Chapter 10

Question 1

What can carbohydrates be covalently linked with?

Answer 1

Proteins and lipids.

Question 2

What functions do carbohydrates fulfill?

Answer 2

Energy source and energy storage.
Structural component of cell walls and exoskeletons.
Informational molecules in cell-cell signaling.
Cellular identification.

Question 3

What are monosaccharides?

Answer 3

Aldehydes or ketones that contain 2 or more alcohol groups.
Smallest monosaccharides are composed of 3 carbons.

Question 4

How are carbohydrates named?

Answer 4

Basic nomenclature:
- Number of carbon atoms in the carbohydrate + -ose.
- Example: three carbons = triose.
Common functional groups:
- All carbohydrates initially had a carbonyl functional group.
- Aldehydes = aldose.
- Ketones = ketose.

Question 5

What are the isomeric forms that monosaccharides exist in?

Answer 5

Isomers: have the same molecular formula but different structures.
- Constitutional isomers: differ in the order of attachment of atoms.
- Stereoisomers: atoms are connected in the same order but differ in spatial arrangement.
— Enantiomers: nonsuperimposable mirror images.
— Diastereoisomers: isomers that are not mirror images.
Epimers: Differ at one of several asymmetric carbon atoms.
- Anomers: isomers that differ at a new asymmetric carbon atom formed on ring closure.
- Diastereoisomers.

Question 6

What are the common carbohydrates in biochemistry?

Answer 6

Ribose: standard five-carbon sugar.
Glucose: standard six-carbon sugar.
Galactose: epimer of glucose.
Mannose: epimer of glucose.
Fructose: ketose form of glucose.

Question 7

How can carbohydrates be stereoisomers?

Answer 7

Enantiomers: stereoisomers that are nonsuperimposable mirror images.
In sugars that contain many chiral centers, only the one that is most distant from the carbonyl carbon is designated as D (right) or L (left).
D and L isomers of a sugar are enantiomers.
Most hexoses in living organisms are D stereoisomers.
Some simple sugars occur in the L form, such as L-arabinose.

Question 8

What are epimers?

Answer 8

D-Mannose and D-galactose are both epimers of D-glucose.
Each epimer differs from D-glucose in the configuration at one chiral center.
D-Mannose and D-galactose vary at more than one chiral center and are diastereomers, but not epimers.

Question 9

Why do many common sugars exist in cyclic forms?

Answer 9

Chemical basis for ring formation is that an aldehyde can react with an alcohol to form a hemiacetal, whereas a ketone can react with an alcohol to form a hemiketal.

Question 10

Explain the cyclization of monosaccharides.

Answer 10

Pentoses and hexoses readily undergo intramolecular cyclization.
Former carbonyl carbon becomes a new chiral center, called the anomeric carbon.

Question 11

Explain the anomeric carbon.

Answer 11

When the former carbonyl oxygen becomes a hydroxyl group, the position of this group determines if the anomer is alpha or beta.
If the hydroxyl group is on the opposite side (trans) of the ring as the CH2OH moiety, the configuration is alpha.
If the hydroxyl group is on the same side (cis) of the ring as the CH2OH moiety, the configuration is beta.

Question 12

Explain the cyclization of monosaccharides for pyranoses and furanoses.

Answer 12

Six-membered oxygen-containing rings are called pyranoses after the pyran ring structure.
Five-membered oxygen-containing rings are called furanoses after the furan ring structure.
The anomeric carbon is usually drawn on the right side.

Question 13

Explain fructose.

Answer 13

Fructose forms both the pyranose form, which predominates when fructose is free in solution, and a furanose form, commonly seen in fructose derivatives.

Question 14

How are sugars reduced?

Answer 14

A solution of glucose contains one-third alpha anomer, two-thirds beta anomer, and about 1% open chain.
Because the two anomeric forms are in equilibrium that passes through an open chain form, the free open chain form reacts with oxidizing agents.
Sugars that react with oxidizing agents are called reducing sugars, whereas those that do not are called nonreducing sugars.

Question 15

Explain how monosaccharides are joined to alcohols and amines through glycosidic bonds.

Answer 15

Bond formed between the anomeric carbon atom and a hydroxyl group of another molecule is called an o-glycosidic bond, and the product is called a glycoside.
Bond formed between the anomeric carbon atom and an amine is called an N-glycosidic bond.
Carbohydrates also form ester linkages to phosphates.
Phosphorylation of sugars occurs routinely during metabolism.

Question 16

Explain how sucrose, lactose, and maltose are the common disaccharides.

Answer 16

Enzymes on the outer surface of intestinal epithelium cleave common disaccharides.
Sucrase cleaves sucrose (table sugar), lactase cleaves lactose (milk sugar), and maltase cleaves maltose.

Question 17

Talk about oligosaccharides.

Answer 17

5 - 15 units typically.
Not commonly found free in cells.
Usually attached to proteins.
Protein with small oligosaccharides attached is called glycoprotein.
About half of mammalian proteins are glycoproteins.
Carbohydrate is attached to amino acids on the protein via its anomeric carbon.
Attachments to proteins may be N-linked (through asparagine) or O-linked (through serine or threonine).
O-linked sugars are added in Golgi; attachment of N-linked sugars starts in ER and then completed in the Golgi.
Function as markers (identity markers, destination markers, etc.)

Question 18

Talk about polysaccharides.

Answer 18

Natural carbohydrates are usually found as polymers.
These polysaccharides can be:
- Homopolysaccharides (one monomer unit).
- Heteropolysaccharides (multiple monomer units).
- Linear (one type of glycosidic bond).
- Branched (multiple types of glycosidic bonds).
Polysaccharides do not have a defined molecular weight.
- This is in contrast to proteins because, unlike proteins, no template is used to make polysaccharides.
- Polysaccharides are often in a state of flux; monomer units are added and removed as needed by the organism.

Question 19

Explain homopolymers of glucose: glycogen.

Answer 19

Glycogen is a branched homopolysaccharide of glucose.
- Glucose monomers form (alpha1 –> 4) linked chains.
- There are branch points with (alpha1 –> 6) linkers every 8 - 12 residues.
- Molecular weight reaches several millions.
- Functions as the main storage polysaccharide in animals.
- Found mostly in muscles and liver.

Question 20

Explain homopolymers of glucose: starch.

Answer 20

Starch is a mixture of two homopolysaccharides of glucose.
- Amylose is an unbranched polymer of (alpha1 –> 4) linked residues.
- Amylopectin is branched like glycogen, but the branch points with (alpha1 –> 6) linkers occur every 24 - 30 residues.
- Molecular weight of amylopectin is up to 200 million.
Starch is the main storage polysaccharide in plants.

Question 21

Explain the metabolism of glycogen and starch.

Answer 21

Glycogen and starch are insoluble due to their high molecular weight and often form granules in cells.
Granules contain enzymes that synthesize and degrade these polymers.
Glycogen and amylopectin have one reducing end but many nonreducing ends.
Enzymatic processing occurs simultaneously in many nonreducing ends.

Question 22

Explain homopolymers of glucose: cellulose.

Answer 22

Cellulose is a linear homopolysaccharide of glucose.
- Glucose monomers form (beta1 –> 4) linked chains.
- Structure is tough and water insoluble.
- Most abundant polysaccharide in nature.
- Cotton is nearly pure fibrous cellulose.

Question 23

Explain cellulose metabolism.

Answer 23

Fibrous structure and water insolubility make cellulose a difficult substrate to act upon.
Most animals cannot use cellulose as a fuel source because they lack the enzyme to hydrolyze (beta1 –> 4) linkages.
Fungi, bacteria, and protozoa secrete cellulase, which allows them to use wood as source of glucose.
Ruminants and terminates live symbiotically with microorganisms that produce cellulase and are able to absorb the free glucose into their bloodstreams.
Beta-1,4 linkages favor straight chains, which are optimal for structural purposes.
Alpha-1,4 linkages favor bent structures, which are more suitable for storage.

Question 24

Explain how bacterial cell walls consist of peptidoglycans.

Answer 24

Linear polysaccharide chains are cross-linked by short peptides, pentaglycines and tetrapeptides.
Bacterial cell walls contain D- amino acids. The cross-links made of them are made by a mechanism different from the regular protein synthesis.
Enzyme involved in the formation of cross-links is Glycopeptide transpeptidase.
Antibiotic Penicillin reacts with the active center of transpeptidase and form an inactive complex, which is indefinitely stable. Penicillin is irreversible suicide inhibitor of transpeptidase.

Question 25

Explain how chitin is a homopolysaccharide.

Answer 25

Chitin is a linear homopolysaccharide of N-acetylglucosamine.
N-acetylglucosamine monomers form (beta1 –> 4)- linked chains.
Forms extended fibers that are similar to those of cellulose.
Found in cell walls in mushrooms and in exoskeletons of insects, spiders, crabs, and other arthropods.