Difference between revisions of "QLanners Week 2"
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I used a combination of my knowledge of DNA transcription and translation from my previous biology classes, along with a little bit of code to do this assignment. | I used a combination of my knowledge of DNA transcription and translation from my previous biology classes, along with a little bit of code to do this assignment. | ||
− | The | + | The original template strand provided was: |
+ | 5’-cgtatgctaataccatgttccgcgtataacccagccgccagttccgctggcggcatttta-3’ | ||
− | The | + | The most efficient way I thought of doing this assignment was making a class called DNA that had functions within the class to find the DNA strand's complementary strands and protein sequences from the various reading frames. |
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− | class DNA | ||
− | + | The first step was to create a function that I called compStrand to take the input template strand and output the complementary strand. In order to make the code easier, I just took off the 5' and 3' of the string. And then after passing the template DNA strand through the function, I was sure to add the end labels back on, being sure to put them on in the reverse 3' to 5' order. | |
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− | + | The complementary strand was found to be: | |
+ | 3'-gcatacgattatggtacaaggcgcatattgggtcggcggtcaaggcgaccgccgtaaaat-5' | ||
− | + | As for the reading frames portion of this assignment, I created another function, called protein, within the class. This function took one argument which was the location of the start of the reading frame. | |
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− | + | For reading frames +1,+2, and +3, I was able to just use the template DNA strand that was already provided, as this string was oriented in the 5' to 3' direction. | |
− | + | Hence I called the function protein() on this strand using the arguments 1, 2 and 3. A _ in the outputted protein sequence indicated a stop codon, so this was also taken into account, as any codons beyond that point would not be translated. I included the full output from the function called as well in parentheses after the shortened protein sequence. I then added the N and C terminus endings to either end of these output strings. | |
− | + | The results were as follows: | |
− | + | +1 N-RMLIPCSAYNPAASSAGGIL-C | |
− | + | +2 N-VC-C (Full outputted protein sequence from function: N-VC_YHVPRITQPPVPLAAF-C) | |
− | + | +3 N-YANTMFRV-C (Full outputted protein sequence from function: N-YANTMFRV_PSRQFRWRHF-C) | |
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+ | As for the -1, -2, and -3 reading frames, I had to assign the complementary strand computed above to a new variable (called compStrand) and then I used the function compStrand[:: | ||
+ | +3 | ||
==Acknowledgements== | ==Acknowledgements== | ||
Revision as of 06:45, 9 September 2017
Contents
The Genetic Code Assignment
Answers to Assignment
Laboratory Notebook
I used a combination of my knowledge of DNA transcription and translation from my previous biology classes, along with a little bit of code to do this assignment.
The original template strand provided was:
5’-cgtatgctaataccatgttccgcgtataacccagccgccagttccgctggcggcatttta-3’
The most efficient way I thought of doing this assignment was making a class called DNA that had functions within the class to find the DNA strand's complementary strands and protein sequences from the various reading frames.
The first step was to create a function that I called compStrand to take the input template strand and output the complementary strand. In order to make the code easier, I just took off the 5' and 3' of the string. And then after passing the template DNA strand through the function, I was sure to add the end labels back on, being sure to put them on in the reverse 3' to 5' order.
The complementary strand was found to be:
3'-gcatacgattatggtacaaggcgcatattgggtcggcggtcaaggcgaccgccgtaaaat-5'
As for the reading frames portion of this assignment, I created another function, called protein, within the class. This function took one argument which was the location of the start of the reading frame.
For reading frames +1,+2, and +3, I was able to just use the template DNA strand that was already provided, as this string was oriented in the 5' to 3' direction.
Hence I called the function protein() on this strand using the arguments 1, 2 and 3. A _ in the outputted protein sequence indicated a stop codon, so this was also taken into account, as any codons beyond that point would not be translated. I included the full output from the function called as well in parentheses after the shortened protein sequence. I then added the N and C terminus endings to either end of these output strings. The results were as follows:
+1 N-RMLIPCSAYNPAASSAGGIL-C +2 N-VC-C (Full outputted protein sequence from function: N-VC_YHVPRITQPPVPLAAF-C) +3 N-YANTMFRV-C (Full outputted protein sequence from function: N-YANTMFRV_PSRQFRWRHF-C)
As for the -1, -2, and -3 reading frames, I had to assign the complementary strand computed above to a new variable (called compStrand) and then I used the function compStrand[::
+3
Acknowledgements
References
Links
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