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?

# 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?

1. A

1

2. B

11

3. C

33

4. D

333

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### Solution:

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

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