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Carbohydrates
Dr Seemal Jelani
Introduction to Biochemistry
1
5/23/2017
Carbohydrates
Most abundant
 Their abundance in human body
is low but constitute about 75%
by mass of dry plant materials
 Green (chlorophyll-containing)
plants produce carbohydrates
via photosynthesis
 A major source of energy from
our diet composed of the
elements C, H and O.
 They also called saccharides,
which means “sugars.”
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Biochemistry
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Carbohydrates
 They are produced by
photosynthesis in plants.
 such as glucose are
synthesized in plants
from CO2 from the air,
H2O from the soil, and
energy from the sun
absorbed in chlorophyll
to form carbohydrates
and O2
 Oxidized in living cells to
produce CO2, H2O, and
energy.
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Biochemistry
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Uses of carbohydrates in plants
 Cellulose (carbohydrates) serve as structural
elements
 Starch (carbohydrates) provide energy
reserves for the plants
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Biochemistry
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 Dietary intake of plant materials is the major
carbohydrate source for humans and animals
 Functions:
 Carbohydrate oxidation provides energy
 Carbohydrate storage in the form of glycogen
provides a short-term energy reserve
 Carbohydrates supply carbon atoms for the
synthesis of other biochemical substances
(proteins, lipids, nucleic acids)
 Carbohydrate form part of the structural
framework of DNA and RNA
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Classification of Carbohydrates
The types of carbohydrates are
• Monosaccharides, the simplest carbohydrates.
• Disaccharides, which consist of two monosaccharides.
• Polysaccharides, which contain many monosaccharides.
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Biochemistry
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General Formula
Cn H2n On
 Carbohydrates are Polyhydroxy aldehydes,
Polyhydroxy ketones
 The carbohydrate glucose is Polyhydroxy aldehyde
and the carbohydrate fructose is Polyhydroxy ketone
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Biochemistry
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CHIRALITY: Handedness in molecules
A chiral object is not superimposable on its
mirror image they do not possess a plane
of symmetry
Two forms of a chiral object are known as
enantiomers
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Mirror image
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Mirror image
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Non superposable
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Chiral HANDS
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Chiral
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Shells
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The plane has the same thing on
both sides for the flask
There is no mirror plane for a hand
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If an object has a plane of
symmetry it is necessarily the
same as its mirror image
The lack of a plane of symmetry
is called “handedness”, Chirality
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Hands, gloves are prime
examples of chiral object
They have a “left” and a
“right” version
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Why this chapter?
Handedness is important in
organic and biochemistry
Molecular handedness makes
possible specific interactions
between enzymes and
substrates
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Chirality in daily life
Most Biomolecules in nature are
chiral (sugars, DNA, proteins, amino
acids, steroids)
Human proteins are exclusively
built from L-amino acids; this
involves receptors which are chiral
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The Reason for Handedness:
Chirality
 Molecules that are not superimposable
with their mirror images are chiral (have
handedness)
 A plane of symmetry divides an entire
molecule into two pieces that are exact
mirror images
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Functional groups in
carbohydrates
 Aldehydes CHO
 Ketone
C=O
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Biochemistry
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Aldose
Ketose
20
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Chirality in carbohydrates
Fischer projections
 Carbohydrates may contain more than one
chiral center
 FP represents a method for giving
molecularity specifications in two dimensions
 FP is a two dimensional notation for showing
the spatial arrangement of groups about
chiral centers in molecule
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Biochemistry
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






Stereoisomers
Enantiomers
Diastereomers
Handedness (Right and left)
D and L
Fischer Projection for 2,3,4-trihydroxybutanal
Epimers
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Biochemistry
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Properties of Enantiomers
 Structural isomers have different properties
 Diastereoisomers have different properties
 Enantiomers have same properties expect
two:
 Their interaction with plane polarized light
 Their interaction with other chiral substances
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Biochemistry
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PLANE-POLARIZED LIGHT
Polarimeter: a device for measuring the extent
of rotation of plane-polarized light
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Chem-241
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Plane-Polarized Light
optical activity
Light vibrating in all planes  to direction of propagation
Plane-polarized light: light vibrating only in parallel planes
Plane-polarized light the vector sum of left
and right circularly polarized light
Optically Activity
Enantiomers (chiral) interact with circularly polarized light
rotating the plane one way with R center
and opposite way with S
result: rotation of plane-polarized light clockwise (+)
or counterclockwise (-)
Plane-Polarized Light
Change in the polarized plane?
(polarimeter)
achiral
sample
no change in the plane
Plane-Polarized Light
Change in the polarized plane?
(polarimeter)
rotates the plane

Classification of Monosaccharides
Monosaccharides consist of
 3-6 carbon atoms typically.
 A carbonyl group (aldehyde or ketone).
 Several hydroxyl groups.
 2 types of monosaccharide structures:
Aldoses and ketoses
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Biochemistry
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Aldoses
O
║
C─H aldose
│
H─ C─OH
│
H─ C─OH
│
CH2OH
Aldoses are monosaccharides
 with an aldehyde group
 with many hydroxyl (-OH)
groups.
triose (3C atoms)
tetrose (4C atoms)
pentose (5 C atoms)
hexose (6 C atoms)
Erythose, an aldotetrose
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Biochemistry
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Ketoses
CH2OH
│
C=O
ketose
│
H─ C─OH
│
H─ C─OH
│
H─C─OH
│
CH2OH
Ketoses are monosaccharides
 with a ketone group
 with many hydroxyl (-OH)
groups.
Fructose, a ketohexose
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Biochemistry
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Learning Check
Identify each as aldo- or keto- and as tetrose, pentose, or
hexose:
O
CH2OH
C H
H C OH
C O
H C OH
HO C H
H C OH
H C OH
H C OH
CH2OH
CH2OH
ketopentose
aldohexose
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Biochemistry
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Structures of Monosaccharides
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Biochemistry
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Fischer Projections
A Fischer projection
 is used to represent carbohydrates.
 places the most oxidized group at the top.
 shows chiral carbons as the intersection of vertical and
horizontal lines.
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Biochemistry
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Chiral Carbon
 Carbon linked to four different groups
CH3CHOHC6H5
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Biochemistry
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D and L Notations
In a Fischer projection, the −OH group on the
 chiral carbon farthest from the carbonyl group determines
an L or D isomer.
 left is assigned the letter L for the L-form.
 right is assigned the letter D for the D-form.
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Biochemistry
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Examples of D and L Isomers of
Monosaccharides
O
O
C H
C H
H
O
OH
HO
H
H
OH
H
OH
CH2OH
D-glucose
Dr Seemal Jelani
Biochemistry
HO
C H
H
OH
H
H
OH
OH
CH2OH
D-ribose
Introduction to
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H
H
OH
H
OH
HO
H
CH2OH
L-galactose
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D-Glucose
 found in fruits, corn
syrup, and honey.
 an aldohexose with the
formula C6H12O6 known
as blood sugar in the
body.
 The monosaccharide in
polymers of starch,
cellulose, and glycogen.
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Biochemistry
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D-Fructose
 is a ketohexose
C6H12O6.
 is the sweetest
carbohydrate.
CH2OH
C O
HO C H
 is found in fruit
juices and honey.
H C OH
H C OH
 converts to glucose
in the body.
CH2OH
D-Fructose
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Biochemistry
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Cyclic Structures
Cyclic structures
 are the widespread form of monosaccharides with 5
or 6 carbon atoms.
O
O
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Biochemistry
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 The cyclic forms of monosaccharides result
from the ability of their carbonyl group to
react intramolecularly with a –OH group
 Cyclic structure is formed when the –OH
group on C-5 reacts with the aldehyde group
or ketone group
 The result is a cyclic hemiacetal or cyclic
hemiketal
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Biochemistry
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Steps for cyclisation
for D-Glucose
 All –OH groups to the
right in the projection
formula appear below the
ring whereas –OH gps to
the left in FP appear
above the ring
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Biochemistry
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Drawing the Cyclic Structure for Glucose
Number the carbon chain and turn clockwise to
linear open chain.
H
O
H H OH H
C
1
H
2C
OH
HO
3C
H
H
4C
OH
H
5C
OH
6CH
Dr Seemal Jelani
Biochemistry
form a
O
HOCH2 C C C C C
H
OH OH H OH
6
5
4
3
2
1
2OH
Introduction to
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 The –OH gp on C-5 adds across then
C=O
to give two stereoisomers
 In aq. Solu of D- glucose, a dynamic equilibrium exists
among the α, β, and open chain forms and there is
continual interconversion among them
 A freshly mixed solution of pure α- D- Glucose slowly
converts to a mixture of both α & β-D-glucose by an
opening and closing of the cyclic structure.
 At equilibrium 63% is β-D-glucose and 37% are α-Dglucose and less than 0.01% in open chain
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Cyclic Structure for Glucose
STEP 2 Fold into a hexagon.
 Bond the C5 –O– to C1.
 Place the C6 group above the
ring.
 Write the –OH groups on C2 and
C4 below the ring.
 Write the –OH group on C3
above the ring.
 Write a new –OH on C1.
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Biochemistry
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CH2OH
6
5
O
4
OH
1
OH
3
2
OH
OH
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Cyclic Structure for Glucose (cont)
STEP 3 Write the new –OH on C1
• down for the  form.
• up for the  form.

CH2OH
O
CH2OH
O
OH
OH
OH
OH

OH
OH
OH
OH
-D-Glucose
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-D-Glucose
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Summary of the Formation of Cyclic
Glucose
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Biochemistry
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Alcohols react with the carbonyl
groups of aldehydes and ketones to
give hemiacetal
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-D-Glucose and β-D-Glucose in Solution
When placed in solution,
 cyclic structures open and close.
 -D-glucose converts to β-D-glucose and vice versa.
 at any time, only a small amount of open chain forms.
CH2OH
CH2OH
O
OH
OH
Dr Seemal Jelani
Biochemistry
O
O
C
OH
H
OH
OH
OH
-D-glucose
(36%)
H
O
OH
OH
CH2OH
D-glucose (open)
(trace)
Introduction to
49
OH
OH
OH
β-D-glucose
(64%)
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Intramolecular Cyclic hemiketal
Structure of Fructose
 Fructose and other ketoses with a long
carbon chain also cyclizes to form hemiketal
 D-fructose and D-ribose form a fivemembered ring
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Cyclic Structure of Fructose
Fructose
 is a ketohexose.
 forms a cyclic structure.
 reacts the —OH on C-5 with the C=O on C-2.
CH2OH
C O
CH2OH
HO C H
CH2OH
OH
H C OH
H C OH
CH2OH
O
O
OH
OH
OH
CH2OH
OH
OH
α-D-fructose
-D-fructose
CH2OH
D-fructose
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Haworth Projection
 This is two dimensional notation that specifies the 3dimensional structure of cyclic form of carbohydrate
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 The D and L form of a monosaccharide is determined
by the position of the terminal CH2OH gp on the
highest-numbered ring carbon atom
 In D-form, this group is positioned above the ring
 In L-form the terminal CH2OH gp is positioned below
the ring (not encountered in biological systems)
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 α and β configurations is determined by the position
of the –OH gp on carbon no 1 relative to the CH2OH
 In β-configuration both of these gps point in the same
direction
 In α-configuration the two gps point in opposites
direction
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Reactions of Monosaccharides
 Oxidation
 Reduction
 Glycoside formation
 Phosphate ester formation
 Amino sugar formation
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Oxidation
 Yield three different types of oxidation products
 Oxidizing agent used to determined the product
Weak oxidizing Agent:
Tollen's Reagent
Benedict Solution
Reducing Sugars
Is a carbohydrate that gives a positive test with TR, FS
and BS
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 Strong oxidizing agent oxidizes both end of
monosaccharide i.e
 Terminal pri-alcohol and carbonyl group to give
dicarboxylic acid
 Such polyhydroxy dicarboxylic acids are known as –
aric acids
 The oxidation of glucose gives Glucaric acid
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Reduction
 Carbonyl group present in a monosaccharide ( aldose,
ketose) can be reduced to a Hydroxyl group using
Hydrogen as a reducing group
 Product is called Sugar Alcohol
 D- Glucitol is also known as D-sorbitol
 These are used as moisturizing agents in foods and
cosmetics because of their affinity for water
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Glycoside Formation
 Cyclic forms of monosaccharides are hemiacetals and
hemiketals react with alcohols to form Acetals and
Ketals
 The general name for monosaccharide acetals and
Ketals is Glycoside
 Glycoside
 It is an Acetal or a Ketal formed from a cyclic
monosaccharide
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



Glycoside produced from glucose is Glycoside
Glycoside produced from galactose is Galactoside
Exist in α and β forms
Named as by listing alkyl or aryl group attached to the
oxygen followed by the name of a monosaccharide
involved with the suffix-ide
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Phosphate Ester formation
 The –OH gp of a monosaccharide can react with
Oxoacids to form Esters
 Phosphate esters are formed from phosphoric acid
and various monosaccharides commonly encountered
in biological system
 Example
 Specific enzymes in the body catalyze the
esterification of the carbonyl group (C1) and the
primary alcohol (C6) of glucose to give
 Glucose-1-Phosphate
 Glucose-6-Phosphate
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 These phosphate esters of glucose are stable
in aqueous solution and play important roles
in the metabolism of carbohydrates
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Amino Sugar formation
 Amino sugars of glucose, mannose and galactose are
common in nature
 They are produced by replacing the –OH group on
carbon 2 on the monosaccharide with an amino group
 Amino sugars and their N-acetyl-derivatives are
important building blocks of polysaccharides such as
cartilage
 The N-acetyl derivatives of D-glucosamine and
D-galactosamine are present in the biochemical markers
on red blood cells, which distinguish the various blood
type
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Disaccharides
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Important Disaccharides
A disaccharide consists of two monosaccharides.
Monosaccharides
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Biochemistry
Disaccharide
glucose + glucose
maltose + H2O
glucose + galactose
lactose + H2O
glucose + fructose
sucrose + H2O
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Disaccharides
 Monosaccharide + Monosaccharide
(Functioning as
(Functioning as
A hemiacetal or
an alcohol)
hemiketal)
Disaccharide + H2O
glycoside
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Maltose
Maltose is






A disaccharide also known as malt sugar.
Composed of two D-glucose molecules.
Obtained from the hydrolysis of starch.
Used in cereals, candies, and brewing.
Found in both the - and β - forms.
The Glycosidic linkage between the two glucose units is called
(1-4) linkage
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Glycosidic Formation
 Cyclic forms of monosaccharides are hemiacetals and
hemiketals, they react with alcohols to form acetals
and Ketals
 The bond that links two monosaccharides of a
disaccharide together is called a Glycosidic linkage
 A Glycosidic linkage is the carbon-Oxygen-carbon
bond that joins the two components of Glycoside
together
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Formation of Maltose
Free α-OH
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Cellobiose
 Produced as an intermediate in the hydrolysis of the
polysaccharide cellulose
 Contains two D-glucose monosaccharide units
 Differ from maltose –must have a β- configuration
 β- (1-4)Glycosidic linkage
 Reducing sugar having three isomeric forms in Aq.
Solu and on hydrolysis produces two D-glucose
molecules
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Difference in maltose and
Cellobiose
 Different in biological behaviors
 Differences are related to stereochemistry of their
glycosidic linkages
 Maltase enzyme which breaks glucose-glucose α (1-4)
linkage present in maltose is present in maltose is
found in human body and in yeast
 That’s why maltose is easily digested by human body
and readily fermented by yeast
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 Both the human body and yeast lacks enzyme
cellobiase which is needed to break the glucoseglucose β (1-4) glycosidic linkage of Cellobiose
 Cellobiose cannot be digested by humans or
fermented by yeast
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Lactose
Lactose
 is a disaccharide of β-Dgalactose and α- Dglucose.
 contains a β -1,4glycosidic bond.
α-form
 is found in milk nearly 49% and milk products.
α-form
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Sucrose
Sucrose or table sugar
• is obtained from sugar cane and sugar beets.
• consists of α-D-glucose and β-D-fructose..
• has an α,β-1,2-glycosidic bond.
α-D-glucose
β -D-fructose
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Sweetness of Sweeteners
Sugars and artificial
sweeteners
 differ in
sweetness.
 are compared to
sucrose (table
sugar), which is
assigned a value of
100.
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Learning Check
Identify the monosaccharides in each of the following:
A. lactose
(1) α-D-glucose
(2) β-D-fructose
(3) β-D-galactose
B. maltose
(1) α-D-glucose
(2) β-D-fructose
(3) β-D-galactose
C. sucrose
(1) α-D-glucose
(2) β-D-fructose
(3) β-D-galactose
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Polysaccharides
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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Polysaccharides
Polysaccharides
 Are polymers of D-glucose.
 Include Amylose and Amylopectin,
starches made of α-D-glucose.
 Include glycogen (animal starch in
muscle), which is made of α-D-glucose.
 Include cellulose (plants and wood),
which is made of β-D-glucose.
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CH2OH
O
OH
OH
OH
OH
α-D-Glucose
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Structures of Amylose and Amylopectin
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Amylose
Amylose is
 a polymer of α-Dglucose molecules.
 linked by -1,4
glycosidic bonds.
 a continuous
(unbranched) chain.
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Amylopectin
Amylopectin
 is a polymer of α-Dglucose molecules.
 is a branched-chain
polysaccharide.
 has α-1,4-glycosidic
bonds between the
glucose units.
 has α-1,6 bonds to
branches.
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Dextrins
 Starches like amylose and amylopectin hydrolyze to
dextrins (smaller polysaccharides)
 Contain 3-8 glucose units
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Glycogen
Glycogen
 is the polysaccharide that
stores α-D-glucose in
muscle.
 is similar to amylopectin,
but is more highly
branched.
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Cellulose
Cellulose
 is a polysaccharide of
glucose units in
unbranched chains.
 has β-1,4-glycosidic
bonds.
 cannot be digested by
humans because
humans cannot break
down β-1,4-glycosidic
bonds.
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