Answer :)
The mRNA sequence is as the following:
5’ GUAUAAGAAGCACUCUACCUCAAUGGGUCCAUGGGAGAAGGUAGGCAUGUGUAUUUGACAAAGGGA 3’
Answer i:)
The amino acid chain of the above mRNA strand (based on the open reading frame mentioned in the second part of the answer):
MET GLY GLU GLY ARG HIS VAL TYR LEU THR LYS GLY
Therefore, the six terminal amino acids are VAL TYR LEU THR LYS GLY
Answer ii:)
The reading frame of the above mRNA is as bellow:
AUGGGAGAAGGUAGGCAUGUGUAUUUGACAAAGGGA
Protein sequence: MET GLY GLU GLY ARG HIS VAL TYR LEU THR LYS GLY
The protein sequence does not contain stop codon; therefore, it is a reading frame. The minimum size of an open reading frame is 100 codons. Therefore, in this reading frame, we suppose that there must be 100 codons until a stop codon does not come.
The other reading frame:
Protein sequence: MET GLY PRO TRP GLU LYS VAL GLY MET CYS ILE STOP
We chose the reading frame from the start codon (AUG). This reading frame is started with AUG codon but contains stop codon and does not reach to minimum 100-codon length.
Protein sequence: MET CYS ILE STOP
This reading frame is too short to consider as an open reading frame.
Answer iii:)
The red circle is showing the possible promoter region i.e. TATA box.

Answer iv:)
The possible amino acid sequence is MET GLY GLU GLY ARG HIS VAL TYR LEU THR LYS GLY in which the terminal amino acid Glycine supports in the flexibility of the protein.

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