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7. Cells are kept in suspension in RPMI, a media for growing cells. You have 30 mL of a suspension at a concentration of Z2x106 cellslmL You need to dilute a portion of these cells such that 100 HL the new suspension will deliver K105 cells when plated in the wells of a 96-we microtitre plate. How would you dilute cells so you could prepare Your answer should account for 5-25% extra volume such wells? as a pipetting allowance (answers may vary) 0.5 x 10 S cells nepa ed mL cells 2.2 x 10 3.1 10 mL RPMI X lo RIU t Total volume of cells prepared

It would be much appreciated if you can help me with the recipe of the buffer above. Thank you.

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Answer #1

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

Consider the formula, C1V1 = C2V2.

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

V1 = Volume of stock to be taken = ?

C2 = Concentration of final volume = 1* 10^5 cells/mL

V2 = Volume of final solution = 125 µL

Now, substitute the values in the given formula.

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

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

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

7. Cells are kept in suspension in RPMI, a media for growing cells. You have 30 mL of a cell suspension at a concentration of

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