BIOMEDE 418 HW 2 Key
Due Feb 8th
15 pts per question; 120 pts total
1. OR-gate logic. Analyze the C1-FFL with OR-logic at the Z promoter. Are there delays following
ON or OFF steps of Sx? What could be the biological use of such a design?
Solution:
After an ON step of Sx, X becomes active X*. On a rapid timescale it binds the Z promoter.
Since Z is regulated by OR-gate logic, X* alone can activate transcription without need of Y.
Therefore there are no delays following an ON step of Sx.
After an OFF step of Sx, X* rapidly becomes inactive X. However, protein Y is still present in
the cell, and if Sy is present, Y is active Y*. Since the Z input function is an OR gate, Y* can
continue to activate transcription of Z even in the absence of X*. Therefore, Z production
persists, until Y degrades/dilutes below its activation threshold for Z. The dynamics of Y are
given by
dY / dt= -αY, (there is no production term because X is inactive following removal of Sx) so
that Y=Ym exp(-α t), where Ym is the level of Y at time t=0. The OFF delay is given by the time
it takes Y to reach its activation threshold for Z, Ky: solving for this time
Y(Tof)= Ym exp(-α Tof)=Ky, yields
Tof= 1/ α log (Ym/Ky)
In summary, the OR-gate C1-FFL shows sign-sensitive delays. It has a delay following OFF
but not ON steps of Sx. The delay depends on the presence of Sy. This behavior is opposite
to that of the C1-FFL with an AND gate, which shows delay upon ON but not OFF steps.
The OR-gate C1-FFL could be useful in systems which need to be protected from sudden
loss of activity of their master regulator X. The OR-gate FFL can provide continued
production during brief fluctuations in which X activity is lost. This protection works for OFF
pulses shorter than Tof. Note that Tof can be tuned by evolutionary selection by adjusting
the biochemical parameters of protein Y such as its expression level Ym and its activation
threshold K
y.
2. A decoration on the FFL. The regulator Y in C1-FFLs in transcription networks is often
negatively auto-regulated. How does this afect the dynamics of the circuit, assuming that it
has an AND input function at the Z promoter? How does it afect the delay times? The Y
regulator in an OR-gate C1-FFL is often positively auto-regulated. How does this afect the
dynamics of the circuit? How does it afect the delay times?
X Y Z
Solution:
The negative auto-regulation on Y speeds its response time. Hence it shortens the time
needed for Y to cross K
yz, the activation threshold for Z. Denoting by Ton the delay in Z
activation following a step of Sx , and assuming strong auto-repression in the negative autoregulation loop, the delay is:
Ton=Kyz/ (see Eq 3.4.7), which is shorter than the delay in an FFL without negative autoregulation. . Following an OFF step of Sx the negative auto-regulation has no efect (see
Exercise 3.5) .
For the C1-FFL with an OR input function at the Z promoter, an ON step of Sx causes
an immediate rise of Z (the auto-regulation of Y has no efect here because one active
regulator input to Z is enough). The positive auto-regulation of Y has an efect upon an OFF
step of Sx. Recall from exercise 3.4, that a positively auto-regulated gene has, in a linear
model, the following dynamics:
(4.3.1) dY / dt = βx+β1 Y –α Y= βx-(-1)Y
where 1Y is the term representing the positive auto-regulation and βx is the efect of X. When Sx
goes to zero, βx=0 and the solution for Y's dynamics is a decay from an initial value Yst:
(4.3.2) Y= Yst exp(-(-1)t)=Yst exp(-'t)
thus regulator Y's levels will exponentially decay with a rate α' = α- β1 , smaller than the rate 
for a gene without positive auto-regulation. Hence, the delay for turning of gene Z will be
longer due to positive auto-regulation. Gene Z production will stop when Y decreases below its
activation threshold:
(4.3.3) Y(Tof)=Kyz  Yst exp(-' Tof)=Kyz  Tof =1/' log (Kyz/Yst)
The delay in the turning of of Z is therefore /' longer than the delay when Y is not autoregulated.