Chapter 7 Questions
Is D-2-deoxygalactose the same chemical as D-2-deoxyglucose? Explain.
Glucose and galactose have different conformations at C4 so they are not the same chemically
16. Cellulose could provide a widely available and cheap form of glucose, but humans cannot digest it. Why not? If you were offered a procedure that allowed you to acquire this ability, would you accept? Why or why not?
Humans cannot digest cellulose because they do not have the enzyme cellulase which breaks it down into glucose. I do not think I would accept a procedure that allowed me to acquire this ability because with this capability, I would also need the larger teeth of animals like cows and sheep to break down the material. However, this would be a good skill to have if you were trying to survive in the wild as there would be food sources readily available.
19. 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. The glucose 1-phosophate is formed by the breakdown of stored muscle glycogen at which glycogen can be broken down. During a "panic attack," the game bird's rate of glycogen breakdown is quite high, approximately 120 μmol/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. (Assume the average molecular weight of a glucose residue in glycogen is 162 g/mol.)
Knowns: rate= 120 umol/min 0.0035 glycogen in muscle MW of glycogen= 162 g/mol ?t bird can fly 1. Find moles of glycogen 3.5x10^-3 g/162g/mol= 2.2x10^-5 moles 2. Convert glycogen breakdown rate 120x10^-6 mol/min x 1 min/60 sec = 2x10^-6 mol/ second 2.2x10^-5 mol/ 2x10^-6 moles/ second= 11 seconds for glycogen breakdown
The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day 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 D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule.
Rate= 3x10^-6 m/s 5x10^-10 m/ residues 1 residue/ 5x10^-10 m x 3x10^-6m/ 1 sec= 6,000 residues per second
· juice or other citrus juices or cream of tartar. 9. The manufacture of chocolates containing a liquid center is an interesting application of enzyme engineering. The flavored liquid center consists largely of an aqueous solution of sugars rich in fructose to provide sweetness. The technical dilemma is the following: the chocolate coating must be prepared by pouring hot melted chocolate over a solid (or almost solid) core, yet the final product must have a liquid, fructose-rich center. Suggest a way to solve this problem. (Hint: sucrose is much less soluble than a mixture of glucose and fructose).
The core should be a semisolid mixture of sucrose and water. Then, the enzyme invertase can be added to the interior mixture, liquefying it. At the same time, quickly coat the exterior with chocolate. Thus, after the exterior chocolate has cooled, the interior will be a liquid center, hydrolyzed by sucrase to form a mixture of fructose, glucose, and sucrose
23. Heparin, a highly negatively charged glycosaminoglycan, is used clinically as an anticoagulant. It acts by binding several plasma proteins, including antithrombin III, an inhibitor of blood clotting. The 1:1 binding of heparin to antithrombin III seems to cause a conformation change in the protein that greatly increases its ability to inhibit clotting. What amino acid residues of antithrombin III are likely to interact with heparin?
The positively charged amino acid (lysine, arginine, and histidine) residues are the most likely to interact with heparin as it is negatively charged.
The almost pure cellulose obtained from the seed threads of Gossypium (cotton) is tough, fibrous, and completely insoluble in water. In contrast, glycogen obtained from muscle or liver disperses readily in hot water to make a turbid solution. Despite their markedly different physical properties, both substances are (1à4)-linked D-glucose polymers of comparable molecular weight. What structural features of these two polysaccharides underlie their physical properties? Explain the biological advantages of their respective properties.
While both substances are linked (1à4), cellulose is linked in a β conformation and glycogen is linked in an α conformation. The β conformation of cellulose forces the structure into extended chains which are linked parallelly by hydrogen bonding between fibers. The linearity of cellulose allows it aggregate and form tough, hydrophobic fibers. Thus, these fibers are perfectly designed for a structural role in plants. · Glycogen is linked through an α conformation, which causes bends in the chain. This causes helical formation instead of long chains; however, the helix is still formed and stabilized through hydrogen bonding. Furthermore, because glycogen is highly branched, its hydroxyls are more exposed and more available to bond to water. Thus, glycogen is highly water soluble and can be separated out in hot water. Glycogen acts, not as structural support, but as storage fuel. Because they are hydrated, hydrogen bonded to a lot water, and have many exposed nonreducing ends, they can be easily broken down and used for energy.
The fructose in honey is mainly in the β-D-pyranose form. This is one of the sweetest carbohydrates known, about twice as sweet as glucose; the β-D-pyranose form of fructose is much less sweet. The sweetness of honey gradually decreases at a high temperature. Also, high-fructose corn syrup (a commercial product in which much of the glucose in corn syrup is converted to fructose) is used for sweetening cold but not hot drinks. What chemical property of fructose could account for both of these observations?
· Fructose can cyclize to form pyranose or furanose, and increasing the temperature increases the direction toward the furanose form, which is less sweet. Thus, the higher the temperature, the less sweet as there will be more furanose over pyranose.
9. The enzyme glucose oxidase isolated from the mold Penicillium notatum catalyzes the oxidation of β-D-glucose to D-glucono-δ-lactone. This enzyme is highly specific for the β anomer of glucose and does not affect the α anomer. In spite of this specificity, the reaction catalyzed by glucose oxidase is commonly used in a clinical assay for total blood glucose- that is, for solutions consisting of a mixture of β- and α-D-glucose. What are the circumstances required to make this possible? Aside from following the detection of smaller quantities of glucose, what advantage does glucose oxidase offer over Fehling's reagent for measuring blood glucose?
· Mutorotation, conversion between the two anomers, is extremely high. Thus, as more of the β anomer is consumed, the α-D-glucose is converted to the β -D-glucose and eventually all of the mixture is oxidized. Fehling's reagent is helpful in detecting all the reducing sugars, whereas glucose oxidase is specific just for glucose.
9. As sweet as sucrose is, an equimolar mixture of its constituent monosaccharides, D-glucose, and D-fructose, is sweeter. Besides enhancing sweetness, fructose has hydroscopic properties that improve the texture of foods, reducing crystallization and increasing moisture. In the food industry, hydrolyzed sucrose is called invert sugar, and the yeast enzyme that hydrolyzes it is called invertase. The hydrolysis reaction is generally monitored by measuring the specific rotation of the solution, which is positive (+66.4o) for sucrose but becomes negative (inverts) as more D-glucose (+52.7o) and D-fructose (-92o) form. From what you know about the chemistry of the glycosidic bond, how would you hydrolyze sucrose to invert sugar nonenzymatically in a home kitchen?
· Sugar could by hydrolyzed nonenzymatically through heating as the glycosidic bond is sensitive to heat. Additionally, the sugar could be hydrolyzed by acidification as the bond is also sensitive to acid. Some home acids include lemon juice or other citrus juices or cream of tartar.
Which bond(s) in α D-glucose must be broken to change its configuration to β D-glucose? Which bond(s) to convert D-glucose to D-mannose? Which bond(s) to convert one "chair" form of D-glucose to other?
· To go from an alpha D-glucose to a beta-D glucose, the bond formed between the C-1 and the hydroxyl on C-5, needs to be broken and reformed oppositely, so that the hydroxyl now points away from C-1 and the hydrogen points down. This is change in configuration. · To convert D-glucose to D-mannose, the hydroxyl or hydrogen substituents on C-2 must be broken and reformed again in the opposite direction. This is change in configuration. To convert between chair conformations, bond breaking is not needed. This is a change in conformation, not configuration
