Math 345 Intro to Math Biology Lecture 19: Models of Molecular Events and Biochemistry

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1 Math 345 Intro to Math Biology Lecture 19: Models of Molecular Events and Biochemistry Junping Shi College of William and Mary, USA

2 Molecular biology and Biochemical kinetics Molecular biology is one of the most important and rapidly developing areas in the life sciences, and now forms the basis of subjects such as physiology, immunology and genetics. Cellular biology is the study of cells, which make up all living creatures, and which occupy an intermediate level of biological complexity between molecules and multicellular organisms. Biochemical kinetics concerns the concentrations of chemical substances in biological systems as functions of time. Biochemical process are often controlled by enzyme catalysts. Enzymes are proteins that catalyze (i.e. accelerate) chemical reactions. Enzymes are biochemical catalysts. In these reactions, the molecules at the beginning of the process are called substrates, and the enzyme converts these into different molecules, the products. Almost all processes in the cell need enzymes in order to occur at significant rates.

3 (Bio)chemical reaction ma + nb k 1 pc + qd

4 (Bio)chemical reaction ma + nb k 1 pc + qd Reaction rate R = 1 m da dt = 1 db n dt = 1 dc p dt = 1 dd q dt

5 (Bio)chemical reaction ma + nb k 1 pc + qd Reaction rate R = 1 m da dt = 1 db n dt = 1 dc p dt = 1 dd q dt Law of mass action: the reaction rate is proportional to the product of the concentrations of the participating molecules so R = k 1 A m B n, (m + n reaction order)

6 (Bio)chemical reaction ma + nb k 1 pc + qd Reaction rate R = 1 m da dt = 1 db n dt = 1 dc p dt = 1 dd q dt Law of mass action: the reaction rate is proportional to the product of the concentrations of the participating molecules so R = k 1 A m B n, (m + n reaction order) A = mk 1 A m B n, B = nk 1 A m B n, C = pk 1 A m B n, D = qk 1 A m B n.

7 (Bio)chemical reaction ma + nb k 1 pc + qd Reaction rate R = 1 m da dt = 1 db n dt = 1 dc p dt = 1 dd q dt Law of mass action: the reaction rate is proportional to the product of the concentrations of the participating molecules so R = k 1 A m B n, (m + n reaction order) A = mk 1 A m B n, B = nk 1 A m B n, C = pk 1 A m B n, D = qk 1 A m B n. Example: Lotka reaction [Lotka, 1920] A + X k 1 2X, X + Y k 2 2Y, Y k 3 P where A is the grass, X is the rabbits, Y is the foxes, and P is the dead foxes. Assuming that A is a constant (infinite amount of grass). X = k 1 AX k 2 XY, Y = k 2 XY k 3 Y

8 Reversible chemical reaction ma + nb k 1 k2 pc + qd

9 Reversible chemical reaction ma + nb k 1 k2 pc + qd Reaction rate R = 1 m da dt = 1 db n dt = 1 dc p dt = 1 dd q dt

10 Reversible chemical reaction ma + nb k 1 k2 pc + qd Reaction rate R = 1 m R = k 1 A m B n k 2 C p D q, da dt = 1 db n dt = 1 dc p dt = 1 dd q dt A = m(k 1 A m B n k 2 C p D q ), B = n(k 1 A m B n k 2 C p D q ), C = p(k 1 A m B n k 2 C p D q ), D = q(k 1 A m B n k 2 C p D q ).

11 Reversible chemical reaction ma + nb k 1 k2 pc + qd Reaction rate R = 1 m R = k 1 A m B n k 2 C p D q, da dt = 1 db n dt = 1 dc p dt = 1 dd q dt A = m(k 1 A m B n k 2 C p D q ), B = n(k 1 A m B n k 2 C p D q ), C = p(k 1 A m B n k 2 C p D q ), D = q(k 1 A m B n k 2 C p D q ). Example: single molecule reaction A k 1 k2 B A = k 1 A + k 2 B, B = k 1 A k 2 B. Solution: A(t) + B(t) = A 0 + B 0, then A = k 1 A + k 2 (A 0 + B 0 A) = k 2 (A 0 + B 0 ) (k 1 + k 2 )A So lim A(t) = k 2 (A 0 + B 0 ), and lim t k 1 + k B(t) = k 1 (A 0 + B 0 ). 2 t k 1 + k 2 The reaction reaches an equilibrium asymptotically. In equilibrium, A B = k 2 k 1.

12 Autocatalytic reaction A chemical reaction is autocatalytic if the reaction product itself is the catalyst for that reaction. ma + nb k 1 (n + p)b A = mk 1 A m B n, B = pk 1 A m B n ma + nb k 1 k2 (n + p)b A = mk 1 A m B n + mk 2 B n+p, B = pk 1 A m B n pk 2 B n+p. Examples: Lotka reaction [Schnakenberg, 1979] 2X + Y 3X, A Y, X B x = x 2 y x + b, y = x 2 y + a. Brusselator [Prigogine, 1980] A X, 2X + Y 3X, B + X Y + D, X E x = a + x 2 y (b + 1)x, y = bx x 2 y. [Gray-Scott, 1983] U + 2V 3V, V k P, and feeding U (amount F ), removing U and V (amount F ) U = UV 2 + F (1 U), V = UV 2 (F + k)v.

13 Simple autocatalytic reactions A + nx k 1 (n + 1)X k2

14 Simple autocatalytic reactions A + nx k 1 (n + 1)X k2 Mass action kinetics: x(t)=amount of X molecules, a(t)=amount of A molecules dx dt = k 1x n a k 2 x n+1, da dt = k 1x n a + k 2 x n+1.

15 Simple autocatalytic reactions A + nx k 1 (n + 1)X k2 Mass action kinetics: x(t)=amount of X molecules, a(t)=amount of A molecules dx dt = k 1x n a k 2 x n+1, x(t) + a(t)=constant, assumed to be 1. Then da dt = k 1x n a + k 2 x n+1. dx dt = k 1x n (1 x) k 2 x n+1 = k 1 x n (k 1 + k 2 )x n+1.

16 Simple autocatalytic reactions A + nx k 1 (n + 1)X k2 Mass action kinetics: x(t)=amount of X molecules, a(t)=amount of A molecules dx dt = k 1x n a k 2 x n+1, x(t) + a(t)=constant, assumed to be 1. Then Assume k 1 = k 2 = 1. da dt = k 1x n a + k 2 x n+1. dx dt = k 1x n (1 x) k 2 x n+1 = k 1 x n (k 1 + k 2 )x n+1. n = 0: Unimolecular reaction A X (uncatalyzed) dx dt = 1 2x, solution x(t) = (x 0 0.5)e 2t.

17 Solution of autocatalytic equations n = 1: Bimolecular reaction A + X 2X (first order) (logistic equation!) dx dt = x x 0 2x2, solution x(t) = 2x 0 + (1 2x 0 )e t.

18 Solution of autocatalytic equations n = 1: Bimolecular reaction A + X 2X (first order) (logistic equation!) dx dt = x x 0 2x2, solution x(t) = 2x 0 + (1 2x 0 )e t. n = 2: Trimolecular reaction A + 2X dx dt = x2 2x 3, solution t = 2 ln 3X (second order) ( x(1 2x0 ) x 0 (1 2x) ) + 1 x 0 1 x.

19 Solution of autocatalytic equations n = 1: Bimolecular reaction A + X 2X (first order) (logistic equation!) dx dt = x x 0 2x2, solution x(t) = 2x 0 + (1 2x 0 )e t. n = 2: Trimolecular reaction A + 2X dx dt = x2 2x 3, solution t = 2 ln 3X (second order) ( x(1 2x0 ) x 0 (1 2x) ) + 1 x 0 1 x.

20 Solution of autocatalytic equations n = 1: Bimolecular reaction A + X 2X (first order) (logistic equation!) dx dt = x x 0 2x2, solution x(t) = 2x 0 + (1 2x 0 )e t. n = 2: Trimolecular reaction A + 2X dx dt = x2 2x 3, solution t = 2 ln 3X (second order) ( x(1 2x0 ) x 0 (1 2x) ) + 1 x 0 1 x. Red: n = 0; Blue: n = 1; Green: n = 2 k 1 = k 2 = 1, x 0 = 0.05

21 Solution of autocatalytic equations Red: n = 0; Blue: n = 1; Green: n = 2 k 2 = 0, x 0 = 0.01, k 1 are different for three curves and k 1 are chosen so x(5) = 0.5 The solutions for n 1 are all sigmoid function with limits 0 and 1 at t = and t =.

22 Sigmoid functions We define a sigmoid function to be y : (, ) (0, ) such that y (t) > 0, y (t)(t t 0 ) < 0 for all t t 0, lim y(t) = 0. In general any solution of y = yf (y) t with decreasing f (y) is a sigmoid function. In many cases, lim y(t) = N (the t carrying capacity when f (N) = 0), and it is also called a generalized logistic curve. One can choose f (y) to fit the data to an appropriate sigmoid function. Typical sigmoid functions: 1 u(t) = 1 + ae k(t t 0) (t = t 0 is the inflection point where u (t 0 ) = 0) 1 u(t) = (1 + ae k(t t0) ) q (t = t 0 is the inflection point where u (t 0 ) = 0) 0.5 u(t) = 0.5[tanh(t) + 1] = 1 + [ e 2t 2 t ] u(t) = 0.5[erf (t) + 1] = 0.5 π e s2 ds (t + 1) [ π ] u(t) = t 2 + 1, u(t) = arctan(t) + 1 In many other problems, sigmoid functions have range ( 1, 1) instead of (0, 1)

23 Sigmoid functions

24 Hill function Hill function [Hill, 1919] is a function in form f (u) = up h p, for u 0, p > 0. The + up point u = h is the half-saturation point since f (h) = 1/2, the half of the asymptotic limit. When p > 1, u = h is also the inflection point where f (u) = 0. When p =, the limit is the step function f (u) = 0 for u < h and f (u) = 1 for u > h.

25 Hill function Hill function [Hill, 1919] is a function in form f (u) = up h p, for u 0, p > 0. The + up point u = h is the half-saturation point since f (h) = 1/2, the half of the asymptotic limit. When p > 1, u = h is also the inflection point where f (u) = 0. When p =, the limit is the step function f (u) = 0 for u < h and f (u) = 1 for u > h. One can define a generalized Hill function by f : [0, ) [0, 1) such that f (0) = 0, f (u) > 0 for u > 0, and lim f (u) = 1. One can obtain a Hill function by restricting u and shifting a sigmoid function.

26 Hill function Hill function [Hill, 1919] is a function in form f (u) = up h p, for u 0, p > 0. The + up point u = h is the half-saturation point since f (h) = 1/2, the half of the asymptotic limit. When p > 1, u = h is also the inflection point where f (u) = 0. When p =, the limit is the step function f (u) = 0 for u < h and f (u) = 1 for u > h. One can define a generalized Hill function by f : [0, ) [0, 1) such that f (0) = 0, f (u) > 0 for u > 0, and lim f (u) = 1. One can obtain a Hill function by restricting u and shifting a sigmoid function. In ecology, people use the Hill function as the predator functional response. [Holling 1959] p = 1 is the type II, and p > 1 is the type III.

27 Hill function Hill function [Hill, 1919] is a function in form f (u) = up h p, for u 0, p > 0. The + up point u = h is the half-saturation point since f (h) = 1/2, the half of the asymptotic limit. When p > 1, u = h is also the inflection point where f (u) = 0. When p =, the limit is the step function f (u) = 0 for u < h and f (u) = 1 for u > h. One can define a generalized Hill function by f : [0, ) [0, 1) such that f (0) = 0, f (u) > 0 for u > 0, and lim f (u) = 1. One can obtain a Hill function by restricting u and shifting a sigmoid function. In ecology, people use the Hill function as the predator functional response. [Holling 1959] p = 1 is the type II, and p > 1 is the type III. In biochemistry, people use the Hill function as the reaction rate. When p = 1, it is called Michaelis-Menton chemical kinetics. [Menten and Michaelis, 1913]

28 Hill functions

29 Example 1. Consider the chemical reactions A + B k C, B + C k A. 1 Set up a system of 3 differential equations by the law of mass action for A(t), B(t) and C(t). 2 Assume that A(0) = 2, B(0) = 1 and C(0) = 1. Show that A(t) + C(t) = 3 for all t > 0. Reduce the system in part a to a system of two equations for A and B. 3 Show that B(t) = e 3kt so lim B(t) = 0. t 4 What are lim A(t) and lim t t 2. [Schnakenberg, 1979] 2X + Y 3X, A Y, X B x = x 2 y x + b, y = x 2 y + a. 1 Find the equilibrium point, and calculate the Jacobian. 2 Under what condition, the equilibrium point is stable or unstable. 3 Fix a = 1/2 and use b > 0 as a bifurcation parameter. Find a Hopf bifurcation point b 1 where a Hopf bifurcation occurs at the positive equilibrium.

30 Nonlinear reaction rates: Michaelis-Menton In a basic enzyme reaction, an enzyme E binds to a substrate S to form a complex ES, which in turn is converted into a product P and the enzyme E. (Enzyme is a biological catalyst.) E + S k 1 k2 ES k 3 E + P

31 Nonlinear reaction rates: Michaelis-Menton In a basic enzyme reaction, an enzyme E binds to a substrate S to form a complex ES, which in turn is converted into a product P and the enzyme E. (Enzyme is a biological catalyst.) E + S k 1 k2 ES k 3 E + P Let u = [S], v = [E], w = [ES] and z = [P]. Then u = k 1 uv + k 2 w, v = k 1 uv + k 2 w + k 3 w, w = k 1 uv k 2 w k 3 w, z = k 3 w.

32 Nonlinear reaction rates: Michaelis-Menton In a basic enzyme reaction, an enzyme E binds to a substrate S to form a complex ES, which in turn is converted into a product P and the enzyme E. (Enzyme is a biological catalyst.) E + S k 1 k2 ES k 3 E + P Let u = [S], v = [E], w = [ES] and z = [P]. Then u = k 1 uv + k 2 w, v = k 1 uv + k 2 w + k 3 w, w = k 1 uv k 2 w k 3 w, z = k 3 w. 1. v + w is a constant, and assume it is E (Quasi steady state approximation) Assume v = 0 and w = 0. Then u = z E 0 k 1 k 3 u = k 2 + k 3 + k 1 u

33 Nonlinear reaction rates: Michaelis-Menton In a basic enzyme reaction, an enzyme E binds to a substrate S to form a complex ES, which in turn is converted into a product P and the enzyme E. (Enzyme is a biological catalyst.) E + S k 1 k2 ES k 3 E + P Let u = [S], v = [E], w = [ES] and z = [P]. Then u = k 1 uv + k 2 w, v = k 1 uv + k 2 w + k 3 w, w = k 1 uv k 2 w k 3 w, z = k 3 w. 1. v + w is a constant, and assume it is E (Quasi steady state approximation) Assume v = 0 and w = 0. Then u = z E 0 k 1 k 3 u = k 2 + k 3 + k 1 u So in a large network of chemical reactions, one may use S P to replace the three equations above, but the reaction rate is the Michaelis-Menton function.

34 Nonlinear reaction rates: Michaelis-Menton In a basic enzyme reaction, an enzyme E binds to a substrate S to form a complex ES, which in turn is converted into a product P and the enzyme E. (Enzyme is a biological catalyst.) E + S k 1 k2 ES k 3 E + P Let u = [S], v = [E], w = [ES] and z = [P]. Then u = k 1 uv + k 2 w, v = k 1 uv + k 2 w + k 3 w, w = k 1 uv k 2 w k 3 w, z = k 3 w. 1. v + w is a constant, and assume it is E (Quasi steady state approximation) Assume v = 0 and w = 0. Then u = z E 0 k 1 k 3 u = k 2 + k 3 + k 1 u So in a large network of chemical reactions, one may use S P to replace the three equations above, but the reaction rate is the Michaelis-Menton function. There are many more complex enzyme reactions which make chemical reaction network (large system of ODEs satisfying mass action law and other laws).

35 Modeling for the CIMA reaction [Lengyel-Epstein, 1991] Modeling of Turing Structures in the Chlorite-Iodide-Malonic Acid-Starch Reaction System. Science. MA + I 2 IMA + I + H +, CIO 2 + I CIO 2 + (1/2)I 2, CIO 2 + 4I + 4H + CI + 2I 2 + 2H 2 O [CIO 2 ], [I 2 ] and [MA] varying slowly, assumed to be constant Let I = X, CIO 2 = Y and I 2 = A. Then the reaction becomes A k 1 X, X k 2 Y, 4X + Y k 3 P Reaction rates k 1, k 2 are constants, and k 3 is proportional to u = a u 4uv 1 + u 2, v = b(u uv 1 + u 2 ). [X ] [Y ] u + [X ] 2 Here [I ] = u(t), [CIO 2 ] = v(t), a, b > 0. Key parameter: a > 0 (the feeding rate)

36 Activator-inhibitor systems In the law of mass action, a gain is represented by a + term, and a loss is represented by a term. In more general case, if the presence of one chemical u increases the generation of another chemical v, then u is called a activator, and if u decreases the generation of another chemical v, then u is an inhibitor. If u is an activator of v, and v is an inhibitor of u, then in u = f (u, v), v = g(u, v), f v < 0 and g u > 0. [Gierer-Meinhardt, 1972] a = ρ a2 h µaa + ρa, h = ρa 2 µ h h + ρ h. Here a is the activator, and h is the inhibitor. µ a and µ h are decay rates of a and h, ρ a and ρ h are the feeding rates of a and h, and ρ is the reaction rate.

37 More models Water-plant model w = a w wn 2, n = wn 2 mn w(t) water, n(t) plant a > 0: rainfall; w: evaporation; wn 2 : water uptake by plants; wn 2 : plant growth; mn: plant loss [Klausmeier, Science, 1999]

38 More models Water-plant model w = a w wn 2, n = wn 2 mn w(t) water, n(t) plant a > 0: rainfall; w: evaporation; wn 2 : water uptake by plants; wn 2 : plant growth; mn: plant loss [Klausmeier, Science, 1999] Epidemic models S = A αsi AS, I = αsi βi AI, R = βi AR assuming that S(0) + I (0) + R(0) = 1 [Kermack and McKendrick, 1927]