Question

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)%0D%0AResponses%0D%0A%0D%0ATCA%0D%0ATCA%0D%0A%0D%0AAGU%0D%0AAGU%0D%0A%0D%0AUCA%0D%0AUCA%0D%0A%0D%0ATCC

Answer 1

To determine the base sequence of the DNA gene that originally produced the mRNA codon AGU, we need to consider the relationship between DNA and mRNA.

1. The mRNA is synthesized from the DNA template strand, where adenine (A) pairs with uracil (U) in RNA (which corresponds to thymine (T) in DNA), cytosine (C) pairs with guanine (G), and vice versa.

2. The mRNA codon AGU translates to:
- A in mRNA corresponds to T in DNA
- G in mRNA corresponds to C in DNA
- U in mRNA corresponds to A in DNA (U is present in RNA; in DNA, it is replaced by T)

So, to derive the DNA codon from the mRNA codon AGU:
- A (mRNA) → T (DNA)
- G (mRNA) → C (DNA)
- U (mRNA) → A (DNA)

Thus, the DNA sequence that codes for the mRNA codon AGU is:

**TCA**

Looking at your provided options, the correct response for the DNA gene that would produce the mRNA codon AGU is **TCA**.