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CHEM 331 Problem Set # 6 - Lehninger 5e, Chapter 7
Due Wednesday, November14, 201 2
ANSWER KEY (56 total pts.)
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 pts.) With reduction of the carbonyl oxygen to a hydroxyl group, the
stereochemistry at C-1 and C-3 is the same; the glycerol molecule is not
chiral.
5
2
) to form bright
yellow crystalline derivatives known as osazones:
The melting temperatures of these derivatives are easily determined and are characteristic
for each osazone. 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:
As the table shows, certain pairs of derivatives have the same melting points, although
the underivatized monosaccharides do not. Why do glucose and mannose, and similarly
galactose and talose, form osazone derivatives with the same melting points?
(3 pts.) The configuration at C-2 of an aldose is lost in its osazone derivative, so
aldoses dif- fering only at the C-2 configuration (C-2 epimers) give the same
derivative, with the same melting point. Glucose and mannose are C-2 epimers
and thus form the same osazone; the same is true for galactose and talose (see Fig.
cellulose and glycogen; (b) D-glucose and D-fructose; (c) maltose and sucrose.
(3 pts.)
a) Both are polymers of D-glucose, but they differ in the glycosidic
linkage: (β 1 à4) for cellulose, (α 1 à4) for glycogen.
b) Both are hexoses, but glucose is an aldohexose, fructose is a
ketohexose.
c) Both are disaccharides, but maltose has two (α 1 à4)-linked D-glucose
units; sucrose has (α 1 à 2 β)-linked D-glucose and D-fructose.
(Glc( 1 2)Fru). Explain why sucrose is not a reducing sugar, even though both
glucose and fructose are.
( 2 pts.)
a) A reducing sugar is one with a free carbonyl carbon that can be
oxidized by Cu
or Fe
b) The carbonyl carbon is C-‐1 of glucose and C-‐2 of fructose. When
the carbonyl carbon is involved in a glycosidic linkage, it is no longer
accessible to oxidizing agents. In sucrose (Glc(α 1 → 2)Fru), both
oxidizable carbons are involved in the glycosidic linkage.
part of this structure that makes the compound a reducing sugar.
(4 pts.)
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 1-‐
phosphate, available for oxidation and energy production.
relationships and connections to one another.
(3 pts.)
A typical proteoglycan consists of a core protein with covalently attached
glycosaminoglycan polysaccharides, such as chondroitin sulfate and keratin sulfate.
The polysaccharides generally attach to a serine residue in the protein via a
trisaccharide (gal–gal–xyl). (See Fig. 7-24, p. 253.)
(3 pts.)
Lectins are proteins that bind to specific oligosaccharides. They interact with specific
cell-surface glycoproteins thus mediating cell-cell recognition and adhesion. Several
microbial toxins and viral capsid proteins, which interact with cell surface receptors,
are lectins.
(4 pts.)
α-‐D-‐galactopyranosyl-‐(1à6)-‐α-‐D-‐glucopyranosyl β-‐D-‐fructofuranoside
(sequential method)
or
[α-‐D-‐Gal p -‐(1à6)-‐α-‐D-‐Glc p -‐(1à2)β-‐D-‐Fru f ]
fructofuranosyl-‐(2 à 1) – α-‐D-‐glucopyranoside. Draw its molecular formula. Is it
a reducing sugar?
(6 pts.)
This sugar, whose common name is melezitose, is NOT a reducing sugar because
all three anomeric carbons are involved in glycosidic bonds.
resulting mass of largely unoriented cellulose fibers into a sheet. Untreated
paper loses most of its strength when wet with water but maintains its strength
when wet with oil. Explain.
(2 pts.)
Water disrupts the intrafiber hydrogen bonds that hold the cellulose
fibers together; oil has no effect on hydrogen bonds.
every 25 residues. How many reducing ends does it have? How many non-‐
reducing ends?
(3 pts.)
There will only be one reducing end. The number of reducing ends will
depend on the degree of branching and the length of the branches. For a
1000 -‐residue polysaccharide, the maximum number of branch points
would be obtained assuming that each branch is only one residue long. If
there is a branch point every 25
th residue you would have 1,000/26 = 38
branch points (using 988 residues) and would have 12 residues left over,
so the main chain would be 962 residues long and there would be 38
branches. The number of non-‐reducing ends would be 38 +1 = 39.
equimolar quantities of 2,3,4,6-‐tetra-‐ O -‐methyl-‐D-‐galactose, 2,3,4-‐tri-‐ O -‐methyl-‐D-‐
mannose, and 2,4,6-‐tri-‐ O -‐methyl-‐D-‐glucose. Treatment of the trisaccharide with
β-‐galactosidase yields D-‐galactose and a disaccharide. Treatment of this
disaccharide with α-‐mannosidase yields D-‐mannose and D-‐glucose. Draw the
structure of the trisaccharide and state its systematic name.
(5 pts.)
D-‐Galactopyranosyl -‐β -‐(1à6)-‐D-‐mannopyranosyl-‐α-‐(1à3)-‐D-‐glucopyranose
Note that the reducing sugar (D-‐glucose, at the reducing end) is not methylated
on the anomeric carbon because although a methyl group was added to that
position in the exhaustive methylation, a methyl ESTER was produced (because
glucose is an aldose instead of an ether. It would be removed by hydrolysis.
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: oligopeptides composed of five
different amino acid residues or oligosaccharides composed of five different
monosaccharide residues? Explain.
(2 pts.)
Oligosaccharides; their monosaccharide 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 α or β.
Furthermore, the polymer can be linear or branched. Oligopeptides are
unbranched polymers, with all amino acid residues linked through
identical peptide bonds.