The amino acid serine is coded by the mRNA codon AGU. What is the base sequence of the DNA gene that originally produced this mRNA codon?(1 point)ResponsesAGUAGUUCAUCATCCTCCTCA

1 answer

To determine the base sequence of the DNA gene that corresponds to the mRNA codon AGU, we need to remember that mRNA is synthesized from the DNA template strand through a process called transcription. During this process:

  • Adenine (A) on DNA pairs with Uracil (U) in RNA (since RNA contains uracil instead of thymine).
  • Thymine (T) on DNA pairs with Adenine (A) in RNA.
  • Cytosine (C) on DNA pairs with Guanine (G) in RNA.
  • Guanine (G) on DNA pairs with Cytosine (C) in RNA.

The mRNA codon "AGU" corresponds to the DNA template strand nucleotide sequence "TCA" (because A pairs with T, G pairs with C, and U pairs with A).

Since DNA is double-stranded and the base sequence of the coding strand is complementary to the template strand, we will reverse the transcription rules.

  1. The mRNA codon is AGU (A, G, U).
  2. The corresponding DNA template strand is TCA (T, C, A).
  3. The DNA coding strand (which has the same sequence as mRNA but with T instead of U) is ACT.

So, the final base sequence of the DNA gene that produced the mRNA codon AGU would be:

ACT (representing the coding strand).

From the options you provided, none match this specific sequence. The correct sequence is ACT.