Three methods of how to calculate the amount of bacteria are illustrated. They will all yield the same results. You should use the method that makes the most sense to you.

The dilution table looks as follows:

Unknown
| Volume Transferred | Dilutent Volume | Total Volume |
---|---|---|---|

Dilution A | 1 ml |
999 ml |
1000 ml |

Dilution B | 1 ml |
99 ml |
100 ml |

Dilution C | 1 ml |
9 ml |
10 ml |

100 of Dilution D is used to inoculate two nutrient agar plates. After incubation the plates show 30 and 31 colonies of bacteria. The average number of bacteria colonies is 31.

**Method 1**

Express the dilution and the inoculation in scientific format. (This value is found by dividing the Volume Transferred by the Total Volume.)

Dilution A | 1x10 ^{-3} |

Dilution B | 1x10 ^{-2} |

Dilution C | 1x10 ^{-1} |

Inoculation | 1x10 ^{-1} |

Calculate the amount of bacteria in the original solution by:

31 | = 3x10^{8} Number of bacteria in the original solution |

(10^{-3} x 10^{-2} x 10^{-1} x 10^{-1}) |

The plate count simulation coefficient is always expressed as a whole number. The answer entered would be 3x10^{8} not 3.1x10^{8}.

**Method 2**

Calculate the ratios of the Total Volume / Volume Transferred and then multiply it by the colonies counted on the plate.

1,000 (Dilution A) = 1000 / 1

100 (Dilution B) = 100 / 1

10 (Dilution C) = 10 / 1

10 (inoculation of 100 or 0.1ml) = 10

__31 (colonies counted on plate)__

300,000,000 = 3x10^{8} Number of bacteria in the original solution

The plate count simulation coefficient is always expressed as a whole number. The answer entered would be 3x10^{8} not 3.1x10^{8}.

**Method 3 - EASY Math**

To use this method three criteria must be met:

- the number of bacteria colonies on the plate must be between 10 and 90
- all dilutions must be done as 1 to 9, 1 to 99 or 1 to 999
- inoculation selected must be 100

The coefficient for this solution will be 3. Use the 10's digit from the plate count.

The exponent for this problem can be determined by counting zeros. Count all the zeros for Dilution A, B, and C and then add 2. The exponent for this problem is 8 (6 zeros in the dilutions +2).

The total number of bacteria in the original sample is 3x10^{8}.

The dilution table looks as follows:

Unknown
| Volume Transferred | Dilutent Volume | Total Volume |
---|---|---|---|

Dilution A | 10 ml |
10 ml |
20 ml |

Dilution B | 1 ml |
5 ml |
6 ml |

Dilution C | 1 ml |
999 ml |
1000 ml |

Dilution D | 1 ml |
9 ml |
10 ml |

100 of Dilution D is used to inoculate two nutrient agar plates. After incubation the plates show 80 and 84 colonies of bacteria. The average number of bacteria colonies is 82.

**Method 1 **

Express the dilution and the inoculation in scientific format. (This value is found by dividing the Volume Transferred by the Total Volume.)

Dilution A 5x10^{-1}

Dilution B 1.667x10^{-1}

Dilution C 1x10^{-3}

Inoculation 1x10^{-1}

Calculate the amount of bacteria in the original solution by:

82 | = 9.84x10^{6} |

(5 x 10^{-1} x 1.667 x 10^{-1} x 10^{-3} x 10^{-1}) |

The plate count simulation coefficient is always expressed as a whole number so round the coefficient to the closest whole number. In this example, 9.84 is rounded to 10 and the resulting answer becomes 10x10^{6}. However, 10x10^{6} is not in standard scientific notation. Convert 10x10^{6} to standard scientific notation by changing the coefficient of 10 to a 1 and increasing the exponent by 1, thus 10 x10^{6} is converted to 1x10^{7}. Therefore, the number of bacteria in the original solution is 1x10^{7}.

**Method 2**

Calculate the ratios of the Total Volume / Volume Transferred and then multiply it by the colonies counted on the plate.

2 (Dilution A) = 20/10 = 2

6 (Dilution B) 6/1 = 6

1000 (Dilution C) = 1000/1 = 1000

10 (inoculation of 100 or 0.1ml) = 10

__82 (results shown on plate) = 82__

98,400,000 Rounds to 100,000,000 = 1x10^{7} Number of bacteria in the original solution

**Method 3 - EASY Math**

To use this solution three criteria must be met:

- the number of bacteria colonies on the plate must be between 10 and 90
- all dilutions must be done as 1 to 9, 1 to 99 or 1 to 999
- inoculation selected must be 100

There is no easy math for this solution because dilutions A & B were not done as 1 to 9, 1 to 99 or 1 to 999.