 # MCQ – Lab calculations

These should be simple and you should be confident in your answers!

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How much NaCl would you need to make 100ml at 150mM – NaCl – FWT 58.44
0.88 g
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
8.8 g
5.84 g
58.4 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
15 ml
M1V1 = M2V2… V required = (200 ml x 0.15 M) / 2M
0.075 ml
75 ml
150 ml
How much agarose would you need to make 50ml at  0.75%?
375 mg
50ml /100ml  x 0.75g = 0.375 g = 375 mg
150 mg
3.75 g
0.75 g
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?
3 μl
5000 units / ml = 5 units / μl. This 2 μl is required per μg DNA. so 2 x 1.5 μg = 3 μl
30 μl
1.5 μl
1.33 μl
You have 500ml of your stock ethanol solution (100 % v/v), how much water should you add to dilute this to 70%
214.3 ml
500 ml x (100 % / 70%) = 714.3 ml total so 500 ml EtOH + 214.3 ml of water
714.3 ml
350 ml
150 ml
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?
9.65 ml
1.035 ml
1 ml
965 μl
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
96.5 ml
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)
218.76 mM
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
8.75 mM
114.3 mM
4.57 M
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?
5 ml
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
1 ml
0.25 ml
500 ml
0.5 ml