concentrations

[Alkendi]

hi all

After i mix the PCR reaction mixture (DNA, primers, Master Mix) i need to add BSA (400ng/ul) how do I calculate how much i need to add of BSA to the mixture.

thanks

[lucaspa]

How many millileters of the mixture do you have? You take the # of ml x .4mg/ml and that gives you the number of mg of BSA you will need.

 

400 ng/microliter (ul) = 0.4 mg/ml

[Alkendi]

ok, but i have the BSA already prepared in solution with a concentration of 400ng/ul. I prepare 25ul of PCR mixture and I did the calculations on it. the result was that 0.025ml*0.4mg/ml=0.01mg of BSA

ok, how many ul of BSA that contain 0.01mg?

 

I am confused

[lucaspa]

ok, but i have the BSA already prepared in solution with a concentration of 400ng/ul. I prepare 25ul of PCR mixture and I did the calculations on it. the result was that 0.025ml*0.4mg/ml=0.01mg of BSA

ok, how many ul of BSA that contain 0.01mg?

 

I am confused

 

You screwed yourself. Your calculation is 25 ul (PCR solution) x 400 ng/ul BSA = 10,000 ng BSA. To get 10,000 ng you need 25 ul of the BSA solution. But if you add 25 ul of BSA solution, you now have 50 ul of PCR mixture that is only 200 ng/ul of BSA. Right?

 

You need to do one of two things:

 

1. Add all your PCR reagents (in solid form) to the BSA solution you have already prepared. That would mean that the solution was 400 ng/ul of BSA.

 

2. Make up a 100x BSA solution. That would be 40 mg/ml BSA solution as your "stock". You would then add 0.25 ul of the stock BSA to your PCR solution to get 400 ng/ul.

[Alkendi]

You screwed yourself. Your calculation is 25 ul (PCR solution) x 400 ng/ul BSA = 10,000 ng BSA. To get 10,000 ng you need 25 ul of the BSA solution. But if you add 25 ul of BSA solution, you now have 50 ul of PCR mixture that is only 200 ng/ul of BSA. Right?

 

You need to do one of two things:

 

1. Add all your PCR reagents (in solid form) to the BSA solution you have already prepared. That would mean that the solution was 400 ng/ul of BSA.

 

2. Make up a 100x BSA solution. That would be 40 mg/ml BSA solution as your "stock". You would then add 0.25 ul of the stock BSA to your PCR solution to get 400 ng/ul.

 

thanks alot for the help:doh: