Question

1. 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	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	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	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	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	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	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	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	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’

2. Now find and circle the promoter region 3’-TATAAAA -5’ in yoursequence above.
3. 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.

4. 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’

5. 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’

6. 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

There are 6 potentialstart codon, TAC is the potential start codon.

mRNA is complimentary to DNA, and mRNA has Uracil(U) in place of Thiamine(T) .

In case of amino acid every codon of mRNA codes for an amino acid.
m- RNA Start - AUG Good bosh DNA -> ТА С ОДОО ДУ АДр) Im-RNAL AAA UAG UGA од АТТІ А ТО) АСТ) U AADUAG Stop (ТАТА ААА АТС TTC TC n AGA пAN C cТТ ТАСТcc TAC TCT стт AnT OC ATTC CAT CC C7 ACCTAC AT TOn A CAA On Сл C n Ch TCT (АСТ) АТС -

DNA“. 5 - ТАС Тсс ТДСТст CTT AGT CGA TTC CAT CGG ACC TAC GAT TGA CAA GCG CGG TCT ACT - 51 im RNA: 5'- AUG AGG AUG AGA GAA UCA GCU AAG GUA GCC UGG AUG GUA ACU GOU CGC GCC AGA UGA - 31 Amino acid segence : Met-Arg - Met - Arg- Glu- ser- Ala-lys - Val - Ala- Top - Met - Leu - This - Val - Arg- Ala- Arg.