At high S, Vo here = k3Eo, = Vmax

1
At high S, Vo here k3Eo, Vmax

So the Michaelis-Menten equation can be written
k3 Eo S
Vmax S
Simplest form
Vo
Vo
Km S
Km S
2
Enzyme inhibition competitive, non-competitive,
and allosteric Competitive
A competitive inhibitor resembles the substrate
3
A competitive inhibitor can be swamped out at
high substrate concentrations
Handout 5-3b
4
Vmax willnot beaffected.
-

Vo
Substrate concentration
Inhibitor looks like the substrate And, like the
substrate, binds to the substrate binding site
5
Biosynthetic pathway to cholesterol
6
Zocor(simvastatin)
CoA
7
½ Vmax w/o inhibitor
½ Vmax withyet more inhibitor
Km remains unchanged. Vmax decreases.
8
Non-competitive inhibitor example Substrate still
binds OK. But an essential participant in the
reaction is blocked (here, by mercury binding a
cysteine sulfhydryl)
9
snapshot
Substrate
Non-competitive inhibitor
Example Hg ions (mercury) binding to SH groups
in the active site
10
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11
Allosteric inhibition
Inhibitor binding site

Active
Inactive
Active
allosteric inhibitor
substrate
Allosteric inhibitor binds to a different site
than the substrate, so it need bear no
resemblance to the substrate
The apparent Km OR the apparent Vmax or both may
be affected. The effects on the Vo vs. S curve
are more complex and will be ignored here
12
Allosteric inhibitors are used by the cell for
feedback inhibition of metabolic pathways
Feedback inhibition of enzyme activity, or End
product inhibition
Linear pathway
Branched pathway
The first step committed to anend-product is
usually inhibited.
13
              Thr deaminaseglucose  ......  --gt
--gt threonine -----------------gt
alpha-ketobutyric acid 
protein
A
Substrate
B
C
isoleucine  (and no other aa)
protein
Allosteric inhibitor Also here Feedback
inhibitor (is dissimilar from substrate)
14
Rich medium provide glucose all 20 amino
acids and all vitamins, etc.
20 minutes!, in a rich medium
60 minutes, in a minimal medium
15
Direction of reactions in metabolism
16
Free
determines the direction of a chemical reagion
17
For the model reaction A B C D,
written in the left-to-right direction
indicated

• Consider the quantity called the change in free
  energy associated with a chemical reaction, or
  ? G
• Such that
• IF ? G IS lt0THEN A AND B WILL TEND TO PRODUCE C
  AND D(i.e., tends to go to the right).
• IF ? G IS gt0THEN C AND D WILL TEND TO PRODUCE A
  AND B.(i.e., tends to go to the left)
• IF ? G IS 0THEN THE REACTION WILL BE AT
  EQUILIBRIUM
• NOT TENDING TO GO IN EITHER DIRECTION IN A NET
  WAY.

18

• ?G ? Go RTln(CD/AB)
• where A, B, C and D are the concentrations of the
  reactants and the products AT THE MOMENT BEING
  CONSIDERED.(i.e., these A, B, C, Ds here are
  not the equilibrium concentrations)
• R UNIVERSAL GAS CONSTANT 1.98 CAL / DEG K
  MOLE (R 2)
• T ABSOLUTE TEMP ( oK ) 0oC 273oK Room temp
  25o C 298o K (T 300)
• ln NATURAL LOG
• ? Go a CONSTANT a quantity related to the
  INTRINSIC properties of A, B, C, and D

19

• Also abbreviated form? G ? Go RTlnQ (Q for
  quotient)
• Where Q (CD/AB)

Qualitative term What molecules are in play
Quantitative term How much of each is present
20
? Go

• STANDARD FREE ENERGY CHANGE of a reaction.
• If all the reactants and all the products
  are present at 1 unit concentration, then
• ? G ? Go RTln(Q) ? Go RTln(11 /
  11)
• ? G ? Go RTln(1)
• or ? G ? Go RT x 0,
• or ? G ? Go,
• when all components are at 1
• .. a special case
• (when all components are at 1)
• 1 usually means 1 M

21

• So ? G and ? Go are quite different,
• and not to be confused with each other.
• ? Go 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.

22
Because,

• at equilibrium, ? G ? Go RTln(Q) 0
• and at equilibrium Q Keq
• (a second special case).
• So at equilibrium, ? G ? Go RTln(Keq) 0
• And so ? Go - RTln(Keq)
• So just measure the Keq,
• Plug in R and T
• Get ?Go, the standard free energy change

23
E.g., lets say for the reaction A B C D,
Keq happens to be

• CeqDeq
• AeqBeq
• Then ? Go -RTlnKeq -2 x 300 x ln(2.5 x 10-3)
• -600
  x -6 3600
• 3600 cal/mole (If we use R2 we are dealing with
  calories)
• Or 3.6 kcal/mole
• 3.6 kcal/mole ABSORBED (positive number)
• So energy is required for the reaction in the
  left-to-right direction
• And indeed, very little product accumulates at
  equilibrium
• (Keq 0.0025)

24
Note

• If ?Go 3.6 for the reaction A B lt --- gtC D
• Then ?Go -3.6 for the reaction C D lt--- gt A
  B
• (Reverse the reaction switch the sign)
• And
• For reactions of more than simple 1 to 1
  stoichiometries
• aA bB lt--gt cC dD,
• ?G ?Go RT ln CcDd                        
   AaBb

25
Some exceptions to the 1M standard condition
Exception 1

• 1) Water 55 M (pure water) is considered the
  unit concentration in this case instead of
  1MThe concentration of water rarely changes
  during the course of an aqueous reaction, since
  water is at such a high concentration.
• So when calulating ?Go, 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 ?Go.

26
Exception 2

• In the same way,
• Hydrogen ion concentration, H 10-7 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 ?Go
• However, I will not bother.
• But it should be understood we are always talking
  about ?Go in this course.

27
Summary

• ?G ?Go RTln(Q)
• This combination of one qualitative and one
  quantitative (driving) term tells the direction
  of a chemical reaction in any particular
  circumstance
• ?Go - RTln(Keq)
• The ?Go for any reaction is a constant that can
  be looked up in a book.

28
Hydrolysis of ATP ATP HOH ? ADP
HPO4--
ATP, a small molecule in the cell that helps in
the transfer of energy from a place where it is
generated to a place where it is needed.
adenine
ribose
e
e
A-R
29
The hydrolysis of ATP
AMP
ADP

• ATP HOH ?? ADP Pi

The ?Go of this reaction is about -7
kcal/mole. Energy is released in this
reaction. This is an exergonic
reaction Strongly to the right, towards
hydrolysis, towards ADP
30
O O O
A-R-O-P-O-P-O-
-O-P-O-
O- O- O-
O O O
A-R-O-P-O-P-O-P-O- HOH
O- O- O-
ATP Adenosine triphosphate
ADP Adenosine diphosphate
Pi Inorganic phosphate
The ?Go of this reaction is about -7
kcal/mole. Energy is released in this
reaction. This is an exergonic
reaction Strongly to the right, towards
hydrolysis, towards ADP
31
High energy bonds

• ?Go of a least -7 kcal/mole is released upon
  hydrolysis
• Designated with a squiggle () often
• ATP A-P-PP
• Rationalized by the relief of electrical
  repulsion upon hydrolysis

?Go -7 kcal/mole
32
Hydrolysis of ADP

• A-R-PPP

Not a high energy bond
ATP HOH
ATPase
? ADP Pi heat
Prob set 4
Keq 100,000
33

• 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 ??

34
E.g., A reaction that requires energy,an
endergonic reactionglucose Pi
glucose-6-phosphate
OP03--
Pi
Keq 2.5 x 10 -3
Glucose Pi --gt glucose-6-P H2O ? Go
3.6 kcal/mole.
ATP         H2O   --gt  ADP     Pi      ? Go  
-7    kcal/mole
Glucose     Pi    --gt  G6P   H2O     ? Go
3.6 kcal/mole
ATP H2O Glucose Pi ? ADP Pi G6P H2O
? Go -3.4 kcal/mole overall
Glucose     ATP   --gt  G6P    ADP
? Go -3.4 kcal/mole overall
net sum of the two considered
reactions
35
? Gos of multiple reactions are additive

• ATP         H2O   --gt  ADP     Pi      ? Go  
  -7    kcal/mole
• Glucose     Pi    --gt  G6P   H2O     ? Go
  3.6 kcal/mole
• Glucose     ATP   --gt  G6P    ADP   ? Go -3.4
  kcal/mole overall net sum of the two
  considered reactions
• Enzymes needed ATPase? Glucose phosphorylase?
• No.
• Just get 7 kcal/mole as heat.
• But Hexokinase
• Glucose AR-P-P-P ??glucose-6-P AR-P-P
• Glucose ATP ?? glucose-6-P04 ADP, ? Go
  -3.4 kcal/mole
• A coupled reaction. A new reaction, ATP not
  simply hydrolyzed.
• 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.

36
ATP
ATP
ATP
ATP
ATP
ATP
ATP
ATP
ATP
37

• So does this solve the direction problem? Only
  for a second
• Where does this ATP come from, if we are E. coli
  growing in minimal medium
• Glucose is the only carbon source.
• Need to make ATP from glucose, and this path
  TAKES energy.
• But need only to regenerate ATP from ADP

Via GLYCOLYSIS, e.g.
Handout 7-1a
38
Regeneration of ATP from ADP

• Two solutions
• 1) Photosynthesis
• 2) Catabolism of organic comounds (e.g., glucose)

39
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.

40
Handout 7-2
41
-OPO3
2 moles of G3P for every mole of glucose
Handout 7-3
42
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43
Glyceraldehyde-3- phosphate
1,3,-diphospho-glyceric acid
Handout 7-4a
44
NAD The acceptor of the electrons (and one
proton)
NAD nicotinamide adenine dinucleotide
Niacin (Vit. B3) ? NAD
Handout 7-4c
45
Loose end 2 NAD Loose end 1 was ATP
Top carbon has been oxidized
46
ATP debt paid in full

• 2 ATP
• 4 ATP
• 2 ATP

glycolysis ends here
Handout 7-2
47

• 1 glucose 2 ADP 2 Pi 2 NAD 2
  pyruvate 2 ATP 2 NADH2
• ? Go -18 kcal/mole
• So overall reaction goes essentially completely
  to the right.

48
(Handout 7-3)
Handout 7-4b
49
pull
pull
Handout 7-4b
50
The second way the cell gets a reaction to go in
the desired direction

• 1) First way was a 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, ? Go
  -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 ? G equation.
• ? G ? Go RTln(products/reactants)

51

• 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.

52
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

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54
Glucose
ATP
ATP
excreted
Handout 7-1b
55
yeast
E. coli
H CO CH3 Acetaldehyde detail
humans
56
Fermentation anaerobiosis (no oxygen)

• Lactate fermentation
• Ethanolic fermentation
• Mutually exclusive, depends on organism
• Other types, less common fermentations, exist
• (e.g., propionic acid fermentation, going on in
  Swiss cheese)

57
The efficiency of fermentation

• glucose--gt 2 lactates,
• without considering the couplings for the
  formation of ATP's (no energy harnessing) 
• ? Go -45 kcal/moleSo 45 kcal/mole to work
  with.
• Out of this comes 2 ATPs, worth 14 kcal/mol.
• So the efficiency is about 14/45 30
• Where did the other 31/45 kcal/mole go?
• Wasted as HEAT.

58
Fermentation goes all the way to the right
glucose--gt 2 lactates, without considering the
couplings for the formation of ATP's

(no energy harnessing) ? Go -45
kcal/mole kcal/moleOut of this comes 2 ATPs,
worth 14 kcal/mol. So the efficiency is about
14/45 30

• Since 2 ATPs ARE produced, taking them into
  account, for the reaction
• Glucose 2 ADP 2 Pi ? 2 lactate 2 ATP
• ?Go -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.

Thats fermentation.
59
Glycerol as an alternative sole carbon and
energy source for E. coli
ATP
NAD
NADH2
DHAP (dihydroxy acetone phosphate)
glycerol phosphate
glycerol
and ADP Pi ? ATP
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61
ATP
glycerol
- O2
Glycerol cannot be fermented. E. coli CANNOT grow
on glycerol in the absence of air These pathways
are real, and they set the rules. Stoichiometry
of chemical reactions must be obeyed. No magic is
involved
62
Energy yield
But all this spewing of lactate turns out to be
wasteful. Using oxygen as an oxidizing agent
glucose could be completely oxidized, to
CO2 That is, burned.
How much energy released then? Glucose 6 O2 ?
6 CO2 6 H2O ?Go -686 kcal/mole ! Compared to
-45 to lactate (both w/o ATP production
considered)

• Complete oxidation of glucose,
• Much more ATP
• But natures solution is a bit complicated.
• The fate of pyruvate is now different