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What is the value of the equilibrium constant of ATP hydrolysis at 37 0C [ADP] [Pi ] Calculate the free enthalpy change ∆G at the ratio ATP/ADP = 100:1

The hydrolysis of ATP (ATP → ADP), a spontaneous process, can be coupled to a non- spontaneous reaction to drive the non-spontaneous reaction forward ATP

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Example: Hydrolysis of Phosphoenolpyruvate (PEP) to drive ATP synthesis ADP + Pi → ATP + H2O; ΔG°' = +55 kJ/mol Need to couple it to another, very

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Creatine Kinase Creatine + ATP Creatine Phosphate + ADP ΔG°' Values of Hydrolysis (kJ/mol) Oh Delta G - the change in Gibbs free energy Can tell

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∆Go' for ATP hydrolysis is more negative than most biochemical reactions and than hydrolysis of phophorylated comounds such as Glc-6-PO4 or Glycerol-3- PO4

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What is the value of the equilibrium constant of ATP hydrolysis at 37 0C [ADP] [Pi ] Calculate the free enthalpy change ∆G at the ratio ATP/ADP = 100:1

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The hydrolysis reaction, ATP + H2O → ADP + phosphate, has a standard free energy, AGoʼ, of -30 5 kJ/mol (a) or (c) Show the formula for the equilibrium

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Hydrolysis Synthesis ∆G = ∆G0 + ln ([ADP][P i]/[ATP]) = -13 7 kcal/mol ~1/500 ATP hydrolysis shifts the equilibria of coupled reactions Coupled reaction:

Calculations in Bioenergetics

Basic Relationships:

Redox potential and standard redox potential (Nernst-Peters equation): E h

´ redox potential at pH 7 (V)

E 0 ´ standard redox potential (both [oxid.] & [red.] are 1 mol/l), at pH 7 (V)

R universal gas constant, R = 8.3143 J K -1

mol -1

T temperature (K)

n number of electrons transferred

F Faraday (Faraday's charge), F = 96 487 J V

-1 mol -1 = 96 487 C mol -1 ln natural logarithm, ln = 2.303 log [oxid.] concentration of the substance in oxidised form (mol/l) [red.] concentration of the substance in reduced form (mol/l)

Free enthalpy change and equilibrium constant:

ΔG free enthalpy change (J mol -1

ΔG 0 standard free enthalpy change (concentrations of all reactants & products are 1 mol/l) (J mol -1

R universal gas constant, R = 8.3143 J K

-1 mol -1

T temperature in (K)

Keq equilibrium constant ln natural logarithm, ln = 2.303 log

[A], [B], and [C],[D] ...actual concentrations of the reactants and products, respectively (mol/l)

[A] eq , [B] eq , and [C] eq , [D] eq ...equilibrium concentrations of the reactants and products, respectively (mol/l)

RT [oxid.]

E h

´ = E

´ + - - - - ln - - - -

n F [red.]

A + B C + D

[C]eq [D] eq K eq [A] eq [B] eq [C] [D]

ΔG = - RT ln K

eq + RT ln - - - - [A] [B] [C] [D] ΔG = ΔG 0 + RT ln - - - - [A] [B] ΔG 0 = - RT ln K eq

2Free enthalpy change and redox potential:

ΔG´ free enthalpy change at pH 7 (J mol

-1 ΔG 0 ´ standard free enthalpy change (concentrations of all reactants are 1 mol/l) at pH 7 (J mol -1 ΔE h ´ difference in redox potentials between two redox systems, at pH 7 (V) ΔE 0

´ difference in standard redox potentials (all reactants are at conc. 1 mol/l) between two redox systems,

at pH 7 (V) n number of electrons transferred

F Faraday (Faraday's charge), F = 96 487 J V

-1 mol -1 = 96 487 C mol -1

Osmotic work:

A osmotic work (J mol

-1

n amount of protons transported against the concentration gradient (numerically equivalent to number of

moles)

R universal gas constant, R = 8.3143 J K

-1 mol -1

T temperature (K)

ln natural logarithm, ln = 2.303 log c 1 concentration of the particles at the original side of the membrane c 2 concentration of the particles at the other side of the membrane E membrane potential difference resulting from uneven distribution of protons (V)

F Faraday (Faraday's charge), F = 96 487 J V

-1 mol -1 = 96 487 C mol -1

Examples:

TASK 1

What is the redox potential E

h´ of the system NAD / NADH against hydrogen electrode, if the standard redox potential E 0

´ is - 0.32 V and ratio NAD

/NADH is 10:1? -1 mol -1 , F = 96 487 J V -1 mol -1

Solution:

Use the Nernst-Peters equation

and take into account that NAD transfers two electrons.

8.3143 x 298.15

E h´ = - 0.32 + - - - - - - - - x ln 10 =

2 x 96 487

= - 0.32 + 0.0128458 x 2.3026 = = - 0.32 + 0.0296 = ≡ - 0.29 V

ΔG´ = - n F ΔE

h ΔG 0

´ = - n F ΔE

0 c 2

A = n RT ln - -

c 1 A

E = - -

R T [NAD

E h

´ = E

´ + - - - - ln - - - -

n F [NADH] 3

TASK 2

What is the value of the equilibrium constant of ATP hydrolysis at 37 0 C [ADP] [P i K eq [ATP] [H 2 O] if ΔG 0

´ = -35 kJ . mol

-1 ? (R = 8.3143 J K -1 mol -1

Solution:

From this equation ln K

eq = ΔG 0 ´/ - RT = - 35 000 / (- 8.3143 x 310.15) = 13.5732 K eq = e

13.5732

≡ 7.8 x 10 5

Or: ln K

eq = 13.5732 log K eq = 13.5732 /2.303 = 5.893 K eq = 10 5.893 ≡ 7.8 x 10 5

TASK 3

ΔG 0 = -35 kJ . mol -1 . Calculate the free enthalpy change ΔG at the ratio ATP/ADP = 100:1. (Temperature 37 0

C, R = 8.3143 J K

-1 mol -1 . Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)

Solution:

ATP + H

O ADP + P

100 1

1 ΔG = - 35 000 + 8.3143 x 310.15 x ln - - - = 100
= - 35 000 - 11 875.26 = = - 46 875.26 = ≡ - 46.9 kJ mol -1 ΔG 0 = - RT ln K eq [ADP]

ΔG = ΔG

0 + RT ln - - - - - [ATP] 4

TASK 4

In the mitochondrial respiratory chain, reduced NAD (NADH) is reoxidised by flavin dehydrogenases:

NADH+H

+ FAD ĺ NAD + FADH 2

The difference in the standard redox potentials

ΔE 0

´ for the oxidation of NADH by FAD is

+0.2 V.

Calculate the standard free enthalpy change ΔG

´ (in kJ mol

-1 (F = 96 487 J V -1 mol -1

Solution:

Take into account that the reaction involves transfer of two electrons. ΔG 0 ´ = - 2 x 96487 x 0.2 = - 38 594.8 ≡ - 38.6 kJ mol -1

TASK 5

What is the osmotic work A for transfer of 1 mole of protons by respiratory chain complexes through the inner mitochondrial membrane, from matrix side (pH = 7) to the inter-membrane space (pH = 6) at 37 0 C ? What is the corresponding difference in membrane potential ΔE (expressed in V) ? (R = 8.3143 J K -1 mol -1 , F = 96 487 J V -1 mol -1

Solution:

n = 1

R = 8.3143 J K

-1 mol -1

T = 310.15 K

c 1 = 10 -7 mol/l c 2 = 10 -6 mol/l A = 1 x 8.3143 x 310.15 x 2.303 x log10 = = 5 937.6 J mol -1 ≡ 5.9 kJ mol -1 ΔE = A / F = 5 937.6 / 96 487 ≡ 0.061 V MUDr. Jan Pláteník, PhD & Prof. MUDr. JiĜí Kraml, DrSc, 2004 c 2

A = n RT ln - -

c 1 ΔG 0

´ = - n F ΔE

0quotesdbs_dbs17.pdfusesText_23

[PDF] [PDF] Calculations in Bioenergetics

Calculations in Bioenergetics

Basic Relationships:

´ redox potential at pH 7 (V)

R universal gas constant, R = 8.3143 J K -1

T temperature (K)

F Faraday (Faraday's charge), F = 96 487 J V

Free enthalpy change and equilibrium constant:

ΔG free enthalpy change (J mol -1

R universal gas constant, R = 8.3143 J K

T temperature in (K)

RT [oxid.]

´ = E

´ + - - - - ln - - - -

A + B C + D

ΔG = - RT ln K

2Free enthalpy change and redox potential:

ΔG´ free enthalpy change at pH 7 (J mol

F Faraday (Faraday's charge), F = 96 487 J V

Osmotic work:

A osmotic work (J mol

R universal gas constant, R = 8.3143 J K

T temperature (K)

F Faraday (Faraday's charge), F = 96 487 J V

Examples:

TASK 1

What is the redox potential E

´ is - 0.32 V and ratio NAD

Solution:

Use the Nernst-Peters equation

8.3143 x 298.15

2 x 96 487

ΔG´ = - n F ΔE

´ = - n F ΔE

A = n RT ln - -

E = - -

R T [NAD

´ = E

´ + - - - - ln - - - -

TASK 2

´ = -35 kJ . mol

Solution:

From this equation ln K

13.5732

Or: ln K

TASK 3

C, R = 8.3143 J K

Solution:

ATP + H

O ADP + P

100 1

ΔG = ΔG

TASK 4

NADH+H

The difference in the standard redox potentials

´ for the oxidation of NADH by FAD is

Calculate the standard free enthalpy change ΔG

´ (in kJ mol

Solution:

TASK 5

Solution:

R = 8.3143 J K

T = 310.15 K

A = n RT ln - -

´ = - n F ΔE