Cells are kept in suspension in RPMI, a media for growing cells. You have 30 mL
ID: 314891 • Letter: C
Question
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.Explanation / Answer
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.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.