Exercise 1

This exercise was generated from a Jupyter notebook. You can download the notebook here.

Problem 1.1: Command line exercises

In this problem, you will play around with the command line on your machine and get more familiar with it.

a) Let's play around with some options for the ls command. First cd into a directory that has some interesting files in it (like ~/bootcamp/command_line_tutorial). Try the following.

ls -F
ls -G    # Might not be as cool with Git Bash on Windows
ls -l
ls -lh
ls -lS
ls -FGLh

You should be able to infer what these different options do, but you should talk with the TAs as well.

Normally, files that begin with a dot (.) are omitted when listing things. They are also generally omitted when you use your OS's GUI-based file handling system (like Finder on Macs). To see them, use ls -a. So, cd into your home directory (you remember how to do that, right?), and then do

ls -a

b) The nuclear option to delete everything in a directory is rm -rf. The r means to delete recursively, and the f means to "force" deletion. I was going to give you an exercise that uses the nuclear option, but I'm not going to do that. So, just forget I said anything. For this part of the problem, I want you to discuss with your neighbor when the nuclear option might be used, and what needs to be in place before exercising it.

c) Try doing this:

ls /

What is /? Try cd-ing there and seeing what's in there. Do not delete anything!

d) Make a copy of the ~/bootcamp directory called ~/bootcamp_sandbox. Make sure to copy over all files. Why might you want do do this?

Problem 1.2: Using string methods

In Lesson 7, we wrote a function to compute the reverse complement of a sequence.

a) Write that function again, still using a for loop, but do not use the built-in reversed() function.

b) Write the function one more time, but without any for loops.

Problem 1.3: Longest common substring

Write a function that takes two sequences and returns the longest common substring. A substring is a contiguous portion of a string. For example:

Substrings of ATGCATAT:


Not substrings of ATGCATAT:

AGCA              # Skipped T
CCATA             # Added another C
Hello, world.     # Has nothing to do with the input sequence

Problem 1.4: RNA secondary structure validator

In this problem, we will write a function that takes an RNA sequence and an RNA secondary structure and decides if the secondary structure is possible given the sequence. Remember, single stranded RNA can fold back on itself and for base pairs. An RNA secondary structure is simply the list of base pairs that are present. We will represent the base pairs in dot-parentheses notation. For example, a sequence/secondary structure pair would be


For convenience of discussion, I have labeled the indices of the bases on the top row. In this case, base 0, a G, pairs with base 9, a C. Base 1 pairs with base 8, and base 2 pairs with base 7. Bases 3, 4, 5, and 6 are unpaired. (This structure is aptly called a "hairpin.")

So, I hope the dot-parentheses notation is clear. An open parenthesis is paired with the parenthesis that closes it. Dots are unpaired.

So, the goal of our function is to check all base pairs present in a secondary structure and see if they are with G-C, A-U, or (optionally) G-U.

a) Write a function to make sure that the number of closed parentheses is equal to the number of open parentheses, a requirement for a valid secondary structure. It should return True if the parentheses are valid and False otherwise.

b) Write a function that converts the dot-parens notation to a tuple of 2-tuples representing the base pairs. We'll call this function dotparen_to_bp(). An example input/output of this function would be:


((0, 9), (1, 8), (2, 7))

Hint: You might find the pop() method of lists useful.

c) Because of sterics, the minimal length of a hairpin loop is three bases. A hairpin loop is a series of unpaired bases that are closed by a base pair. For example, the secondary structure (.(....).) has a single hairpin loop of length 4. So, the structure ((((..)))) is not valid because it has a hairpin loop of only two bases.

Write a function that verifies that a list of base pairs (as outputted by dotparen_to_bp()) satisfies the hairpin requirement.

d) Now write your validator function. The function definition should look like this:

def rna_ss_validator(seq, sec_struc, wobble=True):

It should return True if the sequence is commensurate with a valid secondary structure and False otherwise. The wobble keyword argument is True if we allow wobble pairs (G paired with U). Here are some expected results:

Returns True:

rna_ss_validator('GCAUCUAUGC', '(((....)))')
rna_ss_validator('GCAUCUAUGU', '(((....)))') 
rna_ss_validator('GCAUCUAUGU', '(.(....).)') 

Returns False:

rna_ss_validator('GCAUCUACGC', '(((....)))')
rna_ss_validator('GCAUCUAUGU', '(((....)))', wobble=False) 
rna_ss_validator('GCAUCUAUGU', '(.(....)).') 
rna_ss_validator('GCCCUUGGCA', '(.((..))).')