1. Steady state is not the same as equilibrium
This slide for afternoon class
Molecules are flowing,
but back and forth
but, plenty of net flow

2. Direction of reactions in metabolism
Lec. 8

3. Free energy difference determines the direction of a chemical reaction
atoms
transition state
free energy
catalyzed transition state
free
A + B }
energy
difference
E + F
C + D

4. Change in free energy (DG) Energy that can do work
For: A + B C + D, left-to-right direction as indicated:
Consider the quantity called the change in free energy associated with a chemical reaction, or: Δ G
Δ G < 0 (negative): A AND B WILL TEND TO PRODUCE C AND D (i.e., tends to the right).
Δ G > 0 (positive): C AND D WILL TEND TO PRODUCE A AND B. (i.e., tends to the left)
Δ G = 0: THE REACTION WILL BE AT EQUILIBRIUM:
NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY

5. The change in free energy (DG)
A + B C + D
[C][D]
[A][B]
products
reactants
DG = DGo+ RTln
A, B, C and D = concentrations of the reactants and the products AT THE MOMENT being considered. (i.e., these A, B, C, D’s here are not the equilibrium concentrations)
R = universal gas constant = 1.98 CAL / °K MOLE ( soR =~2)
T = ABSOLUTE TEMP ( oK ) 0oC = 273oK; Room temp = 25o C = 298o K or ~ 300o K
ln = NATURAL LOG
D Go = a CONSTANT: a quantity related to the INTRINSIC properties of A, B, C, and D, to be explained

6. The change in free energy change (DG)
Also abbreviated form: DG = DGo+ RTlnQ (Q for “quotient”)
Where Q = ([C][D]/[A][B])
Qualitative term:
What molecules are in play. This depends on what A, B, C, and D are.
Quantitative term:
How much of each is present, at the moment under consideration
Josiah Willard Gibbs ( )

7. Standard free energy change (DGo)
STANDARD FREE ENERGY CHANGE of a reaction.
If all the reactants and all the products are present at 1 unit concentration, then:
DG = DGo + RTln(Q)
= DGo + RTln([1][1] / [1][1])
= DGo + RTln(1)
= DGo + RT x 0,
or DG = DGo, when all components are at 1
….. a special case
(when all components are at 1)
“1” usually means 1 M

8. Standard free energy change (DGo)
So DG and DGo are quite different,
and not to be confused with each other.
DGo allows us to compare all reactions under the same standard reaction conditions that we all agree to, independent of concentrations.
So it allows a comparison of the stabilities of the bonds in the reactants vs. the products.
It is useful.
AND,
It is easily measured.

9. DG and DGo at equilibrium
Because at equilibrium, DG = DGo + RTln(Q) = 0
and only at equilibrium: Q = Keq =
(a second special case).
So: at equilibrium, DG = DGo + RTln(Keq) = 0
And so: DGo = - RTln(Keq)
So just measure the Keq,
Plug in R and T
Get: DGo, the standard free energy change
Publish it. It gets written in a book.
[C]eq [D]eq
[A]eq [B]eq

10. E.g., let’s say for the reaction A + B C + D, Keq happens to be:
[C]eq[D]eq
[A]eq[B]eq
Then DGo = -RTlnKeq = -2 x 300 x ln(2.5 x 10-3)
= x = +3600
3600 cal/mole (If we use R = 2 we are dealing with calories)
Or: kcal/mole
3.6 kcal/mole ABSORBED (positive number, +3.6))
So energy is required for the reaction in the left-to-right direction
And indeed, very little product accumulates at equilibrium
(Keq = )
= 2.5 x 10-3
so, not very much product

11. (Reverse the reaction: switch the sign)
Note:
If for the reaction A + B C + D, ΔGo = +3.6
Then for the reaction C + D A + B, ΔGo = - 3.6
(Reverse the reaction: switch the sign)
And:
For reactions of more than simple 1 to 1 stoichiometries:
aA + bB <--> cC + dD,
ΔG = ΔGo + RT ln [C]c[D]d                           [A]a[B]b
And RT = 2*300 = 600 cal/deg-mole at room temp., or
RT= 0.002*300 = 0.6 kcal/deg-mole at room temp, more usefully

12. Some exceptions to the 1M standard condition:
1) Water: 55 M (pure water) is considered the “unit” concentration in this case instead of 1M The concentration of water rarely changes during the course of an aqueous reaction, since water is at such a high concentration.
So when calculating DGo, instead of writing in “55” when water participates in a reaction (e.g., a hydrolysis) we write “1.”
This is not cheating; we are in charge of what is a “standard” condition, and we all agree to this: 55 M H20 is unit (“1”) concentration for the purpose of defining DGo.

13. Some exceptions to the 1M standard condition:
In the same way,
Hydrogen ion concentration, [H+]: M is taken as unit concentration, by biochemists.
since pH7 is maintained (buffered) in most parts of the cell despite a reaction that may produce acid or base.
This definition of the standard free energy change requires the designation ΔGo’
However, we will not bother.
But it should be understood we are always talking about ΔGo’ in this course.

14. Summary DG = DGo + RTln(Q)
This combination of one qualitative and one quantitative (driving) term tells the direction of a chemical reaction in any particular circumstance
DGo = - RTln(Keq)
The ΔGo for any reaction is a constant that can be looked up in a book.

15. Energy in a cell: adenosine triphosphate (ATP)
Hydrolysis of ATP: ATP + HOH  ADP + HPO4--
ATP, a cellular small molecule that helps in the transfer of energy from a place where it is made to a place where it is needed.
adenosine
phosphate
Acid anhydride (new functional group: 2 acids joined) OH |
HO - P - OH || O
Dehydration between 2 acids
adenine
ribose
“base”
A-R: a nucleoside
ATP: a nucleotide (a nucleoside triphosphate)

16. The hydrolysis of ATP _ _ _ _ _ _ _ (ADP)
ATP + HOH  ADP + Pi _ _ _ _ _ _ _
(ADP)
The DGo of this reaction is about -7 kcal/mole.
Energy is released in this reaction.
This is an exergonic reaction under standard conditions
Strongly to the right, towards hydrolysis, towards ADP

17. “High energy” bonds
DGo of a least ~ -7 kcal/mole is released upon hydrolysis
Designated with a squiggle (~) often
ATP = A-P-P~P
Rationalized here by the relief of electrical repulsion upon hydrolysis: _
DGo = -7 kcal/mole

18. Hydrolysis of ADP A-R-P~P~P _ AMP _ (ADP) _ DGo = -7 kcal/mole
Not a high energy bond

19. What happens when we dissolve ATP in water?
ATP + HOH  ?
ADP + Pi + heat (7o)
ATP + HOH
+ ATPase
1M ATP, 1 ml water, 1 mmole kcal/mole, cal/mole, 7 cal / mmole, 7 cal:1 ml water: 1o C7o

20. How does the cell harness this energy?
The cell often uses the hydrolysis of ATP to release energy.
The released energy is used to drive reactions that require energy.
How does this work ??

21. An example: an endergonic reaction
Suppose: glucose + Pi glucose-6-phosphate
CH2OH
CH2OPO32-
+ Pi
Glucose + Pi --> glucose-6-P + H2O; Δ Go = +3.6 kcal/mole.
Keq= 2.5 x 10 -3
1) ATP    +     H2O   -->  ADP   +  Pi      Δ Go =  -7    kcal/mole
2) Glucose   +  Pi    -->  G6P   + H2O     Δ Go = +3.6 kcal/mole
ATP + H2O+ Glucose + Pi  ADP + Pi + G6P + H2O
Δ Go = -3.4 kcal/mole overall
Glucose + ATP   G6P + ADP Δ Go = -3.4 kcal/mole overall = net sum of the two considered reactions

22. An endergonic reaction, continued
Our test tube of 1 M ATP in water:
Add glucose + PO4:  nothing
Enzymes needed … ATPase? Glucose phosphorylase?
Add these enzymes: ATP  ADP + Pi, Glucose + Pi  G6P But just get 7 kcal/mole as heat again.
Energy absorbed mostly by raising the temperature of all the water around 7o (25o32o).
Just get warmer glucose.
But:

23. Directly coupled reactions
Answer: Add Hexokinase (instead of ATPase and glucose phosphorylase)
Glucose + AR-P-P-P  glucose-6-P + AR-P-P
Glucose +ATP glucose-6-P04 + ADP, DGo =
A directly coupled reaction. A new reaction, ATP is not simply hydrolyzed.
What a good idea!
Direct coupling of reactions is one of two ways the cell solves the problem of getting a reaction to go in the desired direction. The second: later.
-3.4 kcal/mole

24. ATP: the energy “currency” of the cell

25. Where does this ATP come from?
Glucose is the only carbon source in our minimal medium.
Need to make ATP from glucose; that path TAKES energy.
But need only to regenerate ATP from ADP:
coupled to drive endergonic reactions
ATP . . .
ATP
Glucose
ATP
ADP
biosynthetic pathway to ATP
regeneration of ATP from ADP?
Just need to put that Pi back onto ADP

26. Regeneration of ATP from ADP
Two solutions:
1) Photosynthesis
2) Catabolism of organic compounds (e.g., glucose)
Metabolic breakdown of an organic compound

27. Glucose catabolism overview/preview
1- GLYCOLYSIS (6C  3C)
2- KREBS CYCLE (3C 1C, CO2 release)
3- ELECTRON TRANSPORT CHAIN (oxygen uptake, water release)
Glycolysis, in detail, as:
Basic mechanism of energy metabolism (getting energy by glucose breakdown.)
An example of a metabolic pathway.

28. Glycolytic pathway, Glycolysis Handout 8A glucose oxidative pathways
dihydroxyacetone phosphate
fructose-1,6-diphosphate
glucose-6-phosphate
fructose-6-phosphate
glyceraldehyde-3-phosphate
Glycolytic pathway,
Glycolysis
glucose
1,3-diphosphoglyceric acid
phosphoenol-pyruvic acid
2-phosphoglyceric acid
3-phosphoglyceric acid
yeast
lactic acid
acetaldehyde
ethanol
Handout 8A
pyruvic acid
oxidative pathways

29. The first 5 steps of glycolysis
isomeriz- ation again
phosphorylation
isomerization
phosphorylation
2 moles of G3P for every mole of glucose. And: have spent 2 ATPs

30. Glycolytic pathway Glycolysis Handout 8A front glucose
dihydroxyacetone phosphate
fructose-1,6-diphosphate
glucose-6-phosphate
fructose-6-phosphate
glyceraldehyde-3-phosphate
Glycolytic pathway
Glycolysis
glucose
1,3-diphosphoglyceric acid
phosphoenol-pyruvic acid
2-phosphoglyceric acid
3-phosphoglyceric acid
yeast
lactic acid
acetaldehyde
ethanol
Handout 8A front
pyruvic acid
oxidative pathways

31. Oxidation = loss of electrons
Acid anhydride
functional group (if add HOH get 2 acids)
Need an acceptor for these electrons: an oxidizing agent
Abbreviations for the phosphate group
1,3,-diphospho-glyceric acid
glyceraldehyde-3- phosphate

32. Nicotinamide adenine dinucleotide (NAD)
-2H●
+2H●
“niacin”
or NADred
or “NADH2”
or NADH + H+
or NAD

33. { { Glycolytic pathway Loose end #2 = NAD Loose end #1 was ATPs33
fructose-1,6-diphosphate
glucose-6-phosphate
fructose-6-phosphate
Loose end #2 = NAD
Loose end #1 was ATPs
glyceraldehyde-3-phosphate
Glycolytic pathway
glucose {
1,3-diphosphoglyceric acid
Top carbon has
been oxidized
phosphoenol-pyruvic acid
2-phosphoglyceric acid
3-phosphoglyceric acid {
2 ATP
+ 4 ATP
+ 2 ATP
E. coli,
muscle
yeast
acetaldehyde
ethanol
lactic acid
pyruvic acid
“glycolysis” ends here, at pyruvate

34. Glycolytic pathway ATP debt paid in full 2 ATP + 4 ATP + 2 ATP
dihydroxyacetone phosphate
fructose-1,6-diphosphate
glucose-6-phosphate
fructose-6-phosphate
glyceraldehyde-3-phosphate
Glycolytic pathway
glucose
1,3-diphosphoglyceric acid
ATP debt paid in full
phosphoenol-pyruvic acid
2-phosphoglyceric acid
3-phosphoglyceric acid
E. coli,
muscle
2 ATP
+ 4 ATP
+ 2 ATP
yeast
lactic acid
acetaldehyde
ethanol
pyruvic acid
oxidative pathways
“glycolysis” ends here
Handout 8A

35. Net reaction:
1 glucose + 2 ADP + 2 Pi + 2 NAD pyruvate + 2 ATP + 2 NADH2
DGo = -18 kcal/mole
So overall reaction goes essentially completely to the right.

36. Energy levels in glycolysis
kcal/mole
glucose
2 pyruvates

37. Energy levels in glycolysis
kcal/mole
DG = DGo + RTln[products]
[reactants]
glucose
pull
pull
2 pyruvates

38. The second way the cell gets a reaction to go in the desired direction:
1) First way was: a directly coupled reaction (i.e., a different reaction) .
One of two ways the cell solves the problem of getting a reaction to go in the desired direction Glucose + ATP  glucose-6-P04 + ADP, DGo = -3.4 kcal/mole
2) The second way:
Removal of the product of an energetically unfavorable reaction
Uses a favorable downstream reaction
“Pulls” the unfavorable reaction
Operates on the second term of the DG equation.
DG = DGo + RTln([products]/[reactants])

39. Regeneration of NAD So glucose  pyruvic acid
ADP  ATP, as long as we have plenty of glucose
Are we all set?
No…. What about the NAD?.. We left it burdened with those electrons.
Soon all of the NAD will be in the form of NADH2
Very soon
Glycolysis will screech to a halt !!
Need an oxidizing agent in plentiful supply to keep taking those electron off the NADH2, to regenerate NAD so we can continue to run glucose through the glycolytic pathway.

40. Oxidizing agents around for NAD:
1) Oxygen
Defer (and not always present, actually)
2) Pyruvate, our end-product of glycolysis
In E. coli, humans:
Pyruvate  lactate, NADH2  NAD, coupled
In Yeast:
Pyruvate  ethanol + CO2

41. Glycolytic pathway Lactic acid fermentation glucose oxidative pathways
dihydroxyacetone phosphate
fructose-1,6-diphosphate
glucose-6-phosphate
fructose-6-phosphate
glyceraldehyde-3-phosphate 41
Glycolytic pathway
glucose
1,3-diphosphoglyceric acid
phosphoenol-pyruvic acid
2-phosphoglyceric acid
3-phosphoglyceric acid
E. coli,
muscle
yeast
lactic acid
acetaldehyde
ethanol
pyruvic acid
oxidative pathways
Lactic acid fermentation

42. Overview:
Glucose
ATP
ATP
biosynthetic pathway to NAD
excreted

43. Glycolytic pathway H C=O | CH3 Acetaldehyde detail43
glucose
1,3-diphosphoglyceric acid
phosphoenol-pyruvic acid
2-phosphoglyceric acid
3-phosphoglyceric acid
E. coli,
muscle
yeast
acetaldehyde
ethanol
lactic acid
E. coli
yeast H
C=O |
CH3
Acetaldehyde
detail
pyruvic acid
oxidative pathways
humans
Alcoholic fermentation

44. Fermentation goes all the way to the right
Efficiency: glucose--> 2 lactates,
without considering the couplings for the formation of ATP's (no energy harnessing):  Δ Go = -45 kcal/mole kcal/mole Out of this comes 2 ATPs, worth 14 kcal/mol.
31 kcal/mole output as heat.
So the efficiency is about 14/45 = ~30% Not bad at all.
Since 2 ATPs ARE produced, taking them into account, for the reaction:
Glucose + 2 ADP + 2 Pi  2 lactate + 2 ATP
ΔGo = -31 kcal/mole (45-14)
Very favorable. All the way to the right.
Keep bringing in glucose, keep spewing out lactate,
Make all the ATP you want.