Question

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?

Answer 1

Ans:

a: Assuming logic input function in all parameters:

dX/ dt = β_xθ ( X < K_xx) – α X

dY/ dt = β_xθ ( X < K_xy) θ(Y < Kyy) –αY

The concentration of X initially follows a familiar exponential rise, as long as X < Kxx.

X(t) =βX / α (1-e^-αt)

WhenX(τ1) =K_xy, Y production stops and the concentration of Y exponentially decays from its initial steadystate value of K_yy to 0. The delay is:

X(τ₁) =β_X /α (1-e^-αt) = K_xy

τ₁=1 /α ln (β_X / (β_X - αK_xy)

With strong auto-repression:

β_X /α (1 - (1- ατ₁) ) = K_xy

τ₁ = K_xy / β_X

The time for Y(t) to reach half its steady-state value (K_yy / 2) is:

t_1/2 = τ₁ + ln2 / α

Note that θ (X < K_xy) = 0⇒˙Y= -αY ⇒Y(t) = K_yye^-αt ⇒ ½ K_yy e^-αt ⇒ t = ln 2 / α

b) If X production stops (for example, if its activator becomes inactive) its concentration will exponentially decayfrom its steady state level K_xx towards zero. At delay τ₂ it will cross K_xy

X (τ₂) = K_xx e^-ατ2 = K_xy ⇒ t₂= 1 / α ln K_xx / K_xy

After τ₂ Y is produced at rate τ₂ and will reach half of its steady state level

K_yy aftert t_1/2 = τ₂ + K_yy / 2 β_y