Question

Describe how you would prepare 1L of a 1X TAE solution from a 50X stock solution. Additionally, include how you would prepare 80ml of an agarose gel that is 3% in 1X TAE?

Answer 1

1 L=1000 ml

From, 50 X stock solution has to be 50 times diluted .

So, suppose the stock needed is X ml.

Then the final volume of diluted TAE , will be 50 X.

So the diluent needed will be =50-1 X=49 X.

So, X+49X =1000

X=(1000/50)=20 ml

And the diluent needed =1000-20 =980 ml.

So the stock needed is 20 ml and diluent needed is 980 ml.

3 % agarose means 3 grams of agarose is present in 100 ml of TAE buffer.

So, 100 ml of TAE buffer needs 3 grams of agarose.

1 ml of TAE buffer. Needs (3/100 ) grams of agarose.

80 ml of agarose is needs (3*80)/100 =2.4 grams of agarose.