Question

3. A human gene was initially identified as having three exons and two introns. In order, the exons are 456, 224, and 524 bp, while the introns are 2.3 kb and 4.6 kb.

a. Draw this gene showing the promoter, introns, exons, and the transcription start and stop sites.

b. Surprisingly, it is found that this gene encodes not one but two mRNAs that have only 224 nucleotides in common. The original mRNA is 1204 nucleotides, while the new mRNA is 2524 nucleotides. Use your drawing to show how it is possible for this one region of DNA to encode these two transcripts

Answer 1

And:- prome Transcription starts termination sile (Polyadinylation Exon intron Exen intron zon b) one way to come up with the solution to this problem is to understand that alternative splicing of the primary transcript would result in MRNAs that were only partially identical. So in this case, the two transcripts share 224 nucleotides in common. from the question we can see that this is the exact length of the second Exon. So the answer could be that this Exon is shared by a alternatively- spliced manas. The second transcript contains 2.3 Kb of sequence which is not found in the first transcrupt. so the one which we considered as the first intron could actually be part of the second transcript as that would dresult in the 2524 nucleotides stated as this transcripts length. There can be other combinations too.