• Biochemistry is the study of the chemical substances found in living organisms and the chemical interactions of these
substances with each other.
Mainly cellulose
and starch
oxidation - provides energy
storage, in the form of glycogen, provides a short-term
energy reserve
supply carbon atoms for the synthesis of other
biochemical substances (proteins, lipids, and nucleic
acids)
form part of the structural framework of DNA and
RNA molecules (Chapter 22)
linked to lipids (Chapter 19) are structural components
of cell membranes
linked to proteins (Chapter 20) function in a variety of
Carbohydrates are polyhydroxy aldehydes or ketones or
compounds that produce such substances upon
hydrolysis
Simpler Formula:
▪ CnH2nOn or Cn(H2O)n (hydrates of C) ▪ n= number of atoms
Monosaccharide:
CHOOH H H HO OH H OH H
CH2OH Glucose
CH2OH
O H HO OH H OH H
CH2OH
Oligosaccharides:
▪
Contains ~2-10 monosaccharide units - covalently bonded
to each other
▪
Disaccharides
(contain 2 monosaccharide units) more
common - crystalline water soluble substances
▪ Table sugar (sucrose) and milk sugar (lactose)
▪ Upon hydrolysis they produce monosaccharides
Contains many monosaccharide units covalently bonded
Polymers: May contain 100s of 1000s of monosaccharide
units
Examples:
▪ Cellulose: Paper, cotton, wood
Most monosaccharides exist in two forms: a “left
A molecule (or object) that is not
superimposable on its mirror
image is said to be chiral
Chirality Center
: C atom attached to 4 different groups
A molecule with a chirality center is a chiral molecule
▪
Best way to visualize - look at all C atoms and see if there are
at least two H atoms then that can’t be a chiral center
▪
C atoms with less than one H atoms are worth looking at for
their chirality.
Be careful
as a C atom may look chiral but may not have four
DIFFERENT groups (due to
implicit hydrogens
).
C
CH
3Cl
Cl
H
CH3C
OH
H2C
H CH3
C
CH3
O H2C
CH3
C H
O
CH H3C
CH2
H3C
O
OH
HO
OH
Tartaric Acid * *
* *
Stereoisomers are isomers that have the same
molecular and structural formulas
and
the same
functional group but differ in the orientation of
atoms in space.
Enantiomers
: Stereoisomers that are
nonsuperimposable mirror images of each other
Drawing Enantiomers:
▪
Locate the chiral center
▪
Place two bonds in the plane
▪
One in front of the plane (wedged line), one behind the
plane (dashed line)
▪
Arbitrarily place the groups
▪Draw the mirror image
C C
CH3
H3CH2C H Br
C CH3
H3CH2C H
Br
C CH3
CH2CH3 H
Chiral or Achiral?
H
CH
3H
Enantiomers have same boiling points, melting points
and densities - all these are dependent upon
intermolecular forces and chirality doesn’t depend on
them
Enantiomeric pairs have same solubility in achiral
solvents like ethanol and have different solubility in
chiral solvent like D-2-butanol
Our body responds differently to different
enantiomers:
Body response to D form of hormone epinephrine is 20 times greater than its
L isomer
Natural flavors of spearmint and caraway
Vertical lines represent bonds to groups directed into the
printed page.
Horizontal lines represent bonds to groups directed out of
the printed page
Functional groups of high priority will be written at top
One chiral center: D (
dextro
= right) and L (
levo
= left)
system used to designate the handedness of
glyceraldehyde enantiomers
D-Glyceraldehyde
(-OH on the right-hand side)
L-Glyceraldehyde
Two chiral centers: 2,3,4-trihydroxybutanal
There are four stereoisomers for this compound two pairs of enantiomers
Diastereomers
: Stereoisomers that are
not
Just like constitutional isomers, diastereomers also
differ in most chemical and physical properties.
They also have different boiling points and freezing
points.
In contrast, nearly all the properties of a pair of
enantiomers are the same
except
:
1.
Their interaction with other chiral substances
(receptors, solvents)
Enantiomers are optically active: Compounds
that rotate plane polarized light
Two Types:
▪ Dextrorotatory:
▪ Chiral compound that rotates light towards right (clockwise; +)
▪ Levorotatory:
▪ Chiral compound that rotates light towards left (counterclockwise; -)
▪ There is no correlation between D, L and +,
-▪ In D and L you need to look at the structure
Triose - 3 carbon atoms
Tetrose - 4 carbon atoms
Pentoses - 5 carbon atoms
Hexoses - 6 carbon atoms
Aldoses: monosaccharides with one aldehyde group
Ketoses: monosaccharides with one ketone group
Combined # of C atoms and functional group:
Example:
▪
Aldohexose:
Monosaccharide with
aldehyde
group and 6
C atoms
–
D-glucose
▪
Ketohexose:
Monosaccharide with
ketone
group and 6 C
atoms
–
D-fructose
CHO OH H H HO OH H OH H
CH2OH
CH2OH
O H HO OH H OH H
CH2OH
D-Glucose (aldohexose)
D-Fructose (ketohexose)
D-glyceraldehyde
Dihydroxyacetone
D-glucose
D-galactose
D-fructose
D-ribose
C-4 Isomers: Epimers
CHO OH H H HO OH H OH H
CH2OH D-Glucose (aldohexose) CHO OH H H HO H HO OH H
CH2OH
D-Galactose
(aldohexose)
CH2OH
O H HO OH H OH H
CH2OH
D-Fructose (ketohexose) CHO OH H OH H OH H
CH2OH
D-Ribose
Dominant form of monosaccharides with 5 or more C atoms is cyclic
- cyclic forms are in equilibrium with open chain form
Cyclic forms are formed by the reaction of carbonyl group (C=O)
with hydroxyl (-OH) group on carbon 5
β-form is cis (same side)
OH CH2OH CH2OH
OH
OH O
OH CH2OH
OH OH
O
-D-Fructose -D-Ribose
O
OH OH
OH OH
CH2OH
-D-Glucose
CH2OH group determines D or L series.
▪ CH2OH group pointed up = D series
▪ CH2OH group pointed down = L series
Any —OH group at a chiral center that is to the right in a Fischer
projection formula points down in the Haworth projection formula and any —OH group to the left in a Fischer projection formula
Five important reactions of monosaccharides:
▪ Oxidation to acidic sugars (addition of oxygen)
▪ Reduction to sugar alcohols (addition of hydrogen)
▪ Glycoside formation
▪ Phosphate ester formation
▪ Amino sugar formation
These reactions will be considered with respect to glucose.
CHO OH H H HO OH H OH H
CH2OH
COOH OH H H HO OH H OH H
CH2OH
Oxidation can yield three different types of acidic sugars
depending on the type of oxidizing agent used:
▪ Weak oxidizing agents such as Tollens and Benedict’s solutions oxidize
the aldehyde end to give an aldonic acid.
▪ A reducing sugar is a carbohydrate that gives a positive test with Tollens and Benedict’s solutions.
weak oxidizing
COOH OH H H HO OH H OH H COOH D-glucaric acid CHO OH H H HO OH H OH H
CH2OH
D-glucose
Strong oxidizing agents can oxidize both ends of a
monosaccharide at the same time to produce a
dicarboxylic acid:
▪ Such polyhydroxy dicarboxylic acids are known as aldaric acids.
strong oxidizing
CHO OH H H HO OH H OH H COOH D-gluronic acid
In biochemical systems enzymes can oxidize the
primary alcohol end of an aldose without
oxidation of the aldehyde group =
alduronic acid.
enzymes CHO OH H H HO OH H OH H
CH2OH
The carbonyl group in a monosaccharide (either an aldose or a
ketose) is reduced to a hydroxyl group using hydrogen as the reducing agent.
▪ The product is the corresponding polyhydroxy alcohol - sugar alcohol.
A glycoside is an acetal formed from a cyclic
monosaccharide by replacement of the hemiacetal
carbon
—
OH group with an
—
OR group:
▪ A glycoside produced from glucose - glucoside
▪ A glycoside produced from galactose – galactoside
▪ Glycosides exist in both Alpha and Beta forms
O OH OH OH OH OH
+ CH3CH2OH
O O OH OH OH OH
Phosphate ester formation: The hydroxyl groups of a
monosaccharide can react with inorganic oxyacids to
form inorganic esters.
Phosphate esters of various monosaccharides are stable
An amino sugar - one of the hydroxyl groups of a
monosaccharide is replaced with an amino group
In naturally occurring amino sugars the
carbon 2 hydroxyl
group is replaced by an amino group
• Glycoproteins are used as cell markers for the immune
system
• If incompatible blood is given:
• Causes blood cells to form
clumps (immune respone)
• Can be fatal.
Population Distributions
Type O is the universal
donor
▪ Has no extra sugars like other types.
▪ Cannot accept other types because they have extra sugars.
Type AB is the universal
acceptor
Two monosaccharides can react to form
disaccharide
One monosaccharide acts as a hemiacetal and
other as an alcohol
O OH OH OH OH OH + O O OH OH OH OH O OH OH OH OH OH O OH OH OH OH
Maltose is digested easily by humans because we have
enzymes that can break
α
(1-4) linkages but not
β
(1-4)
linkages of cellobiose. Therefore cellobiose cannot be
digested by humans.
O OH OH OH OH OH + O O OH OH OH OH O OH OH OH OH OH O OH OH OH OH
+ H2O
α (1→4) linkage = maltose O O OH OH OH OH O OH OH OH OH
+ H2O
Lactose - principal carbohydrate in milk.
▪ Human - 7%–8% lactose ▪ cow’s milk - 4%–5% lactose
▪ Lactose intolerance: a condition in which people lack the
enzyme lactase needed to hydrolyze lactose to galactose and glucose.
▪ Lactase hydrolyzes β (1-4) glycosidic linkages.
O O
OH OH OH
OH
O
OH
OH OH
Sucrose (table sugar): The
most abundant of all
disaccharides and found in plants.
It is 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
O
OH OH
OH OH
CH2OH
O OH
OH
CH2OH
CH2OH
OH O OH O OH OH
CH2OH
CH2OH
CH2OH
OH
OH O
-D-Glucose
-D-Fructose
+ (1-2)
Linkage
The Polymer Chain
Many monosaccharide
units bonded with
glycosidic linkages
Two types:
- Linear and branched,
homo- and
A storage form for monosaccharides used as an
energy source for cells
▪
Starch: plants, glucose is the monomeric unit,
α
(1
→
4)
and
α
(1
→
6) linkages
▪
Glycogen: humans and animals, more branched and
Serves as a structural element in plant cell walls and
animal exoskeletons
▪ Cellulose: unbranched, glucose polymer, β (1 4) glycosidic
bond.
▪ Chitin: N-acetyl amino derivative of D-glucose
Repeating
disaccharide
units containing an amino sugar
and negative charges due to the presence of sulfate or
carboxyl groups.
▪ Hyaluronic Acid: Highly viscous - serve as lubricants in the fluid of
joints and part vitreous humor of the eye.
Glycolipids
: a lipid molecule that has one or
more carbohydrate (or carbohydrate
derivative) units covalently bonded to it.
Write the products of the hydrolysis of
lactose
O O
OH OH OH
OH
O
OH
OH OH
OH
