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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