mk_2007 Posted May 24, 2010 Share Posted May 24, 2010 I'm trying to write my lab report up and I'm getting really confused about a certain calculation! I made a 1% agarose gel, and Basically, I added 5 μL of ethidium bromide to a solution (solution= 0.5 g agarose and 50 μL TAE) and the bottle said the ethidium bromide had a concentration of 5 μg/μL. so what concentration of ethidium bromide am I adding to my solution? Help would be much appreciated! Link to comment Share on other sites More sharing options...
jacky Posted May 24, 2010 Share Posted May 24, 2010 5 μg/μL of ethidium bromide in 50 μL TAE so in 100μL ?? by calculation: 100x5/50=10 μL Link to comment Share on other sites More sharing options...
CharonY Posted May 24, 2010 Share Posted May 24, 2010 You do not add a concentration of EtBr, but you add a certain amount into a volume resulting in a given concentration. Your stock of EtBr has 5µg/µl. You took out 5µl which is a total of 5µg. Then divide that amount by your final volume to get the concentration. Link to comment Share on other sites More sharing options...
mk_2007 Posted May 25, 2010 Author Share Posted May 25, 2010 ok. I understand half of that. But why are you taking out 5 μg? I dont understand how you get to that? Link to comment Share on other sites More sharing options...
ewmon Posted May 25, 2010 Share Posted May 25, 2010 Final amount = 5 µL × 5 µg/µL = 25 µg. Final volume = 5 µL + 50 µL = 55 µL. Final concentration = 25 µg / 55 µL = 0.455 µg/µL. But you asked what concentration of ethidium bromide you were adding to the solution, and the answer was already written on the bottle: 5 μg/μL. Link to comment Share on other sites More sharing options...
CharonY Posted May 25, 2010 Share Posted May 25, 2010 My bad. It should have been 25 µg, not 5. Link to comment Share on other sites More sharing options...
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