Q1)
the sequence given, we need to read from 5' to 3' and find where the reading frame starts. That's where atg is found.
5’ agcggg atg agcgcatgtggcgcataactg3’
from here onwards we have to separate the bases into groups of three as these are codons that each code for an amino acid.
5’ agcggg atg agc gca tgt ggc gca taa ctg 3’
Met Ser Ala Cys Gly Ala stop
TAA(UAA in mRNA ) is the stop codon so reading frame stops here
we change base A to T (capitalised)
DNA sequence with amino acids are given
5’ agcggg atg Tgc gca tgt ggc gca taa ctg 3’
N Met Cys Ala Cys Gly Ala stop
after changing the base the amino acid sequence changes from Ser to Cys.
Q2)
the complementary strand of the above strand is as follows
5' cagttatgcgccacatgcgctcatcccgct 3'
start codon starts with atg thats where the reading frame starts
5' cagtt atg cgc cac atg cgc tca tcc cgc t 3'
Met Arg His Met Arg Ser Ser Arg
After changing base from A to T, the complementary strand changes from T to A (capitalised)
5' cagtt atg cgc cac atg cgc Aca tcc cgc t 3'
Met Arg His Met Arg Thr Ser Arg
amino acid changes from Ser to Thr.
Q3)
The sequence with amino acids before inserting a base is
5’ agcggg atg agc gca tgt ggc gca taa ctg 3’
Met Ser Ala Cys Gly Ala stop
We insert a base G shown in capitals
5’ agcggg atg agc Ggca tgt ggc gca taa ctg 3’
This changes the codons of bases after the inserted base
5’ agcggg atg agc ggc atg tgg cgc ata act g 3’
Met Ser Gly Met Trp Arg Ile Thr
the amino acid completely changes from Met Ser Ala Cys Gly Ala
to Met Ser Gly Met Trp Arg Ile Thr
Q4)
the complementary strand before adding a base is
5' cagtt atg cgc cac atg cgc tca tcc cgc t 3'
Met Arg His Met Arg Ser Ser Arg
When we insert a base G, base C is added to the complementary strand
5' cagtt atg cgc cac atg cCgc tca tcc cgc t 3'
this changes the codons
5' cagtt atg cgc cac atg cCg ctc atc ccg ct 3'
Met Arg His Met Pro Leu Ile Pro
With insertion of one base the amino acid sequence changes from
Met Arg His Met Arg Ser Ser Arg
to Met Arg His Met Pro Leu Ile Pro
