Counting Microorganisms

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

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

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

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

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

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?

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

Hemacytometer This slide has 2 independent counting chambers

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

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

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

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?

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

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?

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

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

Dilutions of Bacterial Sample

Dilutions of Bacterial Sample

Dilutions of Bacterial Sample

Dilutions of Bacterial Sample

Dilutions of Bacterial Sample 5672 57 4

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 30-300 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

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

Problem Broths of E.coli and of Micrococcus luteus that contain 109 cells/mL were diluted to 10-6. 0.2mL 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

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

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 -1 -2 -3 -4 -5 -6 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)

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: -3 -4 -5 -6 Positive tubes: 3 2 1 0 Choose the correct combination: 321 and find it in the table

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

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