Chapter 4 Carbohydrate Classification

Question 1

How are Carbohydrates Classified?

Answer 1

Carbohydrates are organized by their number of carbon atoms and functional groups.
Common names are also frequently used when referring to sugars,
such as glucose, fructose, and galactose.

Three-carbon sugars are trioses, four-carbon sugars are tetroses, and so on.

Sugars with aldehydes as their most oxidized group are aldoses;

sugars with ketones as their most oxidized group are ketoses.

Question 2

Nomenclature of Carbohydrate Enantomers

Answer 2

The nomenclature of all sugars is based on the D- and L-forms of glyceraldehyde.
Sugars with the highest-numbered chiral carbon with
the −OH group on the right (in a Fischer projection) are D-sugars;

Those with the −OH on the left are L-sugars. D- and L-forms of the same sugar are enantiomers.

Question 3

What are Diasteriomers?

Answer 3

Diastereomers are nonsuperimposable configurations of molecules
with similar connectivity. ey differ at at least one—but not all—chiral carbons. These also include epimers and anomers.

Question 4

What is an Epimer?

Answer 4

Epimers are a subtype of diastereomers that differ at exactly one chiral carbon.

Question 5

What is an Anomers?

Answer 5

Anomers are a subtype of epimers that differ at the anomeric carbon

Question 6

What is Steroisomers?

Answer 6

Optical isomers, also called stereoisomers, are compounds that have the
same chemical formula; these molecules differ from one another only in
terms of the spatial arrangement of their component atoms. A special type of isomerism exists between stereoisomers that are nonidentical,
nonsuperimposable mirror images of each other.

Question 7

What is cyclicization?

Answer 7

Cyclization describes the ring formation of carbohydrates from their straight-chain forms.

When rings form, the anomeric carbon can take on either an α- or βconformation

Question 8

What are anomeric carbons, alpha and beta anomers>

Answer 8

The anomeric carbon is the new chiral center formed in ring closure; it was the carbon containing the carbonyl in the straightchain form.

α-anomers have the −OH on the anomeric carbon trans to the free − CH2OH group.

β-anomers have the −OH on the anomeric carbon cis to the free − CH2OH group

Question 9

What are Haworth projection?

Answer 9

Haworth projections provide a good way to represent three dimensional structure.

Question 10

What is mutarotation?

Answer 10

Cyclic compounds can undergo mutarotation, in which they shift
from one anomeric form to another with the straight-chain form as an
intermediate.

Question 11

What is Monosaccharides?

Answer 11

Monosaccharides are single carbohydrate units, with glucose as the most commonly observed monomer. They can undergo three main reactions: oxidation–reduction, esterification, and glycoside
formation.

Question 12

Aldoses get Oxidized to ____ and Reduced to ______

Answer 12

Aldoses can be oxidized to aldonic acids and reduced to alditols.

Question 13

Sugars:

Answer 13

Sugars that can be oxidized are reducing agents (reducing sugars)
themselves, and can be detected by reacting with Tollens’ or
Benedict’s reagents.

Sugars with a −H replacing an −OH group are termed deoxy
sugars.

Question 14

Talk about esterification and Phosphorylation of sugars:

Answer 14

Sugars can react with carboxylic acids and their derivatives,
forming esters (esterification). Phosphorylation is a similar
reaction in which a phosphate ester is formed by transferring a
phosphate group from ATP onto a sugar.

Question 15

what is Glycoside formation?

Answer 15

Glycoside formation is the basis for building complex
carbohydrates and requires the anomeric carbon to link to another
sugar

Question 16

What are Disaccharides?

Answer 16

Disaccharides form as result of glycosidic bonding between two monosaccharide subunits; polysaccharides form by repeated. monosaccharide or polysaccharide glycosidic bonding.

Common disaccharides include sucrose (glucose-α-1,2-fructose),
lactose (galactose-β-1,4-glucose), and maltose (glucose-α-1,4-
glucose).

Question 17

types of polysacchrides:

Answer 17

Cellulose, Starches, and Glycogen

Question 18

What is cellulose?

Answer 18

Cellulose is the main structural component for plant cell walls and
is a main source of fiber in the human diet.

Question 19

What are starches (amylose and amylopectin)

Answer 19

Starches (amylose and amylopectin) function as a main energy
storage form for plants

Question 20

What is Glyogen function?

Answer 20

Glycogen functions as a main energy storage form for animals.

Question 21

What is the name for a five-carbon sugar with an aldehyde group?

A six-carbon sugar with a ketone group?

Answer 21

Aldopentose;
ketohexose

Question 22

Explain the relationship between the carbonyl carbon, anomeric carbon, and the
alpha and beta forms of a sugar molecule.

Answer 22

During hemiacetal or hemiketal formation, the carbonyl carbon
becomes chiral and is termed the anomeric carbon. e orientation of
the −OH substituent on this carbon determines if the sugar molecule is
the α- or β-anomer.

Question 23

Explain the difference between esterification and glycoside formation

Answer 23

Esterification is the reaction by which a hydroxyl group reacts with
either a carboxylic acid or a carboxylic acid derivative to form an ester.
Glycoside formation refers to the reaction between an alcohol and a
hemiacetal (or hemiketal) group on a sugar to yield an alkoxy group.

Question 24

What purpose do Tollens’ reagent and Benedict’s reagent serve? How do they differ from each other?

Answer 24

Tollens’ reagent and Benedict’s reagent are used to detect the presence
of reducing sugars. Tollens’ reagent is reduced to produce a silvery mirror when aldehydes are present whereas Benedict’s reagent is
indicated by a reddish precipitate of Cu2O.

Question 25

From a metabolic standpoint, does it make sense for carbohydrates to get oxidized or reduced? What is the purpose of this process?

Answer 25

It makes sense for carbohydrates to become oxidized while reducing
other groups. is is the case because aerobic metabolism requires reduced forms of electron carriers to facilitate processes such as
oxidative phosphorylation. Because carbohydrates are a primary energy source, they are oxidized.

Question 26

Which of the two forms of starch is more soluble in solution. Why?

Answer 26

Amylopectin is more soluble in solution than amylose because of its branched structure. The highly branched structure of amylopectin decreases intermolecular bonding between polysaccharide polymers and increases interaction with the surrounding solution

Question 27

Regarding glycogen and amylopectin, which of these two polymers should
experience a higher rate of enzyme activity from enzymes that cleave side
branches? Why

Answer 27

Glycogen has a higher rate of enzymatic branch cleavage because it
contains significantly more branching than amylopectin

Question 28

When glucose is in a straight-chain formation, it:
A. is an aldoketose.
B. is a pentose.
C. has five chiral carbons.
D. is one of a group of 16 stereoisomers.

Answer 28

D
Glucose is an aldohexose, meaning that it has one aldehyde group and six
carbons. Given this information, (A) and (B) can be eliminated. In aldose
sugars, each nonterminal carbon is chiral. erefore, glucose has four chiral
centers, not five, as mentioned in (C). e number of stereoisomers possible
for a chiral molecule is 2
n
, where n is the number of chiral carbons. Because
glucose has four chiral centers, there are 2
4 = 16 possible stereoisomers.

Question 29

All of the following are true of epimers EXCEPT:
A. they differ in configuration about only one carbon.
B. they usually have slightly different chemical and physical
properties.
C. they are diastereomers (with the exception of glyceraldehyde).
D. they have equal but opposite optical activities.

Answer 29

D
Epimers are monosaccharide diastereomers that differ in their configuration
about only one carbon. As with all diastereomers, epimers have different
chemical and physical properties, and their optical activities have no relation
to each other. Enantiomers have equal but opposite optical activities.
erefore, (D) is the only statement that does not apply to epimers.

Question 30

Aldonic acids are compounds that:
A. can be oxidized, and therefore act as reducing agents.
B. can be reduced, and therefore act as reducing agents.
C. have been oxidized, and have acted as reducing agents.
D. have been oxidized, and have acted as oxidizing agents.

Answer 30

C
Aldonic acids form aer the aldehyde group on a reducing sugar reduces
another compound, becoming oxidized in the process.

Question 31

The formation of α-D-glucopyranose from β -D-glucopyranose is
called:
A. glycosidation.
B. mutarotation.
C. enantiomerization.
D. racemization.

Answer 31

B
Mutarotation is the interconversion between anomers of a compound.
Enantiomerization and racemization, (C) and (D), are related:
enantiomerization is the formation of a mirror-image or optically inverted
form of a compound, whereas racemization is moving a solution toward an
equal concentration of both enantiomers. Glycosidation, (A), is the addition
of a sugar to another compound.

Question 32

Ketose sugars may have the ability to act as reducing sugars.

Which process explains this?
A. Ketose sugars undergo tautomerization.
B. The ketone group is oxidized directly.
C. Ketose sugars undergo anomerization.
D. The ketone group is reduced directly.

Answer 32

A
Ketose sugars undergo tautomerization, a rearrangement of bonds, to
undergo keto–enol shis. is forms an aldose, which then allows them to
act as reducing sugars. A ketone group alone cannot be oxidized.
Anomerization, mentioned in (C), refers to ring closure of a
monosaccharide, creating an anomeric carbon.

Question 33

Which of the following enzymes cleaves polysaccharide chains
and yields maltose exclusively?
A. α-Amylase
B. β-Amylase
C. Debranching enzyme
D. Glycogen phosphorylase

Answer 33

B
β-Amylase cleaves amylose at the nonreducing end of the polymer to yield
maltose exclusively, while α-amylase, (A), cleaves amylose anywhere along
the chain to yield short polysaccharides, maltose, and glucose. Debranching
enzyme, (C), removes oligosaccharides from a branch in glycogen or
starches, while glycogen phosphorylase, (D), yields glucose 1-phosphate.

Question 34

Why is the α-anomer of D-glucose less likely to form than the β -
anomer?
A. The β-anomer is preferred for metabolism.
B. The β-anomer undergoes less electron repulsion.
C. The α-anomer is the more stable anomer.
D. The α-anomer forms more in L-glucose

Answer 34

B
The hydroxyl group on the anomeric carbon of the β-anomer is equatorial,
thereby creating less nonbonded strain than the α-anomer, which has the
hydroxyl group of the anomeric carbon in axial position.

Question 35

Which two polysaccharides share all of their glycosidic linkage
types in common?
A. Cellulose and amylopectin
B. Amylose and glycogen
C. Amylose and cellulose
D. Glycogen and amylopectin

Answer 35

D
Glycogen and amylopectin are the only polysaccharide forms that
demonstrate branching structure, making them most similar in terms of
linkage. Both glycogen and amylopectin use α-1,4 and α-1,6 linkages.
Cellulose uses β-1,4 linkages and amylose does not contain α-1,6 linkages.

Question 36

Which of the following is digestible by humans and is made up of
only one type of monosaccharide?
A. Lactose
B. Sucrose
C. Maltose
D. Cellobiose

Answer 36

C
While maltose and cellobiose both have the same glucose subunits, only
maltose is digestible by humans because the β-glycosidic linkages in
cellobiose cannot be cleaved in the human body.

Question 37

Andersen’s disease (glycogen storage disease type IV) is a
condition characterized by a deficiency in glycogen branching
enzyme. Absence of this enzyme would be likely to cause all of
the following effects EXCEPT:
A. decreased glycogen solubility in human cells.
B. slower action of glycogen phosphorylase.
C. less storage of glucose in the body.
D. glycogen devoid of α-1,4 linkages.

Answer 37

D
In Andersen’s disease, glycogen is less branched than normal, thereby
inducing lower solubility of glycogen. Branches reduce the interactions
between adjacent chains of glycogen and encourage interactions with the
aqueous environment. e smaller number of branches means that glycogen
phosphorylase has fewer terminal glucose monomers on which to act,
making enzyme activity slower than normal overall. Finally, without
branches, the density of glucose monomers cannot be as high; therefore, the
total glucose stored is lower than normal. Glycogen synthase is still
functioning normally, so we would expect normal α-1,4 linkages in the
glycogen of an individual with Andersen’s disease but few (if any) α-1,6
linkages

Question 38

The cyclic forms of monosaccharides are:
I. hemiacetals.
II. hemiketals.
III. acetals.
A. I only
B. III only
C. I and II only
D. I, II, and III

Answer 38

C
Monosaccharides can exist as hemiacetals or hemiketals, depending on
whether they are aldoses or ketoses. When a monosaccharide is in its cyclic
form, the anomeric carbon is attached to the oxygen in the ring and a
hydroxyl group. Hence, it is only a hemiacetal or hemiketal because an
acetal or ketal would require the −OH group to be converted to another
−OR group.

Question 39

A chiral molecule is a type of molecule that [lacks or has] an internal plane of symmetry and has a [superimposable or non-superimposable] mirror image

Answer 39

A chiral molecule is a type of molecule that lacks an internal plane of symmetry and has a non-superimposable mirror image

Question 40

[…] only come in pairs because they are mirror images while there can be many more than two […] depending on the number of stereocenters

Answer 40

Enantiomers only come in pairs because they are mirror images while there can be many more than two diastereomers depending on the number of stereocenters

Question 41

[…] form as a result of a glycosidic bond between two monosaccharide subunits

Answer 41

Disaccharides form as a result of a glycosidic bond between two monosaccharide subunits

Question 42

[…] is the spontaneous shift from one anomeric form to another

Answer 42

Mutarotation is the spontaneous shift from one anomeric form to another

The straight-chain form is an intermediate

Question 43

[…] is the main structural component for plant cell walls

Answer 43

Cellulose is the main structural component for plant cell walls

Question 44

A/an [… anomer] is an anomeric carbon with the group pointing up

Answer 44

A/an β-anomer is an anomeric carbon with the group pointing up

Question 45

[…] is when straight-chain carbohydrates form into rings

Answer 45

Cyclization is when straight-chain carbohydrates form into rings

Question 46

The […] is the convention that is used to designate the configurations of chiral carbons

Answer 46

The D and L system is the convention that is used to designate the configurations of chiral carbons

Based on the D- and L- forms of glyceraldehyde

Nearly all carbohydrates in nature are in the D-configuration

Question 47

[…] are a subtype of diastereomers that differ at exactly one chiral carbon

Answer 47

Epimers are a subtype of diastereomers that differ at exactly one chiral carbon

Question 48

[…] are stereoisomers that are non-superimposable and are not mirror images of each other

Answer 48

Diastereomers are stereoisomers that are non-superimposable and are not mirror images of each other

Question 49

In plants, energy is stored in the form of [type of carbohydrate]

Answer 49

In plants, energy is stored in the form of starch

Question 50

[…] is the basis for building complex carbohydrates and requires the anomeric carbon to link to another sugar

Answer 50

Glycoside formation is the basis for building complex carbohydrates and requires the anomeric carbon to link to another sugar

Question 51

Reducing sugars have a/an […] group attached to the […]

Answer 51

Reducing sugars have a/an -OH group attached to the anomeric carbon

The -OH cannot be attached to any other structures

Reducing sugars:

Sucrose (shown below) is a non-reducing sugar because the anomeric carbon does not have an -OH group attached by itself

Question 52

A/an [… anomer] is an anomeric carbon with the group pointing down

Answer 52

A/an α-anomer is an anomeric carbon with the group pointing down

Question 53

[…] are “hydrates of carbon”

Answer 53

Carbohydrates are “hydrates of carbon”

Usually have the formula Cm(H2O)n

Example: Glucose is C6H12O6

H:O ratio is usually 2:1 due to the “hydration” with H2O

Question 54

[…] are sugars that have had a hydroxyl group replaced with a hydrogen atom

Answer 54

Anomers are a subtype of epimers that differ at the anomeric carbon

Question 55

In a […], a sugar reacts with a carboxylic acid or one of its derivatives to form an ester

Answer 55

In a sugar esterification, a sugar reacts with a carboxylic acid or one of its derivatives to form an ester

Question 56

Polysaccharides are formed by repeated monosaccharide or polysaccharide […] bonding

Answer 56

Polysaccharides are formed by repeated monosaccharide or polysaccharide glycosidic bonding

Question 56

A/an […] is a carbon that, in the acyclic form, is not a stereocenter, but once it takes on the cyclic form, it becomes a stereocenter

Answer 56

An anomeric carbon is a carbon that, in the acyclic form, is not a stereocenter, but once it takes on the cyclic form, it becomes a stereocenter