add solutions to Data_Types and Collection_Data_Types exercises

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 : ::

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

source/Variables.rst

@@ -8,9 +8,9 @@ Variables, Expression and statements
 ************************************
 
 
-Exercices
+Exercises
 =========
 
-exercice
+Exercise
 --------
 blabla
\ No newline at end of file

source/_static/code/sv40.py

+
+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"""
+