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Consider the following mRNA sequence:
5'-UCU ACA GUG AUC GUC ACC -3°
How many codons are present in this mRNA,
if the genetic code is considered non-
overlapping and overlapping respectively ?
(1) 5 and 17 (2) 6 and 16
(3) 6 and 18
(4) 5 and 16
Answers
Answered by
2
i think the ans is option 2
Attachments:
sneh09:
how n why?
beacuse if we consider it non overlaped than it has.6 because a codon is a set of 3 amino acid
but if it overlaps than we have to take the amino acids of other codon also
ya and what about overlapping how 16?
so it is 16
i send u how to solve it
for overlapping sequence
thanku
Answered by
1
Answer:
(2)
Explanation:
if we consider non-overlapping, then there are 6 codons UCU , ACA , GUG , AUC , GUC , ACC . (3 nucleotides = 1 codon) BUT
If we consider non-overlapping , then the out of 3 nucleotides used in coding a codon , 2 nucleotides of it will be used for coding the codo next to it [BUT THE LAST TWO NUCLEOTIDE WILL BE ABLE TO CODE FOR ONLY 1 CODON BECAUSE THE CODONS NEXT TO IT WOULD HAVE 2 AND 1 NUCLEOTIDE RESPECTIVELY]. HENCE 16 codons =>
UCU
CUA
UAC
ACA
CAG
AGU
GUG
UGA
GAU
AUC
UCG
CGU
GUC
UCA
CAC
ACC
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