars Is N-acetyl-–
D
-glucosamine (Fig. 7–9) a reducing sugar? What about
D
-gluconate? Is the disaccharide GlcN(␣1mn1␣)Glc a reducing sugar?
16. Cellulose Digestion Cellulose could provide a widely 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 cellulose to its monosaccharides because they lack cellu-
17. 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, glyco-
gen obtained from muscle or liver disperses readily in hot water to make a turbid solution. Despite
their markedly different physical properties, both substances are (1n4)-linked
D
-glucose 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.
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S-82 Chapter 7 Carbohydrates and Glycobiology
Answer Native cellulose consists of glucose units linked by (b1n4) glycosidic bonds. The
blinkages force the polymer chain into an extended conformation. Parallel series of these ex-
18. Dimensions of a Polysaccharide Compare the dimensions of a molecule of cellulose and a molecule
of amylose, each of M
r
200,000.
19. 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.
Answer First, calculate the growth per second:
20. 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 1-phosphate for energy, in the form
of ATP (Chapter 14). The glucose 1-phosphate 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, approximately 120 mmol/min of glucose 1-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.)
Answer Given the average molecular weight of a glucose residue ⫽162, the amount of us-
able glucose (as glycogen) in 1 g of tissue is
Chapter 7 Carbohydrates and Glycobiology S-83
21. Relative Stability of Two Conformers Explain why the two structures shown in Figure 7–18b are
so different in energy (stability). Hint: See Figure 1–22.
22. 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?
23. 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 anti-
thrombin III are likely to interact with heparin?
24. Permutations of a Trisaccharide Think about how one might estimate the number of possible trisac-
charides composed of N-acetylglucosamine 4-sulfate (GlcNAc4S) and glucuronic acid (GlcA), and draw
10 of them.
25. 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 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.
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S-84 Chapter 7 Carbohydrates and Glycobiology
Answer
sively faster than the form without sialic acid.
26. Information Content of Oligosaccharides The carbohydrate portion of some glycoproteins may
serve as a cellular recognition site. To perform this function, the oligosaccharide moiety 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.
27. Determination of the Extent of Branching in Amylopectin The amount of branching (number
of (␣1→6) glycosidic bonds) in amylopectin can be determined by the following procedure. A sample
of amylopectin is exhaustively methylated—treated with a methylating agent (methyl iodide) that
replaces the hydrogen of every sugar hydroxyl with a methyl group, converting OOH to OOCH
3
. All
the glycosidic bonds in the treated sample are then hydrolyzed in aqueous acid, and the amount of
2,3-di-O-methylglucose so formed is determined.
(a) Explain the basis of this procedure for determining the number of (a1n6) branch points in
amylopectin. What happens to the unbranched glucose residues in amylopectin during the
methylation and hydrolysis procedure?
(b) A 258 mg sample 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 amylopectin contained an (a1n6)
branch. (Assume the average molecular weight of a glucose residue in amylopectin is 162 g/mol.)
Answer
(a) In glucose residues at branch points, the hydroxyl of C-6 is protected from methylation
C
H
3
O
H
O
CH
2
CH
3
O
OH
H
H
OH
H
2,3-Di-O-methylglucose
H
OH
Chapter 7 Carbohydrates and Glycobiology S-85
28. Structural Analysis of a Polysaccharide A polysaccharide of unknown structure was isolated, sub-
jected to exhaustive methylation, and hydrolyzed. Analysis of the products revealed three methylated
sugars: 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,
in the ratio 20:1:1. What is the structure of the polysaccharide?
Answer The polysaccharide is a branched glucose polymer. Because the predominant prod-
Data Analysis Problem
29. Determining the Structure of ABO Blood Group Antigens The human ABO blood group system
was first 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 condi-
tions have not changed its structure? Morgan addressed problem 1 through several methods. One
method is described in his paper (p. 312) as observing “constant analytical values after fractional
solubility tests.” In this case, “analytical 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”?
(b) 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 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)?
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S-86 Chapter 7 Carbohydrates and Glycobiology
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 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.
(d) 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 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 sub-
strate remaining unchanged when the T. foetus enzyme acted on the blood group antigens in the pres-
ence of sugars.
Unchanged substrate (%)
Sugar added A antigen B antigen O antigen
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.
(f) Although the O antigen contains galactose, N-acetylglucosamine, 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 exoglycosidase? (Endoglycosidases cut bonds be-
tween interior residues; exoglycosidases remove one residue at a time from the end 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
Chapter 7 Carbohydrates and Glycobiology S-87
(c) A quantitative assay allows researchers to be sure that none of the activity has been lost
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