Question: Part II – Gene expression Below is a normal sequence on a chromosome, which contains 1 gene. Norm…

Part II – Gene expression

Below is a normal sequence on a chromosome, which contains 1
gene.

Normal DNA:

5′-ATATGCTATAAATAACGAGTATGCTAATAAACGCTTTCGGTAACCGCGAAT-3′

3’-TATACGATATTTATTGCTCATACGATTATTTGCGAAAGCCATTGGCGCTTA-5′

The promoter for the gene is: 5’-TATAAATA-3’ (note that
transcription begins immediately after the promoter) Assume that
transcription ends immediately after this sequence: 5’-ACCGCG-3’
(this is downstream of the polyadenylation signal AAUAAA).

(a) Transcription proceeds from
left to right. Where in the above DNA sequence will the RNA
polymerase bind to initiate transcription?

(b) Assume that the gene begins
immediately after the promoter, and transcribes the DNA to form
mRNA. Show the nucleotide sequence of the pre-mRNA formed (do not
include spaces in your sequence and make sure to identify the 5′
and 3′ ends)

(c) The following intron
5’–UUUCG–3’ is removed during RNA processing. Show the nucleotides
present in theresulting mature mRNA (do not include spaces in your
sequence and make sure to identify the 5′ cap and 3′ poly-A
tail)

(d) The mature mRNA will leave
the nucleus and bind to a ribosome. Show the sequence of amino
acids coded for by this gene

The DNA below is a mutated version of the DNA shown above. This
mutated DNA uses the same promoter and terminator, and has the same
intron as the normal DNA shown above.

Mutated DNA nucleotide sequence (the highlighted bases are
different from the normal DNA):

5′-ATATGCTATAAATAACGAGTATGCATAATAAACGCTTTCGGTAACCGCGAAT-3′

3’-TATACGATATTTATTGCTCATACGTATTATTTGCGAAAGCCATTGGCGCTTA-5′

(e) Show the sequence of amino acids for which this mutated DNA
codes

(f) What type of mutation is shown above? Explain your
answer.