Counting Microorganisms. Methods Turbidity measurements Viable counts Most probable number Direct...
-
Upload
bathsheba-morton -
Category
Documents
-
view
228 -
download
1
Transcript of Counting Microorganisms. Methods Turbidity measurements Viable counts Most probable number Direct...
Counting Microorganisms
Methods
• Turbidity measurements• Viable counts• Most probable number• Direct counts
Most probable Number: MPN
– Based on Probability Statistics– Presumptive test based on given characteristics– Broth Technique
Most Probable Number (MPN)
• Begin with Broth to detect desired characteristic• Inoculate different dilutions of sample to be
tested in each of three tubes
-1 -2 -3 -4 -5 -6Dilution
3 Tubes/Dilution
1 ml of Each Dilution into Each Tube
After suitable incubation period, record POSITIVE TUBES (Have GROWTH and desired characteristics)
MPN - Continued• Objective is to “DILUTE OUT” the organism to zero• Following the incubation, the number of tubes showing the desired
characteristics are recorded• Example of results for a suspension of 1g/10 ml of soil• Dilutions: -1 -2 -3 -4 • Positive tubes: 3 2 1 0
– Choose correct sequence: 321 and look up in table
– Multiply result by middle dilution factor» 150 X 102 = 1.5 X 104/mL» Since you have 1g in 10mL must multiply again by 10» 1.5 X 105/g
Pos. tubesMPN/g (mL)
0.10 0.01 0.001
3 2 1 150
Direct Counts
• The sample to be counted is applied onto a hemacytometer slide that holds a fixed volume in a counting chamber
• The number of cells is counted in several independent squares on the slide’s grid
• The number of cells in the given volume is then calculated
Determining the Direct Count
7
• Count the number of cells in three independent squares– 8, 8 and 5
• Determine the mean– (8 + 8 + 5)/3 =7– Therefore 7 cells/square
Determining the Direct Count (Cont’d)
8
• 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 by the the volume of a square– Therefore 7/ 1 X 10-4 ml = 7 X 104 cells/ml
1mm
1mmDepth: 0.1mm
Problem
• A 500μl sample is applied to a hemacytometer slide with the following dimensions: 0.1mm X 0.1mm X 0.02mm. Counts of 6, 4 and 2 cells were obtained from three independent squares. What was the number of cells per milliliter in the original sample if the counting chamber possesses 100 squares?
Microscopy
Differential Staining
Differential StainingGram Stain
Divides bacteria into two groupsGram Negative & Gram Positive
• Stained Purple – Rods
• Genera Bacillus and Clostridium– Coccus
• Genera Streptococcus, Staphylococcus and Micrococcus
Gram Negative
• Stained Red– Rods:
• Genera Escherichia, Salmonella, Proteus, etc.
– Coccus: • Genera Neisseria, Moraxella and Acinetobacter
Rule of thumb
• If the genus is Bacillus or Clostridium= Gram (+) rod
• If the genus name ends in coccus or cocci (besides 3 exceptions, which are Gram (-))= coccus shape and Gram (+)
• If not part of the rules above, = Gram (-) rods
Cell Wall
15
Peptidoglycanwall
PlasmaMembrane
Lipopolysaccharidelayer
Absent
Gram + Vs Gram -
Method – Primary staining
1. Staining with crystal violet2. Addition of Gram’s iodine (Mordant)
+
Gram positive Gram negative
- - - - - - - - - - - - - - - Plasma membrane - - - - - - - - - - - - - - - Wall:peptidoglycan LPS
Method – Differential step
3. Alcohol wash
Gram positive Gram negative
- - - - - - - - - - - - - - - Plasma membrane - - - - - - - - - - - - - - - Wall: peptidoglycan LPS
+ + + + + + + + + + + + + +
Wall is dehydrated – Stain + iodine complex is trapped
Wall is not dehydrated – Complex is not trapped
Method – Counter Stain
4. Staining with Safranin
Gram positive Gram negative
- - - - - - - - - - - - - - - Plasma membrane - - - - - - - - - - - - - - - Wall:peptidoglycan LPS
+ + + + + + +
+
+ + + + + + + + + + + + + +
18
Summary
19
Fixation
Primary stainingCrystal violet
Counter stainingSafranin
WashDestaining
Acid Fast Staining
• Diagnostic staining of Mycobacterium– Pathogens associated with Tuberculosis and Leprosy– Cell wall has mycoic acid
• Waxy, very impermeable
Method
• Basis: – High level of compounds similar to waxes in their
cell walls, Mycoic acid, makes these bacteria resistant to traditional staining techniques
Method (Cont’d)
• Cell wall is permeabilized with heat• Staining with basic fuchsine
– Phenol based, soluble in mycoic layer– Cooling returns cell wall to its impermeable state
• Stain is trapped
• Wash with acid alcohol– Differential step
• Mycobacteria retain stain• Other bacteria lose the stain
Spore Stain
• Spores:– Differentiated bacterial cell– Resistant to heat, desiccation, ultraviolet, and
different chemical treatments• Thus very resistant to staining too!
– Typical of Gram positive rods • Genera Bacillus and Clostridium
– Unfavorable conditions induce sporogenesis• Differentiation of vegetative cell to endospore
– E.g. Anthrax
Malachite Green Staining
• Permeabilization of spores with heat
• Primary staining with malachite green
• Wash• Counter staining with
safranin
Vegetative cells(actively growing)
Spores(resistant
structures used for survival under
unfavourable conditions.)
Endospore(spore within
cell)
Sporangium(cell +
endospore)
Pathogens
19th Century: Robert Koch
• Studies anthrax disease which kills cows• Grows in pure culture bacteria obtained from
the blood of diseased animals– Bacillus anthracis
• Observations:– Blood of diseased animals transmits the disease– The microorganisms is found only in diseased
animals– The microorganism grown in the lab transmits the
disease to healthy animals26
Robert Koch (Cont’d)
• Conclusion: Microorganisms are responsible of diseases– Pathogens
• These results lead Robert Koch to formulate guidelines to associate a microorganism to a disease– Koch’s postulates
27
Koch’s Postulates
• The microorganism must be present in each diseased case but absent from healthy individuals
• The microorganism must be isolated and grown in pure cultures
• The disease must develop when the isolated microorganism is inoculated in a healthy host
• The same microorganism must be isolated again from the diseased host
28