Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

documents that you may need, Study notes of Biology

useful and details. You can also to take my documents for entertaining and studying

Typology: Study notes

2019/2020

Uploaded on 10/07/2022

mai-hua-2
mai-hua-2 🇻🇳

4 documents

1 / 7

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
CHEM 331 Problem Set #6- Lehninger 5e, Chapter 7
Due Wednesday, November14, 2012
ANSWER KEY (56 total pts.)
!
1. 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 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.
2. Many carbohydrates react with phenyl-hydrazine (C6H5NHNH2) 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.
7–3).
pf3
pf4
pf5

Partial preview of the text

Download documents that you may need and more Study notes Biology in PDF only on Docsity!

CHEM 331 Problem Set # 6 - Lehninger 5e, Chapter 7

Due Wednesday, November14, 201 2

ANSWER KEY (56 total pts.)

  1. 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 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.

  1. Many carbohydrates react with phenyl-hydrazine (C 6

H

5

NHNH

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.

  1. 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.

(3 pts.)

a) Both are polymers of D-glucose, but they differ in the glycosidic

linkage: (β 1 à4) for cellulose, (α 1 à4) 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 à4)-linked D-glucose

units; sucrose has (α 1 à 2 β)-linked D-glucose and D-fructose.

  1. Define "reducing sugar." Sucrose is a disaccharide composed of glucose and 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.

  1. Draw the structural formula for α-D-glucosyl-(1à6)-D-mannosamine and circle the

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.

  1. Sketch the principal components of a typical proteoglycan, showing their

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.)

  1. What are lectins? What are some biological processes which involve lectins?

(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.

  1. The trisaccharide drawn below is named raffinose. What is its systematic name?

(4 pts.)

α-­‐D-­‐galactopyranosyl-­‐(1à6)-­‐α-­‐D-­‐glucopyranosyl β-­‐D-­‐fructofuranoside

(sequential method)

or

[α-­‐D-­‐Gal p -­‐(1à6)-­‐α-­‐D-­‐Glc p -­‐(1à2)β-­‐D-­‐Fru f ]

  1. The systematic name of a sugar is O -­‐α-­‐D-­‐glucopyranosyl-­‐(1à3)-­‐ O -­‐β-­‐D-­‐

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.

  1. Most paper is made by removing the lignin from wood pulp and forming the

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.

  1. A molecule of amylopectin consists of 1000 glucose residues and is branched

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.

  1. Exhaustive methylation of a trisaccharide followed by acid hydrolysis yields

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 -­‐(1à6)-­‐D-­‐mannopyranosyl-­‐α-­‐(1à3)-­‐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.

  1. The carbohydrate portion of some glycoproteins may serve as a cellular

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.