Lab 10 Questions. Microbio

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Hawaii Pacific University *

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Feb 20, 2024

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Lab 10 Exercise Pre-Test 1. One disadvantage of using ultraviolet light to sterilize objects is:  d. all of these 2. Which of the following is the most resistant to ultraviolet radiation?  d. spores 3. Which of the following is the most resistant to destruction from heat?  d. spores 4. An antibiotic disk shows a large zone of inhibition (no growth) when placed on an agar plate covered with test bacteria. This large zone of inhibition would most likely be interpreted as:  a. sensitive or susceptible 5. One factor in getting inaccurate results in testing antibiotic sensitivity on an agar plate is:  d. all of these Critical Thinking A. 1. Why should you avoid looking directly into the ultraviolet (UV) light?  Answer: "Ultraviolet light damages cells in two ways: (1) it triggers mutations in DNA, resulting in thymine dimers, thus preventing successful reproduction, and (2) it causes direct protein damage, as can be seen in anyone suffering from sunburn."  Explanation: Direct exposure to UV light can cause mutations in DNA and damage to proteins, which can have harmful effects on cells and tissues, such as preventing successful reproduction and causing sunburn. 2. Why is the method of testing chemical sensitivity to disinfectants you performed considered somewhat inaccurate?  Answer: "One such test is a disk diffusion method performed on bacteria growing on an agar plate. Although the procedure is rather easy to perform, results are often difficult to interpret because the disinfectant is constantly in contact with the microbe rather than for a short exposure, and the microbe is on a growth medium rather than on a typical environmental surface, which has little nutritional material available."

 Explanation: The method involves constant contact between the disinfectant and the microbe, unlike typical exposure scenarios. Additionally, the microbes are on a growth medium with ample nutrition, which may not accurately represent real-world conditions. 3. Why did Bacillus subtilis show growth from all samples of the Thermal Death Time part of this exercise while the others soon showed no growth at all?  Answer: "Thermal Death Time (TDT) is one of several methods used to explore the relationship between temperature, time, and the death rate of specific microbes. TDT is the time necessary to kill all vegetative cells in a pure broth culture at a predetermined temperature. As temperature goes up, the time necessary to kill microbes goes down."  Explanation: Bacillus subtilis showed growth because it may have been more tolerant to the temperature used in the Thermal Death Time test compared to the other species tested. 4. Why does milk that is pasteurized and then refrigerated have a limited shelf life and eventually “spoil”?  This question is not directly addressed in the provided text. 5. In the procedure used to test bacterial growth against various temperatures (incubator, room, refrigerator, freezer), why should efforts be made to inoculate each tube with the same number of bacteria?  Answer: "The object is to deliver equivalent numbers of bacteria into each tube. Again, using sterile technique, add one calibrated loopful of B. subtilis culture into each of the four tubes previously labeled."  Explanation: Inoculating each tube with the same number of bacteria ensures consistency and accuracy in comparing the growth of bacteria at different temperatures. 6. Why does the Kirby-Bauer procedure require that the concentration of the bacteria be the same, the stage of growth constant, the growth medium the same, and the concentration or amount of drug in each disk constant?  Answer: "Bacteria are standardized by placing them on the test medium in their early stages of growth. This ensures that all cells are equally susceptible to the antimicrobial agent. The concentration of these cells is controlled by comparing the cloudiness of the broth or saline solution that it is in with that of a standard chemical solution, which always displays uniform cloudiness."

 Explanation: Standardization ensures that the results of the Kirby- Bauer procedure are accurate and reliable by controlling various factors that could influence bacterial growth and drug sensitivity. 7. Why are Escherichia coli, Pseudomonas aeruginosa, and Staphylococcus aureus used as standards in the Kirby-Bauer method?  This question is not directly addressed in the provided text. 8. A pure culture was inoculated onto a Mueller-Hinton agar plate. The Kirby-Bauer procedure was performed. One of the drugs tested showed a large zone of inhibition but also had small colonies growing within this zone. Further testing showed that these colonies were not the results of contamination. Why would these colonies be present within this zone of inhibition?  This question is not directly addressed in the provided text. B. MATCHING 1. Disinfectant (d) - a chemical that destroys most or all pathogens on an inanimate object 2. Lawn (a) - solid growth of bacteria across the surface of a plate 3. Sensitive (c) - indicates that an antimicrobial drug would be effective against a specific microbe 4. Ozone (f) - by-product of UV light use 5. Thermal Death Time (e) - amount of time it takes to kill 100% of a bacterial broth culture at a specific temperature 6. Resistant (g) - indicates that an antimicrobial drug would not be effective against a specific microbe 7. Antibiotic (j) - a substance naturally produced by one microbe that kills or inhibits another C. MULTIPLE CHOICE 1. **A disadvantage of using UV light to control microbes is:** - **Answer: d. all of these** - **Reasoning:** UV light has several disadvantages mentioned in the text: - "Ultraviolet light cannot penetrate very well and is only effective on surfaces" (Source: Ultraviolet (UV) light sensitivity section). - "Since bacterial cells have only one chromosome, and even one mutation is often lethal, the large number of mutations caused by UV light often results in the

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death of practically all cells present" (Source: Ultraviolet (UV) light sensitivity section). - "Ultraviolet light penetrates so poorly that regular window glass can prevent the penetration of over 80% of it. If this is the case, why is it so important to have sunglasses made of glass that is 100% UV resistant? Since the tinted glass prevents light from penetrating, the pupils respond by dilating. Without 100% UV protection, these dilated pupils would allow even more UV light to reach the retina." (Source: Ultraviolet (UV) light sensitivity section). 2. **UV light is able to damage bacterial cells by:** - **Answer: b. damaging DNA** - **Reasoning:** The text mentions, "Ultraviolet (UV) light damages cells in two ways: (1) it triggers mutations in DNA, resulting in thymine thymine dimers, thus preventing successful reproduction" (Source: Ultraviolet (UV) light sensitivity section). 3. **Bacillus subtilis and Bacillus cereus most likely showed greater resistance to UV light than Serratia marcescens. This is due to:** - **Answer: b. the Bacillus spp. are spore formers** - **Reasoning:** The text states, "Since bacterial cells have only one chromosome, and even one mutation is often lethal, the large number of mutations caused by UV light often results in the death of practically all cells present. Since most cellular proteins are enzymatic, the few cells that may survive DNA damage will die as a result of enzyme damage" (Source: Ultraviolet (UV) light sensitivity section). 4. **The broth culture of Bacillus subtilis showed growth even after extended heating in a water bath, while Serratia marcescens did not. This is due to:** - **Answer: b. Bacillus subtilis is a spore former** - **Reasoning:** The text mentions, "Some are very sensitive to changes in temperature and are limited to a narrow range of temperature. They are thus obligate in this requirement. In other words, they must be kept at a fairly constant temperature for survival... Even the more temperature tolerant or adaptive (facultative) ones used in this laboratory will have up to a 90% death rate if inoculated from room temperature into media taken directly from a refrigerator" (Source: Thermal Death Time (TDT) section). 5. **In the Kirby-Bauer test, which of the following must be consistent?**

- **Answer: d. all of these** - **Reasoning:** The text states, "Bacteria are standardized by placing them on the test medium in their early stages of growth. This ensures that all cells are equally susceptible to the antimicrobial agent. The concentration of these cells is controlled by comparing the cloudiness of the broth or saline solution that it is in with that of a standard chemical solution, which always displays uniform cloudiness" (Source: The Kirby-Bauer Technique section). 6. **A relatively large zone of inhibition surrounding an antimicrobial disk on a Kirby- Bauer test plate would most likely be interpreted as:** - **Answer: a. sensitive reaction** - **Reasoning:** The text mentions, "The reading is based on the size of the zone of inhibition surrounding each disk. These zones are measured in millimeters (mm), and a difference in size of only 2 to 3 mm can mean the difference between describing an organism as being susceptible or sensitive to the drug, or being resistant, which indicates that the drug would be ineffective" (Source: The Kirby- Bauer Technique section). 7. **A factor in the zone of inhibition size on the Kirby-Bauer plate is:** - **Answer: d. all of these** - **Reasoning:** The text states, "The size of the zone of inhibition is related to the diffusion rate of the chemical placed on the plate as it is in the Kirby-Bauer test" (Source: Chemical Sensitivity section).

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