1. Counting Microorganisms

2. Counting microorganisms
Relative abundance
Turbidity measurements
Direct counts
Absolute counts
Viable counts
Absolute number of growing bacteria
Most probale number (MPN)
Probable number of bacteria with given characteristics

3. Turbidity Measurements
Measures the quantity of light that can go through a sample
The less light that passes the more dense is the population
Measurements of optical density or percent transmission

4. Turbidity Measurements
Spectrophotometer (A600):
Measure of optical density (O.D.)
Light
600nm
Detector…. Reading
Different reading

5. 2.0
1.0
O.D. 600nm
% Transmission
100 50
Cellular density

6. Advantages: Disadvantages: Quick Relative measurement
Does not discriminate between dead and living
Does not discriminate between bacteria and detritus
Does not discriminate between different microbes

7. Problem
Two culture broths have the same optical density. One broth represents a large bacillus (A) whereas the other represents a small bacillus (B).
What can you conclude?

8. Direct Counts
The sample to be counted is applied to a hemacytometer slide which holds a fixed volume in a counting chamber
The number of cells is counted
The number of cells for a given volume is determined

9. Hemacytometer
This slide has 2 independent counting chambers

10. Determining the Direct Count
Count the number of cells in 3 squares
8, 8 and 5
Calculate the average
( )/3 =7
Therefore 7 cells/square

11. Determining the Direct Count
1mm
Depth: 0.1mm
1mm
Calculate the volume of a square:
= 0.1cm X 0.1cm X 0.01cm= 1 X 10-4cm3 or ml
Divide the average number of cells per square by the volume of one square
Therefore 7/ 1 X 10-4 ml = 7 X 104 cells/ml

12. Limits: Advantages: Quick Growth is not required
No information about organism required
Limits:
Does not discriminate between live and dead
May be difficult to distinguish bacteria from detritus

13. Problem
A sample is applied to a hemacytometer slide whose counting chamber has the following dimensions: 0.1mm X 0.1mm X 0.02mm and a total of 100 squares. Counts of 6, 4 and 2 cells were recorded in three independent squares.
What is the volume of the counting chamber?
What is the volume of one square?
What is the number of cells per milliliter in the sample?

14. Solution Volume of counting chamber? Volume of one square?
0.01 X 0.01 X 0.002cm = 2 X 10-7cc or mL
Volume of one square?
2 X 10-7mL/100 squares = 2 X 10-9mL
Number of cells per milliliter?
Avg. Number of cells/square = 4
Therefore 4 cells/2 X 10-9mL = 2 X 109 cells/mL

15. Problem
A sample containing 109 cells/ml is applied to a hemacytometer whose counting chamber has the following dimensions: 0.1mm X 0.2mm X 0.1mm and which has 50 squares. How many cells on average are expected / square?

16. Solution Volume of counting chamber? Volume of one square?
0.01 X 0.02 X 0.01cm = 2 X 10-6cc or mL
Volume of one square?
2 X 10-6mL/50 squares = 4 X 10-8mL
Number of cells for one square?
109 cells/mL X 4 X 10-8mL = 40 cells

17. Viable Counts
A viable cell: a cell which is able to divide and form a population (or colony)
A viable cell count is done by diluting the original sample
Plating aliquots of the dilutions onto an appropriate culture medium
Incubating the plates under appropriate conditions to allow growth
Colonies are formed
Colonies are counted and original number of viable cells is calculated according to the dilution used

18. Dilutions of Bacterial Sample

19. Dilutions of Bacterial Sample

20. Dilutions of Bacterial Sample

21. Dilutions of Bacterial Sample

22. Dilutions of Bacterial Sample
5672 57 4

23. Viable Counts
The total number of viable cells is reported as Colony-Forming Units (CFUs) rather than cell numbers
Each single colony originates from a colony forming unit (CFU)
A plate having colonies is chosen
Calculation:
Number of colonies on plate X ÷ volume plated ÷ dilution = Number of CFU/mL
Ex. 57 CFU X 106 = 5.7 X 107 CFU/mL

24. Advantages: Disadvantage: Determines the number of live organisms
Can discriminate between different microorganisms
Disadvantage:
No universal medium
Requires the growth of the microorganism
Only living cells develop colonies
Clumps or chains of cells develop into a single colony

25. Problem
Broths of E.coli and of Micrococcus luteus that contain 109 cells/mL were diluted to mL were then plated. How many CFUs are expected in each case?
E.coli: 109 X 10-6 X 0.2mL = 200 CFU
M. luteus: (tetrad; therefore 1 CFU = 4 cells)
(109 X 10-6 X 0.2mL)/4 = 50 CFU

26. Most Probable Number: MPN
FondBased on statistical probabilities
Presumptive test based on a given set of characteristics
Broth method

27. Most Probable Number: MPN
Start with broths that allow to detect the desired characteristics
Inoculate different dilutions of the sample to be tested in each of three tubes
Dilution
3 Tubes/Dilution
1 ml of each dilution in each tube
After an appropriate incubation period, record the POSITIVE TUBES (that show growth and the desired characteristics)

28. MPN (Cont’d) The objective is to DILUTE the organism to zero
After the incubation, the number of tubes that have the desired characteristics is recorded
Example of results for a suspension of 1g of soil/10 ml
Dilutions:
Positive tubes:
Choose the correct combination: 321 and find it in the table

29. MPN (Cont’d) Combination chosen: 321
Multiply MPN by the middle dilution factor
150 X 104 = 1.5 X 106 bacteria/mL
Or 1.5 X 106 bacteria/0.1g of soil
Therefore 1.5 X 107 bacteria/g
Pos. Tubes
MPN/g (mL)
0.1
0.01
0.001 3 2 1
150

30. Question
Which method would be the fastest and most accurate to determine the number of bacteria in each of the following cases?
Number of bacteria in a lake
Number of E. coli in 100g of beef
Compare the cell denity of two cultures of B. subtilis
Determine the number of Escherichia and of Lactobacillus in 100mL of milk