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The following segment of DNA encodes for a short polypeptide. Note that DNA triplets encoding for the start codon (AUG) and a stop codon are included in this sequence. What is the polypeptide sequence encoded by this gene?

5’…CCCAGCCTAGCCTTTGCAAGAGGCCATATCGAC…3’
3’…GGGTCGGATCGGAAACGTTCTCCGGTATAGCTG…5’
a) Met-Gly-Arg-Gly-Phe-Ala
b) Met-Ala-Ser-Cys-Lys-Gly
c) Met-Ala-Ser-Cys-Lys-Gly-Ile
d) Met-Gly-Arg-Gly-Phe-Ala-Ile

2 Answers

2 votes

Final answer:

After transcription of the template DNA strand into RNA and translation into amino acids, the polypeptide sequence encoded by the given gene is Met-Ala-Ser-Cys-Lys-Gly. This matches option (b).

Step-by-step explanation:

The student's DNA sequence needs to be transcribed into mRNA and then translated into a polypeptide chain. Transcription involves rewriting the DNA code into RNA, where the DNA base 'A' pairs with 'U' in RNA, 'T' pairs with 'A', 'C' pairs with 'G', and 'G' pairs with 'C'. The DNA sequence given is 5'…CCCAGCCTAGCCTTTGCAAGAGGCCATATCGAC…3', but we should use the template strand (3'…GGGTCGGATCGGAAACGTTCTCCGGTATAGCTG…5') to transcribe into mRNA because transcription uses the template strand of DNA.

Transcription will yield the following mRNA sequence: 5'…CCAGUCCUAGCCUUUGCAAGAGGCCAUAUCGAC…3' (removing the initial 'GGG' since we're told to include only the sequence with start and stop codons). The start codon is AUG, so we look for a sequence that starts with AUG after transcription. This gives us the mRNA sequence AUG (start codon) CAAGAG (Gln, Glu) GCCAUA (Ala, Ile) U (stop codon).

Finally, we translate this mRNA sequence into a polypeptide sequence. According to the genetic code, we get:

  • Met (start codon AUG)
  • Ala (GCU)
  • Ser (UCG)
  • Cys (UGC)
  • Lys (AAA)
  • Gly (GGG)

Therefore, the correct polypeptide sequence encoded by this gene is Met-Ala-Ser-Cys-Lys-Gly, which corresponds to option (b).

User Gregg Duncan
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5.8k points
6 votes

Answer:

So none of the option is correct.

Step-by-step explanation:

As we know the transcription occurs in 5' to 3' direction so 3'-5' work as a template strand for the transcription and the mRNA formed will be :-

5' CCC AGC CUA GCC UUU GCA AGA GGC CAU AUC GAC 3'

Here there is no start codon( AUG) for translation.

So none of the option is correct.

There might be error in the nucleotide sequence.

CCC- pro

AGC - ser

CUA- leu

GCC- ala

UUU- phe

GCA- ala

AGA- arg

GGC- gly

CAU- his

AUC- lle

GAC- asp

User Goper Leo Zosa
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6.0k points