4   Data Types

4.1   Exercices

4.1.1   Exercise

Assume that we execute the following assignment statements:

width = 17
height = 12.0
delimiter ='.'

For each of the following expressions, write the value of the expression and the type (of the value of the expression) and explain.

  1. width / 2
  2. width / 2.0
  3. height / 3
  4. 1 + 2 * 5

Use the Python interpreter to check your answers.

>>> width = 17
>>> height = 12.0
>>> delimiter ='.'
>>>
>>> width / 2
8
>>> # both operands are integer so python done an euclidian division and threw out the remainder
>>> width / 2.0
8.5
>>> height / 3
4.0
>>> # one of the operand is a float (2.0 or height) then python pyhton perform afloat  division but keep in mind that float numbers are aproximation.
>>> # if you need precision you need to use Decimal. But operations on Decimal are slow and float offer quite enought precision
>>> # so we use decimal only if wee need great precision
>>> # Euclidian division
>>> 2 / 3
0
>>> # float division
>>> float(2)/float(3)
0.6666666666666666
>>> # decimal division
>>> from decimal import Decimal
>>> a = Decimal(2)
>>> b = Decimal(3)
>>> a / b
Decimal('0.6666666666666666666666666667')
>>> 1 + 2 * 5
11

4.1.2   Exercise

Write a function which take a radius as input and return the volume of a sphere:

The volume of a sphere with radius r is 4/3 πr3.

What is the volume of a sphere with radius 5?

Hint: π is in math module, so to access it you need to import the math module Place the import statement at the top fo your file. after that, you can use math.pi everywhere in the file like this:

>>> import math
>>>
>>> #do what you need to do
>>> math.pi #use math.pi

Hint: the volume of a spher with radius 5 is not 392.7 !

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import math

def vol_of_sphere(radius):
    """
    compute the volume of sphere of a given radius
    """
    
    vol = float(4)/float(3) * float(math.pi) * pow(radius, 3)
    return vol

python -i volume_of_sphere.py
>>> vol_of_sphere(5)
523.5987755982989

vol_of_sphere.py .

4.1.3   Exercise

Draw what happen in memory when the following statements are executed:

i = 12
i += 2
set 
>>> i = 12
>>> id(i)
33157200
>>> i += 2
>>> id(i)
33157152

and

s = 'gaa'
s = s + 'ttc'
set 
>>> s = 'gaa'
>>> id(s)
139950507582368
>>> s = s+ 'ttc'
>>> s
'gaattc'
>>> id(s)
139950571818896

when an augmented assignment operator is used on an immutable object is that

  1. the operation is performed,
  2. and an object holding the result is created
  3. and then the target object reference is re-bound to refer to the result object rather than the object it referred to before.

So, in the preceding case when the statement i += 2 is encountered, Python computes 1 + 2 , stores the result in a new int object, and then rebinds i to refer to this new int . And if the original object a was referring to has no more object references referring to it, it will be scheduled for garbage collection. The same mechanism is done with all immutable object included strings.

4.1.4   Exercise

how to obtain a new sequence which is the 10 times repetition of the this motif : “AGGTCGACCAGATTANTCCG”::
>>> s = "AGGTCGACCAGATTANTCCG"
>>> s10 = s * 10

4.1.5   Exercise

create a representation in fasta format of following sequence :

Note

A sequence in FASTA format begins with a single-line description, followed by lines of sequence data. The description line is distinguished from the sequence data by a greater-than (“>”) symbol in the first column. The word following the “>” symbol is the identifier of the sequence, and the rest of the line is the description (optional). There should be no space between the “>” and the first letter of the identifier. The sequence ends if another line starting with a “>” appears; this indicates the start of another sequence.

id = "sp|P60568|IL2_HUMAN"

comment = "Interleukin-2 OS=Homo sapiens GN=IL2 PE=1 SV=1"

sequence = """MYRMQLLSCIALSLALVTNSAPTSSSTKKTQLQLEHLLLDLQMILNGINNYKNPKLTRML
TFKFYMPKKATELKHLQCLEEELKPLEEVLNLAQSKNFHLRPRDLISNINVIVLELKGSE
TTFMCEYADETATIVEFLNRWITFCQSIISTLT"""

>>> s = id + comment + '\n' + sequence
or
>>> s = "{id} {comment} \n{sequence}".format(id= id, comment = comment, sequence = sequence)

4.1.6   Exercise

For the following exercise use the python file sv40 in fasta which is a python file with the sequence of sv40 in fasta format already embeded, and use python -i sv40_file.py to work.

how long is the sv40 in bp? Hint : the fasta header is 61bp long. (http://www.ncbi.nlm.nih.gov/nuccore/J02400.1)

pseudocode

write a function fasta_to_one_line that return a sequence as a string without header or any non sequence characters

pseudocode:

function fasta_to_one_line(seq)
header_end_at <- find the first return line character
raw_seq <- remove header from sequence
raw_seq <- remove non sequence chars
return raw_seq
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def fasta_to_one_line(seq):
    header_end_at = seq.find('\n')
    seq = seq[header_end_at + 1:]
    seq = seq.replace('\n', '')
    return seq

fasta_to_one_line.py .

python
>>> import sv40_file
>>> import fasta_to_one_line
>>>
>>> sv40_seq = fasta_to_one_line(sv40_file.sv40_fasta)
>>> print(len(sv40_seq))
5243

Is that the following enzymes:

  • BamHI (ggatcc),
  • EcorI (gaattc),
  • HindIII (aagctt),
  • SmaI (cccggg)

have recogition sites in sv40 (just answer by True or False)?

>>> "ggatcc".upper() in sv40_sequence
True
>>> "gaattc".upper() in sv40_sequence
True
>>> "aagctt".upper() in sv40_sequence
True
>>> "cccggg".upper() in sv40_sequence
False

for the enzymes which have a recognition site can you give their positions?

>>> sv40_sequence = sv40_sequence.lower()
>>> sv40_sequence.find("ggatcc")
2532
>>> # remind the string are numbered from 0
>>> 2532 + 1 = 2533
>>> # the recognition motif of BamHI start at 2533
>>> sv40_sequence.find("gaattc")
1781
>>> # EcorI -> 1782
>>> sv40_sequence.find("aagctt")
1045
>>> # HindIII -> 1046

is there only one site in sv40 per enzyme?

The find method give the index of the first occurence or -1 if the substring is not found. So we can not determine the occurence of a site only with the find method. We will see how to do that when we learn looping and conditions.

4.1.7   Exercise

We want to perform a PCR on sv40, can you give the length and the sequence of the amplicon?

Write a function which have 3 parameters sequence, primer_1 and primer_2

  • We consider only the cases where primer_1 and primer_2 are present in sequence
  • to simplify the exercise, the 2 primers can be read directly on the sv40 sequence.

test you algorithm with the following primers

primer_1 : 5’ CGGGACTATGGTTGCTGACT 3’
primer_2 : 5’ TCTTTCCGCCTCAGAAGGTA 3’

Write the pseudocode before to implement it.

function amplicon_len(sequence primer_1, primer_2)
pos_1 <- find position of primer_1 in sequence
pos_2 <- find position of primer_2 in sequence
amplicon length <- pos_2 + length(primer_2) - pos_1
return amplicon length
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def amplicon_len(seq, primer_1, primer_2):
    """
    given primer_1 and primer_2 
    return the lenght of the amplicon
    primer_1 and primer_2 ar present and in this order 
    in sequence 
    """
    primer_1 = primer_1.upper()
    primer_2 = primer_2.upper()
    sequence = sequence.upper()
    pos_1 = sv40_sequence.find(primer_1)
    pos_2 = sv40_sequence.find(primer_2)
    amplicon_len = pos_2 + len(primer_2) - pos_1
    return amplicon_len

>>> import sv40
>>> import fasta_to_one_line
>>>
>>> sequence = fasta_to_one_line(sv40)
>>> print(amplicon_len(sequence, first_primer, second_primer))
199

amplicon_len.py .

4.1.8   Exercise

reverse the following sequence “TACCTTCTGAGGCGGAAAGA” (don’t compute the complement):

  >>> "TACCTTCTGAGGCGGAAAGA"[::-1]
  or
  >>> s = "TACCTTCTGAGGCGGAAAGA"
  >>> l = list(s)
  # take care reverse() reverse a list in place (the method do a side effect and return None )
  # so if you don't have a object reference on the list you cannot get the reversed list!
  >>> l.reverse()
  >>> print(l)
  >>> ''.join(l)
  or
  >>> rev_s  = reversed(s)
  ''.join(rev_s)

The most efficient way to reverse a string or a list is the way using the slice.

4.1.9   Exercise

The il2_human contains 4 cysteins (C) in positions 9, 78, 125, 145.
We want to generate the sequence of a mutatnt were the cysteins 78 and 125 are replaced by serins (S)
Write the pseudocode, before to propose an implementation:

We have to take care of the string numbered vs sequence numbered:

C in seq -> in string
9 -> 8
78 -> 77
125 -> 124
145 -> 144
generate 3 slices from the il2_human
head <- from the begining and cut between the first cytein and the second
body <- include the 2nd and 3rd cystein
tail <- cut after the 3rd cystein until the end
replace body cystein by serin
make new sequence with head body_mutate tail

il2_human = ‘MYRMQLLSCIALSLALVTNSAPTSSSTKKTQLQLEHLLLDLQMILNGINNYKNPKLTRMLTFKFYMPKKATELKHLQCLEEELKPLEEVLNLAQSKNFHLRPRDLISNINVIVLELKGSETTFMCEYADETATIVEFLNRWITFCQSIISTLT’

head = il2_human[:77]
body = il2_human[77:125]
tail = il2_human[126:]
body_mutate = body.replace('C', 'S')
il2_mutate = head + body_mutate + tail

4.1.10   Exercise

Write a function

  • which take a sequence as parameter
  • compute the GC%
  • and return it
  • display the results readable for human as a micro report like this: ‘the sv40 is 5243 bp length and have 40.80% gc’

use sv40 sequence to test your function.

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def gc_percent(seq):
    gc_pc = float(seq.count('g') + seq.count('c')) / float(len(seq))
    return gc_pc

>>> import sv40
>>> import fasta_to_one_line
>>> import gc_percent
>>>
>>> sequence = fasta_to_one_line(sv40)
>>> gc_pc = gc_percent(sequence)
>>> report = "the sv40 is {0} bp length and have {1:.2%} gc".format(len(sequence), gc_pc)
>>> print report
'the sv40 is 5243 bp length and have 40.80% gc'

gc_percent.py .

gc_pc =  float(sv40_sequence.count('g') + sv40_sequence.count('c')) / float(len(sv40_sequence))
"the sv40 is {0} bp lenght and have {1:.2%} gc".format(len(sv40), gc_pc)