Udine Lectures Lecture #2: Computational Modeling Bud Mishra Professor of Computer Science and Mathematics (Courant, NYU) Professor (Watson School, CSHL) 7 6 2002 02/09/20 Bud Mishra, 2002 Lec.2-1 Modeling 02/09/20 Bud Mishra, 2002 Lec.2-2 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-agents description involves differential equations or update equations. P-agents (representing processes such as transcription, translation, protein binding, proteinprotein interactions, and cell growth.) Inputs of Pagents are concentrations (or numbers) of species and outputs are rates. 02/09/20 Bud Mishra, 2002 Lec.2-3 P-agents and S-agents S1 Process P1 S2 S3 Process P2 S4 02/09/20 Bud Mishra, 2002 Lec.2-4

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 02/09/20 Bud Mishra, 2002 Lec.2-5 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.

02/09/20 Bud Mishra, 2002 Lec.2-6 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 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.

G21(x) 0 02/09/20 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. Bud Mishra, 2002 Lec.2-7 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.. 02/09/20 Bud Mishra, 2002 Lec.2-8 Model of transcription X = concentration of a TF transcription {X,, } = concentration of an mRNA Xm = Cooperativity coefficient Xm = Concentration of X at which transcription of m is halfmaximally activated. (X, Xm, Xm) = X/[ + X] (X, Xm, Xm) = /[ + X]=1 - (X, Xm, Xm) A graph of function = Sigmoid Function 02/09/20 Bud Mishra, 2002

Lec.2-9 Transcription Activation Function x_, _, _ Plot x, 30, 10 , x x Plot3D x, 30, , x, 0, 100 , , 1, 30 , ViewPoint 0, 2, 2

30 x, 0, 100 1 20 0.8 10 0.6 0.4 0.2 20 02/09/20 40 60 80 100 0

20 Bud Mishra, 2002 40 60 80 1 0.75 0.5 0.25 0 100 Lec.2-10 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 achievedleading 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 nonluminescent.. As a symbiont, it lives in high density and is luminescent.. The transcription of the lux genes in this organism controls this luminescence. 02/09/20 Bud Mishra, 2002 Lec.2-11 lux gene + - CRP luxR luxICDABEG LuxR Ai LuxR LuxA

LuxI Ai 02/09/20 + Bud Mishra, 2002 LuxB Lec.2-12 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.

02/09/20 Bud Mishra, 2002 Lec.2-13 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. 02/09/20 Bud Mishra, 2002 Lec.2-14 Biochemical Network LuxA, LuxB

Ai C0 LuxI LuxR luxICDABEG CRP LuxC, LuxD, LuxE luxR 02/09/20 Bud Mishra, 2002 Lec.2-15 Notation

x0 x1 x2 x3 x4 x5 x6 x7 x8 = = = = = = = = = scaled population mRNA transcribed from OL mRNA transcribed from OR protein LuxR protein LuxI

protein LuxA/B protein LuxC/D/E autoinducer Ai complex C0 02/09/20 Bud Mishra, 2002 Lec.2-16 Evolution Equations dx0/dt = kG x0 dx1/dt = Tc[(x8, C0, C0) (cCRP, CRP, CRP)+b] x1/HRNA kG x1 dx2/dt = Tc[(x8, C0, C0) (cCRP, CRP, CRP)+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 02/09/20 Bud Mishra, 2002 Lec.2-17

Parameters Tc Max. transcription rate CRP Cooperativity coef for CRP Tl Max. translation rate CRP Half-max conc for CRP HRNA RNA half-life C0 Cooperativity coef for C0 Hsp

Stable protein half-life C0 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 Rate constant: rC0 02/09/20 dissociates Maximum Population x 0max Bud Mishra, 2002 Lec.2-18 C0 Remaining Questions Simulation: Nonlinearity Hybrid Model (Piece-wise linear)

Stability Analysis Reachability Analysis Robustness 02/09/20 Bud Mishra, 2002 Lec.2-19 Kinetic Equations 02/09/20 Bud Mishra, 2002 Lec.2-20 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: OSullivan & Tompson

02/09/20 Bud Mishra, 2002 Lec.2-21 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 02/09/20 Bud Mishra, 2002 Lec.2-22 Michelis-Mentens Model

(1913) E + A 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 EA) was fast enough for it to be represented by an equilibrium constant for substrate dissociation: Ks = ea/x 02/09/20 ) x = ea/Ks Bud Mishra, 2002 Lec.2-23 M-M Model

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 a0 x= (e0-x)a/Ks ) x[1+(Ks/a)] = e0 ) x = e0/[(Ks/a) + 1] 02/09/20 Bud Mishra, 2002 Lec.2-24 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 02/09/20 Bud Mishra, 2002 Lec.2-25 Van Slyke-Cullen 1914: Van Slyke & Cullen E+A !K1 EA !K2 E + P E (e0-x); A a; EA x; P 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] 02/09/20

Bud Mishra, 2002 Lec.2-26 Steady-State of an Enzyme-Catalyzed Reaction 1925: Briggs-Haldane E + A k-1k1 EA !k2 E + P E e0 x; A a; EA x; P 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 02/09/20 Bud Mishra, 2002 Lec.2-27 The Reversible MichaelisMenten Mechanism E + A k-1k1 EA k-2k2 E + P

E e0-x; A a; EA x; P 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)] 02/09/20 Bud Mishra, 2002 Lec.2-28 More General Model E + A k-1k1 EA k-2k2 EP k-3k3 E + P E e0 x; A a; EA x; EP y; P p v = (kA e0 a kp e0 p)/[1 + (a/kmA) + (p/kmP)] 02/09/20 Bud Mishra, 2002 Lec.2-29 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 + = -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 02/09/20 Bud Mishra, 2002 Lec.2-30 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, K

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 V, p ; Plot p . rulet1, p . rulet2, p . rulet4, p . rulet8 , t, 0, 10 , PlotRange All p 0.1 1. 10. ProductLog 0.11051709180756475812910675254999 t p 0.2 1. 5. ProductLog 0.24428055163203399174476141486524 t p 0.2 2. 5. ProductLog 0.5967298790565082280652996971183 t p 0.2 4. 5. ProductLog 1.7804327427939740340558704534604 t 0.8 x-axis = t y-axis = p 0.6 0.4 0.2

2 02/09/20 4 6 Bud Mishra, 2002 8 10 Lec.2-31 Gated Ionic Channel 02/09/20 Bud Mishra, 2002 Lec.2-32 Example: Gated Ionic Channels Close

d 02/09/20 Ope n 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. Bud Mishra, 2002 Lec.2-33 Gating a Channel with Two States Proteins that switch between an open state and a closed state. Kinetic Model:

C k-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. 02/09/20 Bud Mishra, 2002 Lec.2-34 Law of Mass Action k+ and k- are rate constants. Transition C ! O: J+ = k+[C] Transition C 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 O: j- = k- fO flux C ! O: j+ = k+(1- fO) 02/09/20

Bud Mishra, 2002 Lec.2-35 Final Differential Equation dfO/dt = j+ - j= k- fO + k+(1-fO) = -(k- + k+)[fO k+/(k- + k+)] Take = 1/(k- + k+)], f1 = k+/(k- + k+). dfO/dt = -(fO f1)/ Let Z = -(fO f1); thus, dZ/dt = Z/. Z = Z(0) exp[-t/]. fO(t) = f1 + [fO(0) f1] exp[-t/]. 02/09/20 Bud Mishra, 2002 Lec.2-36 The Family of Solutions: = 2.

The function converges to the equilibrium value f 1 = 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 0.6 5 10 15

20 0.4 0.2 02/09/20 Bud Mishra, 2002 Lec.2-37 Metabolism 02/09/20 Bud Mishra, 2002 Lec.2-38 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. 02/09/20 Bud Mishra, 2002 Lec.2-39 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. 02/09/20 Bud Mishra, 2002

Lec.2-40 ATP Reaction ATP (adenosine triphosphate) ATP +H2O -H2O -P-P-P 02/09/20 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 1:105 Bud Mishra, 2002 Lec.2-41 Energy Released by ATP Energy Potential Barrier: Energy Energy Potenti al Barrier ATP 02/09/20 ADP + iP Energy Release d

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 Bud Mishra, 2002 Lec.2-42 Catalyst/Enzyme Energy enzym e or heat E ATP

E 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 02/09/20 Bud Mishra, 2002 Lec.2-43 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 moleculethis makes the enzyme specific for this reaction. the enzyme also recognizes the product substances, ADP & H3PO4, otherwise the reaction could not be reversibly catalyzed. 02/09/20 Bud Mishra, 2002 Lec.2-44 Unidirectionality of a Reaction Energy enzym e or heat 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. E ATP E ADP + iP 02/09/20 In summary, there are two critical parameters: These two parameters are unrelated (independent) Bud Mishra, 2002 Lec.2-45

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) 02/09/20 Bud Mishra, 2002 Lec.2-46 Problem Such unidirectional processes cannot be easily explained with the classical Michelis-Menten model. Why? 02/09/20

Bud Mishra, 2002 Lec.2-47 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%. 02/09/20 Bud Mishra, 2002 Lec.2-48 Rate of Change: Cooperativity In general, with cooperativity, ( = cooperativity constant), the rate increases rapidly: v = V a/(K + a) and dv/da = a-1 K V/(a+K)2

At the half way point, (dv/da)|a=K = ( (V/K). The rate is 0.1 V at a = K/(91/) and it is 0.9V at a = (91/) K. The needed increase in substrate concentration reduces to 3 fold for = 2. 02/09/20 Bud Mishra, 2002 Lec.2-49 Substrate Concentration and Cooperativity Needed Increase in Substrate Concentration ) Increase in substrate concentration needed to increase the rate from 10% to 90%.--As a function of the cooperativity coefficient : 02/09/20 80 60 40 20 2

3 4 5 , cooperativity coefficient ) Bud Mishra, 2002 6 7 Lec.2-50 Difference in rates: 0.8 Michaeli sMenten v/V 1 0.6

0.4 0.2 -6 02/09/20 -4 -2 Bud Mishra, 2002 Cooperativ e log a 2 4 6 Lec.2-51 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 Michaelis-Menten) The steepest part of the curve is shifted from the origin to a positive concentration typically a concentration within the physiological range for the metabolite. 02/09/20 Bud Mishra, 2002 Lec.2-52 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. 02/09/20 Bud Mishra, 2002 Lec.2-53 Activation & Inhibition 2 1. 2. 1 1 2 1 active site

2 regulatory site 02/09/20 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 siteallosteric regulatory sites. Bud Mishra, 2002 Lec.2-54 Feedback Inhibition

A 1 B 2 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. C 3 D 4 X 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. 02/09/20

Bud Mishra, 2002 Lec.2-55 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 nonintegral. K0.5 = Value of the substrate concentration at which the rate of reaction is half of the maximum achievable. 02/09/20 Bud Mishra, 2002 Lec.2-56 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)] . 02/09/20 Bud Mishra, 2002 Lec.2-57 Adair Equation An enzyme E has two active sites that bind substrate A independently: A + EA A + A+ E Ks1 Ks2 A+EA

Ks2 Ks1 02/09/20 EAA 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 +Ka)/(K2 + Ka + a2) Bud Mishra, 2002 Lec.2-58 Adair Equation In[2]:=

exppp k1 e a K1 a k2 e a K2 a Out[2]= In[10]:= Out[10]= a e k1 a K1 a e k2 a K2

exppp Together Simplify a e K1 k2 a k1 k2 k1 K2 a K1 a K2 02/09/20 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 V a2/(K2 + a2), where V = (k1+k2)e and K = p(Ks1 Ks2) Bud Mishra, 2002 Lec.2-59 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] 02/09/20 Bud Mishra, 2002 Lec.2-60 The Monod-WymanChangeux Model The earliest mechanistic model proposed to account for cooperative effects in terms of the enzymes 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. 02/09/20

Bud Mishra, 2002 Lec.2-61 Example: A protein with two binding sites R2 T2 sk3 2k-3 sk1 2k-1 R1 T1 2sk3 k-3 2sk1 k-1 R0 k2 k-2 02/09/20

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 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 Lec.2-62 Saturation Function R2 T2 sk3 2k-3 sk1 2k-1

R1 T1 2sk3 k-3 2sk1 k-1 R0 k2 k-2 02/09/20 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/kiThen r1 = 2sK1-1r0; r2 = s2K1-2r0; t1 = 2sK1-1t0; t2 = s2K1-2t0 and r0/t0 = K2.

Bud Mishra, 2002 Lec.2-63 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 02/09/20 Bud Mishra, 2002 Lec.2-64 Calculation In[89]:=

MWCy s_, K1_, K2_, K3_, n_ : s K1 1 s K1 ^ n 1 1 s K1 ^ n ; In[90]:= 1 K2 s K3 1 s K3 ^ n 1 1 K2 1 s K3 ^ n ?? MWCy Global`MWCy MWCy s_, K1_ , K2_ , K3_ , n_ 02/09/20

: s n 1 s 1 s n 1 s 1 K1 K3 K1 K2 K3 s n 1 K3 s n 1 K1 K2 Bud Mishra, 2002 Lec.2-65 Calculation In[91]:= Out[91]=

In[92]:= MWCy s, a, 100a, a 100, n n 1 s s 1 n1 s 100 s 1 a a a a2 100 s 1 n s 1 n a a 100 a Plot MWCy s, 2, 200, .02, 1 , MWCy s, 2, 200, .02, 2 , MWCy s, 2, 200, .02, 200 , s, 0, 5 , PlotRange All 1 0.8 0.6

0.4 0.2 1 2 3 4 5 Out[92]= Graphics 02/09/20 Bud Mishra, 2002 Lec.2-66