Stock solutions

Question

Need dilutions to make wells with 1,3 and 10ug of protein per well. That means 1,3, 10 ug per 5uL solution.

prepare samples of each of these proteins from the stock solutions in sufficient amounts to run three gels with two lanes for each protein in each gel.

Note: protein stain used can be in the range of 0.5-50ug of protein per lane.

The following stock solutions will be available: BSA, bovine serum, bovine IgG, each at 10mg/mL

Answer 1

If I understand your question correctly, you will have available to you solutions of BSA, and IgG that are

10 mg/ ml, and you need to dilute them so that you have 1, 3, and 5 ug per 5 ul. If that's correct, then the following should answer your question. If that's not what you meant, then please leave a comment and let us know what you need.

3 gels with 2 lanes for each gel = 6 samples of each dilution.

5 ul for each x 6 samples = 30 uls of diluted samples are needed.

We will prepare 100 uls of each to be sure we have sufficient amounts

10 mg / ml = 10 ug / ul ... stock solutions

To make 5 ug / 5 ul (1 ug / ul) Take 10 ul of stock and dilute add 90 ul diluent

To make 3 ug / 5 ul (0.6 ug / ul) Take 6 ul of stock and add 94 ul diluent

To make 1 ug / 5 ul (0.2 ug / ul) Take 2 ul of stock and add 98 ul diluent

The formula for the above calculations is essentially (x ul)(10 ug/ul) = (100 ul)(1 ug / ul) and x = 10 ul. This is for the 5 ug / 5 ul solution. The others use the same calculation but substituting 3 ug / ul and 1 ug / ul.

Hope this made sense.