Question

Problem 3. Fan-out: Transcription factor X regulates two genes Y1 and Y2. Draw the resulting network, termed a fan- out with two target genes. The activation thresholds for these genes are Ki and K2. The activator X begins to be produced at time t=0 at rate 6, and is degraded/diluted at rate a, and its signal Sx is present throughout. What are the times at which Yn and Y2 reach halfway to their maximal expression? Problem 4. Time-dependent production and decay: A gene Y with simple regulation has a time-dependent production rate B(t) and degradation rate alt). Solve for its concentration as a function of time.

Answer 1

Answer:

Q-4):

• shift, Y has already reached steady state
• at a level Y(t=0)=Yst= 1/ . After the shift, we have:
• dY/dt = 2
• The solution to such an equation is generally Y(t)=C1+C2. e- .t, where the constants C1 and C2 need to
• determined so that Y(t=0)= 1/ , and that Y at long times reaches its new steady state. Verify that the
• function Y(t)= 2/ +( 1/ - 2/ ).e- .t is indeed a solution of the equation, by taking its time derivative.

1):

• The dynamic equation for the concentration of mRNA of gene Y, Ym, is:
• dYm/dt = m mYm
• The dynamical equation for the protein product is due to production of p copies per mRNA and
• degradation/dilution at rate :
• dY/dt = pYm - Y

2):

• In the typical case that mRNA degradation is faster than the degradation/dilution of the protein
• product, we can assume that Ym reaches steady state quickly in comparison to the protein levels. The
• reason is that the typical time for the mRNA to reach steady state is the response time log(2)/ m, which
• is much shorter than the protein response time log(2)/ , because m » .
• The steady-state mRNA level is
• found by setting dYm/dt=0 in the first equation, yielding
• Ym, st = m/ m
• Using this for Ym in the second equation yields the following equation for the protein production rate:
• dY/dt = p m/ m - Y
• The effective protein production rate is therefore
• = p. m/ m

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