Question 2. (9 pts). The following DNA sequence has been used as template for tr

Question

Question 2. (9 pts). The following DNA sequence has been used as template for transcription originating the mRNA that was translated to produce the corresponding polypeptide as follows. DNA RNA polypeptide 5 CAUGGGCUGGAAGGAUUAGAUG 3 N- Met Glu TR Lys Asp C Point mutations have been made to the original DNA template (indicated in bold red color). In each case, identify the type of point mutation with respect to the change in the sequence of bases in the DNA template and the translated polypeptide (use ALL necessary terms to describe the mutation type). a (2 pts) DNA: 3 GTACCCGACTTTCCTAATCTAC 5 What type of point mutation was made in the DNA template above? Will that point mutation affect the polypeptide translated from the mRNA transcribed from the DNA template? EXPLAIN What type of point mutation was made in the DNA template above? Will that point mutation affect the polypeptide translated from the mRNA transcribed from the DNA template? EXPLAIN What type of point mutation was made in the DNA template above? Will that point mutation affect the polypeptide translated from the mRNA transcribed from the DNA template? EXPLAIN d) (3 pts) You will now introduce a TRANSVERSION SILENT point mutation to the original DNA molecule at the heginning of this questinn (in hlue) Write the FNTIRF nOIRI F STRANDFD DNA molecule containing that mutation, the RNA transcribed from it and theesting polypeptide

Explanation / Answer

) There is substitution of thymine between in place of cytosine. Hence, this mutation will be an base substitution point mutation. RNA is complementary to template strand

Template 3’ G TAC CCG ACT TTC   CTA ATC TAC 5’

RNA           5’ C AUG GGC UGA AAG GAU UAG AUG 3’

Peptide        N- Met- Gly- Stop-C

There is premature stop codon created. Hence, it is nonsense mutation,

b). There is a substitution of guanine base by adenine. Hence, it is a base substitution point mutation

Template 3’ G TAC CCT ACC TTC CTA ATC TAC5’

RNA           5’ C AUG GGA UGG AAG GAU UAG AUG 3’

Peptide        N- Met- Gly- Try-   Lys- Asp- Stop-C

There is substitution of Glycine by glycine. Hence, it is a silent mutation.

c) There is a addition of adenine after cytosine. Hence, it is a insertion point mutation.

Template 3’ G TAC CCG ACC   ATT CCT   AAT CTA C 5’

RNA           5’ C AUG GGC UGG UAA GGA UUA GAU G 3’

Peptide          N- Met- Gly- Try-   Stop-C

There is premature termination of polypeptide. Hence, it is an nonsense mutation.

d) Transversion mutation are change of a purine base with a pyrimidine. Silent mutation do not alter the aminoacid sequence of polypeptide.

DNA

5’ C ATG GGA TGG AAG GAT TAG ATG 3’

3’ G TAC CCT ACC TTC   CTA ATC TAC 5’

RNA           5’ C AUG GGU UGG AAG GAU UAG AUG 3’

Peptide        N- Met- Gly- Try-   Lys- Asp- Stop-C

There is a transversion of purine guanine with pyrimidine thymine. However the amino acid remains the same. Hence, it is an transverison silent mutation.

  

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.