Question

Unknown has a molecular weight of 400. An experiment requires an incubation to be carried out with 1 μM of unknown. The incubation will be carried out in a total volume of 500 μL of culture medium. What volume of each of the stock solutions below (a – d) would you need to add to the incubation to give 1 μM of unknown? a. 1 mM b. 0.5% (w/v) c. 20 μg/dL d. 10 μg/mL e. Due to the toxicity of the solvent, the volume added cannot exceed 0.1% (v/v) in the final 500 μL of culture. Which of the stock solutions above COULD actually be used for this incubation?

Answer 1

Data given: Molecular weight of the unknown: 400.

So 400gm per litre gives 1M solution.

Accordingly for 1uM/500ul, 0.2g of unknown needs to be dissolved. So we need 0.2g of unknown in the final volume of 500ul of solution.

Now let's see each of the stock solutions:

A. 1mM

As the stock volume is not mentioned in the question, let's assume 500ul of 1mM of unknown.

To get 1uM from 1mM, we need to divide it by 1000. Hence 0.5ul of 1mM solution needs to be added to 499.5ul of culture medium.

B. 0.5%w/v

That means, 0.5g per 100ml. As molecular weight of the unknown is 400g. Accordingly it becomes 12.5mM. 500ul of this stock contains 0.25×10^-2. But we need 0.2×10^-6 g per 500ul. If we calculate as per then we need to add 0.04 ul of stock to 499.96ul of culture medium.

C. 20ug/dl

That means 20ug/100ml. That means 500ul of stock contains 0.1ug of unknown. But we need 0.2ug per 500ul. As stock solution concentration is lesser then the final required solution we cannot make any dilution.

D. 10ug/ml.

Hence 500ul contains 5ug. As we need 0.2ug/ 500ul, we have to add 20ul of stock solution into 480ul of culture medium.

E. 0.1%v/v means 0.1 ml need to be added to 99.9 ml. Stock solution B, that is 0.5%w/v requires addition of stock volume lesser than 0.1% v/v is a suitable stock solution to avoid toxicity.