Question

An average count of 42 cells/mm² grid is counted in a hemocytometer for a 2.5 ml suspension of cells.

How many cells would have to be actually added to a plate (this is usually called “seed”) to begin a culture with 3.5 x 10⁵ cells/dish, if the line of cells is known to have a plating efficiency of 65%? Show your calculation. (Plating efficiency refers to the % of cells that can survive after seeded in a culture container)

Answer 1

Answer: 5.77 X105 cells (1.819ml of cell suspension) should be added Petri plate for culture

Calculation:

Total no. Of cells = (No. Of average counted cells (N) X Dilution factor (DF) X volume in which cell is suspended; V)/ Area counted X Depth of fluid

Where, Area = 1sqmm            Depth of fluid = 0.1 mm         Sample was not diluted

Total no. Of cells = (N X V)/ (1/10)2 X (0.1/10)

Simplyfy the above calculation;

total No. Of cells = N X V X 104

So the total no. Of cells = 42 X 2.5 X 104

                                    = 105 X104 (10.5 X105)

Plating efficiency = 65% (means out of 100 only 65 cells are able to grow and form colony)

So out of 1 = 65/100

So for 3.5 x105 = (65/100) X 3.5 X 105

                        = 2.27 105

So the total cells required for plating = 3.5 X105 + 2.27 X105

                                                            = 5.77 X105

Since in 2.5 ml of suspension 10.5 X105 cells are present

So the volume of suspended cells require to get 5.77 X105 cells for culture = (10.5 X105)/5.77 X105

                        1.819 ml