4. Consider the following gene structure: 2 3 4 120 150100 190 Exons are number

Question

4. Consider the following gene structure: 2 3 4 120 150100 190 Exons are number (1-4) and sizes are given (e.g. exon 1 is of size 120 nts). The transcript can initiate (begin) at either arrow and exons 2 and/or 3 can be spliced out (noted by dotted splice junctions). (a) (2pt) How many possible isoforms of this gene could exist? (b) (4pt) For each isoform, list the junction spanning RNA-seq read that would support its existence (c) (4pt) It was noted in class that longer read lengths will reduce the need for isoform inference algorithms Assuming only single end reads. what is the shortest read length that would guarantee the ability to unambiguously identify all isoforms of this gene? For this question we require that a junction read must capture at least 5 bp of each adjacent exon

Explanation / Answer

Answer a) There are total 5 possible isoforms of this gene which could exist.

Answer b) Junction spanning RNA-seq read for all isoforms.

1st isoform - 1 2 3 4 (Transcriptions starts from 1st exon and none of the exons is spliced out.)

2nd isoform - 1 3 4 (Transcriptions starts from 1st exon and 2nd exon is spliced out.)

3rd isoform - 1 4 (Transcriptions starts from 1st exon and 2nd,3rd exons are spliced out. )

4th isoform - 2 3 4 (Transcriptions starts from 3rd exon and none of the exon downstream is spliced out. )

5th isoform 2 4 (Transcriptions starts from 2nd exon and 3rd exon is spliced out. )

Answer c) While considering all the 5 potential transcripts, we find that the Exon 4 is the only common DNA region among them. therefore, we may take Exon 4 (from 3" end) as an end read to unambiguously identify all the possible isoforms.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.