Chapter 7 Carbohydrates and the glycoconjugates of cell surfaces
生物化学精品课件(厦大)carbohydrates and glycobiology
Aldoses and Ketoses
Aldehyde vs. Ketone
The smallest sugar--triose
A carbonyl group is a functional group composed of a carbon atom double-bonded to an oxygen atom: C=O.
D or L is determined by the chiral center with highest number
Mutual epimers: differ only around one carbon D-mannose vs. D-glucose D-glucose vs. D-galactose
Stereoisomers with n chiral centers=2n
Four chiral centers can yield 16 isomers of D- and Lforms together.
Ketose with three carbons
Ketose with
No chiral center
Ribose and deoxyribose sugars form part of the structural framework of RNA and DNA.
Polysaccharides are structural elements in the cell walls of bacteria and plants. They are involved in cell-cell recognition.
Key point: One carbon center with four different side groups.
carbohydrates 中国医科大学 糖代谢 生物化学 英文
COOC O ~ P + H2O CH2
PEP
(10) PEP →pyruvate
COOC O~ P CH2 PEP
ADP
ATP COO-
pyruvate kinase
CO CH3
Pyruvate
• Second substrate level phosphorylation • irreversible
1) Glycolytic pathway : G → pyruvate including 10 reactions.
(1) G phosphorylated into glucose 6-phosphate
HO CH2 HH O
OH H
OH
ATP
ADP
H
Mg2+
OH Hexokinase
H OH
bifunctional enzyme
2) Pyruvate kinase
• Allosteric regulation: F-1,6-BP acts as allosteric activator；
ATP and Ala in liver act as allosteric inhibitors;
Mg2+
F- 6-P
F- 1,6-BP
PFK-1
Aldolase
lactate
NAD+
GAP
DHAP
LDH NADH+H+
pyruvate ATP
NAD+ NADH+H+
H3PO4
glyceraldehyde 3-phosphate dehydrogenase
pyruvate kinase Nhomakorabeaglycerate
第7章 糖
苷键的β- 位有吸电子基团者，使α- 位氢活化， 在碱液中与苷键起消除反应而开裂。
CHO OH-H2O
H C
OOH-
CHO
Glc O
Glc O
-H2O
藏红花苦苷
双烯醛 safranol
㈣酶催化水解反应
用酶水解苷键可以获知苷键的构型，可保持苷元
结构不变的真正苷元。
酶专属性高，选择性地催化水解某一构型的苷。
CH2OH OR
OH OH OH
IO4-
BH4-
H+
CH2OH
OH
+
CH2OH
R CHOH CH2OH
IO4-
R CHO +
HCOOH
glc OH
glcO
HO
S
IO4BH4-
OH
glc
glcO
H+
HO
人参皂苷Rb1 20（s）原人参二醇
3%HCl
50 % HA c
O OH
OH
OH
R
HO
glc
glcO
Angyal 用总自由能来分析构象式的稳定性，
比较二种构象式的总自由能差值，能量低的是优
势构象。
如：葡萄糖的二种构象式的比较：
作业：写出 β-D-Glu、α-L-Rha、β-D-Gal、β-D-Xyl的Haworth 投影式和β-D-Glu ，β-D-Gal优势构象式
二 糖的分类
（一）糖的种类
单糖
第 7章
糖
Saccharides and glycosides
本章基本内容
一、概述
二、单糖的立体化学
三、糖和苷分类
四、苷键水解
Carbohydrates
Outline of the section
1. Digestion and absorption of carbohydrates (糖的消化吸收) 2. Anaerobic oxidation (糖的无氧氧化—糖酵解) 3. Tricarboxylate cycle (TCA 循环) 4. Pentose phosphate pathway (PPP, 磷酸戊糖途径) 5. Gluconeogenesis (糖异生) 6. Glycogenesis and glycogenolysis (糖原合成与分解) 7. Blood glucose regulation (血糖的调节)
Glucose(葡萄糖) Galactose(半乳糖)
Monosaccharides （单糖）
O
H
OH
H
OH
H
OH
Glucose（葡萄糖, 己醛糖）
OH
Ribose（核糖, 戊醛糖）
Glucose （葡萄糖）
Enantiomer: 镜像对称（手性 分子）
L-Glucose(葡萄糖)
D-Glucose(葡萄糖) predominant structure
Section 1
Structure, digestion and absorption of carbohydrates
(糖的结构、消化与吸收)
1. What’s carbohydrates/saccharides?
• 糖(carbohydrates)即碳水化合物，其化学本质为多羟醛或多羟酮
类及其多缩聚体化合物，也称saccharides. 结构式Cn(H2O)n，n≥3. – 单糖(monosaccharides)：葡萄糖、果糖、半乳糖
carbohydrate-digetion
Sucrase, Isomaltase, Maltase, Lactase
Sucrose, Dextrin, Maltose, Lactose
Hydrolysis
Fig. 3-7, p. 70
Fig. 3-7a, p. 70
Fig. 3-7b, p. 70
Fig. 3-7c, p. 70
Digestion of disaccharides Virtually no digestion of disaccharides or small oligosaccharides occurs in the mouth and stomach. In human, digestion takes place entirely in the upper small intestine. Disaccharidase activity presents in the microvilli of the intestinal mucosal cells (brush border), rather than in the intestinal lumen. The disaccharidases include lactase, sucrase, maltase, and isomaltase.
• Integral proteins • Each has specific combining site • Undergoes a conformational change upon binding the molecule • Can reverse this change when unbound
Maltase hydrolyze maltose to yield two glucose units.
Carbohydrate(一)
Hexokinase (EC 2.7.1.1.) exerts a key regulatory role in glycolysis. This enzyme has the ability to phosphorylate different hexoses, but phosphorylation of Glc is, by far, its most important function. Four different isozymes of hexokinase have been identified in mammals -Type I, Type II, Type III, and Type IV. As the latter isozyme has a strong specificity for D-Glc, it is also known as glucokinase.
3. Phosphofructokinase-1(磷酸果糖激酶):
fructose-6-phosphate + ATP (FDP) + ADP fructose-1,6-bisphosphate
The Phosphofructokinase-1 reaction is the rate-limiting step of Glycolysis.
4. Four molecules of ATP are produced by substrate-level phosphorylation （底物水平磷酸化） 5. Pyruvate is the end product.
• Substrate-level phosphorylation is the production of ATP from ADP by a direct transfer of a high-energy phosphate group from a phosphorylated intermediate metabolic compound in an exergonic catabolic pathway.
Chapter 7 carbohydrates——食品生物化学(英文)课件PPT
➢Carbohydrates are covalently linked with a variety of other molecules. Carbohydrates linked to lipid molecules, or glycolipids, 醣脂 are mon ponents of biological membranes. Proteins that have covalently linked carbohydrates are called glycoproteins糖蛋白.
Chapter 7 Carbohydrates
碳水化合物
➢What are carbohydrates?
➢Carbohydrates are the single most abundant class of organic molecules found in nature. The name carbohydrate arises from the basic molecular formula (CH2O)n, which can be rewritten (C·H2O)n to show that these substances are hydrates of carbon, where n=3 or more.
OUTLINE
➢7.1 Carbohydrate Nomenclature命名法 ➢7.2 Monosaccharides ➢7.3 Oligosaccharides ➢7.4 Pte Nomenclature
Carbohydrates are generally classified into three groups: monosaccharides (and their derivatives[dɪ'rɪvətɪv]衍生物), oligosaccharides, and polysaccharides.
organic Chapter 7
If the group of highest priority on one carbon is on the same side as the group of highest priority on the other carbon the double bond is Z (zusammen) If the highest priority groups are on opposite sides the alkene is E (entgegen)
The greater the number of attached alkyl groups (i.e. the more highly substituted the carbon atoms of the double bond), the greater the alkene’s stability
This is true only if a small base such as ethoxide is used
Chapter 7
6
The transition state in this E2 reaction has double bond character The trisubstituted alkene-like transition state will be most stable and have the lowest ∆G‡ Kinetic control of product formation: When one of two products is formed because its free energy of activation is lower and therefore the rate of its formation is higher
sugars, carbohydrates and glycobiology (Ch
1. Carbohydrates are aldehyde or ketone compounds with multiple hydroxyl groups or substances that can yield such compounds on hydrolysis (p. 293)
2. Monosacchrides contain one carbonyl group and two or more hydroxyl groups. (p. 294) 2.1 Monosacchrides can be divided into two families: aldoses and ketoses.
1.1 Carbohydrates are the most abundant biomolecules on earth and have multiple roles in all forms of life.
1.1.1 Carbohydrates serve as energy stores (e.g., starch in plants, glycogen in animals), fuels (e.g., glucose), and metabolic intermediates (e.g., ATP, many coenzymes).
2.1.3 The simplest aldose is glyceraldehyde, and the simplest ketose is dihyoxyacetone, both being triose.
2.1.4 Monosacchrides containing four, five, and six carbon atoms in their backbones are called tetroses, pentoses (e.g., ribose and deoxyribose), and hexoses (e.g., glucose and fructose), respectively. Each has both aldoses and ketoses.
TextACarbohydratesManycarbohydrates碳水化合物canbe
Text A CarbohydratesMany carbohydrate s （碳水化合物） can be expressed in the form [Cx (H2O y ], but this is not an accurate description of their structure. Carbohydrates are infact polyhydroxy （多羟基） aldehyde s （醛） or ketone s （酮）and the term also includes any compounds that can be hydrolyzed to form carbohydrates. The simplest carbohydrates are called sugars (or monosaccharide s （单糖） and these may link together to form more complex carbohydrates, starches (or polysaccharide s （多糖） .The diversity of dietary carbohydrates necessitates discussion of several classes of these molecules, ranging from simple sugars to huge, branched polymers.1.MonosaccharidesMonosaccharides or simple sugars are either hexoses (6-carbon （六碳糖）like glucose （葡萄糖）, galactose （半乳糖） and fructose （果糖）, or pentose s (5-carbon （五碳糖）like ribose （核糖）. These are the breakdown products of more complex carbohydrates and can be efficiently absorbed across the wall of the digestive tube （消化道）and transported into blood. Simple sugars are monosaccharides.These are characterised by hydroxyl （羟基）, aldehyde and ketone group s （功能团）. Monosaccharides are either aldose s （醛糖） or ketose s （酮糖）. The aldo- and keto- prefix （前缀）identifies the nature of the carbonyl （羰基，羰酰） group, and the –ose suffix（后缀） designates a carbohydrate.In Fischer projections （费歇尔投影式）, if the hydroxyl group at the lowest stereogenic centre （手相中心）, C5 in this case, points to the right, it may be labelled a D-sugar, if it points to the left, it is an L-sugar.The number of carbon atoms is also indicated, so for example, aldohexose is the six-carbon aldehyde sugar, more commonly referred to as glucose. Intramolecular （分子内）, nucleophilic addition （亲核加成）can take place between a hydroxyl and acarbonyl group of the same molecule, resulting in the formation of a cyclic hemiacetal （半缩醛）.Simple monosaccharides consist of aldehyde or ketone groups and both primary and secondary hydroxyl groups, all of which are available for reaction.An alkaline （碱的，强碱的） solution of copper （铜）(Fehling's solution （斐林试剂） oxidise s （氧化）an aldose to an aldonate （醛糖盐）, and is itself reduced to copper, precipitating （沉淀）as red Cu2O.Hydrogenation （氢化作用） of fructose, and carbohydrates in general, involves addition of hydrogen （氢）to the double bond （双键）of the carbonyl group. The resulting products have an –itol suffix to denote that they are sugar alcohols.Carbohydrate hydroxyl groups form esters on reactions with acid （酸，酸的）chloride s （氯化物） or anhydride s （酐） in the presence of base （碱）. All of the hydroxyl groups react.Ether s （醚，醚类）may be formed by treatment of carbohydrates with alkylhalide s （烷基卤化物） in the presence of base.2.Disaccharides （二糖）Disaccharides are simply two monosaccharides linked together by a glycosidic bond （糖苷键）. The disaccharides most important in nutrition and digestion are:∙lactose （乳糖）or "milk sugar": glucose + galactose ∙sucrose （蔗糖）or "table sugar": glucose + fructose ∙ maltose （麦芽糖）: glucose + glucoseOligosaccharides are relatively short chains of monosaccharides which typically are intermediate s （中间体） in the breakdown of polysaccharides to monosaccharides.3.PolysaccharidesPolysaccharides are the most abundant dietary carbohydrate for all except very young animals. You should be familiar with three important polysaccharides, each of which is a large polymer of glucose:∙ Starch （淀粉） is a major plant storage form of glucose. It occurs in two forms: alpha-amylose （α-直链淀粉）, in which the glucoses are linked together in straight chains, and amylopectin , in which the glucose chains are highly branched. Except for the branch points of amylopectin, the glucose monomer s （单体） in starch are linked via （通过） alpha(1->4 glycosidic bonds （α-1，4-糖苷键）, which, in the digestive tract of mammal s （哺乳动物）, are hydrolyzed by amylase s （淀粉酶）.∙ Cellulose （纤维素） is the other major plant carbohydrate. It is the major constituent of plant cell walls, and more than half of the organic carbon on earth is found in cellulose. Cellulose is composed on unbranched, linear chains of D-glucose molecules, linked to one another by beta(1->4 glycosidic bonds （β-1,4-糖苷键）, which no vertebrate has the capacity to enzymatically digest. Herbivore s(食草动物 subsist largely on cellulose, not because they can digest it themselves, but because their digestive tracts teem with microbes that produce cellulases that hydrolyze cellulose.∙ Glycogen （糖原） is the third large polymer of glucose and is the major animal storage carbohydrate. Like starch, the glucose molecules in glycogen are linked together by alpha(1->4 glycosidic bonds.Text B LipidsLipids are a family of naturally occurring compounds grouped together on the basis of their relative insolubility in water and solubility in nonpolar solvents. These arenaturally occurring molecules that can be isolated from cells and tissues by extraction using nonpolar organic solvents.There are two classes of lipids:i those with ester linkages that can be hydrolyzed, for example, fats and waxesii those without ester linkages and so cannot be hydrolyzed, for example, cholesterol and steroids.Fats and oils are triacylglycerols. They are triesters of glycerol (propane-1,2,3-triol with three long chain carboxylic acids. Reaction with aqueous sodium hydroxide yields glycerol and the three fatty acids.Fatty acids are usually unbranched and contain an even number of carbon atoms, usually from 12 to 20. If there are any double bonds, they tend to have cis geometry. Fats containing one double bond are monounsaturated and those with more than one double bond are polyunsaturated. Saturated fats have a uniform shape that allows them to pack together in a crystal lattice. Unsaturated fats have double bonds that introduce kinks into the hydrocarbon chain making crystal formation more difficult. This is why saturated fats have higher melting points and are solids at room temperature compared to unsaturated fats that tend to be liquids (oils.The most abundant lipid in olive oil is glyceryl trioleate, a triester formed from glycerol and oleic acid, a mono-unsaturated fatty acid with a cis double bond in the middle of the C 18 chain.Waxes are mixtures of esters of long chain carboxylic acids and long chain alcohols. The carboxylic acids usually have an even number of carbon atoms, ranging from 16 to 36, and the long chain alcohols also have an even number of carbon atoms, ranging from 24-36.脂质脂质是一类不溶于水，但溶于非极性溶剂的天然化合物。
生物化学 双语Carbohydrate and Glycobiology练习题
Chapter 2. Carbohydrate and Glycobiology 1. Define the following terms(1)Polysaccharide(2)Furanose(3)Pyranose(4)Chiral carbon(5)Conformation(6)Configuration(7)Epimer(8)Enantiomer(9)Diastereomer(10)Stereoisomer(11)Lectin(12)Anomeric carbon atom(13)Glycosidic bond(14)Glycoform(15)Molecular recognition(16)O-Glycosidic linkage(17)N-glycosidic linkage2. Mono-Choice Questions(1) To possess optical activity, a compound must be:A. a carbohydrate.B. a hexose.C. asymmetric.D. colored.E. D-glucose.(2) Which of the following monosaccharides is not an aldose?A. erythroseB. fructoseC. glucoseD. glyceraldehydeE. ribose(3) The reference compound for naming D and L isomers of sugars is:A. fructose.B. glucose.C. glyceraldehyde.D. ribose.E. sucrose.(4) When two carbohydrates are epimers:A. one is a pyranose, the other a furanose.B. one is an aldose, the other a ketose.C. they differ in length by one carbon.D. they differ only in the configuration around one carbon atom.E. they rotate plane-polarized light in the same direction.(5) Which of following is an anomeric pair?A. D-glucose and D-fructoseB. D-glucose and L-fructoseC. D-glucose and L-glucoseD. α-D-glucose and β-D-glucoseΕ. α-D-glucose and β-L-glucose(6) When the linear form of glucose cyclizes, the product is a(n):A. anhydride.B. glycoside.C. hemiacetal.D. lactone.E. oligosaccharide.(7) Which of the following pairs is interconverted in the process of mutarotation?A. D-glucose and D-fructoseB. D-glucose and D-galactoseC. D-glucose and D-glucosamineD. D-glucose and L-glucoseΕ. α-D-glucose and β-D-glucose(8) Which of the following is not a reducing sugar?A. fructoseB. glucoseC. glyceraldehydeD. riboseE. sucrose(9) From the abbreviated name of the compound Gal(β1 → 4)Glc, we know that:A. C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.B. the compound is a D-enantiomer.C. the galactose residue is at the reducing end.D. the glucose is in its pyranose form.E. the glucose residue is the β anomer.(10) Starch and glycogen are both polymers of:A. fructose.B. glucose1-phosphate.C. sucrose.D. α-D-glucose.Ε. β-D-glucose.(11) Which of the following statements about starch and glycogen is false?A. amylose is unbranched; amylopectin and glycogen contain many (α1 → 6) branches.B. both are homopolymers of glucose.C. both serve primarily as structural elements in cell walls.D. both starch and glycogen are stored intracellularly as insoluble granules.E. glycogen is more extensively branched than starch.(12) Which of the following is a heteropolysaccharide?A. celluloseB. chitinC. glycogenD. hyaluronateE. starch(13) In glycoproteins, the carbohydrate moiety is always attached through the amino acid residues:A. asparagine, serine, or threonine.B. aspartate or glutamate.C. glutamine or arginine.D. glycine, alanine, or aspartate.E. tryptophan, aspartate, or cysteine.(14) The biochemical property of lectins that is the basis for most of their biological effects is their ability to bind to:A. amphipathic molecules.B. hydrophobic molecules.C. specific lipids.D. specific oligosaccharides.E. specific peptides.(15) Carbohydrates areA. polyhydroxy aldehydes.B. polyhydroxy ketones.C. polyhydroxy acids.D. polyhydroxy alcohols.E. A and B.(16) The simplest carbohydrates areA. D- and L-glyceraldehyde.B. dihydroxyacetone and D- and L-glyceraldehyde.C. dihydroxyacetone and glycerate.D. all of the above.E. none of the above.(17) The nutritional storage form(s) of glucose in plants.A. glycogenB. amyloseC. amylopectinD. B and CE. all of the above(18) How do some viruses gain entry into specific cells?A. by attaching to ion channelsB. by cleaving the glycosidic bonds and altering protein shapesC. by binding to glycoproteins on the cell surface that are unique to specific cellsD. all of the above.E. none of the above.(19) Which of the following statements is true about the cyclic (ring) form of D–glucose?A. the cyclic form of D–glucose is a furanose.B. the cyclic form of D–glucose involves the formation of a covalent bond between the carbonyl group and the oxygen of a hydroxyl group along the chain.C. the cyclic form of D–glucose is the result of a bond forming between the C–1 and C–6 carbon atoms.D. in aqueous solution, the cyclic form of D–glucose is rare.(20) Which of the following is true about how a disaccharide is formed from two monosaccharides?A. monosaccharides are linked in a reaction that eliminates water.B. disaccharides form when a hydoxyl group of one monosaccharide reacts with the hydroxyl group of another monosaccharide.C. a disaccharide consists of two monosaccharides linked to each other by an N-glycosidic bond.D. disaccharides are formed by a hydrolysis reaction.(21) The reducing end of a disaccharide or a polysaccharide isA. the end with an anomeric carbon that cannot be oxidized.B. the end that does not have an anomeric carbon.C. the end of a chain with a free anomeric carbon (i. e., not involved in a glycosidic bond).D. the end whose sugar cannot take the linear form.(22) What is the C-4 epimer of glucose?A. alloseB. mannoseC. guloseD. galactoseE. fructose(23) An N-linked glycan would be attached to ______ while an O-linked glycan would be attached to ________.A. Asn; SerB. Lys; TyrC. Gln; ThrD. Arg; SerE. Lys; Ser(24) Which are possible conformations of a furanose molecule?A. envelope and twist.B. chair and boat.C. cis and trans.D. A and B.(25) What type of bond links the monomers of a polysaccharide?A. glucotide bond.B. phosphate ester bond.C. peptide bond.D. glycosidic bond.(26) Who received a Nobel Prize in 1937 for his work on the structures of carbohydrates and the synthesis of Vitamin C?A. Linus Pauling.B. Walter Haworth.C. Max Perutz.D. Emil Fisher.(27) Which statement is false about the sugar units in DNA?A. they are cyclic in DNA.B. it is a deoxy form of ribose.C. it is an epimer of glucose.D. it has a D-configuration.(28) Chitin isA. found in insect and crustacean shells.B. found in fungi cell walls.C. composed of N-acetylglucosamine subunits.D. composed of linear fibrils like cellulose.E. all of the above.(29) A carbohydrate that gives two molecules when it is completely hydrolyzed is known as aA. monosaccharide.B. disaccharide.C. polysaccharide.D. starch.E. trisaccharide.(30) Which group of carbohydrates cannot be hydrolyzed to give smaller molecules?A. monosaccharidesB. disaccharidesC. trisaccharidesD. oligosaccharidesE. polysaccharides(31) In the L- isomer of a Fischer projection of a monosaccharide, the -OH group furthest from the carbonyl is writtenA. on the left of the top chiral carbon.B. on the right of the top chiral carbon.C. on the left of the middle chiral carbon.D. on the left of the bottom chiral carbon.E. on the right of the bottom chiral carbon.(32) One difference between D-glucose and L-glucose isA. the open-chain form of L-glucose does not exist.B. it is not possible to make L-glucose.C. L-glucose has a 5-membered ring , and D-glucose has a 6-membered ring.D. only D-glucose can act as a substrate in metabolic reactions.E. L-glucose cannot form a closed structure.(33) In a disaccharide, two monosaccharides are joined by what kind of bond?A. doubleB. anomericC. alcoholD. glycosidicE. rotational(34) Which of the following contains a β-1,4-glycosidic bond?A. galactoseB. lactoseC. maltoseD. sucroseE. amylose(35) Maltose is aA. monosaccharide.B. disaccharide.C. trisaccharide.D. polysaccharide.E. phosphosaccharide.(36) Amylose is aA. monosaccharide.B. disaccharide.C. trisaccharide.D. polysaccharide.E. phosphosaccharide.(37) Galactose is aA. monosaccharide.B. disaccharide.C. trisaccharide.D. polysaccharide.E. phosphosaccharide.(38) Which of the following contains α-1,6-branches?A. amyloseB. glycogenC. celluloseD. sucroseE. maltose(39) Cellulose is not digestible by humans because it contains glucose units linked by ________-glycosidic bonds.A. α-1,2B. α-1,4C. α-1,6D. β-1,2E. β-1,4(40) Amylose is a form of starch which hasA. only β-1,4-bonds between glucose units.B. only α-1,4-links bonds glucose units.C. both α-1,4-andβ-1,4-bonds between glucose units.D. hemiacetal links joining glucose units.E. carbon-carbon bonds joining glucose units.(41) Amylopectin is a form of starch which hasA. only β-1,4-bonds between glucose units.B. only α-1,4-links bonds glucose units.C. both α-1,4-andα-1,6-bonds between glucose units.D. hemiacetal links joining glucose units.E. carbon-carbon bonds joining glucose units.(42) Cellulose is a polysaccharide which hasA. only β-1,4-bonds between glucose units.B. only α-1,4-links bonds glucose units.C. both α-1,4-andα-1,6-bonds between glucose units.D. hemiacetal links joining glucose units.E. carbon-carbon bonds joining glucose units.(43) Humans cannot digest cellulose because theyA. lack the necessary enzymes to digest β-glycosides.B. are allergic to β-glycosides.C. are poisoned byβ-glycosides.D. have intestinal flora which use up β-glycosides.E. cannot digest chlorophyll.(44) Aspartame® and Saccharin® are two examples ofA. disaccharides.B. polysaccharides.C. chlorosaccharides.D. alcohol sweeteners.E. noncarbohydrate sweeteners.(45) Under acid hydrolysis conditions, starch is converted toA. glucose.B. xylose.C. maltose.D. galactose.E. fructose.(46) Fructose does not undergo hydrolysis because it is aA. aldose.B. hexose.C. reducing sugar.D. monosaccharide.E. disaccharide.(47) Galactose is a product of enzymatic hydrolysis ofA. lactose.B. glucose.C. maltose.D. erythrose.E. sucrose.(48) Which is not a glycoconjugate?A. Proteoglycan.B. Glycolipid.C. Glycoprotein.D. Homoglycan.(49) A saccharide is a ________.A. lipidB. carbohydrateC. nucleic acidD. protein(50) Which is a general term indicating a carbohydrate polymer?A. Glycan.B. Polycarb.C. Multimer.D. Oligosaccharide.(51) Which does not apply to dihydroxyacetone?A. Ketose.B. Triose.C. Chiral.D. Water-soluble.(52) Which structure is NOT a carbohydrate?A.B.C.D.(53) Which is true about naturally occurring monosaccharides?A. The L-isomers predominate.B. The D-isomers predominate.C. The L and D-isomers occur in equal ratios.D. The ratio of L and D-isomers varies widely depending on the source.(54) Examine the Fischer projection below. How is this carbohydrate classified?A. L enantiomer; aldopentose.B. L enantiomer; ketopentose.C. D enantiomer; aldohexose.D. D enantiomer; ketopentose.(55) The Fischer projections of linear D-glucose and D-galactose are shown below. These two molecules are ________.A. epimersB. enantiomersC. anomersD. structural (constitutional) isomers(56) The structures of D-ribose and D-arabinose are shown below. These two molecules are________.A. epimersB. enantiomersC. tautomersD. anomers(57) The intramolecular cyclization reaction of glucose in solution ________.A. generates a chiral centerB. yields a hemiacetalC. usually forms a pyranoseD. All of the above(58) Below is the Fischer projection of D-galactose. Which is the proper Haworth projection of β-D-galactopyranose?A. IB. IIC. IIID. IV(59) Below is the structure for a cyclic D-monosaccharide. Which is the anomeric carbon atom?A. 1B. 2C. 3D. 4E. 5(60) The compounds α-D-fructofuranose and β-D-fructofuranose are ________.A. enantiomersB. mutamersC. anomersD. conformational isomers(61) Anomers can be interconverted ________.A. by rotation about carbon-carbon bondsB. via a linear intermediateC. by an isotopic exchange reactionD. None of the above. Anomers cannot be interconverted.Ans: B(62) The abbreviation for glucose is ________.A. GluB. GcsC. GlcD. Gluc(63) The major proteoglycan in cartilage is ________.A. hyaluronic acidB. cellobioseC. link proteinD. aggrecan(64) Which of the following polysaccharides contains exclusively glucose in (14) glycosidicbonds?A. amyloseB. amylopectinC. glycogenD. celluloseE. chitin(65) Which of the following explains the importance of sucrose as a disaccharide?A. the glycosidic bond is much more stable than a typical glycosidic bondB. the presence of two glucose residues provides a readily available glucose sourceC. β-glycosidic bonds are much more stable than -glycosidic bondsD. since both anomeric carbons are involved in the glycosidic bond, oxidation is limitedE. none of the above(66) How many stereoisomers would be expected from a ketohexose?A. 4B. 8C. 16D. 32E. 64(67) Which of the following formulas would be a carbohydrateA. C5H12O5B. C4H8O6C. C6H10O6D. C3H8O3E. C6H12O63. True/False Questions(1) Most of the carbohydrates on earth are produced by photosynthesis in plants.(2) All carbohydrates have the empirical formula (CH2O)n.(3) The simplest aldose is the chiral molecule glyceraldehyde.(4) All D-enantiomers of carbohydrates are dextrorotatory.(5) The ketohexoses have fewer chiral carbon atoms than the aldohexoses.(6) The reaction between one molecule of alcohol and one molecule of aldehyde yields an acetal.(7) In solution only one anomeric form of a monosaccharide will be present.(8) Furanoses have more three-dimensional conformations than pyranoses.(9) Steric strain is a major factor in determining the conformations of a monosaccharide that predominate in solution.(10) The chair conformation of furanoses is generally preferred to minimize steric repulsion.(11) Vitamin C is an enediol of a lactone and is vital in the synthesis of collagen.(12) The sugar derivative found in DNA is an oxidized form of ribose.(13) Sugar alcohols do not have a carbonyl group.(14) DNA could be referred to as a glycoside.(15) Both nucleotides and nucleosides are glycosides.(16) For cattle, cellulose is a storage homoglycan.(17) The main difference between structures of amylopectin and glycogen is the specific sugar subunit.(18) Homoglycans are used for storage, while heteroglycans are used for cell structure.(19) α-amylases hydrolyze α(1→4) linkages and β-amylases hydrolyze β(1→4) linkages, but neither attack α1→6) linkages.(20) Enzymes that hydrolyze cellulose cannot hydrolyze amylopectin.(21) Cellulose cannot be degraded in cattle.(22) Enzymatic lengthening and degradation of glycogen occurs only at the non-reducing ends.(23) An oligosaccharide nodulation factor from one species can stimulate symbiosis in another species.(24) Unlike proteins, whose primary structures are encoded in DNA, polysaccharides have very little structural variations.(25) Heteroglycan chains may or may not be covalently bound to proteins in proteoglycans.(26) The heteroglycan component of the cell wall is the same in all bacteria, but the cross-linking polypeptide is not.(27) Some glycoproteins may have identical protein components and variable polysaccharide components.(28) The large number of possible oligosaccharide structures in glycoproteins is limited in cells by the kinds of enzymes available for their synthesis.(29) Sucrose is a disaccharide.(30) A reducing sugar gives a precipitate of silver metal with Benedict's reagent.(31) Sucrose is made up of glucose units only.(32) Amylopectin is a straight-chain polysaccharide.(33) Sucrose is a reducing sugar.4. Matching Questions(1) Select the correct carbohydrate for each description.A. sucroseB. glycogenC. galactoseD. maltoseE. glucoseF. celluloseG. fructoseH. lactoseI. amylase1) a carbohydrate that cannot be digested by humans2) a disaccharide that occurs as a breakdown product of starch3) a carbohydrate that stores energy in the human body4) a carbohydrate that is used to build cell walls in plants5) a monosaccharide that combines with glucose to form lactose6) a disaccharide found in milk and milk products7) a disaccharide consisting of glucose and fructose8) a monosaccharide found in fruit juices and honey, the sweetest carbohydrate9) an unbranched carbohydrate that stores glucose in plants10) the product when maltose is hydrolyzed(2) Indicate the monosaccharide(s) produced upon hydrolysis of each carbohydrate.A. glucose + galactoseB. glucose + fructoseC. glucose1) amylopectin2) lactose3) glycogen4) maltose5) sucrose5. Short Answer Questions(1) Draw the Fischer projection structures of all of the trioses and name them.(2) How is the D or L configuration determined?(3) Compare the structures of amylopectin and amylose.(4) How does glycogen differ from starch in structure and function?(5) Draw Howorth projection formulas for dimmers of glucose with the following types of glycosidic linkages:1). A β(1→4) linkage (both molecules of glucose in the β form)2). An α, α(1→1) linkage3). A (1→6) linkage (both molecules of glucose in the β form)(6) Glycogen is highly branched. What advantage, if any, does this provide an animal?(7) No animal is able to digest cellulose. Try to explain this statement with the fact that many animals are herbivores that depend heavily on cellulose as a food source.(8) What are the chemical and structural differences between cellulose and glycogen?(9) List some of the reasons carbohydrates are considered important molecules.(10) Using the structure of D-altrose, shown below, show a Haworth projection for the β anomer.(11) Starch, cellulose, and glycogen are all polysaccharides composed of glucose. What makes each of these polysaccharides unique in spite of the fact that all are composed solely of glucose?(12) List some of the reasons carbohydrates are considered important molecules.6. Fill in The Blanks(1) ________ is the major storage polysaccharide for animals, while ________ is the major storage polysaccharide for plants.(2) Cellulose is a polymer of ________, a common monosaccharide.(3) ________ and ________ are the two kinds of sugars found in bacterial cell walls.(4) The ________ projection illustrates saccharides as linear, whereas the ________ projection suggests the spatial relationships of saccharides.(5) Structurally, starch exists as both unbranched polysaccharide chains called ________ and branched polysaccharide chains called ________.(6) A _______________ is a stereoisomer that is not a mirror image.(7) A _______________ is a five-membered ring formed from a monosaccharide.(8) When an open-chain monosaccharide cyclizes into a ring, it forms a _______________.(9) Plant starch is composed of amylose, a linear polymer of glucose, and a branched polymer of glucose referred to as _______________.(10) Maltose is composed of two molecules of glucose linked together by a(n) ____________ glycosidic bond.(11) _______________ is a galactose joined to a glucose by a β-1,4 glycosidic bond.(12) In N-linked glycoproteins, the carbohydrate portion is attached to a(n)_____________ residue in the protein.(13) A given protein with several potential glycosylation sites can cave many different glycosylated structures called ________________.(14) The influenza virus recognizes _______________ residues of glycoproteins present on cell surface.。
关于碳水化合物的英文作文
关于碳水化合物的英文作文Carbohydrates are an essential source of energy for the body. They are found in a variety of foods, including fruits, vegetables, grains, and dairy products. When consumed, carbohydrates are broken down into glucose, which is used by the body as a primary source of fuel.Carbohydrates can be classified into two main types: simple carbohydrates and complex carbohydrates. Simple carbohydrates, also known as sugars, are found in foods such as candy, soda, and fruit. They are quickly digested and can cause a rapid increase in blood sugar levels. On the other hand, complex carbohydrates, found in foods like whole grains, legumes, and vegetables, take longer to digest and provide a more sustained release of energy.In addition to providing energy, carbohydrates also play a role in maintaining proper immune function and supporting the growth and repair of body tissues. They are also important for brain function, as the brain relies onglucose as its primary source of fuel.It's important to be mindful of the types of carbohydrates consumed, as excessive intake of simple carbohydrates can lead to weight gain and an increased risk of developing chronic diseases such as diabetes and heart disease. Choosing complex carbohydrates and incorporating them into a balanced diet can help support overall health and well-being.。
Glycoconjugate
Diastereomers and epimers
Glucose
Fructose
Epimers? Diastereomers
What is the structure and chemistry of monosaccharides?
4.Modification of one or more groups convert simple sugars into a variety of sugar derivatives.
3. The hexoses are structurally distinguished in configurations at one or more carbons. Diastereomer: differ in configuration at one or more carbons Epimers: differ in configuration at only one carbons.
Isomaltose
(glucose-β-1,4-glucose) (glucose-α-1,6-glucose)
Disaccharides are the simplest oligosaccharides
C. Oxidation of the terminal –CH2OH to –COOH converts the sugar into a-uronic acid, such as glucuronic acid.
What is the structure and chemistry of Oligosaccharides?
Simple sugars may take on several types of structures in solution. 1. simple sugars are classified according to the number of the carbons in the backbone. Eg. Pentose , hexoses 2. most sugars are asymmetric and designated as D or L-in sCteornefoigchueramtiisotnr:y D (dextrorotatory) or L- (levorotatory)
糖和苷(新)
异侧
O CH 2OH CH 2OH O CH 2OH
同侧
D-葡萄糖
O
O
同侧
β
α
异侧
单糖成环后新形成的一个不对称碳原子称为端基碳
糖的绝对构型（D、L）
CHO H C OH HO
CHO C H
--- 以α-OH甘油醛为标准
CH 2OH
CH 2OH
D 型
CHO CHO
L 型 α -OH甘油醛
COOH CHO COOH
Br2 / H2O
稀 HNO3
CH2 OH
CH2OH
COOH
H
伯醇 R C OH
H R' H C R OH
[O] R
H C OH OH
- H2O R
H C O
[O] R
OH C O
醛
OH - H2O R' C O R
羧酸 酮
[O]
R' C R
仲醇
OH
化学反应的活泼性：端基碳原子 > 伯碳 &g：
(1)按苷键原子的不同，酸水解难易程度为：
C-苷>S-苷>O-苷>N-苷 原因：N最易接受质子，而C上无未共享电子对，不能质子化。 (2)呋喃糖苷较吡喃糖苷易水解，水解速率大50-100倍。 原因：呋喃环平面性，各键重叠，张力大。 (3)酮糖较醛糖易水解。 原因：酮糖多呋喃环结构，且端基上接大基团-CH2OH 。图 (4)吡喃糖苷中，吡喃环C5上的取代基越大越难水解, 故有： 五碳糖>甲基五碳糖>六碳糖>七碳糖>5位接-COOH的糖 原因：吡喃环C5上的取代基对质子进攻有立体阻碍。 图
section 1概述
1985年，Feizi等首先提出了“糖分化抗原”的概念，
即在发育过程中细胞糖蛋白和糖脂所携带的糖类抗原 的改变是通过有序地逐个增加或减少糖残基而完成的。 其后发现，每个集团的细胞以不同的方式相互粘附， 细胞和基质之间也存在着相互识别和相互作用，集团 之间又相互识别、相互作用和相互制约，调节和控制 着高等生物沿着固有的空间轴和时间轴井然有序地发 展。这一复杂的发展过程中所需的极其巨大的“生物 信息”是由所含信息量比核酸和蛋白质大几个数量级 的糖链分子来承担。这就导致了“糖生物学”的诞生。
第一章 糖与糖生物学
chapter I Carbohydrate and Glycobiology
• 糖生物学：生命科学中的新前沿
糖类的研究已有百年的历史，研究成果表明，糖类是
生物体内除蛋白质和核酸以外的又一类重要的生物分 子，尤其是一类重要的信息分子。 1900， Landsteiner因发现ABO血型而获得1930年诺 贝尔生理和医学奖。这一发现在第一次世界大战期间 对抢救伤员作出了重大贡献。 1960年Witkins确定了ABO(H)的抗原决定簇是糖类， 并测定了有关糖类的结构。H抗原的前体是糖脂或糖 蛋白质中糖链非还原末端的二糖——半乳糖-N-乙酰氨 基葡萄糖(Gal-N-GlcNAc)。
2. 作为生物体内的主要能源物质
淀粉颗粒
糖原颗粒
肌糖原－能源
动物干重2％
结缔组织－结构糖
韧带－结构糖
3. 作为生物体内的碳源
CO2
光合 作用
分解 代谢
一切含碳物质
糖
合成 代谢
糖－碳来源
4. 作为细胞识别的信息分子
糖蛋白、糖脂、信息分子糖
细胞表面 识别标记
二. 糖的化学本质与分类
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Chapter 7 Carbohydrates and the glycoconjugates of cell surfacesWhat is the structure, chemistry, and biological function of carbohydrates?7.1 Classification of carbohydrates.Carbohydrates are the most abundant biomolecules on earth and have multiple roles in all forms of life..C n(H2O)n.Carbohydrates can be categorized into monosaccharides, oligosaccharides, and polysacchrides. .Monosacchrides are simple sugars consisting of a single polyhydroxyl aldehyde or ketone unit. .Oligosaccharides contain two(disaccharides) or a few monosaccharides joined by glycosidic bonds..Polysaccharides contain long chains of (hundreds to thousands) monosaccharide units joined by glycosidic bonds.7.2 What is the structure and chemistry of Monosacchrides.Carbohydrates are aldehyde or ketone compounds with multiple hydroxyl groups or substances that can yield such compounds on hydrolysis..Monosacchrides can be divided into two families: aldoses and ketoses..aldehyde: the carbonyl groups at the ends of the carbon chains.Ketoses: the carbonyl groups at places other than the ends.The simplest aldose is glyceraldehyde, and the simplest ketose is dihyoxyacetone, both being triose..Monosacchrides containing four, five, and six carbon atoms in their backbones are called.tetroses.pentoses (e.g., ribose and deoxyribose).hexoses (e.g., glucose and fructose respectively ).Each has both aldoses and ketoses.Hexoses are the most common monosacchrides in nature.D-glucose .D-mannose .D-galactose .D-fructose.Stereochemistry is a characteristics of Monosaccharide structure.All the monosaccharides except dihydroxyacetone contain one or more asymmetric (chiral) carbon atoms..Van’t Hoff’s rule : A compound with n asymmetric carbon atoms has a maximum of 2n possible stereoisomers..Stereoisomeric forms: Optical isomer.Enantiomer (mirror image isomer).Diastereomer (non mirror image isomer).Epimer (Diastereomers differ in configuration at a single asymmetric carbon atom, but not the reference carbon).Glyceraldehyde is conventionally used as the standard for defining D and L configurations:.D-glyceraldehyde has the -OH group on the right,.L-glyceraldehyde has the -OH group on the left..For sugars with more than one asymmetric carbon atom, the D- and L- symbols refer to the absolute configuration of the asymmetric carbon farthest to the carbonyl group.D-glucose: the -OH on C-5 has the same configuration as the asymmetric carbon in D-glyceraldehyde, therefore, D- and L- glucoses are not enantiomers but stereoisomers!.Most of the monosaccharides found in living organisms are the D-isomers.D-ribose .D-glucose .D-galactose.D-mannose.D-fructose.Ketoses are often named by inserting an “ul” into the name of the corresponding aldoses.Ribose is for aldopentose.ribulose is for the ketopentoseCyclic structure of monosaccharide.Monosaccharides easily form intramolecular hemiacetals (in aldoses) or hemiketals (in ketoses) in aqueous solutions..Ring formations occur in aqueous solution because aldehydes and ketones react reversibly with hydroxyl groups present in the sugar to form cyclic Hemiacetal..Mutarotation: Mutarotation process produces an equilibrium mixture of α- and β-forms in both Furanose and pyranose ring structure. The open chain form during Mutarotation can participate in oxidation-reduction reactions..Anomer.The carbonyl carbon becomes a new chiral center after cyclic reaction..Up-position β-anomeric form (D), α-anomeric form (L).Down-position α-anomeric form (D), β-anomeric form (L)Monosaccharide derivatives.These derivatives include mainly amino sugars(e.g., glucosamine, galactosamine, and mannosamine, N-acetyl-glucosamine), deoxy sugars(e.g., fucose--deoxygenated galactose, deoxyribose), and acidic sugars (e.g., gluconate--an aldonic acid, glucuronate--an uronic acid)..Amino sugars are often found in structural polysaccharides..Sugar phosphates are common metabolic intermediates in sugar metabolism.Oxidation-reduction reaction.Aldonic acid (oxidation of an aldehyde group), gluconic acid.Uronic acid (oxidation of a terminal CH2OH group but not the aldehyde group), glucuronic acid .Aldaric acid (oxidation of both the aldehyde and CH2OH group), glucaric acid.Lactone (The carbonyl group in both aldonic and uronic acid can react with an OH group in the same molecule to form a cyclic ester.).Alditol (Sugar alcohol after reduction of the aldehyde and ketone groups of monosaccharides, glucitol or sorbitol)Reducing sugar.The free carbonyl carbons (in open chains) in monosaccharides can reduce Cu2+ (cupric ion) to Cu+ (cuprous ion), which in turn forms a red cuprous oxide precipitate. (oxidoreduction reaction) .Such monosaccharides are called reducing sugars..This color reaction is the basis of Fehling’s reaction, a qualitative test for the presence of reducing sugars, was also used to estimate the glucose levels in blood and urine of diabetes patients.Glycoside formation.Hemiacetal reacts with alcohol to form acetal..Glycosidic linkage.Glycoside (glycoside, fructoside, furanoside, pyranoside).Glycosides are stable in basic solutions..糖苷它主要存在于植物的种子、叶子及皮内。