A complete plant gene containing four introns and five exons is carried on a 6.0-kb DNA fragment. DNA sequencing analysis finds that this fragment contains 1000 base pairs that flank the transcribed region of the gene and 5000 base pairs that are transcribed. Four introns contain 3500 base pairs, and five exons contain 1500 base pairs. Northern blot analysis is performed on RNA of this gene using a probe that binds to a portion of one of the exons. RNA isolated from the cytoplasm of cells is compared to RNA isolated from cell nuclei on the northern blot.?

Answer and Explanation:

the total length of the gene is 6,000 base pairs (or) 6.0 kb and it has five exons, four introns and a 1,000 base pair length flanking region for the transcribed region.

Hence, each exon has (1500/5) = 300 by

each intron has (3500/4) = 875 by and

flanking segment (5' and 3') length for the transcribed region would be (1000/2) = 500 bp.

The gene segment is as shown in the 1st diagram attached below

In eukaryotes, transcription takes place in the nucleus which results in the synthesis of hnRNA (heteronuclear ribonucleic acid). This hnRNA (pre-RNA) has both introns and exons, however when it is synthesized or while it is being synthesized the Spliceosomal complex removes introns from hnRNA and becomes converted into mRNA (messenger RNA) in the nucleus itself.  

(second diagram explains further)

Now the synthesized mRNA reaches the cytoplasm. So, if mRNA collected from the nucleus and cytoplasm will have the same length of about 2,500 base pairs only. Therefore, after the commencement of northern blot analysis both mRNA will yield the same kind of result and have unique length.