The exons and introns of a gene are shown below Alternative

The exons and introns of a gene are shown below. Alternative splicing of this gene produces four different mRNA transcripts. All four transcripts utilize the same start codon, which is the first codon in Exon1. All four transcripts utilize the same stop codon, which is the last codon in Exon5 Based on the information provided in the figure below, predict the exon combinations of the four different transcripts and the length (number of amino acids) of each of the four proteins encoded by the four transcripts.

Solution

1. 12345 = 603 nt = 200 amino acid length

2. 1245 = 393 nt = 130 Amino acid length

3. 1235 = 553 nt = 180 Amino acid length

4. 1345 = 563 nt = 187Amino acid length

Each amino acid is encoded by 3 nucleotides and here the last three nucleotides of every splice varient should not consider while calculate amino acids due to their stop codon nature.