580.429 Systems Biology III
1
November 3, 2008
Homework 7 Key
3.3, 4.5 points
Autoregulated cascade: Gene X encodes a repressor that represses gene Y, which also encodes a repressor. Both X and Y negatively regulate their own promoters. a) At time t=0, X begins to be produced at rate β, starting from an initial concentration of X=0. What are the dynamics of X and Y? What are the response times of X and Y? Assume logic input functions, with repression thresholds Kxx , Kxy , for the action of X on its own promoter and on the Y promoter and Kyy for the action of Y on its own promoter. b) At time t=0, production of X stops after a long period of production, and X concentration decays from an initial steady-state level. What are the dynamics of X and Y? What are the response times of X and Y? Diagram of circuit and graph of dynamics: S
Kxx X
Kxy
Y
Kyy
a)Assuming logic input function in all parameters: dX = βX θ(X < Kxx ) − αX dt dY = βY θ(X < Kxy )θ(Y < Kyy ) − αY dt The concentration of X initially follows a familiar exponential rise, as long as X < Kxx . βX (1 − e−αt ) α When X(τ1 ) = Kxy , Y production stops and the concentration of Y exponentially decays from its initial steady state value of Kyy to 0. The delay is: X(t) =
X(τ1 )
=
τ1
=
βX (1 − e−αt ) = Kxy α 1 βX ln α (βX − αKxy )
With strong auto-repression: βX (1 − (1 − ατ1 )) α τ1
= Kxy =
Kxy βX
Kyy ) is: 2 ln2 = τ1 + α
The time for Y (t) to reach half its steady-state value ( t 12
Note that θ(X < Kxy ) = 0 ⇒ Y˙ = −αY ⇒ Y (t) = Kyy e−αt ⇒ 12 Kyy e−αt ⇒ t =
ln 2 α
580.429 Systems Biology III
2
Homework 7 Key
b)If X production stops (for example, if its activator becomes inactive) its concentration will exponentially decay from its steady state level Kxx towards zero. At delay τ2 it will cross Kxy X(τ2 ) = Kxx e−ατ2 = Kxy ⇒ t2 =
1 Kxx ln α Kxy
Kyy 2βy Note: In the dynamics graph above, we assume strong auto-regulation (in which Xss is much smaller than what it β βX βY would be without autorepression Xss = K << ), that Kxx << and that Kyy << α α α After τ2 Y is produced at rate τ2 and will reach half of its steady state level Kyy after t 12 = τ2 +
2
3.4, 4 points
Positive feedback. What is the effect of positive auto-regulation on the response time? Use as a model the following dX linear equation: = β + β1 X − αX. Explain each term and solve for the response time. When might such a dt design be biologically useful?
β represents the basal production rate and β1 X is the positive effect of X on its own production (positive autoregulation). −αX is the degradation/dilution factor, as usual. dX Grouping terms gives us the equation = β − (α − β1 )X, which helps us to see that the degradation rate is dt reduced by positive auto-regulation, to an effective rate of α0 = α − β1 . Assuming that the auto-regulation is not too strong, that is, that β1 < α,0 the term multiplying X is negative and we approach a stable steady- state as described by X(t) = Xss (1 − e−α t ). The response time is defined as the time to reach half of the steady state: T 1par = 2
ln 2 ln 2 > = T 1simple 2 α0 α
So positive auto regulation creates a longer response time than simple regulation does, due to the reduced effective degradation rate. Remember that negative auto-regulation speeds response times. Strong auto-regulation, in which β1 > α, can lead to instability and unchecked growth of X in the model. In real systems, this instability will be limited by other factors (such as saturation of the input function), locking X in an ON state of high expression even after its activating input β vanishes. Therefore this design creates a bi-stable system, where X is either low or high and fixed. This is the paradigm seen in commitment-type biological decisions, such as thouse made during development. Positive feedback characterizes developmental systems that make a switch that is either OFF or locked ON.
3
3.5, 1.5 points
Turning off auto-regulation. What is the dynamics of a negatively-auto-regulated gene once its maximal promoter activity is suddenly reduced from β1 to β2 = 0? What is the response time, and how does it compare to simple regulation? A negatively-auto-regulated gene, the promoter activity of which has reduced to β2 = 0, exponentially decays to ln 2 . The negative auto-regulation has no effect in this case! Assuming an additional zero with a response time of α activator is present which activates transcription at a constant rate β when it is present (the ON step), negative auto-regulation has an asymmetric accelerating effect, decreasing the response time for an ON signal but not affecting the response time for an OFF signal.
580.429 Systems Biology III
4
3
Homework 7 Key
3.6, 2 points
Two-node positive feedback for decision making. During development from an egg to an embryo, cells need to make irreversible decisions to express the genes appropriate to their designated tissue types and to repress other genes. One common mechanism is positive transcriptional feedback between two genes. There are two types of positive feedback made of two transcription factors. The first type is of two positive interactions, X → Y and Y → X. The second type has two negative interactions X a Y and Y a X. What are the stable steady states in each type of feedback? Which type of feedback would be useful in situations where genes regulated by both X and Y belong to the same tissue? Which would be useful when genes regulated by X belong to different tissues than the genes regulated by Y?
X
Y
Positive feedback with two positive interactions has two stable steady-states: X and Y both ON or X and Y both OFF. This is useful when genes regulated by X and Y belong to the same tissue.
X
Y
Positive feedback with two negative interactions has two stable steady-states: X ON and Y OFF or X OFF and Y ON; that is, either X or Y is ON. This is useful when genes regulated by X belong to different tissues or cell fates than the genes regulated by Y.
580.429 Systems Biology III
5
4
Homework 7 Key
4.3, 4.5 points
A decoration on the FFL. The regulator Y in C1-FFLs in transcription networks is often negatively auto-regulated. How does this affect the dynamics of the circuit, assuming that it has an AND input function at the Z promoter? How does it affect the delay times? The Y regulator in an OR-gate C1-FFL is often positively auto-regulated. How does this affect the dynamics of the circuit? How does it affect the delay times?
SX
X Y
AND
Z
OR
Z
Kyz
Kyy SX
X Y
Kyz
Kyy
AND: The negative-auto-regulation on Y speeds its response time, shortening the time needed for Y to cross the activation threshold for ZKyz . If TON is the delay in Z activation following the introduction of X activating signal Kyz SX , and assuming strong auto-repression in the negative-auto-regulation of Y, the delay is TON = , which is β shorter than the delay in a feed forward loop without negative auto regulation. Following the loss of SX , negativeauto-regulation on Y has no effect and the delay is no different than the simple feed forward loop. OR: An ON step of SX causes an immediate rise in Z, and the auto-regulation of Y has no effect because one active input to Z is enough. However, Y’s regulation affects the dynamics in the case of loss of SX . A positively-autoregulted gene has the following dynamics in a linear model: dY = βX + β1 Y − αY = βX − (α − β1 )Y dt where β1 Y is the term representing the positive auto-regulation and BX is the effect of X. When SX goes to zero, βX = 0 and the solution for Y’s dynamics is a decay from an initial value Yss , such that 0
Y = Yss (e−(α−β1 )t ) = Yss e−α t Thus, Y’s levels will exponentially decay with a rate of α0 = α − β1 , which is smaller than the rate α for gene without positive auto-regulation. The delay for turning off gene Z will be longer, and production will stop when Y decreases below its activation threshold, 0
YTOF F = Kyz =⇒ Yss e−α TOF F = Kyz =⇒ TOF F = The delay in turning off Z is therefore when Y is not auto-regulated.
TOF F 0 TOF F
=
α α0
=
α α−β1
1 Yss ln( ) α0 Kyz
simplemodel P.A.R. > 1 =⇒ TOF > TOF , longer than the delay F F
