01/08/2018. Counting Microorganisms. Counting microorganisms. Turbidity Measurements. Relative abundance. Direct counts.

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1 Counting Microorganisms 1 Counting microorganisms 2 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 3 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 1

2 Turbidity Measurements 4 Spectrophotometer (A600): Measure of optical density (O.D.) Light Detector. Reading 600nm Different reading O.D. 600nm 2.0 % Transmission Cellular density 100 Advantages: Quick 6 Disadvantages: Relative measurement Does not discriminate between dead and living Does not discriminate between bacteria and detritus Does not discriminate between different microbes 2

3 Problem 7 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 8 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 9 This slide has 2 independent counting chambers 3

4 Determining the Direct Count 10 Count the number of cells in 3 squares 8, 8 and 5 Calculate the average ( )/3 =7 Therefore 7 cells/square Determining the Direct Count 11 1mm Depth: 0.1mm 1mm Calculate the volume of a square: = 0.1cm X 0.1cm X 0.01cm= 1 X 10-4 cm 3 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 10 4 cells/ml 12 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 4

5 Problem 13 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 14 Volume of counting chamber? 0.01 X 0.01 X 0.002cm = 2 X 10-7 cc or ml Volume of one square? 2 X 10-7 ml/100 squares = 2 X 10-9 ml Number of cells per milliliter? Avg. Number of cells/square = 4 Therefore 4 cells/2 X 10-9 ml = 2 X 10 9 cells/ml Problem 15 A sample containing 10 9 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? 5

6 Solution 16 Volume of counting chamber? 0.01 X 0.02 X 0.01cm = 2 X 10-6 cc or ml Volume of one square? 2 X 10-6 ml/50 squares = 4 X 10-8 ml Number of cells for one square? 10 9 cells/ml X 4 X 10-8 ml = 40 cells Viable Counts 17 A viable cell: a cell which is able to divide and form a population (or colony) 1. A viable cell count is done by diluting the original sample 2. Plating aliquots of the dilutions onto an appropriate culture medium 3. Incubating the plates under appropriate conditions to allow growth Colonies are formed 4. Colonies are counted and original number of viable cells is calculated according to the dilution used Dilutions of Bacterial Sample 18 6

7 Dilutions of Bacterial Sample 19 Dilutions of Bacterial Sample 20 Dilutions of Bacterial Sample 21 7

8 Dilutions of Bacterial Sample Viable Counts 23 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 10 6 = 5.7 X 10 7 CFU/mL Advantages: 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 24 8

9 Problem 25 Broths of E.coli and of Micrococcus luteus that contain 10 9 cells/ml were diluted to mL were then plated. How many CFUs are expected in each case? E.coli: 10 9 X 10-6 X 0.2mL = 200 CFU M. luteus: (tetrad; therefore 1 CFU = 4 cells) (10 9 X 10-6 X 0.2mL)/4 = 50 CFU Most Probable Number: MPN 26 FondBased on statistical probabilities Presumptive test based on a given set of characteristics Broth method Most Probable Number: MPN 27 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 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) 9

10 MPN (Cont d) 28 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 MPN (Cont d) 29 Combination chosen: 321 Pos. Tubes MPN/g (ml) Multiply MPN by the middle dilution factor 150 X 10 4 = 1.5 X 10 6 bacteria/ml Or 1.5 X 10 6 bacteria/0.1g of soil Therefore 1.5 X 10 7 bacteria/g Question 30 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 10