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KEY CONCEPT
1) ∆G is independent
2) ∆G says nothing about the rate!
WHY DO WE WANT TO
KNOW ABOUT ∆G?
KEY CONCEPT
The value ∆G can be used to predict:
1)
2)
∆G = ∆G° + 0.616kcal/M • ln [B]/[A]
A <-----> B
• ∆G = Cellular Change in Free Energy
• ∆Go = Standard Free Energy Change
• 0.616 = Gas/temperature constant in kcal/M
• [A] = concentration of A in M
• [B] = concentration of B in M
SO, WE NEED A WAY TO
GET ∆Go
DO EXPERIMENT IN THE
LAB
Mix equal amounts of A and B in test tube under
STANDARD STATE CONDITIONS and let it go
to equilibrium
Then evaluate the concentration of A and B
0 = ∆G° + 0.616kcal/M • ln [B]/[A]
∆G° = - 0.616kcal/M • ln [B]/[A]
THUS:
WHAT HAPPENS TO ∆G AT
EQUILIBRIUM?
AND:
At equilibrium [B]/[A] is defined as
the k equilibrium (Keq)
THEREFORE:
∆Go = - 0.616kcal/M • ln Keq
SUMMARY
• ∆G = ??
• ∆Go = ??
• Keq = ratio of product [B] to substrate [A]
– Under what condition?
KEY CONCEPT
The correlation between ∆G and ∆G°
depends on the concentration of substrate
and product in the BODY
LETS TRY A
CALCULATION TO SEE
HOW IT WORKS
Glucose <-----> Fructose
A <----------> B
At equilibrium, the ratio of FRU to
GLU is 0.0475 at pH 7 and 23oC
• What is Keq?
• What is ∆Go’?
• What is ∆G?
• What does this info tell us about the reaction?
WHAT IS Keq?
Keq = [B]/[A]
Keq = [FRU]/[GLU]
Keq = 0.0475
WHAT IS ∆Go?
∆G° = -0.616kcal/mol(lnKeq)
∆G° = - 0.616kcal /mol • (ln0.0475)
∆G° = +1.88kcal/mol
WHAT DOES THIS TELL US
ABOUT THE RATE IN THE
BODY?
HOW ABOUT UNDER
STANDARD STATE
CONDITIONS?
WHAT IS ∆G?
∆G = ∆Go + 0.616kcal • ln[B]/[A]
BUT:
What do we need???
So we check cells and find that:
GLU [A] = 2 x 10-4M
FRU [B] = 3 x 10-6M
SO:
∆G = ∆G° + 0.616ln[B]/[A]
∆G = +1.88kcal/mol + 0.616*ln[3 x 10-6]/[2 x 10-4]
∆G = +1.88kcal/mol + (-2.58)
∆G = -0.71kcal/mol
SO WHAT DOES THIS MEAN?
In the body the reaction in which GLU is
converted to FRU is favorable and will
liberate -0.71kcal/mol of free energy.
What is the spontaneity?
SUPER MAJOR
KEY CONCEPT
The interplay between those reactions and
compounds with high levels of free energy
and those without is the whole essence of
metabolism.
NOTES ABOUT ENZYMES
• DO NOT affect the equilibrium, accelerate the
attainment of equilibrium
ENERGY, COUPLING AND
BIOCHEMISTRY
SUPER MAJOR
KEY CONCEPT
ALL reactions have a ∆G and a reaction
that produces a high level of free energy
can be used to drive a reaction that is
unfavorable.
KEY CONCEPT
In biological systems, a spontaneous chemical
reaction that is not at equilibrium is capable
of providing energy as it attempts to attain
equilibrium.
CONSIDER THE REACTION:
A B C D
-1.36 +5.45 -6.82
Total ∆G is additive for this set of reactions
-1.36 + 5.45 + (-6.82) = -2.73
A B C D
-1.36 +5.45 -6.82
Coupling can involve
TWO INDEPENDENT PATHWAYS.
A to B will not occur unless coupled to C to D.
A B
C D
+4.0
-7.3
A + C B + D -3.3
KEY CONCEPT
The overall free energy change for a
chemically coupled reaction is equal to the
sum of the free energy changes of the
individual steps
KEY CONCEPT
A thermodynamically unfavorable reaction
can be driven by a favorable one!
SO UP TO NOW:
Defined free energy
Intrinsic energy within molecules
Obtained as reaction moves to equilibrium
Indicated where it was found
(Chemical bonds, etc)
WHAT IS THE FORM OF
FREE ENERGY
AND
HOW DO WE OBTAIN AND
USE FREE ENERGY?
REMEMBER?
A -------> B = ∆G (-)
QUESTION?
HOW DO WE COLLECT THE FREE
ENERGY FROM THE CHEMICAL
REACTIONS?
OXIDATION
C6H12O6 + O2 -----> H2O + CO2
A -------> B
Oxidation of fuel molecules
generates free energy to drive
reactions
Glucose, Fatty Acids, Amino Acids
KEY CONCEPT
What happens in oxidation?
We need way to collect and transport this energy
Electron Carriers
• NAD +
• FADH +
• NADP +
NADH and FADH2
Transfer electrons to the oxidative
phosphorylation enzymes in the mitochondria
The energy in the electrons are used to
produce a H+ gradient so that ATP can be
synthesized
NADPH
Is the major
electron
donor in
reductive
biosynthesis
SO WE COLLECT ELECTRONS
WITH NADH
Can we use these electrons directly?
KEY CONCEPT
The free energy of electrons are used to
generate ATP
OXIDATIVE PHOSPHORYLATION
KEY CONCEPT
Adenosine triphosphate (ATP)
is the universal currency of free energy
in biological systems
PHOSPHORYL TRANSFER
POTENTIAL
∆G° for the hydrolysis of ATP is
-7.3kcal/mol
ATP + H2O <----> ADP + Pi + H+
SO HOW DOES ATP WORK?
A B
ATP ADP + Pi
+ 4.0
-7.3
A + ATP B + ADP + Pi -3.3
2-59 MBC
KEY CONCEPT
The hydrolysis of ATP changes the Keq by a
very large number and allows
thermodynamically unfavorable reactions to
occur.
ATP HYDROLYSIS
EQUALS AN INPUT OF ENERGY INTO A
REACTION
What will happen to equilibrium?
KEY CONCEPT
• A to B DO NOT have to represent a
chemical reaction!!!!
– Muscle contraction (movement of proteins)
– Ion flux (gradient)
– Change in Protein Conformation
OTHER COMPOUNDS ALSO
HAVE HIGH PHOSPHORYL
TRANSFER POTENTIAL
• Phosphenolpyruvate (PEP) ∆Go -14.3
• Creatine Phosphate ∆Go -10.3
• ATP ∆Go -7.3
• Glucose-1-phosphate ∆Go -5.3
• Glucose-6-phosphate ∆Go -3.3
THERE ARE MANY OTHER
ACTIVATED CARRIERS
COENZYME A
Universal carrier of acyl
groups
Carries 2-24 carbon units
Terminal sulfhydral group is a
reactive site
2-62
MBC
SUMMARY
WE COLLECT THE FREE
ENERGY (∆G) IN TWO BASIC
FORMS
Electrons (e-)
High Energy Phosphate (PO4)
OK
You should know the following concepts:
∆G and its meaning
How enzymes fit into the mix
The form in which free energy is harvested
How free energy is utilized in the form of
ATP.

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L3 energy new_slide30

  • 1. KEY CONCEPT 1) ∆G is independent 2) ∆G says nothing about the rate!
  • 2. WHY DO WE WANT TO KNOW ABOUT ∆G?
  • 3. KEY CONCEPT The value ∆G can be used to predict: 1) 2)
  • 4. ∆G = ∆G° + 0.616kcal/M • ln [B]/[A] A <-----> B • ∆G = Cellular Change in Free Energy • ∆Go = Standard Free Energy Change • 0.616 = Gas/temperature constant in kcal/M • [A] = concentration of A in M • [B] = concentration of B in M
  • 5. SO, WE NEED A WAY TO GET ∆Go
  • 6. DO EXPERIMENT IN THE LAB Mix equal amounts of A and B in test tube under STANDARD STATE CONDITIONS and let it go to equilibrium Then evaluate the concentration of A and B
  • 7. 0 = ∆G° + 0.616kcal/M • ln [B]/[A] ∆G° = - 0.616kcal/M • ln [B]/[A] THUS: WHAT HAPPENS TO ∆G AT EQUILIBRIUM?
  • 8. AND: At equilibrium [B]/[A] is defined as the k equilibrium (Keq) THEREFORE: ∆Go = - 0.616kcal/M • ln Keq
  • 9. SUMMARY • ∆G = ?? • ∆Go = ?? • Keq = ratio of product [B] to substrate [A] – Under what condition?
  • 10.
  • 11. KEY CONCEPT The correlation between ∆G and ∆G° depends on the concentration of substrate and product in the BODY
  • 12. LETS TRY A CALCULATION TO SEE HOW IT WORKS
  • 13. Glucose <-----> Fructose A <----------> B
  • 14. At equilibrium, the ratio of FRU to GLU is 0.0475 at pH 7 and 23oC • What is Keq? • What is ∆Go’? • What is ∆G? • What does this info tell us about the reaction?
  • 15. WHAT IS Keq? Keq = [B]/[A] Keq = [FRU]/[GLU] Keq = 0.0475
  • 16. WHAT IS ∆Go? ∆G° = -0.616kcal/mol(lnKeq) ∆G° = - 0.616kcal /mol • (ln0.0475) ∆G° = +1.88kcal/mol
  • 17. WHAT DOES THIS TELL US ABOUT THE RATE IN THE BODY? HOW ABOUT UNDER STANDARD STATE CONDITIONS?
  • 18. WHAT IS ∆G? ∆G = ∆Go + 0.616kcal • ln[B]/[A] BUT: What do we need???
  • 19. So we check cells and find that: GLU [A] = 2 x 10-4M FRU [B] = 3 x 10-6M SO: ∆G = ∆G° + 0.616ln[B]/[A] ∆G = +1.88kcal/mol + 0.616*ln[3 x 10-6]/[2 x 10-4] ∆G = +1.88kcal/mol + (-2.58) ∆G = -0.71kcal/mol
  • 20. SO WHAT DOES THIS MEAN? In the body the reaction in which GLU is converted to FRU is favorable and will liberate -0.71kcal/mol of free energy. What is the spontaneity?
  • 21. SUPER MAJOR KEY CONCEPT The interplay between those reactions and compounds with high levels of free energy and those without is the whole essence of metabolism.
  • 22. NOTES ABOUT ENZYMES • DO NOT affect the equilibrium, accelerate the attainment of equilibrium
  • 24. SUPER MAJOR KEY CONCEPT ALL reactions have a ∆G and a reaction that produces a high level of free energy can be used to drive a reaction that is unfavorable.
  • 25. KEY CONCEPT In biological systems, a spontaneous chemical reaction that is not at equilibrium is capable of providing energy as it attempts to attain equilibrium.
  • 26. CONSIDER THE REACTION: A B C D -1.36 +5.45 -6.82
  • 27. Total ∆G is additive for this set of reactions -1.36 + 5.45 + (-6.82) = -2.73 A B C D -1.36 +5.45 -6.82
  • 28. Coupling can involve TWO INDEPENDENT PATHWAYS. A to B will not occur unless coupled to C to D. A B C D +4.0 -7.3 A + C B + D -3.3
  • 29. KEY CONCEPT The overall free energy change for a chemically coupled reaction is equal to the sum of the free energy changes of the individual steps
  • 30. KEY CONCEPT A thermodynamically unfavorable reaction can be driven by a favorable one!
  • 31. SO UP TO NOW: Defined free energy Intrinsic energy within molecules Obtained as reaction moves to equilibrium Indicated where it was found (Chemical bonds, etc)
  • 32. WHAT IS THE FORM OF FREE ENERGY AND HOW DO WE OBTAIN AND USE FREE ENERGY?
  • 34. QUESTION? HOW DO WE COLLECT THE FREE ENERGY FROM THE CHEMICAL REACTIONS?
  • 35. OXIDATION C6H12O6 + O2 -----> H2O + CO2 A -------> B
  • 36. Oxidation of fuel molecules generates free energy to drive reactions Glucose, Fatty Acids, Amino Acids
  • 37. KEY CONCEPT What happens in oxidation? We need way to collect and transport this energy
  • 38. Electron Carriers • NAD + • FADH + • NADP +
  • 39. NADH and FADH2 Transfer electrons to the oxidative phosphorylation enzymes in the mitochondria The energy in the electrons are used to produce a H+ gradient so that ATP can be synthesized
  • 40. NADPH Is the major electron donor in reductive biosynthesis
  • 41. SO WE COLLECT ELECTRONS WITH NADH Can we use these electrons directly?
  • 42. KEY CONCEPT The free energy of electrons are used to generate ATP OXIDATIVE PHOSPHORYLATION
  • 43. KEY CONCEPT Adenosine triphosphate (ATP) is the universal currency of free energy in biological systems PHOSPHORYL TRANSFER POTENTIAL
  • 44. ∆G° for the hydrolysis of ATP is -7.3kcal/mol ATP + H2O <----> ADP + Pi + H+
  • 45. SO HOW DOES ATP WORK? A B ATP ADP + Pi + 4.0 -7.3 A + ATP B + ADP + Pi -3.3
  • 47. KEY CONCEPT The hydrolysis of ATP changes the Keq by a very large number and allows thermodynamically unfavorable reactions to occur.
  • 48. ATP HYDROLYSIS EQUALS AN INPUT OF ENERGY INTO A REACTION What will happen to equilibrium?
  • 49. KEY CONCEPT • A to B DO NOT have to represent a chemical reaction!!!! – Muscle contraction (movement of proteins) – Ion flux (gradient) – Change in Protein Conformation
  • 50. OTHER COMPOUNDS ALSO HAVE HIGH PHOSPHORYL TRANSFER POTENTIAL • Phosphenolpyruvate (PEP) ∆Go -14.3 • Creatine Phosphate ∆Go -10.3 • ATP ∆Go -7.3 • Glucose-1-phosphate ∆Go -5.3 • Glucose-6-phosphate ∆Go -3.3
  • 51. THERE ARE MANY OTHER ACTIVATED CARRIERS
  • 52. COENZYME A Universal carrier of acyl groups Carries 2-24 carbon units Terminal sulfhydral group is a reactive site
  • 54. SUMMARY WE COLLECT THE FREE ENERGY (∆G) IN TWO BASIC FORMS Electrons (e-) High Energy Phosphate (PO4)
  • 55. OK You should know the following concepts: ∆G and its meaning How enzymes fit into the mix The form in which free energy is harvested How free energy is utilized in the form of ATP.