Question

I STARTED WRITING THE ANSWERS TO THIS PROBLEM. WE ARE TO COME UP WITH DNA SEQUENCES FROM THE MRNA SEQUENCE. AM I GOING IN THE RIGHT DIRECTION WITH THE DNA SEQUENCES FOR THIS PROBLEM ?

1.The wildtype sequence of part of a protein is:

NH₂-Trp-Trp-Trp-Met-Arg-Glu-Trp-Thr-Met…

            Each mutant in the following table differs from wildtype by a single point mutation (base substitution). Using this information, determine the DNA sequence coding for the wildtype polypeptide. If there is more than one possible DNA sequence, give all possible DNA sequences. Hint: you should only consider the changes that could lead to the mutants listed.

           

           

Mutant	Amino acid sequence of polypeptide
1	Trp-Trp-Trp-Met
2	Trp-Trp-Trp-Met-Arg-Asp-Trp-Thr-Met
3	Trp-Trp-Trp-Met-Arg-Lys-Trp-Thr-Met
4	Trp-Trp-Trp-Met-Arg-Glu-Trp-Met-Met

1. The wildtype sequence of part of a protein is:

NH₂-Trp-Trp-Trp-Met-Arg-Glu-Trp-Thr-Met…

The Mrna transcript of the wildtype sequence is

                                 5’ UGG-UGG-UGG-AUG-AGA-GAG-UGG-ACG-AUG ’3

The Dna sequence would be

                                 5’ ACC-ACC-ACC-TAC-TCT-CTC-ACC-TGC-TAC ‘3

                                 3’ TGG-TGG-TGG-ATG-AGA-GAG-TGG-ACG-ATG ‘5

2. For the 1^st Mutant amino sequence. The 5^th amino acid Arg will convert to a stop codon. AGA to UGA

                               (Mrna) 5’ UGG-UGG-UGG-AUG-UGA ‘3

                                

                                (Dna) 5’ ACC-ACC-ACC-TAC-ACT ‘3

                                           3’ TGG-TGG-TGG-ATG-TGA’5

1. The wildtype sequence of part of a protein is:

NH₂-Trp-Trp-Trp-Met-Arg-Glu-Trp-Thr-Met…

            Each mutant in the following table differs from wildtype by a single point mutation (base substitution). Using this information, determine the DNA sequence coding for the wildtype polypeptide. If there is more than one possible DNA sequence, give all possible DNA sequences. Hint: you should only consider the changes that could lead to the mutants listed.

           

           

Mutant	Amino acid sequence of polypeptide
1	Trp-Trp-Trp-Met
2	Trp-Trp-Trp-Met-Arg-Asp-Trp-Thr-Met
3	Trp-Trp-Trp-Met-Arg-Lys-Trp-Thr-Met
4	Trp-Trp-Trp-Met-Arg-Glu-Trp-Met-Met

1. The wildtype sequence of part of a protein is:

NH₂-Trp-Trp-Trp-Met-Arg-Glu-Trp-Thr-Met…

The Mrna transcript of the wildtype sequence is

                                 5’ UGG-UGG-UGG-AUG-AGA-GAG-UGG-ACG-AUG ’3

The Dna sequence would be

                                 5’ ACC-ACC-ACC-TAC-TCT-CTC-ACC-TGC-TAC ‘3

                                 3’ TGG-TGG-TGG-ATG-AGA-GAG-TGG-ACG-ATG ‘5

2. For the 1^st Mutant amino sequence. The 5^th amino acid Arg will convert to a stop codon. AGA to UGA

                               (Mrna) 5’ UGG-UGG-UGG-AUG-UGA ‘3

                                

                                (Dna) 5’ ACC-ACC-ACC-TAC-ACT ‘3

                                           3’ TGG-TGG-TGG-ATG-TGA’5

Answer 1

Please refer to following explanation for determining the DNA sequence coding for the wild type polypeptide. This explanation will help you clearly decide whether you are going in right direction with the DNA sequences for the problem or not.

During transcription, DNA strand with 3' to 5' polarity is template strand as RNA can be synthesized only in 5' to 3' direction. As mRNA synthesis occurs on template strand, which has 3' to 5' polarity, the base sequence on mRNA molecule is complementary to template strand (3' to 5' polarity) of DNA molecule. The base sequence on other strand of DNA molecule is same to the mRNA transcript. This strand of DNA molecule having base sequence similar to mRNA transcript is called coding strand and its polarity is 5' to 3'. Thymine on coding strand is replaced by Uracil on mRNA transcript.  

You are not going in the right direction with the DNA sequences for the problem as in DNA sequence determined by you for the wild type polypeptide, the base sequence on DNA strand with 3' to 5' polarity is similar to base sequence on mRNA transcript. Instead, the base sequence on DNA strand with 5' to 3' polarity would be similar to base sequence on mRNA transcript. Uracil on mRNA transcript would be replaced by Thymine on DNA strand with 5' to 3' polarity.

So, correct DNA sequence coding for the wild type polypeptide would be

3’ ACC-ACC-ACC-TAC-TCT-CTC-ACC-TGC-TAC 5'

5’ TGG-TGG-TGG-ATG-AGA-GAG-TGG-ACG-ATG 3'

Please note that in the DNA sequence given above the base sequence on DNA strand with 5' to 3' polarity is similar to base sequence on mRNA transcript (except thymine replaces uracil).

Similarly, you are not going in right direction with the DNA sequence for mutant 1. Correct DNA sequence coding for mutant 1 amino acid sequence would be

3’ ACC-ACC-ACC-TAC-ACT 5'

5’ TGG-TGG-TGG-ATG-TGA 3'