how to perform 1: 5 x 10^ 3 dilution ?

Question

For determination of the number of yeast cells in suspension a counting chamber is used.
However, the yeast suspension contains too many cells for direct counting. Hence a 1: 5 x 10^3
dilution is performed

Suggest how to prepare this dilution!
b) 25 μl of this dilution were transferred into the counting chamber. Counting revealed an
average number of 46 cells in the sample. Please give the number of cells per ml in the original
yeast suspension!

Answer 1

To prepare a 1 : 5x10³ dilution, means to dilute the original call suspension by 5000 fold (i.e. 5x10³ = 5000).  So, depending on what final volume of diluted suspension you want, that will dictate the volume to begin with. For example, take 1 ml of original suspension and dilute to a final volume of 5000 ml (5 L). Or, take 10 ul of original suspension and dilute to a final volume of 50000 ul (50 ml), etc.

(b) 46 cells/25 ul = 1.84 cells/ul

1.84 cells/ul x 1000 ul/ml = 1840 cells/ml in the diluted suspension.  Since this is 5000 fold diluted from the original, 

cells/ml in original suspension = 1840 cells/ml x 5000 = 9,200,000 cells/ml = 9.2x10⁶ cells/ml