The following sequence represents the DNA template strand of a gene: 3\' —TAC CG
ID: 7443 • Letter: T
Question
The following sequence represents the DNA template strand of a gene:3' —TAC CGT GTC TCC CTT GAT ATT GAA CTA—5'
nucleotide number 1 5 10 15 20 25
(a) Give the sequence of the mRNA sequence transcribed from this DNA strand and give the amino acid sequence of the protein it encodes.
(b) Give the amino acid sequence of the protein encoded by this sequence after the following mutations have occurred:
• a transition at nucleotide #10
• a transition at nucleotide #12
• a transition at nucleotide #7
• an insertion of a “G” immediately following nucleotide #10
• a three base pair insert (of your choice) starting with base 5
• What kind of mutated (four bases instead of the typical three) “pop” Amino Acid is created by inserting a G between base 14 and 15 of the original RNA strand
Explanation / Answer
a. The DNA strand is:
3' —TAC CGT GTC TCC CTT GAT ATT GAA CTA—5'
The mRNA sequence is:
AUG GCA CAG AGG GAA CUA UAA CUU GAU
The protein sequence is:
Met-Ala-Gln-Arg-Glu-Leu
b. Nucleotide #10 would be T. A transition would convert T to C
So the 4th codon would be affected. AGG would change to GGG. It would code for glycine.
Hence the modified protein sequence would be:
Met-Ala-Gln-Gly-Arg-Glu-Leu
Nucleotide # 12 would be C. A transition would convert C to T.
So the 4th codon would be affected. AGG would change to AGA. It would code for Arginine.
As both AGG and AGA code for arginine there will be no change in the sequence.
Nucleotide #7 would be G. A transition would change it to A. In mRNA, as T is absent, it is seen as U.
So codon CAG would be UAG. this is a stop codon.
the modified protein would be Met-ALa.
With insertion of a G after nucleotide #10, the DNA sequence would be:
3' —TAC CGT GTC TGC CCT TGA TAT TGA ACT A—5'
the mRNA sequence is:
AUG GCA CAG ACG GGA ACU AUA ACU UGA U
THE protein sequence would be:
Met-Ala-Gln-Thr-Gly-Thr-Ile-Thr
A 3bp insert starting at #5.
3' —TAC CCC TGT GTC TCC CTT GAT ATT GAA CTA—5'
the mRNA sequence would be:
AUG GGG ACA CAG AGG GAA CUA UAA CUU GAU
The protein sequence will be:
met-gly-Thr-Gln-Arg-Glu-Leu-
the modified RNA sequence after inserting a G between 14 and 15th nucleotides would be:
AUG GCA CAG AGG GAGA CUA UAA CUU GAU
The ribosome will not recognize a 4-bp codon as it reads only 3 bases at a time. It may have a frameshift and transfer the 4th nucleotide to the next codon. no amino acid may be attached, if frameshift does not occur.