Carbohydrates and Glycobiology PDF

7 Carbohydrates and Glycobiology © 2021 Macmillan Learning  Carbohydrates carbohydrates = aldehydes or ketones with at least two hydroxyl groups, or substances that yield such compounds on hydrolysis many carbohydrates have the empirical formula (CH2O)n Produc...

7 Carbohydrates and Glycobiology © 2021 Macmillan Learning  Carbohydrates carbohydrates = aldehydes or ketones with at least two hydroxyl groups, or substances that yield such compounds on hydrolysis many carbohydrates have the empirical formula (CH2O)n Produced from CO2 and H2O via photosynthesis in plants Range from as small as glyceraldehyde (Mw = 90 g/mol) to as large as amylopectin (Mw > 200,000,000 g/mol) Can be covalently linked with proteins and lipids Fulfill a variety of functions, including:  Classes of Carbohydrates monosaccharides = simple sugars, consist of a single polyhydroxy aldehyde or ketone unit – example: D-glucose disaccharides = oligosaccharides with two monosaccharide units – example: sucrose (D-glucose and D-fructose) oligosaccharides = short chains of monosaccharide units, or residues, joined by glycosidic bonds polysaccharides = sugar polymers with  Clicker Question 1 A(n) is a(n) with two monosaccharide units. A. oligosaccharide; polysaccharide B. polysaccharide; oligosaccharide C. disaccharide; oligosaccharide D. disaccharide; polysaccharide  Clicker Question 1, Response A(n) is a(n) with two monosaccharide units. C. disaccharide; oligosaccharide Oligosaccharides consist of short chains of monosaccharide units, or residues, joined by characteristic linkages called glycosidic bonds. The most abundant are the disaccharides, with two monosaccharide units. 7.1 Monosaccharides and Disaccharides  Principle 1 Carbohydrates can have multiple chiral carbons; the configuration of groups around each carbon atom determines how the compound interacts with other biomolecules. As we saw for L-amino acids in proteins, with rare exceptions, biological evolution selected one stereochemical series (D-series) for sugars.  Stereoisomerism in Sugars sugar stereoisomers arise because many of the carbon atoms to which the hydroxyl groups are attached are chiral centers enzymes that act on sugars are stereospecific The Two Families of Monosaccharides Are Aldoses and Ketoses backbones of monosaccharides: – unbranched carbon chains with single bonds linking all carbon atoms – one of the carbon atoms is double-bonded to an oxygen atom to form a carbonyl group – other carbon atoms are bonded to a hydroxyl group  Aldoses and Ketoses aldose = carbonyl group is at an end of the carbon chain (in an aldehyde group) ketose = carbonyl group is at any other position (in a ketone group)  Clicker Question 2 What chemical feature determines if a sugar is an aldose or a ketose? A. the direction of rotation of polarized light B. the position of the carbonyl carbon C. if the enantiomers resemble either glyceraldehyde or dihydroxyacetone D. if the epimers differ in configuration around one carbon or more than one carbon  Clicker Question 2, Response What chemical feature determines if a sugar is an aldose or a ketose? B. the position of the carbonyl carbon The backbones of common monosaccharides are unbranched carbon chains in which all the carbon atoms are linked by single bonds. In this open-chain form, one of the carbon atoms is double-bonded to an oxygen atom to form a carbonyl group. If the carbonyl group is at an end of the carbon chain, the monosaccharide is an aldose. If the carbonyl group is at any other position, it is a ketose.  Monosaccharide Carbon Backbone 3 C  triose 4 C  tetrose 5C pentose 6 C  hexose 7C heptose 1 3  Trioses C3H6O3 trioses = simplest monosaccharides, three carbon backbone  Tetroses and Pentoses tetroses = four compone compone carbon nt of nt of backbone RNA DNA pentoses = five carbon backbone C5H10O5  Hexoses and Heptoses C6H12O6 hexoses = six carbon backbone heptoses = seven carbon backbone  Clicker Question 3 Glucose is a monosaccharide with a(n): A. aldehyde functional group and six carbons. B. ketone functional group and six carbons. C. aldehyde functional group and seven carbons. D. ketone functional group and seven carbons.  Clicker Question 3, Response Glucose is a monosaccharide with a(n): A. aldehyde functional group and six carbons. Glucose is an aldohexose, a monosaccharide with an aldehyde functional group and six carbons.  What Makes Sugar Sweet? TAS1R2 and TAS1R3 encode sweet- taste receptors binding of a compatible molecule generates a “sweet” electrical signal in the brain – requires a steric match  Clicker Question 4 Why does (R,S)-aspartame NOT stimulate the sweet receptor? A. Site AH+ cannot bind the partially negative oxygen of the carboxylic acid. B. Site B– cannot hydrogen bond with the amine nitrogen. C. Site X cannot accommodate the hydrophobic benzene ring. D. None of the sites can interact with (R,S)- aspartame.  Clicker Question 4, Response Why does (R,S)-aspartame NOT stimulate the sweet receptor? C. Site X cannot accommodate the hydrophobic benzene ring. The steric match must be correct to stimulate the sweet receptor. Site X is oriented perpendicular to AH+ and B– and can accommodate the hydrophobic benzene ring of (S,S)-aspartame. However, Site X cannot accommodate the hydrophobic benzene ring of (R,S)-aspartame. Monosaccharides Have Asymmetric Centers all monosaccharides (except dihydroxyacetone) contain 1+ chiral carbon atom – occur in optically active isomeric forms enantiomers = two different optical isomers that are mirror images a molecule with n chiral centers can have 2n stereoisomers  Fischer Projection Formulas used to represent three-dimensional sugar structures on paper bonds drawn horizontally indicate bonds that project out of the plane of the paper bonds drawn vertically project behind the plane of the paper Enantiomers of Glyceraldehyde  Numbering Carbons of a Sugar carbons are numbered beginning at the end of the chain near the carbonyl group  D Isomers and L Isomers reference carbon = chiral center most distant from the carbonyl carbon two groups of stereoisomers: – D isomers = configuration at reference carbon is the same as D-glyceraldehyde on the right (dextro) in a projection formula most hexoses of living organisms – L isomers = configuration at reference carbon is the same as L-glyceraldehyde on the left (levo) in a projection formula  Non- superimposable mirror images  Non- superimposable mirror images  Clicker Question 5 Which statement is false regarding the enantiomers of glyceraldehyde? A. The enantiomers of glyceraldehyde are mirror images of each other. B. Because glyceraldehyde contains one chiral center, it has two enantiomers. C. In the Fischer projection formula for D- glyceraldehyde, the hydroxyl group is on the right. D. Glyceraldehyde does not occur in optically active isomeric forms.  Clicker Question 5, Response Which statement is false regarding the enantiomers of glyceraldehyde? D. Glyceraldehyde does not occur in optically active isomeric forms. All the monosaccharides except dihydroxyacetone contain one or more asymmetric (chiral) carbon atoms and thus occur in optically active isomeric forms. The simplest aldose, glyceraldehyde, contains one chiral center (the middle carbon atom) and therefore has two different optical isomers, or enantiomers. D-Aldoses D-Ketoses  Epimers epimers = two sugars that differ only in the configuration around one carbon atom D-Mannose and D-galactose are both epimers of D-glucose. D-Mannose and D-galactose vary at more than one chiral center and are diastereomers, but  Structures to Know Glyceraldehyde & dihydroxyacetone are the standard three-carbon sugars Ribose is the standard five-carbon sugar Glucose is the standard six-carbon sugar Galactose is an epimer of glucose Mannose is an epimer of glucose Fructose is the ketose form of glucose The Common Monosaccharides Have Cyclic Structures in aqueous solution, aldotetroses and all monosaccharides with 5+ backbone carbon atoms occur as cyclic structures – covalent bond between the carbonyl group and the oxygen of a hydroxyl group  Clicker Question 6 A monosaccharide with a ketone functional group and seven carbons: A. can never exist in cyclic form. B. is a ketoheptose. C. is the monosaccharide found in RNA but not DNA. D. has a single chiral center.  Clicker Question 6, Response A monosaccharide with a ketone functional group and seven carbons: B. is a ketoheptose. Monosaccharides with seven carbon atoms in their backbones are called heptoses. A ketose of this chain length is a ketoheptose.  Hemiacetals and Hemiketals hemiacetals or hemiketals = derivatives formed by a general reaction between alcohols and aldehydes or ketones – product of the first alcohol molecule addition – a five- or six-membered ring forms if the — OH and carbonyl groups are on the same molecule acetal or ketal = product of the second alcohol molecule addition – forms a glycosidic bond Formation of Hemiacetals and Hemiketals  α and β Stereoisomeric Configurations reaction with the first alcohol molecule creates an additional chiral center (the carbonyl carbon) produces either of two stereoisomeric configurations: α (trans) and β (cis) anomers = isomeric forms of monosaccharides that differ only in their configuration about the hemiacetal or hemiketal carbon atom  Clicker Question 7 What name is given to monosaccharides that differ in configuration about the hemiacetal carbon atom? A. enantiomers B. isomers C. epimers D. anomers  Clicker Question 7, Response What name is given to monosaccharides that differ in configuration about the hemiacetal carbon atom? D. anomers Isomeric forms of monosaccharides that differ only in their configuration about the hemiacetal or hemiketal carbon atom are called anomers. Formation of the Two Cyclic Forms of D-Glucose reaction between the aldehyde group at C-1 and the hydroxyl group at C- 5 forms a hemiacetal linkage mutarotation = the interconversion of α and β anomers  Clicker Question 8 Cyclization of monosaccharides: A. is the reaction of hemiketal or hemiacetal formation. B. is irreversible. C. creates α and β epimers. D. only occurs in hexoses.  Clicker Question 8, Response Cyclization of monosaccharides: A. is the reaction of hemiketal or hemiacetal formation. In aqueous solution, all monosaccharides with five or more carbon atoms in the backbone occur predominantly as cyclic (ring) structures in which the carbonyl group has formed a covalent bond with the oxygen of a hydroxyl group in the same sugar molecule. The formation of these ring structures is the result of a general reaction between alcohols and aldehydes or ketones to form derivatives called hemiacetals or hemiketals.  Pyranoses and Furanoses pyranoses = six- membered ring compounds – form when the hydroxyl group at C- 6 reacts with the keto group at C-2 furanoses = five- membered ring compounds – form when the hydroxyl group at C- 5 reacts with the  Aldonic and Uronic Acids aldonic acids = form following oxidation of the carbonyl carbon of aldoses uronic acids = form following oxidation at C-6 both form stable intramolecular esters called lactones  Phosphorylated Derivatives some sugar intermediates are phosphate esters – example: glucose 6-phosphate stable at neutral pH and bear a negative charge functions to trap sugar inside the cell because most cells do not have membrane transporters for phosphorylated sugars  Sugars That Are, or Can Form, Aldehydes Are Reducing Sugars reducing sugars = undergo a characteristic redox reaction where free aldehyde groups react with Cu2+ under alkaline condition – reduction of Cu2+ to Cu+ forms a brick-red precipitate ketoses that can tautomerize to form aldehydes are also reducing sugars  Chain-Ring Equilibrium and Reducing Sugars The ring forms exist in equilibrium with the open-chain forms Reducing sugars have a free anomeric carbon Aldehyde can reduce Cu2+ to Cu+ (Fehling’s test) Aldehyde can reduce Ag+ to Ag0 (Tollens’ test) Allows detection of reducing sugars, such as glucose (by measuring the amount of oxidizing agent reduced by a sugar The cuprous ion (Cu+) produced forms a red cuprous oxide precipitate.  Clicker Question 10 Which reaction is one that is NOT common of glucose? A. mutarotation B. oxidation to an aldonic acid C. oxidation to an uronic acid D. oxidation by reducing sugars  Clicker Question 10, Response Which reaction is one that is NOT common of glucose? D. oxidation by reducing sugars Glucose does not undergo oxidation by reducing sugars. Glucose itself is a reducing sugar. Free aldehyde groups in glucose undergo a characteristic redox reaction with Cu2+ under alkaline conditions. As glucose is oxidized from aldehyde to carboxylic acid, Cu2+ is reduced to Cu+, which forms a brick-red precipitate. Blood glucose measurements in the diagnosis and treatment of diabetes Untreated diabetes has several long-term consequences: kidney failure, cardiovascular disease, blindness, impaired wound healing, etc. Average [glc]blood over days can be measured because of a nonenzymatic reaction between glc and primary amino groups in Hb The amount of glycated Hb (GHb) reflects the average [glc]blood over the circulating lifetime of RBC (~120 days) Normal levels ~5% of Hb is GHb Diabetic people may have it as high as 13%  O-Glycosidic Bonds O-glycosidic bond = covalent linkage joining two monosaccharides – formed when a hydroxyl group of one sugar molecule reacts with the anomeric carbon of the other – The resulting compound is a  The Reducing End formation of a glycosidic bond renders a sugar nonreducing reducing end = the end of a disaccharide or polysaccharid e chain with a free anomeric Free anomeric carbon carbon  Clicker Question 11 What term is given to carbohydrates linked by their anomeric carbons? A. nonreducing sugars B. glycosides C. anomers D. hemiketals  Clicker Question 11, Response What term is given to carbohydrates linked by their anomeric carbons? A. nonreducing sugars When the anomeric carbon is involved in a glycosidic bond, the easy interconversion of linear and cyclic forms is prevented. Formation of a glycosidic bond therefore renders a sugar nonreducing. Naming Reducing Oligosaccharides step 1: with the nonreducing end on the left, give the configuration (α or β) at the anomeric carbon joining the first unit to the second step 2: name the nonreducing residue using “furano” or “pyrano” step 3: indicate in parentheses the two carbon atoms joined by the glycosidic bond, with an arrow connecting the two numbers step 4: name the second residue and repeat for additional residues  Clicker Question 12 Following the convention for naming reducing oligosaccharides, what is the name of lactose, as shown. A. α-L-glucopyranose- (1→6)-α-L- galactopyranosyl B. β-D-glucopyranose- (1→4)-β-D- galactopyranosyl C. β-D-galactopyranosyl-  Clicker Question 12, Response Following the convention for naming reducing oligosaccharides, what is the name of lactose, as shown. C. β-D-galactopyranosyl-(1→4)-β-D-glucopyranose The image is drawn with the nonreducing end on the left. The anomeric carbon joining the first unit to the second is β and the nonreducing unit is the six-membered ring D-galactose. A (1→4)-glycosidic bond links the two monosaccharides together. The second residue is the six-membered ring D-glucose.  Symbols and Abbreviations for Monosaccharides and Derivatives  Three Common Disaccharides lactose is a reducing disaccharide sucrose and trehalose are nonreducing sugars  Clicker Question 13 Which of these sugars is nonreducing? A. trehalose B. glucose C. maltose D. lactose  Clicker Question 13, Response Which of these sugars is nonreducing? A. trehalose When the anomeric carbon is involved in a glycosidic bond, the easy interconversion of linear and cyclic forms is prevented. Formation of a glycosidic bond therefore renders a sugar nonreducing. Trehalose is a disaccharide of D- glucose that is a nonreducing sugar.  Clicker Question 14 Which item is NOT associated with disaccharides? A. O-glycosidic linkages B. formation of a ketal or acetal C. never having free anomeric carbons D. sucrose  Clicker Question 14, Response Which item is NOT associated with disaccharides? C. never having free anomeric carbons In describing disaccharides and polysaccharides, the end of a chain with a free anomeric carbon (one not involved in a glycosidic bond) is called the reducing end. There are a number of disaccharides with free anomeric carbons, including glucose and lactose. 7.2 Polysaccharides  Principle 2 Monomeric subunits, monosaccharides, serve as the building blocks of large carbohydrate polymers. The specific sugar, the way the units are linked, and whether the polymer is branched determine its properties and thus its function.  Polysaccharides most carbohydrates in nature occur as polysaccharides (Mr > 20,000) also called glycans Video: 1) Carbohydrates & sugars – biochemistry (Osmosis)  Homopolysaccharides and Heteropolysaccharides homopolysaccharides = contain only a single monomeric sugar species – serve as storage forms and structural elements Can be: 1) Storage forms of monosaccharides, used for fuel (starch and glycogen) 2) Structural elements in plant cell walls and animal exoskeletons (cellulose and chitin)  Homopolysaccharides and Heteropolysaccharides heteropolysaccharides = contain 2+ kinds of monomers – provide extracellular support Provide extracellular support for organisms of all kingdoms, E.g. 1) Bacterial cell envelope is composed in part of a heteropolysaccharide 2) Extracellular matrix in animal cells  Clicker Question 15 Which statement is false regarding homopolysaccharides and heteropolysaccharides? A. Homopolysaccharides contain a single monomeric sugar species. B. Some homopolysaccharides serve as structural elements in animal exoskeletons. C. Heteropolysaccharides serve as storage forms of monosaccharides that are used as fuel. D. In animal tissues, the extracellular space is occupied by several types of heteropolysaccharides.  Clicker Question 15, Response Which statement is false regarding homopolysaccharides and heteropolysaccharides? C. Heteropolysaccharides serve as storage forms of monosaccharides that are used as fuel. Some homopolysaccharides serve as storage forms of monosaccharides that are used as fuels; starch and glycogen are homopolysaccharides of this type.  Principle 4 The sequences of complex polysaccharides are determined by the intrinsic properties of the biosynthetic enzymes that add each monomeric unit to the growing polymer. This is in contrast with DNA, RNA, and proteins, which are synthesized on templates that direct their sequence. Polysaccharides Generally Do Not Have Defined Lengths or Molecular Weights this distinction between proteins and polysaccharides is a consequence of the mechanisms of assembly there is no template for polysaccharide synthesis the program for polysaccharide synthesis is intrinsic to the enzymes that catalyze the polymerization of monomer units Some Homopolysaccharides Are Storage Forms of Fuel storage polysaccharides = starch in plant cells and glycogen in animal cells starch and glycogen molecules are heavily hydrated because they have many exposed hydroxyl groups available to hydrogen bond  Starch and Glycogen starch = contains two types of glucose polymers, amylose and amylopectin – amylose = long, unbranched chains of D- glucose residues connected by (α1→4) linkages – amylopectin = larger than amylose with (α1→4) linkages between glucose residues and highly branched due to (α1→6) linkages glycogen = polymer of (α1→4)-linked glucose subunits, with (α1→6)-linked branches Structure of Starch and Glycogen  Clicker Question 16 What O-glycosidic bond is commonly found in amylose, amylopectin, and glycogen? A. (β1⭢4) B. (α1⭢6) C. (β1⭢6) D. (α1⭢4)  Clicker Question 16, Response What O-glycosidic bond is commonly found in amylose, amylopectin, and glycogen? D. (α1⭢4) Amylose, amylopectin, and glycogen all contain (α1→4) linkages joining successive glucose residues.  Clicker Question 17 A glycogen molecule with 28 branches has how many nonreducing and reducing ends? A. 28 nonreducing ends and 28 reducing ends B. 29 nonreducing ends and 1 reducing end C. 1 nonreducing end and 29 reducing ends D. 1 nonreducing end and 1 reducing end  Clicker Question 17, Response A glycogen molecule with 28 branches has how many nonreducing and reducing ends? B. 29 nonreducing ends and 1 reducing end Because each branch of glycogen ends with a nonreducing sugar unit, a glycogen molecule with n branches has n + 1 nonreducing ends, but only one reducing end. Thus, a glycogen molecule with 28 branches has 29 nonreducing ends and 1 reducing end.  Principle 3 Storage of low molecular weight metabolites in polymeric form avoids the very high osmolarity that would result from storing them as individual monomers. If the glucose in liver glycogen were monomeric, the glucose concentration in liver would be so high that cells would swell and lyse from the entry of water by osmosis. Storage of Glucose as Polymers Avoids High Osmolarity hepatocytes in the fed state store glycogen equivalent to a glucose concentration of 0.4 M 0.4 M glucose in the cytosol would elevate the osmolarity – the resulting osmotic entry of water might rupture the cell  Clicker Question 18 Why is it logical for sugars to be added to only one end of glycogen, making them excellent molecules for glucose storage? A. Because the other end of the molecule is anchored to a membrane with a lipid. B. Because it is the reducing end, and a reduction reaction is the mechanism for adding the sugar. C. Because it has the hemiacetal carbon, the one carbon to which more sugars can be attached. D. Because there are many nonreducing ends on one molecule, allowing rapid glucose storage and release.  Clicker Question 18, Response Why is it logical for sugars to be added to only one end of glycogen, making them excellent molecules for glucose storage? D. Because there are many nonreducing ends on one molecule, allowing rapid glucose storage and release. A glycogen molecule with n branches has n + 1 nonreducing ends. When glycogen is used as an energy source, glucose units are removed one at a time from the nonreducing ends. Degradative enzymes that act only at nonreducing ends can work simultaneously on the many branches, speeding the Some Homopolysaccharides Serve Structural Roles cellulose = tough, fibrous, water-insoluble substance – linear, unbranched homopolysaccharide, consisting of 10,000 to 15,000 D-glucose units – glucose residues have the β configuration – linked by (β1→4) glycosidic bonds – animals do not have the enzyme to hydrolyze (β1→4) glycosidic bonds  Clicker Question 19 Even though amylose and cellulose are made of similar homopolysaccharide chains, they have very different properties. Why? A. The β-glycosidic linkage of glucose molecules in cellulose form interchain and intrachain hydrogen bonds that produce straight, stable fibers that exclude water. B. Cellulose is composed of galactose, while amylose is composed of glucose. C. The α-glycosidic linkage of glucose molecules in amylose causes it to form helices that exclude water. D. Amylose makes a linear polymer which  Clicker Question 19, Response Even though amylose and cellulose are made of similar homopolysaccharide chains, they have very different properties. Why? A. The β-glycosidic linkage of glucose molecules in cellulose form interchain and intrachain hydrogen bonds that produce straight, stable fibers that exclude water. The glucose residues in cellulose are linked by (β1→4) glycosidic bonds, allowing for the formation of interchain and intrachain hydrogen bonds. The Challenge of Glycobiology glycobiology = the study of the structure and function of glycoconjugates the challenge is to understand how cells use specific oligosaccharides to encode information about: – intracellular targeting of proteins – cell-cell interactions – cell differentiation and tissue development – extracellular signals  Oligosaccharide Structures Are Information-Dense branched structures, not found in nucleic acids or proteins, are common in oligosaccharides almost limitless variety of oligosaccharides due to differences in: – stereochemistry and position of glycosidic bonds – type and orientation of substituent groups – the number and type of branches  Lectins Are Proteins That Read the Sugar Code and Mediate Many Biological Processes lectins = bind carbohydrates with high specificity and with moderate to high affinity functions: – cell-cell recognition – signaling – adhesion – intracellular targeting of newly synthesized proteins  Selectins selectins = family of plasma membrane lectins that mediate cell-cell recognition and adhesion in a wide range of cellular processes – move immune cells through the capillary wall – mediate inflammatory responses – mediate the rejection of transplanted organs  Clicker Question 30 Which statement about selectins is false? A. They mediate cell-cell recognition. B. They are involved in the movement of immune system cells. C. They can be involved in the process rejection of transplanted organs. D. They are intracellular.  Clicker Question 30, Response Which statement about selectins is false? D. They are intracellular. Selectins are a family of plasma membrane lectins that mediate cell-cell recognition and adhesion in a wide range of cellular processes. Role of Lectin-Ligand Interactions in Leukocyte Movement Binding Site on Influenza Neuraminidase  Lectin-Carbohydrate Interactions Are Highly Specific and Often Multivalent subtle molecular complementarity allows interaction only with the lectin’s correct carbohydrate binding partners  Lectin Multivalency lectin multivalency = single lectin molecule has multiple carbohydrate binding domains – increases effective affinity Interactions of Sugar Residues Due to the Hydrophobic Effect many sugars have a more polar side and a less polar side  Clicker Question 31 Lectins: A. often bind their ligands via multiple weak interactions. B. bind their ligands with relatively low specificity. C. prevent viruses from binding to their target cells. D. are carbohydrates that bind to receptor proteins.  Clicker Question 31, Response Lectins: A. often bind their ligands via multiple weak interactions. Many sugars have a more polar side and a less polar side; the more polar side hydrogen-bonds with the lectin, while the less polar side undergoes interactions with nonpolar amino acid residues through the hydrophobic effect. The sum of all these interactions produces high-affinity binding and high specificity of lectins for their carbohydrate ligands. Biological Interactions Mediated by the Sugar Code