A fragment of M13 phage DNA shown below was mixed with thefollowing primers P1:

Question

A fragment of M13 phage DNA shown below was mixed with thefollowing primers P1: AGTAG AATTGATGCCACCTT; P2: TGT CTG GAA GTTTCA TTC CA and incubated in a thermal cycler in the presence ofdNTPs and Taq DNA polymerase under the conditions of PCR.       AATGCTACTA CTATTAGTAG AATTGATGCC ACCTTTTCAG CTCGCGCCCCAAATGAAAAT ATAGCTAAAC AGGTTATTGA CCATTTGCGA AATGTATCTAATGGTCAAAC TAAATCTACT CGTTCGCAGA ATTGGGAATC AACTGTTACA TGGAATGAAACTTCCAGACA CCGTACTTTA (180)             (a) What is the length of the PCR product? _______
(b) What is the final concentration of PCR product after 10cycles if starting template concentration is 1nM?
I have the answers but I need to know why =( (a) (155NT * 3.4Anstroms=527Anstroms)
and (b) 1nM * 2^9 = 512nM

      AATGCTACTA CTATTAGTAG AATTGATGCC ACCTTTTCAG CTCGCGCCCCAAATGAAAAT ATAGCTAAAC AGGTTATTGA CCATTTGCGA AATGTATCTAATGGTCAAAC TAAATCTACT CGTTCGCAGA ATTGGGAATC AACTGTTACA TGGAATGAAACTTCCAGACA CCGTACTTTA (180)             (a) What is the length of the PCR product? _______
(b) What is the final concentration of PCR product after 10cycles if starting template concentration is 1nM?
I have the answers but I need to know why =( (a) (155NT * 3.4Anstroms=527Anstroms)
and (b) 1nM * 2^9 = 512nM

Explanation / Answer

Answer for 1st part of the question: You Have Two Primers: P1: AGTAG AATTGATGCCACCTT is forward primer meaning from (5' to3') P2: TGT CTG GAA GTT TCA TTC CA is reverse primer, which means (3'to 5') but when it will read it will read (5' to 3') on thecomplement strand, which means it will be (5'--TGGAATGAAACTTCCAGACA--3') 3'--TGT CTG GAA GTT TCA TTC CA--5' 5'--AC CTT ACT TTG AAG GTC TGT--3' (GET THE COMPLEMENT FOR REVERSEPRIMER) 5'--TG GAA TGA AAC TTC CAG ACA--3' In your DNA, PCR will amplify the reigion between the two primers(including the primer), so if you count how many nucleotides thereyou will find them 155.... then multiply by 3.4 A, you will get 527 A AATGCTACTA CTATTAGTAG AATTGATGCC ACCTTTTCAG CTCGCGCCCC AAATGAAAAT ATAGCTAAAC AGGTTATTGA CCATTTGCGA AATGTATCTA ATGGTCAAACTAAATCTACT CGTTCGCAGA ATTGGGAATC AACTGTTACA TGGAATGAAA CTTCCAGACACCGTACTTTA (180) AATGCTACTA CTATTAGTAG AATTGATGCCACCTTTTCAG CTCGCGCCCC AAATGAAAAT ATAGCTAAACAGGTTATTGA CCATTTGCGA AATGTATCTA ATGGTCAAAC TAAATCTACT CGTTCGCAGAATTGGGAATC AACTGTTACA TGGAATGAAACTTCCAGACA CCGTACTTTA (180)

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