Question

ASSIGNMENT

1. For the DNA sequence given below, write the complementary DNA sequence that would complete the double-strand.  

DNA    3’—A   T   T  G   C   T   T   A  C   T   T  G   C   A   T -- 5’

DNA    5’--

2. Does it matter which strand is the ‘code strand’? The following two sequences look identical, except one runs 3’-5’ and the other 5’-3’. For each DNA sequence given below, write the mRNA sequence that would be coded from it. Make sure you indicate the direction of each mRNA strand (i.e. 3’ and 5’ ends).  Use the Universal triplet code to determine the sequence of amino acids that would be generated for each of the mRNA sequences that you generated in question 2. Remember that the reading of mRNA goes in the 5’-3’ direction (see lab notes for examples).

1. To make the mRNA, read from LEFT TO RIGHT → →

To make the amino acid, read triplets from Left to Right.

DNA         3’- T A C C T A C T T T G C C C G A T C C A T– 5’

mRNA  5’---

amino acid

b.  Read letter by letter from RIGHT TO LEFT ←←

      To make the amino acid, read triplets from Left to Right

DNA         5’- T A C C T A C T T T G C C C G A T C C A T– 3’

mRNA  5’---

amino acid

3. Are the amino acid sequences in a and b the same?     YES____          NO____

3. What 3’ to 5’ DNA sequence would correspond to a 5’-AUG-3’ start codon?

RNA 5’— A   U   G—3’

DNA   3’--________--5’

4. Read the DNAsequence given below in the 3’ to 5’ direction and circle all of thepotentialstart codons (i.e. every time you see the DNA version of a start codon- as in answer to question 3).

How many potentialstart codons did you find?_______________

Line	YOUR DNA SEQUENCE (read from left to right)
1	1) 3’- T T C C G T A A C T T C G G C G C A T C T A G C T T G A G C T C C A A T C A G G
2	2) A C T A C T T A T A A A A A T C T T C T C G A G A G A G C T T T A C T C C T A C T C
3	3) T C T T A G T C G A T T C C A T C G G A C C T A C G A T T G A C A A G C G C G G T C
4	4) T A C T A T C T A C T T A T T T A T T T A C G A G C G T T G A T T C T A C C T A C T G
5	5) A C G A C T A G G G C A T T C T A T A G G A T T A A C T C C T T A T T T T A A C T A
6	6) A C T T C C T G G G A A G G C G C C T T T A T C T G A T C C G T A A T C C G T C C G
7	7) A A G G C T C T G A T C G G A T T A C T G G G T C A C T G G A A A G T G A C C G C T
8	8) G T C T A T T A C T G T A T T T C A T C T G A T T G A C T A T T T T A T A G T C G -5’

5. Now find and circle the promoter region 3’-TATAAAA -5’ in yoursequence above.
6. Identify the start codon that is immediately downstream (towards the 5’ end, after the promoter) of the promoter. Indicate the codons that follow the start, and continue until you reach a stop codon.

7. Copy the entire coding sequence that is found in question 4 (from, and including the start codon to the stop codon). Hint: there are 19 codons, each with 3 base pairs, and the coding region is between lines 2 and 4.

DNA:  3’ –

__ __ __    __ __ __   __ __ __    __ __ __    __ __ __   __ __ __    __ __ __    

__ __ __    __ __ __   __ __ __    __ __ __    __ __ __   __ __ __    __ __ __   

__ __ __   __ __ __    __ __ __    __ __ __   __ __ __ -5’

8. Now write down the mRNA sequence that would be generated from the DNA sequence from question 7. Hint: there are 19 codons, each with 3 base pairs.

mRNA: 5’-

__ __ __    __ __ __   __ __ __    __ __ __    __ __ __   __ __ __    __ __ __    

__ __ __    __ __ __   __ __ __    __ __ __    __ __ __   __ __ __    __ __ __   

__ __ __   __ __ __    __ __ __    __ __ __   __ __ __ - 3’

9. Finally write out the sequence of amino acids that the mRNA sequence in question 8 would code for. Hint: there are 18 amino acids (including the start/methionine). Use the Universal Triplet code from the Gene Expression tutorial document.

Amino acid sequence:______-______-_______-_______-______-_____-______-______-______-______-______-______-______-______-______-______-______-______

Answer 1

Template strand 3’- A   T   T  G   C   T   T   A  C   T   T  G   C   A   T - 5’

Coding strand 5’- T    A A C G    A A   T G   A   A C   G T    A -3’

• 5’ to 3’ strand of DNA serve as coding strand.
• 3’ to 5’ strand of DNA act as template strand.
• RNA polymerase recognizes specific sequences in promotes region and binds to template strand.
• RNA is synthesized by complementary base pairing with the template strand. Coding strand serve as reference.
• Thus, the RNA will have same sequence as coding strand (only T is replaced by U) and is oriented in 5’ to 3’ direction.

a.

Template strand:  3’- T A C C T A C T T T G C C C G A T C C A T-5’

Coding strand   : 5’- A T G G A T G A A A CG G G C T A G G T A-3’

mRNA: 5’- AUG GAU GAA ACG GGC UAG G U A-3’

Amino acid:       Met-Asp-Glu- Thr-Ala

b.  

Coding DNA strand: 5’- T A C C T A C T T T G C C C G A T C C A T– 3’

mRNA:          5’- UAC CUA CUU UGC CCG AUC CAU– 3’

Amino acid: Tyr-Leu- Leu- Cys-Pro-Pro-Ile-Leu

No. Amino acid sequences from a and b are not same.

RNA 5’- A  U  G -3’

DNA   3’- T A C -5’