You found that a diluted sample contained 2.3 ng mL^-1 nucleic acid by measuring
ID: 481515 • Letter: Y
Question
You found that a diluted sample contained 2.3 ng mL^-1 nucleic acid by measuring the A_260 of a 1:200 dilution. How much nucleic acid was in the original solution? A 150 mu L solution of RNA (1250 mu g mL^-1) must be diluted prior to analysis on an agrees gel. Two dilutions of the original solution at 1.0 mu g mu L^-1 and 0.10 mu g mu L^-1 are typically run together on the same gel. An aliquot of 10.0 mu L at each dilution is added to 2.0 mu L of 6X gel loading buffer. Show calculations for the preparation of these samples. Keep in mind that you want to preserve as much of the original solution as possible.Explanation / Answer
Ans. 1. Given, concentration of diluted sample = 2.3 ng mL-1
Dilution factor = 1: 200
concentration of original sample = concentration of diluted sample / dilution factor
= (2.3 ng mL-1) / (1/200)
= 4.6 x 102 ng mL-1 ; [1 ng = 10-3 µg]
= 0.46 µg mL-1
Ans. 2. Concentration of stock solution = 1250 µg/ mL
= 1250 µg/ 1000 µL ; [1 mL = 1000 µL]
= 1.250 µg/ µL
Concentration of first desired solution = 1.0 µg/ µL
Dilution factor required = desired concentration / concentration of stock solution
= (1.0 µg/ µL) / (1.250 µg/ µL)
= 1: 1.250
That is, when 1 µL of stock solution (1.250 µg/ µL) is diluted to 1.250 µL, the concertation of resultant desired solution becomes 1.250 µg/ µL. Consider it as follow-
Amount of RNA in 1.0 µL stock solution = 1.250 µg
Final volume of the desired solution = 1.250 µL
Concertation of desired solution = Mass of RNA / volume of final solution
= 1.250 µg / 1.250 µL
= 1.0 µg/ µL
So, the required dilution for our 1st desired solution = 1: 1.250
Volume of stock solution to be diluted = 150 µL
Since, 1.0 µL of stock is to be diluted upto 1.250 µL
Or, 150 µL - - - - (1.250 x 150) µL
= 187.5 µL
Thus, take 150 µL of stock solution in a sterile Eppendorf tube and make the final volume upto 187.5 µL with distilled water. That is -
Volume of desired solution 1 (187.5 µL) = 150 µL (stock soln.) + 37.5 µL distilled water.
Now: Desired solution 2: Desired solution 1 is further diluted to form desired solution 2 as follow-
Concentration of desired solution 2 = 0.1 µg/ µL
Dilution factor required = desired concentration / conc. of desired solution 1
= (0.1 µg/ µL) / (1.0 µg/ µL)
= 1: 10
That is, when 1 µL of desired solution 1 (1.0 µg/ µL) is diluted upto 10 µL, the concertation of resultant desired solution 2 becomes 0.1 µg/ µL.
Let we require to prepare 20 µL (10 µL + 100% excess) of solution 2.
Volume of solution 1 required = volume of solution 2 required x dilution factor
= 20 µL x (1/10)
= 2 µL
Thus, diluted 2 µL of solution 1 upto 20 µL with sterile distilled water to get the desired solution 2.
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