Udine Lectures Lecture 2 Computational Modeling 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 9/9/2020 ©Bud Mishra, 2002 1

Modeling 9/9/2020 ©Bud Mishra, 2002 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-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. 9/9/2020 ©Bud Mishra, 2002 3

P-agents and S-agents S 1 Process P 1 S 2 S 3 Process P 2 S 4 9/9/2020 ©Bud Mishra, 2002 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… 9/9/2020 ©Bud Mishra, 2002 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. 9/9/2020 ©Bud Mishra, 2002 6

Example of a Hybrid System • • G 12(x) ¸ 0 dx/dt =f 2(x) g 2(x) ¸ 0 q 2 dx/dt =f 1(x) g 1(x) ¸ 0 q 1 – dx/dt = f 1(x) in mode q 1 – dx/dt = f 2(x) in mode q 2 • Invariants: Associated with locations q 1 and q 2 are – g 1(x) ¸ 0 and g 2(x) ¸ 0, resp. • G 21(x) ¸ 0 • 9/9/2020 q 1 and q 2 = two discrete modes x = continuous variable evolving as The hybrid system evolves continuously in disc. mode q 1 according to dx/dt = f 1(x) as long as g 1(x) ¸ 0 holds. If ever x enters the “guard set” G 12(x) ¸ 0, then mode transition from q 1 to q 2 occurs. ©Bud Mishra, 2002 7

Generic Equation • Generic formula for any molecular species (m. RNA, protein complex, or small molecule): d. X/dt = synthesis – decay § transformation § transport • Synthesis: – replication for DNA, – transcription of m. RNA, – translation for protein • • Decay: A first order degradation process Transformation: – cleavage reaction – ligand binding reaction • Transport: Diffusion through a membrane. . 9/9/2020 ©Bud Mishra, 2002 8

Model of transcription X = concentration of a TF transcription {X, k, n} F = concentration of an m. RNA • n. Xm = Cooperativity coefficient • k. Xm = Concentration of X at which transcription of m is “half-maximally” activated. • F(X, k. Xm, n. Xm) = Xn/[kn + Xn] • Y(X, k. Xm, n. Xm) = kn/[kn + Xn]=1 - F(X, k. Xm, n. Xm) • A graph of function F = Sigmoid Function 9/9/2020 ©Bud Mishra, 2002 9

Transcription Activation Function 9/9/2020 ©Bud Mishra, 2002 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 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. 9/9/2020 ©Bud Mishra, 2002 11

lux gene + - CRP lux. ICDABEG lux. R Lux. R - Ai Lux. R Lux. A Lux. I Lux. B Ai 9/9/2020 + ©Bud Mishra, 2002 12

Quorum Sensing • The lux region is organized in two transcriptional units: – OL: containing lux. R gene (encodes protein Lux. R = a transcriptional regulator) – OR: containing 7 genes lux. ICDABEG. • Transcription of lux. I produces the protein Lux. I, required for endogenous production of the autoinducer Ai (a small membrane permeable signal molecule (acyl-homoserine lactone). • The genes lux. A & lux. B code for the luciferase subunits • The genes lux. C, lux. D & lux. E code for proteins of the fatty acid reductase, needed for aldehyde substrate for luciferase. • The gene lux. G encodes a flavin reductase. • Along with Lux. R and Lux. I, c. AMP receptor protein (CRP) controls luminescence. 9/9/2020 ©Bud Mishra, 2002 13

Biochemical Network • The autoimmune inducer Ai binds to protein Lux. R to form a complex C 0, which binds to the lux box. • The lux box region (between the transcriptional units) contains a binding site for CRP. • The transcription from the lux. R promoter is activated by the binding of CRP. • The transcription from the lux. ICDABEG is activated by the binding of C 0 complex to the lux box. • Growth in the levels of C 0 and c. AMP/CRP inhibit lux. R and lux. ICDABEG transcription, respectively. 9/9/2020 ©Bud Mishra, 2002 14

Biochemical Network Lux. A, Lux. B Ai C 0 Lux. I Lux. R lux. ICDABEG CRP Lux. C, Lux. D, Lux. E lux. R 9/9/2020 ©Bud Mishra, 2002 15

Notation • • • x 0 = scaled population x 1 = m. RNA transcribed from OL x 2 = m. RNA transcribed from OR x 3 = protein Lux. R x 4 = protein Lux. I x 5 = protein Lux. A/B x 6 = protein Lux. C/D/E x 7 = autoinducer Ai x 8 = complex C 0 9/9/2020 ©Bud Mishra, 2002 16

Evolution Equations… • dx 0/dt = k. G x 0 • dx 1/dt = Tc[Y(x 8, k. C 0, n. C 0) F(c. CRP, k. CRP, n. CRP)+b] – x 1/HRNA –k. G x 1 • dx 2/dt = Tc[F(x 8, k. C 0, n. C 0) Y(c. CRP, k. CRP, n. CRP)+b] – x 2/HRNA –k. G x 2 • dx 3/dt = Tl x 1 –x 3/Hsp-r. Ai. Rx 7 x 3 –r. C 0 x 8 –k. G x 3 • dx 4/dt = Tl x 2 –x 4/Hsp-k. G x 4 • dx 5/dt = Tl x 2 –x 5/Hsp-k. G x 5 • dx 6/dt = Tl x 2 –x 6/Hsp-k. G x 6 • dx 7/dt = x 0(r. All x 4 –r. Ai. Rx 7 x 3+r. C 0 x 8) –x 7/HAi • dx 8/dt = r. Ai. R x 7 x 3 –x 8/Hsp –r. C 0 x 8 -k. Gx 8 9/9/2020 ©Bud Mishra, 2002 17

Parameters Tc Tl HRNA Hsp Max. transcription rate Hup HAi r. All r. Ai. R r. C 0 Unstable protein half-life 9/9/2020 Max. translation rate RNA half-life Stable protein half-life Ai half-life Rate constant: Lux. I ! Ai Rate constant: Ai binds to Lux. I Rate constant: C 0 dissociates n. CRP k. CRP Cooperativity coef for CRP n. C 0 k. C 0 b vb V kg x 0 max Cooperativity coef for C 0 ©Bud Mishra, 2002 Half-max conc for CRP Half-max conc for C 0 Basal transcription rate Volume of a bacterium Volume of solution Growth rate Maximum Population 18

Remaining Questions • Simulation: – Nonlinearity – Hybrid Model (Piece-wise linear) • Stability Analysis • Reachability Analysis • Robustness 9/9/2020 ©Bud Mishra, 2002 19

Kinetic Equations 9/9/2020 ©Bud Mishra, 2002 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: O’Sullivan & Tompson 9/9/2020 ©Bud Mishra, 2002 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 – p. H concept of hydrogen-ion concentration (log scale for p. H) – Precise quantitative parameter for acidity 9/9/2020 ©Bud Mishra, 2002 22

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 9/9/2020 ©Bud Mishra, 2002 23

M-M Model • Facts: Parameters e and a are not directly measurable: e 0 = e + x a 0 = a + x e ¸ 0, a ¸ 0 ) x · e 0 ¿ a 0 a ¼ a 0 {e 0 = e(t 0) {a 0 = a(t 0) x= (e 0 -x)a/Ks ) x[1+(Ks/a)] = e 0 ) x = e 0/[(Ks/a) + 1] 9/9/2020 ©Bud Mishra, 2002 24

Second-Step in M-M EA ! E + P • Simple first order equation • k 2 = Rate constant • Rate Equation: dp/dt = k 2 x = k 2 e 0/[(Ks/a)+1] = k 2 e 0 a/[Ks + a] • v / a, if Ks > a… 9/9/2020 ©Bud Mishra, 2002 25

Van Slyke-Cullen • 1914: Van Slyke & Cullen E+A !K 1 EA !K 2 E + P • E a (e 0 -x); A a a; EA a x; P a p dx/dt = k 1(e 0 -x) a – k 2 x • At equilibrium, dx/dt = 0 ) x = k 1 e 0 a/[k 2+k 1 a] ) v = k 2 x = k 2 e 0 a/[(k 2/k 1) + a] 9/9/2020 ©Bud Mishra, 2002 26

Steady-State of an Enzyme-Catalyzed Reaction • 1925: Briggs-Haldane E + A Àk-1 k 1 EA !k 2 E + P – E a e 0 – x; A a a; EA a x; P a p dx/dt = k 1(e 0 -x)a – k-1 x – k 2 x • At equilibrium: dx/dt = 0 • k 1(e 0 -x)a – k-1 x –k 2 x =0 ) x = k 1 e 0 a/[k-1+k 2+k 1 a] v = k 2 x = k 2 e 0 a/ [ (k-1+k 2)/k 1 + a] = V a/[Km +a] – V = k 2 e 0 = Maximum Velocity – Km = (k-1+k 2)/k 1 = Michelis Constant 9/9/2020 ©Bud Mishra, 2002 27

The Reversible Michaelis-Menten Mechanism E + A Àk-1 k 1 EA Àk-2 k 2 E + P • E a e 0 -x; A a a; EA a x; P a p • dx/dt = k 1(e 0 -x)a + k-2(e 0 -x)p – (k-1+k 2) x = 0, at equilibrium • x = (k 1 e 0 a + k-2 e 0 p)/(k-1 + k 2 + k 1 a + k-2 p) • v = dp/dt =k 2 x – k-2 (e 0 -x) p = (k 1 k 2 e 0 a – k-1 k-2 e 0 p)/(k-1 + k 2 + k 1 a + k-2 p) = (k. A e 0 a – kp e 0 p)/[1 + (a/Km. A) + (p/Km. P)] 9/9/2020 ©Bud Mishra, 2002 28

More General Model E + A Àk-1 k 1 EA Àk-2 k 2 EP Àk-3 k 3 E + P • E a e 0 – x; A a a; EA a x; EP a y; P a p v = (k. A e 0 a – kp e 0 p)/[1 + (a/km. A) + (p/km. P)] 9/9/2020 ©Bud Mishra, 2002 29

Product Concentration • • dp/dt = v = V a/(Km+a) = V(a 0 -p)/(Km+a 0 -p) s V dt = s (Km+a 0 -p)/(a 0 -p) dp s Km dp/(a 0 -p) + s dp = s V dt -Km ln (a 0 -p) + p = Vt + a – a = -Km ln a 0 • Vt = p + Km ln a 0/(a 0 -p) • Vappt = p + Kmapp ln a 0/(a 0 -p) t/[ln a 0/(a 0 -p)] = (1/Vapp) { p /[ln a 0/(a 0 -p)]} + Kmapp/Vapp 9/9/2020 ©Bud Mishra, 2002 30

Simulation • Product concentration: x-axis = t y-axis = p 9/9/2020 ©Bud Mishra, 2002 31

Gated Ionic Channel 9/9/2020 ©Bud Mishra, 2002 32

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 9/9/2020 Open ©Bud Mishra, 2002 33

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. 9/9/2020 ©Bud Mishra, 2002 34

Law of Mass Action k+ and k- are rate constants. Transition C ! O: J+ = k+[C] Transition C Ã O: J- = k-[O] f 0 = [O] = N 0/N = fraction of the open channels = “open” concentration • f. C = [C] = NC/N = fraction of the open channels = “open” concentration f. C = 1 – f. O flux C Ã O: j- = k- f. O flux C ! O: j+ = k+(1 - f. O) • • 9/9/2020 ©Bud Mishra, 2002 35

Final Differential Equation df. O/dt = j+ - j= k- f. O + k+(1 -f. O) = -(k- + k+)[f. O – k+/(k- + k+)] • Take t = 1/(k- + k+)], f 1 = k+/(k- + k+). df. O/dt = -(f. O – f 1)/t • Let Z = -(f. O – f 1); thus, d. Z/dt = Z/t. • Z = Z(0) exp[-t/t]. f. O(t) = f 1 + [f. O(0) – f 1] exp[-t/t]. 9/9/2020 ©Bud Mishra, 2002 36

The Family of Solutions: • • • t = 2. The function converges to the equilibrium value f 1 = 0. 5. The initial values f. O was chosen to be 0. 1, 0. 2, 0. 4, 0. 8 & 1. 0. 9/9/2020 ©Bud Mishra, 2002 37

Metabolism 9/9/2020 ©Bud Mishra, 2002 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. 9/9/2020 ©Bud Mishra, 2002 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. 9/9/2020 ©Bud Mishra, 2002 40

ATP Reaction • ATP (adenosine triphosphate) – The energy currency of the cell – Placed in water ATP splits to form ADP (adenosine diphosphate) and phosphoric acid (i. P for inorganic phosphate) ATP À -H 2 O+H 2 O ADP + H 3 PO 4 • Hydrolysis: unidirectional -P-P-P 9/9/2020 -P-P i. P – This reaction has enormous tendency to proceed from left to right. – At equilibrium the ratio ADP/ATP ¼ 1: 105 ©Bud Mishra, 2002 41

Energy Released by ATP • Energy Potential Barrier: Energy a Energy Potential Barrier ATP ADP + i. P 9/9/2020 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 + H 3 PO 4 ©Bud Mishra, 2002 42

Catalyst/Enzyme Energy a enzyme or heat DE ATP DE ADP + i. P 9/9/2020 • 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 ©Bud Mishra, 2002 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 molecule…this makes the enzyme specific for this reaction. – the enzyme also recognizes the product substances, ADP & H 3 PO 4, otherwise the reaction could not be reversibly catalyzed. 9/9/2020 ©Bud Mishra, 2002 44

Unidirectionality of a Reaction • enzyme or heat In summary, there are two critical parameters: Energy a 1. DE ATP DE ADP + i. P 2. • 9/9/2020 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) ©Bud Mishra, 2002 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) 9/9/2020 ©Bud Mishra, 2002 46

Problem • Such unidirectional processes cannot be easily explained with the classical Michelis-Menten model. • Why? 9/9/2020 ©Bud Mishra, 2002 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%. 9/9/2020 ©Bud Mishra, 2002 48

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. 9 V at a = (91/n) K. – The needed increase in substrate concentration reduces to 3 fold for n = 2. 9/9/2020 ©Bud Mishra, 2002 49

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 ) • 9/9/2020 n, cooperativity coefficient ) ©Bud Mishra, 2002 50

v/V a Difference in rates: Michaelis. Menten Cooperative log a a 9/9/2020 ©Bud Mishra, 2002 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 a typically a concentration within the physiological range for the metabolite. 9/9/2020 ©Bud Mishra, 2002 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 -bisphoglycerate was understood. 9/9/2020 ©Bud Mishra, 2002 53

Activation & Inhibition • • 2 1. 2. 1 • 1 2 1 active site 2 regulatory site 9/9/2020 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 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. ©Bud Mishra, 2002 54

Feedback Inhibition A 1 B • 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. . C – 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. D • 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. 2 3 4 X 9/9/2020 ©Bud Mishra, 2002 55

The Hill equation • Hill (1910): As an empirical description of the cooperative binding of oxygen to haemoglobin. v = V ah/(K 0. 5 h + 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. • K 0. 5 = Value of the substrate concentration at which the rate of reaction is half of the maximum achievable. 9/9/2020 ©Bud Mishra, 2002 56

Hill Plot • Note that v/(V-v) = (a/K 0. 5)h, and log[v/(V-v)] = h log a – h log K 0. 5 • h =Hill coefficient can be determined from a plot of log a vs. log[v/(V-v)]. 9/9/2020 ©Bud Mishra, 2002 57

Adair Equation • An enzyme E has two active sites that bind substrate A independently: A + A+E A + EA Ks 1 Ks 2 A+EA 9/9/2020 Ks 2 Ks 1 EAA – At equilibrium the dissociation constants are Ks 1 and Ks 2 – The rate constants at the two sites are independent and equal k 1 and k 2 • Applying Michaelis-Menten, we have: v = k 1 e 0 a/(Ks 1 +a) + k 2 e 0 a/(Ks 2+a) • Assuming that V/2 = k 1 e 0 = k 2 e 0, we have (2 v/V) = 1/(1+ Ks 1/a) + 1/(1+Ks 2/a) = (1 + K’/a)/(1+K’/a + K 2/a 2) = (a 2 +K’a)/(K 2 + K’a + a 2) ©Bud Mishra, 2002 58

Adair Equation • General Formula with two active sites: (k 1+k 2) e [a 2 + (Ks 1 k 2 +Ks 2 k 2)/(k 1 + k 2) a] [a 2 + (Ks 1+Ks 2) a + Ks 1 Ks 2] • The formula is approximated as ¼ V a 2/(K 2 + a 2), • where V = (k 1+k 2)e and K = p(Ks 1 Ks 2) 9/9/2020 ©Bud Mishra, 2002 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/K 1 + 3 a 2/K 1 K 2 + 3 a 3/K 1 K 2 K 3 + a 4/K 1 K 2 K 3 K 4] [1+4 a/K 1 + 6 a 2/K 1 K 2 + 4 a 3/K 1 K 2 K 3 + a 4/K 1 K 2 K 3 K 4] 9/9/2020 ©Bud Mishra, 2002 60

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. 9/9/2020 ©Bud Mishra, 2002 61

Example: A protein with two binding sites… R 2 sk 1 T 2 sk 3 2 k-1 R 1 2 sk 1 • Consider a protein with two binding sites: • The protein exist in six states: 2 k-3 T 1 k 2 R 0 9/9/2020 k-3 2 sk 3 k-1 T 0 – Ri, i=0, 1, 2; or Ti, i=0, 1, 2, – where i à the number of bound ligands. • Assume that R 1 cannot convert directly to T 1 or viceversa. • Similarly, Assume that R 2 cannot convert directly to T 2 or viceversa. • Let s denote the concentration of the substrate. k-2 ©Bud Mishra, 2002 62

Saturation Function R 2 sk 1 T 2 sk 3 2 k-1 R 1 2 sk 1 2 k-3 T 1 k 2 R 0 9/9/2020 k-3 2 sk 3 k-1 T 0 • Saturation function = Fraction Y of occupied sites: Y= [r 1 + 2 r 2 + t 1 +2 t 2]/2(r 0+r 1+r 2+t 0+t 1+t 2) • Let Ki = k-i/ki…Then – – – r 1 = 2 s. K 1 -1 r 0; r 2 = s 2 K 1 -2 r 0; t 1 = 2 s. K 1 -1 t 0; t 2 = s 2 K 1 -2 t 0 and r 0/t 0 = K 2. k-2 ©Bud Mishra, 2002 63

Final Result • Substituting into the saturation function: Y= s K 1 -1(1+s. K 1 -1)+k 2 -1[s. K 3 -1(1+ s. K 3 -1)] (1+s. K 1 -1)2 + k 2 -1 (1+ s. K 3 -1)2 • More generally, with n binding sites: Y= s K 1 -1(1+s. K 1 -1)n-1+k 2 -1[s. K 3 -1(1+ s. K 3 -1)n-1] (1+s. K 1 -1)n + k 2 -1 (1+ s. K 3 -1)n 9/9/2020 ©Bud Mishra, 2002 64

Calculation 9/9/2020 ©Bud Mishra, 2002 65

Calculation 9/9/2020 ©Bud Mishra, 2002 66