Carbohydrates.pptx

Biochemistry—An Overview
Return to TOC
Section 18.1
Copyright ©2016 Cengage Learning. All Rights Reserved. 1

Return to TOC
Section 18.1
The Study of Living Things
• Biochemistry: Study of the chemical
substances found in living organisms and the
chemical interactions of these substances with
each other
• Biochemical substance: Chemical substance
found within a living organism
• Types of biochemical substances:
– Bioinorganic substances - Water and inorganic salts
– Bioorganic substances - Carbohydrates, lipids,
proteins, and nucleic acids

Return to TOC
Section 18.1
• As isolated compounds, bioinorganic and
bioorganic substances have no life in and of
themselves
– When these substances are gathered together in a
cell, their chemical interactions are able to sustain life
Bioinorganic and Bioorganic Substances

Return to TOC
Section 18.1

Section 18.2
Occurrence and Functions of Carbohydrates
Return to TOC
Plants and Carbohydrates

Section 18.2
Return to TOC
Photosynthesis
• Process in which plants produce carbohydrates
using carbon dioxide, water, and solar energy
Chlorophyll
2 2 2
Plant enzymes
CO + H O+solar energy carbohydrates +O



Section 18.2
Return to TOC
Functions of Carbohydrates in the Human Body
• 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, and nucleic acids)
• Carbohydrates form part of the structural
framework of DNA and RNA molecules

Section 18.2
Return to TOC
Functions of Carbohydrates in the Human Body
• Carbohydrates linked to lipids are structural
components of cell membranes
• Carbohydrates linked to proteins function in a
variety of cell–cell and cell–molecule recognition
processes

Section 18.3
Classification of Carbohydrates
Return to TOC
• Empirical formula of simple carbohydrates -
– CnH2nOn or Cn(H2O)n (hydrate of C)
• n is the number of atoms
• Carbohydrate: Polyhydroxy aldehyde, ketone,
or a compound that produces such substances
upon hydrolysis

Section 18.3
Return to TOC
Monosaccharides
• Contain single polyhydroxy aldehyde or ketone
unit
• Cannot be broken down into simpler substances
by hydrolysis reactions
• Contain 3–7 C atoms
• 5 and 6 carbon species are more common
• Pure monosaccharides - Water soluble white,
crystalline solids
• Monosaccharides - Glucose and fructose

Section 18.3
Return to TOC
Disaccharides
• Contain 2 monosaccharide units covalently
bonded to each other
• Crystalline and water soluble substances
• Common disaccharides - Table sugar (sucrose)
and milk sugar (lactose)
• Upon hydrolysis, they produce 2
monosaccharide units

Section 18.3
Return to TOC
Oligosaccharides
• Contain three to ten monosaccharide units
covalently bonded to each other
• Free oligosaccharides are seldom encountered
in biochemical systems
• Usually found associated with proteins and lipids
in complex molecules
– Serve structural and regulatory functions

Section 18.3
Return to TOC
Polysaccharides
• Contain many monosaccharide units covalently
bonded
• Number of monosaccharide units varies from a
few 100 units to 50,000 units
• Examples:
– Cellulose - Paper, cotton, wood
– Starch - Bread, pasta, potatoes, rice, corn, beans,
and peas

Section 18.4
Chirality: Handedness in Molecules
Return to TOC
Objects and Handedness
• Most biological molecules, including
carbohydrates, exhibit the property of
“handedness” (form of isomerism)
• Most molecules that possess “handedness” exist
in two forms:
– “Left-handed” form
– “Right-handed” form
• Related in the same manner as two hands that are
“mirror images” of each other

Section 18.4
Return to TOC

Section 18.4
Return to TOC
Mirror Images
• Mirror image: Reflection of an object in a mirror
• Classes of objects based on mirror images -
– Superimposable mirror images: Images that
coincide at all points when the images are laid upon
each other
• Achiral molecule
– Nonsuperimposable mirror images: Images where
not all points coincide when the images are laid upon
each other
• Chiral molecule (handedness)

Section 18.4
Return to TOC
Chirality
• Chiral center: C atom attached to 4 different
groups
• A molecule with chiral center is said to be chiral
• A C atom must have four different groups
attached to it in order to be a chiral center
• A chiral C is usually denoted by *
• Bromochloroiodomethane is a chiral organic
molecule

Section 18.4
Return to TOC
Practice Exercise
Indicate whether the circled carbon atom in each
of the following molecules is a chiral center.

Section 18.4
Return to TOC
Practice Exercise
Indicate whether the circled carbon atom in each
of the following molecules is a chiral center.
a. Not a chiral center b. Not a chiral center
c. Chiral center d. Not a chiral center

Section 18.4
Return to TOC
Responses of Left and Right Handed Forms of a
Molecule in a Human Body
• Both forms may be active, one may be more
active, or one may be active and other non-
active
• Example:
– Response of the body to the right-handed hormone
epinephrine is 20 times greater than responses to the
left-handed form
• Almost all monosaccharides are right handed
• Amino acids are always left handed

Section 18.6
Designating Handedness Using Fischer Projection Formulas
Return to TOC
Fischer Project Formula
• Two-dimensional structural notation for showing
the spatial arrangement of groups about chiral
centers in molecules
• According to this formula, a chiral center is
represented as the intersection of vertical and
horizontal lines
• Functional groups of high priority will be written
at the top

Section 18.6
Return to TOC

Section 18.6
Return to TOC
• What are the priority functional groups in
each molecule?
• How do you number the carbon atoms in
each molecule?
• How many chiral centers are there in each
molecule? Indicate the carbon numbers of
each chiral center.

Section 18.6
Return to TOC
Tetrahedral Arrangements
• The four groups attached to the atom at the
chiral center assume a tetrahedral geometry
governed by the following conventions:
– Vertical lines from the chiral center represent bonds
to groups directed into the printed page
– Horizontal lines from the chiral center represent
bonds to groups directed out of the printed page

Section 18.6
Return to TOC
Fischer Project Formulas
• L and D system used to designate the
handedness of glyceraldehyde enantiomers are
shown below

Section 18.6
Return to TOC

Section 18.6
Return to TOC
Fischer Project Formulas
• The D,L system used to designate the
handedness of glyceraldehyde
enantiomers can be extended to other
monosaccharides with more than one
chiral center
– The carbon chain is numbered starting at the
carbonyl group end of the molecule, and the
highest-numbered chiral center is used to
determine D or L configuration

Section 18.6
Return to TOC

Section 18.6
Return to TOC
• Draw the L configuration of
the three monosaccharides
in the previous slide.

Section 18.8
Classification of Monosaccharides
Return to TOC
• Classification based on number of carbon
atoms:
– Triose - 3 carbon atoms
– Tetrose - 4 carbon atoms
– Pentoses - 5 carbon atoms
– Hexoses - 6 carbon atoms
• Classification based on functional groups:
– Aldoses: Monosaccharides with one aldehyde
group
– Ketoses: Monosaccharides with one ketone group
Monosaccharides

Section 18.8
Return to TOC
Monosaccharides
• Combined number of C atoms and functional
group
– Examples:
• Aldohexose - Monosaccharide with aldehyde group
and 6 C atoms
• Ketopentose - Monosaccharide with ketone group and
5 C atoms

Section 18.8
Return to TOC
Exercise
Classify each of the following monosaccharides
according to both the number of carbon atoms and
the type of carbonyl group present.

Section 18.9
Biochemically Important Monosaccharides
Return to TOC
D-Glucose
• Most abundant in nature
• Most important source of human nutrition
• Grape fruit and ripe fruits are good sources of
glucose (20–30% by mass)
– Also named grape sugar
• Other names
– Dextrose
– Blood sugar (70–100 mg/dL)
• Six-membered cyclic form

Section 18.9
Return to TOC
D-Galactose
• Milk sugar
• Synthesized in human beings
• Also called brain sugar
– Part of brain and nerve tissue
• Used to differentiate between blood
types
• Six-membered cyclic form

Section 18.9
Return to TOC
D-Fructose
• Ketohexose
• Sweetest tasting of all sugars
– Found in many fruits and in honey
• Good dietary sugar due to
higher sweetness
• Five-membered cyclic form

Section 18.9
Return to TOC
D-Ribose
• Part of a variety of complex molecules which
include:
– RNA
– ATP
– DNA
• Five-membered cyclic form

Section 18.10
Cyclic Forms of Monosaccharides
Return to TOC
Cyclic Hemiacetal Forms of D-Glucose
• Dominant forms of monosaccharides with 5 or
more C atoms
– Cyclic structures are in equilibrium with open chain
forms
• Cyclic structures are formed by the reaction of
carbonyl group (C=O) with hydroxyl (–OH) group
on carbon 5

Section 18.10
Cyclic Forms of Monosaccharides
Return to TOC
HAWORTH PROJECTION

Section 18.11
Haworth Projection Formulas
Return to TOC
• Haworth projection formula: Two-dimensional
structural notation that specifies the three-
dimensional structure of a cyclic form of a
monosaccharide

Section 18.11
Return to TOC
α and β Configuration
• Determined by the position of the –OH group on
C1 relative to the –CH2OH group that
determines D or L series
– In a β configuration, both of these groups point in the
same direction
– In an α configuration, the two groups point in opposite
directions

Section 18.11
Return to TOC
–OH Group Position
• The specific identity of a monosaccharide is
determined by the positioning of the other –OH
groups in the Haworth projection formula
– Any –OH group at a chiral center that is to the right in a
Fischer projection formula points down in the Haworth
projection formula
– Any –OH group to the left in a Fischer projection formula
points up in the Haworth projection formula

Section 18.11
Return to TOC
Figure 18.17 - The Cyclic Hemiacetal Forms of D-
Glucose

Section 18.11
Return to TOC
• 2 forms of D-Glucose:
– α-form where the –OH of C1 and CH2OH of C5 are
on opposite sides
– β-form where the –OH of C1 and CH2OH of C5 are
on the same side
Cyclic Hemiacetal Forms of D-Glucose
O
OH OH
OH
OH
CH2OH
O
OH
OH
OH
OH
CH2OH
-D-Glucose -D-Glucose
1
2
3
4
5
6
1
2
3
4
5
6
Anomeric
Carbon
Anomeric
Carbon

Section 18.11
Return to TOC
Anomers
• Cyclic monosaccharides that differ only in the
position of the substituents on the anomeric
carbon atom

Section 18.11
Return to TOC
Let us make the Haworth Projection of FRUCTOSE

Section 18.11
Return to TOC

Section 18.11
Return to TOC
Pyranose and Furanose
• Pyranose - Cyclic monosaccharide containing a
six-atom ring
• Furanose - Cyclic monosaccharide containing a
five-atom ring
• Their ring structures resemble the ring structures
in the cyclic ethers pyran and furan, respectively

Section 18.11
Return to TOC
Practice Exercise
Which of the monosaccharides glucose, fructose,
galactose, and ribose has each of the following
structural characteristics? (There may be more than
one correct answer for a given characteristic)
a. It is a pentose.
b. It is a ketose.
c. Its cyclic form has a 6-membered ring.
d. Its cyclic form has two carbon atoms outside the ring.

Section 18.13
Disaccharides
Return to TOC
• Two monosaccharides can react to form a
disaccharide
– One monosaccharide acts as a hemiacetal and the
other as an alcohol
– Resulting ether bond is a glycosidic linkage

Section 18.13
Disaccharides
Return to TOC

Section 18.13
Disaccharides
Return to TOC
Maltose (Malt Sugar)
• Structurally made of 2 D-glucose units, one of
which must be α-D-glucose, linked via an
α(14) glycosidic linkage
• Digested easily by humans because of an
enzyme that can break α(14) linkages
• Baby foods are rich in maltose

Section 18.13
Disaccharides
Return to TOC
Cellobiose
• Produced as an intermediate in the hydrolysis of
the polysaccharide cellulose
– Contains two D-glucose monosaccharide units, one
of which must have a β configuration, linked through
a β(14) glycosidic linkage
• Cannot be digested by humans

Section 18.13
Disaccharides
Return to TOC
Lactose
• Made up of β-D-galactose unit and a D-glucose
unit joined by a β(14) glycosidic linkage
• Milk is rich in the disaccharide lactose
• Lactase hydrolyzes β(14) glycosidic linkages

Section 18.13
Disaccharides
Return to TOC
Lactose Intolerance or Lactase Persistence
• Lactose is the principal carbohydrate in milk
– Human mother’s milk - 7%–8% lactose
– Cow’s milk - 4%–5% lactose
• Lactose intolerance is a condition in which
people lack the enzyme lactase needed to
hydrolyze lactose to galactose and glucose
• Deficiency of lactase can be caused by a
genetic defect, physiological decline with age, or
by injuries to intestinal mucosa

Section 18.13
Disaccharides
Return to TOC
Lactose Intolerance or Lactase Persistence
• When lactose is undigested, it attracts water
causing fullness, discomfort, cramping, nausea,
and diarrhea
• Bacterial fermentation of the lactose further
along the intestinal tract produces acid (lactic
acid) and gas, adding to the discomfort

Section 18.13
Disaccharides
Return to TOC
Sucrose (Table Sugar)
• The most abundant of all disaccharides and
found in plants
• Produced commercially from the juice of sugar
cane and sugar beets
– Sugar cane contains up to 20% by mass sucrose
– Sugar beets contain up to 17% by mass sucrose

Section 18.13
Disaccharides
Return to TOC

Section 18.13
Disaccharides
Return to TOC
What is the linkage that is formed when two
monosaccharides react to form a disaccharide
called?
a.Alcohol group linkage
b.Glycosidic linkage
c.Hemiacetal linkage
d.Acetal linkage

Section 18.13
Disaccharides
Return to TOC
Which disaccharide is a nonreducing sugar?
a.Maltose
b.Lactose
c.Cellobiose
d.Sucrose

Section 18.13
Disaccharides
Return to TOC
Practice Exercise
Which of these disaccharides, i.e., maltose, cellobiose,
lactose, and sucrose, have the following structural or
reaction characteristics? (There may be more than one
correct answer for a given characteristic)
a.Two different monosaccharide units are present.
b.Hydrolysis produces only monosaccharides.
c.Its glycosidic linkage is a “head-to-head” linkage.
d.It is not a reducing sugar.

Section 18.13
Disaccharides
Return to TOC
Practice Exercise
Which of these disaccharides, i.e., maltose, cellobiose,
lactose, and sucrose, have the following structural or
reaction characteristics? (There may be more than one
correct answer for a given characteristic)
a.Two different monosaccharide units are present.
Lactose, sucrose
b.Hydrolysis produces only monosaccharides.
Maltose, cellobiose, lactose, sucrose
c.Its glycosidic linkage is a “head-to-head” linkage.
Sucrose
d.It is not a reducing sugar.
Sucrose

Section 18.14
Oligosaccharides
Return to TOC
• Carbohydrates that contain 3–10
monosaccharide units bonded to each other via
glycosidic linkages
• Generally present in association with other
complex molecules
– Raffinose - Made of 1 galactose, 1 glucose, and 1
fructose
– Stachyose - Made of 2 galactose, 1 glucose, and 1
fructose units
• Commonly found in onions, cabbage, broccoli,
and whole wheat

Section 18.14
Oligosaccharides
Return to TOC
Blood Types and Oligosaccharides
• Human blood is classified into four types
– A, B, AB, and O
– The basis for the difference is the type of sugars
(oligosaccharides) present
– Blood of one type cannot be given to a recipient with
blood of another type
– A transfusion of wrong blood type can cause the
blood cells to form clumps, a potentially fatal reaction
– People with type O blood are universal donors, and
those with type AB blood are universal recipients

Section 18.14
Oligosaccharides
Return to TOC
Blood Types and Oligosaccharides
• In the United States, type O blood is the most
common and type A the second most common
• The biochemical basis for the various blood
types involves oligosaccharides present on
plasma membranes of red blood cells
• The oligosaccharides responsible for blood
groups are D-galactose and its derivatives

Section 18.14
Oligosaccharides
Return to TOC
Other Oligosaccharides
• Solanine, a potato plant toxin, is a
oligosaccharide found in association with an
alkaloid
– Bitter taste of potatoes is due to relatively higher
levels of solanine

Section 18.14
Oligosaccharides
Return to TOC
Which of the following is a toxin produced by
potato plants?
a.Raffinose
b.Stachyose
c.Solanine
d.None of the above

Section 18.15
General Characteristics of Polysaccharides
Return to TOC
The Polymer Chain
• Polysaccharides are polymers of many
monosaccharide units bonded with glycosidic
linkages
• Two types:
– Homopolysaccharide
– Heteropolysaccharide

Section 18.15
Return to TOC
Figure 18.26 - The Polymer Chains of a Polysaccharide

Section 18.15
Return to TOC
Characteristics of Polysaccharides
• Polysaccharides are not sweet and do not show
positive tests with Tollen’s and Benedict’s
solutions, whereas monosaccharides are sweet
and show positive tests
• Limited water solubility
• Examples:
– Cellulose and glycogen - Storage polysaccharides
– Chitin - Structural polysaccharide
– Hyaluronic acid - Acidic polysaccharide

Section 18.15
Return to TOC
A _____ is a polysaccharide in which only one
type of monosaccharide monomer is present.
a.heteropolysaccharide
b.homopolysaccharide
c.heteromonosaccharide
d.homomonosaccharide

Section 18.16
Storage Polysaccharides
Return to TOC
Starch
• Storage polysaccharide: Polysaccharide that is
a storage form for monosaccharides and used
as an energy source in cells
• Starch
– Glucose is the monomeric unit
– Storage polysaccharide in plants

Section 18.16
Return to TOC
Types of Polysaccharides Isolated From Starch
• Amylose
– Unbranched-chain polymer and accounts for 15%–
20% of the starch
– Has α(14) glycosidic bonds
• Amylopectin
– Branched chain polymer and accounts for 80%–85%
of the starch
– Has α(14) and α(16) glycosidic bonds
– Up to 100,000 glucose units are present
– Amylopectin is digested more readily by humans (can
hydrolyze α linkages but not β linkages)

Section 18.16
Return to TOC
Glycogen
• Glucose storage polysaccharide in humans and
animals
• Contains only glucose units
• Branched chain polymer with α(14) glycosidic
bonds in straight chains and α(16) in
branches
• Three times more highly branched than
amylopectin in starch
• Contains up to 1,000,000 glucose units

Section 18.16
Return to TOC
Glycogen
• Excess glucose in blood is stored in the form of
glycogen
Glycogenesis
Glycogenolysis
Glucose glycogen

Section 18.16
Return to TOC
In starch, two different glucose-containing
polysaccharides can be isolated. They are _____
and _____.
a.glycogen; amylose
b.amylose; amylopectin
c.amylose; cellulose
d.glycogen; cellulose

Section 18.17
Structural Polysaccharides
Return to TOC
Cellulose
• Linear homopolysaccharide with β(14)
glycosidic bond
• Contains up to 5000 glucose units with
molecular mass of 900,000 amu
– Cotton has 95% cellulose and wood 50% cellulose
• Humans do not have enzymes that hydrolyze
β(14) linkages and so they cannot digest
cellulose
– Animals also lack these enzymes, but they can digest
cellulose due to the presence of cellulase-producing
bacteria

Section 18.17
Return to TOC
Cellulose
• It serves as dietary fiber in food and readily
absorbs water resulting in softer stools
– 20–35 g of dietary fiber is desired everyday

Section 18.17
Return to TOC
Chitin
• Similar to cellulose structurally and functionally
• Linear polymer with all β(14) glycosidic
linkages
– It has an N-acetyl amino derivative of glucose
• Function is to give rigidity to the exoskeletons of
crabs, lobsters, shrimp, insects, and other
arthropods

Section 18.17
Return to TOC
Figure 18.32(b) - The Structure of Chitin

Section 18.17
Return to TOC
What is the structural component of plant cell
walls?
a.Starch
b.Glycogen
c.Chitin
d.Cellulose

Section 18.18
Acidic Polysaccharides
Return to TOC
Acidic polysaccharides
• Polysaccharides with a repeating disaccharide
unit containing an amino sugar and a sugar with
a negative charge due to a sulfate or a carboxyl
group
• They are heteropolysaccharides, i.e., different
monosaccharides exist in an altering pattern
• Examples:
– Hyaluronic acid
– Heparin

Section 18.18
Return to TOC
Hyaluronic Acid
• Alternating residues of N-acetyl- β-D-
glucosamine and D-glucuronate
• Highly viscous and serve as lubricants in the
fluid of joints as well as vitreous humor of the
eye

Section 18.18
Return to TOC
Heparin
• Polysaccharide with 15–90 disaccharide
residues per chain
• Blood anticoagulant

Section 18.18
Return to TOC
Heparin, a well known acidic polysaccharide, is
best known for its biochemical function as a(n)
_____.
a.coagulant
b.anticoagulant
c.exoskeleton
d.dietary fiber

Section 18.19
Dietary Considerations and Carbohydrates
Return to TOC
Nutrition
• Foods high in carbohydrate content constitute
over 50% of the diet of most people of the world
– Corn in South America
– Rice in Asia
– Starchy root vegetables in parts of Africa
– Potato and wheat in North America
• Balanced dietary food should contain about 60%
of carbohydrate

Section 18.19
Return to TOC
Classes of Dietary Carbohydrates
• Simple carbohydrates: Dietary
monosaccharides or disaccharides
– Sweet to taste and commonly referred to as sugars
– Constitute 20% of the energy in the US diet
• Complex carbohydrates: Dietary
polysaccharides
– Include starch and cellulose, which are normally not
sweet to taste

Section 18.19
Return to TOC
Dietary carbohydrates constitute _____ of the diet
of most people.
a.10%
b.25%
c.35%
d.50%

Section 18.20
Glycolipids and Glycoproteins: Cell Recognition
Return to TOC
• Glycolipid: Lipid molecule that has one or more
carbohydrate (or carbohydrate derivative) units
covalently bonded to it
• Glycoprotein: Protein molecule that has one or
more carbohydrate (or carbohydrate derivative)
units covalently bonded to it
– Such carbohydrate complexes are very important in
cellular functions such as cell recognition

Section 18.20
Return to TOC
A protein molecule that has one or more
carbohydrate units covalently bonded to it is
known as a(n) _____.
a.glycolipid
b.glucoprotein
c.glycoprotein
d.oligosaccharide protein

Chapter 18
Return to TOC
Concept Question 1
D-glucose consists of six carbon atoms. When drawn using the
Fischer projection formula, how do you determine that it is the D
isomer? How many chiral carbon centers does it have and how
many stereoisomers are possible?
a.The –OH on the chiral carbon farthest from the carbonyl group points to the
right. There are five chiral centers providing 32 stereoisomers.
b.The –OH on the chiral carbon farthest from the carbonyl group points to the
right. There are four chiral centers providing 16 stereoisomers.
c.The –OH on the chiral carbon farthest from the carbonyl group points to the
left. There are five chiral centers providing 32 stereoisomers.
d.The –OH on the chiral carbon farthest from the carbonyl group points to the
left. There are four chiral centers providing 16 stereoisomers.

Chapter 18
Return to TOC
Concept Question 2
Prior to a marathon run, an athlete consumes large amounts of
complex carbohydrates to do what is known as “carbohydrate loading.”
What happens in the body to the glucose molecules present in these
complex carbohydrates and why is carbohydrate loading important?
a.The complex carbohydrates are broken down to glucose and any excess glucose, not
used for immediate energy, is stored in the form of glycogen, which can be used later as
a source of stored energy.
b.The complex carbohydrates are broken down to glucose and any excess glucose, not
used for immediate energy, is stored in the form of starch, which can be used later as a
source of stored energy.
c.The complex carbohydrates are broken down to galactose and any excess galactose,
not used for immediate energy, is stored in the form of glycogen, which can be used later
as a source of stored energy.
d.The complex carbohydrates are broken down to glucose and any excess glucose, not
used for immediate energy, is stored in the form of glycogen, which can be used later to
produce muscle tissue needed to complete the marathon.

Chapter 18
Return to TOC

Carbohydrates.pptx

More Related Content

What's hot

Similar to Carbohydrates.pptx

More from ChristianCatacutan2

Recently uploaded

Carbohydrates.pptx