sometimes a mistake occurs in the stranslation of an mRNA strand. Suppose that the reading of the mRNA stand ATGCCCGCTATCCGACGATAA began, by mistake, at the second nucleotide instead of the the first . Write the sequence ot amino acids that would be formed from N-terminus to C-terminus.

Question

sometimes a mistake occurs in the stranslation of an mRNA strand. Suppose that the reading of the mRNA stand ATGCCCGCTATCCGACGATAA began, by mistake, at the second nucleotide instead of the the first . Write the sequence ot amino acids that would be formed from N-terminus to C-terminus.

Answer 1

I DON't CARE

Answer 2

use the universal triplet code to determine the sequence of amino acids that would be generated for mrna sequence 5'atggatgaaacgggctaggua3'

Answer 3

To determine the sequence of amino acids formed from a mRNA strand, we need to consider the genetic code, which relates each mRNA codon to a specific amino acid. In this case, if the reading of the mRNA strand began at the second nucleotide instead of the first, the reading frame would shift by one base.

Let's first write out the mRNA strand, indicating the nucleotides:

ATGCCCGCTATCCGACGATAA

If the reading starts at the second nucleotide, the new codons will be formed as follows:

TGC - CCG - CTA - TCC - GAC - GAT - AA

Now we can use the genetic code to translate these codons into amino acids:

TGC - CCG - CTA - TCC - GAC - GAT - AA
Cys - Pro - Leu - Ser - Asp - Asp - Stop

Therefore, the sequence of amino acids formed from the N-terminus (beginning) to the C-terminus (end) in this scenario would be: Cys - Pro - Leu - Ser - Asp - Asp - Stop.

Answer 4

To determine the sequence of amino acids, we need to consider the correct reading frame of the mRNA strand. In this case, the mistake occurred by starting the reading at the second nucleotide instead of the first. Therefore, we need to shift the reading frame by one nucleotide and then translate the mRNA sequence into amino acids.

The correct reading frame for the mRNA sequence ATGCCCGCTATCCGACGATAA, starting from the first nucleotide, would be:

ATG-CCC-GCT-ATC-CGA-CGA-TAA

Now, let's translate this sequence into amino acids:

ATG - Start codon, codes for methionine (M)
CCC - Codes for proline (P)
GCT - Codes for alanine (A)
ATC - Codes for isoleucine (I)
CGA - Codes for arginine (R)
CGA - Codes for arginine (R)
TAA - Stop codon

So, the sequence of amino acids formed from N-terminus to C-terminus, considering the mistake in the reading frame, would be:

M - P - A - I - R - R

Therefore, the sequence of amino acids would be Methionine (M), Proline (P), Alanine (A), Isoleucine (I), Arginine (R), and Arginine (R).