Microbiology: Enzyme Tests & Coliform Detection Study Guide

#separator:tab #html:true what is ureathe product of the decarboxylation of certain amino acids what is ureasean enzyme present in certain bacteria that can hydrolyze urea into ammonia and CO2 what is the urea hydrolysis reaction equationurea + H2O --urease--> 2NH3 + CO2 several enteric bacteria that reside in the gut possessurease for a bacterium to survive in acidic conditions requiresthat the organism maintains a neutral linternal pH because it is a basic compound, ammonia canneutralize the acidic conditions within the bacterium urea brothmedia used to detect the presence of urease - has a trace amount of yeast extract (which is the only nutrient source in the broth) -urea (the substrate for the enzyme) -potassium phosphate buffer (which is strong enough to maintain a neutral pH UNLESS large amounts of alkaline compounds are present) -phenol red pH indicator urea broth starts off the color ____ and with a pH of _____peach, 6.8 what will the results be if the organism does not produce urease-then the urea in the broth is not broken down, the alkaline product ammonia is not made, and there is no change in pH of the media.  - the organism may die since there are only trace nutrients in the broth - the broth remains peach or turns slightly yellow in color. BACTERIUM IS UREASE NEGATIVE what will the results be if the organism produces urease- then the urea in the broth is broken down, large amounts of the alkaline compound ammonia are made very quickly, so much ammonia produced so rapidly overwhelms the pH buffering capacity of the broth, and the pH of the broth rises and becomes alkaline. - At high pH, the phenol red indicator turns PINK in color BACTERIUM IS UREASE POSITIVE. what are the applications of the urease test?- Helicobacter pylori is a bacterium that can reside in the stomach and has been associated with stomach cancer and stomach ulcers. - in order to survive the acidic pH of the stomach, it produces and secretes a large amount of urease. - to determine if someone is infected w H. pylori an endoscopy is performed, a small gastric sample removed and the sample is tested for the presence of urease what is collagena protein found mostly in fibrous tissues what is gelatina product from the hydrolysis of collagen what is gelatin made ofit is a large protein polymer made up of numerous peptide subunits. -it is an abundant source of amino acids what is the problem with gelatin?it is too large of a molecule to be transported into the bacterial cell.  - for a bacterium to catabolize gelatin, it requires an enzyme that can break down the protein into its amino acid subunits. what is gelatinaseit is a proteolytic enzyme that breaks down gelatin into its individual amino acid components by CLEAVING peptide bonds - it is SECRETED from the bacteria media used in the gelatinase testnutrient gelatin -a simple media that contains gelatin (the substrate for the enzyme) -gelatin is a solidifying agent -control media is FIRM and SOLID in consistency prior to innoculation  if after inoculation it remains solid for the gelatinase testthe organism is GELATINASE NEGATIVE - since media is still solid gelatin, which is the solidifying agent, MUST be present. Gelatin is not broken down if after inolculation it becomes liquified for the gelatinase testthe organism is GELATINASE POSITIVE. - gelatin would solidify the media, so it is no longer present. Gelatin is absent because it was broken down. To break down the gelatin, gelatinase must have been SECRETED. what is starchan extremely large polysaccharide that is made up of glucose molecules linked by glycosidic bonds to form very long chains what is the structure of starchit consists of 2 types of molecules-- amylose and amylopectin what is amyloselong chains of glucose molecules what is amylopectinbranched chains of glucose molecules what is the problem with starch?it is too large to pass through the bacterial cell membrane.  - it may contain thousands of individual glucose units -glucose is a simple monosaccharide that can be taken up by the bacteria and broken down to produce energy (ATP) - Starch could provide a bacterium with a large nutrient source to use starch, bacteria must possessenzymes that can break the glycosidic bonds between glucose molecules - alpha amylase is a SECRETED enzyme that can break down glycosidic bonds  starch --alpha amylase---> glucose media used to determine the presence of alpha-amylaseSTARCH AGAR contains starch (the substrate for alpha-amylase) to determine if a bacterium produces alpha amylase:-the bacteria to be tested is inoculated onto starch plate and grown overnight -if the bacteria has alpha-amylase, it will be secreted from the cell and diffused into the agar around the bacteria -once secreted, alpha amylase can break down the starch in the media to glucose -after overnight, the growth plate is flooded with iodine - IODINE CAN REACT W/ STARCH TO PRODUCE A DARK BROWN/BLACK COLOR -Iodine does NOT react with glucose what does the result of alpha-amylase NEGATIVE look likeplate is flooded with iodine, and the present starch reacts with the iodine and turns the agar dark brown/black. The agar around the bacteria also turns dark brown/black. The agar around the bacteria still contains starch, meaning no substrate breakdown what does the result of alpha-amylase POSITIVE look likethe plate is flooded with iodine. If the present starch reacts with iodine and turns agar dark brown and black, but the agar around the bacteria does NOT turn dark brown/black, NO starch is in the agar around the bacteria. Starch around bacteria was broken down to glucose which does NOT react with iodine. There is a zone of clearing around the bacterium - the bacteria SECRETED alpha-amylase which broke down the starch into glucose what is casein-phosphoproteins -4 different types which accounts for around 80% of the proteins found in cow's milk.  -gives milk its white color -contains between 200 and 220 amino acids what is the problem with caseinwhile casein can provide a source of amino acids for certain bacteria, it is too large to pass through the bacterial cell membrane. - to overcome this problem, certain bacteria possess CASEASE which is an enyzme that can break down casein what is caseasea SECRETED enzyme that can break the peptide bonds between adjacent amino acids. - the breakage of peptide bonds produces smaller peptides and eventually individual amino acids that can be taken up by the bacterium media used in the casease testmilk agar - the most important component is powdered nonfat milk (the source of CASEIN) - casein in the milk provides the substrate that can be used by casease casease negative resultsbacteria grows on the milk agar plate, with NO change in color of the agar plate where the bacteria have grown, since media remains white, NO breakdown of casein, NO casease produced. casease positive resultsif a bacterium has casease, it will be SECRETED.. once secreted, casease will diffuse out through the agar. -casease breaks down the casein in the agar -when casein is broken down, the white color of the agar is lost. -A ZONE OF CLEARING is observed around the bacteria. in casease test, zone of clearing around the bacteria indicatessecretion of casease. Casease has broken down the casein in the media leaving the agar clear. what are coliforms-gram negative rods -can ferment lactose with the production of gas within 48 hours of growth at 37 degrees celsius -generally associated with fecal material -most common coliform is E.coli the presence of coliforms are used to indicatewhether a water sample has become contaminated with fecal material and therefore with potential bacterial pathogens to determine if a water sample is contaminated with coliforms:MULTIPLE TUBE FERMENTATION METHOD, better known as MOST PROBABLE NUMBER (MPN) is performed The MPN procedure allows one to calculate two things...1. total coliform counts in a water sample 2. the E.coli counts in a water samples the MPN procedure requires several days to complete and several different selective media  objective of MPN exerciseto determine whether water from these 2 duck pond sites are contaminated with coliforms and specifically E. coli and if there are differences in the level of contamination between the two sites. All this will be done by calculating the MPN what is lauryl tryptose broth (LTB)LTB is a media SELECTIVE for COLIFORMS and gives a PRESUMPTIVE DETERMINATION of the presence of coliforms What does LTB includeit includes 1. lauryl sulfate- that inhibits the growth of organisms other than coliforms 2. lactose- coliforms can ferment this sugar 3. durham tube- will indicate  after 48 hr incubation, LTB broths are examined for1. growth 2. presence of GAS Tubes that contain probable coliforms should show growth and possibly gas in the durham what is brilliant green lactose bile broth (BGLB)a SELECTIVE media that CONFIRMS the presence of coliforms what does BGLB contain1. lactose- coliforms can ferment this sugar 2. durham tube- will indicate if gas has been produced from the fermentation of lactose 3. 2% bile- inhibits non-coliforms BGLB tubes with gas are consideredPOSITIVE and used to determine total coliform counts E coli brotha media that is SELECTIVE for E.coli when grown at around 45 degrees celsius what does E.coli broth contain1. lactose- coliforms can ferment this sugar 2. durham tube- will indicate if gas has been produced from the fermentation of lactose 3. bile salts- inhibits non-coliforms EC tubes with gas are consideredPOSITIVE and are used to determine the E.coli counts once the numbers of positive BGLB and EC broth tubes have been determined, the results can be applied to the following formula to calculate the MPN: MPN/100ml= 100P/ vVnVa  where P= total number of positive results (BGLB or EC) Vn= combined volume of sample in LTB tubes that produced negative results in BGLB or EC Va= combined volume of sample in all LTB tubes once the numbers of positive BGLB and EC broth tubes have been determined, the results can be applied to the following formula to calculate the MPN:MPN/100ml= 100P/ vVnVa  where P= total number of positive results (BGLB or EC) Vn= combined volume of sample in LTB tubes that produced negative results in BGLB or EC Va= combined volume of sample in all LTB tubes selective media-contains one or more specific compounds that can prevent the growth of certain bacterial species -it allows the growth of some bacterial species while inhibiting the growth of other bacterial species  to achieve selectivity, selective media containsibhibitors. they may adversely effect DNA synthesis, gene expression, enzymatic activity, or membrane permeability, all of which can destroy or inhibit the growth of certain SPECIFIC bacterial species differential media-contains one or more specific compounds that can distinguish between different bacterial species generally, differentail media has 2 important components:1. Substrate(s) such as a carbohydrate that only certain bacterial species can utilize in a specific chemical reaction or a set of chemical reactions 2. Indicator(s) that provide a visible means of showing that a specific chemical reaction has occurred (a color change in the media) selective and differential media is extremely useful when examining mixed bacterial cultures since it provides:i) a simple way to 'screen out' certain bacterial species ii) some biochemical information on the organisms present in the culture (PRESUMPTIVE IDENTIFICATION) true or false: media can be selective AND differentialtrue what bacteria reside in the guts of birds and all mammals (including humans)coliforms general characteristics of coliforms1. gram negative rods 2. non-spore forming bacteria 3. aerobes or facultative anaerobes 4. can FERMENT LACTOSE what kind of media is MacConkey agar and what is it used forselective AND differential, used to identify the presence of coliforms important components in MacConkey Agar1. bile salts (selective) 2. crystal violet dye (selective) 3. neutral red dye (colorless when at pH greater than 6.8, red at pH lower than 6.8) (differential) 4. lactose (differential) why is macconkey agar selectivebile salts and the crystal violet dye INHIBIT the growth of gram positive bacteria... ONLY gram negative bacteria will grow on MacConkey agar why is MacConkey agar differentialnot all gram negative bacteria can ferment the lactose in the media to produce acidic compounds... if acidic products are made, the pH of the media will drop and the neutral red dye will cause the media to turn REDDISH PINK. if lactose cannot be fermented, then NO acidic products, NO drop in pH, and NO change in color. presumed coliform result on MacConkey Agargrowth= gram negative bacteria... reddish/pink bacterial colonies=low pH= presence of acidic products=fermentation of lactose not a coliform result on Macconkey agargrowth=gram negative bacteria... opaque/colorless bacterial colonies=no change in pH=no acidic products present=no fermentation of lactose poor or no growth on macconkey agar?the organism is likely inhibited by crystal violet and/or bile salts. it is presumptively gram positive bacteria EMB agar- what type and used for what?a selective AND differential media used to identify the presence of coliforms (gram negative lactose fermentors) important components in EMB agar:1. eosin Y dye (selective and differential) 2. methylene blue dye (selective and differential) 3. lactose (differential) EMB agar is selective becauseeosin Y and methylene blue dyes inhibit the growth of gram positive bacteria (ONLY gram negative bacteria will grow well) EMB agar is differential becausenot all gram negative bacteria can ferment lactose to produce acidic compounds... not all gram negative bacteria produce the same AMOUNT of acidic compounds if they can ferment lactose poor or no growth on EMB agargram positive bacteria colorless growth on EMB agargram negative bacteria... no change in media pH= no acidic products present=no fermentation of lactose... NOT a coliform possible coliform result on EMB agargrowth=gram negative pink and mucoidy bacterial colonies=slight change in media pH=small amount of acidic products made=slow fermentation of lactose probable coliform in EMB agargrowth=gram negative  dark purple to black bacterial colonies and or GREEN METALLIC SHEEN=drop in media pH=large amounts of acidic products made=vigorous fermentation of lactose what is triple sugar iron agara single media that can be used to determine an organism's ability to ferment three different sugars as well as the ability of an organism to REDUCE SULFIDE what is TSIA used to dohelp to differentiate between enteric bacteria such as salmonella, shigella, and e.coli 3 important points to remember about the fermentation pathway1. glucose is not the only carbohydrate that can be used in glycolysis (sugars such as lactose or sucrose can be used as well) 2. the end products of fermentation are acidic compounds 3. gas can also be an end product of fermentation most enteric bacteria arefacultative anaerobes and can grow under anaerobic environments in the absence of oxygen, certain enteric species can usesulfur as a terminal electron acceptor to produce energy one specific biochemical pathway for sulfur reduction uses _____ as an electron acceptorthiosulfate. -in this pathway under acidic conditions, thiosulfate is reduced by the enzyme THIOSULFATE REDUCTASE to produce SULFITE and HYDROGEN SULFIDE (H2S) which is expelled from the bacterium H2S can be considered a ... that can react with...reducing agent, metal ions to form metal sulfides - one such reaction that occurs in the presence of H2S is the conversion of ferrous sulfate to FERROUS SULFIDE -the product, ferrous sulfide, is an insoluble BLACK metallic compound sulphur reduction rxn equationsthiosulfate --thiosulfate reductase in acidic conditions--> sulfite + H2S H2S + Ferrous sulfate -----> ferrous sulfide (black precipitate) ingredients in TSI agar1. LOW concentration of glucose (0.1%) 2. High concentrations of lactose and sucrose (1% each) 3. sodium thiosulfate (sulfur source) 4. Ferrous sulfate (H2S indicator) 5. phenol red (pH indicator, which is red at neutral pH) preparation of TSI agarTSIA is always prepared as a slant.  - the slant portion of the media provides aerobic condition, while the butt provides anaerobic conditions how TSIA is innoculatedan inoculating needle is used, the butt is first stabbed and then the needle is pulled up the slant on the way out glucose only fermentation result for TSIA testthis means the organism can only ferment the glucose in the media and acidic products are generated.  -these acidic products lower the pH and media turns yellow within a few hours -becuase it is at a low concentration, the glucose is used up within 12 hours and the bacteria now begins to break down amino acids in the media which produces ammonia, an alkaline product -slant part turns red because enough alkaline product is made in the slant to neutralize the pH and cause reversion. So the slant part is red, but the butt is yellow still bc the accumulation of the alkaline product is not enough to change the pH in the butt red slant/yellow butt for TSIA = .....alkaline slant/ acid butt glucose and lactose and/or sucrose fermentation result for TSIA-organism can ferment two or three of the sugars in the media and acidic products are generated -these acidic products lower the pH and the media turns yellow -high concentrations of sucrose and lactose ensure that the sugar supply will not be exhausted -can continue to ferment sugars so no reversion and the media remains yellow yellow slant/yellow butt for TSIA =....acid slant/acid butt no fermentation result for TSIA-bacteria cannot ferment any of the 3 sugars present in the media -bacteria instead breaks down the amino acids in the media which produces ammonia which raises the pH -turns the media redder in color than the uninoculated control media gas production for TSIA test-gas may be produced from the fermentation of glucose, sucrose, OR lactose -lifting of the media from the bottom of the tube and cracks or bubbles in the butt portion of the media is indicative of gas production sulfur reduction in TSIA test-bacteria CAN ferment at least one of the three sugars -fermentation generates acidic prudcts which lowers the pH and media turns yellow - under acidic conditions IF thiosulfate reductase present will reduce thiosulfate in media to produce H2S which will react with ferrous sulfate in the media to produce ferrous sulfide BLACK PRECIPITATE=H2S PRODUCED=SULFUR REDUCTION for TSIA, if fermentation by glucose only, the slant will appearred due to reversion. if fermentation by 2 or 3 sugars, the slant will appear YELLOW if black precipitate obscurs the color of the media butt,it is acidic bc H2S production requires acidic environment Columbia CNA with 5% sheeps blood agar is used tospecifically grow Gram positive organisms such as staphylococci, streptococci, and enterococci what is in Columbia CNA blood agar-digested casein -digested animal tissue -beef extract -yeast extract -corn starch -sheep blood EXTREMELY nutrient rich What does CNA mean in the Columbia CNA blood agarC= COLISTIN NA= NALIDIXIC ACID these are antibiotics that act as SELECTIVE AGENTS against gram negative organisms the insertion of colistin into the outermembrane...disrupts the integrity of the outer membrane, which can lead to bacterial lysis nalidixic acid inhibitsDNA gyrase/topoisomerase -gyrase and topoisomerase allow supercoiled DNA to be relaxed and reformed and is necessary for DNA replicaton... therefore, nalidixic acid inhibits DNA synthesis -Gram negative bacteria are MORE SENSITIVE to nalidixic acid than gram positive the sheep blood in Columbia CNA agar also makes the media...differential. -different bacteria will show different red blood cell hemolysis patterns when grown on agar that contains sheep's blood beta hemolysiscomplete lysis of RBCs alpha hemolysispartial lysis of RBCs gamma hemolysisno lysis of RBCs mannitol salt agar useto help identify pathogenic staphylococcus species (i.e.,S. aureus) from non-pathogenic staphylococcus species (i.e., S. epidermidis) 3 important components in MSA7.5% salt (high concentration) (selective) mannitol (a carbohydrate) (differential) phenol red (pH indicator) red at neutral pH (differential) mannitol salts agar is selective becausethe high salt concentration only allows staphylococcus species to grow... most bacteria dehydrate at the high salt concentration and die mannitol salts agar is differential because-some staphylococcus species CAN ferment the mannitol in the media, which produces acidic products -acidic products in the media will cause the pH to drop and the phenol red indicator to turn yellow -non pathogenic staphylococcus that CANNOT ferment mannitol will NOT produce acidic products so no change in pH and the media will remain RED poor or no growth in mannitol salt agar=organism is inhibited by the high salt concentration and therefore not staphylococcus good growth in mannitol salt agar=staphylococcus yellow growth or halo around growth in MSA=low pH=acid=mannitol fermentation... possibly pathogenic S. aureus red growth or no color change around growth in MSAno pH change=no acid=no mannitol fermentation... non-pathogenic staphylococcus blood agar =tryptic soy agar+ 5% sheeps blood blood agar provides bacteria withnutrient rich environment and is useful when attempting to culture fastidious organisms (organisms that require strict physiological conditions) hemolysins are able tohemolyze/lysis RBCs and to destroy hemoglobin beta hemolysis produces acomplete zone of clearing around a bacterial colony.  -toxin secreted by the bacteria diffuses through the agar lysing all RBCs it comes into contact with, destroying the hemoglobin within the RBC  alpha hemolysis producesa green coloring around a bacterial colony... hydrogen peroxide secreted by certain bacteria diffuses through the agar and oxidized the hemoglobin in RBCs to produce methemoglobin, which is green in color.  S. epidermidis is part of the normal human ....skin flora and is generally not considered pathogenic s. aureus is ....an opportunistic pathogen that can cause minor to major skin infections, pneumonia, TSS, etc differentiation between S. epidermidis and S. aureusboth share many same metabolic pathways and biochemical properties, but two tests can be used to differentiate the two.. 1. mannitol fermentation 2. presence of coagulase coagulase is only present ins. aureus -protein which binds to PROTHROMBIN- which is involved in blood coagulation and the generation of fibrin clots prothrombin is inactive andmust be cleaved in order for it to be converted into the active enzyme THROMBIN coagulase +prothrombin =staphylothrombin... -binding of coagulase to inactive prothrombin ALONE (no cleavage required) produced the active enzyme STAPHYLOTHROMBIN which can bind to and act on fibrinogen in s. aureus, coagulase can be present in 2 forms1. bound coagulase 2. Free coagulase bound coagulase -attached to the bacterial cell wall -binds to and activates prothrombin in blood plasma -also binds to fibrinogen in blood plasma -bound fibrinogen cleaved by coagulase activates prothrombin to produce fibrin clots free coagulasesecreted by the bacteria into the surrounding environment -binds to and activates prothrombin in blood plasma -coagulase activated prothrombin acts on fibrinogen in the blood plasma to produce fibrin clots two types of coagulase testscoagulase slide test coagulase tube test what is plasmathe yellow colored liquid portion of the blood that the red blood cells are suspended in what is 90% of plasmawater plus compounds such as clotting factors like prothrombin and fibrinogen coagulase tube test-detects both bound and free coagulase -bacteria is mixed with rabbit plasma in a test tube.... if the bacteria possesses coagulase it will bind to prothrombin in the rabbit plasma, prothrombin will become activated and will convert fibrinogen to fibrin and a clot will form when the fibrin links together role of coagulase in S. auerus virulence-possible role in allowin gthe bacteria to evade the host immune response -production of a fibrin clot around the bacteria may protect it from phagocytosis by macrophages -bacteria cannot be 'cleared' from the host and the infection proceeds "ribose sugars and phosphates are linked by"covalent bonds and make up the sugar phosphate backbone of the molecule deoxyribonuclease (DNase)-enzyme secreted by certain bacterial species -considered a bacterial virulence factor that contributes to infection  -DNase breaks the covalent bonds between phosphate and ribose sugar molecules, causing DEPOLYMERIZATION of DNA into short nucleotide chains one type of DNase can... while the other can....break the bond between the 5' carbon of the ribose sugar and the phosphate the other type of DNase can cleave the bond between the phosphate and the 3' carbon atom of ribose sugar DNase agar components1. DNA (substrate) 2. methyl green dye because dye binds to polymerized, uncleaved DNA media is blue/green in color dye will NOT remain bound to small DNA fragments or short nucelotide chains dnase negative resultbacteria has grown but the agar surrounding the bacteria growth remains green/blue -methyl green dye has remained bound to the DNA -methyl green dye ONLY binds to the polymerized DNA -DNA has not been cleaved DNase positive resultzone of clearing -methyl green dye only binds to polymerized DNA -DNA must have been cleaved to remove bound dye -DNase must have been secreted and cleaved the DNA around the bacteria  two main goals in antimicrobial susceptibility testing1. finding the best antimicrobial drug to use that can kill the disease causing bacteria but NOT the normal bacterial flora 2. finding the appropriate therapeutic dose... want enough to kill the disease causing bacterium but that has little toxicity or causes few side effects to humans several different approaches to determine the susceptibility of a bacterium to an antimicrobial drug:1. kirby-bauer method 2. e-test 3. tube dilution advantages/disadvantages of kirby baueradvantages: 1. quick and easy 2. many different antimicrobial drugs tested simultaneously 3. can measure zone of clearing and see where minimal inhibitory concentration (MIC) of the drug is  disadvantages: 1. does not provide info on a therapeutic dose 2. cannot determine the exact concentration of the drug at different places in the zone of clearing advantages/disadvantages of E-testadv: 1. quick and easy 2. many diff antimicrobial drugs can be tested simultaneously 3. can give approximate MIC dis: 1. cannot provide definitive MIC E-strip placed on innoclulated plate, contains different concentrations of the drug noted at various points on the strip... MIC is the area where bacterial growth begins adv/disadv to tube dilution testadv: -can give precise MIC -test can be automated dis: -prior to automation more complicated and time consuming -prior to automation could only test a single microbial drug at a time tube dilution test method and theory1. prepare dilutions of the chosen antimicrobial drug 2. add equal volumes of bacteria to be tested into each tube of broth media 3. incubate tubes overnight 4. following day observe tubes for growth 5. first dilution tube that shows no growth is the MIC automated antimicrobial susceptibility testingcan incubate, read, and report the results of up to 100 antimicrobial susceptibility plates/panels in a total of 18 hours

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