Identification Methods

Question 1

Catalase

Answer 1

detects the production of catalase enzyme, which decomposes hydrogen peroxide into water and oxygen, resulting in bubbles; false positives can occur from using an iron loop, blood agar media, and reading the result too late

Question 2

Coagulase

Answer 2

a protein that has prothrombin-like activity (can convert fibrinogen into fibrin), resulting in clotting; There are two types of coagulase tests based on detecting bound (slide) or free (tube) coagulase

Question 3

Latex agglutination

Answer 3

kits used to detect clumping factor A (a STAPAU cell wall antigen) instead of performing coagulase tests; MRSA can give false negatives

Question 4

DNase

Answer 4

an enzyme that cleaves DNA into nucleotide subunits; a media with toluidine blue is used because it will turn pink if DNase is present

Question 5

Modified oxidase test

Answer 5

detects the enzyme cytochrome oxidase, which oxidizes tetramethyl-p-phenylenediamine to form indophenol; dimethyl sulfoxide (DMSO) is added in the modified test to make bacterial cells more permeable; blue is positive

Question 6

PYR

Answer 6

detects the enzyme L-pyroglutamyl aminopeptidase, which cleaves B-naphthylamine from L-pyrrolidonyl-B-naphthylamide (PYR); a red color is formed when N,N-dimethylaminocinnamaldehyde is added and free B-naphthylamine is present

Question 7

CAMP Test

Answer 7

detects CAMP factor, which is a heat-stable protein produced by bacteria as a result of a synergetic association with STAPAU

Question 8

Hippurate Hydrolysis

Answer 8

detects the enzyme hippuricase, which cleaves hippurate into sodium benzoate and glycine; A precipitate is formed when ferric chloride is added and sodium benzoate is present or a purple color forms from adding ninhydrin when glycine is present

Question 9

Bile Esculin Test

Answer 9

determines if an organism can hydrolyze esculin to esculetin and glucose in the presence of bile; esculetin combines with ferric ions in the agar to produce a black color

Question 10

6.5% NaCl Test

Answer 10

tests if an organism can grow in a broth tube with a high salt concentration; turbidity or an acidic pH level = growth

Question 11

Bile Solubility Test

Answer 11

bile salts can lyse S. pneumoniae when added to actively growing bacteria in agar or broth, leading to colonies dissolving (clearing of colonies)

Question 12

Gelatin Hydrolysis

Answer 12

tests the presence of the gelatinase enzyme that can liquify gelatin

Question 13

Leucine Aminopeptidase (LAP) Test

Answer 13

LAP enzymes hydrolyze leucine-p-naphthylamide to release leucine and B-naphthylamide. Free B-naphthylamide reacts with cinnamaldehyde reagent to form a red color

Question 14

Lecithinase Test

Answer 14

assesses a bacteria’s ability to produce lecithinase (toxic, phospholipase enzyme); lecithinase degrades lecithin into diglycerides leading to a white, opaque zone on a egg-yolk agar; used for Clostridium spp.

Question 15

Nagler test

Answer 15

Lecithinase-positive clostridia can be divided based on Nagler test results; an antitoxin reagent is swabbed over one half of an egg yolk plate and a single streak of a lecithinase positive organism is made across the untreated and treated halves of the plate. A Nagler positive organism shows opacity in the untreated half and no opacity in the treated half while a negative one has opacity on both halves (antitoxin did not affect lecithinase activity)

Question 16

Loeffler’s Stain/Medium

Answer 16

used to identify Corynebacterium species

Question 17

Lysozyme Test

Answer 17

to test if an organism is resistant to lysozyme (will grow in a tube containing lysozyme if resistant)

Question 18

MRS Broth

Answer 18

Used to cultivate lactobacilli

Question 19

Starch Hydrolysis

Answer 19

used to determine the ability of an organism to produce amylase and utilize starch as a carbon source

Question 20

Cystine trypticase agar (CTA)

Answer 20

traditional method for identifying Neisseria spp.; results depend on the ability of an organism to grow with specific carbohydrates checked every 24 hours for 72 hours

Question 21

Rapid carbohydrate tests

Answer 21

check for preformed enzymes (not growth dependent), results done after 1-4 hours and based on the commercially prepared kit

Question 22

Oxidase Test

Answer 22

tests the enzymes cytochrome oxidase or indophenol oxidase; the test reagent turns blue when it is oxidized

Question 23

Chromogenic substrate tests

Answer 23

detects enzymes leading to colored endproducts being formed when a specific substrate is hydrolyzed by the enzyme (Ex: B-galactosidase)

Question 24

Penicillin disk test

Answer 24

a penicillin disk is placed onto CHOC overnight and then a gram stain is prepared of the bacteria at the edge of the zone of inhibition; actual rods will elongate while actual cocci remain round

Question 25

Nitrate reduction test

Answer 25

In the anaerobic respiration process, bacteria derive their required oxygen from nitrate (NO3–) or can use NO3– as a terminal electron acceptor. In this process, nitrate needs to be reduced to nitrite. Nitrite reacts with acetic acid and forms nitrous acid. See image for full reactions. Positive = red. Some bacteria can reduce nitrite to nitrogen gas, giving a negative result. To differentiate a true negative and denitrification from each other, zinc powder is added. If nitrate is present, it will be reduced with the zinc (turning red), which indicates a true negative.

Question 26

X and V Paper disk Test

Answer 26

This test is used to differentiate Haemophilus spp. from each other based on their need for hemin (X factor) and NAD (V factor) for growth. Species that require only the X-factor will grow around X-disk and XV-disk but not around V-disk. Species that require only the V-factor will grow around the V-disk and XV-disk but not around X-disk. Species that require both X-factor and V-factor will grow around the XV-disk but not around the X-disk and V-disk.

Question 27

Porphyrin test

Answer 27

determines if an isolate requires X factor; those that do not require X possess enzymes that convert aminolevulinic acid (ALA) to hemin with porphyrins as an intermediate; porphyrins are detected by the reddish-orange fluorescence when exposed to UV light

Question 28

o-Nitrophenyl-B-D-galactopyranoside (ONPG)

Answer 28

The ONPG test is used to differentiate lactose fermenters, late lactose fermenters, and non lactose fermenters from each other. Two enzymes are involved in the fermentation of lactose into glucose and galactose (a permease and B-galactosidase) Some bacteria only have the enzyme B-galactosidase, resulting in a slower lactose fermentation than those that have both enzymes. B-galactosidase will cleave ONPG into galactose and o-nitrophenol (yellow). Clear = - ; yellow = +

Question 29

What is the difference between Kliger’s iron agar and Triple Sugar Iron Agar?

Answer 29

Triple Sugar Iron Agar: contains 3 sugars (glucose, lactose, and sucrose), so if the organism can ferment glucose or one of the other sugars it will be A/A Kliger’s Iron Agar: only contains glucose and lactose, so if the organism can only ferment glucose, then it will be K/A

Question 30

Voges-Proskauer Test

Answer 30

Tests to see if a bacteria follows the butylene glycol pathway to produce acetylmethylcarbinol (acetoin) and butanediol after glycolysis

Question 31

Urease

Answer 31

Urease is an enzyme that can hydrolyze urea into ammonia and carbon dioxide, resulting in an alkaline pH. The change in pH is detected using the indicator phenol red (positive/basic = red and negative/acidic = yellow)

Question 32

Triple Sugar Iron Test

Answer 32

Used to differentiate bacteria based on their ability to ferment glucose, lactose, and sucrose, and release acid and hydrogen sulfide gas. Same process as the KIA slant but it also has sucrose.

Question 33

Satellitism Test

Answer 33

Used to distinguish Haemophilus influenzae from other Haemophilus species. H. influenzae alone can’t grow in a blood agar medium, but it can grow (satellite) around S. aureus growing on BAP plate because it will hydrolyze blood cells, releasing V factor

Question 34

Phenylalanine Deaminase

Answer 34

Bacteria capable of producing phenylalanine deaminase enzyme can deaminate phenylalanine resulting in phenyl pyruvic acid and ammonia. Only Proteus, Morganella, and Providencia genera have the deaminase enzyme needed for this reaction. A green color develops if positive due to the addition of ferric chloride that reacts with the phenyl pyruvic acid.

Question 35

Oxidative-Fermentation Test

Answer 35

Saccharolytic microorganisms degrade glucose either fermentatively or oxidatively. Fermenters: can produce acid in both aerobic and anaerobic conditions. Oxidizers: can produce acid in only aerobic conditions (needs oxygen as a terminal electron acceptor). Yellow color = acidic/positive and Green color = basic/negative

Question 36

Motility Test

Answer 36

Motility by the bacterium is demonstrated in a semi-solid agar medium, where motile bacteria will migrate readily through the media causing cloudiness. Nonmotile bacteria will not cause cloudiness because they are not spreading throughout the media.

Question 37

Methyl Red (MR) Test

Answer 37

It identifies bacterial ability to produce stable acid end products by means of a mixed-acid fermentation of glucose. Organisms metabolizing pyruvic acid by the mixed acid pathway will produce more acid end products, such as lactic acid and acetic acid, and maintain an acidic environment (red color). Organisms that follow the VP pathway will decrease the pH, leading to a yellow color.

Question 38

Malonate Test

Answer 38

To test the ability of the organism to utilize malonate as sole source of carbon. Organisms which simultaneously utilize malonate and ammonium sulfate produce sodium hydroxide which thereby results in an alkaline reaction and changes the indicator from its original green color to light blue.

Question 39

MUG (4-methylumbelliferyl-beta-D-glucuronide) Test

Answer 39

MUG is a fluorogenic substrate of β-glucuronidase, which is an enzyme produced by most strains of E. coli and other Enterobacteriaceae. The end product of hydrolysis, 4-methylumbelliferyl (4-MU), fluoresces blue under long-wavelength ultraviolet light; however, verotoxin-producing strains of E. coli do not produce MUG.

Question 40

Lysine Iron Agar Test

Answer 40

To differentiate gram-negative bacilli based on decarboxylation and deamination of lysine and the formation of hydrogen sulfide (H2S). When glucose is fermented, the butt will become acidic (yellow), causing lysine decarboxylase to be triggered to neutralize acids changing the butt back to an alkaline pH (purple). If there is no lysine decarboxylase, it will remain acidic and yellow = negative. If deamination of lysine occurs in the slant, a burgundy color forms on the slant. If deamination does not occur, the LIA slant remains purple.

Question 41

Litmus Milk Medium Test

Answer 41

To differentiate microorganisms based on various metabolic reactions in litmus milk, including fermentation, reduction, clot formation, digestion, and the formation of gas

Question 42

Kligler’s Iron Agar (KIA) Test

Answer 42

Test to assess the different abilities of bacteria to either ferment glucose and/or lactose in a different way or lack of ability to ferment them at all together with the ability to produce H2S gas. Bacteria capable of fermenting glucose will initially ferment the glucose, releasing metabolic acids = yellow broth from decrease in pH. If the bacteria are lactose non-fermenter, they won’t be able to use lactose as a carbon source, and the depletion of glucose will trigger oxidative metabolism of peptone in the slant = increase in pH of the slant (red color). There is not enough oxygen in the butt to change the color of the butt so it will be K/A

Question 43

Indole Test

Answer 43

Detects a bacteria’s ability to produce indole from tryptophan, using the enzyme tryptophanase. The final products are indole, pyruvic acid, and ammonium. Indole, if present, combines with the aldehyde in the reagent to produce a pink to red-violet quinoidal compound (benzaldehyde reagent- tube test) or a blue to green color (cinnamaldehyde reagent- spot test).

Question 44

Hydrogen Sulfide (H2S) Test

Answer 44

Some bacteria can metabolically reduce sulfur-containing compounds to hydrogen sulfide and derive energy during the process. Microbial production of H2S is detected by allowing it to react with ferric ions or lead acetate to produce black-colored ferrous sulfide or lead sulfide respectively

Question 45

Growth at 42°C Test

Answer 45

The growth at 42°C test is a presumptive identification test used to differentiate between the species of the fluorescent pseudomonas group as well as other non-fermentative bacteria. P. aeruginosa can grow at 42°C.

Question 46

Flagellar Stain

Answer 46

Motile bacteria (mostly) possess flagella, but they have different amounts, shapes, and positions.

Question 47

Esculin Hydrolysis Test

Answer 47

Based on the hydrolysis of esculin (a glucoside) into glucose and esculetin by a microorganism that has constitutive β-glucosidase or esculinase enzyme. If esculin is hydrolyzed, there will be a loss of fluorescence and ferric ammonium citrate can be added to the tube. If esculetin is produced, it will react with ferric ions and result in a black color (positive)

Question 48

Decarboxylases

Answer 48

This is a group of enzymes that are capable of reacting with the carboxyl portion of amino acids, forming alkaline-reacting amines. Usually testing lysine, argininine, or ornithine. A positive result is purple and negative is yellow/no color change

Question 49

Citrate Utilization

Answer 49

Some bacteria can use citrate (from sodium citrate) as a sole source of carbon. Sodium citrate and ammonium phosphate (sole source of nitrogen) and put together with bacteria. If the bacteria is able to use citrate, then the production of alkaline by-products will occur (ammonia). Bromthymol blue is used as an indicator (basic = blue and acidic = green)

Question 50

Cetrimide Agar Test

Answer 50

Performed to identify or differentiate Pseudomonas aeruginosa from other microorganisms based on its ability to grow with cetrimide (quaternary ammonium detergent that is toxic to most bacteria)

Question 51

Butyrate disk test

Answer 51

Used to ID Moraxella catarrhalis based on the presence of the enzyme butyrate esterase. There are two substrates that can be used: bromochloro-indolyl butyrate and 4- methylumbelliferyl butyrate. The hydrolysis of bromochloro- indolyl substrate by butyrate esterase releases indoxyl, which in the presence of oxygen forms indigo (blue color). The hydrolysis of 4-methylumbelliferyl substrate produces a fluorescent compound visible under UV light.

Question 52

Beta (β) Lactamase Test

Answer 52

Used to detect beta-lactamase enzymes produced by different bacteria.

Question 53

Aerotolerance

Answer 53

determines if an isolate can grow aerobically, anaerobically, or both by incubated a CDC blood plate in anaerobic conditions and a CHOC plate in aerobic conditions and observing growth

Question 54

Disk tests for anaerobic bacteria

Answer 54

The most commonly used disks are vancomycin, colistin, and kanamycin on a CDC blood plate in anaerobic conditions for 2-3 days; 10mm or more is susceptible and less is resistant

Question 55

Sodium polyanethol sulfonate (SPS)

Answer 55

used to identify anaerobic GPC; it is the same as the disk test but a zone of inhibition is 12mm instead of 10mm

Question 56

Nitrate test for anaerobic bacteria

Answer 56

A paper disk saturated with nitrate is placed onto CDC blood for 24-72 hours of incubation instead of the tube test used for other bacteria

Question 57

Bile tolerance test for anaerobic bacteria

Answer 57

Bile disk- any zone = susceptible
Bile tube- inoculate two anaerobic broth tubes (THIO) one with bile and another without and then compare growth within the two tubes; equal growth in both tubes = resistant
Bile-containing agar- growth = resistant

Question 58

Catalase for anaerobic bacteria

Answer 58

Same as aerobic bacteria but it uses 15% instead of 3%

Question 59

Lipase test

Answer 59

Lipase degrades triglycerides into glycerol and free fatty acids, leading to a pearly, iridescent sheen on egg yolk agar surface (“oil on water” appearance)

Question 60

Spore test (ethanol method)

Answer 60

bacteria that produce spores are resistant to ethanol’s antibacterial effects; two CDC plates are incubated (one with ethanol and one without), spore forming bacteria will grow on both plates