Question

Your TA is going to have to make a lot of 0.5X TAE for the amount of agarose gels we are going to be running. 0.5X TAE consists of 0.02 M Tris-base, 9 mM acetic acid and 500 mM EDTA. To make his job simple he'll be preparing a 10x solution of TAE and diluting it to 0.5X TAE. Could you please tell him the amount, in grams, of each component he must add prior to adding water to reach one liter to make the 10x solution? and finally what fold-dilution must she make to go from 10X to 0.5X?

Answer 1

PREPARATION OF 10 X SOLUTION OF TAE= 1000 ml

10X TAE BUFFER CONTAINS

1] 0.02 M Tris-base,=

121.14 gm of tris base -----------1000 ml distilled water--------1 M

? gms of tris base ------------ 1000 ml D.w--------------0.02 M

= 121.14 X 0.02

= 2.42 gms.

2] 9 mM acetic acid

molarity is 17.4

N1V1 =N2V2

17.4 X V1 = 0.009 X 1000

V1 = 0.009X 1000/17.4

= 9/17.4

= 0.52 ml

require 0.52 ml of acetis acid

3]500 mM EDTA

292.24 gm EDTA ------------- 1000 ml-----------1 M or 1000 mM

? ---------------------------------1000 ml of dw --------500mM

= 146 gms of EDTA

he should add 2.42 gms of tris base ,0.52 ml of acetic acid & 146.12 gm of EDTA dissolved in small amount of water then make final volume upto 1000 ml with distilled water.

final concentration is 10X

preparation of 0.5 X buffer

first make 1X Tae buffer from stock of 10X buffer - take 1 ml of 10X bufer add 9 ml of distilled water

then from 1X buffer take 1:1 dilutions for making 0.5X buffer

for making 1000 ml of 0.5 X buffer take 500 ml of 1 X buffer & add 500 ml of distilled water to it.