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Home , Anomer, Pyranose

Class 7: Outline and Objectives

l l , ; ; l , Welcome to Class 7 l , l and glycosidic bonds l l , , , l Bacterial cell walls (peptidoglycans) l Glycoconjugates: Proteoglycans and glycoproteins

Introductory l Bioenergetics: ATP and coupled reactions l Phosphoryl group transfers l Concentration dependence of ∆G

Monosaccharides Stereoisomers of terminal carbon (C1) is carbonyl () second carbon (C2) is carbonyl () Monosaccharides are chiral. A molecule with n chiral centers can have 2n possible stereoisomers.

The chiral center most distant from the carbonyl carbon defines D- and L-forms.

L- and D- isomers of the same compound are mirror images (enantiomers).

Enantiomers of compounds with more than one chiral center have all chiral centers reversed. figure 7-1 figure 7-2 The most common monosaccharides

1 D-aldoses () D-ketoses ()

(achiral)

figure 7-3 The more commonly occurring aldoses are shown in red boxes figure 7-3 The more commonly occurring ketoses are shown in red boxes

Hemiacetals and hemiketals Epimers of

If two differ only in the configuration around one Hemiacetals and hemiketals are molecules with hydroxyl and ether groups carbon atom, they are called epimers. on the same carbon. They result from the reaction between aldehyde or keto groups and alcohol. The reaction is freely reversible. D- and D- are both epimers of D-Glucose. D-Mannose and D-Galactose are not epimers of one another.

Although epimers are isomeric, they are not mirror images (enantiomers) figure 7-5 and in general they have different chemical and physical properties. figure 7-4

2 Cyclic forms of monosaccharides The actual conformation of a ring is not flat, but assumes a chair-like shape

Monosaccharides contain both aldehyde or keto groups and hydroxyl groups. In aqueous solutions, most monosaccharides occur as cyclic structures. They result from or hemiketal formation between aldehyde or keto groups and hydroxyl groups on the same 1% molecule. The reaction is freely reversible.

A new asymmetric C atom (anomeric carbon) is formed in the process of forming a cyclic D-Glucose is the that hemiacetal, making two most commonly occurs in isomeric forms (anomers) nature as a . possible, designated α and β.

33% (at equilibrium) 66% figure 7-6 figure 7-7, 7-8

Why more beta than alpha D-glucopyranose? Haworth Perspectives of Cyclic Sugars ● Substituents that appear on the right side in Fischer projections are below D-Glucopyranose adopts only one of the two the plane of the ring in Haworth perspectives. possible chair forms where all pyranose ● If the hydroxyl group of anomeric carbon is on the same side of the ring as substituents are arranged equatorially. the hyrdoxyl group of the highest numbered asymmetric carbon (e.g., C5 α-D-Glucopyranose has 4 equatorial and 1 axial of a ), the is defined as α (opposite side ≡ β anomer). But, substitutions on the pyranose ring whereas β-D- this is not always easy to see. Glucopyranose has 5 equatorial substituents on ● A practical rule, which works for both D- and L-pyranoses and furanoses, the pyranose ring. Minimization of steric hindrance 33% is that if the hydroxyl group on the anomeric carbon is trans to the terminal favors equatorial positions for the highest number CH2OH in the Haworth perspective drawing, the is an α anomer; if it of pyranose substituents. The is cis to the terminal CH2OH, it is a β anomer. involving stabilization of the axial configuration of 5 HO the hydroxyl group on the anomeric carbon α-D-Fructo- β-D-Ribo- 4 1 β through molecular orbital overlap of the oxygen 2 α furanose lone pairs and the anomeric carbon bond with its 3 2 OH group is not enough to stabilize the alpha form and therefore in the case of D- glucopyranose sterics trumps the anomeric HO OH H 66% or β or β effect. OH H HO

α-D-Glucopyranose β-D-Glucopyranose

3 Mutarotation Pyranoses and furanoses

! Although anomers are isomeric, they are not mirror images (enantiomers). In general, they have different physical and chemical properties. Anomers rotate polarized light differently.

! Interconversion between α and β anomers occurs via the linear Glucose: almost (aldehyde or ketone) form of the respective monosaccharide until exclusively pyranose : 67% pyranose, equilibrium between the two forms is reached. This is called 33% furanose mutarotation. Their equilibrium ratio need not be 1:1! Because anomers rotate polarized light differently, the of the solution changes in the process.

! At equilibrium, the linear (aldose or ) form is present only in minute amounts.

figure 7-7

Sugars as reducing agents Sugars as reducing agents Hemiacetals are easily converted to aldehydes; Hemiacetals are easily converted to aldehydes; aldehydes are easily oxidized to acids. The oxidation of the aldehyde involves aldehydes are easily oxidized to acids. The oxidation of the aldehyde involves transfer of two electrons to an acceptor, which becomes reduced. transfer of two electrons to an acceptor, which becomes reduced. Therefore, monosaccharides are reducing sugars. (Ketones, as well as aldehydes, Therefore, monosaccharides are reducing sugars. react with oxidants, but ketones react more slowly, and the products of ketose Reducing sugars can be detected in solution by adding some colorless substance, oxidation include , derived from C1 and C2). such as AgNO3, which is reduced to a colored product, such as Ag↓.

+ H2O + H2O

+ 3H+ + 3H+

figure 7-10 figure 7-10

4 Chemical oxidation products of glucose Blood glucose determination

Oxidized glucose (gluconate) has a strong tendency to internally + OH– esterify >> lactone formation. This helps to drive the reaction by lowering [product].

Assay: a peroxidase reaction uses the H2O2 produced by glucose oxidase to convert a colorless compound into a colored one, which absorbs light at a particular wavelength.

figure 7-3 figure 7-9

Oxidation at other carbons is more difficult, but Hemiacetals and hemiketals can be esterified such oxidation products do occur in nature with alcohols to form acetals and ketals

C6

C1

(the oxidized carbon is shown in color) In contrast to hemiacetals and hemiketals, acetals and ketals are relatively stable. figure 7-9 figure 7-5

5 Formation of the acetal Common disaccharides

Reducing sugars have a free anomeric carbon. Formation of an acetal from a hemiacetal and an alcohol (hydroxyl group).

Dehydration Non-reducing sugars have no free anomeric carbons. Wavy lines: Anomer not specified (could be α or β)

Non-reducing sugars are named pyranosides or furanosides.

O- figure 7-10 figure 7-11

Naming Conventions Polysaccharides (glycans)

Reducing are named ending with the sugar that has the reducing anomeric carbon .

Non-reducing oligosaccarides can be named beginning from either end sugar.

H α or O β-D-fructofuranosyl α-D-glucopyranoside Fru(β2↔1α)Glc

Raffinose α-D-galactopyranosyl-(1→6)-α-D-glucopyranosyl β-D-fructofuranoside Gal(α1→6)Glc(α1↔2β)Fru or β-D-fructofuranosyl α-D-glucopyranosyl-(6→1)-α-D-galactopyranoside figure 7-12 figure 7-11 Fru(β2↔1α)Glc(6→1α)Gal

6 Some polysaccharides Starch

l Starch (plants) l : α1→4 l : α1→4, α1→6 Glucose l Glycogen (animals, bacteria): α1→4, α1→6 Maltose (more branched than starch) l Cellulose: β1→4

Starch and cellulose both consist of recurring units of D-glucose. Their different properties result from different types of glycosidic linkage.

l Peptidoglycans (bacterial cell walls) l Chitin (exoskeletons, cell walls): N-acetyl-D-glucosamine β1→4

figure 7-13 a,b,c

Structure of starch Starch

Maltose

Starch granules

figure 7-13a,b,c figure 7-19a,b What is the advantage of storing glucose as a polymer?

7 Starch Starch

Maltose Maltose

figure 7-13a,b,c figure 7-13a,b,c What is the advantage of having only one reducing end? What is the advantage of having many non-reducing ends (branching)?

Chitin Cellulose

180° flip Cellulose accounts for over half of the carbon in the biosphere.

The disaccharide unit of cellulose is called . N-acetyl-D-glucosamine: β1→4

Chitin is the principal structural component of the exoskeletons of arthropods (crustaceans, insects, and spiders) and is present in the cell walls of fungi and some algae. After cellulose, from which it only differs in the acetylated amino figure 7-14, 7-20 group at C2, chitin is the next most abundant in the biosphere. figure 7-16a

8 Peptidoglycans in Proteoglycans bacterial cell walls (more than protein)

Glycosaminoglycans ≡ unbranched polysaccharides of alternating uronic acid (oxidized at C6) and GlcNAc or GalNAc residues (often sulfated)

Penicillin interferes with cell wall formation by preventing the synthesis of cross-links. (Alexander Fleming)

Core proteins + covalently linked ≡ proteoglycans

Proteoglycans form the ground substance of connective tissue (cartilage, tendon, skin, blood vessel walls). They have a slimy, mucuslike consistency. figure 20-30 figure 7-22 What is the advantage of having unusual (D-) amino acids?

Glycoproteins Glycoproteins (more protein than carbohydrate) (more protein than carbohydrate)

GlcA-GlcNS

GlcA-GalNAc

Immunoglobin Plasma membrane protein Immunoglobin Plasma membrane protein

Almost all secreted and membrane-associated Almost all secreted and membrane-associated proteins of eukaryotic cells are glycosylated. figures 5-22b, 7-26 proteins of eukaryotic cells are glycosylated. figures 5-21b, 7-26

9 Glycoproteins Glycoproteins

figure 7-30 What is the advantage of having so much potential variation?

The equilibrium constant for a reaction,

Introduction to Bioenergetics K'eq, is mathematically related to ∆G' º

A + B C + D

Standard free energy change (1 M concentrations, etc.):

[C][D] = [A][B]

[A], [B], [C], [D] are the molar concentrations of the reaction components at equilibrium.

If [C][D] > [A][B] at equilibrium, then lnK'eq is positive, and therefore ∆G' º is negative. This means if initially all reactants are present at 1 M concentration, the reaction would go from A + B to C + D before and until equilibrium is reached.

10 The actual ∆G of a reaction depends on reactant Standard free energy changes are additive and product concentrations as well as ∆G'º

A + B C + D

If the reactants are initially present not at 1 M, but at different concentrations (nonstandard conditions):

The criterion for the direction of net spontaneous reaction is ∆G, not ∆G' º.

A reaction with a positive ∆G' º can go forward as long as ∆G is negative. If the two reactions can be effectively coupled, a reaction with a large This is the case when becomes negative ([C][D] < [A][B]), for negative ∆G' º can “drive” a reaction with a positive ∆G' º.

The pathway in a coupled reaction from A to C is different from the example when products C and D are constantly removed as soon as they are individual reactions A to B (1) and B to C (2). formed.

Standard free energy changes are additive Energy coupling Example: glucose phosphorylation

Energy coupling occurs

Glucose + Pi → Glucose 6-P + H2O ΔG' º = 13.8 kJ/mol through shared intermediates

(Pi in this case). ATP + H2O → ADP + Pi ΔG' º = –30.5 kJ/mol

Glucose + ATP → ADP + Glucose 6-P ΔG' º = –16.7 kJ/mol

Glucose phosphorylation with Pi is endergonic.

ATP hydrolysis to ADP and Pi is highly exergonic. ATP hydrolysis coupled to glucose phosphorylation is exergonic.

figure 1-27b

11 Nucleotides and nucleosides Adenosine triphosphate (ATP)

Hydrolysis of the γ- and β-phosphates is highly exergonic. Adenine

γ β α

D-

Nucleoside (phosphate groups are Nucleotide usually complexed with Mg2+) = Nucleoside-P Nucleoside-diP figures 1-26, 13-12 Nucleoside-triP

ATP hydrolysis

Pi ≡ inorganic phosphate

Factors favoring hydrolysis: 1. Relief of electrostatic repulsion

2. Pi is stabilized by resonance 3. Mass action favors hydrolysis

(high [H2O])

figure 13-11

12 In intact cells, ∆G for ATP hydrolysis is often much more negative than ∆G' º (—30.5 kJ/mol), ranging from —50 to —65 kJ/mol. This is because [ATP]/[ADP][Pi] > 1.0 in cells

Energy released by hydrolysis of biological phosphate compounds Hydrolysis of phosphocreatine

Phosphocreatine has a high phosphoryl group transfer potential. It can drive the formation of ATP from ADP.

figure 13-19 figure 13-15

13 ATP can provide energy by group transfer even when there is no net transfer of P

Derivation of energy from ATP hydrolysis generally involves covalent participation of ATP in the reaction.

Formation of glutamine by

condensation of glutamate with NH3 is endergonic (positive ΔG' º).

Formation of γ-glutamyl P by transfer of P from ATP is exergonic (negative ΔG' º).

Formation of glutamine by displacement of P from γ-glutamyl P by NH3 is exergonic (negative ΔG' º).

The net coupled reaction is exergonic (negative ΔG' º). figure 13-18

14