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Bioenergetics 2nd lecture

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Standard free energy change ∆Go

and equilibrium constant Keq

Keq is obtained by dividing [products] by [reactants]when the reaction reaches equilibrium

Keq =

At equilibrium ∆G = 0

[Products][Reactants]

Standard free energy change ∆Go

and equilibrium constant Keq

∆G= ∆Go + RT ln

0 = ∆Go + RT ln Keq

∆Go = - RT ln Keq

[Products]

[Reactants]

At equilibrium∆G =0

[Products] =Keq

[Reactants

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∆Go and equilibrium constant Keq

∆G= ∆Go + RT 2.3 log

• At equilibrium

0 = ∆Go + RT ln Keq

∆Go= - RT ln Keq

• At standard conditions

∆G= ∆Go + RT 2.3 log 1∆G= ∆Go

[Products]

[Reactants]

Glucose 6- phosphate Fructose 6- phosphate

0.66 mol/L 0.33 mol/L

∆G= ∆Go + RT 2.3 log 0.33/ 0.66

∆Go = + 0.4 kcal/mol

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Glucose 6- phosphate Fructose 6- phosphate

∆Go = + 0.4 kcal/mol

∆G= ∆Go + RT 2.3 log 0.09/0.9

∆G= - 0.96

Glucose 6- phosphate Fructose 6- phosphate

1 mol/L 1 mol/L

∆G= ∆Go + RT 2.3 log 1/1

∆G= ∆Go

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∆Gº and Keq

Keq ∆Gº

1 0

101 - 1.36

102 - 2.72

103 - 4.08

10-1 1.36

10-2 2.72

10-3 4.08

If Keq = 1, then ∆Gº = 0

If Keq > 1, then ∆Gº < 0

If Keq < 1, then ∆Gº > 0

K

1041 10310110-4 10-3 10-1

4.1 2.7 1.3 0 -1.36 -2.7 -4.1

∆Gο

K very large Reaction goes to completion

K very small Reaction goes hardly at all

More reactants than products More products than reactants

∆Gº and Keq

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Exergonic reactions in Biochemistry• Complex structures simple structures

Proteins → amino acids

Starch → n glucose

glucose + O2 → CO2 + H2O

• More specificallyHydrolysis reactions

Decarboxylation reactions (release of CO2 )

pyruvate ( C3 ) → acetyl- (C2) +CO2

Oxidation with O2

How can Endergonic reactions proceed?Endergonic reaction can be driven by an

exergonic reaction if the two can be

coupled

A → B ∆G = + 5 kcal/mol

C → D ∆G = - 9 kcal/mol

When the two are coupled

A + C → B + D ∆G = - 4 kcal/mol

A + C → I → B + D

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Example of coupling an exergonic process with an endergonic one

Endergonic and exergonic reactions are indirectly coupled through ATP/ ADP cycle

ATP + H2O → ADP + Pi ∆Gº = - 7.3

ADP + Pi → ATP + H2O ∆Gº = + 7.3

A B ∆Gº = - 2.3

ATP ADP + Pi

D C ∆Gº = -1.7

??

?

?

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Energy released in exergonic reactions

Chemical energy in the products of endergonic reactions

Heat

Adenine + ribose

adenosine

+ 3 phosphate

ATP

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∆Gº for hydrolysis of some phosphorylated compounds

Compound +H2O Product + phosphate ∆Gº

Phosphoenol pyruvate Pyruvate -14.8

1,3 bisphosphoglycerate 3 phosphoglycerate -11.8

Creatine phosphate Creatine - 10.3

ATP ADP - 7.3

Glucose 1- phosphate Glucose -5.0

Glucose 6- phosphate Glucose -3.3

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Phosphorylation of glucose

Glucose + Pi → Glc. 6 P + H20 ∆Gº= +3.3

ATP + H2O → ADP + Pi ∆Gº = -7.3

Glucose + ATP → Glc. 6 P + ADP ∆Gº = -4.0

O2

CO2

ATP is not a long term storage form of energy

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Approximate glucose uptake and ATP turnover by various tissues

Tissue Approximate O2

consumption mole/ day

Equivalent glucose

mole/ day

ATP turnover mole/day

Brain 3.4 0.57 20.4

Heart 1.9 0.32 11.4

Kidney 2.9 0.49 17.4

Liver 3.6 0.6 21.6

Muscle 3.3 0.54 19.8

Total 15.1 2.52 90.6

• Based on O2 consumption

• Assuming glucose is the fuel used

C6H12O6 + 6O2 6CO2 + 6H2O + 36ATP

Other nucleotide triphosphates

• GTP, UTP, CTP

• Synthesized from ATP

ATP + GDP ADP + GTP– GTP in protein synthesis

– UTP in polysaccharide synthesis

– CTP in phospholipids synthesis

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UDP- is a carrier of activated sugar