Question

1. A sequence of a eukaryotic gene (coding strand) is shown below, RNA polymerase recognizes the sequence ‘TATAAT’ and initiates transcription six nucleotides downstream of the sequence. The in tron splice sites are CUU (5’ splice site) and AAG (3’ splice site), poly -A tails are added following the sequence AGUUGG. The poly- A tails are 20 nucleotides.

a. Predict the sequence of mature mRNA and denote 5’ and 3’ ends.

b. If this is an oncogene that is elevated in cancer cells, design two siRNAs to knock down the mRNA, list the sequences of the siRNAs and highlight the targeted regions on the mRNA.

c. If you plan to study whether the gene is regulated by miRNA(s), you will search for potential miRNA target sequences on the mRNA, highlight the region for miRNA targeting.

1 GATGAGTTAT AATATTTCTC TCCAGGCATG GAGTATTCCG GTGTGCGATC GCCGTTATCG ATCGATCGAT

71 ATCGATCGAA TCCCCCTTTG GACCACCCTG GGTTGCCCTC TAAGCATAAT TCGTCGTCGT ACAACGCCAT

141 GTTTCTGT AATTAAAATT TGTTTGCCTC AGTTGGATGT

I would like all parts answered since it all technically counts as one question. Thanks !!

Answer 1

GIVEN THAT:-

(1) Question :-

sequence of a eukaryotic gene (coding strand) is shown below, RNA polymerase recognizes the sequence ‘TATAAT’ and initiates transcription six nucleotides downstream of the sequence. The in tron splice sites are CUU (5’ splice site) and AAG (3’ splice site), poly -A tails are added following the sequence AGUUGG. The poly- A tails are 20 nucleotides.

Coding strand is given:

1 GATGAGTTAT AATATTTCTC TCCAGGCATG GAGTATTCCG GTGTGCGATC GCCGTTATCG ATCGATCGAT

71 ATCGATCGAA TCCCCCTTTG GACCACCCTG GGTTGCCCTC TAAGCATAAT TCGTCGTCGT ACAACGCCAT

141 GTTTCTGT AATTAAAATT TGTTTGCCTC AGTTGGATGT

(a):-

TATAAT is the recognition sequence recognised by RNA polymerase enzyme and the transcription initiation site in this sequence is six nucleotides downstream hence counting six nucleotides downstream from the sequence TATAAT is CTC sequence which is the initiation site. So now during transcription process, RNA polymerase will make a copy of the gene from the DNA to RNA as needed. all of the nucleotide sequences will be copied as it is in mRNA except A where RNA polymerase incorporates U in place of it. thus unprocessed mRNA from the above sequence will be from CTC site:

CUCCAGGCAUGGAGUAUUCCGGUGUGCGAUCGCCGUUAUCGAUCGAUCGAUAUCGAUCGAAUCCCCCUUUGGACCACCCUGGGUUGCCCUCUAAGCAUAAUUCGUCGUCGUACAACGCCAUGUUUCUGUAAUUAAAAUUUGUUUGCCUCAGUUGGAUGU

(b):

the highlighted sequence is intron and will be spliced out thus the mRNA sequence after splicing is:

CUCCAGGCAUGGAGUAUUCCGGUGUGCGAUCGCCGUUAUCGAUCGAUCGAUAUCGAUCGAAUCCCCCAUAAUUCGUCGUCGUACAACGCCAUGUUUCUGUAAUUAAAAUUUGUUUGCCUCAGUUGGAUGU

(c):

mRNA sequence after poly a tailing is:

5'CUCCAGGCAUGGAGUAUUCCGGUGUGCGAUCGCCGUUAUCGAUCGAUCGAUAUCGAUCGAAUCCCCCAUAAUUCGUCGUCGUACAACGCCAUGUUUCUGUAAUUAAAAUUUGUUUGCCUCAGUUGGAAAAAAAAAAAAAAAAAAAA3'

Thus ,.a,b,c are explained..