5. You will need to start with DNA that was extracted from tissue or bacteria. T
ID: 3164858 • Letter: 5
Question
5. You will need to start with DNA that was extracted from tissue or bacteria. This DNA will likely be either genomic or plasmid. Remember that one human cell has 46 chromosomes containing 3.2 billion base pairs (or 6.4 billion nucleotide bases). That is a lot of DNA to Took through to find a specific segment. Let's suppose your target sequence is located within this segment of DNA below. You will use a pair of primers to identify your segment of interest. Write the sequence and polarity of the primers below: Forward Primer 5'GTATA TTTCT 5' Reverse Primer T2 5, A G T T C A G T A T A C C T G C G A A G C A A A A G A A C C T A T A C G 3. 6. In cycle 1 and all subsequent cycles of the PCR reaction, a copy of each of the two original template strands will be made in a 5' to 3' direction. Write the sequence of the copies that are made from the strands of T1 and T2. T1 3TCAAGTCATATGGACGCTTCGTTTTCTTGGATATGC5 C1F S GTATA 3 C1R 3 TTTCT5 T2 S'AGTTCAGTATACCTGCGAAGCAAAA GAACCTATACG 3 7. During cycle 2, new DNA strands will be synthesized from T1,T2, CIF, and CIR. Show the sequences of the new strands from CIF and CIR below: (label 5' and 3' of newly synthesized strands) C1F 5 GTATACCTGCGAAGCAAAAGAACCTATACG3 C2R ????? C2F GTATAExplanation / Answer
5). THE COMPLETE SEQUENCE IS:
T1 31 TCAAGTCATATGGACGCTTCGTTTTCTTGGATATGC 51
Forward primer 51GTATA
TTTCTT Reverse primer
T2 51AGTTCAGTATACCTGCGAAGCAAAAGAACCTATACG31
|
|
T1 31 TCAAGTCATATGGACGCTTCGTTTTCTTGGATATGC 51
T2 51AGTTCAGTATACCTGCGAAGCAAAAGAACCTATACG31
+
T1 31 TCAAGTCATATGGACGCTTCGTTTTCTTGGATATGC 5`1
T2 51AGTTCAGTATACCTGCGAAGCAAAAGAACCTATACG31
6).
T1 31 TCAAGTCATATGGACGCTTCGTTTTCTTGGATATGC 51
C1F 51GTATA
C1R31 TTTCT51
T2 51AGTTCAGTATACCTGCGAAGCAAAAGAACCTATACG31
|
|
T1 31 TCAAGTCATATGGACGCTTCGTTTTCTTGGATATGC 51
C1F 51AGTTCAGTATACCTGCGAAGCAAAAGAACCTATACG31
+
C1R 31 TCAAGTCATATGGACGCTTCGTTTTCTTGGATATGC 5`1
T2 51AGTTCAGTATACCTGCGAAGCAAAAGAACCTATACG31
7). C1F 51GTATACCTGCGAAGCAAAAGAACCTATACG31
C2R TTTCT
C2F GTATA
C1R 31 TCAAGTCATATGGACGCTTCGTTTTCTTGG 5`1
|
|
C1F 51GTATACCTGCGAAGCAAAAGAACCTATACG31
C2R 31 TCAAGTCATATGGACGCTTCGTTTTCTTGG 51
+
C2F 51AGTTCAGTATACCTGCGAAGCAAAAGAA31
C1R31 TCAAGTCATATGGACGCTTCGTTTTCTT 5`1
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