Compute steady states of gene expression

[cascadeless]

Homework Statement

A bacteria that normally divides every 20 minutes express gene X. The production rate of protein X is 5nM/min. The protein is stable and does not degrade.

1. What is the concentration of X in the steady state?
2. The same bacteria enter into a stress state at t=0 for 3 hours (before t=0 the level of protein X is
   at the steady state you found in a). In this state of X increases to 20nM/min but the cell division time changes to 60 min. Calculate the dynamic level of protein X starting t=0. What is the steady state in this case?
3. At t=3 hours the bacteria gets a signal that cases the protein X to become unstable with half-life of 10 min. Calculate now the dynamic level of protein X and the steady state. How would the maximal level of X change if the bacteria got this signal after only 1 hour?
   

Homework Equations

$$\beta ... \text{production rate}$$

• Degradation/dilution rate in units of 1/time
$$x\alpha = \alpha_{dil} + \alpha_{deg}$$

• Change in concentration of Y
$$dY/dt = \beta - \alpha Y$$

• Steady state (solving for dY/dt=0)
$$Y_{ST} = \beta / \alpha$$

Further information on "An Introduction to Systems Biology" by Uri Alon.

The Attempt at a Solution

[/B]
For the first, I have to determine alpha and beta. Beta is obviously 5, and I thought alpha is 1/20, thus 0.05. I am not sure if this is true since the assignment states that the protein does not degrade. Forgive me, I am not that into biology. So the steady state should be 5/0.05 = 100.

I am not sure how to do that with the time shift. I tried to plot it and see how it is looking, so I let MatLab solve the differential equation stated above with initial condition 100. When t is below 180 (3 hours) I set beta to 20 and alpha to 1/60. And when t is higher than 180 then I set the parameters as before. However, that looks like the following:

Unfortunately, I don't know how to calculate the dynamic level of protein X starting with t=0, that's the main problem which I also need for 3.

For 3 I've no idea currently.
I appreciate any help!

 

[epenguin]

I guess this is about some kind of bacteriostatic device like a Chemiostat keeps the bacterial population density constant, hremoving them at the same rate the population increases? More explanation would have been helpful to understand what the problem is.
I agree with your 100 nM
You could at least calculate the nM for the second steady-state and third steady states in the same kind of way independent of your simulations as a check.
What is Y? Is it what you previously called X?
Will try and come back when I when I have more time.

 

[cascadeless]

Thanks a lot for your reply. Oh yes, Y is X, sorry. The book denotes it as Y and the assignment as X. The second steady state would then be 1200, however, I'm not sure if my simulation is ok, looks weird to me. To be honest, I am not really able to provide you more background, it is just a "simple transcriptional process". Perhaps you may like to take a look at the book (pdf), ~page 18: https://www.google.com/url?sa=t&rct...cedownload=1&usg=AOvVaw0vBzFeIJ3J9W7Lbe2sU4BK

I really appreciate your help!

 

[epenguin]

It will be the weekend before I have the time to come back.
Please tidy up your notation.
Maybe a more arithmetic rather than formulaic reasoning would tell you if your steady states are right and reasonable.
I see the model is different from what I imagined, It is essential to know what the model is!
Now I know that your second steady state looks right to me.
As well as Matlab, your text at that point tells you the solution of this standard and elementary differential equation, so you can also calculate it that way.

 

[cascadeless]

Unfortunately, I am not able to edit my post above, and I've noticed another bug in the degradation/dilution rate formula, I am sorry. Here are the correct equations:

Homework Equations

$$\beta ... \text{production rate}$$

• Degradation/dilution rate in units of 1/time
$$\alpha = \alpha_{dil} + \alpha_{deg}$$

• Change in concentration of X
$$dX/dt = \beta - \alpha X$$

• Steady state (solving for dX/dt=0)
$$X_{ST} = \beta / \alpha$$

I think the model is about gene regulation, thus we have a transcription factor that regulates a gene and produces a protein. And since the level of transcription changes with time, we can describe it with differential equations. So, the model can be basically described as $$\frac{dX}{dt} = \beta - \alpha X$$.
I am still confused with 3, however, I will try harder tomorrow. I'd still be happy for any further help.

 

[cascadeless]

I am still confused with the description: "A bacteria that normally divides every 20 minutes express gene X. The production rate of protein X is 5nM/min. The protein is stable and does not degrade."

When a bacteria B express a gene X, then is B a repressor of X, no? Shouldn't then be the production rate equal to zero? And the plot would then show a decreasing function?

And when there is no degradation, how can then alpha be the degradation rate 1 over time?

 

[Ygggdrasil]

I am still confused with the description: "A bacteria that normally divides every 20 minutes express gene X. The production rate of protein X is 5nM/min. The protein is stable and does not degrade."

When a bacteria B express a gene X, then is B a repressor of X, no? Shouldn't then be the production rate equal to zero? And the plot would then show a decreasing function?

And when there is no degradation, how can then alpha be the degradation rate 1 over time?

When a cell divides it splits its contents between two cells. So it there were 100 molecules in the parental cell, each daughter cell will have 50 molecules. Alternatively, you can think of cell division as dilution; the cells are growing in volume, such that the total volume doubles every 20 min. If the initial concentration of Xwas 100nM, and no X is being synthesized, the concentration of X will be 50nM after 20min.

Essentially, you can treat the cell division time as a halflife.

 

[cascadeless]

Okay, thank you very much.

Now I have another question due to 3:
"At t=3 hours the bacteria gets a signal that cases the protein X to become unstable with half-life of 10 min."

I would say that alpha changes when the protein becomes unstable. Does that mean, that alpha is now (1/20 was the dilution rate before):
$$\alpha = \frac{1}{20} + \frac{1}{10}$$

Or does it mean that alpha changes from 1/20 to 1/10?

 

[Ygggdrasil]

• Degradation/dilution rate in units of 1/time
$$\alpha = \alpha_{dil} + \alpha_{deg}$$

As you indicated in the relevant equations you listed in your previous post, you would add the rate of dilution to the rate of degradation to determine the total alpha.

It is important to note that a half life of 20 min does not correspond to a degradation rate of 1/20. You should look for a formula that relates halflife to decay rate.

 

[epenguin]

I am still confused with the description: "A bacteria that normally divides every 20 minutes express gene X. The production rate of protein X is 5nM/min. The protein is stable and does not degrade."

When a bacteria B express a gene X, then is B a repressor of X, no? Shouldn't then be the production rate equal to zero? And the plot would then show a decreasing function?

And when there is no degradation, how can then alpha be the degradation rate 1 over time?

The first sentence of the bolded section is rather confused. The bacterium is a repressor??
All you are asked to do for the questions about steady-states is to calculate at what rate the protein is produced that maintains its concentration constant whilst the volume it is in is expanding, as explained by Ygggdrasil. An essentially arithmetical calculation.

(Then to be sure it is a wonder that the bacteria and can actually do this, and surely has a mechanism to enable it involving a repressor. I doubt this mechanism is responding to the amount of enzyme, but probably to the amount of a product produced by the catalysis by the enzyme. However you are not yet asked anything at all about this so don't worry about it for now.)

I could not make out your last sentence but when there is no degradation it is just the α_deg that is 0.

I have not checked your calculation of the dynamics of the change from one state to another, but if you were to do the calculation I suggested earlier and get the mathematical expression for the degradation, then comparing it with your computation would either increase your confidence or alert you to need for correction. (This is basic stuff for all biomathematical modelling.)

 

[cascadeless]

Thank you guys so much, you helped me a lot!

 

[cascadeless]

I have just a stupid other question, which is not worth to create a new thread.
It is just a notation issue and I'm not quite sure, it says:
"The degradation rate of an activator is $$\alpha = 10min^{-1}$$"

So, do I have to calculate further with alpha=10 or alpha=1/10?

Actually I was pretty sure that I should take 1/10, however, there is an example on the lecture notes which says that the degradation rate is $$\alpha = 0.1min^{-1}$$" and he calculated the model with alpha=0.1, but maybe it is a mistake.

 

[Ygggdrasil]

The statement that α = 10 min^-1 pretty clearly means that α = 10 min^-1.

Here it's useful to take notes of the units. A value with units of time^-1 will be a rate whereas a value of time would more likely to be a halflife.

 

[cascadeless]

Meaning that I should calculate further with α = 10, right?

 

[Ygggdrasil]

You should calculate further with α = 10 min^-1. It is important to consider the units when performing the calculations.

 

[cascadeless]

Ah, now I see, later I have $$x min^{-1} \cdot y min$$ and it cancels to x*y.
Thank you!