Frameshift Insertion Creating Premature UAG Stop Codon

Steps:

• Transcribe the DNA strand to mRNA, replacing T with U, yielding four codons for the amino acids.
• Determine insertion positions A, B, C, D along the DNA, corresponding to adding a U in mRNA at those points.
• For each position, shift the reading frame by inserting U and re-read codons from the mutation point onward.
• Check each new frame for the UAG stop codon; the position creating UAG earliest terminates translation prematurely.

Why B is correct:

• Insertion at B adds U after the first codon, shifting the frame so the next three nucleotides form UAG, immediately stopping translation per the universal genetic code.

Why the others are wrong:

• A: Insertion before the start shifts frame but forms codons without UAG, allowing full translation.
• C: Insertion mid-sequence creates non-stop codons, extending translation beyond four amino acids.
• D: Insertion at the end adds U after the last codon, not shifting the original frame to produce UAG.

Final answer: B