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Q.In an mRNA strand having the sequence of CGGAUCGAUG, if mutation removes all cytosine - (1+1+1=3)

(a) How will the codon be read ?
(b) What will be the function of AUG present in the middle of mRNA ?
(c) Write the complementary base pairing of the rule.
Mizoram MbseMizoram Board of School Education HSSLC 2024Subjective· 3mImportance★★★★★
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Deleting the two cytosines shifts the reading frame; the new codon sequence runs into a premature stop codon (UGA), truncating the protein, and the internal AUG loses any start-codon role.

Original mRNA: 5'-C G G A U C G A U G-3' (10 bases). Removing all cytosine (positions 1 and 6) leaves: G G A U G A U G (8 bases).

  1. How will the codon be read? This deletion of nucleotides shifts the reading frame downstream of the mutation — a frameshift mutation. Reading the new sequence in triplets from the 5' end: GGA – UGA – UG (last codon incomplete). GGA codes for glycine, but the very next codon, UGA, is a stop codon. This means translation would terminate immediately after just one amino acid, producing a severely truncated, non-functional polypeptide — none of the codons after the deletion point match the original reading frame.
  2. Function of the AUG present in the middle of the mRNA: …

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