Question

Linda has asked Brenn to prepare a 1% agarose gel for a gel electrophoresis analysis. But before he could do that, he needs to prepare 600 ml of 1x TAE from 95x TAE. How much 95x TAE and water should he add to create the 1x TAE solution? Afterwards, how much agarose does he need to weigh out, in grams, and add into 50 ml of 1x TAE?

Answer 1

To prepare 600 mL of 1x TAE:

Stock = 95x TAE

Working = 1x TAE (600mL)

Calculations:

Stock Concentration                 Working Concentration

C1V1    =   C2V2

(95x TAE) X V1          =            (1x TAE) X (600mL)

                     V1          =              (1x TAE) X (600mL) / (95x TAE)

                     V1         =          6.316 mL

                     V1          =         6.32 mL

Volume of 95x TAE used = 6.32 mL

Volume of water = Total volume of 1xTAE to be prepared - Volume of 95x TAE

Volume of water = 600mL – 6.32mL

Volume of water = 593.68 mL

So 6.32 mL of 95x TAE and 593.68 mL of water should be added to create the 1x TAE solution.

To weigh-out agarose in grams:

1 % agarose = 1 gram of agarose in 100 mL of 1x TAE

For 50 mL 1x TAE

Weight of agarose = (50/100) X 1 gram

= 0.5 gram

Weight of agarose for 50 mL of 1x TAE = 0.5gram

i.e. to prepare 1% agarose gel 0.5 gram of Agarose is added to 50 mL 1xTAE.