← previous - home - next →

Table of contents

1. Introduction
2. Variables
3. Data structures
4. Conditional statements and loops
5. Some exercises ← (Notebook)
6. Introduction to functions
   1. File manipulation
7. From 0D to 1D
   1. Adding lateral diffusion
8. From 1D to 2D
9. Playing with the model

4. Some exercises

Now, you should have enough to go through a small batch of exercises 🥳 🥳 🥳.

They might be a bit hard, but that’s normal, please don’t hesitate to ask us if you have any trouble understanding something.

The answers are still accessible using the answer function with the number of the question as an argument. Also, for some of the following questions, hints are available, run the command hint(<question_number>) to access the hint. For example, for a hint for question 3, you can run hint(3).

Moreover, you will need a bit of help for the next exercises so we wrote some functions that will be useful. We will describe how they work and how to use them. To load the said function, please run the line below.

Note: If you feel like it, you can try to implement these functions yourself.

from Resources.UsefulFunctions import *
from Resources.Answers import answer, hint

Moreover, we need the parameters described earlier, even though it might not be necessary, we rewrite them here for commodity

mu_a = 2.8e-4
mu_i = 5e-3
tau = .1
k = -.005
size = 100
dx = dy = 2. / size
T = 9.0
dt = .001
n = int(T / dt)

“Small” exercise that does not count

Write a loop that computes the sum of the even numbers between 0 and 30.

# Write the loop here

Before creating a working 2D Turing pattern, let work out how to make the concentration within a given cell change over time according to the function we defined earlier.

Exercise 3

Write a loop that increments the concentration value of the concentration a according to the function da_alone.

The function da_alone implements the change of concentration ($\delta a$) according to $a$, $\delta t$ and $k$:

$\delta a = \delta t (a - a^3 +k)$

da_alone takes as parameters the initial concentration value $a$ (a), the $\delta t$ parameter (dt) and the $k$ parameter (k) and outputs the differential of concentration $\delta a$ for that specific initial concentration.

a = 0.1
# Write here the code necessary
print(f'{a                            = }')
print( 'Expected value (for a = 0.1) : 0.9974896544606241')

Exercise 3 (bonus)

If you feel like it, you can write the function da_alone.

For a solution you can type da_alone?? to have access to its implementation.

# da_alone??

Exercise 4

Now we have access to the last value of the concentration but we would like to be able to access all the values in order to plot them.

To do so, write a piece of code that stores all the intermediary results in a list.

A = [.01]
print(f'Last value of A (A[-1])           --> {A[-1]}')
print( 'Expected value (for A[-1] = 0.01) --> 0.9971706727639877')

Given the list of concentrations, one can plot its evolution over time using the function plot_concentration_1cell:

# plot_concentration_1cell(A)

Exercise 5

Now that we know how to compute the evolution of a given concentration, we want to compute the co-evolution of an activator and its inhibitor.

To do so we can use the functions da and di that implements the following equations:

$\delta a = \delta t(a-a^3-i+k)$

$\delta i = \frac{\delta t}{\tau}(a -i)$

Note that the function da is slightly different to da_alone since it incorporates the inhibitor action $-i$.

The function da takes as an input the original concentration $a$ (a), the time increment parameter $\delta t$ (dt), the constant $k$ (k) as for the function da_alone but also the original inhibitor concentration $i$ (i).

The function di takes as an input the original concentration $i$ (i), the time increment parameter $\delta t$, the constant $\tau$ (tau) and the original activator concentration $a$ (a).

Write code to store in two lists the evolution of the concentrations of the activator and the inhibitor.

A = [0.4]
I = [0.15]
# Write the code here
print('For the starting values of A[0]=0.4 and I[0]=0.15:')
print(f'Last value of A --> {A[-1]}')
print( 'Expected value  --> 0.17217946292184916')
print(f'Last value of I --> {I[-1]}')
print( 'Expected value  --> 0.1733148395515316')

As before, it is possible to plot the values of the concentrations over time using the function plot_concentration_1cell.

Side note !

If you were not able to solve the previous question, or if you cannot wait before looking at what the graph looks like, the function answer_results is there for you!

You can call it the following way:

answer_results(<question_number>, <param_name>=<param_value>, ...)

For example, to get the values of A and I from question 4 you can call answer_results the following way:

A, I = answer_results(4, A=0.4, I=0.15, dt=dt, k=k, tau=tau, n=n)

If you don’t know what parameters to give, you can just call answer_results(<question_number>) and hopefully it will help …

Now, you can call the function plot_concentration_1cell with input A and I to see the concentration evolution!

# Uncomment the following line if you have not computed yet A and I
A, I = answer_results(4, A=0.4, I=0.15, dt=dt, k=k, tau=tau, n=n)
plot_concentration_1cell(A, I)

← previous - home - next →