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.
As others have said, that mass is too small to measure with a standard analytical scale. There are two options you could use:
- Make a larger volume of your 200 ug/mL solution.
- 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).
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.