diff --git a/source/Collection_Data_Types.rst b/source/Collection_Data_Types.rst
index a73ea4b2d4ee72cf555eebedba71a4fff3df67f7..9170f1492ba83c7df6946df865e6f061a7e579f6 100644
--- a/source/Collection_Data_Types.rst
+++ b/source/Collection_Data_Types.rst
@@ -8,10 +8,10 @@
Collection Data Types
*********************
-Exercices
+Exercises
=========
-Exercice
+Exercise
--------
| Draw the representation in memory of the following expressions.
@@ -29,8 +29,38 @@ Exercice
:alt: set
:figclass: align-center
+::
+
+ x = [1, 2, 3, 4]
+ x += [5, 6]
+
+.. figure:: _static/figs/augmented_assignment_list.png
+ :width: 400px
+ :alt: set
+ :figclass: align-center
+
+::
+
+ >>> x = [1, 2, 3, 4]
+ >>> id(x)
+ 139950507563632
+ >>> x += [5,6]
+ >>> id(x)
+ 139950507563632
+
-Exercice
+compare with the exercise on string and integers:
+
+Since list are mutable, when ``+=`` is used the original list object is modified, so no rebinding of *x* is necessary.
+We can observe this using *id()* which give the memory adress of an object. This adress does not change after the
+``+=`` operation.
+
+.. note::
+ even the results is the same ther is a subtelty to use augmented operator.
+ in ``a operator= b`` python looks up ``a`` ’s value only once, so it is potentially faster
+ than the ``a = a operator b``.
+
+Exercise
--------
wihout using python shell, what is the results of the following statements:
@@ -51,14 +81,14 @@ all the digits after the periods are discarded.
In python3 we will obtain the expected result (see :ref:``)
-Exercice
+Exercise
--------
How to compute safely the average of a list? ::
float(sum(l)) / float(len(l))
-exercise
+Exercise
--------
Draw the representation in memory of the following expressions. ::
@@ -98,20 +128,30 @@ Draw the representation in memory of the following expressions. ::
x = [1, ['a','b','z'], 3, 4]
+Exercise
+--------
+
+from the list l = [1, 2, 3, 4, 5, 6, 7, 8, 9] generate 2 lists l1 containing all odd values, and l2 all even values.::
+
+ l = [1, 2, 3, 4, 5, 6, 7, 8, 9]
+ l1 = l[::2]
+ l2 = l[1::2]
+
-exercise
+Exercise
--------
generate a list containing all codons. ::
-
+
+ bases = 'acgt'
codons = []
- for a in 'acgt':
- for b in 'acgt':
- for c in 'acgt':
+ for a in bases:
+ for b in bases:
+ for c in bases:
codon = a + b + c
codons.append(codon)
-exercice
+Exercise
--------
From a list return a new list without any duplicate, regardless of the order of items.
@@ -127,7 +167,7 @@ solution ::
-exercice
+Exercise
--------
let the following enzymes collection: ::
@@ -221,7 +261,7 @@ If we want also the position, for instance to compute the fragments of dna. ::
-exercice
+Exercise
--------
From a list return a new list without any duplicate, but keeping the order of items.
For example: ::
@@ -242,7 +282,7 @@ solution ::
>>> uniq_items = set()
>>> l_uniq = [x for x in l if x not in uniq_items and not uniq_items.add(x)]
-exercice
+Exercise
--------
list and count occurences of every 3mers in the following sequence ::
@@ -301,7 +341,7 @@ solution bonus ::
print kmer, " = ", occurence
-exercice
+Exercise
--------
given the following dict : ::
diff --git a/source/Data_Types.rst b/source/Data_Types.rst
index 0ac0e094fd9b2a5c3d1163574210d20fb2df2a7d..23cdd154c74c58f65402ea5b3f587f30a4442b68 100644
--- a/source/Data_Types.rst
+++ b/source/Data_Types.rst
@@ -6,3 +6,305 @@
**********
Data Types
**********
+
+Exercices
+=========
+
+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
+
+
+Exercise
+--------
+
+Practice using the Python interpreter as a calculator:
+
+| The volume of a sphere with radius r is 4/3 πr\ :sup:`3`. 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 ::
+
+ >>> import math
+ >>> math.pi
+
+ Hint: 392.7 is wrong! ::
+
+ >>> import math
+ >>> float(4)/float(3) * float(math.pi) * pow(5, 3)
+
+
+Exercise
+--------
+
+Draw what happen in memory when the following statements are executed: ::
+
+ i = 12
+ i += 2
+
+.. figure:: _static/figs/augmented_assignment_int.png
+ :width: 400px
+ :alt: set
+ :figclass: align-center
+
+::
+
+ >>> i = 12
+ >>> id(i)
+ 33157200
+ >>> i += 2
+ >>> id(i)
+ 33157152
+
+
+and ::
+
+ s = 'gaa'
+ s = s + 'ttc'
+
+.. figure:: _static/figs/augmented_assignment_string.png
+ :width: 400px
+ :alt: set
+ :figclass: align-center
+
+::
+
+ >>> 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
+
+#. the operation is performed,
+#. and an object holding the result is created
+#. 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.
+
+Exercise
+--------
+
+how to obtain a new sequence which is the 10 times repetition of the this motif : "AGGTCGACCAGATTANTCCG"::
+ >>> s = "AGGTCGACCAGATTANTCCG"
+ >>> s10 = s * 10
+
+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)
+
+
+Exercise
+--------
+
+For the following exercise use the python file :download:`sv40 in fasta <_static/code/sv40.py>` which is a python file with the sequence of sv40 in fasta format
+already embeded, and use python -i sv40.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)
+
+::
+
+ >>> # remove the header
+ >>> sv40_sequence = sv40[61:]
+ >>> # remove the carriagge return character \n
+ >>> sv40_sequence = sv40_sequence.replace('\n' , '')
+ >>> # then compute the lenght
+ >>> len(sv40_sequence)
+ 5243
+
+Is that the following enzymes:
+* BamHI (ggatcc),
+* EcorI (gaattc),
+* HindIII (aagctt),
+* SmaI (cccggg)
+have recogition sites in sv40? ::
+
+ >>> "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.
+
+
+Exercise
+--------
+
+We want to perform a PCR on sv40, can you give the length and the sequence of the amplicon?
+to simplify the 2 primers are given in 5'3'
+| CGGGACTATGGTTGCTGACT
+| TCTTTCCGCCTCAGAAGGTA
+(write the peudocode before coding)
+
+| find the position of the first primer in sv40
+| find the position of the 2nd primer in sv40
+| # the position of primer are position of the begining of the primer
+| lenght of amplicon = position 2nd primer + len(2nd primer) - position 1rst primer
+
+::
+
+ first_primer = "CGGGACTATGGTTGCTGACT".lower()
+ second_primer = "TCTTTCCGCCTCAGAAGGTA".lower()
+ pos_1 = sv40_sequence.find(first_primer)
+ pos_2 = sv40_sequence.find(second_primer)
+ amplicon_len = pos_2 + len(second_primer) - pos_1
+ print amplicon_len
+ 199
+
+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 obect 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.
+
+Exercise
+--------
+
+| il2_human = 'MYRMQLLSCIALSLALVTNSAPTSSSTKKTQLQLEHLLLDLQMILNGINNYKNPKLTRMLTFKFYMPKKATELKHLQCLEEELKPLEEVLNLAQSKNFHLRPRDLISNINVIVLELKGSETTFMCEYADETATIVEFLNRWITFCQSIISTLT'
+| 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 pseudo code, before to propose an implementation:
+
+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
+
+::
+
+ 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
+
+Exercise
+--------
+
+# use again the sv40 sequence and compute the gc%
+# generate a "micro" report like this 'the sv40 is 5243 bp lenght and have 40.80% gc'
+
+::
+
+ 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)
\ No newline at end of file
diff --git a/source/Variables.rst b/source/Variables.rst
index 7c7ab3eff342e962d0de68a103193592ad0f191a..d4037b7019f545bc8af484766b7ca455a729684d 100644
--- a/source/Variables.rst
+++ b/source/Variables.rst
@@ -8,9 +8,9 @@ Variables, Expression and statements
************************************
-Exercices
+Exercises
=========
-exercice
+Exercise
--------
blabla
\ No newline at end of file
diff --git a/source/_static/code/sv40.py b/source/_static/code/sv40.py
new file mode 100644
index 0000000000000000000000000000000000000000..c20535d15121890063fc76d6a9e351ac73a211c2
--- /dev/null
+++ b/source/_static/code/sv40.py
@@ -0,0 +1,78 @@
+
+sv40 = """>gi|965480|gb|J02400.1|SV4CG Simian virus 40 complete genome
+GCCTCGGCCTCTGCATAAATAAAAAAAATTAGTCAGCCATGGGGCGGAGAATGGGCGGAACTGGGCGGAG
+TTAGGGGCGGGATGGGCGGAGTTAGGGGCGGGACTATGGTTGCTGACTAATTGAGATGCATGCTTTGCAT
+ACTTCTGCCTGCTGGGGAGCCTGGGGACTTTCCACACCTGGTTGCTGACTAATTGAGATGCATGCTTTGC
+ATACTTCTGCCTGCTGGGGAGCCTGGGGACTTTCCACACCCTAACTGACACACATTCCACAGCTGGTTCT
+TTCCGCCTCAGAAGGTACCTAACCAAGTTCCTCTTTCAGAGGTTATTTCAGGCCATGGTGCTGCGCCGGC
+TGTCACGCCAGGCCTCCGTTAAGGTTCGTAGGTCATGGACTGAAAGTAAAAAAACAGCTCAACGCCTTTT
+TGTGTTTGTTTTAGAGCTTTTGCTGCAATTTTGTGAAGGGGAAGATACTGTTGACGGGAAACGCAAAAAA
+CCAGAAAGGTTAACTGAAAAACCAGAAAGTTAACTGGTAAGTTTAGTCTTTTTGTCTTTTATTTCAGGTC
+CATGGGTGCTGCTTTAACACTGTTGGGGGACCTAATTGCTACTGTGTCTGAAGCTGCTGCTGCTACTGGA
+TTTTCAGTAGCTGAAATTGCTGCTGGAGAGGCCGCTGCTGCAATTGAAGTGCAACTTGCATCTGTTGCTA
+CTGTTGAAGGCCTAACAACCTCTGAGGCAATTGCTGCTATAGGCCTCACTCCACAGGCCTATGCTGTGAT
+ATCTGGGGCTCCTGCTGCTATAGCTGGATTTGCAGCTTTACTGCAAACTGTGACTGGTGTGAGCGCTGTT
+GCTCAAGTGGGGTATAGATTTTTTAGTGACTGGGATCACAAAGTTTCTACTGTTGGTTTATATCAACAAC
+CAGGAATGGCTGTAGATTTGTATAGGCCAGATGATTACTATGATATTTTATTTCCTGGAGTACAAACCTT
+TGTTCACAGTGTTCAGTATCTTGACCCCAGACATTGGGGTCCAACACTTTTTAATGCCATTTCTCAAGCT
+TTTTGGCGTGTAATACAAAATGACATTCCTAGGCTCACCTCACAGGAGCTTGAAAGAAGAACCCAAAGAT
+ATTTAAGGGACAGTTTGGCAAGGTTTTTAGAGGAAACTACTTGGACAGTAATTAATGCTCCTGTTAATTG
+GTATAACTCTTTACAAGATTACTACTCTACTTTGTCTCCCATTAGGCCTACAATGGTGAGACAAGTAGCC
+AACAGGGAAGGGTTGCAAATATCATTTGGGCACACCTATGATAATATTGATGAAGCAGACAGTATTCAGC
+AAGTAACTGAGAGGTGGGAAGCTCAAAGCCAAAGTCCTAATGTGCAGTCAGGTGAATTTATTGAAAAATT
+TGAGGCTCCTGGTGGTGCAAATCAAAGAACTGCTCCTCAGTGGATGTTGCCTTTACTTCTAGGCCTGTAC
+GGAAGTGTTACTTCTGCTCTAAAAGCTTATGAAGATGGCCCCAACAAAAAGAAAAGGAAGTTGTCCAGGG
+GCAGCTCCCAAAAAACCAAAGGAACCAGTGCAAGTGCCAAAGCTCGTCATAAAAGGAGGAATAGAAGTTC
+TAGGAGTTAAAACTGGAGTAGACAGCTTCACTGAGGTGGAGTGCTTTTTAAATCCTCAAATGGGCAATCC
+TGATGAACATCAAAAAGGCTTAAGTAAAAGCTTAGCAGCTGAAAAACAGTTTACAGATGACTCTCCAGAC
+AAAGAACAACTGCCTTGCTACAGTGTGGCTAGAATTCCTTTGCCTAATTTAAATGAGGACTTAACCTGTG
+GAAATATTTTGATGTGGGAAGCTGTTACTGTTAAAACTGAGGTTATTGGGGTAACTGCTATGTTAAACTT
+GCATTCAGGGACACAAAAAACTCATGAAAATGGTGCTGGAAAACCCATTCAAGGGTCAAATTTTCATTTT
+TTTGCTGTTGGTGGGGAACCTTTGGAGCTGCAGGGTGTGTTAGCAAACTACAGGACCAAATATCCTGCTC
+AAACTGTAACCCCAAAAAATGCTACAGTTGACAGTCAGCAGATGAACACTGACCACAAGGCTGTTTTGGA
+TAAGGATAATGCTTATCCAGTGGAGTGCTGGGTTCCTGATCCAAGTAAAAATGAAAACACTAGATATTTT
+GGAACCTACACAGGTGGGGAAAATGTGCCTCCTGTTTTGCACATTACTAACACAGCAACCACAGTGCTTC
+TTGATGAGCAGGGTGTTGGGCCCTTGTGCAAAGCTGACAGCTTGTATGTTTCTGCTGTTGACATTTGTGG
+GCTGTTTACCAACACTTCTGGAACACAGCAGTGGAAGGGACTTCCCAGATATTTTAAAATTACCCTTAGA
+AAGCGGTCTGTGAAAAACCCCTACCCAATTTCCTTTTTGTTAAGTGACCTAATTAACAGGAGGACACAGA
+GGGTGGATGGGCAGCCTATGATTGGAATGTCCTCTCAAGTAGAGGAGGTTAGGGTTTATGAGGACACAGA
+GGAGCTTCCTGGGGATCCAGACATGATAAGATACATTGATGAGTTTGGACAAACCACAACTAGAATGCAG
+TGAAAAAAATGCTTTATTTGTGAAATTTGTGATGCTATTGCTTTATTTGTAACCATTATAAGCTGCAATA
+AACAAGTTAACAACAACAATTGCATTCATTTTATGTTTCAGGTTCAGGGGGAGGTGTGGGAGGTTTTTTA
+AAGCAAGTAAAACCTCTACAAATGTGGTATGGCTGATTATGATCATGAACAGACTGTGAGGACTGAGGGG
+CCTGAAATGAGCCTTGGGACTGTGAATCAATGCCTGTTTCATGCCCTGAGTCTTCCATGTTCTTCTCCCC
+ACCATCTTCATTTTTATCAGCATTTTCCTGGCTGTCTTCATCATCATCATCACTGTTTCTTAGCCAATCT
+AAAACTCCAATTCCCATAGCCACATTAAACTTCATTTTTTGATACACTGACAAACTAAACTCTTTGTCCA
+ATCTCTCTTTCCACTCCACAATTCTGCTCTGAATACTTTGAGCAAACTCAGCCACAGGTCTGTACCAAAT
+TAACATAAGAAGCAAAGCAATGCCACTTTGAATTATTCTCTTTTCTAACAAAAACTCACTGCGTTCCAGG
+CAATGCTTTAAATAATCTTTGGGCCTAAAATCTATTTGTTTTACAAATCTGGCCTGCAGTGTTTTAGGCA
+CACTGTACTCATTCATGGTGACTATTCCAGGGGGAAATATTTGAGTTCTTTTATTTAGGTGTTTCTTTTC
+TAAGTTTACCTTAACACTGCCATCCAAATAATCCCTTAAATTGTCCAGGTTATTAATTCCCTGACCTGAA
+GGCAAATCTCTGGACTCCCCTCCAGTGCCCTTTACATCCTCAAAAACTACTAAAAACTGGTCAATAGCTA
+CTCCTAGCTCAAAGTTCAGCCTGTCCAAGGGCAAATTAACATTTAAAGCTTTCCCCCCACATAATTCAAG
+CAAAGCAGCTGCTAATGTAGTTTTACCACTATCAATTGGTCCTTTAAACAGCCAGTATCTTTTTTTAGGA
+ATGTTGTACACCATGCATTTTAAAAAGTCATACACCACTGAATCCATTTTGGGCAACAAACAGTGTAGCC
+AAGCAACTCCAGCCATCCATTCTTCTATGTCAGCAGAGCCTGTAGAACCAAACATTATATCCATCCTATC
+CAAAAGATCATTAAATCTGTTTGTTAACATTTGTTCTCTAGTTAATTGTAGGCTATCAACCCGCTTTTTA
+GCTAAAACAGTATCAACAGCCTGTTGGCATATGGTTTTTTGGTTTTTGCTGTCAGCAAATATAGCAGCAT
+TTGCATAATGCTTTTCATGGTACTTATAGTGGCTGGGCTGTTCTTTTTTAATACATTTTAAACACATTTC
+AAAACTGTACTGAAATTCCAAGTACATCCCAAGCAATAACAACACATCATCACATTTTGTTTCCATTGCA
+TACTCTGTTACAAGCTTCCAGGACACTTGTTTAGTTTCCTCTGCTTCTTCTGGATTAAAATCATGCTCCT
+TTAACCCACCTGGCAAACTTTCCTCAATAACAGAAAATGGATCTCTAGTCAAGGCACTATACATCAAATA
+TTCCTTATTAACCCCTTTACAAATTAAAAAGCTAAAGGTACACAATTTTTGAGCATAGTTATTAATAGCA
+GACACTCTATGCCTGTGTGGAGTAAGAAAAAACAGTATGTTATGATTATAACTGTTATGCCTACTTATAA
+AGGTTACAGAATATTTTTCCATAATTTTCTTGTATAGCAGTGCAGCTTTTTCCTTTGTGGTGTAAATAGC
+AAAGCAAGCAAGAGTTCTATTACTAAACACAGCATGACTCAAAAAACTTAGCAATTCTGAAGGAAAGTCC
+TTGGGGTCTTCTACCTTTCTCTTCTTTTTTGGAGGAGTAGAATGTTGAGAGTCAGCAGTAGCCTCATCAT
+CACTAGATGGCATTTCTTCTGAGCAAAACAGGTTTTCCTCATTAAAGGCATTCCACCACTGCTCCCATTC
+ATCAGTTCCATAGGTTGGAATCTAAAATACACAAACAATTAGAATCAGTAGTTTAACACATTATACACTT
+AAAAATTTTATATTTACCTTAGAGCTTTAAATCTCTGTAGGTAGTTTGTCCAATTATGTCACACCACAGA
+AGTAAGGTTCCTTCACAAAGATCAAGTCCAAACCACATTCTAAAGCAATCGAAGCAGTAGCAATCAACCC
+ACACAAGTGGATCTTTCCTGTATAATTTTCTATTTTCATGCTTCATCCTCAGTAAGCACAGCAAGCATAT
+GCAGTTAGCAGACATTTTCTTTGCACACTCAGGCCATTGTTTGCAGTACATTGCATCAACACCAGGATTT
+AAGGAAGAAGCAAATACCTCAGTTGCATCCCAGAAGCCTCCAAAGTCAGGTTGATGAGCATATTTTACTC
+CATCTTCCATTTTCTTGTACAGAGTATTCATTTTCTTCATTTTTTCTTCATCTCCTCCTTTATCAGGATG
+AAACTCCTTGCATTTTTTTAAATATGCCTTTCTCATCAGAGGAATATTCCCCCAGGCACTCCTTTCAAGA
+CCTAGAAGGTCCATTAGCTGCAAAGATTCCTCTCTGTTTAAAACTTTATCCATCTTTGCAAAGCTTTTTG
+CAAAAGCCTAGGCCTCCAAAAAAGCCTCCTCACTACTTCTGGAATAGCTCAGAGGCCGAGGCG"""
+
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