Question

If you want to reproduce the shuttling experiment as performed during week 3. you will need to purchase Cycloheximide (CHX) from one of three potential suppliers. CHX has a molar mass of 281.35 g/mol and can be dissolved in DMSO, water, or EtOH but with different solubilities. You will need to use CHX at a final working concentration of 17 μg/mL in a 150 mm dish with a total media volume of 14 mL.

Determine how much of the stock solution purchased from each company you would need (in μL) to add to the media in the 150 mm dish to obtain the final working concentration of CHX. Give your answers in μL using the scientific notation.

a. Company A supplies CHX dissolved in DMSO at 100 mg/ mL:

Volume needed in μL

b. Company A supplies CHX dissolved in water in 2% (w/v) solution

Volume needed in μL:

c. Company A supplies CHX dissolved in EtOH at 50 mM

Volume needed in μL:

Answer 1

In general for calculating required volumes of stock solutions to get a specific working concentration, e use the formula

C1 * V1 = C2 * V2

Where, C1 and V1 are the concentration and volume of the stock solution respectively.

C2 and V2 are concentration and volume of the working solution respectively.

The concentrations could either be in molarity (M) or percentage (%) or Normality (N). One has to remember that both the side should have identical units of volume and concentration for the above equation to be use correctly.

Conversions used,

1g = 1000mg = 1000000 µg

1mg = 1000 µg or 1 µg = 0.001 mg

1ml = 1000µl or 1 µl = 0.001 ml

1mM = 0.001 M

Molarity = no. of moles of a solute dissolved in one litre of a solvent.

No. moles of a solute = ‘grams of the solute taken’/ molecular weight (MW)of the solute.

In the given question, MW of CHX = 281.35

1. C1 = 100mg/ml = 100 * 1000 µg/ml, V1 = ‘x’ ml; C2 = 17 µg/ml, V2 = 14 ml

The equation becomes, 100000 * x = 17 * 14

X = 238 x 10^-5 ml = 2.38 µl

2. Here the concentration of stock solution is given in % and we need to convert it into µg/ml

2% (w/v) solution means 2g of solute in 100ml of solution or 0.02g of solute in 1 ml of solution.

0.02g/ml = 0.02 * 10⁶ µg/ml = 20000 µg/ml

So our C1 = 20000 µg/ml

Equation becomes, 20000 µg/ml *x = 17 µg/ml * 14

X = (238/2) * 10^-4 ml = 11.9 µl

3. In this case the concentration of stock is in molarity. We can convert the working concentration of 17 µg/ml to molarity

17 µg/ml = 17 µg in 1 ml of solution

Or 17000 µg in 1L of solution (1L = 1000ml)

Or 0.017 g in 1L of solution (converting micrograms to grams)

No. moles of a solute = ‘grams of the solute taken’/ molecular weight (MW)of the solute.

Molarity = 0.017/281.35 in 1L = 6.042 * 10^-5 M = 0.06 mM

Now using the equation,

50 mM * x = 0.06 mM * 14

X = (0.06 * 14)/50 ml = 0.0168 ml = 16.8 µl