metabolism overview
TRANSCRIPT
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Energy is not Cycled
CO2 GlucoseCO2
Fat, muscle, etc.
Anabolism
Energy (light) Energy (heat)
Energy (heat)
Catabolism
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Catabolism Drives Analobolism
Catabolism
Breakdown of nutrients (e.g.
glucose)
Releases energy for anabolic
reactions
Releases heat
Anabolism (Biosynthesis) Synthesis of macromolecules
Requires energy
Releases heat
Intermediary Metabolism
Metabolic pathways involving low
molecular weight (
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Catabolism Driving Anabolism
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Separation of Anabolic and MeabolicPathways
Cellular Compartmentalization
Fatty acid catabolismin mitochondria
Fatty acid synthesisin cytoplasm
Different concentrations of products, reactants and regulators
Unique Cofactors
NAD/NADH for catabolism
NADP/NADPH for anabolism
When it goes wrong .
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Classes of Biochemical Reaction
Oxidation-reduction
Lactate dehydrogenase
Cleavage of carbon bonds
Adol condensations (aldose)
Claison condensations (citrate synthase)
Decarboxylations (acetoacetate decarboxylase)
Internal rearrangements, isomerizations and eliminations
Phosphohexose Isomerase
Group transfers (eg acyl,glucosyl, phosphoryl)
Hexokinase
Free radical reactions
Ribonucleotide reductase
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Oxidation-Reduction
Dehydrogenases
Dehydrogenations - loss of 2 electrons & 2 hydride ions
Oxidases
Oxygen becomes bonded to carbon
Oxygenases
Oxidases that use molecular oxygen
Oxidation
Reduction
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Phosphate and PhosphateTransfer
Transient Intermediate
Nucleophilic Attack (Glucokinase)
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Consider a Typical Enzymatic Reaction
ABCD
The reaction is reversible
Forward and backward reactions occur at the same time
The rate of each reaction is dependent on the concentration
of reactants
As A and B are used up, the forward rate decreases, C and D
increase and the rate of the reverse reaction increases
At the steady state, the forward and backward reaction rates
are the same
For the forward reaction;A and B are substrates
C and D are products
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Consider a Typical Enzymatic Reaction
ABCD
ForwardRate
Proportional to [A][B] = k1[A][B]
ReverseRate
Proportional to [C][D] = k2[C][D]
At Equilibrium: forward rate = reverse rate
k1and k2
are
constants
k1 [C][D]
k2 [A][B]
=
k1[A][B] = k2[C][D]Kd [C][D]
[A][B]=
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Equilibrium Constant
For the reaction:
At equilibrium:
Where Keq = the equilibrium constant
[A] = the molar concentration of A etc.
----------------------------------------------------------------------
K'eq
pH 7 (10-7
M H+) 55.5 M H2O
ABCD
Keq
[C][D][A][B]
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Equilibrium Constant
All enzymatic
reactions are
reversible
ABCD
Keq [C][D][A][B]
Enzymatic Reactions are reversible
AorB ForwardCorD
AorB
CorDBackwards
}
}
At dynamicequilibrium
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Change in Gibbs Free Energy G
The free energy change drives a reaction
A negativeG drives the reaction forward(as written)
A positiveG drives the reaction backwards(as written)
The G depends on the concentrations of reactants andproducts, temperature and pressure
Standard G(G)
One molar reactants and products, 298 K (25 C), 1 atm
Standard Biological G(G)
As G+ pH 7 (10-7M H+), 55.5 M H2O, 1 mM Mg2+
The standard free energy change of a reaction cannot
used to reliably predict the net direction of a reaction in
vivo since the reactants and products are not at 1 M
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Enzymes Lower the Activation Energy
Enzymes work by lowering the activation energy
Therefore:
Enzymes increase the dynamics of a reaction
Enzymes may increase the net rate of product formation
Enzymes do not change the Keq
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Free Energy Change
G = free energy change
G = standard G
G' = biological standard G
1 M, 298 K, 1 atm
10-7
M H+, 55.5 M H2O
ABCD
1 mM Mg2+
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Relationship between G and Keq
For the reaction
From thermodynamic principles:
At equilibrium
Therefore:
R= the gas constantT= the temperature in Kelvin
GG0' RTln[C][D][A][B]
ABCD
G
o'
RTlnKeq'
G 0
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Law of Mass Action
GG0' RTln[C][D][A][B]
G 0G0' RTln[C][D]
[A][B]
At equilibrium
ABCD
Remove some of the productsCandD
G is zero:
G becomes negative G 0G0' RTln[C][D]
[A][B]
Remove some of the substratesAandB
G becomes positive G 0G0' RTln[C][D]
[A][B]
A +B C +D
A +B C +D
Forwards
Backwards
Keq
[C][D]
[A][B]
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Gs are Additive
Glucose + Pi G-6-P + H20
ATP + H20 ADP + Pi
Since only starting and final states are considered:
Glucose + ATP G-6-P + ADP
+13.8 kJ/mol
G'
-30.5 kJ/mol
-16.7 kJ/mol
The hydrolysis of ATP is used to drive the reaction forward
Consider the reaction:
Energetically unfavorable reactions are coupled to favorable ones
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Keqs are Multiplicative
Glucose + Pi G-6-P + H20
ATP + H20 ADP + Pi
Since only starting and final states are considered:
Glucose + ATP G-6-P + ADP
3.9 x10-3M
Keq
2.0 x 105M
7.8 x 102M
The hydrolysis of ATP is used to drive the reaction forward
Consider the reaction:
Energetically unfavorable reactions are coupled to favorable ones
Overall G = G1+ G2where G1= RTln Keq1
Overall G = RTln Keq1 + RTln Keq2 (adding logs = multiplication)
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Operational Gs
ATP + H20 ADP + Pi
G'
-30.5 kJ/mol
Consider the reaction:
Observed versus biological standard Gs
Conditions vary, but in general the true G of ATP
hydrolysis in vivo is more negative than standard G
In human erythrocytes at 25 C
[ADP] = 0.25 x 10-3
[Pi] = 1.65 x 10-3
[ATP] = 2.25 x 10-3
Gp = G + RT ln [ADP][Pi]
[ATP]
Gp = -30.5 + 21 kJ/mol
G of ATP hydrolysis = 51.5 kJ/mol
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Metabolic Pathways & Law of Mass Action
Products are moved through pathways by:Increasing the level of substrate
Decreasing the level of product
Especially useful for energetically unfavorable
reactions
ABCD Keq [C][D][A][B]
GG0' RTln[C][D][A][B]
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Relationship Between Keqand G
Keq G (kJ/mol)
0.001 17.1
0.01 11.4
0.1 5.7
1 0.0
10 -5.7
100 -11.41,000 -17.1
Why is G 0 when Keq
is 1?
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ATP
Adenosine
ATP
AMP
ADP
Phosphoester
bonds
Phosphoanhydride
bonds
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ATP Hydrolysis
Hydrolysis with relief of charge repulsion
Resonance stabilization Ionization
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ATP Forms a Complex With Mg2+
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ATP is a High-Energy Compound
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1,3-BPG Hydrolysis
H+
1,3-Bisphosphoglycerate4- + H2O
3-phosphoglycerate3- Pi2- + H+
G = -49.3 kJ/mol
ionization
Resonance Stabilization and Ionization
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TransphosphorylationsPhosphocreatine is an Energy Reservoir
ADP + phosphocreatine ATP + creatine G = -12.5 kJ/mol
2ADP ATP + AMP G = ~ 0
ATP + NDP ADP + NTP G = ~ 0
Creatine Kinase
Other Transphosphorylations
Adenylate Kinase
Nucleoside diphosphate Kinase
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Acetyl-CoA HydrolysisResonance Stabilization and Ionization
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Thioesters Versus Oxygen EstersNo Resonance Stabilization in Thioesters
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Comparison of Gs of Hydrolysis
Standard Free Energies of Hydrolysis
G kJ/mol(approx)
Phospoenolpyruvate - 62
1,3-bisphoshoglycerate to 3-BPG - 49
Phosphocreatine - 43ATP to ADP
ATP to AMP + PPi
ADP to AMP
PPi
AcetylCoA - 31
Hexose phosphates - 13 to - 21
AMP - 14
Gycerol-1-phosphate - 9.2
{~ - 31
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ATP Coupling is Multi-StepGlutamine Synthetase
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Na+K+ ATPaseATP Hydrolysis Powers the Pump
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ATPase Pumping MechanismPhosphorylation changes the conformation of the pump
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Palmitoyl-CoA SynthesisAMP is conjugated to palmitate
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RNA ElongationTwo High-Energy Bonds are Used
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Oxidation
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Food for WorkComparison to an electrical motor
BatteryElectric
MotorWORK
e.g. movement
Food Mitochondria WORKe.g. muscle use
Electrons
Physical
Coupling
Electrons ATP
Electromotif Force (emf)
Proton Motive Force
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Redox Pair
Fe 2+ Cu2+ Fe3+ Cu++ +
Fe2+
Cu2+Fe
3+
Cu+++
e-e
-
Reducing agent(reductant,
Oxidizing agent(oxidant, electron
acceptor)electron donor)
Ferrous FerricCupric Cuprous
Oxidation
loss of electrons
Reduction
gain of electrons
T f f El t i O i C d
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Transfer of Electrons in Organic CompoundsOxidation of a reducing sugar by cupric ions
CR
O
H
+ 4OH - + 2Cu2+ Cu2O+ 2H 2OCR
O
OH
+
CR
O
H
+ 2OH - + H2OCR
O
OH
+
2OH - + 2Cu2+ Cu2O H2O++2e -
2e -
Oxidation - loss of electrons
Reduction - gain of electrons
cupric ion
T f f El t i O i C d
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Transfer of Electrons in Organic CompoundsOxidation of a reducing sugar by cupric ions
CR
O
H
+ 2OH - + H2OCR
O
OH
+
2OH - + 2Cu2+ Cu2O H2O++2e -
2e -
2OH-give H2O plus O2-
2Cu2++ 2e-give 2Cu+
2Cu
+
+ O
2-
give Cu2O
OH replaces H yields 1e-
H combines with OH-
to give H2O plus 1e-
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Oxidation States of CarbonOwnership of Electrons H< C < S < N < O
Electronsbelongingto
carbon
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Methods of Electron Flow
Fe 2+ + Cu 2+ Fe 3+ + Cu+
Direct Electron Transfer
Use of Hydride Ions
Use of Hydrogen Atoms
Direct Oxidation by Oxygen to Give
a Covalently Bound Oxygen
AH 2 + B A + BH 2
AH 2 A + 2H + 2e -
NAD + + H - + H + e - NADH + H+
NAD + + AH 2 NADH + H +
R-CH 3 +1 /2O2 R-CH 2 -OH
Electron acceptor
Electron acceptor
Electron acceptor
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Nernst Equation
E = E0+RT
nFln
[electron acceptor]
[electron donor]
__ _______________
E = E0+n
ln[electron acceptor]
[electron donor]_______________0.026 V
n =no. of electrons/moleculeF = Faraday Const.
T= temp (K)
R = gas constant
Relates Standard Reduction Potential to True Reduction Potential
E is the biological reduction
potential (pH 7)
f G
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Relationship of G to E
G = -n F E
or
G 0' = -n F E0'
F = Faraday Const.
Th H d Q i h E
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The Hardest Question on the Exam
G = -n F E
or
G 0' = -n F E0'F = Faraday Const.
Given E for two reactions, predict thefavorable direction the reaction; what
becomes oxidized and what reduced
Preparation
Work through the example in Lehninger pages 510/1
NAD A t 2 El t
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NAD+ Accepts 2 ElectronsTransferred as a Hydride Ion (H-)
E l f NAD ifi it
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Examples of NAD specificity
NADH Ab b t 340
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NADH Absorbs at 340 nmCan be used to measure concentration
Pellagra
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PellagraDermatitis, diarrhea, dementia, death
Maize is deficient in tryptophan and niacin
Alcoholics have reduced absorption of niacin
QuickTime and a
TIFF (Uncompressed) decompressorare needed to see this picture.
P ll hi t
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Pellagra, a history
Once common in the Southern
US, pellagra was thought to becontagious
1914 - Dr Joseph Goldberg wentto the S. to find a cure
Came to believe it was related to
nutrition
Controlled trials in orphanagesand a convict camp (inducedpellagra using a restricted diet)
In 1937 vitamin B3 was identified
Native Americans did not sufferfrom pellagra because they treatedcorn with lime which made niacinavailable
QuickTime and a
TIFF (Uncompressed) decompressorare neede d to see this picture.
Fl i d i di l tid (FAD)
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Flavin adenine dinucleotide (FAD)Accepts One or Two Electrons
Fl t i
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Flavoproteins