PCR Math Jim Burnette Given conversions: •Average molecular weight (MW) of a base pair = 650 daltons -18 •1 dalton (atomic mass unit) = 1.66 x 10 µg 23 •Avogadro’s number = 6.022x10 molecules gram 1 -6 µg 10 -9 ng 10 -12 pg 10 Given experiment: •lambda genome = 48,502 bp (base pairs) I used lambda because I could double check my work. You can use any genome size. •Primer pair has 1 target per genome •PCR amplimer (band) size is 450 bp (again, you can use any size) •Start with 1 picogram (pg) of DNA in the reaction •Assume doubling during all cycles of PCR. (This not true for the first 5 and last 5 or so cycles. But the assumption is ok.) Given the above, answer these three questions. As you answer, describe the logic you are using. All units must be shown. Use scientific notation and rounding when necessary. (I’m a stickler for scientific notation. It must be in the proper format or I will not accept it. Same for units.) 1. How many genomes of lambda are in 1 pg of lambda DNA? 2. Find the number of copies of the target after 30 cycles of PCR. 3. Find the mass of DNA generated after 30 cycles. 4. Find the moles of DNA generated after 30 cycles. 1 Find the number of copies of the target after 30 cycles of PCR. Need number of starting molecules (genomes) in 1 pg of lambda DNA. To get this we first need to know the mass of 1 genome. Since we were given daltons, we will calculate MW/genome in daltons and then convert to pg. A. Daltons per genome. ⎛ 48,502bp ⎞ ⎛ 650daltons ⎞ 7 ?daltons /genome = ⎜ ⎟ ⋅ ⎜ ⎟ = 3.1x10 daltons /genome bp ⎝ genome ⎠ µg. ⎝ ⎠ B. Convert MW in daltons to MW in −18 ⎛1.66x10 µg ⎞ −11 daltons #µg /genome = 3.1x10 7 genome ⋅ ⎜ ⎟ = 5.1x10 µg /genome ⎝ dalton ⎠ ( € € ) C. Convert MW in µg to MW in pg ⎛ 5.11x10−11µg ⎞ ⎛1x10 6 pg ⎞ −5 # pg /genome = ⎜ ⎟ ⋅ ⎜ ⎟ = 5.1x10 pg /genome genome ug ⎝ ⎠ ⎝ ⎠ D. calculate number of genomes € ⎛ 1genome ⎞ 4 # genomes = 1pg⎜ ⎟ ≅ 2x10 genomes −5 ⎝ 5.1x10 pg ⎠ Therefore, 1 pg of lambda DNA contains approximately 20,000 genomes. 2. Find the number of copies of the target after 30 cycles of PCR. € In 20,000 genomes there are 20,000 targets if the primers amplify only 1 target per genome. (This is easy but may not be for a student.) Student needs to know that since PCR doubles the number N of targets at every cycle the total doubling at the end is 2 where N=number of cycles. 20,000 ⋅ 2 30 = 2.1x1013 copies € 3. Find the mass of DNA generated after 30 cycles. 13 To find the mass of 2.1x10 copies of the target we need to know the mass of 1 target. The target is 450 bp in length. A. Find MW in daltons of target massof t arget = 450bp ⋅ 650daltons /bp = 2.9x10 5 daltons /t arget B. Convert MW in daltons to MW in µg. € ⎛1.66x10−18 µg ⎞ −13 # daltons /t arget = (2.9x10 5 daltons /t arget ) ⋅ ⎜ ⎟ = 4.8x10 µg /t arget dalton ⎝ ⎠ C. Calculate total mass ⎛ 4.8x10−13 µg ⎞ total mass DNA after 30cycles = 2.1x10 copies ⋅ ⎜ ⎟ = 10µg DNA copy ⎝ ⎠ 13 € Therefore 10µg of DNA will be made after 30 cycles of PCR. 4. Find the moles of DNA generated after 30 cycles. € Simply divide the number of molecules generated by Avogadro’s number. Of course the student will have to remember that 1 mole of anything is 6.022x1023 molecules. ⎛ ⎞ 1mole −11 # moles = 2.1x1013 molecules ⋅ ⎜ ⎟ = 3.5x10 moles 23 ⎝ 6.022x10 molecules ⎠ Since 3.5x10-11 moles is a difficult number, it would be reported as 35 pmol. There is another way to do this. The student must convert µg DNA to grams DNA. € € # moles = grams DNA MW in daltons # moles = 10x10−6 g = 3.5x10−11 moles 5 2.9x10 daltons Extensions: •Give a sequence and a total number of A, G, C, and Ts for 1 strand. Ask the student how many moles of each nucleotide will be consumed in the PCR. You would need to provide the MW of each base (or make them memorize it ). They would have to remember that for every A in the top strand they would need the equivalent number of Ts for the bottom, etc. See http://www.neb.com/nebecomm/tech_reference/general_data/default.asp for physical properties of nucleic acids and amino acids. •Have the student plot a portion of the reaction on semi-log paper. •other?