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If i needed to prepare 200µg/ml of proteinase K, and the proteinase K was in a solid powder form, would I have to weight out 200 µg using an analytical balance, and if so, is it possible with a balance that only goes up to four decimal places? (i.e. 0.0000) I assumed 1 microgram is 1×10-6 gram and therefore 0.000001 gram.

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    $\begingroup$ How is this a biology question? If you are interested in the finer points of using a balance, you might enjoy reading this: A question of balance $\endgroup$ – Roland Mar 25 '15 at 10:12
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As others have said, that mass is too small to measure with a standard analytical scale. There are two options you could use:

  1. Make a larger volume of your 200 ug/mL solution.
  2. Make a concentrated stock such as 20 mg/mL. This a 100X stock; you can make 1 mL of 200 ug/mL proteinase K with 10 uL of stock and 990 uL of buffer. Small volumes are easier to measure than small masses (with a micropipette).
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  • $\begingroup$ I guess i still am having issues with the concepts of concentrations. It is just such a hard concept to wrap my head around, i wish i could have practice with it. For your first option, making a larger volume, what would that entail? The recipe requires 200ug/ml , so i wouldnt know how to deal with anything else. For the second option, how would i go about making the 20mg/ml stock? Would I measure out 20miligrams of the powder = 0.020grams on the scale? And then mix that with 1 ml of water? But even with that more concentrated stock, i dont understand how i go from the 20mg/ml to the 200ug/ul $\endgroup$ – Ro Siv Mar 25 '15 at 1:13
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    $\begingroup$ @RoSiv You should practice. I imagine your teacher could help you and there are many resources on the internet. $\endgroup$ – canadianer Mar 25 '15 at 1:54
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In this case make a large quantity and store it as stock - this is a general lab practice. Preparing 10 ml of your protK solurion would you need 2000ug (=2mg) of protK powder.

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