Chap. 7. Problem 2.
Epimers are stereoisomers that differ in the configuration about
only one carbon. The epimers of these sugars at carbons 2, 3,
and 4 therefore are:
(a) D-allose: D-altrose (C-2); D-glucose (C-3); D-gulose (C-4).
(b) D-gulose: D-idose (C-2); D-galactose (C-3); D-allose (C-4).
(c) D-ribose: D-arabinose (C-2); D-xylose (C-3).
Chap. 7. Problem 6.
(a) Cellulose and glycogen: Both of these compounds are
homopolysaccharides of D-glucose. Cellulose is a linear
polymer, whereas glycogen is a branched polymer. Oglycosidic linkages in cellulose are exclusively (ß14). Oglycosidic linkages in glycogen are (14) in the main chains
and (16) at branch points.
(b) D-glucose and D-fructose: Both of these monosaccharides are
hexoses. D-fructose is a ketose, and D-glucose is an aldose.
(c) Maltose and sucrose: Both of these sugars are disaccharides.
Maltose contains two (14) linked D-glucose units. Sucrose
contains (1 2ß) linked D-glucose and D-fructose units.
Maltose is a reducing sugar; sucrose is not.
Chap. 7. Problem 9.
Straight-chain fructose can cyclize to either the pyranose or
furanose forms. The observations in the problem can be explained
if heating converts more of the fructose to its furanose form,
which is less sweet than the pyranose form.
Chap. 7. Problem 10.
Although glucose oxidase is specific for the ß anomer of Dglucopyranose, the enzyme can ultimately oxidize all of the
glucose in solution because the ß and  anomers are in equilibrium
via mutarotation. Glucose oxidase is more accurate than Fehling’s
reagent for measuring glucose in the blood, because the enzyme
is specific for glucose and does not detect other reducing sugars
(e.g., galactose) that react with Fehling’s reagent.
Chap. 7. Problem 13.
Lactose (Gal(ß14)Glc) exists in two anomeric forms because the
free anomeric carbon (C-1) in the glucose residue can undergo
mutarotation. In sucrose (Glc(1 2ß)Fru), the anomeric carbons
of both monosaccharides are linked via an O-glycosidic bond.
Thus, sucrose lacks a free anomeric carbon that can intercovert
between  and ß forms via mutarotation.
Chap. 7. Problem 15.
N-acetyl-ß-D-glucosamine is a reducing sugar because it contains
a free anomeric carbon at C-1 that can open to the straightchain form and therefore can be oxidized. D-gluconate is not a
reducing sugar because its anomeric carbon at C-1 is already
oxidized to the level of a carboxylic acid. The disaccharide
GlcN(1 1)Glc is not a reducing sugar because it lacks a free
anomeric carbon. The anomeric carbons of both glucose units in
this compound are tied up in an O-glycosidic linkage and cannot
open to the straight-chain forms required for oxidation.
Chap. 7. Problem 17.
In glycogen, the (14)
linkages in the main chains
produce bends in the chains and
limit the formation of long
fibers. Branching also favors
the formation of a globular,
granular structure. Many of the
hydroxyl groups of glucose units
in the polymer are exposed to
water and are hydrated, which explains why glycogen can be
dispersed in hot water to make a turbid solution. In cellulose,
glucose units are linked via (ß14) linkages. This allows the
polymer to adopt an extended conformation in which parallel chains
are held together via numerous interchain hydrogen bonds. Water
is mostly excluded from cellulose which forms insoluble, tough
fibers. Cellulose therefore is well suited to take on a structural,
supportive role in plants. Glycogen, due to branching and hydration,
is well suited to serve as an energy repository from which glucose
units can readily be liberated by enzymatic cleavage.
Chap. 7. Problem 22.
Chondroitan sulfate contains a large number of negatively charged
carboxylate and sulfate functional groups. In solution, these
negative charges repel one another and force the molecule into an
extended conformation. Chondroitan sulfate also is extensively
hydrated due to the prevalence of polar and charged groups, and
this increases the volume occupied by this molecule in solution.
The dehydrated solid form of chondroitan sulfate is produced by
removal of water molecules and addition of positively charged
counterions such as sodium ion which masks the negative charges
of the polymer. In this form, the volume of the molecule is
greatly reduced from that observed in solution.
Chap. 7. Problem 26.
Oligosaccharides composed of five different monosaccharide
residues actually can produce a greater variety of structures than
oligopeptides composed of five different amino acid residues.
Oligopeptides are unbranched polymers in which every amino acid is
linked via a simple peptide bond. In oligosaccharides, O-glycosidic
linkages can be formed using several different hydroxyl groups in
the monomer units, and each glycosidic bond can be either  or ß.
In addition, branched structures are possible. Overall,
monosaccharide units in oligosaccharides can be combined in more
ways than the amino acids of an oligopeptide.