LAB REPORT NUMBER TWO DATE: 3/25/2010 inal attachment Lab Experiment number 11 PURPOSE: To learn the Gram stain technique, the reason for the stain, and how to identify the results of the organisms stained. MATERIALS: Bunsen burner, inoculating loop, staining tray, glass slides, bibulous paper, lens paper, oil, and microscope METHODS: Apply Crystal Violet (Primary stain) for 1 minute. Rinse with D-water Apply Iodine (Mordant) for 1 minute. Rinse with D-water. Apply Alcohol (Decolorize) for 30 seconds. Rinse with D-water. Apply Safarin (Counterstain) for 1 minute.
Blot dry with bibulous paper.
MICROORGANISMS USED: E. coli, B. cereus, S. aureus & E. coli (mixture) RESULTS/DATA USED: E. coli cell shape was bacilli (rod) with a diplobaccillus arrangement. The color was pink because it was Gram negative. B. cereus cell shape was bacilli (rod) with a diplobacillus arrangement. The color was purple because it was Gram positive.
S. aureus & E. coli (mixture) cell shape was cocci (spherical) with a staphylococcus arrangement. The color was mostly purple with some noticeable pink but the mixture was Gram positive. CONCLUSIONS E. coli is Gram negative, B. ereus is Gram positive, S. aureus & E. coli mixture is Gram positive. REVIEW QUESTIONS: Question 1: What are the advantages of differential staining procedures over the simple staining technique? Answer: Simple stains are used to just give color to microbes on slides. Differential stains tell the chemical composition of organisms. Source: http://www. bmb. psu. edu/courses/micro107/notes/staining. htm Question 2: Cite the purpose of each of the following reagents in a differential staining procedure. Answer: a. Primary stain: Passes the color of the stain to all of the cells. b.
Counterstain: Used to stain red the cells that have been decolorized (Gram – cells). c. Decolorizing agent: removes the primary stain so that the counterstain can be absorbed. d. Mordant: Increases the cells’ affinity for a stain by binding to the primary stain. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ Pages 73 & 74 Question 3: Why is it essential that the primary stain and the counterstain be of contrasting colors? Answer: Cell types or their structures can be distinguished from one another on the basis of the stain that is retained.
Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ Pages 73 Question 4: which is the most crucial step in the performance of the Gram staining procedures? Explain. Answer: Decolorization is the most crucial step of the Gram stain. Over-decolorization will result in lost of the primary stain causing Gram positive organisms to appear Gram negative. Under-decolorization will not completely remove the CV-I (crystal-violet-iodine) complex, causing Gram negative organisms to appear Gram positive. Source: Microbiology – A Laboratory Manual 4th Edition/ James G.
Cappuccino, Natalie Sherman/ 2008/ Pages 74 Question 5: Because of a snowstorm, your regular laboratory session was cancelled and the Gram staining procedure was performed on cultures incubated for a longer period of time. Examination of the stained Bacillus cereus slides revealed a great deal of color variability, ranging from an intense blue to shades of pink. Account for this result. Answer: The organisms lost their ability to retain the primary stain and appear to be gram-variable. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ Pages 74
LAB EXPERIMENT NUMBER 12 PURPOSE: The purpose of the Acid fast stain is to identify the members of the genus Mycobacterium, which represent bacteria that are pathogenic to humans. Mycobacteria has a thick, waxy wall that makes penetration by stains extremely difficult so the acid fast stain is used because once the primary stain sets it cannot be removed with acid alcohol. This stain is a diagnostic value in identifying these organisms. MATERIALS: * Bunsen burner * Hot plate * Inoculating loop * Glass slides * Bibulous paper * Lens paper * Staining tray * Microscope METHODS: 1.
Prepared a bacterial smear of M. smegmatic, S. aureus, & a mixture of M. smegmatic & S. aureus 2. Allowed 3 bacterial slides to air dry & then heat fixed over Bunsen burner 8 times. 3. Set up for staining over the beaker on hot plate, flooded smears with primary stain-crystal fuchsin and steamed for 8 minutes. 4. Rinsed slides with water 5. Decolorized slides with acid alcohol until it runs clear with a slight red color. 6. Rinsed with water 7. Counterstained with methylene blue for 2 minutes 8. Rinsed slides with water. 9. Blot dry using bibulous paper and examine under oil immersion
MICROORGANISMS USED: * Mycobacterium smegmatic * S. aureus * A mixture of S. aureus & M. smegmatic RESULTS AND DATA USED: 1. M. smegmatic, a bacilli bacteria that colored pink resulting in acid fast. 2. S. aureus, a cocci bacteria that colored blue resulting in non acid fast. 3. M. smegmatic & S. aureus resulted in both acid fast & non acid fast. CONCLUSION The conclusion to the acid fast stain is that S. aureus lacks a cellular wax wall causing the primary stain to be easily removed during decolorization, causing it to pick up the counterstain-methylene blue.
This results in a non acid fast reaction, meaning it is not in the genus Mycobacterium. M. smegmatic has a cellular wax wall causing the primary stain to set in and not be decolorized; this results in an acid fast reaction meaning it is in the genus Mycobacterium. REVIEW QUESTIONS Question 1: Why must heat or a surface-active agent be used with application of the primary stain during acid-fast staining? Answer: It reduces surface tension between the cell wall of the myobacteria and the stain. Source: Microbiology – A Laboratory Manual 4th Edition/ James G.
Cappuccino, Natalie Sherman/ 2008/page 79 Question 2: Why is acid-alcohol rather than ethyl alcohol used as a decolorizing agent? Answer: Acid-fast cells will be resistant to decolorization since the primary stain is more soluble in the cellular waxes than in the decolorizing agent. Ethyl alcohol would make the acid fast cells non-resistant to the decolorization. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ page 79 Question 3: What is the specific diagnostic value of this staining procedure? Answer: Acid-fasting staining represents bacteria that is athogenic to humans Question 4: Why is the application of heat or a surface-active agent not required during the application of the counter stain in acid-fast staining? Answer: The counter stain methylene blue is only needed to give the stain its color. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 79 Question 5: A child presents symptoms suggestive of tuberculosis, namely a respiratory infection with a productive cough. Microscopic examination of the child’s sputum reveals no acid-fast rods.
However, examination of gastric washings reveals the presence of both acid-fast and non-acid fast bacilli. Do you think the child has active tuberculosis? Explain. Answer: Yes, the child may have active tuberculosis. Although, acid-fast microorganisms are not easily removed and non-acid fast are. Tuberulosis represents bacteria that are pathogenic to humans, the stain is of diagnostic value identifying these organisms. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 79 LAB EXPERIMENT NUMBER 13 PURPOSE:
The purpose of this experiment is to identify the difference between the bacterial spore and vegetative cell forms. The vegetative cells are highly resistant, metabolically inactive cell types. The endospore is released from the degenerating vegetative cell and becomes an independent cell. MATERIALS: * Bunsen burner * hot plate * staining tray * inoculating loop * glass slides * bibulous paper * lens paper * microscope METHODS: 1. The spore stain (Schaeffer-Fulton Method) is performed on a microscopic slide by making an individual smear of the bacteria on slide and heat fixing until dry. . Flood the smears with malachite green and place on top of a beaker of warm water on a hot plate, allowing it to steam for 5 minutes. 3. Remove the slide and rinse with water. 4. Add counter stain safranin for 1 minute then rinse again with water and blot dry with bibulous paper. MICROORGANISMS USED: * S. aureus * S. aureus & B. cereus mix RESULTS/DATA USED 1. B. cereus- green spores, pink vegetative cells, endospore located in center of cell 2. B. cereus & S. aureus- green spores, pink vegetative cells, endospore located in center of cell CONCLUSION:
An endospore is a special type of dormant cell that requires heat to uptake the primary stain. To make endospores readily noticeable, a spore stain can be used. In using a microscope, under oil immersion, you will be able to identify the color of the spores, color of the vegetative cells and be able to locate the endospore in certain bacteria like S. aureus and B. cereus. REVIEW QUESTIONS Question 1: Why is heat necessary in spore staining? Answer: The heat dries the dye into the vegetative cell of the spore. Source: Microbiology Lab Manual, 8th edition, Cappuccino & Sherman, p. 5 Question 2: Explain the function of water in spore staining. Answer: The water removes the excess primary stain, while the spores remain green the water rinses the vegetative cells that are now colorless. Source: Microbiology Lab Manual, 8th edition, Cappuccino & Sherman, p. 85 Question 3: Assume that during the performance of this exercise you made several errors in your spore-staining procedure. In each of the following cases, indicate how your microscopic observations would differ from those observed when the slides were prepared correctly.
Answer: a. ) You used acid-alcohol as the decolorizing agent. The alcohol would wash out all coloring from the bacteria. Source: Microbiology Lab Manual, 8th edition, Cappuccino & Sherman, p. 85 b. ) You used safranin as the primary stain and malachite green as the counterstain. Safranin will absorb to vegetative cells and not endospores since you need heat for endospores to form and malachite green will not absorb without heat but it will to vegetative cells. Source: Microbiology Lab Manual, 8th edition, Cappuccino & Sherman, p. 85 c. You did not apply heat during the application of the primary stain. Without heat, the endospores will not form and it will not penetrate the spore to color the vegetative cell. Source: Microbiology Lab Manual, 8th edition, Cappuccino & Sherman, p. 85 Question 4: Explain the medical significance of a capsule. Answer: The capsule protects bacteria against the normal phagocytic activities of the host cells. Source: Microbiology Lab Manual, 8th edition, Cappuccino & Sherman, p. 87 Question 5: Explain the function of copper sulfate in this procedure.
Answer: It is used as a decolorizing agent rather than water, washes the purple primary stain out of the capsular material without removing the stain bound to the cell wall, the capsule absorbs the copper sulfate and will appear blue. Source: Microbiology Lab Manual, 8th edition, Cappuccino & Sherman, p. 88 LAB EXPERIMENT NUMBER 44A PURPOSE: The purpose of this experiment is to identify the best chemotherapeutic agents used for infectious diseases. S. aureus is the infectious disease used for this experiment. MATERIALS: * Sensi-disc dispensers or forceps * Bunsen burner * sterile cotton swabs * glassware marking pencil millimeter ruler METHODS: Using the Kirby-Bauer antibiotic sensativity test method is used. This method uses an Antibiotic Sensi-disc dispenser, which placed six different types of antibiotics on an Mueller-Hinton agar plate, infected with S. aureus. The antibiotics are in the form of small, round disc, approximately 5mm in diameter. The anitbiotics are placed evenly away from each other on the S. aureus infected Mueller-Hinton agar plate and incubated at 37 degrees Celcius for up to 48 hours. After the completed incubation time, any area surrounding the antibiotic disc which shows a clearing or an “area of inhibition” is then measured.
Measurements are taken from the diameter of each antibiotic area of inhibition. This measurement will determine which of the antibiotics is best to be used against the specific organism. (In this case, S. aureus) MICROORGANISMS USED: S. aureus ANTIBIOTICS USED: Ticarcilin Erythomycin Clindamycin Gentamicin Vancoymycin Lmipenem RESULTS/DATA USED: A chart showing the measurements of each antibiotic is used to determine its effectiveness. The three different types of ranges are: Resistant (Least useful) Intermediate (Medium useful) Susceptible (Most useful) The following results are: Zone Size Ticarcilin – 25mm (Susceptible)
Erythomycin – 20mm (Intermediate) Clindamycin – 20mm (Intermediate) Gentamicin – 15mm (Susceptible) Vancoymycin – 13 mm (Susceptible) Lmipenem – 21 mm (Susceptible) CONCLUSION: 4 of the 6 antibiotics above can be effectively used against inhibiting this organism (S. aureus). This information would be passed on to the provider of the infected patient, so the patient can be given the antibiotic chosen by their provider and recover from this infection. LAB EXPERIMENT NUMBER 46B PURPOSE: The purpose of this experiment is to evaluate the effectiveness of antiseptic agents against selected test organisms.
MATERIALS: The materials used are five Tryticase soy agar plates. METHODS: 24-48 hours Trypticase soy broth cultures of E. coli, B. cereus, S. aureus and M. spegmatic. MICROORGANISMS USED: The microorganisms used were E. coli, B. cereus, S. aureus and M. spegmatic. RESULTS/DATA USED: The data collected in this experiment shows chlorine bleach having the broadest range of microbial activity because it has the strongest ingredients. Tincture of iodine and hydrogen peroxide seems to have the narrowest range because the contents aren’t as strong. CONCLUSION:
The Agar Plate-Sensitivity Method shows the effectiveness of antiseptic agents against selected test organisms. The antiseptic exhibited microbicidal activity against each microorganism. REVIEW QUESTIONS: Question 1: Evaluate the effectiveness of a disinfectant with a phenol coefficient of 40. Answer: A disinfectant with a phenol coefficient of 40 indicates the chemical agent being more effective than the phenol. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 302 Question 2: Can the disinfection period (exposure time) be arbitrarily increased? Explain.
Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 302 Answer: Yes, the longer the disinfection period the greater the antimicrobial activity. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 302 Question 3: A household cleanser is labeled germicidal. Explain what this means to you. Answer: A household cleaner labeled germicidal means something that would and the killing of germs on a surface. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page`302
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