Your favorite patient has just had surgery. After surgery Bactroban was ap Show more I need help on 13 please Your favorite patient has just had surgery.
After surgery Bactroban was applied. After a few days you notice an infection around the incision. Even though more Bactroban is applied the infection persists. You want to identify the bacteria causing the infection to determine what treatment is best. First you need to collect a sample. 1. What kind of medium do you place your swab into to grow your bacteria? Mueller-Hinton agar medium 2. After incubating for 24 hours will you have a pure culture? Give an explanation for your answer! No. Pure culture is not seen. This is due to the fact that the pure culture is infected by the microbes. After incubation the plates were examined for zone of inhibition. Zone of inhibition 22 mm was considered as sensitive and such isolates were excluded from the study whereas 21 was considered as resistant and reported as a mupirocin (bactroban) resistant MRSA (methillin resistant staphylococcus aureus) and included in the study. 3. What is the next step and what is the purpose of it? Perform the Disc diffusion method or Kirby Bauer Test to determine the minimum inhibitory concentration for mupirocin (Bactroban). After isolating different species of bacteria you see two distinct types of colonies. -One has medium-sized shiny colonies that are mucoid with entire margins convex elevation and a whitish translucent appearance. This colony type is the dominant type on the plate. -The other has circular pinhead colonies which are convex with entire margins and have a yellowish-white color. There are only 2 colonies that appear this way. You make a pure culture of each type of colony using half of one colony labeled with its colony description. You take the other half of the colony you used to make your pure culture and perform a simple stain to determine the size and shape of the different bacteria. 4. What kind of simple stain would you perform and why? What color will the cells appear when looking through the microscope and why? I would use methylene blue to stain my bacterial cells. I placed loopful of water; added a small amount of the organisms in the water and heat fixed it until the water evaporated. I did this for each type of colony. Methylene blue is a basic stain that will stain the bacterial cells because of the negative charge on their cell walls. Therefore the cells will appear blue when viewed under the microscope. The negative charge comes from the presence of peptidoglycan which is located just outside of the cytoplasmic membrane. 5. Do you think the stain has affected the cells shape? Why or why not? When heat fixing onto the slide the cells shape may have been distorted. 6. You determine that the cells from the whitish translucent colony are rod shaped. What term do you use to describe this cell type? Bacillus You find that the cells from the smaller yellowish-white colonies are circular and seem to appear in clumps. What cell type is this? Micrococcus 7. After you incubate your pure culture for 24 hours you perform a gram stain. You find that your circular cells have stained purple. What does this result mean? Dont just answer gram+ or gram-. Give a brief explanation what this reaction is based on! My circular cells that stained purple are considered to be Gram positive. The Gram-positive cell wall has many interconnecting layers of peptidoglycan (composed of NAG and NAM) and makes up about 60-90% of the cell. At the second step Grams iodine is added (mordant) and forms this CV-I (Crystal violet-iodine) complex so that the dye cannot be removed easily. At this step both Gram-positive bacteria and Gram-negative will remain purple. Next is the decolorizer step which is an important step in the Gram staining procedure. At this step the Gram-positive cells will retain its CV-I complex and will not be decolorized by the alcohol therefore appearing purple. It will remain purple even after the last step (counterstain). This is because in the Gram-positive bacteria the Crystal violet and the iodine combine to create a larger molecule that precipitates within the cell. The alcohol will then cause the peptidoglycan layer to dehydrate in the gram-positive cell wall and therefore decrease the space between the molecules causing the cell to trap the CV-I complex within the cell. 8. You find that your rod shaped cells have stained pink. What does this result mean? Dont just answer gram+ or gram-. Give a brief explanation what this reaction is based on! This means my rod-shaped cells are Gram-negative. The Gram-negative cell wall has a much thinner single layer of peptidoglycan that is surrounded by an outer membrane (phospholipids lipopolysaccharides and lipoprotein). Only 10-20% of the Gram-negative cell wall is peptidoglycan. At the second step Grams iodine is added (mordant) and forms this CV-I (Crystal violet-iodine) complex so that the dye cannot be removed easily. At this step both Gram-positive bacteria and Gram-negative will remain purple. Next is the decolorizer step which is an important step in the Gram staining procedure. At this step the Gram-negative cells will not retain its CV-I complex and will be decolorized by the alcohol therefore being colorless. At the last step (counterstain) using Safranin the Gram-negative bacteria that are colorless will now be directly stained by the Safranin therefore appearing pink. In the case of the Gram-negative bacteria the alcohol dissolves the outer lipid layer from the cells. The thin layer of peptidoglycan is unable to retain the CV-I complex and the cell is decolorized Summarize your results: Colony Appearance Cell Shape Gram Stain Medium-sized shiny colonies; mucoid; entire margins; convex; whitish translucent appearance Bacillus Gram-negative Circular pinhead colonies; convex; entire margins; yellowish-white color Micrococcus Gram-positive 9. Based on the gram stain results which bacteria do you think is likely causing the infection? Give your reason. The Gram-negative is more likely to be causing the infection. This is due to the Gram-negative cell wall containing an outer membrane that is composed of phospholipids and liopolysaccahrdies. The lipopolysaccahrides are resombble for the toxicity of the Gram-negative bacteria ebcase when these bacteria die and their cell wall lyse these liposoolyscashides will be relased and will thus be called endotoxins. The effect of these endotoxins is the same in the human body: tjeu will stimulate macrophages to release ctyokines in very high concentrations. 10. If you plate the bacteria causing the infection will it grow on a. LB agar? Yes. It will grow on Luria Broth agar because it is a rich medium. b. GMS agar? c. PE agar? It will not grow on PE agar because it is a selective medium used to grow Gram-positive organisms therefore the Gram-negative will not grow on this type of medium. 11. What tests can you now perform to help identify the problem bacteria? a. You decide to start with fermentation tests. You incubate your tubes for 24 hours. In both the PRB glucose tube and PRB lactose tube you see yellow broth and an air bubble in the Durham tube. What does this mean? Phenol red in Phenol red broth is a pH indicator. This indicator will turn yellow in the presence of acid production during glucose and lactose fermentation. The gas produced is trapped as a bubble. Thus it can ferment both glucose and lactose. Summarize your results: Glucose Appearance Result Glucose Appearance Result Lactose Appearance Result Lactose Appearance Result yellow and gas production Positive Yellow and gas production Positive Now that you know the fermenting capabilities of your bacteria you decided to run the IMViC tests. The first test you try is the Citrate Utilization test. a. When making the Simmons citrate media you added glucose along with the citrate. Will your test yield accurate results if you use this media? Why or why not? b. If you dont add glucose to the Simmons citrate media will your test results be accurate? Why or why not? c. After running the test properly you find that the media is a deep blue color and there is growth on the slant. What does this result mean? d. Next you perform the Indole test. You find out that your bacteria cannot degrade tryptophan. What does this tube look like? e. You inoculate your bacteria into 2 MRVP tubes and incubate them for 72 hours. When you perform the methyl red test with one of the 2 MRVP tubes the media retains a yellow color. What does this result mean? Give 2 possible explanation for this result! i. Now you perform the Voges-Proskauer test using the second MRVP tube and the media develops a rose color. What does this result mean and does this result help to interpret the result of the methyl red test? Give a detailed explanation!!! a) If we use glucose then citrate non-utilizers will also grow which will give a false positive result as we have seen in the preceding test they can ferment glucose. b) If we do not add glucose we will be able to get an accurate result. c) A blue test indicates citrate positive. It may be a coliform. d) Indole test is performed in presence of tryptophan in the medium (peptone water). The final product is indole which is detected by adding Kovacs Reagent which will result in a pink layer on the surface of the media. If it does not degrade tryptophan then the color of the top will remain unchanged. As this is does not degrade tryptophan it may be Klebsiella pneumoniae or enterobacter aerogens. e) Positive methyl red test are indicated by the development of red color after the addition of methyl red reagent. Methyl red is a pH indicator which turns red below pH 4. This low pH will be prevalent if the microbe performs mixed acid fermentations. A no color change indicates a negative result which indicates the bacterium may be Klebsiella pneumoniae or enterobacter aerogens. i) The VP test uses alpha-naphthol and potassium hydroxide to test for the presence of acetoin produced by Klebsiella spp Enterobacter spp Hafnia spp and Serratia spp. A rose color indicates VP positive (acetoin producer). It only substantiates the results of the MR test as the question still remains the same as both Klebsiella pneumoniae and enterobacter aerogens are positive for VP test. Summarize your results: Methyl Red Appearance Result Methyl Red Appearance Result Voges-Proskauer Appearance Result Voges-Proskauer Appearance Result Indole Appearance Result Indole Appearance Result Citrate Appearance Result Citrate Appearance Result 12. Knowing all these different test results can you clearly identify the bacterium that is causing the infection? If not what additional test(s) would you do to identify the bacterium? Give a detailed explanation! (Hint: You can use your results from class to help answer this. You may need to do some extra searches outside of the book.) The identification is still nor clear. As both Klebsiella pneumoniae and enterobacter aerogens are the possibility. To differentiate them motility test can be used. Soft agar stabs of the two bacteria on Sulfite-Indole-Motility (SIM) agar will result spreading of the bacterium from the stab. Otherwise hanging drop test can be used to view them directly under microscope. E. aerogens is motile and K. pneumoniae is not. 13. You decide to perform the litmus test as your final test. You inoculate a tube containing litmus milk medium and incubate the tube for 24hrs. What would you expect to see when you look at the tube the next day? Give a detailed explanation!!!!! 14. Based on what you know so far what antibiotic might you recommend? Why? Based on the above results carbapenems will be the best choice of antibiotics. Imipenem doripenem and meropenem act against both E. aerogens and K. pneumoniae. 15. Can you transform these cells following the protocol in Chapter 19 of your lab manual? Why or why not? In order to conclude this series of experiments you would like to extract genomic DNA from your identified bacterium. Your supervisor tells you to get 5 X 10 9 healthy bacterial cells for the extraction. What would you do to get these cells? Give a detailed explanation!!! Show less
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