Bistability in Boolean network models Matthew Macauley Department - PowerPoint PPT Presentation

Bistability in Boolean network models Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4500, Spring 2016 M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 1 / 18

Motivation Weaknesses of previous Boolean models All processes take a single timestep. Only assumed high and low levels in intracellular lactose. Medium levels are needed for bistability to exist. No time-delays incorporated. In this section, we’ll see how to add these types of features to Boolean models. In particular, we’ll see how to incorporate features such as: loss of concentration due to dilution and degradation; time-delays due to cellular processes M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 2 / 18

Dilution and degradation Suppose Y regulates the production of X . Assume Y ♣ t q ✏ 1 implies X ♣ t � 1 q ✏ 1. (activation takes 1 step). Generally, the loss of X due to dilution and degradation takes several steps. Introduce new variables X old ♣ 1 q , X old ♣ 2 q , . . . , X old ♣ n q . Properties (i) If Y ♣ t q ✏ 0 and X ♣ t q ✏ 1, then X old ♣ 1 q ♣ t � 1 q ✏ 1. (“ X has been reduced once by dilution & degradation.”) (ii) If Y ♣ t q ✏ 0 and X old ♣ i ✁ 1 q ♣ t q ✏ 1, then X old ♣ i q ♣ t � 1 q ✏ 1. (“ X has been reduced i times by dilution & degradation.”) (iii) The number of “old” variables is determined by the number of timesteps required to reduced r X s below the discretation threshold. Thus, X ♣ t � 1 q ✏ 1 when either of the following holds: Y ♣ t q ✏ 1 (new amount will be produced by t � 1), X ♣ t q ❫ X old ♣ n q ♣ t q ✏ 1 (previous amounts of X still available). ✁ ✠ X ♣ t � 1 q ✏ Y ♣ t q ❴ X ♣ t q ❫ X old ♣ n q ♣ t q M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 3 / 18

Other features Medium levels of lactose Introduce a new variable L high so that L high ✏ 1 implies L ✏ 1. High lactose: L ✏ 1, L high ✏ 1. Medium lactose: L ✏ 1, L high ✏ 0. Low lactose: L ✏ 0, L high ✏ 0. We can ignore any state for which L ✏ 0, L high ✏ 1. Previously, we introduced a variable L ℓ to denote “ at least low levels of lactose ,” so ♣ L , L ℓ q ✏ ♣ 1 , 0 q described “medium lactose.” This is an equally valid way to acheive the same goal. Time delays Say R regulates production of X , delayed by time τ ( n steps). Introduce new variables R 1 , R 2 , . . . , R n , with transition functions: R 1 ♣ t � 1 q ✏ R ♣ t q R 2 ♣ t � 1 q ✏ R 1 ♣ t q . . . R n ♣ t � 1 q ✏ R n ✁ 1 ♣ t q X ♣ t � 1 q ✏ R n ♣ t q M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 4 / 18

Estimating constants for our Boolean model 3-variable ODE model of the lac operon (Yildirim and Mackey, 2004) Let M ♣ t q ✏ mRNA, B ♣ t q ✏ β -galactosidase, and A ♣ t q ✏ allolactose (concentrations), respectively. 1 � K 1 ♣ e ✁ µτ M A τ M q n dM dt ✏ α M K � K 1 ♣ e ✁ µτ M A τ M q n ✁ ⑨ γ M M dB dt ✏ α B e ✁ µτ B M τ B ✁ ⑨ γ B B dA L A dt ✏ α A B K L � L ✁ β A B K A � A ✁ ⑨ γ A A We need to estimate these rate constants and time delays from the literature. Time delays: τ M ✏ . 10 min, τ B ✏ 2 . 00 min. Degradtion rates are harder to determine experimentally, and they vary widely in the literaure. Sample values: ✩ γ A ✏ . 52 min ✁ 1 , . 0135 min ✁ 1 , . 00018 min ✁ 1 ✬ ✬ ✫ γ B ✏ . 00083 min ✁ 1 , γ M ✏ . 411 min ✁ 1 , ✬ ✬ ✪ µ P ♣ . 0045 , . 0347 q M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 5 / 18

Estimating constants for our Boolean model Approach We’ll select “middle of range” estimates for the rate constants: µ ✏ . 03 min ✁ 1 , γ A ✏ . 014 min ✁ 1 ù ñ γ A ✏ γ 1 � µ ✏ . 044, ⑨ γ B ✏ . 001 min ✁ 1 ù ñ γ B ✏ γ B � µ ✏ . 031, ⑨ γ M ✏ . 411 min ✁ 1 ù ñ γ M ✏ γ M � µ ✏ . 441. ⑨ Degradation is assumed to be exponential decay: x ✶ ✏ ✁ kx implies x ♣ t q ✏ Ce ✁ kt . The half-life is the time t such that: t ✏ ln 2 x ♣ t q ✏ Ce ✁ kt ✏ . 5 C e ✁ kt ✏ . 5 ✁ kt ✏ ln 1 ù ñ ù ñ ù ñ 2 k Half-lives ⑨ h A ✏ ln 2 γ A ✏ 15 . 753 ( approx. 1 time-step to decay ) ⑨ h B ✏ ln 2 ⑨ γ B ✏ 22 . 360 ( approx. 2 time-steps to decay ) ⑨ ⑨ h M ✏ ln 2 γ M ✏ 1 . 5 ( approx. 0 time-steps to decay ) ⑩ M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 6 / 18

A Boolean model incorporating dilution and degradation Model assumptions Variables are M , B , A . Glucose absent. Intracellular lactose present, two parameters: L and L high . Time-step ✓ 10 min. ⑨ Ignore (all ✦ 10): τ M ✏ . 10 min, τ B ✏ 2 min, h M ✏ 1 . 572 min. Introduce variables for dilution and degradation: (since ⑨ A old h A ✓ 15 . 8 ✓ 1 timestep) (since ⑨ B old , B old ♣ 2 q h B ✓ 22 . 4 ✓ 2 timesteps) Proposed model ✁ ✠ f M ✏ A f B ✏ M ❴ B ❫ B old ♣ 2 q ✁ ✠ f A ✏ ♣ B ❫ L q ❴ L high ❴ A ❫ A old ❫ B f B old ♣ 1 q ✏ M ❫ B ✁ ✠ f A old ✏ ♣ B ❴ L q ❫ L high ❫ A f B old ♣ 2 q ✏ M ❫ B old ♣ 1 q Most of the functions should be self-explanatory. M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 7 / 18

A Boolean model incorporating dilution and degradation Justification for f A ✁ ✠ f A ✏ ♣ B ❫ L q ❴ L high ❴ A ❫ A old ❫ B There are 3 ways for allolactose to be available at t � 1: (i) β -galactosidase and lactose are present; (ii) high levels of lactose (assume basal concentrations of β -galactosidase); (iii) Enough allolactose is present so that it’s not degraded below the threshold, and no β -galactosidase is present. Let’s write our model into polynomials form, with parameters ♣ L , L h q and variables ♣ x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q ✏ ♣ M , A , A old , B , B old ♣ 1 q , B old ♣ 2 q q : f M ✏ A f 1 ✏ x 2 ✁ ✠ f A ✏ ♣ B ❫ L q ❴ L high ❴ A ❫ A old ❫ B f 2 ✏ x 2 ♣ 1 � x 3 q♣ 1 � x 4 q � ♣ Lx 4 � L h � x 4 LL h q � x 2 ♣ 1 � x 3 q♣ 1 � x 4 q♣ Lx 4 � L h � x 4 LL h q ✁ ✠ f A old ✏ ♣ B ❴ L q ❫ L high ❫ A f 3 ✏ ♣ 1 � x 4 L q♣ 1 � L h q x 2 ✁ ✠ f B ✏ M ❴ B ❫ B old ♣ 2 q f 4 ✏ x 1 � x 4 ♣ 1 � x 6 q � x 1 x 4 ♣ 1 � x 6 q f B old ♣ 1 q ✏ M ❫ B f 5 ✏ ♣ 1 � x 1 q x 4 f B old ♣ 2 q ✏ M ❫ B old ♣ 1 q f 6 ✏ ♣ 1 � x 1 q x 5 M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 8 / 18

Using Sage to compute the fixed points (high lactose) Conclusion: There is a unique fixed point, ♣ M , A , A old , B , B old ♣ 1 q , B old ♣ 2 q q ✏ ♣ x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q ✏ ♣ 1 , 1 , 0 , 1 , 0 , 0 q This is exactly what we expected: the lac operon is ON. M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 9 / 18

Using Sage to compute the fixed points (low lactose) We need to backsubstitute. Recall that x k i ✏ x i for all k . The last equation: x 6 6 � x 4 6 � x 3 6 ✏ 0 implies x 6 ✏ 0. Plug this into the previous equation: x 5 � x 4 6 � x 6 ✏ 0 (with x 6 ✏ 0) implies x 5 ✏ 0. And so on. We get a unique fixed point: ♣ M , A , A old , B , B old ♣ 1 q , B old ♣ 2 q q ✏ ♣ x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q ✏ ♣ 0 , 0 , 0 , 0 , 0 , 0 q This is exactly what we expected: the lac operon is OFF. M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 10 / 18

Using Sage to compute the fixed points (medium lactose) The last (7th) equation implies x 6 ✏ 0. The 6th one then implies x 5 ✏ 0. The 5th equation gives no information ( x 4 can be anything), as does the 4th ( x 2 4 � x 4 ✏ 0). The 3rd equation says x 3 ✏ 0. The 2nd equation says x 2 ✏ x 4 , and the 1st equation says x 1 ✏ x 4 . We get two fixed points: ♣ M , A , A old , B , B old ♣ 1 q , B old ♣ 2 q q ✏ ♣ x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q ✏ ♣ 0 , 0 , 0 , 0 , 0 , 0 q , or ♣ 1 , 1 , 0 , 1 , 0 , 0 q . M. Macauley (Clemson) Bistability in Boolean network models Math 4500, Spring 2016 11 / 18