Unlike inorganic catalysts, enzymes are specific

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Unlike inorganic catalysts, enzymes are specific
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Unlike inorganic catalysts, enzymes are specific

•                                       succinic
  dehydrogenase
• HOOC-HCCH-COOH lt-------------------------------gt
  HOOC-CH2-CH2-COOH
  
  
  2H
• fumaric acid                                  
                     succinic acid
• NOT a substrate for the enzyme
• 1-hydroxy-butenoate 
    HO-CHCH-COOH
• (simple OH instead of one of
  the carboxyl's)
• Maleic acid
• 
  
• 
  
  maleic
  acid
• Platinum will work with all of these,
  indiscriminantly

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• Enzymes work as catalysts for two reasons
• They bind the substrates putting them in close
  proximity.
• They participate in the reaction, weakening the
  covalent bonds
• of a substrate by its interaction with the
  enzymes amino acid residue side groups (e.g.,
  by stretching).

Movie http//www.columbia.edu/cu/biology/courses/
c2005/images/dhfr_movie2.gif
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Chemical kinetics

• Substrate ? Product
• (reactants in enzyme catalyzed reactions are
  called substrates)
• S ? P
• Velocity V ?P/ ? t
• So V also -?S/ ?t (disappearance)
• From the laws of mass action
• ?P/ ?t - ?S/ ?t k1S k2P
• For the INITIAL reaction, P is small and can be
  neglected
• ?P/ ?t - ?S/ ?t k1S
• So the INITIAL velocity Vo k1S

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Vo ?P/ ? t
P vs. t Slope Vo
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Effect of different initial substrate
concentrations
0.6
S4
S3
0.4
P
S2
0.2
S1
0.0
t
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Vo the slope in each case
Effect of different initial substrate
concentrations
0.6
S4
S3
0.4
P
S2
0.2
S1
0.0
t
Vo k1S
Slope k1
Considering Vo as a function of S (which wil be
our usual useful consideration)
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Now, with an enzyme
We can ignore the rate of the non-catalyzed
reaction
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Enzyme kinetics (as opposed to simple chemical
kinetics)
Vo independent of S
Vo proportional to S
Can we understand this curve?
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Michaelis and Menten mechanism for the action of
enzymes (1913)
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Michaelis-Menten mechanism
X

• Assumption 1. E S lt--gt ES this is how enzymes
  work, via a complex
• Assumption 2. Reaction 4 is negligible, when
  considering INITIAL velocities (Vo, not V).
• Assumption 3. The ES complex is in a
  STEADY-STATE, with its concentration unchanged
  with time during this period of initial rates. 
• (Steady state is not an equilibrium condition, it
  means that a compound is being added at the same
  rate as it is being lost, so that its
  concentration remains constant.)

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E S
ES
E P
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Michaelis-Menten Equation(s)
See handout at your leisure for the derivation
(algebra, not complicated, neat)
k3EoS
Vo
(k2k3)/k1 S
If we let Km (k2k3)/k1, just gathering 3
constants into one, then
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All the ks are constants for a particular enzyme
Rate is proportional to the amount of enzyme
At high S (compared to Km), Rate is constant Vo
k3Eo
At low S (compared to Km), rate is proportional
to S Vo k3EoS/Km
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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
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• Now, Vmax k3Eo
• So k3 Vmax/Eo
• the maximum (dP/dt)/Eo, the maximum
  (-dS/dt)/Eo
• k3 the TURNOVER NUMBER
• the maximum number of moles of substrate
  converted to product per mole of enzyme per
  second
• max. no. of molecules of substrate converted to
  product per molecule of enzyme per second
• Turnover number then is a measure of  the
  enzyme's catalytic power.

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Some turnover numbers (per second)

• Succinic dehydrogenase 19 (below average)
• Most enzymes 100 -1000
• The winner
• Carbonic anhydrase (CO2 H20 H2CO3)
• 600,000
• Thats 600,000 molecules of substrate, per
  molecule of enzyme, per second.
• Picture it!
• You cant.

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Km ?
Vmax/2 is achieved at a S that turns out to be
numerically equal to Km
So Km is numerically equal to the concentration
of substrate required to drive the reaction at ½
the maximal velocity Try it Set Vo ½ Vmax and
solve for S.
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Another view of Km
Consider the reverse of this reaction (the
DISsociation of the ES complex)
The equilibrium constant for this dissociation
reaction is
Kd ES / ES k2/k1
(Its the forward rate constant divided by the
backward rate constant. See the Web lecture if
you want to see this relationship derived)
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Consider in reverse
Kd k2/k1
Km (k2k3)/k1 (by definition)
IF k3 ltlt k2, then Km k2/k1 But k2/k1 Kd
(from last graphic) so Km Kd for the
dissociation reaction (i.e. the equilibrium
constant)
(and 1/Km the association constant)
So the lower the Km, the more poorly it
dissociates. That is, the more TIGHTLY it is held
by the enzyme
And the greater the Km, the more readily the
substrate dissociates, so the enzyme is binding
it poorly
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Km ranges

• 10-6M is good
• 10-4M is mediocre
• 10-3M is fairly poor

So Km and k3 quantitatively characterize how an
enzyme does the job as a catalyst
k3, how good an enzyme is in facitiating the
chemical change (given that the substrate is
bound) Km, how well the enzyme can bind the
substrate in the first place
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Enzyme inhibition competitive, non-competitive,
and allosteric
A competitive inhibitor resembles the substrate
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Competitive inhibitor can be swamped out at high
substrate concentrations
Handout 5-3b
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-
Vo
Substrate concentration
Inhibitor looks like the substrate and like the
substrate binds to the substrate binding site
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Biosynthetic pathway to cholesterol
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Zocor(simvastatin)
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½ Vmax w/o inhibitor
½ Vmax withyet more inhibitor
Km remains unchanged. Vmax decreases.
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Substrate
Non-competitive inhibitor
Example Hg ions (mercury) binding to SH groups
in the active site
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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)
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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 affects on the Vo vs. S curve
are more complex and 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
First committed step is usually inhibited
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              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)
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Rich medium provide glucose all 20 amino
acids and all vitamins, etc.
60 minutes, in a minimal medium
20 minutes !, in a rich medium
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Direction of reactions in metabolism
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Free
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
  ? 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.

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• ?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

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

Qualitative term
Quantitative term
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? 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(Q) ? 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

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• 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.

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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

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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)

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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

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Some exceptions to the 1M standard condition
Exception 1

• 1) Water 55 M (pure water) is considered unit
  concentration 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.

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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 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.

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Summary

• ?G ?Go RTln(Q)
• This combination of one qualitative and one
  quantitative (driving) term tell 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.

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• GOT THIS FAR

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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.
e
e
A-R
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The hydrolysis of ATP
ADP
AMP

• ATP HOH ?? ADP Pi
• A adenine (2 rings with Ns)
• R ribose (5-carbon ring sugar)

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-
-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
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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
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• A-R-PPP

Not a high energy bond
ATP HOH
ATPase
? ADP Pi heat
Prob set 4
Keq 100,000
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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 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
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? 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 7kcal/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.
• 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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ATP
ATP
ATP
ATP
ATP
ATP
ATP
ATP
ATP
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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 TAKES
  energy.
• But need only to regenerate ATP from ADP

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

• Two solutions
• 1) Photosynthesis
• 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
• Basic mechanism of energy metabolism(getting
  energy by glucose breakdown.)
• An example of a metabolic pathway.

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Handout 7-2