Dilution, degradation, and time delays in algebraic models Matthew - PowerPoint PPT Presentation

Dilution, degradation, and time delays in algebraic models Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Algebraic Biology M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 1 / 1

Motivation We’ve seen how to incorporate the following features into ODE models: dilution of protein concentation due to cellular growth; degradation (or decay) of protein concentration; time-delays due to cellular processes. In this section, we’ll see how to add these types of features to Boolean models. Our Boolean models will be derived from the 3-variable and 5-variable ODE models from the previous lecture. M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 2 / 1

Dilution and degradation Suppose Y regulates the production of X . Assume Y p t q “ 1 implies X p t ` 1 q “ 1. (activation takes 1 step). Generally, the loss of X due to dilution and degradation takes n timesteps. Introduce new variables X old p 1 q , X old p 2 q , . . . , X old p n ´ 1 q . Properties (i) If Y p t q “ 0 and X p t q “ 1, then X old p 1 q p t ` 1 q “ 1. (“ X has been reduced once by dilution & degradation.”) (ii) If Y p t q “ 0 and X old p i ´ 1 q p t q “ 1, then X old p i q p 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 reduce r X s below the discretation threshold. Thus, X p t ` 1 q “ 1 when either of the following holds: Y p t q “ 1 (new amount will be produced by t ` 1), X p t q ^ X old p n ´ 1 q p t q “ 1 (previous amounts of X still available). ´ ¯ X p t ` 1 q “ Y p t q _ X p t q ^ X old p n q p t q M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 3 / 1

Other features 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 p t ` 1 q “ R p t q R 2 p t ` 1 q “ R 1 p t q R 3 p t ` 1 q “ R 2 p t q . . . R n ´ 1 p t ` 1 q “ R n ´ 2 p t q X p t ` 1 q “ R n p t q Medium levels of lactose Introduce a new variable L m meaning “at least medium levels” of lactose. Clearly, L “ 1 implies L m “ 1. High lactose: L “ 1, L m “ 1. Medium lactose: L “ 0, L m “ 1. Low lactose levels: L “ 0, L m “ 0. We can ignore any state for which L “ 1, L m “ 0. M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 4 / 1

Estimating constants for our Boolean model 3-variable ODE model of the lac operon (Yildirim and Mackey, 2004) Let M p t q “ mRNA, B p t q “ β -galactosidase, and A p t q “ allolactose (concentrations), respectively. 1 ` K 1 p e ´ µτ M A τ M q n dM dt “ α M K ` K 1 p e ´ µτ M A τ M q n ´ Ă γ M M dB dt “ α B e ´ µτ B M τ B ´ Ă γ B B dA L A dt “ α A B K L ` L ´ β A B K A ` A ´ Ă 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 p . 0045 , . 0347 q M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 5 / 1

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 A ` µ “ . 044, Ă γ B “ . 001 min ´ 1 ù ñ γ B “ γ B ` µ “ . 031, Ă γ M “ . 411 min ´ 1 Ă ù ñ γ M “ γ M ` µ “ . 441. Degradation is assumed to be exponential decay: x 1 “ ´ kx implies x p t q “ Ce ´ kt . The half-life is the time t such that: t “ ln 2 x p t q “ Ce ´ kt “ . 5 C e ´ kt “ . 5 ´ kt “ ln 1 ù ñ ù ñ ù ñ 2 k Half-lives h A “ ln 2 Ă γ A “ 15 . 753 ( approx. 1 time-step to decay ) Ă Ă h B “ ln 2 γ B “ 22 . 360 ( approx. 2 time-steps to decay ) Ă h M “ ln 2 Ă γ M “ 1 . 5 ( approx. 0 time-steps to decay ) Ą M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 6 / 1

A Boolean model incorporating dilution and degradation Model assumptions Variables are M , B , A . Glucose absent. Intracellular lactose present, two parameters: L and L m . Time-step « 12 min. Ă Ignore (all ! 12): τ M “ . 10 min, τ B “ 2 min, h M “ 1 . 572 min. Introduce variables for dilution and degradation: (since Ă A old h A « 15 . 8 « 1 timestep) (since Ă B old , B old p 2 q h B « 22 . 4 « 2 timesteps) Proposed model ´ ¯ f M “ A f B “ M _ B ^ B old p 2 q ´ ¯ f A “ p B ^ L m q _ L _ A ^ A old ^ B f B old p 1 q “ M ^ B ´ ¯ f A old “ p B _ L m q ^ L ^ A f B old p 2 q “ M ^ B old p 1 q Most of the functions should be self-explanatory. M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 7 / 1

A Boolean model incorporating dilution and degradation Justification for f A ´ ¯ f A “ p B ^ L m q _ L _ A ^ A old ^ B There are 3 ways for allolactose to be available at t ` 1: (i) β -galactosidase and at least medium levels of 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 p L , L m q and variables p x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q “ p M , A , A old , B , B old p 1 q , B old p 2 q q : f M “ A f 1 “ x 2 ´ ¯ f A “ p B ^ L m q _ L _ A ^ A old ^ B f 2 “ x 2 p 1 ` x 3 qp 1 ` x 4 q ` p L m x 4 ` L ` x 4 LL m q ` x 2 p 1 ` x 3 qp 1 ` x 4 qp L m x 4 ` L ` x 4 LL m q ´ ¯ f A old “ p B _ L m q ^ L ^ A f 3 “ p 1 ` x 4 L m qp 1 ` L q x 2 ´ ¯ f B “ M _ B ^ B old p 2 q f 4 “ x 1 ` x 4 p 1 ` x 6 q ` x 1 x 4 p 1 ` x 6 q f B old p 1 q “ M ^ B f 5 “ p 1 ` x 1 q x 4 f B old p 2 q “ M ^ B old p 1 q f 6 “ p 1 ` x 1 q x 5 M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 8 / 1

Using Sage to compute the fixed points (high lactose) Conclusion: There is a unique fixed point, p M , A , A old , B , B old p 1 q , B old p 2 q q “ p x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q “ p 1 , 1 , 0 , 1 , 0 , 0 q This is exactly what we expected: the lac operon is ON. M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 9 / 1

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: p M , A , A old , B , B old p 1 q , B old p 2 q q “ p x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q “ p 0 , 0 , 0 , 0 , 0 , 0 q This is exactly what we expected: the lac operon is OFF. M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 10 / 1

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: p M , A , A old , B , B old p 1 q , B old p 2 q q “ p x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q “ p 0 , 0 , 0 , 0 , 0 , 0 q , or p 1 , 1 , 0 , 1 , 0 , 0 q . M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 11 / 1

Fixed points of our model and bistability Here is a table showing the fixed points of our model, depending on whether extracellular lactose levels are low, medium, or high. Inducer level L L m M A A old B B old p 1 q B old p 2 q operon Low lactose 0 0 0 0 0 0 0 0 OFF High lactose 1 1 1 1 0 1 0 0 ON Medium lactose 0 1 0 0 0 0 0 0 OFF Medium lactose 0 1 1 1 0 1 0 0 ON Suppose lactose concentration is low ( L “ L m “ 0), and so the operon is OFF. The current state is p M , A , A old , B , B old p 1 q , B old p 2 q q “ p x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q “ p 0 , 0 , 0 , 0 , 0 , 0 q , Now, let’s change L m from 0 to 1, increasing the lactose level to medium. We are now in the 3rd fixed point above, and so the operon is still OFF. Conversely, suppose lactose concentration is high ( L “ L m “ 1), and so the operon is ON. The current state is p M , A , A old , B , B old p 1 q , B old p 2 q q “ p x 1 , x 2 , x 3 , x 4 , x 5 , x 6 q “ p 1 , 1 , 0 , 1 , 0 , 0 q , Now, let’s change L from 1 to 0, reducing the lactose level to medium. This takes us to the 4th fixed point above, and so the operon is still ON. M. Macauley (Clemson) Dilution, degradation, & time-delays Algebraic Biology 12 / 1