If an RNA sequence has 999 bases and codes fo | vedclass

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(C) The genetic code is read in groups of $3$ bases,known as codons.Since the deletion occurs at position $901$,all codons from this point onwards wil

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$33$

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(C) The genetic code is read in groups of $3$ bases,known as codons.Since the deletion occurs at position $901$,all codons from this point onwards will be affected.The total number of bases is $999$.The deletion occurs at position $901$.The number of bases remaining after the deletion point is $999 - 901 + 1 = 99$ bases.Since each codon consists of $3$ bases,the number of codons affected is $99 / 3 = 33$ codons.Therefore,$33$ codons will be altered due to the frame-shift mutation.

If an $RNA$ sequence has $999$ bases and codes for a protein with $333$ amino acids,and the base at position $901$ is deleted,resulting in a sequence of $998$ bases,how many codons will be altered?

Similar Questions

According to the genetic code,some amino acids are encoded by more than one codon. This characteristic of the genetic code is . . . . . . .

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

The property of a codon for always coding for a specific amino acid is called:

Which of the following pairs is correctly matched with respect to the codon and the amino acid it codes for?

In the genetic code,there are $64$ codons.

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