If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901…
- 1
- 11
- 33
- 333
Solution
Total bases present in RNA = 999
Bases left after deletion of 901st base
In RNA = 999 - 901 = 98
3 codon= 1 amino acid
Therefore 98/3= approx 33
A number of codons present in 98 bases= 33 (Approximately as three codons code for one amino acid).
Asked in: NEET 2017
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