Western Blot - How to Load Equal Amounts and Volumes of Protein?

[shawna82]

Hi there,
I am new to the area of biochemistry and am having trouble understanding how to obtain equal amounts and volumes of protein in sample buffer for different protein samples. I have obtained the protein concentrations (ug/ul) for each of my test samples using the Bradford assay method. I have read previous posts in the Western Blot forum discussing how to go about doing calculations, but I am left more confused than ever!

My goal is to load 40ug of protein in volumes of 10ul for each test sample into separate wells (40ug protein/10ul loading buffer). I am using a 4X sample buffer. My lowest protein concentration is 9.64ug/ul for unknown “A” and 19.92ug/ul for unknown “B.” After having conducted an assay, 10ul of these protein samples were each digested with 30ul of digestion buffer, diluting my protein samples by a factor of 4 (total volume 40ul/10ul of protein in solution = 4, the dilution factor). Because I want to achieve 40ug of protein in my loading buffer (10ul), I have divided this amount by the determined concentration of protein for unknown “A” and then multiplied this value by my dultion factor (4):

Unknown “A”: (40ug÷9.64ug/ul)(4) = (4.15ul)(4) = 16.6ul,
Unknown “B”: (40ug÷19.64ug/ul)(4) = (2.04ul)(4) = 8.2ul

From this point forward, I am confused about 1) how to obtain 40ug of unknown A for a volume of 10ul for loading. I was thinking to go ahead and increase my total loading volume to 20ul since my wells are capable of taking this maximum volume. I also confused about 2) how much of my 4X sample buffer and water to add to achieve a total volume of 10ul? Would anyone be so kind to explain as simple as possible how to go about doing such calculations? I greatly appreciate any help I can get with understanding how to achieve the correct ug of protein within a set volume.

Shawna