Q The given nucleotide sequence on mRNA is shown below:
5' AUGUCAUGGGAGUGA GUUGGGCUA AAAUAG 3'
a) how many amino acids will be inserted in
polynucleotide chain under normmal conditions ?
(6) How many acids will be inserted in
polynucleotide chain in a mutated situation by
the deletion of 9Th nucleotide in the cistron
part of the DNA ?
Answers
Answer:
(a) 4
(b) 7
Explanation:
Part (a):
There is terminator codon at position 13, 14, 15 i e UGA. Since there are 12 bases between initiator and terminator (including initiator and excluding terminator) so (12/3) i.e. 4 amino acids will be added.
Part (b):
On removal of 9th nucleotide, the chain will become:5' AUG UCA UGG AGU GAG UUG GGC UAA AAUAG 3 (Since, genetic code is coma less, spaces have been introduced to make it easier to read and have no actual significance) Now terminator codon is present at 22, 23, 24 position i e It can clearly be seen that there are 7 triplets between initiator and terminator (including initiator and excluding terminator) so 7 amino acid polynucleotide chain will be inserted. Note:- There are three terminator codons viz. UAA, UAG and UGA.
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