You are a scientist working for a paternity testing center. You spent all Friday isolating dsDNA...

Ans. #A. An absorbance of 1.00 means dsDNA (double stranded DNA) concentration of 50 ng/ μL OR 20 ug/ mL

#A1. [DNA] unknown = Absorbance of unknown x (50 ug/ mL)

Or, [DNA] unknown = 0.15 x (50 ug/ mL)

Or, [DNA] unknown = 7.50 ug/ mL

Therefore, [DNA] in original sample 1 = 7.50 ug/ mL

#A2. [DNA] unknown = Absorbance of unknown x (50 ug/ mL)

Or, 5 ug/ mL = Abs x (50 ug/ mL)

Or, Abs = (5 ug/ mL) / (50 ug/ mL)

Hence, Abs = 0.10

Therefore, Abs of sample 2 = 0.10

#A3. [DNA] unknown = Absorbance of unknown x (50 ug/ mL)

Or, [DNA] unknown = 0.06 x (50 ug/ mL)

Or, [DNA] unknown = 3.0 ug/ mL

Therefore, [DNA] in original sample 3 = 3.0 ug/ mL

#B. Given, volume of sample 3 = 100.0 uL = 0.100 mL

Amount of DNA in 100.0 uL original sample =

[DNA] in original sample x Volume of solution in mL

= (3.0 ug/ mL) x 0.100 mL

                                    = 0.30 ug

#C. Mass of dsDNA in one cell = 6.0 x 10^-12 g                             ; [1 g = 10⁶ ug]

                                                = 6.0 x 10^-6 ug

Hence, DNA content in cell = 6.0 x 10^-6 ug / cell

Now,

Number of cells in sample 3 = Total DNA content of sample 3 / DNA content of cell

                                                = 0.30 ug / (6.0 x 10^-6 ug / cell)

                                                = 5.00 x 10⁴ cells