Cells are kept in suspension in RPMI a media for growing cel

Cells are kept in suspension in RPMI, a media for growing cells. You have 30 mL of a cell suspension at a concentration of 7.2 times 10^6 cells/mL. You need to dilute a portion of these cells such that 100 mu L of the new suspension will deliver 1 times 10^5 cells when plated in the wells of a 96-well microstate plate. How would you dilute some cells so you could prepare 48 such wells? You answer should account for 5-25% extra volume as a pipeline allowance (answers may vary). ________ mL cells ________ mL RPMI _______ Total volume of cells prepared It would be much appreciated if you can help me with the recipe of the buffer above. Thank you.

Solution

As it is given to take 25% of the extra volume, the final volume to be considered at each dilution is, 100 µL.

From the formula, C1V1 = C2V2.

Where, C1 = Concentration of the stock = 7.2* 10^6 cells/mL

V1 = Volume of stock to be taken = ?

C2 = Final volume concentration; given as 1* 10^5 cells/mL

V2 = Final solution volume (per each well) = 100 µL + 25 µL = 125 µL

By substitute the values in the given formula, we get,

7.2* 10^6 * VI = 1* 10^5 * 125 µL

V1 = 125/72 = 1.736 µL

Thus, take 1.736 µL from the stock solution and make the volume up to 125 µL. Take 100 µL from this solution, the resulting cell volume is, 1* 10^5 cells/mL

The total volume of stock (RPMI) needed is = 1.736*48 = 83.328 or approximately 83 µL or 0.083 ml.

Number of cells present in 83 µL of stock is, = 7.2* 10^6* 0.083/ 30 = 19.2* 10^3 cells/ml

Thus, 48 wells require, 48*125 = 6000 µL or 6 ml of solution.