In article <[email protected]>, [email protected] says... > > Hello Guys > I have a trivial query, I have a protein stock 1000 µg per ml. Now I would > like to load on each well 0.25, 0.5, 1, 2, 4, 8, 16 and 32 µgs. > For 0.25 µg, 0.25 µl from stock + 0.25 µl 2x sample buffer , total 0.5 µl > load on first lane. If I want to load higher amount like 16 µg and 32 µgs > more volume to be loaded but I could load maximum16 µl in each well. In > that case, should I use C1 V1 = C2 V2 formula, 1000 µg * ? = 16 µg * 8 µl > V1 = 0.128 micro liter from the stock, 7.8 ul milli q water, to that add 8 > µl 2x sample buffer and load 16 ul to the well, Am I right doing things > here?????
Your sample is 1 µg/µl. If your wells can hold only 16 µl and you still have to add the sample buffer, then the maximum sample size is 8 µl, which is equivalent to 8 µg of protein. The alternative would be to use gels with larger wells, so that you can load higher sample volumes. What you should do in any case is to load the same volume into each well, otherwise the gel will run unevenly. So for example pipett 10, 8, 6, 4, 2 µl sample into vials, make to 10 µl total volume with the buffer you have dissolved your protein in (i.e., 0, 2, 4, 6, 8 µl of buffer) and then add 10 µl of SDS-sample buffer to each sample. Mix and load your 16 µl per well. _______________________________________________ Methods mailing list [email protected] http://www.bio.net/biomail/listinfo/methods
