4) Consider the following sequence of a DNA double helix. (20 poines) A) Using t

Question

4) Consider the following sequence of a DNA double helix. (20 poines) A) Using the genetic code table below, A) determine the mRNA sequence resulting from transcription. (5 points) Seond letter UAA Stop UGA Stop UAG Stop UOG TrpG ??? CCA Pra Lys AGG B) Depending on a combination of intracellular and extracelilir conditions, the mRNAttained in part (A) is used to either produce a 20 amino acid prodein (Protein D or an 11 amino acid protein (Protein II, What are the sequences of these two prodeins? (5 points) (0 ame (Il oreure scds C) Consider the following point mutation, where a cytosine in the 3'-' sequence is substituted with a thymine as i How would this mutation affect Protein I and Protein II. Write down the new amino acid sequences for both proteins that result from these mutations. (Hint: Determine the mRNA sequence with the T incorporated in place of the C first). (10 points)

Explanation / Answer

A.We know that transcription proceeds in 5' -3' direction, that is, 5' end is formed first and proceeds in 3' direction. So we will take 3'-5' sequence of DNA into consideration

First of all, the given sequence of DNA is

3'- GTTTACGCTGCGCCCGGCAGTACGACAACCGAAAGGCAT GAGCACTCTGTGTGATCCTTVTAAATCCAA -5'

So the resulting m-RNA should be

5'-CAAAUGCGACGCGGGCCGUCAUGCUGUUGGCUUUCCGUACUCGUGAGACACACUAGGAAGAUUUAGGUU -3'.

B. The process of protein formation always starts at the codon AUG. We can say, AUG is start codon, every cycle will only begin with AUG. No other codon can replace it.

Now, the 2 protiens secreted from this mRNA are-

i) Protein I- 20 amino acids chain

CAAAUGCGACGCGGGCCGUCAUGCUGUUGGCUUUCCGUACUCGUGAGACACACUAGGAAGAUUUAGGUU

So the resulting amino acid sequence will be

Met- Arg- Arg- Gly- Pro- Ser- Cys- Cys- Trp- Leu- Ser- Val- Leu- Val- Arg- His- Thr- Arg- Lus- Ile- STOP

ii). Protein II- 11 amino acids chain

Here, the initiation will start at AUG

CAAAUGCGACGCGGGCCGUCAUGCUGUUGGCUUUCCGUACUCGUGAGACACACUAGGAAGAUUUAGGUU

So, the resulting sequence of amino acids will be

Met- Leu- Leu- Ala- Phe- Arg- Thr- Arg- Glu- Thr- His- STOP.

C. Due to point mutation, the resulting DNA sequence will be

3'-GTTTACGCTGCGCCCGGCAGTACGACAATCGAAAGGCATGAGCACTCTGTGTGATCCTTVTAAATCCAA -5'

Therefore the mRNA will be

5'- CAAAUGCGACGCGGGCCGUCAUGCUGUUAGCUUUCCGUACUCGUGAGACACACUAGGAAGAUUUAGGUU -3'.

Now, the protein sequence will be

i). Protein I

5'- CAAAUGCGACGCGGGCCGUCAUGCUGUUAGCUUUCCGUACUCGUGAGACACACUAGGAAGAUUUAGGUU -3'

The 20 amino acid will reduce to 9 amino acid chain

Met- Arg- Arg- Gly- Pro- Ser- Cys- Cys- Trp- STOP

ii). Protein II

5'- CAAAUGCGACGCGGGCCGUCAUGCUGUUAGCUUUCCGUACUCGUGAGACACACUAGGAAGAUUUAGGUU - 3'

The 11 amino acid chain will remain 11 amino acids chain.

Met- Leu- Leu- Ala- Phe- Arg- Thr- Arg- Glu- Thr- His- STOP.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

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