MA508 – Worksheet 1

MA508 – Worksheet 1
In 1965, the Nobel prize for physiology and medicine was awarded to Jaques Monod
and Fran¸cois Jacob for their study of the lac operon. This is a series of genes that produces
proteins involved in lactose metabolism in bacteria. Some of these proteins bind to and
regulate the genes in the lac operon. Since the lac operon is complex, here we will analyze
a simpler system – a single gene that is regulated by its protein product. You will see that,
even in this simple system, complex behaviors can arise.
As you probably know, a gene is a section of DNA that codes for a single protein. To
turn the DNA into protein, an enzyme called RNA polymerase binds to a region of DNA
called the promoter, and then transcribes the following DNA into RNA. Then, this RNA is
translated into a protein (see Fig. 1).
DNA
RNA polymerase
transcription
RNA
translation
Protein
regulation
breakdown
Figure 1: A cartoon showing the experimental system
Proteins can bind to the promoter region and affect how easily RNA polymerase binds.
In this way, the formation of the protein can be regulated. Proteins, once made, do not last
forever but are eventually broken down and destroyed. Thus, by regulating the formation
of protein, the amount of protein can be controlled.
Here, we will study a single gene that is regulated by its own protein product. In
particular, RNA polymerase cannot bind to the promoter unless two of the protein molecules
are bound (see Fig. 1). Mathematically, the system can be expressed with the following
equation
p˙ = −kp + A
p
2
p
2 + B2
where k, A, and B are positive constants that determine how quickly the protein is destroyed, how quickly it is formed, and how strongly the protein affects its formation, respectively. For simplicity, assume k = B = 1.
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Answer the following questions.
Here, again, is the governing equation
p˙ = −p + A
p
2
p
2 + 1
1a) Draw the phase line for the case where A = 6 (i.e., plot ˙p vs. p, indicate any fixed
point(s) and indicate the stability of each. On the horizontal axis (p) indicate the flow
directions).
1b) Without solving the equation, sketch p(t) as a function of t for several initial conditions for A = 6.
1c) Draw the phase line for the case where A = 1.
1d) Without solving the equation, sketch p(t) as a function of t for several initial conditions for A = 1.
1e) Suppose that you wanted to maintain a non-zero steady-state concentration of protein in one of your cells. Would you choose A = 1 or A = 6? What else would you have to
do to ensure a non-zero steady-state?
I have uploaded a couple movies showing simulations of this system. In each movie, on
the left, the protein molecules “dance” randomly due to Brownian motion. When they are
near either of two binding sites on the DNA molecule, they bind. When one protein molecule
is bound to DNA, the protein turns yellow. If a second protein molecule then binds, they
turn red and more protein is formed. All the while, the protein is being destroyed.
Watch the two movies (N10 start.mov, and N60 start.mov), which show what happens
when we initially have 10, or 60 protein molecules. On the right is a plot of the number
of molecules as a function of time, p(t). Time is measured in milliseconds, so the entire
experiments occurs over five seconds.
2) Based on these movies, and assuming that the system at least approximately obeys the
differential equation, what can you say about the parameter A in the system? This is maybe
a hard question, but think about what makes the system behave differently for A = 1 than
for A = 6, and then think about what would be required to get the behavior you observe
in the movies.
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