ch 16 carbohydrates

49. Some athletes eat diets high in carbohydrates before an event. Suggest a biochemical basis for this practice.

Athletes try to increase their stores of glycogen before an event. The most direct way to increase the amount of this polymer of glucose is to eat carbohydrates.

29. What is the main structural difference between cellulose and starch?

Both cellulose and starch are polymers of glucose. In cellulose, the monomers are joined by a beta glycosidic linkage, whereas in starch they are joined by an alpha-glycosidic linkage

27. How does chitin differ from cellulose in structure and function?

Chitin is a polymer of N- acetyl-b-d-glucosamine, whereas cellulose is a polymer of d-glucose. Both polymers play a structural role, but chitin occurs in the exoskeletons of invertebrates and cellulose primarily in plants.

47. Why is the polysaccharide chitin a suitable material for the exoskeleton of invertebrates such as lobsters? What other sort of material can play a similar role?

Chitin is a suitable material for the exoskeleton of invertebrates because of its mechanical strength. Individual polymer strands are cross-linked by hydrogen bonding, accounting for the strength. Cellulose is another polysaccharide cross-linked in the same way, and it can play a similar role

6. Why are furanoses and pyranoses the most common cyclic forms of sugars?

Furanoses and pyranoses have five-membered and six-membered rings, respectively. rings of this size are the most stable and the most readily formed.

4.What is the difference between an enantiomer and a diastereomer?

Enantiomers are nonsuperimposable, mirror-image stereoisomers. Diastereomers are nonsuperimposable, nonmirror- image stereoisomers.

35. All naturally occurring polysaccharides have one terminal residue, which contains a free anomeric carbon. Why do these polysaccharides not give a positive chemical test for a reducing sugar?

The concentration of reducing groups is too small to detect.

39. No animal can digest cellulose. Reconcile this statement with the fact that many animals are herbivores that depend heavily on cellulose as a food source.

The digestive tract of these animals contains bacteria that have the enzyme to hydrolyze cellulose. (tricohelylase bacteria)

23. What is the metabolic basis for the observation that many adults cannot ingest large quantities of milk without developing gastric difficulties?

In some cases the enzyme is missing. In other cases, the enzyme isomerizes galactose to glucose for further metabolic breakdown.

26. What are some of the main differences between the cell walls of plants and those of bacteria?

The cell walls of plants consist mainly of cellulose, whereas those of bacteria consist mainly of polysaccharides with peptide crosslinks.

7. How many chiral centers are there in the open-chain form of glucose? In the cyclic form?

There are four chiral centers in the open-chain form of glucose (carbons two through five). Cyclization introduces another chiral center at the carbon involved in hemiacetal formation, giving a total of five chiral centers in the cyclic form.

17. What is unusual about the structure of N-acetylmuramic acid (Figure 16.18) compared with the structures of other carbohydrates?

This compound contains a lactic acid side chain.

12. How does the cyclization of sugars introduce a new chiral center?

Two different orientations with respect to the sugar ring are possible for the hydroxyl group at the anomeric carbon. The two possibilities give rise to the new chiral center.

22. Draw a Haworth projection for the disacchabride gentibiose, given the following information: (a) It is a dimer of glucose. (b) The glycosidic linkage is b(1 -->6). (c) The anomeric carbon not involved in the glycosidic linkage is in the alpha configuration.

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24. Draw Haworth projection formulas for dimers of glucose with the following types of glycosidic linkages: (a) A b(1--> 4) linkage (both molecules of glucose in the b form) (b) An a,a(1 -->> 1) linkage (c) A b(1 -->6) linkage (both molecules of glucose in the b form)

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8. Following are Fischer projections for a group of five-carbon sugars, all of which are aldopentoses. Identify the pairs that are enantiomers and the pairs that are epimers. (The sugars shown here are not all of the possible five-carbon sugars.) book

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13. Convert the following Haworth projections to a Fischer projection. Name the monosaccharides you have drawn.

book D-allose and D-altrose

14. Convert each of the following chair conformations to an open-chain from and to a Fischer projection. Name the monosaccharides you have drawn.

book D-allose and D-galactose

2. Name which, if any, of the following are epimers of d-glucose: d-mannose, d-galactose, d-ribose.

d-Mannose and d-galactose are both epimers of d-glucose, with inversion of configuration around carbon atoms 2 and 4, respectively; d-ribose has only five carbons, but the rest of the sugars named in this question have six.

An oligosaccharide

is a compound formed by the linking of several simple sugars (monosaccharides) by glycosidic bonds.

furanose

is a cyclic sugar that contains a five-membered ring similar to that in furan.

pyranose

is a cyclic sugar that contains a six-membered ring similar to that in pyran.

Polysaccharides

is a polymer of simple sugars, which are compounds that contain a single carbonyl group and several hydroxyl groups.

ketose

is a sugar that contains a ketone group

an aldose

is a sugar that contains an aldehyde group;

A glycosidic bond

is the acetal linkage that joins two sugars.

A glycoprotein

the covalent bonding of sugars to a protein.

43. How would you expect the active site of a cellulase to differ from the active site of an enzyme that degrades starch?

A cellulase (an enzyme that degrades cellulose) needs an active site that can recognize glucose residues joined in a beta-glycosidic linkage and hydrolyze that linkage. An enzyme that degrades starch has the same requirements with regard to glucose residues joined in an alpha-glycosidic linkage.

19. Define the term reducing sugar.

A reducing sugar is one that has a free aldehyde group. The aldehyde is easily oxidized, thus reducing the oxidizing agent.

38. Glycogen is highly branched. What advantage, if any, does this provide an animal?

Because of the branching, the glycogen molecule gives rise to a number of available glucose molecules at a time when it is being hydrolyzed to provide energy. A linear molecule could produce only one available glucose at a time.

11. Two sugars are epimers of each other. Is it possible to convert one to the other without breaking covalent bonds?

Converting a sugar to an epimer requires inversion of configuration at a chiral center. This can be done only by breaking and reforming covalent bonds.

34. Explain how the minor structural difference between alpha and beta-glucose is related to the differences in structure and function in the polymers formed from these two monomers.

Differences in structure: cellulose consists of linear fibers, but starch has a coil form. Differences in function: cellulose has a structural role, but starch is used for energy storage.

21. Name two differences between sucrose and lactose. Name two similarities.

Differences: sucrose contains fructose, whereas lactose contains galactose. Sucrose has an a,b(1-> 2) glycosidic linkage, whereas lactose has a b(1 ->4) glycosidic linkage. Similarities: sucrose and lactose are both disaccharides, and both contain glucose.

28. How does glycogen differ from starch in structure and function?

Glycogen and starch differ mainly in the degree of chain branching. Both polymers serve as vehicles for energy storage, glycogen in animals and starch in plants.

30. What is the main structural difference between glycogen and starch?

Glycogen exists as a highly branched polymer. Starch can have both a linear and a branched form, which is not as highly branched as that of glycogen.