Question

.           Describe in detail, how you would prepare a 0.8% agarose gel, which is 1X TAE and 0.1ug/ml Ethidium bromide. Your gel box holds 50 ml. TAE is 50X, Ethidum bromide is 10 mg/ml.

Answer 1

0.8% of agarose means that 0.8gms of agarose dissolved in 100ml of 1X TAE.

For 50 ml we will need 0.4 gm of agarose.

The stock solution of TAE is 50X. We need to make 50 ml of 1X TAE.

So we will need X ml of 50X TAE.

C1V1 = C2V2

C1 stock concentration

V1 volume of stock solution

C2 working concentration

V2 volume of working solution.

50 *X = 1 * 50

X = 1ml.

So we need to add 1ml 0f 50X TAE in 49 ml of distilled water to make 50ml of 1X TAE.

0.1ug/ml Ethidium bromide. Ethidium bromide is 10 mg/ml

We need X ml of Ethidium bromide to make 0.1microgram/ml solution

1mg = 1000 microgram

C1V1 = C2V2

10*1000 *X = 0.1*50

X = 0.1*50/10000

X= 5/10000

X= 5*1000/10000

X= 1/2

So we need 0.5 microlitre of 10mg/ml of Ethidium bromide