Stoker7e_PPT_CH18.pptx about carbohydrates

Carbohydrates

1. Copyright ©2016 Cengage Learning. All Rights Reserved. 1

2. Copyright ©2016 Cengage Learning. All Rights Reserved. 2 18.1 Biochemistry—An overview 18.2 Occurrence and functions of carbohydrates 18.3 Classification of carbohydrates 18.4 Chirality: Handedness in molecules 18.5 Stereoisomerism: Enantiomers and diastereomers 18.6 Designating handedness using Fischer projection formulas 18.7 Properties of enantiomers 18.8 Classification of monosaccharides 18.9 Biochemically important monosaccharides 18.10 Cyclic forms of monosaccharides 18.11 Haworth projection formulas 18.12 Reactions of monosaccharides 18.13 Disaccharides 18.14 Oligosaccharides 18.15 General characteristics of polysaccharides 18.16 Storage polysaccharides 18.17 Structural polysaccharides 18.18 Acidic polysaccharides 18.19 Dietary considerations and carbohydrates 18.20 Glycolipids and glycoproteins: Cell recognition

3. Copyright ©2016 Cengage Learning. All Rights Reserved. 3 The Study of Living Things Biochemistry : Study of the chemical substances found in living organisms and the chemical interactions of these substances with each other Biochemical substance : Chemical substance found within a living organism Types of biochemical substances: Bioinorganic substances - Water and inorganic salts Bioorganic substances - Carbohydrates, lipids, proteins, and nucleic acids

4. Copyright ©2016 Cengage Learning. All Rights Reserved. 4 As isolated compounds, bioinorganic and bioorganic substances have no life in and of themselves When these substances are gathered together in a cell, their chemical interactions are able to sustain life Bioinorganic and Bioorganic Substances

5. Biochemistry, the study of the chemical substances of living organisms, includes the study of: carbohydrates, lipids, proteins, and salts. carbohydrates, lipids, proteins, and organic salts. carbohydrates, lipids, proteins, and nucleic acids. carbohydrates, lipids, and proteins. Copyright ©2016 Cengage Learning. All Rights Reserved. 5

6. Biochemistry, the study of the chemical substances of living organisms, includes the study of: carbohydrates, lipids, proteins, and salts. carbohydrates, lipids, proteins, and organic salts. carbohydrates, lipids, proteins, and nucleic acids. carbohydrates, lipids, and proteins. Copyright ©2016 Cengage Learning. All Rights Reserved. 6

7. Plants and Carbohydrates It is estimated that more than half of all organic carbon atoms are found in the carbohydrate materials of plants Human uses for carbohydrates of the plant kingdom extend beyond food Carbohydrates in the form of cotton and linen are used as clothing Carbohydrates in the form of wood are used for shelter and heating and in making paper Copyright ©2016 Cengage Learning. All Rights Reserved. 7

8. Copyright ©2016 Cengage Learning. All Rights Reserved. 8 Most of the matter in plants, except water, is carbohydrate material Carbohydrates account for 75% of dry plant material and are produced by photosynthesis Cellulose - Structural element Starch/glycogen - Energy reservoir Carbohydrates

9. Photosynthesis Process in which plants produce carbohydrates using carbon dioxide, water, and solar energy Copyright ©2016 Cengage Learning. All Rights Reserved. 9

10. Functions of Carbohydrates in the Human Body Carbohydrate oxidation provides energy Carbohydrate storage, in the form of glycogen, provides a short-term energy reserve Carbohydrates supply carbon atoms for the synthesis of other biochemical substances (proteins, lipids, and nucleic acids) Carbohydrates form part of the structural framework of DNA and RNA molecules Copyright ©2016 Cengage Learning. All Rights Reserved. 10

11. Functions of Carbohydrates in the Human Body Carbohydrates linked to lipids are structural components of cell membranes Carbohydrates linked to proteins function in a variety of cell–cell and cell–molecule recognition processes Copyright ©2016 Cengage Learning. All Rights Reserved. 11

12. Identify the functions of carbohydrates in the human body. They provide energy during their oxidation and a short-term energy reserve. They form part of the framework for nucleic acids and supply carbon atoms for synthesis of other biomolecules. When linked to proteins, they are involved in the cell–cell and cell–molecule recognition process, and they function as structural components of cell membranes when linked to lipids. All the above. Copyright ©2016 Cengage Learning. All Rights Reserved. 12

13. Identify the functions of carbohydrates in the human body. They provide energy during their oxidation and a short-term energy reserve. They form part of the framework for nucleic acids and supply carbon atoms for synthesis of other biomolecules. When linked to proteins, they are involved in the cell–cell and cell–molecule recognition process, and they function as structural components of cell membranes when linked to lipids. All the above. Copyright ©2016 Cengage Learning. All Rights Reserved. 13

14. Empirical formula of simple carbohydrates - C n H 2n O n or C n (H 2 O) n (hydrate of C) n is the number of atoms Carbohydrate : Polyhydroxy aldehyde, ketone, or a compound that produces such substances upon hydrolysis Copyright ©2016 Cengage Learning. All Rights Reserved. 14

15. Monosaccharides Contain single polyhydroxy aldehyde or ketone unit Cannot be broken down into simpler substances by hydrolysis reactions Contain 3–7 C atoms 5 and 6 carbon species are more common Pure monosaccharides - Water soluble, white crystalline solids Monosaccharides - Glucose and fructose Copyright ©2016 Cengage Learning. All Rights Reserved. 15

16. Disaccharides Contain 2 monosaccharide units covalently bonded to each other Crystalline and water soluble substances Common disaccharides - Table sugar (sucrose) and milk sugar (lactose) Upon hydrolysis, they produce 2 monosaccharide units Copyright ©2016 Cengage Learning. All Rights Reserved. 16

17. Oligosaccharides Contain three to ten monosaccharide units covalently bonded to each other Free oligosaccharides are seldom encountered in biochemical systems Usually found associated with proteins and lipids in complex molecules Serve structural and regulatory functions Copyright ©2016 Cengage Learning. All Rights Reserved. 17

18. Polysaccharides Contain many monosaccharide units covalently bonded Number of monosaccharide units varies from a few 100 units to 50,000 units Examples: Cellulose - Paper, cotton, wood Starch - Bread, pasta, potatoes, rice, corn, beans, and peas Copyright ©2016 Cengage Learning. All Rights Reserved. 18

19. Carbohydrates are classified as: monosaccharides, disaccharides, oligosaccharides, and polysaccharides. polyhydroxy aldehydes or ketones. monosaccharides and disaccharides. monosaccharides, oligosaccharides, polysaccharides, and polyhydroxy aldehydes or ketones. Copyright ©2016 Cengage Learning. All Rights Reserved. 19

20. Carbohydrates are classified as: monosaccharides, disaccharides, oligosaccharides, and polysaccharides. polyhydroxy aldehydes or ketones. monosaccharides and disaccharides. monosaccharides, oligosaccharides, polysaccharides, and polyhydroxy aldehydes or ketones. Copyright ©2016 Cengage Learning. All Rights Reserved. 20

21. Objects and Handedness Most biological molecules, including carbohydrates, exhibit the property of “handedness” (form of isomerism) Most molecules that possess “handedness” exist in two forms: “Left-handed” form “Right-handed” form Related in the same manner as two hands that are “mirror images” of each other Copyright ©2016 Cengage Learning. All Rights Reserved. 21

22. Mirror Images Mirror image : Reflection of an object in a mirror Classes of objects based on mirror images - Superimposable mirror images : Images that coincide at all points when the images are laid upon each other Ex. Dinner plate with no design Achiral molecule Nonsuperimposable mirror images : Images where not all points coincide when the images are laid upon each other Ex. Human hands Chiral molecule (handedness) Copyright ©2016 Cengage Learning. All Rights Reserved. 22

23. Copyright ©2016 Cengage Learning. All Rights Reserved. 23 Chirality Chiral center : C atom attached to 4 different groups A molecule with chiral center is said to be chiral A C atom must have four different groups attached to it in order to be a chiral center A chiral C is usually denoted by * Bromochloroiodomethane is a chiral organic molecule

24. Practice Exercise Indicate whether the circled carbon atom in each of the following molecules is a chiral center. Copyright ©2016 Cengage Learning. All Rights Reserved. 24

25. Practice Exercise Indicate whether the circled carbon atom in each of the following molecules is a chiral center. Copyright ©2016 Cengage Learning. All Rights Reserved. 25 a. Not a chiral center b. Not a chiral center c. Chiral center d. Not a chiral center

26. Responses of Left and Right Handed Forms of a Molecule in a Human Body Both forms may be active, one may be more active, or one may be active and other non-active Example: Response of the body to the right-handed hormone epinephrine is 20 times greater than responses to the left-handed form Almost all monosaccharides are right handed Amino acids are always left handed Copyright ©2016 Cengage Learning. All Rights Reserved. 26

27. A carbohydrate molecule whose mirror image is not superimposable is said to be a _____ molecule. achiral chiral nonidentical nonsuperimposable Copyright ©2016 Cengage Learning. All Rights Reserved. 27

28. A carbohydrate molecule whose mirror image is not superimposable is said to be a _____ molecule. achiral chiral nonidentical nonsuperimposable Copyright ©2016 Cengage Learning. All Rights Reserved. 28

29. Copyright ©2016 Cengage Learning. All Rights Reserved. 29 Stereoisomers Isomers that have the same molecular and structural formulas but differ in the orientation of atoms in space Two types: Enantiomers : Stereoisomers whose molecules are non-superimposable mirror images of each other Molecules with chiral center Diastereomers : Stereoisomers whose molecules are not mirror images of each other Example: Cis-trans isomers

30. What feature does a carbohydrate possess that generates stereoisomerism? Presence of a chiral center Presence of structural rigidity Presence of three or more carbon atoms Presence of an achiral center Copyright ©2016 Cengage Learning. All Rights Reserved. 30

31. What feature does a carbohydrate possess that generates stereoisomerism? Presence of a chiral center Presence of structural rigidity Presence of three or more carbon atoms Presence of an achiral center Copyright ©2016 Cengage Learning. All Rights Reserved. 31

32. Copyright ©2016 Cengage Learning. All Rights Reserved. 32 Fischer Project Formula Two-dimensional structural notation for showing the spatial arrangement of groups about chiral centers in molecules According to this formula, a chiral center is represented as the intersection of vertical and horizontal lines Functional groups of high priority will be written at the top

33. Copyright ©2016 Cengage Learning. All Rights Reserved. 33 Tetrahedral Arrangements The four groups attached to the atom at the chiral center assume a tetrahedral geometry governed by the following conventions: Vertical lines from the chiral center represent bonds to groups directed into the printed page Horizontal lines from the chiral center represent bonds to groups directed out of the printed page

34. Copyright ©2016 Cengage Learning. All Rights Reserved. 34 Fischer Project Formulas L and D system used to designate the handedness of glyceraldehyde enantiomers are shown below

35. Copyright ©2016 Cengage Learning. All Rights Reserved. 35 Fischer Project Formulas We now consider Fischer projection formulas for the compound 2,3,4-trihydroxybutanal, a monosaccharide with four carbons and two chiral centers There are four stereoisomers for this compound and two pairs of enantiomers

36. Copyright ©2016 Cengage Learning. All Rights Reserved. 36 Fischer Project Formulas The D,L system used to designate the handedness of glyceraldehyde enantiomers can be extended to other monosaccharides with more than one chiral center The carbon chain is numbered starting at the carbonyl group end of the molecule, and the highest-numbered chiral center is used to determine D or L configuration Epimers : Diastereomers whose molecules differ only in the configuration at one chiral center

37. What formula is used to show the two-dimensional structure of groups around chiral centers in a molecule? Fischer projection Haworth projection Line angle Expanded Copyright ©2016 Cengage Learning. All Rights Reserved. 37

38. Next meeting Copyright ©2016 Cengage Learning. All Rights Reserved. 38

39. What formula is used to show the two-dimensional structure of groups around chiral centers in a molecule? Fischer projection Haworth projection Line angle Expanded Copyright ©2016 Cengage Learning. All Rights Reserved. 39

40. How can you identify the D isomer of a carbohydrate by looking at the Fischer projection formula? The –OH group on the chiral carbon farthest from the carbonyl group is pointing to the right. The –OH group on the chiral carbon farthest from the carbonyl group is pointing to the left. The –OH group on the chiral carbon closest to the carbonyl group is pointing to the right. The –OH group on the chiral carbon closest to the carbonyl group is pointing to the right. Copyright ©2016 Cengage Learning. All Rights Reserved. 40

41. How can you identify the D isomer of a carbohydrate by looking at the Fischer projection formula? The –OH group on the chiral carbon farthest from the carbonyl group is pointing to the right. The –OH group on the chiral carbon farthest from the carbonyl group is pointing to the left. The –OH group on the chiral carbon closest to the carbonyl group is pointing to the right. The –OH group on the chiral carbon closest to the carbonyl group is pointing to the right. Copyright ©2016 Cengage Learning. All Rights Reserved. 41

42. Copyright ©2016 Cengage Learning. All Rights Reserved. 42 Constitutional Isomers and Diastereomers Constitutional isomers differ in most chemical and physical properties Have different boiling points and melting points Diastereomers also differ in most chemical and physical properties Have different boiling points and freezing points In contrast, nearly all the properties of a pair of enantiomers are the same Two differences: Their interaction with plane polarized light Their interaction with other chiral substances

43. Interaction of Enantiomers with Plane-Polarized Light Properties of light: Ordinary light waves- Vibrate in all directions Plane polarized light waves - Vibrate only in one direction Plane-polarized light is rotated clockwise (to right) or counterclockwise (to left) when passed through enantiomers Direction and extent of rotation will depend upon the concentration of the enantiomer Same concentration of two enantiomers rotates light to same extent but in opposite directions Copyright ©2016 Cengage Learning. All Rights Reserved. 43

44. Figure 18.10 - Vibrational Characteristics of Ordinary Light and Plane-Polarized Light Copyright ©2016 Cengage Learning. All Rights Reserved. 44

45. Copyright ©2016 Cengage Learning. All Rights Reserved. 45 Dextrorotary and Levorotatory Compounds Enantiomers are optically active, i.e., they are compounds that rotate the plane of polarized light Dextrorotatory compound : Chiral compound that rotates light towards right (clockwise; +) Levorotatory compound : Chiral compound that rotates light towards left (counterclockwise; -) There is no correlation between D, L and +, - In D and L system, the structure is viewed + and – can be determined using a polarimeter

46. Copyright ©2016 Cengage Learning. All Rights Reserved. 46 Interactions Between Chiral Compounds Right- and left-handed baseball players cannot use the same glove (chiral) but can use the same hat (achiral) Two members of the enantiomer pair (chiral) react differently with other chiral molecules Enantiomeric pairs have same solubility in achiral solvents like ethanol and have different solubility in chiral solvent like D-2-butanol

47. Copyright ©2016 Cengage Learning. All Rights Reserved. 47 Interactions Between Chiral Compounds Enantiomers have same boiling points, melting points, and densities All these are dependent upon intermolecular forces, whereas chirality doesn’t depend on such forces Our body responds differently to different enantiomers One may give higher rate or one may be inactive Example: Body response to D isomer of hormone epinephrine is 20 times greater than its response to L isomer

48. What is meant when it is said that carbohydrates are optically active compounds? A solution of carbohydrate when placed in a light beam reflects the light back to the observer. A solution of carbohydrate will rotate the plane of polarized light. A solution of carbohydrate will not rotate the plane of polarized light. Being optically active, they do not possess handedness. Copyright ©2016 Cengage Learning. All Rights Reserved. 48

49. What is meant when it is said that carbohydrates are optically active compounds? A solution of carbohydrate when placed in a light beam reflects the light back to the observer. A solution of carbohydrate will rotate the plane of polarized light. A solution of carbohydrate will not rotate the plane of polarized light. Being optically active, they do not possess handedness. Copyright ©2016 Cengage Learning. All Rights Reserved. 49

50. Copyright ©2016 Cengage Learning. All Rights Reserved. 50 Classification based on number of carbon atoms: Triose - 3 carbon atoms Tetrose - 4 carbon atoms Pentoses - 5 carbon atoms Hexoses - 6 carbon atoms Classification based on functional groups: Aldoses : Monosaccharides with one aldehyde group Ketoses : Monosaccharides with one ketone group Monosaccharides

51. Monosaccharides Combined number of C atoms and functional group Examples: Aldohexose - Monosaccharide with aldehyde group and 6 C atoms Ketopentose - Monosaccharide with ketone group and 5 C atoms Copyright ©2016 Cengage Learning. All Rights Reserved. 51

52. What structural feature distinguishes an aldose and a ketose? An aldose possesses an aldehyde group, and a ketose possesses a ketone group. An aldose contains 5 carbon atoms, and a ketose contains 6 carbon atoms. An aldose possesses only –OH groups, and a ketose possesses a carbonyl group. An aldose possesses a ketone group, and a ketose possesses an aldehyde group. Copyright ©2016 Cengage Learning. All Rights Reserved. 52

53. What structural feature distinguishes an aldose and a ketose? An aldose possesses an aldehyde group, and a ketose possesses a ketone group. An aldose contains 5 carbon atoms, and a ketose contains 6 carbon atoms. An aldose possesses only –OH groups, and a ketose possesses a carbonyl group. An aldose possesses a ketone group, and a ketose possesses an aldehyde group. Copyright ©2016 Cengage Learning. All Rights Reserved. 53

54. Exercise Classify each of the following monosaccharides according to both the number of carbon atoms and the type of carbonyl group present. Copyright ©2016 Cengage Learning. All Rights Reserved. 54

55. Exercise Classify each of the following monosaccharides according to both the number of carbon atoms and the type of carbonyl group present. Copyright ©2016 Cengage Learning. All Rights Reserved. 55 Aldopentose Ketohexose Aldo Ketopentose

56. D-Glucose Most abundant in nature Most important source of human nutrition Grape fruit and ripe fruits are good sources of glucose (20–30% by mass) Also named grape sugar Other names Dextrose Blood sugar (70–100 mg/dL) Six-membered cyclic form Copyright ©2016 Cengage Learning. All Rights Reserved. 56

57. D-Galactose Milk sugar Synthesized in human beings Also called brain sugar Part of brain and nerve tissue Used to differentiate between blood types Six-membered cyclic form Copyright ©2016 Cengage Learning. All Rights Reserved. 57

58. D-Fructose Ketohexose Sweetest tasting of all sugars Found in many fruits and in honey Good dietary sugar due to higher sweetness Five-membered cyclic form Copyright ©2016 Cengage Learning. All Rights Reserved. 58

59. D-Ribose Part of a variety of complex molecules which include: RNA ATP DNA Five-membered cyclic form Copyright ©2016 Cengage Learning. All Rights Reserved. 59

60. Which monosaccharide is the most important from a human nutritional standpoint? D-Galactose D-Mannose D-Glucose D-Fructose Copyright ©2016 Cengage Learning. All Rights Reserved. 60

61. Which monosaccharide is the most important from a human nutritional standpoint? D-Galactose D-Mannose D-Glucose D-Fructose Copyright ©2016 Cengage Learning. All Rights Reserved. 61

62. Cyclic Hemiacetal Forms of D-Glucose Dominant forms of monosaccharides with 5 or more C atoms Cyclic structures are in equilibrium with open chain forms Cyclic structures are formed by the reaction of carbonyl group (C=O) with hydroxyl (–OH) group on carbon 5 Copyright ©2016 Cengage Learning. All Rights Reserved. 62

63. Figure 18.17 - The Cyclic Hemiacetal Forms of D-Glucose Copyright ©2016 Cengage Learning. All Rights Reserved. 63

64. Copyright ©2016 Cengage Learning. All Rights Reserved. 64 2 forms of D-Glucose: α -form where the –OH of C1 and CH 2 OH of C5 are on opposite sides β -form where the –OH of C1 and CH 2 OH of C5 are on the same side Cyclic Hemiacetal Forms of D-Glucose

65. Anomers Cyclic monosaccharides that differ only in the position of the substituents on the anomeric carbon atom Copyright ©2016 Cengage Learning. All Rights Reserved. 65

66. Cyclic Forms of Other Monosaccharides Intramolecular cyclic hemiacetal formation and the equilibrium between various forms are not restricted to glucose All aldoses with five or more carbon atoms establish similar equilibria, but with different percentages of the alpha, beta, and open-chain forms Fructose and other ketoses with a sufficient number of carbon atoms also cyclize Copyright ©2016 Cengage Learning. All Rights Reserved. 66

67. Copyright ©2016 Cengage Learning. All Rights Reserved. 67 Pyranose and Furanose Pyranose - Cyclic monosaccharide containing a six-atom ring Furanose - Cyclic monosaccharide containing a five-atom ring Their ring structures resemble the ring structures in the cyclic ethers pyran and furan, respectively

68. Monosaccharides in nature have cyclic forms because they can form: internal acetals that are stable. internal hemiacetals that are stable. internal acetals that are rapidly converted to stable hemiacetals. internal hemiacetals that are rapidly converted to stable acetals. Copyright ©2016 Cengage Learning. All Rights Reserved. 68

69. Copyright ©2016 Cengage Learning. All Rights Reserved. 69

70. Monosaccharides in nature have cyclic forms because they can form: internal acetals that are stable. internal hemiacetals that are stable. internal acetals that are rapidly converted to stable hemiacetals. internal hemiacetals that are rapidly converted to stable acetals. Copyright ©2016 Cengage Learning. All Rights Reserved. 70

71. Haworth projection formula : Two-dimensional structural notation that specifies the three-dimensional structure of a cyclic form of a monosaccharide Copyright ©2016 Cengage Learning. All Rights Reserved. 71

72. Copyright ©2016 Cengage Learning. All Rights Reserved. 72 α and β Configuration Determined by the position of the –OH group on C1 relative to the –CH 2 OH group that determines D or L series In a β configuration, both of these groups point in the same direction In an α configuration, the two groups point in opposite directions

73. Copyright ©2016 Cengage Learning. All Rights Reserved. 73 –OH Group Position The specific identity of a monosaccharide is determined by the positioning of the other –OH groups in the Haworth projection formula Any –OH group at a chiral center that is to the right in a Fischer projection formula points down in the Haworth projection formula Any –OH group to the left in a Fischer projection formula points up in the Haworth projection formula

74. Haworth projection formula for a monosaccharide is a formula showing: a line structure of a stable internal acetal. a two-dimensional notation that specifies the three-dimensional structure of a cyclic monosaccharide. a line structure of a stable internal hemiacetal. a two-dimensional notation that specifies the three-dimensional structure of a stable cyclic acetal. Copyright ©2016 Cengage Learning. All Rights Reserved. 74

75. Haworth projection formula for a monosaccharide is a formula showing: a line structure of a stable internal acetal. a two-dimensional notation that specifies the three-dimensional structure of a cyclic monosaccharide. a line structure of a stable internal hemiacetal. a two-dimensional notation that specifies the three-dimensional structure of a stable cyclic acetal. Copyright ©2016 Cengage Learning. All Rights Reserved. 75

76. Practice Exercise Which of the monosaccharides glucose, fructose, galactose, and ribose has each of the following structural characteristics? (There may be more than one correct answer for a given characteristic) It is a pentose. It is a ketose. Its cyclic form has a 6-membered ring. Its cyclic form has two carbon atoms outside the ring. Copyright ©2016 Cengage Learning. All Rights Reserved. 76

77. Practice Exercise Which of the monosaccharides glucose, fructose, galactose, and ribose has each of the following structural characteristics? (There may be more than one correct answer for a given characteristic) It is a pentose. Ribose It is a ketose. Fructose Its cyclic form has a 6-membered ring. Glucose, galactose Its cyclic form has two carbon atoms outside the ring. Fructose Copyright ©2016 Cengage Learning. All Rights Reserved. 77

78. Five important reactions of monosaccharides: Oxidation to acidic sugars Reduction to sugar alcohols Glycoside formation Phosphate ester formation Amino sugar formation Glucose will be used as the monosaccharide reactant Other aldoses, as well as ketoses, undergo similar reactions Copyright ©2016 Cengage Learning. All Rights Reserved. 78

79. Oxidation to Produce Acidic Sugars The redox chemistry of monosaccharides is closely linked to the alcohol and aldehyde functional groups Oxidation can yield three different types of acidic sugars depending on the type of oxidizing agent used Aldonic acid - Formed when weak oxidizing agents such as Tollens and Benedict’s solutions oxidize the aldehyde end Reducing sugar : Carbohydrate that gives a positive test with Tollens and Benedict’s solutions Copyright ©2016 Cengage Learning. All Rights Reserved. 79

80. Copyright ©2016 Cengage Learning. All Rights Reserved. 80 Strong oxidizing agents can oxidize both ends of a monosaccharide at the same time (the carbonyl group and the terminal primary alcohol group) to produce a dicarboxylic acid Such polyhydroxy dicarboxylic acids are known as aldaric acids In biochemical systems, enzymes can oxidize the primary alcohol end of an aldose such as glucose, without oxidation of the aldehyde group, to produce an alduronic acid Oxidation to Produce Acidic Sugars

81. Copyright ©2016 Cengage Learning. All Rights Reserved. 81 Reduction to Produce Sugar Alcohols The carbonyl group in a monosaccharide (either an aldose or a ketose) is reduced to a hydroxyl group using hydrogen as the reducing agent The product is the corresponding polyhydroxy alcohol called sugar alcohol or alditol Sorbitol - Used as a moisturizing agent in foods and cosmetics and as a sweetening agent in chewing gum

82. Next meeting 8/31 Copyright ©2016 Cengage Learning. All Rights Reserved. 82

83. Glycloside Formation Glycoside : Acetal formed from a cyclic monosaccharide by replacement of the hemiacetal carbon –OH group with an –OR group Glucoside - Glycoside produced from glucose Galactoside - Glycoside produced from galactose Exist in both α and β forms Copyright ©2016 Cengage Learning. All Rights Reserved. 83

84. Copyright ©2016 Cengage Learning. All Rights Reserved. 84 Phosphate Ester Formation Hydroxyl groups of a monosaccharide can react with inorganic oxyacids to form inorganic esters Phosphate esters of various monosaccharides are stable in aqueous solution and play important roles in the metabolism of carbohydrates

85. Copyright ©2016 Cengage Learning. All Rights Reserved. 85 Amino Sugar Formation Amino sugar - Formed when one of the hydroxyl groups of a monosaccharide is replaced with an amino group In naturally occurring amino sugars, the C2 hydroxyl group is replaced by an amino group Amino sugars and their N -acetyl derivatives are important building blocks of polysaccharides such as chitin and hyaluronic acid

86. Which of the following sugars can be oxidized by Tollens and Benedict’s reagents in basic conditions? D-glucose, an aldohexose D-fructose, a ketohexose D-mannose, an aldohexose All the above Copyright ©2016 Cengage Learning. All Rights Reserved. 86

87. Which of the following sugars can be oxidized by Tollens and Benedict’s reagents in basic conditions? D-glucose, an aldohexose D-fructose, a ketohexose D-mannose, an aldohexose All the above Copyright ©2016 Cengage Learning. All Rights Reserved. 87

88. Copyright ©2016 Cengage Learning. All Rights Reserved. 88 Two monosaccharides can react to form a disaccharide One monosaccharide acts as a hemiacetal and the other as an alcohol Resulting ether bond is a glycosidic linkage

89. Copyright ©2016 Cengage Learning. All Rights Reserved. 89 Maltose (Malt Sugar) Structurally made of 2 D-glucose units, one of which must be α -D-glucose, linked via an α(1 4) glycosidic linkage Digested easily by humans because of an enzyme that can break α (1  4) linkages Baby foods are rich in maltose

90. Copyright ©2016 Cengage Learning. All Rights Reserved. 90 Cellobiose Produced as an intermediate in the hydrolysis of the polysaccharide cellulose Contains two D-glucose monosaccharide units, one of which must have a β configuration, linked through a β (1  4) glycosidic linkage Cannot be digested by humans

91. Copyright ©2016 Cengage Learning. All Rights Reserved. 91 Lactose Made up of β - D-galactose unit and a D-glucose unit joined by a β (1  4) glycosidic linkage Milk is rich in the disaccharide lactose Lactase hydrolyzes β (1  4) glycosidic linkages

92. Lactose Intolerance or Lactase Persistence Lactose is the principal carbohydrate in milk Human mother’s milk - 7%–8% lactose Cow’s milk - 4%–5% lactose Lactose intolerance is a condition in which people lack the enzyme lactase needed to hydrolyze lactose to galactose and glucose Deficiency of lactase can be caused by a genetic defect, physiological decline with age, or by injuries to intestinal mucosa Copyright ©2016 Cengage Learning. All Rights Reserved. 92

93. Lactose Intolerance or Lactase Persistence When lactose is undigested, it attracts water causing fullness, discomfort, cramping, nausea, and diarrhea Bacterial fermentation of the lactose further along the intestinal tract produces acid (lactic acid) and gas, adding to the discomfort Copyright ©2016 Cengage Learning. All Rights Reserved. 93

94. Sucrose (Table Sugar) The most abundant of all disaccharides and found in plants Produced commercially from the juice of sugar cane and sugar beets Sugar cane contains up to 20% by mass sucrose Sugar beets contain up to 17% by mass sucrose Copyright ©2016 Cengage Learning. All Rights Reserved. 94

95. What is the linkage that is formed when two monosaccharides react to form a disaccharide called? Alcohol group linkage Glycosidic linkage Hemiacetal linkage Acetal linkage Copyright ©2016 Cengage Learning. All Rights Reserved. 95

96. What is the linkage that is formed when two monosaccharides react to form a disaccharide called? Alcohol group linkage Glycosidic linkage Hemiacetal linkage Acetal linkage Copyright ©2016 Cengage Learning. All Rights Reserved. 96

97. Which disaccharide is a nonreducing sugar? Maltose Lactose Cellobiose Sucrose Copyright ©2016 Cengage Learning. All Rights Reserved. 97

98. Which disaccharide is a nonreducing sugar? Maltose Lactose Cellobiose Sucrose Copyright ©2016 Cengage Learning. All Rights Reserved. 98

99. Practice Exercise Which of these disaccharides, i.e., maltose, cellobiose, lactose, and sucrose, have the following structural or reaction characteristics? (There may be more than one correct answer for a given characteristic) Two different monosaccharide units are present. Hydrolysis produces only monosaccharides. Its glycosidic linkage is a “head-to-head” linkage. It is not a reducing sugar. Copyright ©2016 Cengage Learning. All Rights Reserved. 99

100. Practice Exercise Which of these disaccharides, i.e., maltose, cellobiose, lactose, and sucrose, have the following structural or reaction characteristics? (There may be more than one correct answer for a given characteristic) Two different monosaccharide units are present. Lactose, sucrose Hydrolysis produces only monosaccharides. Maltose, cellobiose, lactose, sucrose Its glycosidic linkage is a “head-to-head” linkage. Sucrose It is not a reducing sugar. Sucrose Copyright ©2016 Cengage Learning. All Rights Reserved. 100

101. Carbohydrates that contain 3–10 monosaccharide units bonded to each other via glycosidic linkages Generally present in association with other complex molecules Raffinose - Made of 1 galactose, 1 glucose, and 1 fructose Stachyose - Made of 2 galactose, 1 glucose, and 1 fructose units Commonly found in onions, cabbage, broccoli, and whole wheat Copyright ©2016 Cengage Learning. All Rights Reserved. 101

102. Blood Types and Oligosaccharides Human blood is classified into four types A, B, AB, and O The basis for the difference is the type of sugars (oligosaccharides) present Blood of one type cannot be given to a recipient with blood of another type A transfusion of wrong blood type can cause the blood cells to form clumps, a potentially fatal reaction People with type O blood are universal donors, and those with type AB blood are universal recipients Copyright ©2016 Cengage Learning. All Rights Reserved. 102

103. Blood Types and Oligosaccharides In the United States, type O blood is the most common and type A the second most common The biochemical basis for the various blood types involves oligosaccharides present on plasma membranes of red blood cells The oligosaccharides responsible for blood groups are D-galactose and its derivatives Copyright ©2016 Cengage Learning. All Rights Reserved. 103

104. Other Oligosaccharides Solanine, a potato plant toxin, is a oligosaccharide found in association with an alkaloid Bitter taste of potatoes is due to relatively higher levels of solanine Copyright ©2016 Cengage Learning. All Rights Reserved. 104

105. Which of the following is a toxin produced by potato plants? Raffinose Stachyose Solanine None of the above Copyright ©2016 Cengage Learning. All Rights Reserved. 105

106. Which of the following is a toxin produced by potato plants? Raffinose Stachyose Solanine None of the above Copyright ©2016 Cengage Learning. All Rights Reserved. 106

107. The Polymer Chain Polysaccharides are polymers of many monosaccharide units bonded with glycosidic linkages Two types: Homopolysaccharide Heteropolysaccharide Copyright ©2016 Cengage Learning. All Rights Reserved. 107

108. Figure 18.26 - The Polymer Chains of a Polysaccharide Copyright ©2016 Cengage Learning. All Rights Reserved. 108

109. Characteristics of Polysaccharides Polysaccharides are not sweet and do not show positive tests with Tollen’s and Benedict’s solutions, whereas monosaccharides are sweet and show positive tests Limited water solubility Examples: Cellulose and glycogen - Storage polysaccharides Chitin - Structural polysaccharide Hyaluronic acid - Acidic polysaccharide Copyright ©2016 Cengage Learning. All Rights Reserved. 109

110. A _____ is a polysaccharide in which only one type of monosaccharide monomer is present. heteropolysaccharide homopolysaccharide heteromonosaccharide homomonosaccharide Copyright ©2016 Cengage Learning. All Rights Reserved. 110

111. A _____ is a polysaccharide in which only one type of monosaccharide monomer is present. heteropolysaccharide homopolysaccharide heteromonosaccharide homomonosaccharide Copyright ©2016 Cengage Learning. All Rights Reserved. 111

112. Starch Storage polysaccharide : Polysaccharide that is a storage form for monosaccharides and used as an energy source in cells Starch Glucose is the monomeric unit Storage polysaccharide in plants Copyright ©2016 Cengage Learning. All Rights Reserved. 112

113. Types of Polysaccharides Isolated From Starch Amylose Unbranched-chain polymer and accounts for 15%–20% of the starch Has α (1  4) glycosidic bonds Amylopectin Branched chain polymer and accounts for 80%–85% of the starch Has α (1  4) and α (1  6) glycosidic bonds Up to 100,000 glucose units are present Amylopectin is digested more readily by humans (can hydrolyze α linkages but not β linkages) Copyright ©2016 Cengage Learning. All Rights Reserved. 113

114. Glycogen Glucose storage polysaccharide in humans and animals Contains only glucose units Branched chain polymer with α (1  4) glycosidic bonds in straight chains and α (1  6) in branches Three times more highly branched than amylopectin in starch Contains up to 1,000,000 glucose units Copyright ©2016 Cengage Learning. All Rights Reserved. 114

115. Glycogen Excess glucose in blood is stored in the form of glycogen Copyright ©2016 Cengage Learning. All Rights Reserved. 115

116. In starch, two different glucose-containing polysaccharides can be isolated. They are _____ and _____. glycogen; amylose amylose; amylopectin amylose; cellulose glycogen; cellulose Copyright ©2016 Cengage Learning. All Rights Reserved. 116

117. In starch, two different glucose-containing polysaccharides can be isolated. They are _____ and _____. glycogen; amylose amylose; amylopectin amylose; cellulose glycogen; cellulose Copyright ©2016 Cengage Learning. All Rights Reserved. 117

118. Cellulose Linear homopolysaccharide with β (1  4) glycosidic bond Contains up to 5000 glucose units with molecular mass of 900,000 amu Cotton has 95% cellulose and wood 50% cellulose Humans do not have enzymes that hydrolyze β (1  4) linkages and so they cannot digest cellulose Animals also lack these enzymes, but they can digest cellulose due to the presence of cellulase-producing bacteria Copyright ©2016 Cengage Learning. All Rights Reserved. 118

119. Cellulose It serves as dietary fiber in food and readily absorbs water resulting in softer stools 20–35 g of dietary fiber is desired everyday Copyright ©2016 Cengage Learning. All Rights Reserved. 119

120. Chitin Similar to cellulose structurally and functionally Linear polymer with all β (1  4) glycosidic linkages It has an N -acetyl amino derivative of glucose Function is to give rigidity to the exoskeletons of crabs, lobsters, shrimp, insects, and other arthropods Copyright ©2016 Cengage Learning. All Rights Reserved. 120

121. Figure 18.32(b) - The Structure of Chitin Copyright ©2016 Cengage Learning. All Rights Reserved. 121

122. What is the structural component of plant cell walls? Starch Glycogen Chitin Cellulose Copyright ©2016 Cengage Learning. All Rights Reserved. 122

123. What is the structural component of plant cell walls? Starch Glycogen Chitin Cellulose Copyright ©2016 Cengage Learning. All Rights Reserved. 123

124. Acidic polysaccharides Polysaccharides with a repeating disaccharide unit containing an amino sugar and a sugar with a negative charge due to a sulfate or a carboxyl group They are heteropolysaccharides, i.e., different monosaccharides exist in an altering pattern Examples: Hyaluronic acid Heparin Copyright ©2016 Cengage Learning. All Rights Reserved. 124

125. Hyaluronic Acid Alternating residues of N -acetyl- β -D-glucosamine and D-glucuronate Highly viscous and serve as lubricants in the fluid of joints as well as vitreous humor of the eye Copyright ©2016 Cengage Learning. All Rights Reserved. 125

126. Heparin Polysaccharide with 15–90 disaccharide residues per chain Blood anticoagulant Copyright ©2016 Cengage Learning. All Rights Reserved. 126

127. Heparin, a well known acidic polysaccharide, is best known for its biochemical function as a(n) _____. coagulant anticoagulant exoskeleton dietary fiber Copyright ©2016 Cengage Learning. All Rights Reserved. 127

128. Heparin, a well known acidic polysaccharide, is best known for its biochemical function as a(n) _____. coagulant anticoagulant exoskeleton dietary fiber Copyright ©2016 Cengage Learning. All Rights Reserved. 128

129. Copyright ©2016 Cengage Learning. All Rights Reserved. 129 Nutrition Foods high in carbohydrate content constitute over 50% of the diet of most people of the world Corn in South America Rice in Asia Starchy root vegetables in parts of Africa Potato and wheat in North America Balanced dietary food should contain about 60% of carbohydrate

130. Classes of Dietary Carbohydrates Simple carbohydrates : Dietary monosaccharides or disaccharides Sweet to taste and commonly referred to as sugars Constitute 20% of the energy in the US diet Complex carbohydrates : Dietary polysaccharides Include starch and cellulose, which are normally not sweet to taste Copyright ©2016 Cengage Learning. All Rights Reserved. 130

131. Dietary carbohydrates constitute _____ of the diet of most people. 10% 25% 35% 50% Copyright ©2016 Cengage Learning. All Rights Reserved. 131

132. Dietary carbohydrates constitute _____ of the diet of most people. 10% 25% 35% 50% Copyright ©2016 Cengage Learning. All Rights Reserved. 132

133. Glycolipid : Lipid molecule that has one or more carbohydrate (or carbohydrate derivative) units covalently bonded to it Glycoprotein : Protein molecule that has one or more carbohydrate (or carbohydrate derivative) units covalently bonded to it Such carbohydrate complexes are very important in cellular functions such as cell recognition Copyright ©2016 Cengage Learning. All Rights Reserved. 133

134. A protein molecule that has one or more carbohydrate units covalently bonded to it is known as a(n) _____. glycolipid glucoprotein glycoprotein oligosaccharide protein Copyright ©2016 Cengage Learning. All Rights Reserved. 134

135. A protein molecule that has one or more carbohydrate units covalently bonded to it is known as a(n) _____. glycolipid glucoprotein glycoprotein oligosaccharide protein Copyright ©2016 Cengage Learning. All Rights Reserved. 135

136. Concept Question 1 D-glucose consists of six carbon atoms. When drawn using the Fischer projection formula, how do you determine that it is the D isomer? How many chiral carbon centers does it have and how many stereoisomers are possible? The –OH on the chiral carbon farthest from the carbonyl group points to the right. There are five chiral centers providing 32 stereoisomers. The –OH on the chiral carbon farthest from the carbonyl group points to the right. There are four chiral centers providing 16 stereoisomers. The –OH on the chiral carbon farthest from the carbonyl group points to the left. There are five chiral centers providing 32 stereoisomers. The –OH on the chiral carbon farthest from the carbonyl group points to the left. There are four chiral centers providing 16 stereoisomers. Copyright ©2016 Cengage Learning. All Rights Reserved. 136

137. Concept Question 1 D-glucose consists of six carbon atoms. When drawn using the Fischer projection formula, how do you determine that it is the D isomer? How many chiral carbon centers does it have and how many stereoisomers are possible? The –OH on the chiral carbon farthest from the carbonyl group points to the right. There are five chiral centers providing 32 stereoisomers. The –OH on the chiral carbon farthest from the carbonyl group points to the right. There are four chiral centers providing 16 stereoisomers. The –OH on the chiral carbon farthest from the carbonyl group points to the left. There are five chiral centers providing 32 stereoisomers. The –OH on the chiral carbon farthest from the carbonyl group points to the left. There are four chiral centers providing 16 stereoisomers. Copyright ©2016 Cengage Learning. All Rights Reserved. 137

138. Concept Question 2 Prior to a marathon run, an athlete consumes large amounts of complex carbohydrates to do what is known as “carbohydrate loading.” What happens in the body to the glucose molecules present in these complex carbohydrates and why is carbohydrate loading important? The complex carbohydrates are broken down to glucose and any excess glucose, not used for immediate energy, is stored in the form of glycogen, which can be used later as a source of stored energy. The complex carbohydrates are broken down to glucose and any excess glucose, not used for immediate energy, is stored in the form of starch, which can be used later as a source of stored energy. The complex carbohydrates are broken down to galactose and any excess galactose, not used for immediate energy, is stored in the form of glycogen, which can be used later as a source of stored energy. The complex carbohydrates are broken down to glucose and any excess glucose, not used for immediate energy, is stored in the form of glycogen, which can be used later to produce muscle tissue needed to complete the marathon. Copyright ©2016 Cengage Learning. All Rights Reserved. 138

139. Concept Question 2 Prior to a marathon run, an athlete consumes large amounts of complex carbohydrates to do what is known as “carbohydrate loading.” What happens in the body to the glucose molecules present in these complex carbohydrates and why is carbohydrate loading important? The complex carbohydrates are broken down to glucose and any excess glucose, not used for immediate energy, is stored in the form of glycogen, which can be used later as a source of stored energy. The complex carbohydrates are broken down to glucose and any excess glucose, not used for immediate energy, is stored in the form of starch, which can be used later as a source of stored energy. The complex carbohydrates are broken down to galactose and any excess galactose, not used for immediate energy, is stored in the form of glycogen, which can be used later as a source of stored energy. The complex carbohydrates are broken down to glucose and any excess glucose, not used for immediate energy, is stored in the form of glycogen, which can be used later to produce muscle tissue needed to complete the marathon. Copyright ©2016 Cengage Learning. All Rights Reserved. 139