2. Sequencing of PCR templates (lengths of double-stranded DNA to be amplified b
ID: 184642 • Letter: 2
Question
2. Sequencing of PCR templates (lengths of double-stranded DNA to be amplified by the polymerase chain reaction) is a job often sent out to university or commercial centers where the latest robotic capillary electrophoresis equipment can make short work of a tricky analysis. Sequencing centers are picky about sample preparation, setting strict limits on the volume and concentration of sample submitted and even the type of tube it's submitted in. The required concentration of template is proportional to the size of the molecule in the sample submitted-for a 1800 base-pair section of DNA, the ideal sample concentration is 180 ng/. Fortunately, the sample we're about to submit meets these requirements exactly. a) Given that the molecular weight of DNA is about 660 daltons per base pair in double stranded DNA, what would be the molecular weight of the 1800 bp fragment? (10 Points) How many molecules ofthe double-stranded DNA template would be of sample used in each sequencing reaction? (Hint: Avogadro Number is Naw 6.022 103 molecules/mol) (10 Points) b) und in the 5 c) The sequencing reaction also uses a single-stranded DNA "primer" molecule to bind to the template DNA and give the DNA polymerase a starting point. If the primer is a 20- base sequence, and 0.5 1 of a 2 pmol/ primer solution is used in the sequencing reaction, which molecule is in molar excess at the start of the reaction: the primer or the template? (5 Points)Explanation / Answer
Answer for 2(a):
The molecular weight of a molecule is always expressed as (weight /moles). The molecular weight of standard dna is approximately 660 daltons/bp for double stranded dna. Thus for a sample of 1800 bp the molecular weight of the sample would be approximately = 660*1800 daltons = 1188000 daltons = (1188000 * 1.661 x 10^-24) gm = 1.97*10^-18 gm
as 1 dalton = 1.661×1027 kg = 1.661 x 10^-24 gm
Answer for 2(b):
One unified atomic mass unit (U) or dalton is approximately the mass of one nucleon and is numerically equivalent to 1 g/mol. The concentration of DNA in the solution is 180 ng/ microlitre. So in 1.5 microlitre the concentration or amount of DNA is 1.5*180 ng = 270 ng = 2.7 x 10^-7 gm. Thus the number of moles of the sample fragment present in the solution is (1.97*10^-18)/(2.7*10^-7) = 7.29 * (10 ^-12) .
So the no of molecules of the fragment present is = N(avogardo) * 7.29 * 10 ^-12 = 4.39*(10^12)
Answer for Question 2(c):
At the start of the reaction, the mix has 7.29 * (10 ^-12) moles while the primer present is [ (2*10^-12) /0.5 ] moles = 0.5 *10^-12 moles. thus the sample is present in excess
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