Shown below is the sequence of a short fictitious eukaryotic gene. Both strands

Question

Shown below is the sequence of a short fictitious eukaryotic gene.

Both strands of DNA are shown. After exhaustive studies, you have determined the following:

• Transcription in this organism always starts at the sixth nucleotide after the TATAA. That is, given the following sequence, the first nucleotide of the mRNA would be X: TATAAnnnnnX (where n can be any nucleotide).

• The intron splice sites have been well defined. Introns always begin with GUAUGU and end with CAG or UAG. That is, given the following sequences, the bases in bold will be in the mature mRNA while the italicized nucleotides will be spliced out as an intron:
XXXXGUAUGU(X)nCAGXXXXXX
XXXXGUAUGU(X)nUAGXXXXXX

• The intron-splicing machinery processes the RNA from 5' to 3'. It finds the first GUAUGU sequence and the first subsequent (moving 5' to 3') CAG or UAG and removes the intron between them. (Note that these rules and sequences are similar to those of real eukaryotic organisms.)

• This organism adds poly(A) tails immediately after the sequence GAAUAAAU. Poly(A) tails are usually about seven nucleotides long.

• RNA is processed in this sequence: transcription, then splicing, then polyadenylation and capping.

a) The promoter sequence is TATAA. Why wouldn’t the sequence TATA (or even TATATA, for that matter) work as a promoter sequence? (Hint: remember that a promoter is not just a place for RNA polymerase to bind; a promoter must indicate which direction RNA polymerase must read the DNA.)

b) What is the sequence of the mature mRNA from this gene?

c) What is the sequence of the protein produced from this gene?

For the following mutations, describe the changes to the mRNA sequence (either list the new mRNA sequence or just list that of the altered areas) and give the sequence of the protein produced by the mutant gene. Base pairs are numbered and “C/G base pair” means: C on the top strand, G on the bottom. (The base pairs in question are highlighted in bold.) Consider each mutation separately.

d) T/A base pair 52 is changed to A/T.

e) G/C base pair 41 is changed to T/A.

f) T/A base pair 86 is changed to G/C.

g) T/A base pair 127 is changed to A/T.

h) A/T base pair 24 is deleted.

5' -GAAGCTAGAGGTCAAT ACCTGTATAAATGAAAAGGCGCTGGTATGTCCGAAT AGCATGCA. 1 60 3'-c TTCGATC TCCAGTTATGGACATATTTAC TTTTCCGCGACCATACAGGcTTATCGTACGT TGcCTCTGTATGTATTACTGTAGCTTTAAGGTACTACGTATGTCCGTATGTAATA. 61 120 CTTGTACGGAGACATACATAATGACATCGAAATTCCATGATGCATACAGGCATACATTAT AATAACTGTACAGTAACTAATGATGGTTGACGATACCCTCGGAATAAATGCGCATACGTA-3 121 180 TTATTGACATGTCATTGATTACTACCAA CTGCTATGGGAGCCTTATTTACGCGTATGCAT-5

Explanation / Answer

If polymerase recognises TATA or TATATA then this will be read same on both the strands from 5’-3’ direction. Whereas we need transcription of DNA where genes are coded, this TATAA give direction to RNA polymerase to transcribe the genes downstream.

AUG is the start codon thus the protein sequence from the above RNA would be

AAGGCGCUGUUAUGUCCGAAUAGCAUGCAGAACAUGCCUCUCUUUAAGGUACUACUAACUAAUGAUGGUUGACGAUACCCUCGGAAUAAAUAAAAAAA

h) A/T base pair 24 is deleted.

If this pair is deleted then TATAA box would be non-functional and ultimately no mRNA would be transcribed.

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