Download Bioenergetics - Biochemistry - Lecture Notes and more Study notes Biochemistry in PDF only on Docsity!
Bioenergetics
Basis of Thermodynamics
Every living cell and organism must perform work to stay alive, to grow and to reproduce. The ability to harvest energy from nutrients or photons of light and to channel it into biological work is the miracle of life. Living organisms carry out a remarkable variety of energy transductions. The biological energy transductions obey the physical laws that govern all natural processes, including the laws of thermodynamics.
1 st^ Law of Thermodynamics
The energy of the universe remains constant.
2 nd^ Law of Thermodynamics
All spontaneous processes increase the entropy of the universe.
State functions depend only on the initial and final conditions not on path taken between the initial and final conditions. They are independent of path. The important state functions for the study of biological systems are: G, the Gibbs free energy which is equal to the total amount of energy capable of doing work during a process at constant temperature and pressure.
- If ∆G is negative, then the process is spontaneous and termed exergonic.
- If ∆G is positive, then the process is nonspontaneous and termed endergonic.
- If ∆G is equal to zero, then the process has reached equibrium. H , the Enthalpy which is the heat content of the system.
- When ∆H is negative the process produces heat and is termed exothermic.
- When ∆H is positive the process absorbs heat and is termed endothermic. S , the Entropy is a quantitative expression of the degree of randomness or disorder of the system.
- When ∆S is positive then the disorder of the system has increased.
- When ∆S is negative then the disorder of the system has decreased.
The conditions of biological systems are constant temperature and pressure. Under such conditions the relationships between the change in free energy, enthalpy and entropy can be described by the expression where T is the temperature of the system in Kelvin.
∆G = ∆H − T∆S
Equilibrium Constants
All spontaneous processes proceed until equilibrium is reached. Consider the following chemical reaction.
A + B C + D
k (^1)
k (^2)
The forward rate of product formation is = k 1 [A][B] The reverse rate of reactant formation is = k 2 [C][D]
At equilibrium the concentrations of products and reactants are such that forward and reverse rates are equal k 1 [Aeq ][Beq ] = k2 [Ceq ][Deq ]. A little algebra and presto
[ ][ ]
[ ][ ]
2
1 eq eq
eq eq eq A B
C D
k
k K = =
At equilibrium ∆G = 0. The biochemist standard state the concentration of reactants and products are initially set at 1 M, the temperature is 298 oK, the pressure is 1 atm, the pH is 7.0 and the concentration of water is 55 M. The biochemists constants are written as ∆Go^ ’ and K’ (^) eq. This is the only standard state we will work with in this class so forgive if I occasionally drop the prime. ∆G o^ ’ is a constant characteristic for each reaction just as K’eq is a constant characteristic for each reaction. These two constants have a simple relationship.
eq ∆ G o^ ' =− RT ln K ' or
−∆ = RT
G
eq
o
K e
' '
The actual free energy change depends on the reactant and product concentrations.
[ ][ ]
[ ][ ]
' ln A B
C D
G Go RT
Reactions can be coupled together. The standard free energy changes are additive. Cool feature of state functions. Multiply the equilibrium constants Ie. (1)Glucose + Pi ^ glucose-6-phosphate + H 2 O^ ∆Go^ ’ = 13.8 kJ/mol; K’^ eq = 3.9^ X^10 -3^ M^ -^. (2)ATP + H 2 O ADP + Pi ∆Go^ ’ = -30.5 kJ/mol; K’eq = 2 X 105 M.
(Sum)ATP + glucoseglucose-6-phosphate + ADP ∆G o’ = 13.8 kJ/mol + -30.5 kJ/mol = -16.7 kJ/mol K’ (^) eq = (3.9 X 10 -3^ M-1^ ) X(2 X 10 5 M)=7.8 X 10 2
Given (1) Phosphoenolpyruvate (PEP) + H 2 0 pyruvate + Pi ∆Go^ ’ = -61.9 kJ/mol (2) ATP + H 2 O ADP + Pi ∆Go^ ’ = -30.5 kJ/mol What is the free energy change for: PEP + ADP ATP + pyruvate? (1) PEP + H 2 0 pyruvate + Pi ∆Go^ ’ = -61.9 kJ/mol (2) ADP + Pi ATP + H 2 O ∆Go^ ’ = 30.5 kJ/mol *
(Sum) PEP + ADP ATP + pyruvate ∆Go^ ’ = -61.9 kJ/mol + 30.5 kJ/mol = -31.4 kJ/mol *Note: change the direction of the reaction change the sign of the ∆Go^ ’, invert equilibrium constant. 1/K’ (^) eq.
Thermodynamics of ATP Hydrolysis ATP is the principle energy currency of the cell that links catabolism to anabolism. ATP has a large negative standard free energy change of hydrolysis. ATP + H 2 O ADP + Pi ∆Go^ ’ = -30.5 kJ/mol What is the chemical basis of the large, negative free energy change?
- The hydrolytic cleavage of the γ-phosphate anhydride bond relieves electrostatic repulsion in ATP.
Other High Energy Phosphorylated compounds.
Phosphoenolpyruvate – PEP PEP3-^ + H 2 O Pyruvate -^ + Pi2-^ ∆Go^ ’ = -61.9 kJ/mol
O P O -
O -
C O
CH 2
C
O
H O H
O
P O- O -
C HO
CH 2
C
O
C
H 2 C
C
O
H
Phosphoenolpyruvate contains one phosphate ester bond that can under go hydrolysis to yield the enol form of pyruvate which immediately tautomerizes to the more stable keto form of pyruvate. The reactant PEP has only one stable form while the product pyruvate has two possible forms. This extra stabilization of the product is the greatest contributor to the high standard free energy of hydrolysis.
1,3-Bisphosphoglycerate 1,3-Bisphosphoglycerate 2-^ + H 2 O 3-Phosphoglycerate3-^ + Pi 2-^ + H+^ ∆Go^ ’ = -49.3 kJ/mol
This high energy compound contains one phosphoanhydride bond.
The product 3-phosphoglyceric acid immediately ionizes to produce a carboxylate anion.
The removal of the 3-phosphoglyceric acid and the resonance stabilized phosphate favor the forward reaction.
O
O C H 2 C C
OH
H
O O P
O O - O- H O H
O
O CH 2 C C
OH
H
O OH P
O O - O-
O
O CH 2
C C
OH
H
O O-
H+
Phosphocreatine Phosphocreatine 2-^ + H 2 O Creatine + Pi2-^ ∆Go^ ’ = -49.3 kJ/mol
O
O -
N H C N CH 3
CH 2
H
O H
H
NH 2
C
O O -
P
O O - O
H O
H 2 N C N CH 3
CH 2
NH 2 H
C
O O -
H 2 N C^ N^ CH 3
CH 2
NH 2 H
C
O O -
The release of Pi and the resonance stabilized creatine favor the forward reaction.
Note for all of these phosphate releasing reactions, the phosphate formed is stabilized by resonance favoring product formation.
Standard Free Energies of Phosphate Ester Hydrolysis of Some Biological Compounds Compound (^) ∆Go^ ’ (kJ/mol)
Phosphoenolpyruvate (PEP) -61. 1,3-Bisphosphoglycerate -49. Acetyl phosphate -43. Phosphocreatine -43. ADPAMP + Pi -35. PPi -33. ATPAMP + PPi -32. ATPADP + Pi -30. Glucose-1-phosphate -20. Fructose-6-phosphate -15. Glucose-6-phosphate -13. Glycerol-3-phosphate -9. AMPadenosine + Pi -9. The compounds with more negative free energy of phosphate ester hydrolysis than ATP can phosphorylate ADP to form ATP. Ie. PEP + ADP ATP + pyruvate ∆Go^ ’ = -31.4 kJ/mol
ATP provides energy by group transfers, not simple hydrolysis. Ie. (1)Glucose + Pi glucose-6-phosphate + H 2 O ∆Go^ ’ = 13.8 kJ/mol; K’ (^) eq = 3.9 X 10 -3^ M -^. (2)ATP + H 2 O ADP + Pi ∆Go^ ’ = -30.5 kJ/mol; K’eq = 2 X 105 M.
(Sum)ATP + glucoseglucose-6-phosphate + ADP ∆G o’ = 13.8 kJ/mol + -30.5 kJ/mol = -16.7 kJ/mol K’ (^) eq = (3.9 X 10 -3^ M-1^ ) X(2 X 10 5 M)=7.8 X 10 2
Most of the group transfer reactions of ATP are SN 2 nucleophilic substitutions. In the examples above the nucleophile is an oxygen of an alcohol. Each of the three phosphates of ATP are susceptible to nucleophilic attack. Nucleophilic attack at the γ-phosphate results in ADP and the transfer of phosphate to the nucleophile. Nucleophilic attack at the β-phosphate results in AMP and the transfer of a pyrophosphate group to the nucleophile. Nucleophilic attack at the α-phosphate results in pyrophosphate and the adenylylation of the nucleophile.
Inorganic Pyrophosphatase The nucleophilic attack at the α-phosphate of ATP results in an adenylylated nucleophile and pyrophosphate. The ubiquitous enzyme inorganic pyrophosphatase provides an additional thermodynamic push for the adenylylation reaction by catalyzing the hydrolysis of pyrophosphate into two molecules of phosphate (∆Go^ ’ = -33.5 kJ/mol). This enzyme makes adenylylation reactions very favorable thermodynamically.
Example: Fatty acyl-CoA synthesis.
ATPAMP + PPi ∆Go^ ’ = -32.2 kJ/mol PPi2Pi ∆Go^ ’ = -33.5 kJ/mol
ATPAMP + 2Pi ∆Go^ ’ = -65.7 kJ/mol
CoA—SH + PalmitatePalmitoyl-CoA ∆Go^ ’ = 31.4 kJ/mol
Palmitate + ATP + CoASHpalmitoyl-CoA + AMP + 2 Pi ∆Go^ ’ = −65.7 kJ/mol +31.4 kJ/mol= −34.3 kJ/mol
N N N
N
NH (^2)
O OH OH
H H H H
O P O O-
O P
O
O-
P O
O
O-
N N N
N
NH 2
O OH OH
H H H H
P O O-
O
Fatty acyl-CoA synthetase
O C
O-
CH 3 (CH 2 ) 14
C
O CH 3 (CH 2 ) 14 O
P O-
O
O-
P O
O
O-
H H O
P O
O
O-
2 - O^ H
inorganic pyrophosphatase
C
O CH 3 (CH 2 ) 14 S^ CoA
N N N
N
NH 2
O OH OH
H H H H
P O O-
O
N N N
N
NH (^2)
O
OH OH
H H H H
O P O O-
O P
O
O-
P O
O
O-
γ β α
:OR :OR^ :OR
RO P
O
O-
O - RO P
O
O-
O P
O O - O-
ADP AMP
N N N
N
NH (^2)
O
OH OH
H H H H
O P O O-
O R
PPi
Three Positions on ATP for Nucleophilic Attack
O
OH OH
H H H H
O P O O-
O P
O
O-
P O
O
O-
N N N
N
NH (^2)
O
OH OH
H H H H
P O O-
O
O
Rna
:
N N
NH 2
O
N N N
N
NH (^2)
O
O OH
H H H H
P O O-
O
O
Rna
N
NH 2
N O O
OH OH
H H H H
O P O O-
P O -
O
O-
P O
O
O-
P OH
O
O-
H 2 O
2
RNA Synthesis
N N N
N
NH 2
O
OH OH
H H H H
O P O O-
O P
O
O-
P
O
O -
H 2 N CH C CH (^3)
O-
O
H 2 N CH C CH 3
O
N N N
N
NH 2
O
OH OH
H H H H
O P O O-
O
P O-
O
O-
P
O
O-
P
O
O-
H 2 O
2
Inorganic phosphatase
Activation of an aminoacid for aminoacyl-tRNA synthesis
