Question

Gel Electrophoresis and Bioinformatics Analyses of wild type and mutant Arabidopsis 1. The procedure for making 30 mL of 1.2% agarose in 1x TAE is: g agarose mL lX TAE 2. Briefly explain the roles of each of the following in electrophoresis. Agarose - Tris - EDTA- Power supply - 3. What is the stain we used to view DNA and how does it work? 4. Which samples amplified and which ones did not?

Answer 1

1) 1.2% agarose means 1.2g agarose dissolved in 100ml solution.

1ml solution contain = 1.2/100

30ml solution contain=1.2*30/100

=0.36g

0.36g of agarose dissolved in 100ml 1*TAE

2) tris : a strong buffer which neutralised or stabilized the pH.

EDTA: DNAase enzyme require divalent cation as a cofactor for the activation of the enzyme. EDTA chelates the divalent cation and does not allow the enzyme to function.

power supply: As the DNA is negatively charged so it start moving towards the positive charge. In gel electrophoresis two electrode are present anode and cathode which work when the power supply start.

3) ethidium bromide (Ebr) is used as a dye . It intercalate the DNA and give orange color in the UV spectrometer.

4) only those regions are ampilied for which the primers have been added to the solution . Ampilification occur thermocycler process called PCR.