The following DNA sequence represents a eukaryotic gene. Indicate which is the t
ID: 87967 • Letter: T
Question
The following DNA sequence represents a eukaryotic gene. Indicate which is the template and which is the coding strand, determine where transcription begins (assume it ends at the end of the sequence presented), and write out the nucleotide sequence of the initial transcript. You need to process the initial transcript to make an mRNA (there is an intron in the gene: using the information about splicing on pp. 306-307 of your textbook, locate and remove the intron) and write out the sequence of the completely processed mRNA. Finally, translate the mRNA and write out the amino acid sequence of the encoded polypeptide. 5'- ACCGGCCTCTCTCTCACACACAGTACGGCGTTTATTGGGACTTCAAATCATAT GGCCATTACAAGCCCTGCAGAAAGGGAAGGCGAAGCTCAGGATAAGTA... 3 ' - TGGCCGGAGAGAGAGTGTGTGTCATGCCGCAAATAACCCTGAAGTTTAGTATA CCGGTAATGTTCGGGACGTCTTTCCCTTCCGCTTCGAGTCCTATTCAT... ..CAGGTGACACTCTACTCACCGGATAAGTATTGTCATCTCCCAAGGGGGTG AATATCCGCGCATCCGCTTTATATCGCCTGATTAACGCAGTCCTACACAAT- 3' ..GTCCACTGTGAGATGAGTGGCCTATTCATAACAGTAGAGGGTTCCCCCAC TTATAGGCGCGTAGGCGAAATATAGCGGACTAATTGCGTCAGGATGTGTTA- 5'Explanation / Answer
A DNA which is a eukaryotic is nature has many unique features, from the sequence given above we are told to perform and identify the "Central Dogma" of life, which is The DNA transcribed to an mRNA and then to a Protein sequence. Here the given coding sequence we can identify from the characters of it.
It has to be start form the 5' end as the sequence always gets replicated from the direction of 5' to 3'. Now looking at both the sequences the matched version of it with a corresponding sequence as the nitrogenous nucleotides A T G and C, to their corresponding to whom they bind is T A C and G.
Looking at the first sequence it starts from 5' and then ACCGGC, just down to that second sequence is the corresponding sequence which matches also, as it is starting from the 3' end.
Then the replication takes place from the 3rd sequence, it starts with the ... which is to show the continuation from the 1st sequence. Here look into the final sequence which corresponds to the 3rd sequence.
So, we claim the 1st and 3rd sequence as the template strand and the 2nd and 4th to be the codon sequence.
Taking this replicated sequence into consideration we can perform the mRNA building up process.
A RNA sequence always have nucleotide bases as the U A G C because it replaces U with T.
So our codon sequence now has to be transcribed, so the matched corresponding sequence would be containing,
A T G C from the DNA and in mRNA it is U A C G.
DNA starts from 5' TGGCCG........GTGTTA 3'
mRNA transcript would be 3' ACCGGCCUCUCUCUCACACACAGUACGGCGUUUAUUGGGACUUCAAAUCAUAUGGCCAUUACAAGCCCUGCAGAAAGGGAAGGCGAAGCUCAGGAUAAGUA......CAGGUGACACUCUACUCACCGGAUAAGUAUUGUCAUCUCCCAAGGGGGUGAAUAUCCGCGCAUGGCGUUUAUAUCGCCUGAUUAACGCAGUCCUACACAAU- 5'
Next step is to translate this sequence, this part is simple as you know that intron sequence has to be removed from here. We just need to take the non-coding sequence out of here, that is to say mRNA code for amino acid in triplicate, like take the first two codons- ACC and GGC, now looking at the amino acids chart we can make out that this two sequence codes for ACC- threonine, GGC- Glycine. Like wise continue and just exclude the intron by looking that which codons codes for nothing.