These should be simple and you should be confident in your answers!
Keep practicing until you get it!
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How much NaCl would you need to make 100ml at 150mM – NaCl – FWT 58.44
m (in g) = FWT (in g/mol) x conc (in M) x vol (in L) = 58.44 x 0.15 M x 0.1 L = 0.8766 g
You have a 2M stock NaCl stock solution, you need to make a 200ml of 150mM, how much of your stock solution do you need?
It can’t be done, you need to use powder
M1V1 = M2V2… V required = (200 ml x 0.15 M) / 2M
How much agarose would you need to make 50ml at 0.75%?
50ml /100ml x 0.75g = 0.375 g = 375 mg
Your restriction enzyme is supplied at a concentration of 5000 units / ml, your protocol calls for 10 units per μg of DNA. You want to digest 1.5 μg of plasmid. How much enzyme do you need?
5000 units / ml = 5 units / μl. This 2 μl is required per μg DNA. so 2 x 1.5 μg = 3 μl
You have 500ml of your stock ethanol solution (100 % v/v), how much water should you add to dilute this to 70%
500 ml x (100 % / 70%) = 714.3 ml total so 500 ml EtOH + 214.3 ml of water
Your media comes already supplemented to 0.07 mM CaCl2, to induce differentiation you need to supplement this to a final concentration of 2 mM. How much of your 1M CaCl2 stock solution would you add to 500 ml of media?
What do we need? 2 mM – 0.07 mM = 1.93 mM. What have we got? 1M = 1000 mM. So volume needed = (1.93 / 1000) x 500 ml = 0.000965 ml = 965 μl
You are writing up your methods section and your protocol just said dissolve 5.3 g of Tris in 200 ml of water, what concentration was your solution? (Tris FWT 121.14)
5.3 g / 121.14 = 0.044 this would be the conc in M if it was made up in 1L, however we have used 200 ml so it is 5 times more concentrated than this: 0.044 x 5 = 218.76 mM
You are making up your transfer buffer for a western blot. You have a 5x solution of transfer which needs to be supplemented with 10 % Methanol (MeOH, assume 100% stock) before use. Which of the following options would be appropriate for your 800 ml transfer tank?
560 ml water, 160 ml 5X buffer, 80 ml MeOH
10% of 800 ml = 80 ml MeOH, 800/5 = 160 ml of buffer, 800-80-160=560 ml of water
240 ml water, 160 ml 5X buffer 160 ml MeOH
632 ml water, 160 ml 5X buffer, 8 ml MeOH
704 ml water, 16 ml 5X buffer, 8 ml MeOH
What dilution(s) would you need to do from a stock solution of 20 mM to give a solution of 10 µM?
1 in 100 then 1 in 20
20 mM = 20 000 µM therefore you need a 1 in 2000 dilution. 1 in 100 then 1 in 20 = 100×20 = 1 in 2000. IF you were making up a large volume you could do this directly i.e a 1 in 200 dilution, but if you were wanting small final volume your pipetting would be more accurate if you did a serial dilution, hence the two stops here.
1 in 500
1 in 100 then 1 in 2
1 in 2 then 1 in 250
You need to supplement the diet of a mouse with zinc . You have a solution of zinc chloride which has a concentration of 100 µg/ml. A mouse eats 5g of food per day. How much of this solution would you need to add to 100 g of laboratory animal food to supplement the mouse with 25 µg of zinc chloride per day?
You could do this in different orders depending on how you like to break down these sort of questions. If we start with the food, 5g of food/day 100g total therefore we need 20 days work of ZnCl2. 20 days x 25 µg per day = 500 µg. Our stock is at 100 µg/ml therefore we need 5 ml. The other way would be to say 25 µg is 0.25 ml of the stock x 20 days = 5 ml