Bklein7 Week 2

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Electronic Lab Notebook

[Note: After class on 9/17, I copied over this section directly from my user page. Pardon any redundancy. For future weeks, I will merge the electronic lab notebook entries with my individual journal entires.]

  • Write out the complementary strand of DNA below the strand shown and be sure to label the 5’ and 3’ ends of the complementary strand.
    • In writing the complimentary strand...
      1. Begin with the 3' and end with the 5' labels, as this strand will run antiparallel to the existing DNA strand.
      2. Use the rules of complimentary base pairing for DNA (A pairs with T; C pairs with G) to write the complimentary nucleotide sequence.
      3. The final product can be seen below:
5’-cgtatgctaataccatgttccgcgtataacccagccgccagttccgctggcggcatttta-3’
3'-gcatacgattatggtacaaggcgcatattgggtcggcggtcaaggcgaccgccgtaaaat-5'
  • Using the genetic code, translate all possible reading frames of this DNA sequence.
    • In order to translate the DNA sequence, we must first write out the mRNA transcripts of each DNA strand:
      1. Simply copy the DNA sequences, but replace every "T" in the sequence with a "U".
      2. The final product can be seen below:
5’-cguaugcuaauaccauguuccgcguauaacccagccgccaguuccgcuggcggcauuuua-3’
3'-gcauacgauuaugguacaaggcgcauauugggucggcggucaaggcgaccgccguaaaau-5'
  • There are six possible reading frames that can be used to translate these mRNA strands.
    1. For the +1 reading frame, take the top strand (which is read 5' to 3' as is) and divide it into trinucleotides-treating the first three nucleotides as the first codon
      • +1: 5’-cgu aug cua aua cca ugu ucc gcg uau aac cca gcc gcc agu ucc gcu ggc ggc auu uua-3’
      • Next, use the genetic code to translate each trinucleotide into an amino acid. In writing the polypeptide, begin with the N-terminus and end with the C-terminus. The result is as follows: N-ter-R M L I P C S A Y N P A A S S A G G I L-C-ter
    2. For the +2 reading frame, ignore the first nucleotide in the sequence, and then divide the remaining sequence into trinucleotides
      • +2: 5’-c gua ugc uaa uac cau guu ccg cgu aua acc cag ccg cca guu ccg cug gcg gca uuu ua-3’
      • Translate the polypeptide, stopping transcription when a "stop codon" is reached: N-ter-V C (stop)-C-ter
    3. For the +3 reading frame, ignore the first two nucleotides in the sequence, and then divide the remaining sequence into trinucleotides
      • +3: 5’-cg uau gcu aau acc aug uuc cgc gua uaa ccc agc cgc cag uuc cgc ugg cgg cau uuu a-3’
      • Translation: N-ter-Y A N T M F R V (stop)-C-ter
    4. For the -1 reading frame, begin by reversing the bottom mRNA strand so that it reads 5' to 3'. Then divide it into trinucleotides:
      • -1: 5'-uaa aau gcc gcc agc gga acu ggc ggc ugg guu aua cgc gga aca ugg uau uag cau acg-3'
      • Translation: No protein synthesized; first codon within this reading frame is a stop codon
    5. For the -2 reading frame, redivide the -1 reading frame so that the first nucleotide in the sequence is ignored:
      • -2: 5'-u aaa aug ccg cca gcg gaa cug gcg gcu ggg uua uac gcg gaa cau ggu auu agc aua cg-3'
      • Translation: N-ter-K M P P A E L A A G L Y A E H G I S I-C-ter
    6. For the -3 reading frame, redivide the -2 reading frame so that the first two nucleotides in the sequence are ignored:
      • -3: 5'-ua aaa ugc cgc cag cgg aac ugg cgg cug ggu uau acg cgg aac aug gua uua gca uac g-3'
      • Translation: N-ter-K C R Q R N W R L G Y T R N M V L A Y-C-ter
  • Which of the reading frames (if any) of the reading frames you translated is an open reading frame, i.e., does not contain a stop codon?
    • The +1, -2, and -3 reading frames did not contain any stop codons. Therefore, these reading frames are considered "open".

Assignment Answers

The Genetic Code

DNA & mRNA Sequences

  • DNA Strands
5’-cgtatgctaataccatgttccgcgtataacccagccgccagttccgctggcggcatttta-3’
3'-gcatacgattatggtacaaggcgcatattgggtcggcggtcaaggcgaccgccgtaaaat-5'
  • mRNA Strands
5’-cguaugcuaauaccauguuccgcguauaacccagccgccaguuccgcuggcggcauuuua-3’
3'-gcauacgauuaugguacaaggcgcauauugggucggcggucaaggcgaccgccguaaaau-5'

Reading Frames

  • +1
mRNA: 5’-cgu aug cua aua cca ugu ucc gcg uau aac cca gcc gcc agu ucc gcu ggc ggc auu uua-3’
Translation: N-ter-R M L I P C S A Y N P A A S S A G G I L-C-ter
  • +2
mRNA: 5’-c gua ugc uaa uac cau guu ccg cgu aua acc cag ccg cca guu ccg cug gcg gca uuu ua-3’
Translation: N-ter-V C (stop)-C-ter
  • +3
mRNA: 5’-cg uau gcu aau acc aug uuc cgc gua uaa ccc agc cgc cag uuc cgc ugg cgg cau uuu a-3’
Translation: N-ter-Y A N T M F R V (stop)-C-ter
  • -1
mRNA: 5'-uaa aau gcc gcc agc gga acu ggc ggc ugg guu aua cgc gga aca ugg uau uag cau acg-3'
Translation: No protein synthesized; first codon within this reading frame is a stop codon
  • -2 
mRNA: 5'-u aaa aug ccg cca gcg gaa cug gcg gcu ggg uua uac gcg gaa cau ggu auu agc aua cg-3'
Translation: N-ter-K M P P A E L A A G L Y A E H G I S I-C-ter
  • -3
mRNA: 5'-ua aaa ugc cgc cag cgg aac ugg cgg cug ggu uau acg cgg aac aug gua uua gca uac g-3'
Translation: N-ter-K C R Q R N W R L G Y T R N M V L A Y-C-ter
  • The following reading frames were open: +1, -2, and -3.

Links

Assignments Pages

Individual Journal Entries

Shared Journal Entries

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