Question

Plasmid isolation

1. Write a set of instructions for preparing Solutions I, II, and III, using the stock solutions/powders provided below.

Solution	Volume (mL)	Stock solutions/powders available
Solution I, the resuspension solution: 15 mM Tris-HCl, pH 8.0, 10 mM EDTA, 100 μg/mL RNase	50	1 M Tris pH 8.0, 0.5 M EDTA 1%(w/v) RNase (boiled for 20 min to remove any DNase activity)
Solution II, the alkaline lysis solution: 0.2 N NaOH, 1%(w/v) SDS	25	5 M NaOH 10%(w/v) SDS
Solution III, the neutralization solution, 3 M Potassium Acetate, pH 5.2	100	Powder KOAc (m.wt 98.14) Glacial Acetic Acid (density 1.05 g/mL, m. wt. 60)

Answer 1

Preparation of solution I- 50ml

Tris HCL- 50/(1000/15)= 50/66.67= 0.75ml or 750ul

EDTA: 50/(500/10)= 50/50= 1ml

RNase: 50/(10000/100)= 0.5ul

Water = (50-0.75-1)= 48.25ml

Preparation of solution II 25ml

NaOH:  25/(5/0.2)= 1ml

SDS- 25/(10/1)= 2.5gm

Preparation of solution 100ml

Pottasium acetate-

3=( wt/98.14)x10=> wt.= 30x98.14/10= 294.42gm

Acetic acid=

3=(gm/ molwt)X10= > gm =3x 60= 180gm

So volume of acetic acid required= gmx density= 180x1.05= 189ml

Hope it's clear..thanks