BIOTECHNOLOGY & GENETIC ANALYSIS - Essay Example

This is mainly attributed to accumulation of pollutant along the path of river and hence bacterial population need to evolved catabolic capability to survive and hence more plasmid bearing bacterial population were found in second sample. Similarly most of the plasmid was found to be in size range more than 35 KB clearly indicates most of them are Conjugative plasmids as this group of plasmid has more number of genes compared to non-conjugative plasmid to carryout conjugation process and hence larger the size.

Here re-suspension solution constitutes of glucose, EDTA and Tris each have its own role. Glucose provides osmotic stress and EDTA as chelating agents which binds to heavy metals and helps in disintegration of cell wall, Tris act as buffering agent and maintains pH of over all reactions to avoid any pH dependent side reaction. In this stage cell become highly fragile and some are break open. This solution is mixture of SDS and NaOH. Here NaOH provides alkaline condition which helps in cell lyses and denaturation of DNA while SDS dissolves cell wall constituents and induces extensive cell lyses.

It also helps in proteins denaturation and precipitation. In this stage most of cell constituents get denatured including genomic DNA, But as plasmid is in its CCC (covalently closed circular) forms will not denatured completely and most of them remains in its native configuration.Step 4: Neutralisation Solution Here potassium acetate and acetic acid act as neutralizing agent to bring back the pH to normal. Similarly it induces the renaturation of DNA. Because of larger size most of the Genomic DNA remains denatured and mingled with proteins remains with cell debris while plasmid being smaller molecule except out to supernatant .

Step 5: centrifugation at high speed; During this stage all cell derbies along with genomic DNA settled at the bottom of tube and being smaller in size plasmid remains in supernatant. Which subsequently used for further purification and transformation.Ans 3 protocol 6: Here we have two different observation 1) colonies from tube 2 grown as blue colored colonies 2) while from tube 3 there is mixture of blue and white. This can be explained as follows. In case of tube 2 there is only vector pGEM3Z used for transformation.

The plasmid pGEM3Z have lacZ gene as marker which code for enzyme called beta glycosidase. After transformation cells where plated on LA supplemented with X-gal and IPTG. Now in presence of IPTG expression of lac Z induces and leads to synthesis of beta-glycosidase which subsequently acts on X-gal and cleaved it to chromogenic intermediate give rise to blue color. While in case of tube 3 there was plasmid vector along with insert gene (ligation mixture) and plated on similar plate after transformation.

As vector pGEM3Z having MCS (multiple cloning sites) in side the lacZ gene any insertion or recombination leads to inactivation of lacZ (insertional inactivation). Inactive lacZ will not code for functional beta glycosidase and hence colonies having insertion give rise to white colors. In another scenario where cut plasmid re-ligated without any insertion during