Question

You would like to make a special lysis buffer for single worm PCR. The buffer contains 25 mM KCl, 50 mM Tris HCl (pH 8.3), 5 mM MgCl₂ and 0.090% NP40, 0.9% Tween20 and 0.01% (w/v gelatin). 50 ug/mL Proteinase K is added immediately before use. You have powder KCl, 1M Tris HCl (pH 8.3), 250 mM MgCl₂, 100% NP40 and 100% Tween-20 and powder gelatin. The Proteinase K stock solution is at a concentration of 1 mg/mL.

a] If you want to prepare 10 mL of lysis buffer without the Proteinase K, describe how much KCl you would need.

b] If you want to prepare 10 mL of lysis buffer without the Proteinase K, describe how much Tris HCl you would need.

c] If you want to prepare 10 mL of lysis buffer without the Proteinase K, describe how much MgCl₂ you would need.

d] If you want to prepare 10 mL of lysis buffer without the Proteinase K, describe how much NP-40 you would need.

e] If you want to prepare 10 mL of lysis buffer without the Proteinase K, describe how much Tween-20 you would need.

f] If you want to prepare 10 mL of lysis buffer without the Proteinase K, describe how much gelatin you would need.

g] Once you have prepared the lysis buffer, you are ready to take an aliquot to make your working solution to collect lysates. If you wanted to prepare 1 mL of the lysis buffer with the proteinase K added, how much of the 1 mg/ml stock would you need to achieve the final concentration of 50 ug/mL (micrograms/mL)?

one question with many parts.... this problem is my biggest weakness... please help

Answer 1

a. The molar mass of KCl is 74.55.

Hence to make 25mM solution, 10ml we need  18.63 mg of KCl powder.

b. We need 0.5 ml of 1M stock solution to make 50mM, 10 ml solution. C1V1 =C2V2 where C1 is the stock concentration and V1 is the volume required of stock solution. C2 is required concentration and V2 is the volume of required concentration. Hence V1 = C2V2/C1

c. As per the above formula, we need 0.2 ml of MgCl2 from 250 mM stock.

d. 9ul of NP 40

as the stock is 100 per cent we need to know 0.09% of 10 ml that would be = (0.09/100)*10 ml = 0.009 ml or 9ul

e. From the percentage formula + (0.9/100)*10ml = 0.09 ml or 90 ul

f. for 0.01 % of gelatin in 10 ml solution = (0.01/100)* 10= 0.001 gram or 1 mg

g. As C1V1= C2v2

C1= 1mg/ml or 1000 ug/ml as there is 1000 microgram in one milligram.

V1 = required volume of this stock, to be found out

C2= 50 ug/ml (the required concentration)

V2 = 1 ml (the required volume of 50 ug /ml

Hence V1 = (50 ug/ml *1ml )/ 1000ug/ml = 0.05 ml or 50 ul.

While solving these problems the units conversion must be kept in mind all the units of measurements should be the same when multiplying or dividing.