21 31 GGCGTATAAA GCGACGACTG TAGACTGACG AGCCTATCCA 41 TGGACGCGCC ATGGCCCTGT AAGCG

Question

21 31 GGCGTATAAA GCGACGACTG TAGACTGACG AGCCTATCCA 41 TGGACGCGCC ATGGCCCTGT AAGCGGTGCG ATGCAATAAA 81 ACGCGTATCA GTCATTCAGC GTAGTCTGAT GCCAGTCGAC 121 TGC 51 61 7 1 91 101 Please answer the following question(s): 1. For the eukaryotic DNA sequence shown, what region of the mRNA contains the open reading frame that will be translated into protein? a. 27-35 O b. 41-101 c. 51-59 Hint First find the TATA box and the polyadenylation sequences, then think about the location of the sequences with respect to the final length of the mRNA. Next step is to use Kozak's rule to find the optimal AUG

Explanation / Answer

TATA box- TATAAA (present in the region 1 to 10)

consensus kozak sequence- GCCGCCAUGG in mRNA. the underlined is the start codon. the 3rd upstream 'G' (in bold) and immediate downstream 'G' (in bold) are important for a kozak sequence.

a similar sequence is present in the region 41 to 61. (CGCGCCATGG)

polyadenylation sequence- AAUAAA (present in 71 to 80 region)

therefore, 41 to 80 region would be transcribed.

The question asks about protein. Proteins would be translated from a start codon to stop codon. Start codon is usually AUG. Stop codon could be UAG, UAA and UGA.

The codon at 51 (ATG)- AUG in mRNA Would act as start codon.

The codon at 60 (TAA)- UAA in mRNA would act as stop codon.

Therefore, the region from nucleotide 60 would not be translated.

The region that is translated- 51 to 59. (Option C)

this region that is transcribed- 41 to 101.

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