MATERIALS AND METHODS
OBJECTIVE:
1) The present study was carried out to enumerate and identify lipolytic
Microorganisms from Milk samples collected from milk collection centres.
GENERAL METHODS:
1) Lipolytic activity can be measured by a clear zone formation around Each colony due to hydrolysis of Tributyrin as a substrate in the Lipase reagent when colonies grown on Tributyrin agar.
2) Lipolytic activity can be measured by a Greenish blue zone formation Around colonies due to lipase producing microorganisms present in the Sample when grown on butter fat medium.
3) Production of lipase from bacteria can be tested by the following methods.
1) Growth on Egg - Yolk Agar
2) Tween- Hydrolysis
3) Inoculation on to a nutrient agar medium containing a lipid.
MATERIALS REQUIRED:
1) Butter fat agar medium (PH- 7.8), 2) Ringer solution - full strength,
3) Ringer solution - quarter strength, 4) CUSO4 Standard solution,
5) Hot water bath, 6) Sterile 500ml beaker, 7) Sterile dilution botteles
8) Sterile 1ml, 10ml, 25ml pipettes, 9) sterile spreader
10) 2 Milk samples collected from milk Local Dairy Units and named as
Sample-1, sample-2.
PRINCIPLEl :
The development of greenish blue zone around the growth / colony is due to insoluble copper salts of fatty acids set free on lipolysis by lipase producing microorganisms.
MEDIA AND MATERIALS REQUIRED:
1) Butter fat agar medium preparation:
Butter fat : 5.0 gm
Yeast extract : 3.0 gm
Peptone : 5.0 gm
Agar : 15.0 gm
Distilled Water: 1000ml
Milk sample 1 : 5ml
Milk sample 2 : 5ml
Dissolved the constituents in Distilled water and sterilize at 1210 C for 10 min .
2) Ringer Solution Full strength :
Nacl : 9.0 Kg
Kcl : 0.42 gm
Anhydrous cacl2: 0.48 gm
Sodium Bicaronate: 0.20 gm
Dis.H20 : 1000ml
Dissolved the ingredients in Distilled water. Dispensed in flasks and sterile at 1210c for 10 min. In the present work we measured the constituents for 125 ml.
3)Ringer Solution - Quarter strength :
1 Part of Full strength Ringer’s solution and 3 parts of Distilled water.
Autoclaved at 1210c for 10 min. In Present work we Prepared Ringer Full strength solution 32 ml and mixed with 168 ml Autoclaved at 1210c for 10 min.
4)Cuso4 Solution ( Aqueous standard Solution):
Nacl - 58 .44 gm
Kcl - 74.55 gm
CaCl2 - 110.99 gm
NaHCO3 - 84.01 gm
Distilled Water - 1000 ml
Dissolve all the ingredients in required amount and preserve in amber
Coloured bottle in dark environment .
PROTOCOL:
1) Prepared butterfat agar medium plates 24 hrs before the start of the Experiment.
2) Warmed the Ringer’s solution at 40-450c for 15 min in a hot water bath.
3) Milk sample was taken in a sterile beaker and warmed at 40-450c for 15 min.
4) Pipetted 25 ml of gently mixed, melted Milk samples in to 125ml of warm (400 c) Ringer’s solution in a dilution bottle and shacked the mixture well.
5) Prepared sterile dilutions of the above mixture in Quarter strength Ringer’s Solution.
6) Poured 1ml (or) 0.1 ml two milk samples from the different dilutions (10-4 to 10-6) on butter fat agar plates.
7) Spreader the suspension over the media to ensure uniform distribution of Cells on the Butter fat agar medium to get isolated colonies.
8) Incubated the inverted plates at 21± 10c for 3 days and 32±10c for 2 days.
9) After incubation the plates were observed for the appearance of colonies of microorganisms
10) The plates were flooded with saturated CUSO4 Solution for 10-15 min.
11) The CUSO4 Solution was drained and the agar plate was kept under running tap water to remove the excess CUSO4 Solution.
COLONY MORPHOLOGY :
The number of lipolytic organisms per gram was estimated by standard
formula Dilution factor (1: 10 ) :
No . Of Lipolytic Organisms / gram ═ No. of Colonies × Dilution Factor Volume of sample added |
Dilution of the sample was made by adding 1ml of the sample to 9ml of Distilled water to make a sample solution of 1:10
OBSERVATION RESULTS :
Sample-1 : 1) The No. of colonies was counted 10-4 in all the 6 plates kept for analysis . Plate No 1( LP1) : 74 Colonies Plate No.2 (LP2) : 70 Colonies Plate No.3(LP3) : 68 Colonies Plate No.4(LP4) : 76Colonies Plate No.5(LP5) : 80 Colonies Plate No.6 (LP6) : 72Colonies Average ═ ( 74 + 70+ 68+ 76 +80+72)/6 ═ 73 2) The No. of colonies was counted 10-5 in all the 6 plates kept for analysis. Plate No 1( LP1) : 78 Colonies Plate No.2 (LP2) : 78 Colonies Plate No.3(LP3) : 75 Colonies Plate No.4(LP4) : 75Colonies Plate No.5(LP5) : 82 Colonies Plate No.6 (LP6) : 72Colonies Average ═ (78 + 78 + 75+ 75+82+72 )/6═ 76 3) The No. of colonies was counted 10-6 in all the 6 plates kept for analysis. Plate No 1( LP1) : 76Colonies Plate No.2 (LP2) : 70 Colonies Plate No.3(LP3) : 72 Colonies Plate No.4(LP4) : 74Colonies Plate No.5(LP5) : 76Colonies Plate No.6 (LP6) : 76Colonies Average ═ (76 + 70 + 72+ 74 +76+76 )/6 ═ 74 The number of colonies in the dilutions of 10-4 to 10-6 was counted 73, 76, 74 respectively. |
Sample -2: 1) The No. of colonies was counted 10-4 in all the 6 plates kept for analysis. Plate No 1( LP1) : 44 Colonies Plate No.2 (LP2) : 30 Colonies Plate No.3(LP3) : 35 Colonies Plate No.4(LP4) : 42Colonies Plate No.5(LP5) : 48 Colonies Plate No.6 (LP6) : 43Colonies Average ═ ( 44 + 30+ 35+ 42 +48+43 )/6═ 40 2) The No. of colonies was counted 10-5 in all the 6 plates kept for analysis. Plate No 1( LP1) : 48 Colonies Plate No.2 (LP2) : 41 Colonies Plate No.3(LP3) : 35 Colonies Plate No.4(LP4) : 42Colonies Plate No.5(LP5) : 46 Colonies Plate No.6 (LP6) : 42Colonies Average ═ (48 + 41 + 35+ 42+46+42)/6 ═ 42 3) The No. of colonies was counted 10-6 in all the 6 plates kept for analysis. Plate No 1( LP1) : 76Colonies Plate No.2 (LP2) : 70 Colonies Plate No.3(LP3) : 72 Colonies Plate No.4(LP4) : 74Colonies Plate No.5(LP5) : 76Colonies Plate No.6 (LP6) : 76Colonies Average ═ (48 + 41 + 44+ 42 +46+45 )/6═ 44 The number of colonies in the dilutions of 10-4 to 10-6 was counted 40, 42, 44 respectively. |
COLONY MORPHOLOGY:
Sample: 1
The colonies were large to very large, gray-white, blistery and dry in appearance
Sample: 2:
The colonies were Small , Creamy white , round , discrete colonies
C) OBSERVATION OF ZONE FORMATION:
After adding CUSO4 Solution the colonies appears blue in colour due to insoluble copper salts of fatty acids set free on lipolysis by lipase producing micro organisms . The washed plates were observed for the presence of zones around the Colonies. The presence of greenish- Blue zones around the colonies was Observed . This indicates the presence of the lipase production by the microorganisms present in the butter sample .
GRAM STAINING TECHNIQUE:
The test was originally developed by Christian Gram in 1884, but was modified by Hucker in 1921. The modified procedure provided greater reagent stability and better differentiation of organisms.
The Gram stain is used to classify bacteria on the basis of their forms, sizes, cellular morphologies, and Gram reactions; in a clinical microbiology laboratory, it is additionally a critical test for the rapid presumptive diagnosis of infectious agents and serves to assess the quality of clinical specimens.
Gram stain permits the separation of all bacteria into two large groups, those which retain the primary dye (gram-positive) and those that take the color of the counterstain (gram-negative).
The primary dye is crystal violet and the secondary dye is usually either safranin O or basic fuchsin. Some of the moreCommon
Formulations include: Saturated crystal violet (approximately 1%), Hucker’s crystal violet, and 2% alcoholic crystal violet.
PROTOCOL:
1. Deparaffinize sample and hydrate to distilled water
2. Place slides of staining rack
3. Add 1ml of crystal violet solution to the sample. Alternativley gentian violet can be used
4. Add 250ul of 5% sodium biocarbonate
5. Gently mix solutions for one minute
6. Rinse with water
7. Cover sample with Gram's iodine solution and incubate for 1 minute
8. Rinse with water
9. Blot with filter paper until dry
10. Incubate breifly in acetone. This step should be short as longer incubation will cause some gram positives to become gram negative. The time of incubation is usually determined by the decolorization of the background tissue. Repeat if necessary.
11. Cover sample with basic fuchsin solution for 3 minutes
12. Rinse with water
13. Blot excess water with filter paper
14. Dip slide in acetone
15. Dip slide in 0.1% picric acid in acetone
16. Rinse in acetone
17. Rinse in 50% acetone - 50% histolene
18. Rinse in 100% histolene. Repeat if necessary
19. Mount slide
Gram positive bacteria stain blue and gram negative bacteria stain red.
PROCEDURE:
1)The gram-positive cell envelope consists of a thick layer of peptidoglycan embedded with techoic acids and a plasma membrane comprised of phospholipids with integral membrane proteins traversing the bilayer.
2)The cells are flooded with crystal violet dye. Crystal violet is a water-soluble, basic dye. In solution, basic dyes produce dye particles with positive charges (cations). Sometimes the crystal violet dye particle is abbreviated CV+.
3)The individual crystal violet ions penetrate the thick peptidoglycan layer of the cell as well as the plasma membrane, making their way through the matrix created by the crosslinking of polysaccharides and proteins within the peptidoglycan layer.
4)Gram's iodine solution is added. This solution consists of a mixture of iodine and potassium iodide. The active constituent in this solution is the iodide ion or an iodine-iodine complex.
5) Like the crystal violet dye particles, the iodide ions are also able to penetrate the thick peptidoglycan layer of the cell. Here, the iodide ions mix with the crystal violet dye particles that were added in the previous step.
6) The crystal violet and iodide ions react, forming a crystal violet-iodine complex. This complex is insoluble in water and produces particles much larger than either the iodide ions or the crystal violet ions individually.
7) The alcohol/acetone mixture displaces water in the peptidoglycan layer, resulting in dehydration. This loss of water causes the thick peptidoglycan layer to shrink, tightening the matrix created by the crosslinking of polysaccharides and proteins.
8) It is important to know that this decolorizing step is a critical step in the Gram stain protocol. Exposure to the alcohol for too long can cause cells that are gram-positive to lose too much of the dye complex due to damage to the peptidoglycan layer. These cells will not appear gram-positive when the staining procedure is complete.
9) The counterstain, normally safranin, is added. Like crystal violet, safranin is a weakly water-soluble, basic dye that produces cationic stain particles in solution that bind negatively charged moieties such as the techoic acids, peptides and phospholipid heads within the envelope and in the cytoplasm.
10) Safranin, because of its small size, is able to penetrate the dehydrated peptidoglycan layer and bind to negatively charged moieties. Because the safranin is much lighter in color than the crystal violet-iodine complex.
11) When viewed under a microscope, gram-positive cells appear purple due to the crystal violet-iodine complex retained inside.
12) The gram-negative cell envelope consists of a thin layer of peptidoglycan surrounded by two phospholipid membranes, one interior and one exterior. Polysaccharide chains are bound to the phosphate heads of the outer membrane to form lipopolysaccharides. Both the membranes contain integral membrane proteins. Place cursor over each membrane for ID.
13) The cells are flooded with crystal violet dye. Crystal violet is a water-soluble, basic dye. In solution, basic dyes produce dye particles with positive charges (cations). Sometimes the crystal violet dye particle is abbreviated CV+.
14) The individual crystal violet ions penetrate the thin peptidoglycan layer of the cell as well as the plasma membrane, making their way through the matrix created by the crosslinking of polysaccharides and proteins within the peptidoglycan layer.
15) Gram's iodine solution is added. This solution consists of a mixture of iodine and potassium iodide. The active constituent in this solution is the iodide ion or an iodine-iodine complex.
16) Gram's iodine solution is added. This solution consists of a mixture of iodine and potassium iodide. The active constituent in this solution is the iodide ion or an iodine-iodine complex.
17) Gram's iodine solution is added. This solution consists of a mixture of iodine and potassium iodide. The active constituent in this solution is the iodide ion or an iodine-iodine complex.
18) A decolorizing solution, normally consisting of a mixture of ethyl alcohol and acetone, is added. Numerous variations of the decolorizing solution formula are used in labs.
19) The alcohol/acetone mixture displaces water in the peptidoglycan layer, resulting in dehydration. This loss of water causes the thin peptidoglycan layer to shrink slightly, tightening the matrix created by the crosslinking of polysaccharides and proteins. The alcohol/acetone mixture also disrupts and dissolves the outer membrane, exposing the peptidoglycan layer to the environment.
20) Safranin, because of its small size, is able to penetrate the dehydrated peptidoglycan layer and bind to negatively charged moieties. Because
Safranin is the only stain present, the cells will have a pink or red Colour.
GRAM’S STAINING:
Sample-1:
Rod shaped cells was observed with purple colour. Therefore it was identified as Gram positive bacteria.
Sample-2:
Rod shaped cells was Observed with purple colour. Therefore it was also identified As Gram positive bacteria.
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