Lecture 4: Carbohydrates Reference: Lieberman and Marks Chapter 5 and Lecture Notes I. Explain the basic nomenclature of carbohydrates. A. Carbohydrates are classified into three groups: monosaccharides (and their derivatives), oligosaccharides, and polysaccharides 1. Monosaccharides (simple sugars) cannot be broken down into simpler sugars under mild conditions a.) Simple monosaccharides consist of a linear chain of three or more carbon atoms one of which forms a double bond with oxygen 2. Oligosacchardies : oligo, meaning “few,” and consist of from two to ten simple sugar molecules 3. Polysaccharides are polymers of the simple sugars – starch or glycogen a.) Most sugars end with the suffix “–ose” : glucose b.) Most sugars are in the D-configuration -For sugars, the D or L designation refers to the highest numbered chiral carbon atom - D rotates the polarization around the chiral carbon clockwise and L rotates the polarization counter-clockwise II. Describe the basic structural features of carbohydrates in terms of aldose or ketose grouping, numbers of carbons and stereochemistry A. Aldose Grouping -Contains an aldehyde functional group and if it contains 3C or more it is chiral B. Ketose Grouping -Contains a ketone functional group and if it contains 4C or more it is chiral III. Draw the structures of dihydroxyacetone and D-glyceraldehyde (this will serve you well during metabolism). -See above IV. Define the terms “pyranose” and “furanose”. A. Pyranose: Cyclic form of aldose (but can also be a furanose) B. Furanose: Cyclic form of ketose (but can also be a pyranose) V. Explain ring formation by carbohydrates. A. Identify the anomeric carbon in a ring structure. -The “anomeric carbon” is the carbonyl carbon. Anytime an anomeric carbon has an –OH group attached to it, it will be a free anomeric carbon. However, if the –OH group is replaced by another atom (say C or N) or the –OH group is changed to an –O-R group, then the anomeric carbon cannot reversilbly form the ring and linear structures. In this case, the anomeric carbon is no longer “free”. B. Identify whether the anomeric carbon is in an α versus β configuration and how this occurred during ring formation. Carbohydrate molecules can form cyclic structures when an –OH group attacks the carbonyl carbon, forming a new covalent bond. Because of ring geometry and bond strain, only 5-member (furanose) and 6-member (pryanose) rings will be formed. The carbonyl carbon (the anomeric carbon) and its substituents form a planar bond structure because of the double bonded C=O. Because of this, the attacking –OH can add to the anomeric carbon from the top of the plane, causing the carbonyl oxygen to become an OH group that now points down below the original plane (this is the alpha configuration), or the attacking –OH can add to the anomeric carbon from the bottom of the plane, causing the carbonyl oxygen to become an -OH group that now points up above the original plane (this is the beta configuration). There is an equal probability of the two states forming so both alpha and beta forms will exist simultaneously C. Explain how pyranose versus furanose rings are formed and which form will dominate based on whether the carbohydrate is an aldose or ketose. -Ring formation read above - Bond strain, ring substitutions, etc, will favor one form over the other. Aldoses “prefer” the pyranose form whereas ketoses “prefer” the furnanose form. -Not exclusive, but preferred. These configurations are interchangeable VI. Define the term “reducing sugar” and explain how a reducing sugar can reduce another molecule or ion. A. Reducing sugars are sugars with free anomeric carbons - they will reduce oxidizing agents, such as peroxide, ferricyanide and some metals (Cu and Ag) B. In glucose the aldehyde group contains significant electron density because of the carbonyl group and the simple hydrogen atom attached to the carbonyl carbon. Because of this, the aldehyde group can give an electron to another atom. In the process the aldehyde gets oxidized (i.e. it loses an electron = oxidized) and in the process becomes a carboxylic acid. The other atom gets reduced (i.e., gains an electron). VII. List and describe some common monosaccharide derivatives. • Sugar alcohols: result from reduction of sugars • Deoxy sugars: constituents of DNA, etc. • Sugar esters: phosphate esters like ATP are important • Amino sugars contain an amino group in place of a hydroxyl group • Acetals, ketals and glycosides: basis for oligo- and poly-saccharides VIII. Define the term “acetylated” amino sugar. -The acetyl group is a carbonyl with a methyl group on it. -CO-CH3 IX. List the common amino sugar disaccharides found in glycosaminoglycans X. Explain the basic structural features of oligosaccharides and the basis of their nomenclature. A. A saccharide polymer containing a small number (typically two to ten) of component sugars, also known as simple sugars (monosaccharides) B. Individual units are called “residues” -be able to identify anomeric carbons and reducing and non-reducing ends. -Note carefully the nomenclature of links! Be able to recognize alpha(1,4), beta(1,4), etc Naming of oligosaccharides proceeds from left to right (from the reducing end to the terminal) In parentheses, between the names of the monosaccharide residues, the number of the anomeric carbon atom, an arrow symbol, and the number of the carbon atom bearing the connecting oxygen of the next monosaccharide unit are listed. Appropriate symbols are used to indicate the stereochemistry of the glycosidic bonds (α or β), the configuration of the monosaccharide residue (D orL), and the substitutions at oxygen atoms XI. Describe the glycosidic bond and define the α versus β configuration. A. Glycosidic bonds join sugars B. Alpha is below the plane of the ring and beta is above the plane XII. Explain why sucrose does not have a reducing end, but lactose does. A. Sucrose does not have a free anomeric carbon and therefore is unable to participate in oxidation/reduction reactions. Lactose has an anomeric carbon on the glucose end (This is known as the reducing end). XIII. Explain that starch and glycogen are glucose storage systems, and that individual glucose molecules can be removed as needed. A. Starch (plants) 1. Two forms: amylose (linear chain) and amylopectin (branched structure). 2. Most starch is 10-30% amylose and 70-90% amylopectin B. Glycogen (animals) 1. Glycogen constitutes up to 10% of liver mass and 1-2% of muscle mass 2. Very similar to amylopectin: except much greater number of branches C. Both are long glucose chains and the glucose molecules can be cleaved off and used as needed XIV. Explain that α-1,4-linked glucose is found in glycogen and that β-1,4-linked glucose is found in cellulose. Explain some structural differences that arise from this difference. A. The α-1,4-linked glucose is flexible and forms twisting branched chains that can be easily broken down in the body. The β-1,4-linked glucose is stiff and forms strong tightly packed sheets that become extended ribbons, and our bodies are unable to break them down.