SCCC Micro Lab Final – Flashcards
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Unlock answers| (2-9) Bacteria can survive at temperatures as low as ____ and as high as _____. |
| -10°C, 110°C |
| (2-9) Define the 3 cardinal temperatures: |
-minimum - organism will not survive below this temp -maximum - organism will not survive above this temp -optimum - temperature at which an organism shows its highest growth rate |
| (2-9) Define the optimum temperature range for - Psychrophile |
| below 20°C |
| (2-9) Define the optimum temperature range for - Psychrotroph |
| between 0°C - 30°C |
| (2-9) Define the optimum temperature range for - Mesotroph |
| between 15°C - 45°C |
| (2-9) Define the optimum temperature range for - Thermophile |
| above 40°C |
| (2-9) Define the optimum temperature range for - Obligate Thermophile |
| will not grow below 40°C |
| (2-9) Define the optimum temperature range for - Faculative Thermophile |
| will grow below 40°C |
| (2-9) Define the optimum temperature range for - Extreme Thermophile |
| grow best betwen 65°C - 110°C |
| (2-9) Thermal Classifications of Bacteria |
| [image] |
| (2-9) Why do different temperatures produce different growth rates? |
| Because different bacteria are adapted to different habitats. Different enzymes dictate what temperature organisms can thrive. |
| (2-14) Define - Germicide |
| substances designed to reduce the number of pathogens |
| (2-14) Define - Disinfectant |
| germicides used on surfaces & liquids |
| (2-14) Define - Antiseptic |
| germicides used on or in living tissue |
| (2-14) Name the test used to determine the effectiveness of a germicide |
| Use-Dilution Test |
| (2-14) Write the genus of each of the 3 organisms usually tested |
| Staphylococcus, Salmonella, Pseudomonas |
| (2-14) Glass ______ are coated with _________. |
| beads, living bacteria |
| (2-14) The coated beads are then dipped in varying ____________ of a test germicide |
| dilutions |
| (2-14) They are then _______________. |
| transferred to a growth medium |
| (2-14) After incubation, we look for _______________ |
| microbial growth |
| (2-14) To meet the standards, the solution must be able to inhibit growth ________% of the time |
| 95 |
| (2-14) Which germcide was the most effective and at what concentration? Which was the least effective? |
Lysol was the most efffective, worked at all concrentrations on S. aureus and E. coli. Hydrogen peroxide was the least effective, did not work on E. coli at any concentration, only worked at 3% concentration to kill S. aureus. Hydrogen peroxide will not work on organisms that produce catalase. |
| (2-14) Which organism seemed to be most resistant to the most germicides? |
| E. coli |
| (2-14) Explain the purpose of the control tubes. |
-control tube #1 was to make sure the bacteria are alive -control tube #2 was to make sure the broth is sterile -control tube #3 was to make sure the bacteria was not washed of the glass beads -control tube #4 was to make sure the beads and the sterile water were sterile |
| (7-3) Define antibiotic |
| natural antimicrobial agents produced by microorganisms |
| (7-3) What is the current term for an agent that is used to treat a bacterial infection? |
| anti-microbials |
| (7-3) The disc diffusion test is also called the: |
| Kirby-Bauer test |
| (7-3) Paper discs are impregnated with |
| antimicrobics |
| (7-3) They are then placed on a large plate that has been covered with a _____ of bacteria |
| lawn |
| (7-3) Explain the 2 things that occur during incubation |
-growth of the bacteria -diffusion of antimicrobial into the agar |
| (7-3) The concentration of the antimicrobial substance is ____________ close to the disc |
| higher |
| (7-3) The size of the "zone of inhibition" is dependant on: |
-sensitivity of the bacteria to the specific antimicrobial agent -the point at which the chemical's minimum inhibitory concentratio (MIC) is reached |
| (7-3) Define bactericidal |
| a drug that kills the organism |
| (7-3) Define bacteriostatic |
| a drug that stops the organism's growth but does not kill it |
| (7-3) The name of the type of agar used is |
| Mueller-Hinton |
| (7-3) The pH of the media is |
| 7.2 - 7.4 |
| (7-3) The media is ______ deep in the plate |
| 4 mm |
| (7-3) What effect will thick agar have on zone size? |
| slows lateral diffusion and produces smaller zones |
| (7-3) The bacterial broth must have a _____________ turbidity standard |
| 0.5 McFarland turbidity standard |
| (7-3) The chemical used to prepare McFarland standards is called |
| barium sulfate |
| (7-3) Chloramphenicol - Cellular target |
| prevents peptide bond formation during translation |
| (7-3) Ciprofloxacin - cellular target |
| interferes with DNA replication |
| (7-3) Trimethoprim - cellular target |
| inhibits purine and pyrimidine synthesis |
| (7-3) Penicillin - cellular target |
| inhibits cross-linking of the cell wall's peptidoglycan |
| (7-3) What might be the consequence of pouring the plates 2 mm deep instead of 4 mm deep? |
| The antimicrobial will diffuse faster and produce larger zones than the standard |
| (7-3) The Mueller-Hinton plates are supposed to be used within a specific time after their preparation and should be free of visible moisture. Why do you think this is so? |
-Dry plates will slow lateral diffusion producing smaller zones -Moisture will dilute the bug |
| (7-3) In clinical applications of the Kirby-Bauer test, diluted cultures (for the McFarland standard comparison) must be used within 30 minutes. Why is this important? |
| Bugs may keep growing in the broth causing too much turbidity |
| (7-3) E. coli and S. aureus were chosen to represent Gram-negative and Gram-positive bacteria. For penicillin, is there a difference in susceptibility between the Gram-positive and Gram-negative bacteria? If so, why? |
| Penicillin is a cell wall inhibitor. Penicillin breaks down the peptidoglycan in the bacterial cell wall. Gram positive bacteria have cell walls made up of peptidoglycan, but gram negative bacteria have a cell wall made of a phospholipid bilayer and not much peptidoglycan. Also, gram negative cell walls have pumps to remove antibiotics. |
| (7-3) You record zone diameters of 25 mm for chloramphenicol and penicillin disks. Which antibiotic would be more effective against this organism? What does this tell you about comparing zone diameters to each other and the importance of the zone diameter chart? |
| The chloramphenicol would be more effective against this organism because the susceptible standard is ; 18. The resistant standard for penicillin is ; 28. You cannot go by the diameter of the clearing without using the interpretive chart. |
| (7-3) How does the antibiotic get from the disk into the agar? |
| The antibiotic diffuses into the agar |
| (7-3) Does the agar have an antibiotic beyond the zone of inhibition? |
| Yes, but it is not at a concentration that will inhibit visible growth of the bacteria |
| (7-3) What was the purpose of inoculating NA plates with samples taken from the zones of inhibition? |
| to see if the antibiotic killed the bacteria (bactericidal) or if it stopped the growth but did not kill the microbe (bacteriostatic) |
| (7-6) Define epidemiology |
| the study of the causes, occurrence, transmission, distribution, and prevention of diseases in a population |
| (7-6) Define fomite |
| contaminated object |
| (7-6) Define vector |
| biting insect |
| (7-6) Define common source epidemic |
| when a disease is transmitted from an area or from water that infects many people at once |
| (7-6) Define propagated transmission |
| a disease transmitted from person to person |
| (7-6) Define index case |
| first case of the disease |
| (7-6) Twelve ways infectious diseases are transmitted |
| ingestion, inhalation, direct skin contact, open wounds or lesions in skin, animal bites, direct blood to blood contact, sexual contact, people ; animal, biting insects, aerosols, contaminated objects, water contaminated with feces |
| (7-6) Equation for incidence rate |
number of new cases in a; time period ------------------------------------------- x K size of at-risk population at midpoint of time period |
| (7-6) Equation for point prevalence |
number of existing cases at a point in time ------------------------------------------- x K size of at-risk population |
| (7-6) What is "K" in the two equations and why do we use it? |
| K is some power of 10 to get its value up to a number bigger than one |
| (7-6) Descriptive epidemiology |
| Looks at the cause and source of the disease |
| (7-6) Analytical epidemiology |
| Compares two groups of pepole, one that had contact with the injurious agent and one that did not. Looks for factors that may have preceded the disease |
| (7-6) Experimental epidemiology |
| begins with a hypothesis and uses experimentation |
| (7-6) Case reporting |
-used to determine the chain of transmission of a particular disease -once the chain of transmission is discovered, measures can be taken to stop the spread of the disease -case reporting use used to approximate the incidence and prevalence of a disease -these are also called "Morbidity Measures" |
| (7-6) Incidence & Prevalence Calculations |
week - # of new cases 1 - 2 2 - 5 3 - 4 4 - 3 5 - 5 6 - 2
Assume population 1,200 & illness lasts 3 weeks, K=1000
week 1 incidence 2/1,200 = 0.0001666 x 1000 = 1.7 cases per 1000 people week 1 prevalence 2/1,200 = 0.0001666 x 1000 = 1.7 cases per 1000 people
week 2 incidence 5/1,200 = 0.00416 x 1000 = 4.2 cases per 1000 people week 2 prevalence 2+5/1,200 = 0.00583 x 1000 = 5.8 cases per 1000 people
week 3 incidence 4/1200 = 0.0033 x 1000 = 3.3 cases per 1000 people week 3 prevalence 2+5+4/1200 = 0.0091 x 1000 = 9.1 cases per 1000 people
week 4 incidence 3/1200 = 0.0025 x 1000 = 2.5 cases per 1000 people week 4 prevalence 5+4+3/1200 = 0.01 x 1000 = 10 cases per 1000 people
week 5 incidence 5/1,200 = 0.00416 x 1000 = 4.2 cases per 1000 people week 5 prevalence 4+3+5/1200 = 0.01 x 1000 = 10 cases per 1000 people
week 6 incidence 2/1,200 = 0.0001666 x 1000 = 1.7 cases per 1000 people week 6 prevalence 3+5+2/1200 = 0.0083 x 1000 = 8.3 cases per 1000 people |
| (7-6) Explain how you determined the index case |
| this was the first time growth appeared |
| (7-6) Suggest possible reasons for cases showing no growth between cases that exhibited growth |
| They had a subclinical infection so they did not have enough bug to cause disease or only a mild case of the disease, but they may have been carriers to spread the disease |
| (12-1) Candida albicans - what is it called when it infects the oral cavity |
| thrush |
| (12-1) Candida albicans - what is it called when it infects female genitals |
| vulvovaginitis |
| (12-1) Candida albicans - what is it called when it infects the skin |
| cutaneous candidiasis |
| (12-1) Candida albicans - name three other organs it can infect |
| lungs, bronchi, kidneys |
| (12-1) Candida albicans - who is most susceptible |
| diabetics, immunocompromised (AIDS), catheterized patients, antibiotic therapy |
| (12-1) Candida albicans is unique in that cells form ___________ when grown at 37°C |
| pseudohyphae |
| (12-1) Candida albicans - pseudohyphae help them to ______ to tissue |
| adhere |
| (12-1) Aspergillis - what is the general term for infection with this organism |
| aspergillosis |
| (12-1) Aspergillis - what is caused by invasive aspergillosis |
| necrotizing pneumonia |
(12-1) septate or nonseptate hyphae? [image] |
| septate hyphae |
(12-1) septate or nonseptate hyphae? [image] |
| nonseptate hyphae |
| (12-1) [image] |
| Rhizopus sporangiophoes |
| (12-1) [image] |
| Aspergillus condiophore |
| (12-1) [image] |
| Penicillum condiophore |
| (8-12) Fecal contamination is evidenced by the presence of a group of bacteria called: |
| Enterobacteriaceae |
| (8-12) Most Enterobacteriacea can ferment ______ and produce ________ & ________. |
| lactose, acid, gas |
| (8-12) These organisms are called ________ |
| coliforms |
| (8-12) The pores in the membrane filter are ____ wide. |
| 0.45 μm wide |
| (8-12) The agar that is used is called |
| Endo agar |
| (8-12) Endo agar selects against Gram _____ bacteria |
| positive |
| (8-12) It contains the sugar ___________ |
| lactose |
| (8-12) A dye in the media changes in response to changes in ____ |
| pH |
| (8-12) Coliform colonies appear: |
| red and/or mucoid with a gold or green metallic sheen |
| (8-12) Does Endo agar only support the growth of coliforms |
| No |
| (8-12) What is the color of the colonies we should count? |
| red or metallic |
| (8-12) a countable plate has between ___ & ___ coliform colonies |
| 20 & 80 |
| (8-12) In order to be safe to drink (________), a sample must contain less than __________ |
| potable, one coliform per 100 ml |
| (8-12) Formula to find total coliforms/100 ml |
total coliforms/100 ml = coliform colonies counted x 100 ml/volume of original sample in ml
ex. 2 colonies counted in 10 ml sample = 2 x 100ml/10 ml = 2 x 10ml = 20 colonies per 100 ml (not potable) |
| (11-7) ELISA - a positive result is: |
| color change |
| (11-4) Define antigen |
| any substance that causes antibody formation |
| (11-4) define antibody |
| a protein produced by the body in response to a specific antigen |
| (11-4) when large antigens and antibodies combine it is called: |
| agglutination |
| (11-4) A direct agglutination test involves combining antibodies and naturally particulate antigens. What does "particulate" mean? |
| can be seen with the naked eye |
| (11-4) Why did they have to combine the antigens with the red beads before adding the antibody? |
| The antigens are too small to be seen even when bound to antibodies |
| (11-4) Application |
| agglutination reactions may be used to detect the presence of either antigen or antibody in a sample. Direct agglutination reactions are used to diagnose some diseases, determine if a patient has been exposed to a certain pathogen, and are involved in blood typing. Indirect agglutination is used in some pregnancy tests as well as in diagnosing disease. |
| (11-4) [image] |
| Direct agglutination |
| (11-4) Indirect agglutination |
| [image] |
| (11-6) Mononucleosis Hemagglutination Test |
-Some IgM antibodies are not specific -They react with more than one antigen -They are called heterophile antibodies -To test for mono, we look for antibodies (IgM) against the Epstein-Barr Virus (EBV) -Patient serum is 1st mixed iwith guinea pig kidney. It binds all the IgM except IgM directed against EBV -Next mix the above mixture with horse RBCs -If IgM against EBV is in the serum, the horse RBCs will agglutinate |
