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Carbohydrates (saccharides) Organic biomolecules: 1.Carbohydrates 2.Lipids 3.Proteins 4.Nucleic acids.

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Presentation on theme: "Carbohydrates (saccharides) Organic biomolecules: 1.Carbohydrates 2.Lipids 3.Proteins 4.Nucleic acids."— Presentation transcript:

1 Carbohydrates (saccharides) Organic biomolecules: 1.Carbohydrates 2.Lipids 3.Proteins 4.Nucleic acids

2 Functions of carbohydrates Energy: metabolism and storage Structural and protective Precursors for other biomolecules Cell recognition and regulation Components of DNA and RNA units Look for these functions in the chapter: - which molecules that we discuss fit which category?

3 Terminology Basic building block: monosaccharides –Polyhydroxyaldehydes (figure 18.1) –Polyhydroxyketones (figure 18.2) Bigger units constructed of monosaccharides: polysaccharides –Disaccharides –Oligosaccharides –Polysaccharides

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6 Important points of monosaccharides: Biologically important sugars are generally 3-6 C’s in length Number from nearest the carbonyl end The stereochemistry of important hexoses in biology is ____ at the next-to-terminal carbon Carbohydrate = carbo + hydrate

7 Sugar terminology: -ose Aldoses = Ketoses = Triose, tetrose, pentose, hexose: Note: sugars have several identities so what terminology you use depends on what identity you are addressing

8 Important monosaccharides: Hexoses: –D-glucose = blood sugar, dextrose –D-fructose = levulose –D-galactose: –D-mannose: Pentoses: –D-ribose (deoxyribose –at C2 - in DNA) –D-xylose

9 Construct sugar model Furanose = 5 membered ring Pyranose = 6 membered ring Construct D-glucose as open chain, then convert to cyclic form

10 Conversion of D-glucose to its cyclic forms: Pyranose or furanose? Anomers: differ in configuration at the hemiacetal carbon (C1 here) (Fischer projection) (Haworth projections)

11 Anomers: Hexoses and pentoses form stable hemiacetals which are 5 or 6 membered rings Each hemiacetal structure is one of two possible anomers (alpha or beta) Practice converting Fischer projections into Haworth projections

12 Do alpha or beta anomers matter? Yes they do! (stay tuned)

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14 Monosaccharide properties Sweetness Very water soluble Crystalline at RT Not soluble in nonpolar solvents –Slightly soluble in EtOH, MeOH Undergo mutarotation to rapid equilibration in water between anomeric forms –Each sugar has its own characteristic equilibrium ratio; we have the example of glucose –Positive Benedict’s test because ___________

15 Check your lab results: Monosaccharides: –Solubility in water –Benedict’s test –Crystalline at RT Diastereomers with characteristic m.p.’s –Optical rotation Diastereomers with unique optical activity

16 Disaccharides Dehydration reaction between a hemiacetal and an alcohol forms a glycoside ( = an acetal of a carbohydrate) This linkage is made through a stereospecific enzyme-catalyzed reaction in biological systems –The difference in starch vs. cellulose: digestibility by humans is an example of structure → function

17 Maltose: malt sugar or corn sugar; From action of amylase on starch Benedict’s test results:

18 Cellobiose = from partial hydrolysis of cellulose, by enzyme cellulase Reducing sugar = Benedict’s test results: _____

19 Lactose = milk sugar Reducing sugar?____

20 Sucrose = table sugar; sugar cane Reducing sugar?_____ (check your lab results!)

21 Polysaccharides Storage –Starch –glycogen Structural –Cellulose –Chitin Recognition –Glycolipids –glycoproteins Essentially nonreducing sugars: Benedicts ____ mutarotation____

22 Starch: shown here is a small unit What type of glycosidic linkage is shown between glucose units?_________ Amylose (10-30%) linear, long chains

23 Amylopectin: 70-90%, branched Glycogen: even more branched Starch continued:

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25 Check your lab results: Iodine test:____

26 Cellulose: Glycosidic linkage is _________ Planar formation of chains Need cellulase to cleave this linkage: provided by microorganisms of ruminants Fiber for humans

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30 Cell recognition The ABO blood groups are major sugars, expressed on cell membranes, called antigens Antibody response to these particular sugar attachments is expressed early in life, presumably due to similar environmental immunogens exposures; so everybody has the antibodies appropriate to their blood type.

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33 Frequency of ABO phenotype in US population WhiteBlackAsian Native American O 45494379 A 4127 16 B 1019254 AB 445<1

34 Frequency of ABO phenotypes worldwide About 21% of worldwide population has A allele. Distribution in native populations worldwide

35 Type B: 16% of worldwide population has the allele Distribution in native populations worldwide

36 About 63% of world population is type O Distribution in native populations worldwide http://anthro.palomar.edu/vary/images/map_of_A_blood_allele.gif

37 Other antigens D-antigens: a different sugar chain attached to the cell membrane If you have D-antigen, you are Rh+ If you don’t express D-antigen, you are Rh- Ratio of people who are Rh+ : Rh- –White Americans: 85:15 –Black Americans: 92:8

38 More on antigens Duffy antigens (Fya and Fyb) provide a recognition site for parasite plasmodium vivax to dock and enter cell White Americans: 65% Fya+; 80% Fyb+ (Or <1% Fya- and Fyb- ) Black Americans: 10% Fya+; 23% Fyb+ (Or 68% Fya- and Fyb- ) So who is genetically more resistant to this particular malarial parasite?

39 This end is an inositol sugar A malarial parasitic glycosylphosphatidylinositol (GPI); a parasitic sugar that helps initiate malarial infection. 1.Identify the first four sugars 2.What are the glycosidic linkages between the sugar units?


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