If there are 999 bases in an RNA that codes 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?

33 codons will be altered if the base atposition 901 is deleted from an mRNA having 999 bases. This can be calculated as
Total number of bases in the given mRNA = 999
Number of codons for this mRNA $=\frac{999}{3}=333$
If one base is deleted at position 901 then the number bases unchanged are 900 which codes for $\frac{900}{3}$= 300amino acids
Thus, number of altered codons will be
333 - 300= 33