An Eukaryotic gene has four exons and three introns. The 1st, 2nd, 3rd and 4th e

Question

An Eukaryotic gene has four exons and three introns. The 1st, 2nd, 3rd and 4th exons are 538, 321 and 416 and 384 bp long respectively, while the 1st, 2nd and 3rd introns are 3.6 kbp, 1.9kbp and 2.2 kbp respectively.

It was discovered that this gene encodes not one, but three mature mRNA types, each producing a different polypeptide (Longest = polypeptide A, Second longest = Polypeptide B and the Smallest = Polypeptide C), due to alternative splicing. The largest mRNA is 1659 bp long and produces a 480 amino acid long polypeptide. The
start codon is located 100 bases from the 5’end of the mRNA (consider that the start codon is AUG and it begins at the 101 base position).

> If it was discovered that Exon 1 and 4 are constitutive exons and the polypeptides B and C are produced by the mutual exclusion of Exon 2 and Exon 3 respectively;
I. What would be the amino acid similarity between Polypeptide A and B? (Give as a % of Polypeptide A) ___________________
Show your calculations...........

II. What would be the amino acid similarity between Polypeptide A and C? (Give as a % of Polypeptide A) ___________________
Show your calculations............

Explanation / Answer

[Note: Introns are spliced from mature mRNA and thus are not translated.]

Total nucleotide length is 1559, which means that if all the mRNA is translated then 553 amino acid polypeptide will be formed (1659 / 3 = 553 a.a ; i codon represents 3 bases which code for 1 amino acid). However, we find that this mRNA codes for 480 amino acid long polypeptide.

If we look carefully, start codon is located 100 b.p downstream from 5' end. This reduces the mRNA sequence to 1559 b.p (start site 101) which will code for 520 amino acids. This implies that 120 b.p from 3' end are untranslated because 520 - 480 = 40 amino acids = 120 b.p.

So, actual nucleotide base pairs used for translation in each exon are: 438, 321, 416 and 264 b.p.

Now let's see whether the nucleotide base pairs in each codon is divisible by 3, to find out whether the codons are in frame. You can see, that they are in frame.

I. amino acid similarity between Polypeptide A and B = 100% - (321 b.p / 480 a.a)%

= 100% - (107 a.a / 480 a.a)%

= 100% - 22.3 %

= 77.7 %

II. amino acid similarity between Polypeptide A and C = 100% - (416 b.p / 480 a.a) %

= 100% - (139 a.a / 480 a.a) %

= 100% - 30.0 %

= 70.0 %

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