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

Carbohydrates(saccharides)

Organic biomolecules:

1. Carbohydrates

2. Lipids

3. Proteins

4. Nucleic acids

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?

Terminology

• Basic building block: monosaccharides– Polyhydroxyaldehydes (figure 18.1)– Polyhydroxyketones (figure 18.2)

• Bigger units constructed of monosaccharides: polysaccharides– Disaccharides– Oligosaccharides– Polysaccharides

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

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

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

Construct sugar model

• Furanose = 5 membered ring

• Pyranose = 6 membered ring

• Construct D-glucose as open chain, then convert to cyclic form

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)

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

Do alpha or beta anomers matter?

Yes they do!(stay tuned)

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 ___________

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

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

Maltose: malt sugar or corn sugar; From action of amylase on starch

Benedict’s test results:

Cellobiose = from partial hydrolysis of cellulose, by enzyme cellulase

Reducing sugar = Benedict’s test results: _____

Lactose = milk sugar

Reducing sugar?____

Sucrose = table sugar; sugar cane

Reducing sugar?_____ (check your lab results!)

Polysaccharides• Storage

– Starch– glycogen

• Structural– Cellulose– Chitin

• Recognition– Glycolipids– glycoproteins

Essentially nonreducing sugars:

Benedicts ____mutarotation____

Starch: shown here is a small unit

What type of glycosidic linkage is shown between glucose units?_________

Amylose (10-30%) linear, long chains

Amylopectin: 70-90%, branchedGlycogen: even more branched

Starch continued:

Check your lab results:

Iodine test:____

Cellulose: Glycosidic linkage is _________

Planar formation of chainsNeed cellulase to cleave this linkage: provided by microorganisms of ruminantsFiber for humans

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.

Frequency of ABO phenotype in US population

White Black AsianNative American

O 45 49 43 79

A 41 27 27 16

B 10 19 25 4

AB 4 4 5 <1

Frequency of ABO phenotypes worldwide

About 21% of worldwide population has A allele.

Distribution in native populations worldwide

Type B: 16% of worldwide population has the allele

Distribution in native populations worldwide

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

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

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?

This end is an inositol sugar

A malarial parasitic glycosylphosphatidylinositol (GPI); a parasitic sugar that helps initiate malarial infection.

1. Identify the first four sugars2. What are the glycosidic linkages

between the sugar units?

OH

OH

H

O

OH

HH

OH

O P

O

OH

O

NH2

OH

OH

H

O

HH

OH

CH2OH

H

OH

OH

H

O

OH

HH

OH

H

OH H

O

H

NH2H

OH

CH2OH

H

H OH

H

OH

HH

O

HOH

H OH6

5 4

3

21

PHO

OO

SH