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At high S, Vo here = k3Eo, == Vmax
So the Michaelis-Menten equation can be written:
Vo =
k3 [Eo] [S]
Km + [S]
Vo = Vmax [S]
Km + [S]
Simplest
form
Enzyme inhibition:
competitive, non-competitive, and allosteric
Competitive:
A competitive inhibitor resembles the substrate
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3
A competitive inhibitor can be swamped out at high
substrate concentrations
Handout 5-3b
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Vmax will
not be
affected.
+
Vo
Apparent (measured) Km increases
Substrate concentration
Inhibitor looks like the substrate
And, like the substrate, binds to the substrate binding site
Biosynthetic pathway to cholesterol
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Zocor
(simvastatin)
=“CoA”
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½ Vmax
w/o inhibitor
½ Vmax with
yet more
inhibitor
Km remains unchanged. Vmax decreases.
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Non-competitive inhibitor example
Substrate still binds OK.
But an essential participant in the reaction is blocked
(here, by mercury binding a cysteine sulfhydryl)
Hg++
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snapshot
Substrate
Non-competitive inhibitor
Example: Hg ions (mercury) binding to –SH groups in the active site
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Allosteric inhibition
Inhibitor
binding
site
+
Active
Active
= allosteric inhibitor
Inactive
= 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
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Allosteric inhibitors are used by the cell for
feedback inhibition of metabolic pathways
Feedback inhibition of enzyme activity, or “End product inhibition”
Linear pathway:
PQRSTUV
Branched pathway:
End product
End product
The first step committed to anend-product is usually inhibited.
End
product
protein
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Thr deaminase
glucose ...... --> --> threonine -----------------> alpha-ketobutyric acid
A
Substrate
B
C
protein
isoleucine
(and no other aa)
Allosteric inhibitor
Also here: Feedback inhibitor
(is dissimilar from substrate)
Rich medium = provide glucose + all 20 amino acids and all vitamins, etc.
20 minutes!, in a rich medium
60 minutes, in a minimal medium
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Direction
Flow
of glucose
in E. coli
of
reactions
in metabolism
Macromolecules
Polysaccharides
Lipids
Nucleic Acids
Proteins
yn
th
e
tic
pa
t
hw
ay
monomers
bi
os
intermediates
glucose
Each arrow = a specific chemical reaction
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Energy
}Freedifference
determines the direction
of a chemical reagion
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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: Δ G
Such that:
• IF Δ G IS <0:
THEN A AND B WILL TEND TO PRODUCE C AND D
(i.e., tends to go to the right).
• IF Δ G IS >0:
THEN C AND D WILL TEND TO PRODUCE A AND B.
(i.e., tends to go to the left)
• IF Δ G IS = 0:
THEN THE REACTION WILL BE AT EQUILIBRIUM:
NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY.
ΔG = Δ
Go+
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RTln([C][D]/[A][B])
• 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, D’s 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
• Δ Go = a CONSTANT:
a quantity related to the INTRINSIC properties of A, B, C, and D
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Also abbreviated form:
Δ G = Δ Go+ RTlnQ (Q for “quotient”)
Where Q = ([C][D]/[A][B])
Qualitative term:
What molecules
are in play
Quantitative
term:
How much of
each is present
Josiah
Willard
Gibbs
(1839 1903)
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Δ Go
STANDARD FREE ENERGY CHANGE of a reaction.
If all the reactants and all the products are present at 1 unit
concentration,
then:
Δ G = Δ Go + RTln(Q) = Δ Go + RTln([1][1] / [1][1])
Δ G = Δ Go + RTln(1)
or Δ G = Δ Go +RT x 0,
or Δ G = Δ Go,
when all components are at 1
….. a special case
(when all components are at 1)
“1” usually means 1 M
So Δ G and Δ Go are quite different,
and not to be confused with each other.
Δ 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.
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Because,
• at equilibrium, Δ G = Δ Go + RTln(Q) = 0
and at equilibrium Q = Keq =
(a second special case).
[C]eq [D]eq
[A]eq [B]eq
• So: at equilibrium,
Δ G = Δ Go + RTln(Keq) = 0
• And so: Δ Go = - RTln(Keq)
• So just measure the Keq,
• Plug in R and T
• Get: ΔGo, the standard free energy change
E.g., let’s say for the reaction A + B
Keq happens to be:
[C]eq[D]eq
[A]eq[B]eq
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C + D,
= 2.5 x 10-3
Then Δ Go = -RTlnKeq = -2 x 300 x ln(2.5 x 10-3)
= -600
x
-6
= +3600
3600 cal/mole (If we use R=2 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)
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Note:
If ΔGo = +3.6 for the reaction A + B < --- >C + D
Then ΔGo = -3.6 for the reaction C + D <--- > A + B
(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
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 1M
The 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.
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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 ΔGo’
However, I will not bother.
But it should be understood we are always talking about ΔGo’ in
this course.
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Summary
ΔG = Go + RTln(Q)
This combination of one qualitative and one
quantitative (driving) term tells the direction of a
chemical reaction in any particular circumstance
ΔGo = - RTln(Keq)
The ΔGo for any reaction is a constant that can be looked up
in a book.
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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.
Acid anhydride (new functional group)
adenine
ribose
e
A-R
e
H
H
O
O
|
|
HO - P - OH HO - P - OH
||
||
O
O
Dehydration between 2 acids
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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
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O O O
||
||
||
A-R-O-P-O-P-O-P-O- + HOH
|
|
|
O- O- O-
ATP
Adenosine triphosphate
O O
||
||
A-R-O-P-O-P-O- +
|
|
O- O-
ADP
Adenosine
diphosphate
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
O
||
-O-P-O|
O-
Pi
Inorganic
phosphate
“High energy” bonds:
• Go of a least ~ -7 kcal/mole is released upon
hydrolysis
• Designated with a squiggle (~) often
• ATP = A-P-P~P
• Rationalized by the relief of electrical repulsion upon
hydrolysis:
ΔGo = -7 kcal/mole
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Hydrolysis of ADP
AMP
ΔGo = -7 kcal/mole
ADP
A-R-P~P~P
Not a high energy bond
ATP + HOH + ATPase
Keq ~ 100,000
 ADP + Pi + heat
Prob set 4
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• 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 ??
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E.g., A reaction that requires energy,
an endergonic reaction:
glucose + Pi
glucose-6-phosphate
OP03--
+Pi
Glucose + Pi --> glucose-6-P + H2O;
ATP
+
Δ Go = +3.6 kcal/mole.
H2O --> ADP + Pi
Glucose + Pi
--> G6P + H2O
Δ Go = -7
Keq= 2.5 x 10 -3
kcal/mole
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
Δ Go’s of multiple reactions are additive
ATP + H2O --> ADP + Pi
Δ Go = -7 kcal/mole
Glucose + Pi --> G6P + H2O Δ Go = +3.6 kcal/mole
Glucose + ATP --> G6P + ADP Δ 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, Δ 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.
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Flow of glucose in E. coli
Macromolecules
Polysaccharides
Lipids
Nucleic Acids
Proteins
ATP
ATP
monomers
ATP
ATP
ay
ATP
ATP
yn
th
e
ATP
tic
pa
t
ATP
hw
ATP
bi
os
intermediates
glucose
Each arrow = a specific chemical reaction
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• 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:
ATP
ATP
Via GLYCOLYSIS, e.g.
Handout 7-1a
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Regeneration of ATP from ADP
Two solutions:
• 1) Photosynthesis
Metabolic breakdown of
an organic compound
• 2) Catabolism of organic comounds (e.g.,
glucose)
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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:
A) Basic mechanism of energy metabolism
(getting energy by glucose breakdown.)
B)
An example of a metabolic pathway.
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Handout 7-2
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The first 5 steps of glycolysis
dihydroxyacetone
pnosphate
-OPO3
5
PO 3
e
l as
o
Ald OH,
+ATP
+ATP
Fructose-6-P
G-6-P
Glucose
Fructose- +H OH
-H
Isomerization
Phosphorylation 1,6-di-P
Phosphorylation
1
2
3
Handout 7-3
4
gyceraldehyde
-3-phosphate
2 moles of
G3P for every
mole of
glucose
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Acid anhydride
functional group
(if add HOH
get 2 acids)
Glyceraldehyde-3- phosphate
1,3,-diphospho-glyceric acid
Abbreviations for the phosphate group
Need an acceptor
for these electrons:
An oxidizing agent.
Handout 7-4a
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NAD: The acceptor of the electrons (and one proton)
NAD = nicotinamide adenine dinucleotide
Niacin (Vit. B3)  NAD
Handout 7-4c
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Loose end #2 = NAD
Loose end #1 was ATP
Top carbon has
been oxidized
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ATP debt
paid in full
-2 ATP
+ 4 ATP
= + 2 ATP
“glycolysis” ends here
Handout 7-2
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1 glucose + 2 ADP + 2 Pi + 2 NAD
Δ Go =
2 pyruvate + 2 ATP + 2 NADH2
-18 kcal/mole
So overall reaction goes essentially completely to the right.
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(Handout 7-3)
Handout 7-4b
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•Δ G = Δ Go +
RTln([products]
[reactants])
pull
pull
Handout 7-4b
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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, Δ 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 Δ G equation.
• Δ G = Δ Go + RTln([products]/[reactants])
• 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.
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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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Glucose
GAL-3-P
ATP
1,3-Di-PGA
ATP
Glucose
NAD
NADH2
Biosynthetic pathway to NAD
Lactate
Pyruvate
excreted
Handout 7-1b
yeast
humans
E. coli
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H
C=O
|
CH3
Acetaldehyde
detail
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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)
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The efficiency of fermentation
glucose--> 2 lactates,
without considering the couplings for the formation of
ATP's (no energy harnessing):
Δ Go = -45 kcal/mole
So 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.
Fermentation goes all the way to the right
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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.
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
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.
That’s fermentation.
Gl;ycerol as an alternative sole carbon and energy source for E. coli
ATP
glycerol
+NAD
glycerol
phosphate
- O2
?
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+ NADH2
DHAP
(dihydroxy
acetone
phosphate)
glycolysis
+O2
CO2 + H2O
and ADP + Pi  ATP
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Glycerol + ATP → glycerol phosphate → DHAP
NAD → NADH2
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+NAD
ATP
glycerol
glycerol
phosphate
- O2
+ NADH2
DHAP
(dihydroxy
acetone
phosphate)
glycolysis
+O2
Glycerol cannot be fermented.
CO2 + H2O
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
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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
ΔGo = -686 kcal/mole !
Compared to -45 to lactate (both w/o ATP production considered)
Complete oxidation of glucose,
Much more ATP
But nature’s solution is a bit complicated.
The fate of pyruvate is now different