Gene X encodes a repressor that represses gene Y, which also encodes a repressor

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 , 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?

Explanation / Answer

Ans:

a: Assuming logic input function in all parameters:

dX/ dt = x ( X < Kxx) – X

dY/ dt = x ( X < Kxy) (Y < Kyy) –Y

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

X(t) =X / (1-e-t)

WhenX(1) =Kxy, Y production stops and the concentration of Y exponentially decays from its initial steadystate value of Kyy to 0. The delay is:

X(1) =X / (1-e-t) = Kxy

1=1 / ln (X / (X - Kxy)

With strong auto-repression:

X / (1 - (1- 1) ) = Kxy

1 = Kxy / X

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

t1/2 = 1 + ln2 /

Note that (X < Kxy) = 0Y= -Y Y(t) = Kyye-t ½ Kyy e-t t = ln 2 /

b) If X production stops (for example, if its activator becomes inactive) its concentration will exponentially decayfrom its steady state level Kxx towards zero. At delay 2 it will cross Kxy

X (2) = Kxx e-2 = Kxy t2= 1 / ln Kxx / Kxy

After 2 Y is produced at rate 2 and will reach half of its steady state level

Kyy aftert t1/2 = 2 + Kyy / 2 y

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