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Berg JM, Tymoczko JL, Stryer L. Biochemistry. 5th edition. New York: W H Freeman; 2002.

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Biochemistry. 5th edition.

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Problems

1. Word origin. Account for the origin of the term carbohydrate.

Carbohydrates were originally regarded as hydrates of carbon because the empirical formula of many of them is (CH2O)n.

2. Diversity. How many different oligosaccharides can be made by linking one glucose, one mannose, and one galactose? Assume that each sugar is in its pyranose form. Compare this number with the number of tripeptides that can be made from three different amino acids.

Three amino acids can be linked by peptide bonds in only six different ways. However, three different monosaccharides can be linked in a plethora of ways. The monosaccharides can be linked in a linear or branched manner, with α or β linkages, with bonds between C-1 and C-3, between C-1 and C-4, between C-1 and C-6, and so forth. Consequently, the number of possible trisaccharides greatly exceeds the number of tripeptides.

3. Couples. Indicate whether each of the following pairs of sugars consists of anomers, epimers, or an aldose-ketose pair:

(a) Aldose-ketose; (b) epimers; (c) aldose-ketose; (d) anomers; (e) aldose-ketose; (f) epimers.

(a) d-glyceraldehyde and dihydroxyacetone

(b) d-glucose and d-mannose

(c) d-glucose and d-fructose

(d) α-d-glucose and β-d-glucose

(e) d-ribose and d-ribulose

(f) d-galactose and d-glucose

4. Tollen's test. Glucose and other aldoses are oxidized by an aqueous solution of a silver-ammonia complex. What are the reaction products?

Aldoses are converted into aldonic acids; the aldehyde group of the sugar is oxidized to a carboxylate.

5. Mutarotation. The specific rotations of the α and β anomers of d-glucose are +112 degrees and +18.7 degrees, respectively. Specific rotation, [α]d, is defined as the observed rotation of light of wavelength 589 nm (the d line of a sodium lamp) passing through 10 cm of a 1 g ml-1 solution of a sample. When a crystalline sample of α-d-glucopyranose is dissolved in water, the specific rotation decreases from 112 degrees to an equilibrium value of 52.7 degrees. On the basis of this result, what are the proportions of the α and β anomers at equilibrium? Assume that the concentration of the open-chain form is negligible.

The proportion of the α-anomer is 0.36, and that of the β anomer is 0.64.

6. Telltale adduct. Glucose reacts slowly with hemoglobin and other proteins to form covalent compounds. Why is glucose reactive? What is the nature of the adduct formed?

Glucose is reactive because of the presence of an aldehyde group in its open-chain form. The aldehyde group slowly condenses with amino groups to form Schiff-base adducts.

7. Periodate cleavage. Compounds containing hydroxyl groups on adjacent carbon atoms undergo carbon-carbon bond cleavage when treated with periodate ion (IO4-). How can this reaction be used to distinguish between pyranosides and furanosides?

A pyranoside reacts with two molecules of periodate; formate is one of the products. A furanoside reacts with only one molecule of periodate; formate is not formed.

8. Oxygen source. Does the oxygen atom attached to C-1 in methyl α-d-glucopyranoside come from glucose or methanol?

From methanol.

9. Sugar lineup. Identify the following four sugars.

Image ch11fb1.jpg

(a) β-d-Mannose; (b) β-d-galactose; (c) β-d-fructose; (d) β-d-glucosamine.

10. Cellular glue. A trisaccharide unit of a cell-surface glycoprotein is postulated to play a critical role in mediating cell-cell adhesion in a particular tissue. Design a simple experiment to test this hypothesis.

The trisaccharide itself should be a competitive inhibitor of cell adhesion if the trisaccharide unit of the glycoprotein is critical for the interaction.

11. Mapping the molecule. Each of the hydroxyl groups of glucose can be methylated with reagents such as dimethylsulfate under basic conditions. Explain how exhaustive methylation followed by compete digestion of a known amount of glycogen would enable you to determine the number of branch points and reducing ends.

Reducing ends would form 1,2,3,6-tetramethylglucose. The branch points would yield 2,3-dimethylglucose. The remainder of the molecule would yield 2,3,6-trimethylglucose.

12. Component parts. Raffinose is a trisaccharide and a minor constituent in sugar beets.

(a) Not a reducing sugar. No open-chain forms are possible. (b) d-Galactose, d-glucose, d-fructose. (c) d-Galactose and sucrose (glucose = fructose).

(a) Is raffinose a reducing sugar? Explain.

(b) What are the monosaccharides that compose raffinose?

(c) β-Galactosidase is an enzyme that will remove galactose residues from an oligosaccharide. What are the products of β-galactosidase treatment of raffinose?

Image ch11fb2.jpg

13. Anomeric differences. α-d-Mannose is a sweet-tasting sugar. β-d-Mannose, on the other hand, tastes bitter. A pure solution of α-d-mannose loses its sweet taste with time as it is converted into the β anomer. Draw the β anomer and explain how it is formed from the α anomer.

Image ch11fb3.jpg

Image app5fb18.jpg

The hemiketal linkage of the α anomer is broken to form the open form. Rotation about the C-1 and C-2 bonds allows the formation of the β anomer, and a mixture of isomers results.

14. A taste of honey. Fructose in its β-d-pyranose form accounts for the powerful sweetness of honey. The β-d-furanose form, although sweet, is not as sweet as the pyranose form. The furanose form is the more stable form. Draw the two forms and explain why it may not always be wise to cook with honey.

Heating converts the very sweet pyranose form to the more stable but less sweet furanose form. Consequently, it is difficult to accurately control the sweetness of the preparation, which also accounts for why honey loses sweetness with time. See Figure 11.6 for structures.

15. Making ends meet. (a) Compare the number of reducing ends to nonreducing ends in a molecule of glycogen. (b) As we will see in Chapter 21, glycogen is an important fuel storage form that is rapidly mobilized. At which end—the reducing or nonreducing—would you expect most metabolism to take place?

(a) Each glycogen molecule has one reducing end, whereas the number of nonreducing ends is determined by the number of branches, or α-1,6 linkages. (b) Because the number of nonreducing ends greatly exceeds the number of reducing ends in a collection of glycogen molecules, all of the degradation and synthesis of glycogen takes place at the nonreducing ends, thus maximizing the rate of degradation and synthesis.

16. Carbohydrates and proteomics. Suppose that a protein contains six potential N-linked glycosylation sites. How many possible proteins can be generated, depending on which of these sites is actually glycosylated? Do not include the effects of diversity within the carbohydrate added.

64. Each site either is or is not glycosylated so that there are 26 = 64 possible proteins.

Chapter Integration Problem

17. Stereospecificity. Sucrose, a major product of photosynthesis in green leaves, is synthesized by a battery of enzymes. The substrates for sucrose synthesis, d-glucose and d-fructose, are a mixture of α and β anomers as well as acyclic compounds in solution. Nonetheless, sucrose consists of α-d-glucose linked by its carbon-1 atom to the carbon-2 atom of β-d-fructose. How can the specificity of sucrose be explained in light of the potential substrates?

As discussed in Chapter 9, many enzymes display stereo-specificity. Clearly, the enzymes of sucrose synthesis are able to distinguish between the isomers of the substrates and link only the correct pair.

Image ch11f6

By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.

Copyright © 2002, W. H. Freeman and Company.
Bookshelf ID: NBK22366

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