Question

The RNA sequence below depicts the part of a 330 nucleotides long mRNA and it encodes the C-terminal portion of a protein.
5'-... ... ... AAC ACC ACG ACC CAU GUG GCG AGA CGG UAG-3'
A mutation was identified in this RNA denoted as \(322A \rightarrow U\). This nucleotide change is represented by which of the following class(es) of mutation?

• A. Missense mutation
• B. Non-sense mutation
• C. Transversion mutation
• D. Silent mutation

Hint:

Count codons backward from the stop codon UAG to find nucleotide 322, then check both the amino acid change and the base class change.

Answer 1

The Correct Option is B, C

Step 1: Locate nucleotide 322 in the given sequence.
The mRNA is 330 nucleotides long, and the stretch shown ends at the very last nucleotide, position 330, right before the 3' end.
The last codon shown is \(UAG\), the stop codon, so its three bases occupy positions 328, 329 and 330 (U = 328, A = 329, G = 330).
Counting backward in triplets from there: \(CGG\) sits at 325-327, and \(AGA\) sits at 322-324.
So nucleotide 322 is the first base of the codon \(AGA\), which is A.

Step 2: Find the codon before and after the mutation.
Before the mutation, the codon at position 322-324 is \(AGA\), which codes for arginine (Arg).
The mutation changes the base at position 322 from A to U, so the codon becomes \(UGA\).
\(UGA\) is one of the three stop codons (along with \(UAA\) and \(UAG\)), so translation now stops at this position instead of continuing.

Step 3: Classify the mutation by its effect on the protein.
A mutation that turns a codon for an amino acid into a stop codon is called a nonsense mutation, because it cuts the protein short.
This rules out missense mutation (that would swap in a different amino acid, not a stop) and silent mutation (that would keep the same amino acid).
So option (B), Non-sense mutation, is correct.

Step 4: Classify the mutation by nucleotide chemistry.
Purines are A and G; pyrimidines are C and U.
A transition is a substitution within one group, purine to purine or pyrimidine to pyrimidine. A transversion is a substitution that crosses between the two groups.
Here the original base A is a purine and the new base U is a pyrimidine, so this change crosses groups and is a transversion.
So option (C), Transversion mutation, is also correct.

Final Answer:
The mutation \(322A \rightarrow U\) turns codon \(AGA\) (Arg) into the stop codon \(UGA\), which makes it a nonsense mutation, and since A (purine) changes to U (pyrimidine), it is also a transversion.
\[ \boxed{\text{(B) Non-sense mutation and (C) Transversion mutation}} \]