You are a scientist working for a paternity testing center. You spent all Friday
ID: 543044 • Letter: Y
Question
You are a scientist working for a paternity testing center. You spent all Friday isolating dsDNA from 100 hair samples from children and their possible dads. In order to get a quick start on your Southern Blot for Monday, you took absorbance readings for all samples using a spectrophotometer. On Monday morning you realize that you have misplaced some info for 3 of the samples. The bad news is that that spectrophotometer is broken, so you can’t take new measurements – but you need the information right away.
The missing info is indicated in the Table below:
SAMPLE
A260 nm
Concentration
1
0.15
?
2
?
5 µg/ml
3
0.06
?
Some useful information: The path length is 1 cm, and for dsDNA = 20 g-1cm-1L (i.e. 1 A260 O.D. unit for dsDNA = 50 µg/ml). The total volume for each sample is 100µl.
a) Calculate the missing values for each sample (reporting concentrations in µg/ml), and show your work. (3 marks)
b) How much total DNA do you have in sample 3? (1 mark)
c) If one diploid cell has ~ 6 picograms of dsDNA, how many cells worth of DNA do you have in sample 3 (total volume)? (1 mark) N.B. 1pg = 1 x 10-12 g
SAMPLE
A260 nm
Concentration
1
0.15
?
2
?
5 µg/ml
3
0.06
?
Explanation / Answer
Ans. #A. An absorbance of 1.00 means dsDNA (double stranded DNA) concentration of 50 ng/ L 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
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