ch07_selProbs-key - Carbohydrates and Glycobiology 1 Sugar...

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Unformatted text preview: Carbohydrates and Glycobiology 1. Sugar Alcohols 1n the monosaccharide derivatives known as sugar alcohols, the carbonyl oxygen is reduced to a hydroxyl group. For example, D—glyceraldehyde can be reduced to glycerol. However, this sugar alcohol is no longer designated D or L. Why? Answer With reduction of the carbonyl oxygen to a Mdroxyl group, the stereochendstry at (3—1 and (3—3 is the same; the glycerol molecule is not chiral. 2. Recognizing Epimers Using Figure 17—3, identify the epiIners of (a) D-allose, (h) D—gulose, and (c) D-n‘hose at [3—2, ICE-3. and (3—4. Answer Epimers differ by the configuration about only one carbon. (:1) maltrose ((3—2). D—glucose [(3.3), ogulose [(3—4) (h) oidose ((3—2). D—galactose ((3—3), mallose ((3—4) (c) o—arabinose ((3-2), D—xylose [(3—3) 5. Configuration and Conformation Which bond(a) in a—o—gluooae must be broken to change its config- uration to {Lo—glucose? Which bond(s} to convert n—glucose to D—rnannose? Win-ch bond [5} to convert one “chair” form of Irv-glucose to the other? Answer 'Ib convert wry—glucose to {Lo—glucose, the bond between [5—1 and the hydroxafl on (3—5 must be broken and refijrmed in the opposite configuration [as in Fig. T—fi). 'Ib convert D—ghlcose to D—Inannoae, either the —H or the —DH on [3—2 must be broken and reformed in the opposite configuration Conversion between chair conformations does not require bond breakage; this is the critical distinction between configurafion and (Information. 8. Reducing Sugars Draw the structural formula for u—D-glucuyl-[l—}6)—D-Inarmosa.1njne and circle the part of this structure that makes the compound a reducing sugar. Answer GHBOH H H H HO OH H |—CH2 H 0H H 1—] OH OH reducing Ho H 3 g H H 9. Hemiacetal and Glycosidic Linkages Explain the difference between a heIniacetal and a glycoside. Answer A hemiaoetal is formed when an aldose or ketose condenses with an alcohol; a glyco— side is formed when a herniacetal condenses with an alcohol (see Fig. T—Er, p. 238). 21. 22. Growth Rate of Bamboo The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 113 mfday under optimal conditions. Given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth, calculate the number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate. Each o—glucose unit contributes «0.5 nm to the length of a cellulose molecule. Answer First, calculate the growth per second: as mfday (a4 hiday) (so IIIiIiJh}(E-D em) Given that each glucose residue increases the length of the cellulose chain by D.5 nm (5 X 10“” m), the number of residues added per second is s x 1e—6 ms 5 x 10—1” mfresidue = a x iii—Ems. = EJJDD residuesa's Glycogen as Energy Storage: How Long Can a Game Bird Fly? Since ancient times it has been observed that certain game birds, such as grouse, quail, and pheasants, are easily fatigued. The Greek historian Xenophon wrote, “The bustards. . . can be caught if one is quick in starting them up, for they will fly only a short distance, like partridges, and soon tire; and their flesh is delicious.“ The flight muscles of game birds rely almost entirely on the use of glucose 1—phosphate for energy, in the form of ATP (Chapter 14}. The glucose 1—phosphate is formed by the breakdown of stored muscle glyco— gen, catalyzed by the enzyme glycogen phosphorylase. The rate of ATP production is limited by the rate at which glycogen can be broken down During a “panic flight,“ the game bird‘s rate of glycogen breakdown is quite high, approximately 120 amoli'min of glucose 1-phosphate produced per gram of fresh tissue. Given that the flight muscles usually contain about 0.35% glycogen by weight, calculate how long a game bird can fly. (Assmne the average molecular weight of a glucose residue in glycogen is 162 gfmol.) Answer Given the average molecular weight of a glucose residue = 162, the amount of use able glucose (as glycogen) in 1 g of tissue is 3.5 x 10'3 g 162 gfmol In 1 min, 121] pmol of glucose 1—phosphate is produced, so 121] pmol of glucose is hydrolyzed. Thus, depletion of the glycogen would occur in [2.2 X 111‘5 mol) [6G stnJh) 120 x 10‘“ mot'rrun = 2.2 x 10—5 mol =11s ...
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