In a 125 amino acid sequence,if the codon for | vedclass

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Question

(A) $UAA$ is one of the three stop codons (nonsense codons) which do not code for any amino acid.If the codon for the $25^{	ext{th}}$ amino acid is m

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a polypeptide of $24$ amino acids is formed

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(A) $UAA$ is one of the three stop codons (nonsense codons) which do not code for any amino acid.If the codon for the $25^{	ext{th}}$ amino acid is mutated to $UAA$,the translation process will be terminated at that specific position because no $tRNA$ molecule can recognize the stop codon to bring the next amino acid.Consequently,the ribosome will release the growing peptide chain after the $24^{	ext{th}}$ amino acid.Therefore,a polypeptide chain consisting of $24$ amino acids will be formed.

In a $125$ amino acid sequence,if the codon for the $25^{\text{th}}$ amino acid is mutated to $UAA$,then:

Similar Questions

The possibility of a three-letter genetic code was first suggested by:

$A$: Triplet genetic code can be confirmed by frame shift mutations.
$R$: Frame shifting involves the change in protein product coded by triplet codons.

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