Transcript Udine Lectures Lecture #2: Computational Modeling ¦ Bud Mishra Professor of Computer Science and Mathematics (Courant, NYU) Professor (Watson School, CSHL) 7 ¦ 6 ¦ 2002 11/7/2015 ©Bud Mishra,

Udine Lectures
Lecture #2:
Computational Modeling
¦
Bud Mishra
Professor of Computer Science and Mathematics (Courant, NYU)
Professor (Watson School, CSHL)
7 ¦ 6 ¦ 2002
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©Bud Mishra, 2002
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Modeling
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Modeling Biomolecular
Networks
• Agents and Modes:
– Species and Processes: There are two kinds of agents:
– S-agents (representing species such as proteins, cells and
DNA): S-agents are described by concentration (i.e.,
their numbers) and its variation due to accumulation or
degradation. S-agent’s description involves differential
equations or update equations.
– P-agents (representing processes such as transcription,
translation, protein binding, protein-protein
interactions, and cell growth.) Inputs of P-agents are
concentrations (or numbers) of species and outputs are
rates.
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P-agents and S-agents
S1
Process P1
S2
S3
Process P2
S4
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Agents & Modes
• Each agent is characterized by a state x 2 Rn and
• A collection of discrete modes denoted by Q
• Each mode is characterized by a set of differential
equations (qi 2 Q & z 2 Rp is control)
dx/dt = fqi(x,z),
– and a set of invariants that describe the conditions under
which the above ODE is valid…
– these invariants describe algebraic constraints on the
continuous state…
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Mode Definition
• Modes are defined by the transitions among its
submodes.
• A transition: specifies source and destination
modes, the enabling condition, and the associated
discrete update of variables.
• Modes and submodes are organized hierarchically.
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Example of a Hybrid System
• q1 and q2 = two discrete modes
• x = continuous variable evolving as
G12(x) ¸ 0
dx/dt =f2(x)
g2(x) ¸ 0
q2
dx/dt =f1(x)
g1(x) ¸ 0
q1
G21(x) ¸ 0
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– dx/dt = f1(x) in mode q1
– dx/dt = f2(x) in mode q2
• Invariants:Associated with locations
q1 and q2 are
– g1(x) ¸ 0 and g2(x) ¸ 0, resp.
• The hybrid system evolves
continuously in disc. mode q1
according to dx/dt = f1(x) as long as
g1(x) ¸ 0 holds.
• If ever x enters the “guard set”
G12(x) ¸ 0, then mode transition
from q1 to q2 occurs.
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Generic Equation
• Generic formula for any molecular species (mRNA,
protein, protein complex, or small molecule):
dX/dt = synthesis – decay § transformation § transport
• Synthesis:
– replication for DNA,
– transcription of mRNA,
– translation for protein
• Decay: A first order degradation process
• Transformation:
– cleavage reaction
– ligand binding reaction
• Transport: Diffusion through a membrane..
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Model of transcription
X = concentration of a TF
transcription
{X,k, n}
F = concentration of an mRNA
• nXm = Cooperativity coefficient
• kXm = Concentration of X at which transcription of m is “halfmaximally” activated.
• F(X, kXm, nXm) = Xn/[kn + Xn]
• Y(X, kXm, nXm) = kn/[kn + Xn]=1 - F(X, kXm, nXm)
• A graph of function F = Sigmoid Function
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Transcription Activation
Function
x_,
Plot
x, 30, 10 ,
_,
x
_
Plot3D
x, 30,
,
x, 0, 100 ,
, 1, 30 ,
ViewPoint
0, 2, 2
x
30
x, 0, 100
1
20
0.8
10
0.6
0.4
1
0.2
20
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40
60
80
100
0
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20
40
60
80
0.75
0.5
0.25
0
100
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Quorum Sensing in V. fischeri
• Cell-density dependent gene expression in prokaryotes
– Quorum = A minimum population unit
• A single cell of V. fischeri can sense when a quorum of
bacteria is achieved—leading to bioluminescence…
• Vibrio fiscehri is a marine bacterium found both as
– a free-living organism, and
– a symbiont of some marine fish and squid.
• As a free-living organism, it exists in low density is non-luminescent..
• As a symbiont, it lives in high density and is luminescent..
• The transcription of the lux genes in this organism controls this
luminescence.
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lux gene
+
-
CRP
luxR
luxICDABEG
LuxR
-
Ai
LuxR
LuxA
LuxI
LuxB
Ai
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Quorum Sensing
• The lux region is organized in two transcriptional units:
– OL: containing luxR gene (encodes protein LuxR = a transcriptional
regulator)
– OR: containing 7 genes luxICDABEG.
• Transcription of luxI produces the protein LuxI, required for
endogenous production of the autoinducer Ai (a small membrane
permeable signal molecule (acyl-homoserine lactone).
• The genes luxA & luxB code for the luciferase subunits
• The genes luxC, luxD & luxE code for proteins of the fatty acid
reductase, needed for aldehyde substrate for luciferase.
• The gene luxG encodes a flavin reductase.
• Along with LuxR and LuxI, cAMP receptor protein (CRP) controls
luminescence.
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Biochemical Network
• The autoimmune inducer Ai binds to protein LuxR
to form a complex C0, which binds to the lux box.
• The lux box region (between the transcriptional
units) contains a binding site for CRP.
• The transcription from the luxR promoter is
activated by the binding of CRP.
• The transcription from the luxICDABEG is activated
by the binding of C0 complex to the lux box.
• Growth in the levels of C0 and cAMP/CRP inhibit
luxR and luxICDABEG transcription, respectively.
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Biochemical Network
LuxA, LuxB
Ai
C0
LuxR
LuxI
luxICDABEG
CRP
LuxC, LuxD, LuxE
luxR
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Notation
•
•
•
•
•
•
•
•
•
x0 = scaled population
x1 = mRNA transcribed from OL
x2 = mRNA transcribed from OR
x3 = protein LuxR
x4 = protein LuxI
x5 = protein LuxA/B
x6 = protein LuxC/D/E
x7 = autoinducer Ai
x8 = complex C0
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Evolution Equations…
• dx0/dt = kG x0
• dx1/dt = Tc[Y(x8, kC0, nC0) F(cCRP, kCRP, nCRP)+b]
– x1/HRNA –kG x1
• dx2/dt = Tc[F(x8, kC0, nC0) Y(cCRP, kCRP, nCRP)+b]
– x2/HRNA –kG x2
• dx3/dt = Tl x1 –x3/Hsp-rAiRx7 x3 –rC0x8 –kG x3
• dx4/dt = Tl x2 –x4/Hsp-kG x4
• dx5/dt = Tl x2 –x5/Hsp-kG x5
• dx6/dt = Tl x2 –x6/Hsp-kG x6
• dx7/dt = x0(rAll x4 –rAiRx7 x3+rC0x8) –x7/HAi
• dx8/dt = rAiR x7 x3 –x8/Hsp –rC0x8-kGx8
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Parameters
Tc
Tl
HRNA
Hsp
Hup
HAi
rAll
rAiR
rC0
Max. transcription rate
Max. translation rate
RNA half-life
Stable protein half-life
Unstable protein half-life
Ai half-life
Rate constant: LuxI ! Ai
Rate constant: Ai binds to LuxI
Rate constant: C0 dissociates
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nCRP
kCRP
nC0
kC0
b
vb
V
kg
x0max
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Cooperativity coef for CRP
Half-max conc for CRP
Cooperativity coef for C0
Half-max conc for C0
Basal transcription rate
Volume of a bacterium
Volume of solution
Growth rate
Maximum Population
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Remaining Questions
• Simulation:
– Nonlinearity
– Hybrid Model (Piece-wise linear)
• Stability Analysis
• Reachability Analysis
• Robustness
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Kinetic Equations
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Early Examples
• Enzymes for fermentation:
– Hydrolysis of Sucrose
– Enzyme = Invertase
Sucrose + Water ! glucose + fructose
– Acidity of the mixture, an important parameter
– At optimal value of acidity,
rate of reaction / amount of enzyme
– 1890: O’Sullivan & Tompson
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History
• 1892: Brown
– Ideas of “Enzyme-Substrate Complex” and “Law of Mass Action”
• 1902-3: Henri
– More precise in terms of chemical and mathematical models
– Equilibrium between:
• Free Enzyme
• Enzyme-Substrate Complex
• Enzyme-Product Complex
• 1909: Sorensen
– pH concept of hydrogen-ion concentration (log scale for pH)
– Precise quantitative parameter for acidity
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Michelis-Menten’s Model (1913)
E + A À EA ! E + P
– E = Enzyme,
• e= instantaneous enzyme concentration
– A = Substrate,
• a = instantaneous substrate concentration
– EA=Enzyme-Substrate Complex,
• x = instantaneous EA concentration
– P = Product,
• p = instantaneous concentration
• Assumption:
– The reversible first step (E + A À EA) was fast enough for it to be
represented by an equilibrium constant for substrate dissociation:
Ks = ea/x
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) x = ea/Ks
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M-M Model
• Facts: Parameters e and a are not directly
measurable:
e0 = e + x
a0 = a + x
e ¸ 0, a ¸ 0 )
a ¼ a0
{e0 = e(t0)
{a0 = a(t0)
x · e0 ¿ a0
x= (e0-x)a/Ks
) x[1+(Ks/a)] = e0
) x = e0/[(Ks/a) + 1]
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Second-Step in M-M
EA ! E + P
• Simple first order equation
• k2 = Rate constant
• Rate Equation:
dp/dt = k2 x = k2 e0/[(Ks/a)+1]
= k2 e0 a/[Ks + a]
• v / a, if Ks > a…
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Van Slyke-Cullen
• 1914: Van Slyke & Cullen
E+A !K1 EA !K2 E + P
• E a (e0-x); A a a; EA a x; P a p
dx/dt = k1(e0-x) a – k2 x
• At equilibrium, dx/dt = 0
) x = k1e0a/[k2+k1a]
) v = k2 x = k2 e0 a/[(k2/k1) + a]
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Steady-State of an EnzymeCatalyzed Reaction
• 1925: Briggs-Haldane
E + A Àk-1k1 EA !k2 E + P
– E a e0 – x; A a a; EA a x; P a p
dx/dt = k1(e0-x)a – k-1x – k2x
• At equilibrium: dx/dt = 0
• k1(e0-x)a – k-1x –k2x =0
) x = k1 e0 a/[k-1+k2+k1 a]
v = k2 x = k2 e0 a/ [ (k-1+k2)/k1 + a]
= V a/[Km +a]
– V = k2 e0 = Maximum Velocity
– Km = (k-1+k2)/k1 = Michelis Constant
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The Reversible MichaelisMenten Mechanism
E + A Àk-1k1 EA Àk-2k2 E + P
• E a e0-x; A a a; EA a x; P a p
• dx/dt = k1(e0-x)a + k-2(e0-x)p – (k-1+k2) x
= 0, at equilibrium
• x = (k1 e0 a + k-2 e0 p)/(k-1 + k2 + k1 a + k-2 p)
• v = dp/dt =k2 x – k-2 (e0-x) p
= (k1 k2 e0 a – k-1 k-2 e0 p)/(k-1 + k2 + k1 a + k-2 p)
= (kA e0 a – kp e0 p)/[1 + (a/KmA) + (p/KmP)]
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More General Model
E + A Àk-1k1 EA Àk-2k2 EP Àk-3k3 E + P
• E a e0 – x; A a a; EA a x; EP a y; P a p
v = (kA e0 a – kp e0 p)/[1 + (a/kmA) + (p/kmP)]
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Product Concentration
•
•
•
•
dp/dt = v = V a/(Km+a) = V(a0-p)/(Km+a0-p)
s V dt = s (Km+a0-p)/(a0-p) dp
s Km dp/(a0-p) + s dp = s V dt
-Km ln (a0-p) + p = Vt + a
– a = -Km ln a0
• Vt = p + Km ln a0/(a0-p)
• Vappt = p + Kmapp ln a0/(a0-p)
t/[ln a0/(a0-p)]
= (1/Vapp) { p /[ln a0/(a0-p)]} + Kmapp/Vapp
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Simulation
• Product concentration:
tpFun t_, a0_, K_, V_ :
NSolve t
Log a0
a0 p
1 V p
Log a0
a0 p
rulet1
rulet2
rulet4
rulet8
rulet1,
tpFun t, 0.1, 1, 1 ;
tpFun t, 0.2, 1, 1 ;
tpFun t, 0.4 , 1, 1 ;
tpFun t, 0.8 , 1, 1 ;
rulet2, rulet4, rulet8
K V, p ;
Plot p . rulet1, p . rulet2, p . rulet4,
p . rulet8 , t, 0, 10 , PlotRange
All
p 0.1 1. 10. ProductLog 0.11051709180756475812910675254999
p 0.2 1. 5. ProductLog 0.24428055163203399174476141486524
p 0.2 2. 5. ProductLog 0.5967298790565082280652996971183
p 0.2 4. 5. ProductLog 1.7804327427939740340558704534604
0.8
x-axis = t
y-axis = p
0.6
0.4
0.2
2
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4
6
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8
10
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t
t
t
t
Gated Ionic Channel
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Example:
Gated Ionic Channels
Closed
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Open
• Example from Fall et al.
• Gated Aqueous Channel: An
aqueous pore selective to
particular types of ions.
• Portions of a transmembrane
protein form the “gate” & is
sensitive to membrane potential
• Based on the membrane
potential, the pore can be in
OPEN or CLOSED states.
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Gating a Channel
with Two States
• Proteins that switch between an “open” state and a “closed”
state.
• Kinetic Model:
C Àk-k+ O.
• C = Closed State, O = Open State.
– These states represent a complex set of underlying molecular states
in which the pore is either permeable or impermeable to ionic
charge.
– The transitions between C and O are unimolecular process because
they involve only one (the channel) molecule.
– The transitions between molecular states are reversible.
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Law of Mass Action
k+ and k- are rate constants.
Transition C ! O: J+ = k+[C]
Transition C Ã O: J- = k-[O]
f0 = [O] = N0/N
= fraction of the open channels = “open” concentration
• fC = [C] = NC/N
= fraction of the open channels = “open” concentration
fC = 1 – fO
flux C Ã O: j- = k- fO
flux C ! O: j+ = k+(1- fO)
•
•
•
•
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Final Differential Equation
dfO/dt = j+ - j= k- fO + k+(1-fO)
= -(k- + k+)[fO – k+/(k- + k+)]
• Take t = 1/(k- + k+)], f1 = k+/(k- + k+).
dfO/dt = -(fO – f1)/t
• Let Z = -(fO – f1); thus, dZ/dt = Z/t.
• Z = Z(0) exp[-t/t].
fO(t) = f1 + [fO(0) – f1] exp[-t/t].
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The Family of Solutions:
•
t = 2.
• The function converges to the equilibrium value f1 = 0.5.
• The initial values fO was chosen to be 0.1, 0.2, 0.4, 0.8 & 1.0.
ff t_, f0_, finf_, _ :
finf
f0
finf Exp t
N;
Plot ff t, 0.1, 0.5, 2 , ff t, 0.2, 0.5, 2 ,
ff t, 0.4, 0.5, 2 , ff t, 0.8, 0.5, 2 ,
0.8
ff t, 1.0, 0.5, 2 , , t, 0, 20 ,
PlotRange
All
5
10
15
20
0.6
0.4
0.2
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Metabolism
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Metabolism
• The complex of chemical reactions that
– convert foods into cellular components
– provide the energy for synthesis,
– and get rid of used up materials.
• Metabolism is (artificially) thought to consist of
– ANABOLISM:
• Building up activities
– CATABOLISM:
• Breaking down activities.
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Anabolism
• Requirements for synthesis of macromolecules:
– Varieties of organic compounds (e.g., amino acids)
– Precursor molecules of nucleic acids
– Energy.
• External supply of precursors consists of molecules that can
enter the cells and may have to be drastically altered by
chemical reactions inside the cells.
• Energy needed for metabolism is used up or made available
in the reshuffling of chemical bonds.
• In order for these processes (construction or destruction) to
proceed stably, metabolisms need to be unidirectional.
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ATP Reaction
• ATP (adenosine triphosphate)
ATP À
-H2O
+H2O
-P-P-P
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ADP + H3PO4
-P-P iP
– The energy currency of the cell
– Placed in water ATP splits to form
ADP (adenosine diphosphate) and
phosphoric acid (iP for inorganic
phosphate)
• Hydrolysis: unidirectional
– This reaction has enormous tendency to
proceed from left to right.
– At equilibrium the ratio
ADP/ATP ¼ 1:105
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Energy Released by ATP
• Energy Potential Barrier:
Energy
Potential
Barrier
ATP
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ADP + iP
Energy
Released
– In a population of molecules of
ATP, there is a distribution of
energies…
– A small percentage of molecules
has enough energy to make the
transition over the barrier
– To undergo a chemical reaction,
a molecule must be activated
(distorted to a transition state)
from where it glides into the new
structure … ADP + H3PO4
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Catalyst/Enzyme
enzyme
or heat
DE
ATP
DE
ADP + iP
• A catalyst such as an enzyme
forms with the substrate
molecules a complex that
distorts the molecules forcing
them into a state close to the
activated transition state at the
top of the energy barrier…
– An enzyme does not change the
difference in energy between the
substrate and the product;
– It reduces effective potential
barrier
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ATP Reaction
• In the hydrolysis of ATP,
– the enzyme ATPase operates primarily on the P-O-P
bond of ATP and on the HO-H bond of the water
molecule.
– the surface groups of the enzyme recognizes the whole
ATP molecule…this makes the enzyme specific for this
reaction.
– the enzyme also recognizes the product substances, ADP
& H3PO4, otherwise the reaction could not be reversibly
catalyzed.
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Unidirectionality of a
Reaction
•
enzyme
or heat
In summary, there are two
critical parameters:
1.
The energy difference
between substrate and
product, which determines
in which direction the
reaction will proceed;
2. The potential barrier, which
controls the rate of the
reaction.
DE
ATP
DE
ADP + iP
•
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These two parameters are
unrelated (independent)
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Futile Cycles
• All living organisms require a high degree of control over
metabolic processes so as to permit orderly change without
precipitating catastrophic progress towards thermodynamic
equilibrium.
• Examples: Processes such as glycolysis and gluconeogenesis
are
– essentially reversal of each other
– but cannot occur simultaneously
– as it would simply result in continuous hydrolysis of ATP resulting
in eventual death.
• These complementary processes are either in different segregated
populations of cells or in different compartments of the same cell (e.g.
glycolysis and gluconeogenesis in liver tissues)
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Problem
• Such unidirectional processes cannot be easily
explained with the classical Michelis-Menten
model.
• Why?
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Rate of Change:
Michaelis-Menten
• In Michalis-Menten Equation:
v = Va/(K+a), and dv/da = KV/(a+K)2
• At the half way point,
(dv/da)|a=K = 1/4 (V/K)
– The rate is 0.1 V at a = K/9 and it is 0.9 V at a = 9 K.
– An enormous increase in substrate concentration (81
fold) is needed to increase the rate from 10% to 90%.
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Rate of Change:
Cooperativity
• In general, with cooperativity, (n = cooperativity
constant), the rate increases rapidly:
v = V an/(Kn + an) and dv/da = n an-1 Kn V/(an+Kn)2
• At the half way point,
(dv/da)|a=K = (n/4) (V/K).
– The rate is 0.1 V at a = K/(91/n) and it is 0.9V at a = (91/n)
K.
– The needed increase in substrate concentration reduces
to 3 fold for n = 2.
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Substrate Concentration and
Cooperativity
• Increase in substrate concentration needed to increase the rate from
10% to 90%.--As a function of the cooperativity coefficient n:
80
60
40
20
2
3
4
5
6
7
n, cooperativity coefficient )
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Difference in rates:
1
0.8
MichaelisMenten
0.6
0.4
Cooperative
0.2
log a a
-6
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-4
-2
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2
4
6
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Cooperativity
• A property arising from “cooperation” among
many active sites from polymeric enzymes…
• The main role in metabolic regulation:
– Property of responding with exceptional sensitivity to
change in metabolic concentrations.
– Shows a characteristic S-shaped (sigmoid) response curve
(as opposed to a rectangular hyperbola of MichaelisMenten)
– The steepest part of the curve is shifted from the origin
to a positive concentration a typically a concentration
within the physiological range for the metabolite.
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Allosteric Interaction
• To permit inhibition or activation by metabolically
appropriate effectors, many regulated enzymes have evolved
sites for effector binding that are separate from catalytic
sites.
• These sites are called “allosteric sites”
– (Greek for different shape or another solid):
– Monod, Cahngeux and Jacob: 1963.
– Enzymes possessing allosteric sites are called “allosteric enzymes.”
• Many allosteric enzymes are cooperative and vice versa.
– Haemoglobin was known to be cooperative for more than 60 years
before the allosteric effect of 1,2-bisphosphoglycerate was
understood.
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Activation & Inhibition
•
2
1
1
2
•
1.
2.
•
1 active site
•
2 regulatory site
•
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In cells, not only the amounts, but also the activity
of enzyme is regulated (via allosteric regulatory
sites).
The regulator substances fall into two categories:
Activators &
Inhibitors.
Inhibitors decrease enzyme activity, either by
competing with the substrate for the active sites, or
by changing the configuration of the enzyme…
Activators increase the activity of enzymes when they
combine with them…
These kinds of regulation occurs by the regulators
combining with the enzyme at a site other than the
active site…allosteric regulatory sites.
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Feedback Inhibition
A
1
B
2
C
3
D
4
X
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• In the example shown, an amino acid (say X) is
made by a pathway in which each arrow indicates
an enzyme reaction.
• The level of X in the cell controls the activity of
enzyme 1 of the pathway..
– The more X is present, the less active the enzyme is.
– When external X decreases the enzyme becomes active
again.
– Example of negative feedback accomplished by
allosteric sites and competition for the active sites.
• Unidirectionality of A !1 B reaction is very
important as a small amount of X should have
large effect on the rate of the reaction.
©Bud Mishra, 2002
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The Hill equation
• Hill (1910): As an empirical description of the cooperative
binding of oxygen to haemoglobin.
v = V ah/(K0.5h + ah)
• h = Hill coefficient = cooperativity coefficient.
– Based on a limiting physical model of substrate binding, h takes an
integral value.
– Experimentally determined h coefficient is often non-integral.
• K0.5 = Value of the substrate concentration at which the rate
of reaction is half of the maximum achievable.
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Hill Plot
• Note that
v/(V-v) = (a/K0.5)h, and
log[v/(V-v)] = h log a – h log K0.5
• h =Hill coefficient can be
determined from a plot of
log a vs. log[v/(V-v)] .
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Adair Equation
• An enzyme E has two active sites that bind
substrate A independently:
A
+
A+E
A
+
EA
Ks1
Ks2
A+EA
Ks2
Ks1
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E AA
– At equilibrium the dissociation constants are
Ks1 and Ks2
– The rate constants at the two sites are
independent and equal k1 and k2
• Applying Michaelis-Menten, we have:
v = k1 e0 a/(Ks1 +a) + k2 e0 a/(Ks2+a)
• Assuming that V/2 = k1 e0 = k2 e0, we have
(2v/V) = 1/(1+ Ks1/a) + 1/(1+Ks2/a)
= (1 + K’/a)/(1+K’/a + K2/a2)
= (a2 +K’a)/(K2 + K’a + a2)
©Bud Mishra, 2002
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Adair Equation
• General Formula with two active
sites:
(k1+k2) e [a2 + (Ks1 k2 +Ks2 k2)/(k1 + k2) a]
[a2 + (Ks1+Ks2) a + Ks1 Ks2]
• The formula is approximated as
¼ V a2/(K2 + a2),
• where
V = (k1+k2)e and K = p(Ks1 Ks2)
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Adair Equation for
Haemolobin
• An equation of the same general form with four
binding sites, as haemoglobin can bind upto four
molecules of oxygen:
y = (v/V) =
[a/K1 + 3a2/K1K2 + 3a3/K1K2K3 + a4/K1K2K3K4]
[1+4a/K1 + 6a2/K1K2 + 4a3/K1K2K3 + a4/K1K2K3K4]
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The Monod-WymanChangeux Model
• The earliest mechanistic model proposed to account for
cooperative effects in terms of the enzyme’s conformation:
(1965).
• Assumptions:
– Cooperative proteins are composed of several identical reacting units,
called protomers, that occupy equivalent positions within the protein.
– Each protomer contains one binding site per ligand.
– The binding sites within each protein are equivalent.
– If the binding of a ligand to one protomer induces a conformational
change in that protomer, an identical change is induced in all
protomers.
– The protein has two conformational states, usually denoted by R and T,
which differ in their ability to bind ligands.
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Example: A protein with two
binding sites…
R2
T2
sk3 2k-3
sk1 2k-1
R1
2sk1
T1
2sk3 k-3
k-1
R0
k2
k-2
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T0
• Consider a protein with two binding
sites:
• The protein exist in six states:
– Ri, i=0,1,2; or Ti, i=0,1,2,
– where i à the number of bound ligands.
• Assume that R1 cannot convert
directly to T1 or viceversa.
• Similarly, Assume that R2 cannot
convert directly to T2 or viceversa.
• Let s denote the concentration of the
substrate.
©Bud Mishra, 2002
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Saturation Function
R2
T2
sk3 2k-3
sk1 2k-1
R1
2sk1
T1
2sk3 k-3
k-1
R0
k2
k-2
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T0
• Saturation function = Fraction Y of
occupied sites:
Y=
[r1 + 2 r2 + t1 +2t2]/2(r0+r1+r2+t0+t1+t2)
• Let Ki = k-i/ki…Then
–
–
–
–
–
r1 = 2sK1-1r0;
r2 = s2K1-2r0;
t1 = 2sK1-1t0;
t2 = s2K1-2t0
and r0/t0 = K2.
©Bud Mishra, 2002
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Final Result
• Substituting into the saturation function:
Y=
s K1-1(1+sK1-1)+k2-1[sK3-1(1+ sK3-1)]
(1+sK1-1)2 + k2-1 (1+ sK3-1)2
• More generally, with n binding sites:
Y=
s K1-1(1+sK1-1)n-1+k2-1[sK3-1(1+ sK3-1)n-1]
(1+sK1-1)n + k2-1 (1+ sK3-1)n
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Calculation
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Calculation
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