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Shown below is the structure of an Arabidopsis gene, divided into 10 segments, designated A-J. The gene contains three exons,
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As per Chegg Q&A Guidelines, I am answering the first four subquestions.

Here we have an Arabidopsis gene. The gene is divided into ten segments A-J as shown in the question above.

The segment C has the promoter, segments D, F and H have exons while segments E and G have Introns and the segment I have polyA addition site.

a) The DNA is acted upon by the RNA polymerase which recognizes the promoter and the transcription is initiated from the transcription start site until the terminator is met. Since the terminator is not specified in the given sequence, the entire sequence after the promoter will be copied by the RNA polymerase and will be a part of the primary transcript. Hence the primary transcript will have segments D, E, F, G, H, I and J.

b) This primary transcript will then undergo further processing wherein there will be capping to the 5' end, polyA tailing to the 3' end and the splicing out of the introns and the joining of the exons. Thus the mature transcript will lose segments E, H and J (E and H due to splicing and J due to polyadenylation). Thus the completely processed transcript or the mRNA will have segments D, F, H and I.

c) The addition of nucleotides happens at the 5' end in the form of a methyl guanosine cap and at the 3' end in the form of a polyA tail. Thus, new nucleotides will be added to the segments D and I.

d) The translation begins from the start codon AUG which is also called the translation initiation codon. This codon is present on the first exon and determines the frame in which the ribosomal machinery would translate the mRNA transcript. The first exon is present on sequence D. Hence the translation initiation codon will be present on segment D.

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