Skip to content

Latest commit

 

History

History
145 lines (115 loc) · 14.2 KB

Python_03_problemset.md

File metadata and controls

145 lines (115 loc) · 14.2 KB

Python 3 Problem Set -- Strings

  1. What data types are considered sequences in Python?
  2. What is the length of the following DNA string? Is this DNA string a Python sequence?
GATGGGATTGGGGTTTTCCCCTCCCATGTGCTCAAGACTGGCGCTAAAAGTTTTGAGCTTCTCAAAAGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCGGGGACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGCCCCCTCTGAGTCAGGAAACATTTTCAGACCTATGGAAACTACTTCCTGAAAACAACGTTCTGTCCCCCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTCCCCGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTCCCCCCGTGGCCCCTGCACCAGCAGCTCCTACACCGGCGGCCCCTGCACCAGCCCCCTCCTGGCCCCTGTCATCTTCTGTCCCTTCCCAGAAAACCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTCCCCTGCCCTCAACAAGATGTTTTGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACACCCCCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGCCCCCACCATGAGCGCTGCTCAGATAGCGATGGTCTGGCCCCTCCTCAGCATCTTATCCGAGTGGAAGGAAATTTGCGTGTGGAGTATTTGGATGACAGAAACACTTTTCG

  1. Make sure to commit your changes along the way. You can wait until the end to push them to your remote repo, or you can do it now. It is probably smart to commit after each problem set question.

  2. In the interpreter:

    • Create a variable named 'DNA' which contains the sequence above.
    • Count the number of A's
    • Count the number of T's
    • Count the number of G's
    • Count the number of C's

  3. In the interpreter:

    • Create a variable named 'bird' with the contents 'chicken'
    • Convert the contents of 'bird' to be uppercase and print
    • Exit the interpreter

  4. Create a script with vi that counts the number of A's, T's, C's, and G's regardless of case: a. Create a short string with a known composition of nucleotides, for example, 'ATTGGGCCCC' A=1, T=2, G=3, C=4. This is a postive control. b. Test your script by running with the postive control. c. Run with this unknown/experimental sequence:

GATGGGATTggggttttccccTCCCATGTGCTCAAGACTGGCGCTaaaaGttttGAGCTTCTCaaaaGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCggggACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGccccCTCTGAGTCAGGAAACAttttCAGACCTATGGAAACTACTTCCTGaaaaCAACGTTCTGTccccCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTccccGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTccccCCGTGGccccTGCACCAGCAGCTCCTACACCGGCGGccccTGCACCAGccccCTCCTGGccccTGTCATCTTCTGTCCCTTCCCAGaaaaCCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTccccTGCCCTCAACAAGATGttttGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACAccccCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGccccCACCATGAGCGCTGCTCAGATAGCGATGGTCTGGccccTCCTCAGCATCTTATCCGAGTGGAAGGAAATTTGCGTGTGGAGTATTTGGATGACAGAAACACttttCG
  1. Find and replace all instances of 'T' with 'U' in this DNA sequence.
  • Start with this small test sequence (i.e., postive control): ATGCATGC
  • The test should return AUGCUGC
  • Now run your code with this DNA sequence:
GATGGGATTGGGGTTTTCCCCTCCCATGTGCTCAAGACTGGCGCTAAAAGTTTTGAGCTTCTCAAAAGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCGGGGACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGCCCCCTCTGAGTCAGGAAACATTTTCAGACCTATGGAAACTACTTCCTGAAAACAACGTTCTGTCCCCCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTCCCCGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTCCCCCCGTGGCCCCTGCACCAGCAGCTCCTACACCGGCGGCCCCTGCACCAGCCCCCTCCTGGCCCCTGTCATCTTCTGTCCCTTCCCAGAAAACCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTCCCCTGCCCTCAACAAGATGTTTTGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACACCCCCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGCCCCCACCATGAGCGCTGCTCAGATAGCGATGGTCTGGCCCCTCCTCAGCATCTTATCCGAGTGGAAGGAAATTTGCGTGTGGAGTATTTGGATGACAGAAACACTTTTCG

the result should be:

GAUGGGAUUGGGGUUUUCCCCUCCCAUGUGCUCAAGACUGGCGCUAAAAGUUUUGAGCUUCUCAAAAGUCUAGAGCCACCGUCCAGGGAGCAGGUAGCUGCUGGGCUCCGGGGACACUUUGCGUUCGGGCUGGGAGCGUGCUUUCCACGACGGUGACACGCUUCCCUGGAUUGGCAGCCAGACUGCCUUCCGGGUCACUGCCAUGGAGGAGCCGCAGUCAGAUCCUAGCGUCGAGCCCCCUCUGAGUCAGGAAACAUUUUCAGACCUAUGGAAACUACUUCCUGAAAACAACGUUCUGUCCCCCUUGCCGUCCCAAGCAAUGGAUGAUUUGAUGCUGUCCCCGGACGAUAUUGAACAAUGGUUCACUGAAGACCCAGGUCCAGAUGAAGCUCCCAGAAUUCGCCAGAGGCUGCUCCCCCCGUGGCCCCUGCACCAGCAGCUCCUACACCGGCGGCCCCUGCACCAGCCCCCUCCUGGCCCCUGUCAUCUUCUGUCCCUUCCCAGAAAACCUACCAGGGCAGCUACGGUUUCCGUCUGGGCUUCUUGCAUUCUGGGACAGCCAAGUCUGUGACUUGCACGUACUCCCCUGCCCUCAACAAGAUGUUUUGCCAACUGGCCAAGACCUGCCCUGUGCAGCUGUGGGUUGAUUCCACACCCCCGCCCGGCACCCGCGUCCGCGCCAUGGCCAUCUACAAGCAGUCACAGCACAUGACGGAGGUUGUGAGGCGCUGCCCCCACCAUGAGCGCUGCUCAGAUAGCGAUGGUCUGGCCCCUCCUCAGCAUCUUAUCCGAGUGGAAGGAAAUUUGCGUGUGGAGUAUUUGGAUGACAGAAACACUUUUCG
  1. Find and replace all instances of 'T' with 'U' in this DNA sequence. Notice that there are both uppercase and lowercase characters. Start with a small test sequence that contains both upper and lower cased characters ('AtGcaTgC'). Confirm that you are generating the correct resulting sequence ('AUGCAUGC').
GATGGGATTggggttttccccTCCCATGTGCTCAAGACTGGCGCTaaaaGttttGAGCTTCTCaaaaGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCggggACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGccccCTCTGAGTCAGGAAACAttttCAGACCTATGGAAACTACTTCCTGaaaaCAACGTTCTGTccccCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTccccGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTccccCCGTGGccccTGCACCAGCAGCTCCTACACCGGCGGccccTGCACCAGccccCTCCTGGccccTGTCATCTTCTGTCCCTTCCCAGaaaaCCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTccccTGCCCTCAACAAGATGttttGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACAccccCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGccccCACCATGAGCGCT

the result should be:

GAUGGGAUUGGGGUUUUCCCCUCCCAUGUGCUCAAGACUGGCGCUAAAAGUUUUGAGCUUCUCAAAAGUCUAGAGCCACCGUCCAGGGAGCAGGUAGCUGCUGGGCUCCGGGGACACUUUGCGUUCGGGCUGGGAGCGUGCUUUCCACGACGGUGACACGCUUCCCUGGAUUGGCAGCCAGACUGCCUUCCGGGUCACUGCCAUGGAGGAGCCGCAGUCAGAUCCUAGCGUCGAGCCCCCUCUGAGUCAGGAAACAUUUUCAGACCUAUGGAAACUACUUCCUGAAAACAACGUUCUGUCCCCCUUGCCGUCCCAAGCAAUGGAUGAUUUGAUGCUGUCCCCGGACGAUAUUGAACAAUGGUUCACUGAAGACCCAGGUCCAGAUGAAGCUCCCAGAAUUCGCCAGAGGCUGCUCCCCCCGUGGCCCCUGCACCAGCAGCUCCUACACCGGCGGCCCCUGCACCAGCCCCCUCCUGGCCCCUGUCAUCUUCUGUCCCUUCCCAGAAAACCUACCAGGGCAGCUACGGUUUCCGUCUGGGCUUCUUGCAUUCUGGGACAGCCAAGUCUGUGACUUGCACGUACUCCCCUGCCCUCAACAAGAUGUUUUGCCAACUGGCCAAGACCUGCCCUGUGCAGCUGUGGGUUGAUUCCACACCCCCGCCCGGCACCCGCGUCCGCGCCAUGGCCAUCUACAAGCAGUCACAGCACAUGACGGAGGUUGUGAGGCGCUGCCCCCACCAUGAGCGCU
  1. Write a script that calculates the AT content in the DNA string below.
    AT content is the proportion of bases that are either A or T.
    It is ALWAYS a good idea to test your code with test data.
    For example, the below sequence is long and you don't know for sure how many As and Ts are present.
    Test your code with a DNA string that you KNOW the correct answer.
    In 'AATTGGCCA' you know you have 3 As and 2 Ts.
    Add code to calculate the GC content in addition to the AT content.
GATGGGATTGGGGTTTTCCCCTCCCATGTGCTCAAGACTGGCGCTAAAAGTTTTGAGCTTCTCAAAAGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCGGGGACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGCCCCCTCTGAGTCAGGAAACATTTTCAGACCTATGGAAACTACTTCCTGAAAACAACGTTCTGTCCCCCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTCCCCGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTCCCCCCGTGGCCCCTGCACCAGCAGCTCCTACACCGGCGGCCCCTGCACCAGCCCCCTCCTGGCCCCTGTCATCTTCTGTCCCTTCCCAGAAAACCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTCCCCTGCCCTCAACAAGATGTTTTGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACACCCCCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGCCCCCACCATGAGCGCTGCTCAGATAGCGATGGTCTGGCCCCTCCTCAGCATCTTATCCGAGTGGAAGGAAATTTGCGTGTGGAGTATTTGGATGACAGAAACACTTTTCG

Verify your code is correctly reporting AT and GC content:

The above sequence has these counts:

nt count
T 187
G 218
A 167
C 270
content type %
AT content 42%
GC content 58%
  1. Extract and print the substring from nucleotide postion 100 (not the same as its index) to nucleotide position 200 in this DNA sequence. Start with a small test sequence to make sure you are getting the correct output.
GATGGGATTGGGGTTTTCCCCTCCCATGTGCTCAAGACTGGCGCTAAAAGTTTTGAGCTTCTCAAAAGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCGGGGACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGCCCCCTCTGAGTCAGGAAACATTTTCAGACCTATGGAAACTACTTCCTGAAAACAACGTTCTGTCCCCCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTCCCCGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTCCCCCCGTGGCCCCTGCACCAGCAGCTCCTACACCGGCGGCCCCTGCACCAGCCCCCTCCTGGCCCCTGTCATCTTCTGTCCCTTCCCAGAAAACCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTCCCCTGCCCTCAACAAGATGTTTTGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACACCCCCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGCCCCCACCATGAGCGCTGCTCAGATAGCGATGGTCTGGCCCCTCCTCAGCATCTTATCCGAGTGGAAGGAAATTTGCGTGTGGAGTATTTGGATGACAGAAACACTTTTCG

The resulting sub-sequence should be:

GCTGGGCTCCGGGGACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTG

Test other sequence inputs and the resulting subsequences with the Sequence Manipulation Site

  1. Now use this DNA sequence and calculate the GC content in your substring (from 100 to 200, like above) but it should be regardless of case:
GATGGGATTggggttttccccTCCCATGTGCTCAAGACTGGCGCTaaaaGttttGAGCTTCTCaaaaGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCggggACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGccccCTCTGAGTCAGGAAACAttttCAGACCTATGGAAACTACTTCCTGaaaaCAACGTTCTGTccccCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTccccGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTccccCCGTGGccccTGCACCAGCAGCTCCTACACCGGCGGccccTGCACCAGccccCTCCTGGccccTGTCATCTTCTGTCCCTTCCCAGaaaaCCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTccccTGCCCTCAACAAGATGttttGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACAccccCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGccccCACCATGAGCGCT

The GC content for the correct subsequence is 65%

  1. Reverse Complements:

Reverse Complement Overview

A sequence and its reverse complement represent the two strands of double stranded DNA or RNA. To calculate the reverse complement of DNA the nucleotides need to be translated into the complementary nucleotide, A -> T, T->A, G->C, and C->G, and the entire sequence needs to be reversed.

To read more about reverse complements and to test your output see Qiagen's page on reverse complements

 Original Sequence  5'ATGCAGGGGAAACATGATTCAGGAC 3'  
 Complement         3'TACGTCCCCTTTGTACTAAGTCCTG 5'    
 Reverse Complement 5'GTCCTGAATCATGTTTCCCCTGCAT 3'

Write a script to generate and print the reverse complements of a DNA sequence. Hint for reverse. Use string formating for printing.
Use this sequence:

GATGGGATTggggttttccccTCCCATGTGCTCAAGACTGGCGCTaaaaGttttGAGCTTCTCaaaaGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCggggACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGccccCTCTGAGTCAGGAAACAttttCAGACCTATGGAAACTACTTCCTGaaaaCAACGTTCTGTccccCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTccccGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTccccCCGTGGccccTGCACCAGCAGCTCCTACACCGGCGGccccTGCACCAGccccCTCCTGGccccTGTCATCTTCTGTCCCTTCCCAGaaaaCCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTccccTGCCCTCAACAAGATGttttGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACAccccCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGccccCACCATGAGCGCT

The resulting reverse complement sequence should be:

AGCGCTCATGGTGGGGGCAGCGCCTCACAACCTCCGTCATGTGCTGTGACTGCTTGTAGATGGCCATGGCGCGGACGCGGGTGCCGGGCGGGGGTGTGGAATCAACCCACAGCTGCACAGGGCAGGTCTTGGCCAGTTGGCAAAACATCTTGTTGAGGGCAGGGGAGTACGTGCAAGTCACAGACTTGGCTGTCCCAGAATGCAAGAAGCCCAGACGGAAACCGTAGCTGCCCTGGTAGGTTTTCTGGGAAGGGACAGAAGATGACAGGGGCCAGGAGGGGGCTGGTGCAGGGGCCGCCGGTGTAGGAGCTGCTGGTGCAGGGGCCACGGGGGGAGCAGCCTCTGGCGAATTCTGGGAGCTTCATCTGGACCTGGGTCTTCAGTGAACCATTGTTCAATATCGTCCGGGGACAGCATCAAATCATCCATTGCTTGGGACGGCAAGGGGGACAGAACGTTGTTTTCAGGAAGTAGTTTCCATAGGTCTGAAAATGTTTCCTGACTCAGAGGGGGCTCGACGCTAGGATCTGACTGCGGCTCCTCCATGGCAGTGACCCGGAAGGCAGTCTGGCTGCCAATCCAGGGAAGCGTGTCACCGTCGTGGAAAGCACGCTCCCAGCCCGAACGCAAAGTGTCCCCGGAGCCCAGCAGCTACCTGCTCCCTGGACGGTGGCTCTAGACTTTTGAGAAGCTCAAAACTTTTAGCGCCAGTCTTGAGCACATGGGAGGGGAAAACCCCAATCCCATC
  1. Write a script to find the starting nucleotide position of an EcoRI GAATTC site in the below DNA sequence. Remember DNA sequences start with a 1 and a python string starts with an index of 0. Run with test data first.
GATGGGATTGGGGTTTTCCCCTCCCATGTGCTCAAGACTGGCGCTAAAAGTTTTGAGCTTCTCAAAAGTCTAGAGCCACCGTCCAGGGAGCAGGTAGCTGCTGGGCTCCGGGGACACTTTGCGTTCGGGCTGGGAGCGTGCTTTCCACGACGGTGACACGCTTCCCTGGATTGGCAGCCAGACTGCCTTCCGGGTCACTGCCATGGAGGAGCCGCAGTCAGATCCTAGCGTCGAGCCCCCTCTGAGTCAGGAAACATTTTCAGACCTATGGAAACTACTTCCTGAAAACAACGTTCTGTCCCCCTTGCCGTCCCAAGCAATGGATGATTTGATGCTGTCCCCGGACGATATTGAACAATGGTTCACTGAAGACCCAGGTCCAGATGAAGCTCCCAGAATTCGCCAGAGGCTGCTCCCCCCGTGGCCCCTGCACCAGCAGCTCCTACACCGGCGGCCCCTGCACCAGCCCCCTCCTGGCCCCTGTCATCTTCTGTCCCTTCCCAGAAAACCTACCAGGGCAGCTACGGTTTCCGTCTGGGCTTCTTGCATTCTGGGACAGCCAAGTCTGTGACTTGCACGTACTCCCCTGCCCTCAACAAGATGTTTTGCCAACTGGCCAAGACCTGCCCTGTGCAGCTGTGGGTTGATTCCACACCCCCGCCCGGCACCCGCGTCCGCGCCATGGCCATCTACAAGCAGTCACAGCACATGACGGAGGTTGTGAGGCGCTGCCCCCACCATGAGCGCTGCTCAGATAGCGATGGTCTGGCCCCTCCTCAGCATCTTATCCGAGTGGAAGGAAATTTGCGTGTGGAGTATTTGGATGACAGAAACACTTTTCG
  • Also find the ending nucleotide position of the EcoRI site?
  • Use f"" string formating to print out these two values like this:
EcoRI startPos:yourStartPos endPos:yourEndPos

With the above sequence you should get:

startPos:396 endPos:402
  1. ADD/COMMIT/PUSH