Question at position 64 Translate the coding DNA sequence into protein as a ribosome would. I recommend working this out on scrap paper, then finding the answer. Coding DNA: 5'-GCACTATGCCCCTGCGGGGATGAGTA-3'N-Met-Pro-Leu-Arg-Gly-CN-Ala-Leu-Cys-Pro-Cys-Gly-Asp-Glu-CN-Met-Pro-Leu-Arg-Gly-Trp-CNo protein is madeN-Met-Ser-Arg-Gly-Val-Pro-Val-C单项选择题
A
N-Met-Pro-Leu-Arg-Gly-C
B
N-Ala-Leu-Cys-Pro-Cys-Gly-Asp-Glu-C
C
N-Met-Pro-Leu-Arg-Gly-Trp-C
D
No protein is made
E
N-Met-Ser-Arg-Gly-Val-Pro-Val-C

Question

Question at position 64 Translate the coding DNA sequence into protein as a ribosome would. I recommend working this out on scrap paper, then finding the answer. Coding DNA: 5'-GCACTATGCCCCTGCGGGGATGAGTA-3'N-Met-Pro-Leu-Arg-Gly-CN-Ala-Leu-Cys-Pro-Cys-Gly-Asp-Glu-CN-Met-Pro-Leu-Arg-Gly-Trp-CNo protein is madeN-Met-Ser-Arg-Gly-Val-Pro-Val-C单项选择题

A

N-Met-Pro-Leu-Arg-Gly-C

B

N-Ala-Leu-Cys-Pro-Cys-Gly-Asp-Glu-C

C

N-Met-Pro-Leu-Arg-Gly-Trp-C

D

No protein is made

E

N-Met-Ser-Arg-Gly-Val-Pro-Val-C

BlackTom AI · Accepted Answer

Carefully restating the problem helps. The coding DNA sequence given is 5'-GCACTATGCCCCTGCGGGGATGAGTA-3'. We need to translate this as a ribosome would, using standard codon-to-amino acid rules, with T replaced by U for the mRNA and reading from the start codon in the correct reading frame.

First, convert to codons from the 5' end: GCA | CTA | TGC | CCC | TGC | GGG | GAT | GAG | TA. The last two bases 'TA' are incomplete for a full codon, so translation stops after the eighth codon.

Now map each codon to an amino acid using the universal genetic code:
- GCA codes for Alanine (Ala).
- CTA codes for Leucine (Leu).
- TGC codes for Cysteine (Cys).
- CCC codes for Proline (Pro).
- TGC codes for Cysteine (Cys).
- GGG codes for Glycine (Gly).
- GAT codes for Aspartic acid (Asp).
- GAG codes for Glutamic acid (Glu).

Putting these together gives Ala-Leu-Cys-Pro-Cys-Gly-Asp-Glu as the peptide sequence produced by the first eight codons. The trailing two nucleotides 'TA' do not form a complete codon, so they do not contribute to the amino acid sequence; there is no stop codon indicated within the eight codons here, so translation would produce the chain up to Glu.

Now evaluate the answer options:
- Option A: N-Met-Pro-Leu-Arg-Gly-C. This starts with Met and includes different amino acids (Met vs Ala) and a different sequence overall, so it does not match the translation above.
- Option B: N-Ala-Leu-Cys-Pro-Cys-Gly-Asp-Glu-C. This matches the sequence derived above exactly in order and identity of amino acids; the trailing '-C' corresponds to the terminal C-terminus annotation rather than an extra codon, so this aligns with the expected peptide.
- Option C: N-Met-Pro-Leu-Arg-Gly-Trp-C. This contains Met at the start and a different series of residues, including Trp, which do not appear in the translation above, so it is incorrect.
- Option D: No protein is made. The sequence does produce a peptide through the first eight codons, so this statement is false.
- Option E: N-Met-Ser-Arg-Gly-Val-Pro-Val-C. This sequence has entirely different residues (Ser, Val, etc.) and does not fit the codon-derived translation, so it is incorrect.

In summary, the only option that aligns with the codon-by-codon translation of the given coding DNA is the second option, which matches Ala-Leu-Cys-Pro-Cys-Gly-Asp-Glu (with the terminal C indicating the chain terminus rather than an additional residue).