Question

1) After taking an aliquot of your cells and diluting them with Trypan Blue in a 1:1 dilution, you load the cell/Trypan mixture onto a hemocytometer to count the cells.

You obtain the following counts:

Quadrant	Transparent cells	Blue cells
1	65	2
2	71	4
3	68	1
4	73	3
5	67	2

What is the average number of Trypan-diluted cells per quadrant?

2) Now, take into consideration the Trypan Blue dilution. What is the more concentrated number of cells per quadrant?

3) What is the volume in each quadrant of a hemocytometer?

4) What is the cell density (concentration) of the cells (cells/mL)?

5) If your original cell culture (the one you took an aliquot from to count) has 13 mL of cell suspension, what is the total number of cells you have in culture?

Answer 1

1. Trypan blue dye is used in the dye exclusion test to determine the viable cells in a given culture. The viable or intact cells exclude or do not take up the dye and apeear to be transparent cells while the dead cells take up the dye and appear in blue colour.

Average no of cells per quadrant= total no of cells in all quadrant/no of quadrants

So, average no of transparent cells= (65+71+68+73+67)/5 =344/5= 68.8

average no of blue cells = (2+4+1+3+2)/5 = 2.4

2. Dilution of trypan blue= 1:1 (1 part of 0.4% trypan blue is mixed with 1 part cell suspension)

No of cells per quadrant= average no of cells/quadrant x dilution of trypan blue (1)

Transparent cells= 68.8 x 1 = 68.8cells

Blue cells= 2.4 x 1 = 2.4

3. The hemocytometer is divided into 9 squares, each of size: 1mm x 1mm.

Afetr covering the sample with the cover slip, the resulting height of chamber= 0.1mm

Volume of the haemocytometer chamber= Length x width x height

=1 mm x 1 mm x 0.1 mm = 0.1 mm³ = 10^-4 ml

4. Now, the no of cells in the total volume of chamber = average no of cells x 10⁴ ml

No of cells per ml of suspension= concentrated no of cells per quadrant x volume of chamber

Transparent cells/ml = 68.8 x 10⁴ = 6.88 x 10⁵ cells/ml

Blue cells/ml = 2.4 x 10⁴ cells/ml

5. Volume of cell suspension in original cell culture= 13ml

Total no of cell in the original cell culture= No of cells/ml x volume of cell suspension

Total no of transparent cells= 6.88 x 10⁵ x 13 = 8.94 x 10⁷cells

Total no of blue cells= 2.4 x 10⁴ x 13 = 3.12 x 10⁵ cells