Viable Counts: Dilution Arithmetic - Basics

The answer to the dilution problem is in scientific notation:

ax10b

The variable a is the coefficient. Variable b is the exponent. Remember that:

4x105 = 400,000
3x106 = 3,000,000
2x107 = 20,000,000
1x108 = 100,000,000


Dilution

Dilution Example 1

If a sample has 4x108 (or 400,000,000) bacteria and it is diluted as follows:

1ml sample
9ml diluent
10ml resulting dilution

The resulting dilution has 4x107 bacteria.

Dilution Example 2

If a sample has 4x108 (or 400,000,000) bacteria and it is diluted as follows:

1ml sample
99ml diluent
100ml resulting dilution

The resulting dilution has 4x106 bacteria.

Dilution Example 3

If a sample has 4x108 (or 400,000,000) bacteria and it is diluted as follows:

1ml sample
999ml diluent
1,000ml resulting dilution

The resulting dilution has 4x105 bacteria.

Dilution Example 4

If a sample has 4x108 (or 400,000,000) bacteria and it is diluted as follows:

1ml sample
19ml diluent
20ml resulting dilution

The resulting dilution has 2x107 bacteria (=400,000,000 x (1/20))

Dilution Example 5

If a sample has 4x108 (or 400,000,000) bacteria and it is diluted as follows:

1ml sample
399ml diluent
Total 400ml

The resulting dilution has 1x106 bacteria (=400,000,000 x (1/400))

Inoculation and Incubation

After diluting the sample plates are inoculated. Typically, 100 Ál is inoculated which is 0.1 ml.

If the dilution has 4x103 (or 4,000) bacteria then the inoculation will have an average of 4x102 (or 400) bacteria. In the Plate Count Simulation the resulting incubation would end up with TNTC meaning To Numerous To Count.

If the dilution has 5x102 (or 500) bacteria then the inoculation will have an average of 5x101 (or 50) bacteria. In this simulation the resulting incubation would end up with numbers between 46 and 54. Some variation is included in the plating computation to simulate the variations in actual bacteria growth.

 

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