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?

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If there are $999$ bases in an $RNA$ that code for a protein with $333$ amino acids,and the base at position $901$ is deleted such that the length of the $RNA$ becomes $998$ bases,how many codons will be altered?

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