Question

Below is a dog gene with four exons and three introns. You examined the expression of this gene in the liver with northern blotting. An arrow in Exon 1 is the transcription initiation site. Exon 1 Exon 2 Exon 3 Exon 4 (200 bp) (150 bp) (170 bp) (300 bp) = When you did northern blot analysis with Exon 4 as a probe, you recognized two bands on a blot. Their sizes are 820 and 670 bp long. When you did it with Exon 2 as a probe, you only recognized the 820 bp long band. How can this be possible? Explain. You are allowed to write ONE short sentence or TWO.

please help :)

Answer 1

This is possible only when the RNA is processed in two different ways that give rise to two different types of transcripts from a single gene, this is known as alternative splicing. In the first splicing events, only introns are removed and all exons are joined together, hence we got a band of 820 bp (sum of all exons fragment). In the second case exon, 2 is also removed with introns during splicing, as a result of this one fragment of size 670 bp (820-150) was observed. Now if we do the northern blotting with exon 4 probe two bands will be observed, since exon 4 is present in both processed RNA but northern blotting with exon 2 will give only one band because exon two is present in only first processed RNA.