Answer Happy

The following eukaryotic DNA sequence is a made up gene. It contains a promoter (green), a coding sequence (yellow to grey), a terminator (grey). Within the coding sequence there are 2 exons (uppercase black) and 2 introns (lower case black). With this information please answer the questions below. 5 ATATTATagclasatcogicag GCATGGCGAGGCC gcgtactcag ACCTAACCTATA 3 TATAATA galagonagle CGTACCGCTCCOG ogcatgage TOGATTOGATAT 5 tattagcgcatg A GACATCAG AATAAATAAADCTTTOOGCTTTT 3 atsategogtacTCTGTAGTCTTATTTATTICOAAACCOOAAAA Question 3 The nucleotides from position to would be transcribed to RNA. -22 to +80 +1 to +66 -22 to +66 +1 to +80 1/1 pts
1/1 pts Question 4 The transcription factors would recognize and bind to nucleotides between the positions +1 to +25 -22 to -1 +1 to +55 -22 to -15
Question 5 10/10 pts What would the sequence of the immature mRNA be? Place the sequence of ALL the transcript that would be synthesized from the gene. In each of the available lines you should write in order. The sequence is divided by sections. WRITE THE NUCLEOTIDES IN CAPITAL LETTERS. Exon 1 AGCAUGGCGAGG Intron 1 GCGUACUCAG Exon 2 ACCUAACCUAUA Intron 2 UAUUAGCGCAUG Exon 3 no terminator AGACAUCAGAAU Terminator AGCUUUGGGCUU
The following eukaryotic DNA sequence is a made up gene. It contains a promoter (green), a coding sequence (yellow to grey), a terminator (grey). Within the coding sequence there are 2 exons (uppercase black) and 2 introns (lower case black). With this information please answer the questions below. 5' ATATTATagctaaatcggtcag GCATGGCGAGGCC gcgtactcag ACCTAACCTATA 3' TATAATAtogatttagccagtc CGTACCGCTCCGG cgcatgagtc TGGATTGGATAT -22 +1 +30 5' tattagcgcatg A GACATCAG AATAAATAAAGCTTTGGGCTTTT 3' ataatcgcgtacTCTGTAGTCTTATTTATTICGAAACCCGAAAA +60 -80