Question

Assuming the average weight of a deoxynucleotide monophosphate (dNMP) is 327.0 g/mol, how many picomoles of DNA are present in 500ng of a 1500bp DNA fragment?

Answer 1

Ans. 1 base pair (pb) means two nucleotides are paired with complementary H-bonding.

So, Total number of dNMP in 1500 bp = (2 dNMP/ base pair) x number of base pairs

                                                            = (2 dNMP/ base pair) x 1500 base pairs

                                                            = 3000 dNMP

Total molar mass of 1500 bp DNA = (Molar mass / dNMP) x No. of dNMP in DNA

                                                            = (327.0 g mol^-1 / dNMP) x 3000 dNMP

                                                            = 981000 g/ mol

# Given,

            Mass of DNA sample = 500 ng                                          ; [1 ng = 10^-9 g]

                                                = 500 x 10^-9 g

                                                = 5.00 x 10^-7 g

Now,

            Moles of DNA = Mass / Molar mass

                                                = 5.00 x 10^-7 g / (981000 g/ mol)

                                                = 5.0968 x 10^-13 mol                         ; [1 mol = 10¹² picomole]

                                                = 5.0968 x 10^-13 x 10¹² pmol

                                                = 0.51 pmol

Hence, amount of 1500bp DNA in sample = 0.51 pmol