1. Hemocytometer
 For hematology, microbiology, cell culture, and many applications that require use of suspensions of cells it is necessary to determine cell concentration.
 A device used for determining the number of cells per unit volume of a suspension is called a counting chamber.
 The most widely used type of chamber is called a hemocytometer, since it was originally designed for performing blood cell counts.
 A hemacytometer is a microscope chamber slide with a small (3mm x 3mm) square etched onto the surface.
 The slide has a coverslip, which rests exactly 0.1 mm above the slide.
 Cells in suspension are introduced into this area, and then counted.
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2. A B 1 2 3 5 4 9 6 7 8
1/5*4= 1/20=0.05
¼=0.25
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The main divisions separate the grid into 9 large squares.

3. Thus the entire counting grid lies under a volume of 0.9 mm-cubed.
Each square has a surface area of one square mm, and the depth of the chamber is 0.1 mm.
Thus the entire counting grid lies under a volume of 0.9 mm-cubed.
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4. For large cells this may mean counting the four large corner squares and the middle one.
For a dense suspension of small cells you may wish to count the cells in the four 1/25 sq. mm corners plus the middle square in the central square.
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5. It is essential to be extremely careful with higher power objectives, since the counting chamber is much thicker than a conventional slide.
 The chamber or an objective lens may be damaged if the user is not not careful. One entire grid on standard hemacytometers with Neubauer rulings can be seen at 40x (4x objective).
 Suspensions should be dilute enough so that the cells or other particles do not overlap each other on the grid, and should be uniformly distributed.
 To perform the count, determine the magnification needed to recognize the desired cell type.
 Sometimes you will need to dilute a cell suspension to get the cell density low enough for counting. In that case you will need to multiply your final count by the dilution factor.
 For example, suppose that for counting you had to dilute a suspension of Chlamydomonas 10 fold.
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6. Spore suspension
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7. Counted=200 Spore
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8. 0.04 x 0.1= 0.004 mm3 =The total volume in each square mm-cubed
5 x = 0.02 mm3 = Combined volume of five squares
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9. 200 spore were countes 200 / 0.02= 10000 spore/ mm3
10000 x 1000= = 1 x 107 spore/ml
Total counted spore *50000*D=200*50000*1= spore/ml
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10. C2= that we wanted=1 x 10 5 spore/ ml V2= that we wanted= 2 liters
1000
N x D
n x 16
N= Counted Spore
D= diluted suspension
n= Number of squers
Spore/ml= 4000
1000 = = 1 x spore/ ml
200 x 1
5 x 16
1 x spore/ ml Stock
C1V1=C2V2
C1= 1 x spore/ ml
C2= that we wanted=1 x spore/ ml
V2= that we wanted= 2 liters
V1= ?
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11.  Suppose that you counted 125 cells (total) in the five squares.
C1V1=C2V2= (1x 10 7) x V1= (1x 10 5) x 2000
V1=2x 10 8 / 1x10 7= V1= 20 ml
20 ml
water
2000 ml
1x10 5 spore/ml
 Cells are often large enough to require counting over a larger surface area.
 For example, you might count the total number of cells in the four large corner squares plus the middle combined.
 Each square has surface area of 1 mm-squared and a depth of 0.1 mm, giving it a volume of 0.1 mm-cubed.
 Suppose that you counted 125 cells (total) in the five squares.
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12. Again, multiply by 1000 to determine cell count per ml (250,000).
You then have 125 cells per 0.5 mm-cubed, which is 250 cells/mm cubed.( 5 Squers * 0.1 mm-cubed= 0.5 mm-cubed)
125/0.5=250 cells/mm-cubed
Again, multiply by 1000 to determine cell count per ml (250,000).
 Sometimes you will need to dilute a cell suspension to get the cell density low enough for counting. In that case you will need to multiply your final count by the dilution factor.
 For example, suppose that for counting you had to dilute a suspension of Chlamydomonas 10 fold.
Toatal spore per 5 squares * 2000* D=125* 2000*D= 125*2000*10=
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