1. 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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11/9/2018

2. Modeling
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3. 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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4. P-agents and S-agents ©Bud Mishra, 2002 S1 Process P1 S2 S3 Process P2
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5. Agents & Modes dx/dt = fqi(x,z), ©Bud Mishra, 2002
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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6. Mode Definition ©Bud Mishra, 2002
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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11/9/2018

7. Example of a Hybrid System
q1 and q2 = two discrete modes
x = continuous variable evolving as
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.
dx/dt =f1(x)
g1(x) ¸ 0 q1
dx/dt =f2(x)
g2(x) ¸ 0 q2
G12(x) ¸ 0
G21(x) ¸ 0
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11/9/2018

8. dX/dt = synthesis – decay § transformation § transport
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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11/9/2018

9. Model of transcription
{X,k, n}
F = concentration of an mRNA
X = concentration of a TF
nXm = Cooperativity coefficient
kXm = Concentration of X at which transcription of m is “half-maximally” 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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10. Transcription Activation Function
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11. 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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12. lux gene + - ©Bud Mishra, 2002 luxR luxICDABEG CRP LuxR Ai LuxA LuxI
LuxB + -
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13. Quorum Sensing ©Bud Mishra, 2002
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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14. Biochemical Network ©Bud Mishra, 2002
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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15. Biochemical Network ©Bud Mishra, 2002 Ai C0 luxR LuxA, LuxB LuxI LuxR
luxICDABEG
CRP
LuxC, LuxD, LuxE
luxR
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16. Notation ©Bud Mishra, 2002 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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17. Evolution Equations… ©Bud Mishra, 2002 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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18. Parameters ©Bud Mishra, 2002 Tc nCRP Tl kCRP HRNA nC0 Hsp kC0 Hup b
Max. transcription rate
nCRP
Cooperativity coef for CRP Tl
Max. translation rate
kCRP
Half-max conc for CRP
HRNA
RNA half-life
nC0
Cooperativity coef for C0
Hsp
Stable protein half-life
kC0
Half-max conc for C0
Hup
Unstable protein half-life b
Basal transcription rate
HAi
Ai half-life vb
Volume of a bacterium
rAll
Rate constant: LuxI ! Ai V
Volume of solution
rAiR
Rate constant: Ai binds to LuxI kg
Growth rate
rC0
Rate constant: C0 dissociates
x0max
Maximum Population
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19. Remaining Questions ©Bud Mishra, 2002 Simulation: Stability Analysis
Nonlinearity
Hybrid Model (Piece-wise linear)
Stability Analysis
Reachability Analysis
Robustness
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20. Kinetic Equations
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21. Early Examples ©Bud Mishra, 2002 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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22. History ©Bud Mishra, 2002 1892: Brown 1902-3: Henri 1909: Sorensen
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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23. 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 ) x = ea/Ks
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24. M-M Model ©Bud Mishra, 2002 x= (e0-x)a/Ks ) x[1+(Ks/a)] = e0
Facts: Parameters e and a are not directly measurable:
e0 = e + x {e0 = e(t0)
a0 = a + x {a0 = a(t0)
e ¸ 0, a ¸ 0 ) x · e0 ¿ a0
a ¼ a0
x= (e0-x)a/Ks
) x[1+(Ks/a)] = e0
) x = e0/[(Ks/a) + 1]
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11/9/2018

25. Second-Step in M-M EA ! E + P ©Bud Mishra, 2002
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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26. Van Slyke-Cullen ©Bud Mishra, 2002 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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11/9/2018

27. Steady-State of an Enzyme-Catalyzed 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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11/9/2018

28. The Reversible Michaelis-Menten 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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11/9/2018

29. More General Model ©Bud Mishra, 2002
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)]
©Bud Mishra, 2002
11/9/2018

30. 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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11/9/2018

31. Simulation ©Bud Mishra, 2002 Product concentration: x-axis = t
y-axis = p
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32. Gated Ionic Channel
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33. Example: Gated Ionic Channels
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.
Closed
Open
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34. 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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35. = fraction of the open channels = “open” concentration
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
fC = 1 – fO
flux C Ã O: j- = k- fO
flux C ! O: j+ = k+(1- fO)
©Bud Mishra, 2002
11/9/2018

36. 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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37. The Family of Solutions:
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.
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38. Metabolism
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39. Metabolism ©Bud Mishra, 2002 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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40. 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.
©Bud Mishra, 2002
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41. ATP Reaction ©Bud Mishra, 2002 ATP (adenosine triphosphate)
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
ATP À -H2O+H2O ADP + H3PO4
-P-P-P
-P-P iP
©Bud Mishra, 2002
11/9/2018

42. Energy Potential Barrier
Energy Released by ATP
Energy Potential Barrier:
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
ATP
Energy a
ADP + iP
Energy Potential Barrier
Energy Released
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43. Catalyst/Enzyme ©Bud Mishra, 2002
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
ATP
Energy a
ADP + iP
D E
enzyme or heat
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11/9/2018

44. ATP Reaction ©Bud Mishra, 2002 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.
©Bud Mishra, 2002
11/9/2018

45. Unidirectionality of a Reaction
In summary, there are two critical parameters:
The energy difference between substrate and product, which determines in which direction the reaction will proceed;
The potential barrier, which controls the rate of the reaction.
These two parameters are unrelated (independent)
ATP
Energy a
ADP + iP
D E
enzyme or heat
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46. Futile Cycles ©Bud Mishra, 2002
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)
©Bud Mishra, 2002
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47. Problem
Such unidirectional processes cannot be easily explained with the classical Michelis-Menten model.
Why?
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48. 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%.
©Bud Mishra, 2002
11/9/2018

49. 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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50. 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:
Needed Increase in Substrate Concentration )
n, cooperativity coefficient )
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51. Difference in rates: ©Bud Mishra, 2002 Michaelis-Menten Cooperative
v/V a
Michaelis-Menten
Cooperative
log a a
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52. Cooperativity ©Bud Mishra, 2002
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 Michaelis-Menten)
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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53. 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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54. Activation & Inhibition
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. 1 2
active site
regulatory site
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55. Feedback Inhibition ©Bud Mishra, 2002
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. A 1 B 2 C 3 D 4 X
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56. The Hill equation ©Bud Mishra, 2002 v = V ah/(K0.5h + ah)
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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57. log[v/(V-v)] = h log a – h log K0.5
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)] .
©Bud Mishra, 2002
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58. Adair Equation ©Bud Mishra, 2002
An enzyme E has two active sites that bind substrate A independently:
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) A +
A+E A + EA
Ks1
Ks2
Ks2
A+EA
Ks1
EAA
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59. (k1+k2) e [a2 + (Ks1 k2 +Ks2 k2)/(k1 + k2) a]
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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11/9/2018

60. 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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61. The Monod-Wyman-Changeux 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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11/9/2018

62. Example: A protein with two binding sites…
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. R0 T0 T1 T2 R1 R2 k2
k-2
k-3
2k-3
k-1
2k-1
sk1
2sk1
2sk3
sk3
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11/9/2018

63. [r1 + 2 r2 + t1 +2t2]/2(r0+r1+r2+t0+t1+t2)
Saturation Function
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. R0 T0 T1 T2 R1 R2 k2
k-2
k-3
2k-3
k-1
2k-1
sk1
2sk1
2sk3
sk3
©Bud Mishra, 2002
11/9/2018

64. Final Result ©Bud Mishra, 2002
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:
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
©Bud Mishra, 2002
11/9/2018

65. Calculation
©Bud Mishra, 2002
11/9/2018

66. Calculation
©Bud Mishra, 2002
11/9/2018