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Lecture 4: Carbohydrates
Reference: Lieberman and Marks Chapter 5 and Lecture Notes
I. Explain the basic nomenclature of carbohydrates.
A. Carbohydrates are classified into three groups: monosaccharides (and their
derivatives), oligosaccharides, and polysaccharides
1. Monosaccharides (simple sugars) cannot be broken down into simpler
sugars under mild conditions
a.) Simple monosaccharides consist of a linear chain of three or
more carbon atoms one of which forms a double bond with oxygen
2. Oligosacchardies : oligo, meaning “few,” and consist of from two to ten
simple sugar molecules
3. Polysaccharides are polymers of the simple sugars – starch or glycogen
a.) Most sugars end with the suffix “–ose” : glucose
b.) Most sugars are in the D-configuration
-For sugars, the D or L designation refers to the highest
numbered chiral carbon atom
- D rotates the polarization around the chiral carbon
clockwise and L rotates the polarization counter-clockwise
II. Describe the basic structural features of carbohydrates in terms of aldose or ketose
grouping, numbers of carbons and stereochemistry
A. Aldose Grouping
-Contains an aldehyde functional group and if it contains 3C or more it is
B. Ketose Grouping
-Contains a ketone functional group and if it contains 4C or more it
is chiral
III. Draw the structures of dihydroxyacetone and D-glyceraldehyde (this will serve you
well during metabolism).
-See above
IV. Define the terms “pyranose” and “furanose”.
A. Pyranose: Cyclic form of aldose (but can also be a furanose)
B. Furanose: Cyclic form of ketose (but can also be a pyranose)
V. Explain ring formation by carbohydrates.
A. Identify the anomeric carbon in a ring structure.
-The “anomeric carbon” is the carbonyl carbon. Anytime an anomeric
carbon has an –OH group attached to it, it will be a free anomeric carbon. However, if
the –OH group is replaced by another atom (say C or N) or the –OH group is changed to
an –O-R group, then the anomeric carbon cannot reversilbly form the ring and linear
structures. In this case, the anomeric carbon is no longer “free”.
B. Identify whether the anomeric carbon is in an α versus β configuration and how
this occurred during ring formation.
Carbohydrate molecules can form cyclic structures when an –OH group attacks
the carbonyl carbon, forming a new covalent bond. Because of ring geometry and bond
strain, only 5-member (furanose) and 6-member (pryanose) rings will be formed. The
carbonyl carbon (the anomeric carbon) and its substituents form a planar bond structure
because of the double bonded C=O. Because of this, the attacking –OH can add to the
anomeric carbon from the top of the plane, causing the carbonyl oxygen to become an OH group that now points down below the original plane (this is the alpha configuration),
or the attacking –OH can add to the anomeric carbon from the bottom of the plane,
causing the carbonyl oxygen to become an -OH group that now points up above the
original plane (this is the beta configuration). There is an equal probability of the two
states forming so both alpha and beta forms will exist simultaneously
C. Explain how pyranose versus furanose rings are formed and which form will
dominate based on whether the carbohydrate is an aldose or ketose.
-Ring formation read above
- Bond strain, ring substitutions, etc, will favor one form over the other. Aldoses
“prefer” the pyranose form whereas ketoses “prefer” the furnanose form.
-Not exclusive, but preferred. These configurations are interchangeable
VI. Define the term “reducing sugar” and explain how a reducing sugar can reduce
another molecule or ion.
A. Reducing sugars are sugars with free anomeric carbons - they will reduce
oxidizing agents, such as peroxide, ferricyanide and some metals (Cu and Ag)
B. In glucose the aldehyde group contains significant electron density because of
the carbonyl group and the simple hydrogen atom attached to the carbonyl carbon.
Because of this, the aldehyde group can give an electron to another atom. In the process
the aldehyde gets oxidized (i.e. it loses an electron = oxidized) and in the process
becomes a carboxylic acid. The other atom gets reduced (i.e., gains an electron).
VII. List and describe some common monosaccharide derivatives.
Sugar alcohols: result from reduction of sugars
Deoxy sugars: constituents of DNA, etc.
Sugar esters: phosphate esters like ATP are important
Amino sugars contain an amino group in place of a hydroxyl group
Acetals, ketals and glycosides: basis for oligo- and poly-saccharides
VIII. Define the term “acetylated” amino sugar.
-The acetyl group is a carbonyl with a methyl group on it. -CO-CH3
IX. List the common amino sugar disaccharides found in glycosaminoglycans
X. Explain the basic structural features of oligosaccharides and the basis of their
A. A saccharide polymer containing a small number (typically two to ten[1]) of
component sugars, also known as simple sugars (monosaccharides)
B. Individual units are called “residues”
-be able to identify anomeric carbons and reducing and non-reducing ends.
-Note carefully the nomenclature of links! Be able to recognize alpha(1,4),
beta(1,4), etc
Naming of oligosaccharides proceeds from left to right (from the reducing
end to the terminal) In parentheses, between the names of the monosaccharide residues,
the number of the anomeric carbon atom, an arrow symbol, and the number of the carbon
atom bearing the connecting oxygen of the next monosaccharide unit are listed.
Appropriate symbols are used to indicate the stereochemistry of the glycosidic bonds (α
or β), the configuration of the monosaccharide residue (D orL), and the substitutions at
oxygen atoms
XI. Describe the glycosidic bond and define the α versus β configuration.
A. Glycosidic bonds join sugars
B. Alpha is below the plane of the ring and beta is above the plane
XII. Explain why sucrose does not have a reducing end, but lactose does.
A. Sucrose does not have a free anomeric carbon and therefore is unable to
participate in oxidation/reduction reactions. Lactose has an anomeric carbon on the
glucose end (This is known as the reducing end).
XIII. Explain that starch and glycogen are glucose storage systems, and that individual
glucose molecules can be removed as needed.
A. Starch (plants)
1. Two forms: amylose (linear chain) and amylopectin (branched
2. Most starch is 10-30% amylose and 70-90% amylopectin
B. Glycogen (animals)
1. Glycogen constitutes up to 10% of liver mass and 1-2% of muscle mass
2. Very similar to amylopectin: except much greater number of branches
C. Both are long glucose chains and the glucose molecules can be cleaved off and
used as needed
XIV. Explain that α-1,4-linked glucose is found in glycogen and that β-1,4-linked glucose
is found in cellulose. Explain some structural differences that arise from this difference.
The α-1,4-linked glucose is flexible and forms twisting branched chains that can
be easily broken down in the body. The β-1,4-linked glucose is stiff and forms strong
tightly packed sheets that become extended ribbons, and our bodies are unable to break
them down.