5 Jan 2012

ISOLATION OF MICROORGANISMS WITH LIPOLYTIC ACTIVITY FROM DAIRY PRODUCTS




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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