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Announcements / Reminders! • Midterm TA led Review Sessions! Welcome to Class 7! Sunday, February 23 from 8-10pm! Location: Science Center Main Room (315)! • Office Hours! Prof Salomon: SFH 270 on Thursday Feb 13 and Feb 20, 2:30-4:30! TAs: by appointment! Introductory Biochemistry! • Sapling problem set 3 due Monday, February 24, 5pm! • Conference Sections on Monday and Tuesday except Mon 12-1 will not be held due to university holiday! • Midterm 1 is Tuesday February 25 at 1 pm! Location: Last names beginning with A-M Macmillan 117! Last names beginning N-Z BERT 130! (BERT= Building for Enviromental Research and Teaching, with greenhouse on top. Previously called Hunter labs)! Active transport! Two types of active transport! l Energy is required to transport a solute against a gradient:! !∆Gt = RT ln(C2 /C1) ! ! !(C2=destination of transported molecules, C1=origin of transported !molecules)! l Primary active transport: Solute accumulation (endergonic, “uphill”) is ! !coupled directly to an exergonic chemical reaction (ATP hydrolysis).! l Secondary active transport: Solute accumulation (“uphill”) is coupled! !to the exergonic (“downhill”) flow of a different solute.! Xout + ATP + H2O → Xin + ADP + Pi Xin + ATP + H2O → Xout + ADP + Pi! Xout + Sout → Xin + Sin! Sout + ATP + H2O → Sin + ADP + Pi! figure 11-35! 1 Secondary active transport Primary active transport:! Sodium/potassium ATPase! figures 11-37, 11-38! Primary active transport: Sodium/potassium ATPase! Primary active transport: Sodium/potassium ATPase figures 11-37, 11-38 2 Secondary active transport: Glucose transport, driven ! by ion gradients, into intestinal epithelial cells ! Transport ATPases! Three types of ATPases:! ! P-type! V-type! F-type! ! All three transport cations. Example: H+ ATPase ! (ATP synthase in the direction of ATP formation) figure 11-43 Example: Na+K+ ATPase! Example: H+ ATPase ! (creates a low-pH compartment) From Lehninger 3rd Ed., figure 12-31! Active transport Solute transport across membranes l Energy is required to transport a solute against a gradient:! !∆Gt = RT ln(C2 /C1)! l Primary active transport: Solute accumulation (endergonic, “uphill”) is ! !coupled directly to an exergonic chemical reaction (ATP hydrolysis).! l Secondary active transport: Solute accumulation (“uphill”) is coupled! !to the exergonic (“downhill”) flow of a different solute.! l When the solute is an ion and its movement is not accompanied! !by a counterion, an electric potential is produced. Such a transport! !process is electrogenic. Otherwise it is electroneutral.! ! l The electrical potential difference affects the energetics of electrogenic transport.! Chemical potential gradient! (concentration difference) Electrochemical potential gradient (electrical potential and concentration difference) figure 11-27 3 A few words about electrochemical potentials Class 7: Outline and Objectives ∆Gt = ∆G`0 + RT ln(C2 /C1) = RT ln(C2 /C1) ; ∆G`0 = 0 because the molecule is unchanged It’s mathematically clear, as well as intuitively obvious, that at equilibrium, when ∆Gt = 0, that RT ln(C2 /C1) also = 0, and this means that (C2 /C1)eq = 1, which in turn means that C2 = C1.! But, what happens if the solute is charged, and there is an electrical potential difference (E2 – E1 = ∆ψ = Vmembrane) between the two regions?! In this case, the expression includes an additional term ! l Monosaccharides! l Aldoses, ketoses; hemiacetals; epimers ! l Pyranoses, furanoses! l Mutarotation, anomers! l Disaccharides and glycosidic bonds! l Polysaccharides ! l Starch, glycogen, cellulose, chitin! l Bacterial cell walls (peptidoglycans)! l Glycoconjugates: Proteoglycans and glycoproteins! ∆Gt = RT ln(C2 /C1) + ZF∆ψ! Z is the charge of the solute, F is the Faraday constant: F = 96.5 kJ/mol-V.! When Z = 0 (uncharged solute), or when ψ = 0 (no electrical potential across the membrane), the expression reduces to ∆Gt = RT ln(C2 /C1).! l Bioenergetics: ATP and coupled reactions ! l Phosphoryl group transfers! l Concentration dependence of ∆G! An important consequence is that at equilibrium ∆Gt =0, C2 does not necessarily = C1.! ln(C2 /C1)eq = –ZF∆ψ/RT ! !(C2 /C1)eq = e–ZF∆ψ/RT! Monosaccharides terminal carbon (C1) is carbonyl! (aldehyde)! Stereoisomers ! of glyceraldehyde second carbon (C2) is carbonyl! (ketone)! The most common monosaccharides! figure 7-1! 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 7-2! 4 D-aldoses (aldehydes) D-ketoses (ketones) (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 Glucose If two sugars differ only in the configuration around one! carbon atom, they are called epimers.! ! D-Mannose and D-Galactose 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 7-4! and in general they have different chemical and physical properties.! Hemiacetals and hemiketals are molecules with hydroxyl and ether groups on the same carbon. They result from the reaction between aldehyde or keto groups and alcohol. The reaction is freely reversible.! figure 7-5! 5 Cyclic forms of monosaccharides The actual conformation of a pyranose 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 hemiacetal or hemiketal formation between aldehyde or keto groups and hydroxyl groups on the same molecule. The reaction is freely reversible.! 1% A new asymmetric C atom (anomeric carbon) is formed in the process of forming a cyclic hemiacetal, making two isomeric forms (anomers) possible, designated α and β.! 33% (at equilibrium) 66% D-Glucose is the aldose that most commonly occurs in ! nature as a monosaccharide. figure 7-6! figure 7-7, 7-8! Haworth Perspectives of Cyclic Sugars! Why more beta than alpha D-glucopyranose? D-Glucopyranose adopts only one of the two possible chair forms where all pyranose substituents are arranged equatorially. α-D-Glucopyranose has 4 equatorial and 1 axial substitutions on the pyranose ring whereas β-DGlucopyranose has 5 equatorial substituents on the pyranose ring. Minimization of steric hindrance favors equatorial positions for the highest number of pyranose substituents. The anomeric effect involving stabilization of the axial configuration of the hydroxyl group on the anomeric carbon through molecular orbital overlap of the oxygen lone pairs and the anomeric carbon bond with its OH group is not enough to stabilize the alpha form and therefore in the case of Dglucopyranose sterics trumps the anomeric effect.! ! ! 33% ● Substituents that appear on the right side in Fischer projections are below the plane of the ring in Haworth perspectives.! ● If the hydroxyl group of anomeric carbon is on the same side of the ring as the hyrdoxyl group of the highest numbered asymmetric carbon (e.g., C5 of a hexose), the anomer is defined as α (opposite side ≡ β anomer). But, this is not always easy to see.! ● A practical rule, which works for both D- and L-pyranoses and furanoses, is that if the hydroxyl group on the anomeric carbon is trans to the terminal CH2OH in the Haworth perspective drawing, the sugar is an α anomer; if it is cis to the terminal CH2OH, it is a β anomer.! 5 HO α-D-Fructofuranose! 2 α! 4 1 3 HO 66% or! OH α-D-Glucopyranose! OH β! H β! β-D-Ribofuranose! 2 or! H β! HO β-D-Glucopyranose! 6 Pyranoses and furanoses Mutarotation ! 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 (aldehyde or ketone) form of the respective monosaccharide until equilibrium between the two forms is reached. This is called mutarotation. Their equilibrium ratio need not be 1:1! Because anomers rotate polarized light differently, the optical rotation of the solution changes in the process. ! ! ! At equilibrium, the linear (aldose or ketose) form is present only in minute amounts.! Glucose: almost exclusively pyranose Fructose: 67% pyranose, 33% furanose figure 7-7! Sugars as reducing agents Sugars as reducing agents Hemiacetals are easily converted to aldehydes; aldehydes are easily oxidized to acids. The oxidation of the aldehyde involves transfer of two electrons to an acceptor, which becomes reduced. Therefore, monosaccharides are reducing sugars. (Ketones, as well as aldehydes, react with oxidants, but ketones react more slowly, and the products of ketose oxidation include glycolaldehyde, derived from C1 and C2). + H2O Hemiacetals are easily converted to aldehydes; aldehydes are easily oxidized to acids. The oxidation of the aldehyde involves transfer of two electrons to an acceptor, which becomes reduced. Therefore, monosaccharides are reducing sugars. Reducing sugars can be detected in solution by adding some colorless substance, such as AgNO3, which is reduced to a colored product, such as Ag↓. + H2O + 3H+ figure 7-10! + 3H+ figure 7-10! 7 Chemical oxidation products of glucose Blood glucose determination! Oxidized glucose (gluconate) has ! a strong tendency to internally ! esterify >> lactone formation. This helps to drive the reaction by lowering [product]. + OH–! 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 7-9! figure 7-3! Oxidation at other carbons is more difficult, but such oxidation products do occur in nature Hemiacetals and hemiketals can be esterified with alcohols to form acetals and ketals C6! C1! (the oxidized carbon is shown in color)! figure 7-9! In contrast to hemiacetals and hemiketals, acetals and ketals are relatively stable.! figure 7-5! 8 Formation of the acetal disaccharide maltose Formation of an acetal from! a hemiacetal and an alcohol (hydroxyl group). Dehydration Common disaccharides Reducing sugars have a free! anomeric carbon. 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-glycosidic bond figure 7-10! Naming Conventions figure 7-11! Polysaccharides (glycans) Reducing oligosaccharides are named ending with the sugar that has the reducing anomeric carbon . Non-reducing oligosaccarides can be named beginning from either end sugar.! H or β-D-fructofuranosyl α-D-glucopyranoside Fru(β2↔1α)Glc figure 7-11! α O 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 Fru(β2↔1α)Glc(6→1α)Gal figure 7-12! 9 Some polysaccharides Glucose Starch l Starch (plants)! l Amylose: α1→4 ! l Amylopectin: α1→4, α1→6! l Glycogen (animals, bacteria): α1→4, α1→6 ! !(more branched than starch)! l Cellulose: β1→4 ! Maltose 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 7-19a,b! figure 7-13a,b,c! What is the advantage of storing glucose as a polymer? 10 Starch Starch Maltose Maltose figure 7-13a,b,c! What is the advantage of having only one reducing end? figure 7-13a,b,c! 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 cellobiose. N-acetyl-D-glucosamine: β1→4! figure 7-14, 7-20! 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 group at C2, chitin is the next most abundant polysaccharide in the biosphere. figure 7-16a! 11 Proteoglycans! (more carbohydrate than protein) Peptidoglycans in ! bacterial cell walls Glycosaminoglycans ≡ unbranched polysaccharides of alternating uronic acid (oxidized at C6) and GlcNAc or GalNAc residues (often sulfated)! ! ! ! ! ! ! ! ! ! ! Core proteins + covalently linked glycosaminoglycans ≡ proteoglycans! Penicillin interferes with cell wall! formation by preventing the synthesis! of cross-links.! (Alexander Fleming) figure 20-30! Proteoglycans form the ground substance of connective tissue! (cartilage, tendon, skin, blood vessel walls). They have a slimy, mucuslike! consistency. figure 7-22! What is the advantage of having unusual (D-) amino acids? Glycoproteins! (more protein than carbohydrate) Glycoproteins! (more protein than carbohydrate) GlcA-GlcNS GlcA-GalNAc Immunoglobin! Plasma membrane protein! Almost all secreted and membrane-associated ! proteins of eukaryotic cells are glycosylated.! figures 5-22b, 7-26! Immunoglobin! Almost all secreted and membrane-associated ! proteins of eukaryotic cells are glycosylated.! Plasma membrane protein! figures 5-21b, 7-26! 12 Glycoproteins Glycoproteins figure 7-30! What is the advantage of having so much potential variation? Introduction to Bioenergetics The equilibrium constant for a reaction, 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. ! 13 The actual ∆G of a reaction depends on reactant and product concentrations as well as ∆G'º! Standard free energy changes are additive C + D A+B 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. ! ! This is the case when becomes negative ([C][D] < [A][B]), for ! ! example when products C and D are constantly removed as soon as they are formed. ! Standard free energy changes are additive Glucose + Pi → Glucose 6-P + H2O! !ΔG' º = 13.8 kJ/mol! ATP + H2O → ADP + Pi !ΔG' º = –30.5 kJ/mol! ! ! Glucose + ATP → ADP + Glucose 6-P If the two reactions can be effectively coupled, a reaction with a large 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! individual reactions A to B (1) and B to C (2).! Energy coupling! Example: glucose phosphorylation! Energy coupling occurs! through shared intermediates! (Pi in this case). !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! 14 Nucleotides and nucleosides Adenosine triphosphate (ATP) Hydrolysis of the γ- and β-phosphates is highly exergonic. Adenine γ β α D-Ribose Nucleoside Nucleotide = Nucleoside-P Nucleoside-diP Nucleoside-triP (phosphate groups are usually complexed with Mg2+) figures 1-26, 13-12! 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! 15 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! 16 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! 17