4. The exons and introns of a gene are shown below. Alternative splicing of this
ID: 320056 • Letter: 4
Question
4. 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.
ATG (Start codon) Exon 1 180 nt Exon 2 40 nt Exon 3 210 nt Exon 4 50 nt TAG (Stop codon) Exon 5 123 nt (nt nucleotides)Explanation / Answer
since exon 1 and exon 5 are always involved
the other combinations will be
exon 1 +exon3+ exon4+ exon5 = 180 +210+50+123=563/3= 187 amino acids -1 stop codon = 186aa
exon 1+ exon2 + exon4+ exon5 =180 +40+50+123=393/3= 131- i stop codon = 130 aa
exon 1+ exon2 +exon3+exon5 = 180 +40+210+123=553/3=184 - stop codon= 183 aa
exon 1+ exon2 +exon3+ exon4+ exon5 =180 +40+210+50+123= 603/1 =201 - stop codon = 200 aa
i divide nucleotides by 3 because 3 nucleotides code 1 amino acid .... and i subtract 1 because stop codon would not contribute aminoacid
Hope it helps
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