3. You want to determine the concentration of viable cells in a suspension of ce

Question

3. You want to determine the concentration of viable cells in a suspension of cells using a hemocytometer. You dilute the cell suspension 1:4 with trypan blue and load the hemocytometer. You count two large nonadjacent squares and find 57 "clear" cells/3 "blue" cells in one square and 60 clear" cells/5 "blue" cells in the second square What is the concentration of viable cells in the cel suspension in cells/ml? b. What is the % viability of the cell suspension. 4. You want to deliver 1.5 x 105 cells in 0.3 ml per well of an 8-well chamber slide. What will be the final concentration of cells in each well? 5. Assuming that you have a cell suspension at 1.25 x 106 cells/ml, cell medium (diluent) and volu want to plate all 8 wells (1.5 x 105 cells in 0.3 ml per well), how would you proceed? Show your calculation

Explanation / Answer

3. Trypan blue stain dead cell, so the cells which are blue are not viable. Then the no of viable cells

= (57-3)+(60-5)/2= 54+55/2= 54.5

Thus the no of cells /ml

= 54.5 X 104 ( required to get the cell no /ml)

=5.45 X 106

now the cells were diluted to 1:4 with trypan blue, then the total no of cells

= 5.45 X 4 ( dilution factor) X 106

= 21.8 X 106   or 22 X 106 / ml

4. If 1.5 X 105 cells were plated with 0.3 ml medium / well in 8 well plate, then the concentration of cells in each well will be-

0.3 ml contains 1.5 X 105 cells

thus 1 ml contains 1.5 X 105 X 1 / 0.3 = 5 X 105 cells / ml

5. The cell soluiton I have is 1.25 X 106

now I need to seed1.5 X 105 cells in 0.3 ml / ml of a 8 well plate.

so, for 1 well, 1.5 X 105

      now 8 well requires = 8 X 1.5 X 105 = 12 X 105 cells.

   and 0.3 X 8 = 2.4 ml of cell culture medium

Now, the stock cell solution is 1.25 X 106 / mland the working cell count is 12 X 105 or 1.2 X 106

I will take 1.4 ml of medium and then will add 1 ml of cell suspecion from the stock ( which has the concentration1.25 X 106 / ml)

thus the current cell suspension will contain 1.25 X 106 cell in 2.4 ml and when 0.3 ml from this will be added to 1 well it will contain almost 1.5 X 105 cells / well.

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