BIOCHEMICAL TEST TO IDENTIFY BACTERIA IN THE LABORATORY Details of the test include: Stating the test State the use/aim of the test Define the test Mention the main reagents State the principle of the test and methods State the quality control of the test A: Enzyme identification test: They include the following test; a) Coagulase test b) Catalase test c) Urease test d) Phosphatase test e) DNA-ase test f) Oxidase test g) Lethicinase test 1. UREASE TEST Aim: To differentiate the enterobacteria strains that produces the urease enzyme. Ways of doing the test which may include; the Protease, Klebsiella and the Yesinia species and also to identify the Cryptococcus species, Brucella, Helicobacter pylori Using the Rosco urease identification tablet Using modified Christensen’s urea broth Materials: Rosco urease tablets 0.25ml physiological saline Test tube Incubator/water bath Bijou bottle Inoculating loop Christensen’s modified urea broth Principle of the test The test organism is cultured in a medium which contains urea, when the strain is urease-producing , the enzymes urease will 1
Transcript
BIOCHEMICAL TEST TO IDENTIFY BACTERIA IN THE LABORATORYDetails of the test include:
Stating the test State the use/aim of the test Define the test Mention the main reagents State the principle of the test and methods State the quality control of the test
A: Enzyme identification test:They include the following test;
a) Coagulase testb) Catalase testc) Urease testd) Phosphatase teste) DNA-ase testf) Oxidase testg) Lethicinase test
1. UREASE TESTAim:To differentiate the enterobacteria strains that produces theurease enzyme. Ways of doing the test which may include; the Protease,Klebsiella and the Yesinia species and also to identify theCryptococcus species, Brucella, Helicobacter pylori
Using the Rosco urease identification tablet Using modified Christensen’s urea broth
Principle of the testThe test organism is cultured in a medium which contains urea,when the strain is urease-producing , the enzymes urease will
1
break down the urea (by hydrolysis) to give ammonia and thebicarbonate released as carbondioxide, with the release of theammonia, the medium becomes alkaline showing a pink red colour ofthe indicator.
Method for the Rosco urease tablet:a) Prepare the dense milky suspension in 0.25ml of
physiological saline in a test tubeb) Add a urease tablet, close the tube and incubate at 35-37o C
in an incubator or water bath for 4 hours or overnightc) Observe for colour change within the 4hours
Results: Red/purple colour shows positive urease test Yellow/orange colour shows negative urease test
The urease test using Christensen’s (modified) urea broth:a) Inoculate heavily the test organism in a bijou bottle
containing 3ml sterile Christensen’s modified urea broth b) Incubate at 35-37oc for3-12hours in a water bathc) Look for the colour change in the medium
Results: Pink colour shows positive urease test No pink colour shows a negative urease test
2. OXIDASE TEST (Cytochrome oxidase test)Aim:Is to assist in the identification of Pseudomonas, Nesseria,Vibrio Brucella and Postuerella species all of which produce theCytochrome oxidase enzyme
Materials/requirements: Oxidase reagent Filter paper Petri dish Pipette Applicator stick
2
Principle:A piece of filter a paper is soaked in a few drops of oxidasereagent, a colony of the organism is then smeared on to thefilter paper and a blue-purple colour shows a positive oxidasetest and a negative oxidase test I shown by no blue-purple colourwithin 10 seconds
Method of the test:a) Place a piece of filter in a clean Petri-dish and add 2-3
drops of freshly prepared oxidase reagentb) Using a piece of stick or glass rod remove a colony of the
test organism and smear on the filter paperc) Look for the development of the blue-purple colour within
few secondsResults:
Blue-purple colour shows positive oxidase test within 10seconds
No blue-purple colour shows negative oxidase test within 10seconds
Controls: Positive oxidase control is Pseudomonas aeroginosa Negative oxidase control is Escherichia coli
3. DNA-ase test:Aim: Identification of the S. aureus which produces thedeoxyribonuclease (DNA-ase) enzymes and its used only when plasmaisn’t there to do coagulase test or when the coagulase test isdifficult to interpretMaterials/requirements:
Principle:The deoxyribonuclease hydrolyzes the deoxyribonucleic acid (DNA);the test organism is cultured on the medium which contains DNA.After overnight incubation, the colonies are tested for DNA-aseproduction by flooding the plate with a weak hydrochloric acid
3
solution. The acid precipitates unhydrolyzied DNA. DNA-aseproducing colonies are therefore surrounded by clear areas due toDNA hydrolysisMethods:
a) Divide the DNA-ase plate into the required number of stripsby making the underside of the plate
b) Using a sterile loop or swab, spot inoculate the test andcontrol organism. Make sure each test area is labeledclearly
c) Incubate the plate at 35-37oC f overnightd) Cover the surface of the plate with 1 mol/l hydrochloric
acid solution. Tip off excess acide) Look for the clearing around the colonies within 5minutes of
adding the acidResults:
Clearing around the colonies shows DNA-ase positive test No clearing around the colonies shows DNA-ase negative
strainsControl:
Positive control is S. aureus Negative control is S.epidemidis4. PHOSPHATASE
Aim: The test determines the inactivation of the enzyme phosphatase byheat that naturally occurs in milk. Correct degree of heating inpasteurization kills this enzyme. However, its presence shows poorpasteurization.
Principle This enzyme liberates P-nitrophenol from di-sodium P-nitrophenylphosphate and thereby producing a yellow colour at PH of 10.0 that canbe quantified with a colorimeter.
Method Add 1ml of the buffer substrate in test tubes. Stopper and
incubate at 37°C for 5 minutes.
Remove from a water bath and add 1ml of the milk and mix.
Re-stopper, invert for several times and re-incubate for 2hours.
Set controls using 1ml of boiled milk and treated as the test andread after 2 hours.
Read the result by comparing the colour produced using acolorimeter
Results:
A yellow colour produced shows a phosphatase test positive hencephosphatase was liberated.
Negative test , no yellow colour produced
5. LECITHINASE TEST
Aim:
It’s to identify the ability of the microbes to produce the enzymelecithinase and the organism that produces the enzymes are theBacillus cerulus which is also a strongly hemolytic on BA and motile,B. anthracis which is non-motile and non-hemolytic, B. thoringiensisand the Clostridium spp
Requirements: Two nutrient broth tubes sub-cultured with suspected Bacillus
colonies (1st have Bacillus cerulus and the 2nd has Bacillussubtilis)
Inoculating loop Incubator (35-
+10C) Manitol –egg yolk polymycin (MYP) agar Thiogylcolate broth with inoculated Clostridia spp Clostridium perfrigens type A antitoxins Anaerobic gas pak jar system Anaero GRO egg yolk agar
5
Principle Clostridium perfrigens produces an enzyme Lethicinase which breaksdown lipoproteins in egg York medium into phosphocoline and glycerideswhich results into a formation of a precipitate in the medium givingthe medium opacity/opalescence. If lipase is present, there will behydrolysis which is seen as a fatty layer covering the colony andlactose fermentation will be seen as reddening in the medium onexposure to air. Protenase activity will be seen as an area ofclearing around the colony due to breakdown of casein in the media.
The bacillus produces the enzyme lecithinase which will hydrolyze thelecillin to form a zone of white precipitate around the coloniesshowing positive test and yellow colonies showing negative lecithinasetest
Method for the Bacillus
a) Transfer aseptically the MYP into a Petri-dishb) Divide the agar into three equal partsc) Transfer a loopful of culture from the 1st t nutrient broth
and inoculate a segment on the MYP agar surface and streakon the medium to obtain isolated colonies
d) Do the same for the 2nd nutrient brothe) Inoculate the uninoculated broth (control) on one side of
the MYP agarf) Incubate the plate at 34-36 0C for 24-48 hoursg) Examine the colour change and opacity zones around the
coloniesResults:
1st plate appears pink-red without manitol fermentation andclear zones of opacity conforming lecithinase activity hencepositive result
Colonies in the 2nd plate fermented the manitol appearingyellow with no opacity showing a negative lecithinase test
Methods for the clostridium:a) Divide three plate into two halvesb) To one halve add 60microlitres of clostridium perfrigens
type A antitoxin and spread evenlyc) Allow to absorb and dry then mark the side inoculated
6
d) Take a loopful of the test organism from the thiogylcolatebroth and strike it as a straight line from the antitoxinfree side to the antitoxin side
e) Inoculate the control organism on the same plate in the sameway
f) Incubate anaerobically in a gas pak jar after streaking at35-37 OC for 24-48 hours
g) Examine Results:
Opacity shows a positive lecithinase test Clearance around the colony indicates protease activity
No opacity shows lecithinase test negative
6. Coagulase test:Aim: The test is used to identify staphylococcus aureus which producesthe enzyme coagulase form other strains of the staphylococcuswhich don’t produce the coagulase enzymes and then coagulase iscapable of coagulating the EDTA-treated plasma to produce clumpsTypes of coagulase test enzymes:
a) Free coagulase: these convert the fibrinogen to fibrin byactivating the coagulase reacting factor presenting plasmawhich is detected by the appearances of the fibrin in thetube.
b) Bound coagulase (clumping factor): this converts thefibrinogen directly to fibrin without requiring a coagulasefactor reacting and it’s detected by clumping of thebacterial cells in the rapid slide method.
Materials: EDTA antcoagulated human plasma or rabbit plasma Oxalate or heparin plasma can also be used but not citrated
plasma because citrate utilizes bacteria example theEnterococci which may cause clotting of the plasma usingknown coagulase positive S.aueus
Principle:Coagulase produced by the bacteria will cause the plasma to clotby converting fibrinogen to fibrin forming visible clumps.Methods:
a) Tube method
7
b) Slide methodNB: - A tube method is used after the slide test has proved to benegative or not clear and staphylococcus has been isolated from aserious infection. A tube test method can also be required to detect some strains ofthe staphylococcus which are resistant to the methicilin (MRSA)Slide test method for bound coagulase:
a) Place a drop of distilled water on each end of a slide or ontwo separate slides
b) Emulsify a colony of the test organism in each of the dropsto make two thick suspension
c) Add a loop full of the plasma on one of the suspension andmix gently
d) Look for clumping of the organism within 10 secondsResults:
Clumping within 10 seconds shows staphylococcus aureus sobound coagulase
No clumping within 10 seconds shows no bound coagulase so nostaphylococcus aureus
Tube test method for free coagulase:a) Take three small test tubes and label them; T-test organism,
(18-24 hr Broth culture), POS-Positive control (18-24 hr S.aureus Broth culture) and NEG-Negative control (sterilebroth culture)
b) Pipette 0.2ml of the plasma into each tubec) Add 0.8ml of the test broth culture to the test tube labeled
Td) Add 0.8 ml of the S. aureus culture and add to the test tube
labeled POSe) Add 0.8 ml of the sterile broth to the tube labeled NEGf) After mixing all gently, incubate the three tubes at 35-37
oC g) Examined for clotting after 1 hour if no clotting, reexamine
after 3 hours, if still no clotting leave to stand overnightat RTP and re-examine.
Results: Clotting of the tube content or fibrin clot in the tube
shows S. aureus positive
8
No clotting or fibrin clot shows negative control tube or S.aureus negative
Control: Positive coagulase control is S. aureus Negative coagulase control is Escherichia coli or S.
epidermidis
7. CATALASE TEST:Aim: To differentiate those bacteria that produce the enzymes catalasesuch as the Staphylococci from the non catalase producing enzymeslike the streptococci and also to differentiate the bacilluswhich are catalase positive from the clostridium which arecatalase negative.Requirement:
Hydrogen peroxide, 3% Sterile applicator stick Colony of the Test organism in a medium (not a blood agar
medium because blood contains the enzyme catalase Test tube
Method of the test tube:a) On a clean glass test tube, pipette 2-3 ml of the hydrogen
peroxide (H2O2) solutionb) Using a sterile applicator stick or glass rod, pick a colony
of organism from the culture media and mix with the 3% H2O2 c) Observe for reaction immediately
Results: Appearance of air bubbles means catalase positive and if
it’s a gram positive cocci then it’s Staphylococcus No air bubbles appearance means catalase negative bacteria
and if gram cocci then it’s streptococcus not staphylococcusMethod for slide test:
a) Place a slide in a Petri dish and coverb) Using an inoculating loop, pick a small portion of the test
organism from a well isolated 18-24 hrs colony and place onthe slide
c) Pipette 1 drop of 3% H2O2 and add to the slide on theculture, don’t mix
d) Cover the Petri dish to limit the aerosols
9
e) Observe for the air bubbles formingResults:
Positive air bubbles shows catalase positive organism Negative air bubbles shows catalase negative organism
Control: Positive catalase test is staphylococcus spp Negative control is streptococcus species
Aim: The Indole test is a biochemical test performed on bacterial speciesto determine the ability of the organism to split Indole from theamino acid tryptophan. This division is performed by a chain of anumber of different intracellular enzymes, a system generally referredto as "Tryptophanase." This is basically used to differentiate E. colifrom Klebsiella, Yesinia, Serratia and other enterobacteria.
Materials: Inoculating loop Tryptophan water/ peptone water Bijou bottle Incubator 0.5ml of Kovac’s reagents/Ehrlich’s
Principle
10
Indole is generated by reductive deamination from tryptophan via theintermediate molecule indolepyruvic acid. Tryptophanase catalyzes thedeamination reaction, during which the amine (-NH2) group of thetryptophan molecule is removed. Final products of the reaction areIndole, pyruvic acid, ammonia (NH3) and energy and the colour changeof red produced due to the effect of the Kovac’s or Ehrlich’s reagentwhich contains 4, P-dimethylaminobenzaldehyde which reacts with theIndole.
Method:a) Inoculate the test organism in a bijou bottle which contains
3mls of peptone water or tryptophan waterb) Incubate at 35-37 OC for 48hoursc) Test for Indole by adding o.5ml of the Kovac’s reagent,
shake gently and examine for Red colour in the surface layerwith 10 minutes
Results: Red surface layer shows positive Indole test hence the
bacteria that produces the Indole is present No red surface layer within 10 minutes hence absence of the
bacteria that produces the Indole Controls:
Positive control is Escherichia coli Negative control is lactobacillus spp
2. BILE ESCULIN TEST:Aim: To differentiate Enterococcus and group D streptococci which areBile tolerant and can hydrolyze Esculin from non group D viridiangroups streptococci which grows poorly on the BileRequirements:
Bile- Esculin agar medium (agar slant or plate) Inoculating wire loop Incubator Timer
Principle of the test:Bile –Esculin test is based on the ability of certain bacterianotably the group D streptococci and Enterococcus species tohydrolyze Esculin in the presence of BileProcedure:
11
a) With an inoculating wire, touch two or three morphologicallysimilar streptococcal colonies and inoculate the slant ofthe bile Esculin medium with an S- shaped motion or streakthe surface of the bike Esculin plate for isolation
b) The inoculation tube is incubated at 35-37 OC for 24 hoursc) Read results after 24hrs inoculation
Results: Diffused blackening of more than half of the slant within
24- 48 hrs indicate esculin hydrolysis On plates, black haloes will be observed around isolated
colonies and any blackening is considered positive All group D streptococcus will be bile –esculin positive
within 48 hrsControl:
Positive control Enterococcus spp Negative control is viridians Streptococcus not of group D
Limitation of the test: 3% of some of the streptococcus may also hydrolyze esculin
in presences of the bile
3. Methyl red test:Aim:Methyl red test determine whether the microbes perform mixed acidfermentation when supplied with glucose.Requirements:
Methyl red indicator Inoculating loop Incubator Barrits reagent Methyl red Voges- Proskauer broth (MR-VP) Pure culture of microorganism
Principle:In mixed acid fermentation, three acids formed in significantamounts (acetic acid, lactic and succinic) when glucose isfermented and the large amount of acid results into a significantdecrease in the Ph of the medium below 4.4 and this is visualizedby using pH indicator, methyl red which is yellow above Ph 5.1and red at ph 4.4 so a colour change to red on addition of methyl
12
red shows a positive result and a yellow colour shows a negativetest.Methods:
Inoculate two tubes containing MR-VP broth with a pureculture of the microorganism under investigation using ainoculating loop
Incubate at 35 OC for 4 days Add 5 drops of the methyl red indicator solution to the
first tube To the 2nd tube add Barrits reagent Observe the medium for colour change
Results: MR Test positive shows a red colour after adding the MR MR Test negative shown when Colour remained yellow
Quality control: Positive control is Escherichia coli Negative control is Enterobacter aerogenes
4. SATELLITISM TEST:Aim:Used to identify the Haemophilus influenzae that can’t grow onthe 5% sheep blood agar since it contains hemin (factor X) butLuck the factor V (NAD)Requirement:
Sterile physiological saline or sterile peptone water Chocolate agar medium with Haemophilus influenza suspect Sterile swab Nutrient agar or tryptic soya agar plate A plate of blood agar Pure culture of S. aureus Incubator Carbondioxide jar
Principle:Satillism test is based on the ability of the Haemophilusinfluenza to grow on sheep blood agar very close to the coloniesof S. aureus as the S. aureus produces the NAD (factor V) as ametabolic bi-product when grown in culture responsible for the
13
growth of the Haemophilus influenzae a phenomenon called thesatelliting.Method:
a) Mix a loopful of the suspected colonies of Haemophilusinfluenzae in about 2mls of the sterile physiological saline
b) Using a sterile swab, inoculate the organism suspensionabove in a;i. Plate of nutrient agarii. Plate of blood agar
c) Streak a pure culture of S. aureus across each of theinoculate plate
d) Incubate both plates in a carbondioxide jar enrichedatmosphere at 35-370C for 18-24 hours
e) Examine the culture for growth and satellite coloniesResults:The suspected colonies are of H. influenza if;
Growth is seen in the blood agar but not in the nutrientagar plate
The colonies near the column of S. aureus growth are largerthan those from it
If satellites colonies are present are both blood are nutrientagar plates than the organism is probably a Haemophilus speciesthat requires only factor V even in the absence of factor X suchas H. parainfluenzae
5. CARBOHYDRATE/SUGAR/GLUCOSE UTILIZATION Aim:Bacteria produce acidic products when they ferment certaincarbohydrates/sugar. The carbohydrate utilization tests are designedto detect the change in pH which would occur if fermentation of thegiven carbohydrate/sugar occurred.
Principle:Acids lower the pH of the medium which will cause the pH indicator(phenol red) to turn yellow. If the bacteria do not ferment thecarbohydrate then the media remains red. If gas is produced as a bi-product of fermentation, then the Durham tube will have a bubble init.
Requirements:
14
Basal medium which contains; peptone, meat extract, sodiumchloride, and indicator in which carbohydrates are added in afinal concentration of 0.5-1.0% water, put Durham’s tube in aninverted manner.
Method Add one drop of pure culture of organism to each test tube of
carbohydrate broth.
Incubate overnight at 37°C.
Either Andrade’s or phenol red indicator is used.
Results Using Andrade’s indicator
Alkaline: - No colour change (negative), Acid: - Red (positive) Using Phenol red indicator
Red (negative), Yellow (positive) If there is sugar fermentation, acid will be produced turning the
broth indicator pink. If no sugar fermentation, the media will remain unchanged and
there will be no colour change
6. EJIKMAN TEST:Aim:Some spore forming bacteria give false positive reactions withpresumptive coliform count and resist chlorinated water. Eijkmantest detects true Coliform in a positive presumptive test i.e.for Escherichia coliRequirement:
Tryptophan water Autoclave Incubator Inoculating loop Positive control Negative control Kovac’s reagent Tubes Lactose bile broth
Principle
15
The test is done by incubating subcultures from presumptive testsat 44°C and 37°C in lactose containing medium which is inhibitoryto spore forming bacteria e.g. Lauryl tryptose broth or Brilliantgreen lactose bile broth. Coliform produce gas from lactose at37°C while Escherichia coli (E. coli) produce gas and Indole at44°CMethod
a) Prepare 5-10ml of Lauryl tryptose (lactose) broth or brilliantgreen lactose bile broth plus Durham’s tubes and other tubescontaining 5-10ml of Tryptone water. Autoclave at 121°C for 15minutes.
b) Incubate lactose media at 44°C and at 37°C and Tryptone water at44°C.
c) Inoculate a loop-full or drop of each of the presumptive culturesin the above mentioned media and re-incubate immediately at theright temperature
d) Set positive (E. coli) and negative controls (Klebsiella)alongside the test.
e) Remove the tubes incubated at 44°C after 24 hours and add fewdrops of Kovac’s or Ehrlich’s reagent to Tryptone water cultures.Check for the reactions of the control organisms.
f) Remove the tubes incubated at 37°C after 48 hours and examine forgas production then record.
Results Combination of positive and negative results of gas production at
37°C, gives the most probable number of coliform bacilli per100mls of water. This value is the confirmed coliform count.
Combination of positive and negative results for gas and Indoleproduction at 44°C gives the most probable number of E. coli per100ml of water. This value is the confirmed E. coli count.
Control:
Positive control is E.coli
Negative control is Klebsiella spp
7. POTASSIUM CYANIDE (KCN) Aim:
16
The test is to demonstrate the ability of the organism to grow in thepresence of cyanide; it is required to differentiate enterobacteriasuch as Klebsiella, pneumonia and Pseudomonas aeroginosa from otherorganisms.
Principle:When KCN broth is inoculated with a few drops of heavily growing brothculture of the organism and incubated, the enterobacteria like theklebsiella will utilize the cyanide and grow which is shown byturbidity in the culture plate showing utilization of the cyanide.
Method a) Inoculate a tube of KCN broth with a few drops of a heavily
growing broth culture. b) Incubate at 37°C and observe at 24 hours.
Results A positive result is the presence of turbidity in the KCN broth
i.e. growth
8. BILE SOLUBILITY TEST Aim To differentiate Streptococcus pneumoniae which is soluble in bile andbile salts from other Streptococcus species such as Streptococcusviridian
Principle A heavy inoculum of the test organism is emulsified in physiologicalsaline to give a turbidity suspension. A bile salt sodiumdeoxychollate is then added. The test can be performed by adding thebile salt to a broth culture of the organism and the bile saltsdissolve streptococcus pneumoniae as shown by clearing of theturbidity within 10-15 minutes. Streptococcus viridian is notdissolved and therefore there is no clearing.
17
Method for tube method Emulsify 2g colonies of the suspected/test organism in 2ml
physiological saline in a tube
Divide the organism suspension between two tubes
To the first tube, add 2drops of the sodium deoxychollate reagentand mix
To the other tube ( negative control) add 2drops of the steriledistilled water and mix
Incubate both tubes at 35-370C for 10-15 minutes
Examine for clearance in turbidity in the tube of sodiumdeoxychollate
Results Clearance of turbidity indicates Streptococcus pneumoniae No clearance of turbidity-Viridian organism probably not S.
pneumoniae.
Control: Bile solubility positive organism is Streptococcus pneumonia Bile solubility negative organism is Enterococcus faecalis
9. AESCULIN HYDROLYSIS TEST Aim:To determine the ability of the bacteria to hydrolyze the glycosideaesculin to aesculinetin and glucose in the presence of 10% -40% bilewhich inhibit the growth of the other gram positive organisms and onlyallow the growth of the Streptococci and the Enterococci hencedifferentiating Enterococci from the Streptococci
Principle:Aesculin is incorporated into agar with ferric citrate and bile salts(bile aesculin agar). Hydrolysis of the aesculin forms aesculinetin(6, 7-dihydroxycoumarin) and glucose. The aesculin forms dark brown orblack complexes with ferric citrate, allowing the test to be read. Thepresence of a dark brown or black halo indicates that the test ispositive and no colour change indicates a negative aesculin test.
Requirement: Nutrient broth aesculin
18
Discrete bacterial colony Bile aesculin agar plate Wire loop 1% ferric citrate
Method a) Inoculate a tube of Aesculin broth with a test organism from the
discrete bacterial colonyb) Incubate at 37 0C for 24 – 48 hours.c) Add a drop of 1% ferric citrate to the broth cultured) Examine for colour change
Results: Positive results shows dark or black colour showing aesculin has
been split Negative colour change show aesculin test negative
Control: Positive control is Enterococcus faecalis NCTC 12697 Negative control is Streptococcus agalactiae NCTC 8181
10. ELEK’S PLATE TEST:Aim:An ELEK test is used to differentiate Corynebacterium diphtheriaeorganisms from normal flora organisms also called diphtheroids and toidentify the toxicity of the C. diphtheriae
Requirement: Sterile filter paper Diphtheriae antitoxin Petri-dish Incubator Protease peptone water Inoculating loop
Principle The test organism is inoculated on a serum agar plate containing afilter paper soaked with diphtheria antitoxin. The antigens andantibodies are allowed to diffuse towards each other and a line ofwhite precipitate occurs when the antigen and antibody meet in optimalproportions which are observed under dark background using transmittedlight.
19
Method a) Using sterile forceps soak strip of sterile filter paper 1.5 x
7cm in diphtheria antitoxin diluted to 750 units per ml andallows the strip to drain.
b) Lay the strip in a sterile 9cm diameter Petri-dish. Place thePetri-dish at 37 0C in the incubator for 20 minutes until thestrip is dry.
c) Prepare the serum culture medium by adding 3ml of clear sterileserum e.g. horse, rabbit, calf or 0.3% potassium tellurite to15ml of sterile cooled at 48 0C Columbia or protease peptoneagar.
d) Pour the serum agar medium into the Petri-dish containing theantitoxin strip and allow the medium to set firmly.
e) Using a wax pencil, draw lines at right angles to the strip across the base of the Petri-dish to indicate where to inoculatethe test and control organisms.
f) Using a sterile wire loop, heavily inoculate the test and controlorganisms (the toxigenic and non-toxigenic Corynebacteriumdiphtheriae strains).
g) Invert the plate and incubate at 37oC overnight
ELEK's Test Agar Plate Horizontal lines: toxin-producing bacteria to be tested. Vertical rectangle: filter paper soaked with antitoxin
antibodies. White lines: place where the filter paper and bacteria meet
are lines of precipitation
Results Read the results using transmitted light with the culture plate heldagainst a dark background.
White line of precipitates indicate toxin production (positivetest)
No precipitation shows a negative test (No toxinproduction).
20
11. Optochin:Aim:Is used for the presumptive identification of the alpha-hemolyticstreptococci as S.pneumoniaeMaterials:
Inoculating loop 5% sheep blood agar plate pure culture of S.pneumoniae Optochin disk (P) incubator candle jar (for Co2)
Principle:The Optochin test is dependent on the sensitivity of theStreptococcus pneumonia strains of the bacteria when the “P” diskis applied on the blood agar inoculated with the S.pneumoniaestrain microbe which is an alpha-hemolytic streptococcus andpositive results are measured in terms of the zones of inhibitionin mm showing Optochin sensitivity.Method:
a) using a inoculating loop, streak two or three suspectedcolonies of a pure culture to be tested on the 5% sheepblood agar plate
b) place a P-disk within the streaked area of the plate c) incubate the blood agar at 35-370C with 5% Co2 for 18-24
hoursd) Observe for zones of inhibition of growth.
Results: a zone of 14mm or more indicates sensitivity and allows for
presumptive identification of pneumonia (positive forOptochin test)
no zone of inhibition indicates no sensitivity and negativefor Optochin test
a zone of inhibition of less than 14mm or no zone indicatespneumonia in question and a solubility test should be doneto confirm pneumonia, if Bile soluble then pneumoniapositive.
Quality control:
21
Test all new lots of the Optochin with a positive and negativecontrol.
Positive control: growth of S. pneumonia strains ATCC4961(inhibited by Optochin)
Negative control: growth of S. mitis strains ATCC49456 (notinhibited by Optochin)
12. POTASSIUM HYDROXIDE TEST:Aim:To identify gram positive and gram negative organism in a culturemedium.Requirement:
Test organism in a culture (solid medium/agar slant) KOH solution 3% Glass slide
Principle:The KOH will dissolve the gram negative thin wall ofpeptidoglycan off forming visible clumps when the test organismis mixed in KOH solution but the gram positive thick wall ofpeptidoglycan won’t be dissolved off, so no clumps.Method:
To the glass slide is added 3% KOH solution drop Add an inoculum of the test organism from the plate Mix well and observe for reaction/clumps
Results: Large visible clumps shows a positive KOH test meaning a
gram negative organism No large visible clumps shows a negative KOH test meaning
gram positive organismControl:
Positive: E. coli Negative: S. epidermidis
13. Citrate utilization test:Aim:Is to identify the Enterobacterium, it’s based on the ability oforganism to utilize glucose as the only source of carbon.Requirements:
Bacterial suspension
22
Sterile physiological saline Test tube Citrate tablet Incubator
Principle of the test:When an organic acid such as citrate is used as a carbon andenergy source, alkaline carbonates and bicarbonates ultimatelyare produced. The visible presence of growth on the medium andthe change in pH indicator color due to the increased pH are thesigns that an organism can import citrate and use it as a solecarbon and energy source; such organisms are considered to becitrate positive.Procedures: using the Rosco citrate
a) Prepare a dense bacterial suspension of the test organism in0.25 ml sterile saline in a sterile tube
b) Add a citrate tabletc) Incubate overnight at 35-370C
Results: Red colour shows a positive citrate utilization test, hence
citrate was utilized Yellow-orange colour shows a negative citrate utilization
test hence citrate wasn’t utilizedControl:
Positive control is Klebsiella pneumoniae Negative control is Escherichia coli
Method using the Simmon’s citrate agara) Prepare slopes of medium in bijou bottles as recommended by
the manufacturer b) Using a sterile straight wire loop, first streak the slope
with a saline suspension of the organism and then stab thebutt
c) Incubate at 350C for48 hoursd) Look for bright blue colour in the medium
Results: Bright blue colour shows a positive citrate utilization test No colour change shows a negative citrate utilization test
Control:
23
The same as for the citrate tablet
14. Voges-Proskauer test:Aim:It is used to test the production of acid from dextrose phosphatebroth to assist in differentiation of Enterobacteria such asKlebsiella pneumoniae and Vibrio cholera
Principle The organism produces acetyl methyl carbinol and it is reductionproduct when they break down dextrose. This is oxidized byaddition of alkaline to give di-acetyl (pink).
Method:a) To a 2 day broth culture, add 1 ml of 1% KOH. b) Leave at room temperature for 1 hour. c) And add 3 ml of O’Meara’s reagent. d) Observe for colour change
Results Pink colour indicates positive (Klebsiella and Enterobacter) No colour indicates negative (E. coli and Citrobacter)
C. SEROLOGICAL TESTThese include the following test;
1. Germ tube test2. Lancefield grouping test
The details include:1. GERM TUBE TEST:
Aim:It is used in examination and identification of Candida albicansfrom cultures which gives a germ tube test positive from otherCandida species
Requirement: Test sera Positive pure culture of yeast (C. albicans) Incubator Carbondioxide jar Microscope slide
24
Cover slip Test tube Inoculating loop
Principle:Candida albicans when grown on sabround agar medium had theability to form germ tube (sporeting yeast cells) when inoculatedon a sterile test sera and incubated in a 10% Carbondioxide jarwhich will be seen as small filament projecting from the cellsurface. Method:
a) Set up two tubes in a rack and label them A and Bb) Add 0.5ml of the test sera in each tubec) To tube A, add using a sterile loop a small portion of the
pure colony of the C. albicans, mix and incubate I 10%Carbondioxide jar for 2 hours
d) Wait for 10 minutes and at interval pick a drop of thesample put on the slide and cover using a cove slip andexamine by X10 and X40 objectives
Results: Small filament projecting from the cell surface shows
positive Germ tube test hence sporeting yeast. No small filament observed shows Germ tube test negative
hence no sporeting.
2. LANCEFIELD GROUPING TESTAim: For the quantitative identification a typing of the streptococci,using the streptococci antisera to identify the group A, B, C, G,and L streptococci and serotype group B streptococci and S.suisMaterials for used:
5% blood agar plate Glucose broth Inoculating loop Centrifuge Pipette 0.006N, 0.1N and 0.2N HCl Water bath (1000C) Phenol red indicator 0.2N NaOH
25
Capillary tubesPrinciple of the test:This test detect the carbohydrate in the cell wall of theLancefield group D streptococcus and when an acid antigen extractmixed with a specific antiserum was directed against bacteriumsurface components, the cells bound together through antigen-antibody bonds to form aggregates (precipitation) in thecapillary tube and this is usually visible to the naked eye.Method of the test:
a) The streptococci is grown in 5% blood agar plateb) Pick colonies of the bacteria and add to 6ml glucose broth
and incubate at 35-370C overnightc) Centrifuge the suspension for 10 minutes at 3000rpm and
remove the supernatantd) Add 0.1 ml of either an 0.06N, 0.1N or O.2N HCl to the
bacteria pellete) The acid suspension is placed in a water bath (1000C) for 15
minutesf) Cool the acid suspension under tab waterg) The Ph-value is adjusted to approximately 7 by adding
droplets of 0.2N NaOH until the colour is brown/orange(phenol red, red at ph >8.2 and yellow at ph <6.4)
h) Centrifuge the suspension for 10 minutes at 3000 rpm andtransfer the supernatant to a new tube
i) Suck equally the antiserum first into the capillary tube andthen an equal volume of acid antigen extract (the antiserumshould be on top and the acid antigen below to allow thediffusion of the antiserum through the acid extract)
j) Observe for precipitation in the capillary tube against alight source
Results: Precipitation in the tube shows positive Lancefield test No precipitation in the tube shows negative Lancefield test
Controls used: Positive control are the Lancefield group D streptococcus
strains Negative control is normal saline solution
26
REFERENCES:-1. www.microbeonline.com/bacteriology 2. Lancefield, R.C. Serological differentiation of specific
types of bovine hemolytic streptococci (group B) J.Exp Med1934; 59: 441-458
3. Stotved HC, S, konradsen HB False negative results in typinggroup B Streptococci by the standard Lancefield a higherExtraction method J CLIN Microbiol 2002 May; 40(5): 1882-3
4. Odds, F.C. (1979), Candida and candidiasis (12th Ed)Baltimore University Press, London, pp 29-40
5. Casanova, M, Cervera, A.M., Gozalbo, D &Martinez, J.P.(1997) Hemin inducers germ tube formation in Candidaalbicans, infect, immune Oct. 65 (10) 4360-4
6. Monica Cheesbrough, district laboratory Practice Part 2,Biochemical test, chapter 7, subunit 7.5, page: 62-70
