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| Are lipases extracelluar or intracellular enzymes? Why? | Extracellular, because lipids are macromolecules and are too big to be transported into the cell, so they need to be broken down. |
| What is the agar made of in the Lipid hydrolysis test? Why is it used? | Tributyrin, because it is the simplest triglyceride found in natural fats. |
| Describe the process of the lipid hydrolysis test. And state how to interpret a (+) and (-) result. | Get a Tributyrin agar plate and inoculate each side with a lipase positive and lipase negative organism. Incubate for 24 hrs./ A (+) result will be seen as clearing around the organism. A negative result will not having any clearing and does NOT have lipase present. |
| T/F.? Spirit blue dye binds only to intact triglycerides.? | True |
| What is the enzyme, metabolic process, and end products in the lipid hydrolysis test? | Lipase/ Lipolysis (lipid hydrolysis)/ Fats hydrolyzed into glycerol (which goes through glycolysis) and 3 fatty acids (which go through B-oxidation to produce Acetyl CoA, which then goes through the Krebs cycle). |
| What is the purpose of the Lipid hydrolysis test? | To determine if the organism can produce Lipase in order to break down fats to use as an energy source. |
| What is the purpose of the Starch hydrolysis test? | To determine if the organism can produce amylase and/or Oligo-1,6- glucosidase, to break down starches to use as an energy source. |
| What are the 2 enzymes that can be produced by bacteria in the starch hydrolysis test? | Amylase and Oligo-1,6-glucosidase. |
| The 2 forms that Starch exists in is ____________ (linear) and ______________ (branched). | Amylose and Amylopectin. |
| What enzyme breaks down Amylopectin (branched starch)? | Oligo-1,6-glucosidase. |
| What is the enzyme, metabolic process, and end products in the starch hydrolysis test? | Amylase and Oligo-1,6-glucosidase/ hydrolysis/ (Many) alpha-D-Glucose |
| Describe the process of the starch hydrolysis test. | Get a starch agar plate, inoculate the plate with a (+) and (-) organism on each side of the plate. Incubate. After incubation add Iodine to the growth and observe for a halo around the growth, which would indicate a (+) result. |
| What does a (+)/(-) result look like in the starch hydrolysis test, and what do they mean? | (+) has a halo around the growth- which means that the Iodine reacted with Glucose/ (-) result has no halo/ clearing around the growth, which means that the Iodine reacted with the starch. |
| The Lipid hydrolysis test uses a plate of ____________ agar. | Tributyrin. |
| The Antimicrobial susceptibility test uses a plate of ________________ agar. | Mueller Hinton. |
| The Antimicrobial susceptibility test is also called the _______________ test. | Kirby-Bauer test |
| Describe the process of the Antimicrobial susceptibility test. | Get a big Mueller Hinton agar plate, inoculate the plate completely with a sterile cotton swab with full streaks and covering all of the agar with the bacteria completely, Apply the antibiotic disks by spacing them out (Tetracycline, Penicillin, Streptomycin, and Chloramphenicol), Incubate 16-18 hours. Observe the plates and measure the "Zone of inhibition" in millimeters, use the table to determine the bacterias susceptibility to the drug measured. |
| What does SIM stand for? | Sulfur reduction, Indole production, and Motility. |
| Describe the processes of the S-I-M Tests and the interpretation of the results. | You grab 4 SIM tubes, You aseptically STAB inoculate 3 of the tubes with 3 different bacteria., incubate. After incubation observe the tubes for Sulfur reduction and Motility 1st!! Sulfur reduction- if there is black in the medium then it is (+) for sulfur reduction (sulfur gains a Hydrogen=reduced) It is either reduced by Cysteine desulfurase or Thiosulfate reductase... Motility- If there is growth (not black growth, just growth) radiating outward from the stabline, then it is (+) for motility./After recording data, add Kovac's reagent to all three tubes, if there is a red precipitate at the top of the tube, then the result is (+) and it means that Tryptophan was broken down by tryptophanase into Indole and pyruvate. |
| A (+) result for Indole production in the SIM test is indicative of, What? | That tryptophan was broken down by tryptophanase into Indole and Pyruvate./ That Tryptophanase is present, which produces a red color. |
| Name the two enzymes that reduce Sulfur, in the SIM test. | Cysteine desulfurase and Thiosulfate reductase. |
| A (+) result for the Sulfur reduction test in the SIM media is indicative of, What? | That either Cysteine desulfurase or Thiosulfate reductase is present and they reduced sulfur to Hydrogen Sulfide (H2S). |
| What does KIA stand for? | Kligler Iron Agar |
| What does the edge of the zone of inhibition represent? Does this edge represent the end of diffusion of the drug? | The MIC (minimum inhibitory concentration)/ No, the drug still diffuses into the agar, but the edge of the zone of inhibition is where the drugs concentration becomes too low to be lethal to the bacteria. |
| _____________ is the hydrogen sulfide (H2S) indicator in the KIA test./ ______________ is the pH indicator in the test. | Ferrous sulfate/ Phenol red |
| Describe the process of the KIA test. | Get 4 or 5 KIA slants and inoculate them with different bacteria that your testing. Streak the slants with the loop and stab the butts with the needle.. Incubate and then observe. |
| What does the KIA test, test for? | Fermentation of Glucose, Lactose and Sucrose., Sulfur reduction, and Gas from fermentation. |
| Why is a black butt considered to be a yellow butt in the KIA test?... How can you determine what sugar was fermented in the process? | A yellow butt indicates Acid production, But black indicates that Hydrogen sulfide was produced by the thiosulfate. BUT There has to be Acidic conditions In order for Thiosulfate reduction to take place. Therefore, a black butt is indicative of Sulfur reduction AND fermentation./ If the butt is black, then it HAS TO HAVE ACID, which means that it used fermentation. In order to find out what it fermented you have to look at the slant. If the slant is RED= glucose was fermented., If the slant is Yellow= Glucose AND Lactose and/or Sucrose was fermented. |
| In the KIA test, An organism that does Not ferment any of the carbohydrates but utilizes the animal proteins will ___________ the medium and turn it ______________. | Alkalinize/ Red |
| What reaction is taking place in the KIA test when black appears in the medium? | Ferrous sulfate (apart of the medium) is reacting with hydrogen sulfide (H2S). |
| Describe/explain (give symbols) these results from the KIA test; A tube with a yellow slant/yellow butt...... Tube with Red slant/yellow butt..... Red slant/Red Butt..... Red slant/No change in Butt.... Black precipitate in Agar.. | (A/A)= Glucose and Lactose was fermented with acidic end products/ (K/A)= Proteins catabolized Aerobically with alkaline end products, Glucose was fermented with acid end products(Reversion)/(K/K) No fermentation, proteins catabolized anaerobically and aerobically with alkaline end products (indicative of a strict aerobe)./ (K/NC)= No fermentation, proteins catabolized aerobically with alkaline end products./ (H2S)= Sulfur reduction. |
| What is Reversion? What test can this be seen in? | When an organism switches from fermentation of a product to deamination. |
| T/F.? In the KIA test, a tube with a red slant and yellow butt, means that the organism fermented only Glucose and NOT lactose or Sucrose, But also catabolized proteins.? What is this process called? | True/ Reversion. |
| ___________ are Exotoxins that destroy red blood cells and hemoglobin. | Hemolysin's. |
| What are hemolysins? | They are exotoxins that are secreted by Gram positive cocci that destroy RBC's and hemoglobin. |
| What are the 3 types of hemolysis and what do they stand for? | B-hemilysis= (Hemolysin is present) complete destruction of RBC's and hemoglobin, which results in clearing around the colonies./ (Alpha) a-hemolysis= (Hemolysin is present) Partial RBC destruction and is apparent by greenish discoloration around the colonies/ (gamma) y-hemolysis= No hemolysis, the organism does NOT hemolyze RBC's. |
| Describe the process of the Blood Agar test. And describe the results after Incubation. | Get a Blood agar plate. Divide the plate into 3 quadrants and label for each bacteria being tested. Aseptically streak each quadrant with its bacteria and then stab the agar on the streak 3 times. Incubate and observe./ Beta hemolysis- Complete destruction of RBC's and hemoglobin by the presence of hemolysin., Alpha hemolysis- partial destruction of RBC's and hemoglobin, seen by greenish color.., Gamma hemolysis- No RBC's hemolyzed. |
| The stabs in the Blood Agar test encourages _____________ activity because of the reduced oxygen concentration. | Streptolysin |
| What test uses the streak-stab technique? | The Blood Agar test. |
| What are the 2 types of coagulase enzymes? | Free coagulase and bound coagulase. |
| How does coagulase benefit a bacterium? | Coagulase causes blood plasma to clot, which increases bacterial resistance to phagocytosis by surrounding the infecting bacteria with a clot. |
| Describe the process of the Coagulase test and what the results mean. | Get small rabbit plasma tubes and label with correct bacterium. Aseptically inoculate each with the metal loop and cover with foil. Then incubate, and observe./ A (+) result indicates that coagulase is present and the medium is solid (plasma clumped). A (-) result indicates that coagulase is absent and the medium is liquid (no plasma clumping). |
| What test uses TTC to make interoperation of results easier? What is TTC and how does it work? | The Motility test/ Tetrazolium Salt- when it is oxidized it is colorless, but when it is reduced (gains an electron), then it turns red. This makes motility easier to detect. |
| Describe the process for the motility test and describe the results. | You get some motility test media tubes and aseptically inoculate each tube with different bacteria by stabbing them once down the middle. Incubate and observe. A (+) (motile) result will have red growth away from the stab line. A (-) result will only have red growth in the stab-line. |
| A ___________ is used on non living surfaces. | disinfectant |
| A ___________ is used on living tissue. | antiseptic. |
| Describe the process of the "Effectiveness of disinfectants" test. | Control #1- Aseptically inoculate control tube #1 with culture/ Control #2- Sterile broth and nothing done to it. Get 5 sterile beads. Alcohol flame forceps and aseptically add 4 of the beads into the assigned culture tube for 1 minute.. The 5th bead is to be placed in a petri dish with sterile water and then transferred to a sterile control broth., After 1 minute the broth is decanted and the 4 beads are to be transferred to a petri dish with filter paper.. After this 3 of the beads are to be aseptically transferred (alcohol flamed loop) to their own petri dishes with different dilution concentrations of the disinfectant being tested. The 4th bead is to be aseptically transferred to a petri dish with sterile water.( this bead is a control to test that the bacteria wasn't removed by the water sloshing, therefore the disinfectant actually worked and that it wasn't from sloshing around). Let the beads sit for 10minutes in the disinfectants/ water and then aseptically remove them with forceps into their own sterile broth tubes. Then Incubate. Then observe for growth. |
| How many controls were used in the "Effectiveness of disinfectants" test? What are their purposes? | 4 controls/ Control 1- To make sure that the bacteria was viable., Control 2- to make sure that the broth was sterile., Control 3- to make sure that the bacteria was not washed off of the bead by sloshing of broth or water., Control 4- Was to make sure that the beads and sterile water were actually sterile. |
| Describe the process of the "Effectiveness of disinfectants" test. | Control #1- Aseptically inoculate control tube #1 with culture/ Control #2- Sterile broth and nothing done to it. Get 5 sterile beads. Alcohol flame forceps and aseptically add 4 of the beads into the assigned culture tube for 1 minute.. The 5th bead is to be placed in a petri dish with sterile water and then transferred to a sterile control broth., After 1 minute the broth is decanted and the 4 beads are to be transferred to a petri dish with filter paper.. After this 3 of the beads are to be aseptically transferred (alcohol flamed loop) to their own petri dishes with different dilution concentrations of the disinfectant being tested. The 4th bead is to be aseptically transferred to a petri dish with sterile water.( this bead is a control to test that the bacteria wasn't removed by the water sloshing, therefore the disinfectant actually worked and that it wasn't from sloshing around). Let the beads sit for 10minutes in the disinfectants/ water and then aseptically remove them with forceps into their own sterile broth tubes. Then Incubate. Then observe for growth. |
| The multiple tube fermentation technique for total coliform determination uses a technique called ______________. | Most probable number. |
| What is a coliform? | A Gram (-) rod shaped bacteria that ferments lactose to produce acid/gas and is found in the digestive tracts of warm blooded animals. |
| Describe the process of the multiple tube fermentation technique. And describe the results. | Grab 15 LTB tubes Label 5 Group A, 5 Group B and 5 Group C/ Get 2 more tubes with 9.0 mL of water and Label one 1:10 and the other 1:100.. Use a pipet and add 1.0 mL of the original contaminated water to the tube 1:10 tube (mix well) and then use a new pipet and from the 1:10 tube add 1.0 mL to the 1:100 tube (mix well).. (discard pipets).. After this you will use a new pipet and add 1.0mL of the original sample to all 5 tubes in Group A (mix well), Use a new pipet and add 1.0mL to all 5 tubes in group B from the 1:10 solution (mix well), Use a new pipet and add 1.0mL to all 5 tubes in Group C from the 1:100 solution.(mix well)/ Incubate tubes @ 37 C for 48 hours and then observe for growth and /or gas in the inverted durham tubes. (Gas is (True +) result) (Growth is a Potential (+)) If the durham tube has gas and/or growth, then it is to be used in the BGLB tubes and EC tubes. Aseptically inoculate 1 BGLB tube and 1 EC tube for each (+) LTB (mix well) and Incubate BGLB tubes @ 37 C for 48 hours and EC tubes @ 42 C (water bath) for 48 hours. Observe each tube for gas and count them. |
| What does LTB stand for? BGLB? EC? | Lauryl Tryptose Broth/ Brilliant Green Lactose Bile/ E. Coli Broth |
| What is the purpose of the LTB tubes? What does it contain? | To presumptively determine the presence or non-presence of coliforms./ Contains Lactose and Lauryl sulfate- inhibits gram (+) bacteria. |
| What media in the multiple tube fermentation technique, is confirmatory media? | BGLB and EC |
| What does BGLB stand for and how does it work? | Brilliant green Lactose bile- The brilliant green is a pH indicator, the Lactose is the fermentable sugar and the Bile inhibits gram (+) bacteria from growing. |
| Describe the process of the multiple tube fermentation technique. And describe the results. | Grab 15 LTB tubes Label 5 Group A, 5 Group B and 5 Group C/ Get 2 more tubes with 9.0 mL of water and Label one 1:10 and the other 1:100.. Use a pipet and add 1.0 mL of the original contaminated water to the tube 1:10 tube (mix well) and then use a new pipet and from the 1:10 tube add 1.0 mL to the 1:100 tube (mix well).. (discard pipets).. After this you will use a new pipet and add 1.0mL of the original sample to all 5 tubes in Group A (mix well), Use a new pipet and add 1.0mL to all 5 tubes in group B from the 1:10 solution (mix well), Use a new pipet and add 1.0mL to all 5 tubes in Group C from the 1:100 solution.(mix well)/ Incubate tubes @ 37 C for 48 hours and then observe for growth and /or gas in the inverted durham tubes. (Gas is (True +) result) (Growth is a Potential (+)) If the durham tube has gas and/or growth, then it is to be used in the BGLB tubes and EC tubes. Aseptically inoculate 1 BGLB tube and 1 EC tube for each (+) LTB (mix well) and Incubate BGLB tubes @ 37 C for 48 hours and EC tubes @ 42 C (water bath) for 48 hours. Observe each tube for gas and count them. |
| What does the P stand for in the MPN calculation? ... V(n)? ... V(a)? | P= total number of (+) results (BGLB or EC)/ V(n)= total volume of sample that in LTB tubes that were (-)/.. V(a)= Volume of sample in all LTB tubes. |
| Describe the process of the multiple tube fermentation technique. And describe the results. | Grab 15 LTB tubes Label 5 Group A, 5 Group B and 5 Group C/ Get 2 more tubes with 9.0 mL of water and Label one 1:10 and the other 1:100.. Use a pipet and add 1.0 mL of the original contaminated water to the tube 1:10 tube (mix well) and then use a new pipet and from the 1:10 tube add 1.0 mL to the 1:100 tube (mix well).. (discard pipets).. After this you will use a new pipet and add 1.0mL of the original sample to all 5 tubes in Group A (mix well), Use a new pipet and add 1.0mL to all 5 tubes in group B from the 1:10 solution (mix well), Use a new pipet and add 1.0mL to all 5 tubes in Group C from the 1:100 solution.(mix well)/ Incubate tubes @ 37 C for 48 hours and then observe for growth and /or gas in the inverted durham tubes. (Gas is (True +) result) (Growth is a Potential (+)) If the durham tube has gas and/or growth, then it is to be used in the BGLB tubes and EC tubes. Aseptically inoculate 1 BGLB tube and 1 EC tube for each (+) LTB (mix well) and Incubate BGLB tubes @ 37 C for 48 hours and EC tubes @ 42 C (water bath) for 48 hours. Observe each tube for gas and count them. |
