Relationship of Capsules with Virulence and Capsule Staining Techniques Lab

Discuss the relationship of capsules with virulence. Bacteria that has a capsule is more virulent and less susceptible to antibiotics/antiseptics Recognize the steps of a capsule stain and compare with a negative stain. A negative stain stains the background so the bacteria show up white against it. In capsule staining, the organism is stained and the background is stained and the capsule shows up as a clear halo around the organism. Procedure for Negative Stain 1. Place drop of India in stain onto slide near one end 2. Use toothpick to get sample of plaque from teeth. Mix sample into drop of India ink. Do not spread with loop or toothpick 3. To spread the mixture evenly use another slide. a. Take second slide and back up to the drop. Only back up into the drop until you contact the drop. b. Now draw the drop across with the slide to spread. 4. Let the preparations air dry. DO NOT heat fix 5. Examine under 100x oil immersion lens Procedure of Capsule Staining 1. Use toothpick to scrape between your teeth for plaque. 2. Smear on slide, air dry and heat fix. 3. Flood with safranin for approx 1.5 - 2 min 4. Rinse with water 5. Blot dry with bilbulous paper 6. Put a couple drops of india ink on the stained smear 7. Spread the india ink with a second slide (as done in negative stain protocol) 8. Air dry for 5 mins 9. Examine slide under 100x oil immersion lens Perform and list steps of a spore stain. Note the results of positive vs negative. Endospore stain procedure: 1. Prepare smears of organisms to be tested for endospores. 2. Heat fix the smears. Heat fix more than usual but let cool and then repeat heat fix 3 times. 3. Place slide on staining tray 4. Flood smear with malachite green. 5. Time for 20 mins, replenish if evaporates too quickly, slide needs to remain flooded for full 20 mins 6. Carefully remove slide 7. Rinse well with water 8. Drain the slide and counterstain with safranin for 2 mins 9. Wash, blot, and examine. 10. The vegetative cells will appear red and the spores will appear green. Positive- spores are present (you see green spores and pink vegetative cells) Negative- no spores are present (you only see pink vegetative cells) What genera are endospore formers? Bacillus and Clostridium; endospores contain the cell’s DNA, ribosomes and Dipicolonic acid. Dipicolonic acid is a spore specific chemical that aid endospores ability to remain dormant Perform and list the steps in an acid-fast stain. Note positive vs negative results. Procedure: 1. Prepare smear, break up clumps so that organism is not clustered 2. Air dry and heat fix 3. Flood with carbolfuschsin 4. Keep flooded for 10 mins, do not allow to dry out 5. Rinse with water 6. Decolorize with acid alcohol, about 5 seconds of one dropper running over smear 7. Rinse with water 8. Counterstain by flooding with methylene blue for 1 min. 9. Rinse with water, blot with bibulous paper. Positive acid-fast will show up as bright pink Negative acid-fast shows up as blue What genera are acid fast positive? What makes them stain this way? Mycobacterium, Norcardia. Postitive results are pink bacteria (from the Carbol Fuschia). They stain this way because of the mycolic acid in their cell walls which prevents the acid-alcohol wash from decolorizing them. Give an example of a pathogenic acid fast and a pathogenic spore forming bacterial species. Give disease and organisim. Acid fast – tuberculosis, Mycobacterium Tuberculosis and Leprosy caused by Mycobacterium Leprae Spore-forming: Anthrax; Bacillus anthracis Use the spectrophotometer to estimate growth of broth cultures, note how to use a blank to zero, and read OD or absorbance. To set the spectrophotometer to zero: 1. Use down arrow to select OD600 and hit enter 2. Hit enter again 3. Place a clean blank in the holder, arrow/notch facing you 4. Close lid and press 0.00 (zero) button. Repeat steps 3 & 4 before putting in a sample to be tested. Compare the growth curve by plate count and spectrophotometry. Growth curve plots the increase in cell numbers versus length of incubation. This allows measurement of the rate of growth and shows its generation time. For the growth curve, serial dilutions are performed and a spread plate is done for each dilution at 30 minute intervals in order to calculate the number of colony forming units at a given time. For the spectrophotometer, growth measurements are also taken at 30 minute intervals. This was done with the large flask that was being shaken in the incubator room. The growth curve by plate count was more effective than the spectrophotometer because it counts live bacteria while the other (known as indirect method or turbidity) counts all cells which may be dead or alive. The disadvantage of the growth curve would be that it takes more time (because the cells have to grow). *Be able to make bacteria counts from plates with known amounts of solution. I think this is serial dilutions and using the log to figure out the number of colonies in the original Be able to explain how to make a serial dilution of a sample and read a plate with growth from a diluted sample to determine the concentration of the original sample. Compare the effectiveness of hand sanitizer, soap and the alcohol wipes. Explain and be able to interpret the raw data, see lab report. Note what percent reduction refers to. Most effective: alcohol wipes Least effective: soap *Percent reduction refers to the amount of bacteria killed or removed from a sample, e.g. 0.9 = 90%. **Compare the effectiveness of the antiseptics we used with the different organisms we tested. Note the methods used to evaluate the antiseptics. (Mariah, please update with the results you guys have) The only results I have are written in blue, but I will add out groups information for the antibiotics below as well -Mariah P.aeruginosa - resistant to all antibiotics; resistant to antiseptics & disinfectants tested (hand sanitizer, lysol, bleach, cavicide; betadine was minimally effective) B.cereus - sensitive to hydrogen peroxide and lysol (16 mm) Resistant to hand soap Intermediate to hand sanitizer B.cereus- sensitive to caviside Resistant to Hand soap Intermediate to bleach *S.epidermis - sensitive to lysol, hydrogen peroxide, greatest zone of inhibition (35mm) Intermediate to hand sanitizer and bleach E.coli - sensitive to betadine, hand sanitizer, lysol Define optimum, minimum and maximum temperatures for growth. Optimum temperature is the temp that the bacteria like best and at which they will have the most growth Minimum and maximum temps are the min/max that they are able to grow at. Compare ecology of an organism with optimum temperature and pH in lab. The ecology of an organism is where is typically is found, the environment it lives in. This environment will have the optimum temperature and pH for this organism. Explain two ways to grow obligate anaerobes. Fluid Thioglycollate Medium consumes oxygen and permits the growth of obligate anaerobes. It is a differential media that has a redox indicator in it which makes oxygen show up red. This medium also allows for microbes to grow at different depths to determine what type of aerobe they are. *Anaerobic jar, it depletes the oxygen from the environment so that only obligate anaerobes and facultative anaerobes can grow Recognize oxygen requirements from FTM tubes. Oxygen requirements for FTM tubes: 1. Obligate aerobes require oxygen for growth so they can only grow at the surface. They use oxidative respiration 2. Obligate anaerobes cannot survive in the presence of oxygen because they lack the enzymes superoxide dismutase (SOD) and catalase or peroxidase which detoxify damaging oxygen by-products such as peroxides and superoxides. They use anaerobic respiration or fermentation 3. Facultative anaerobes can grow with or without oxygen. They don’t care. 4. Microaerophiles require small amounts of oxygen for respiration but can not survive in atmospheric oxygen due to limited ability to carry out respiration 5. Aerotolerant anaerobes can survive in oxygen due to SOD but do not use oxygen as a terminal electron acceptor. They use fermentation or anaerobic respiration Give 2 essential enzymes needed to live in an oxygen environment. superoxide dismutase (SOD) Catalase or perioxidase Compare the growth of Serratia marcescens and Pseudamonas at different temperatures. When serratia marcescens is grown atv 25 degrees C it produces a red pigment (called prodigiosin) when grown at 37 degrees Celsius it is white. Pseudamonas emits a blue-green color that diffuses into the agar; this happens at any temp. Define mesophile, psychorophiles and thermophiles and give examples. Mesophile - optimum growth temp between 20-50 degrees Celsius; pathogens for humans Psychrophiles - op temp -5 - 20 degrees Celsius; found in arctic and antarctic Thermophiles - like it hot, 50-80 degrees Celsius Compare TDP and TDT as terms and for the organisms used in lab. TDP - thermal death point, this is the temperature at which a culture is killed in 10 minutes. TDT - thermal death time, this is the amount of time required to kill a suspension of cells or spores at a given temperature Compare the lethal effects of temperature on Staphylococcus, Bacillus species and Escherichia coli and discuss why there is a difference. Bacillus requires higher heat at a longer time to ensure death because it is spore-forming. Compare the effects of pH on the growth of the organisms that we used in lab. Optimal pH for species tested: S. epidermidis: pH 5.5 E. coli: pH 7 A.Faecalis: pH 7 C. glabrata: pH 5.5 Graph the growth using percent transmittance against pH or temperature, see worksheet for the lab. Optimal temperature for species tested: S. epidermidis: 37 C B. cereus: 37 C E. coli: 37C B. stearothermophilus: 55 C Describe how UV effect our test organisms and how it differed from what we expected. Explain why. Bacillus species were the least susceptible because they are spore forming Define transformation and briefly summarize the procedure. Transformation: modification of a cell/bacterium by the uptake and incorporation of extracellular DNA. Procedure: 1. Take 2 microcentrifuge tubes: 1 labeled P+ and 1 labeled P-. These tubes will have 10 mL of either plasmid (P+) or (P-). The “P-” tube will act as a negative control as it will not be exposed to the plasmid. 2. Take tube containing ice-cold Transformation solution and place on ice. 3. Take 250mL of TS and add to P- tube. Take 250mL of the TS and add to the P+ tube, keep both on ice. 4. Pick 3 small colonies of E. coli from the culture plate with a sterile loop and mix it into each of the tubes and place on ice for 10 mins. (This treatment stresses the bacterium, rendering its cell membrane and cell wall permeable to the plasmid. This process will take the recipient E. Coli “competent” to uptake the plasmid) 5. After 10 mins, walk with your ice bath and tuves to the water bath. Transfer both the tubes from the ice container into the water bath at 42C for 50 sec. The transfer from the ice to the water bath should be immediate. When incubating the tubes in the water bath, be sure that the uve bottoms ar all the way to the water below the rack. (this is called heat shock) 6. After 50 sec, immediately place the tubes on ice for 2 mins 7. Move to room temp, add 250Ml of recovery media to both tubes, mix gently, and leave at room temp for 15 mins 8. Using lab marker label 4 plates #1-4 9. Transfer 100mL from P+ tube to the LB/Amp/Ara plates. Top sprwad evenly. 10. 10Transfer 100mL from P- tube to the LB and LB/AMP plates, Top spread evenly. 11. Allow liquid to absorb (15 min). Invert the plates and incubate. Compare the results and the purpose of the plates: LB P- : Massive growth. How come? We put bacteria on the plate but it didn't have the plasmid. But there was no need for it because it was just regular broth medium. There was nothing to stop it from growing. There was no antibiotic, so it grew. This show that the procedure of the Transformation solution--the more of ice--didn't kill them or change them. LB/amp P-: This was our control. The procedure did not make them amp resistant. So it showed that it did not obtain the ability to grow in amp. There was nothing on the plate. LB/amp P+: We were supposed to get several white colonies. The reason is because they were transformed and they now picked up the plasmid so they could grow in amp. (tech might have accidentally dumped arabinose. This was a mistake) LB/amp/ara P+: Should have gotten red ones and the reason why is bc it picked up the plasmid and were able to grow in amp. However, they should have turned red bc they had arabinose-the inducer--to turn on the red fluorescent protein. Note: -Satellite colonies: mean that the ones growing around the red colonies are not transformed. The reason they do this is because they are living off the goodwill of the red colonie in the middle. -If there is a non-resistant cell near the transformed cell on the plate producing amp, it can survive and make a small colonie or a satellite colony. Explain what pTOM (red version of pGLO) is and what RFLP (GFP) is. See web site: pTOM (red version of pGLO): Allows us to easily explore mechanisms of gene regulation and genetic selection. We are allowed to do so by the pGLO plasmid, which carries a gene for the production of Green Fluorescent Protein. The PGLO system incorporates an arabinose promoter which controls expression of the GFP in transformed cells. Expression of GFP gene can be turned on by including arabinose (a sugar) in the growth medium. Resistance to the antibiotic AMP is used as the selection mechanism for transformation. Cells transformed with pGLO can grow on AMP-plates. Transformed cells will appear white on Amp plates not contained arabinose, and fluorescent green when arabinose is included (for us it was pink). The process is observable with a UV lamp. The pGLO system also helps us experiment purification of the recombinant protein, GFP. what RFLP (GFP): ???? Someone pls help Explain the function of the following in transformation: Transformation solution- TS LB-Broth medium for colonies to grow on Ampicillin-The antibiotic--to select for transformations by plating on a medium containing amp. Arabinose-The inducer--allows to express the gene of interest by growing cells on a medium containing arabinose. Explain how to use the filter paper disk method of antiseptic evaluation. 1. Divide plate into 4 quadrants and label each quadrant with one of the antiseptics/disinfectants 2. Transfer 0.1mL of you test organism to each plate and top spread (using the hockey stick) 3. Using alcohol-flamed forceps, pick up a sterile paper disk and dip it HALFWAY into a disinfectant/antiseptic, then place the disk on the proper quadrant of the inoculated media. The chemical will wick up into the paper disk. No need to push in, the liquid should adhere to the agar. 4. Repeat procedure with other chemicals. Use 4. 5. Incubate plates at proper temp (which temp was this?) Compare effectiveness of different antiseptics on the different organisms use in class. most vs least E. coli: H202/Bleach P. aeruginosa: Betadine,Sanitizer/Everything else was 0 B. cereus: Lysol/hand soap S. epidermidis: Lysol/bleach Define zone of inhibition and what causes it. Also explain why it may differ with diffusion. -If an antibiotic stops the bacteria from growing or kills the bacteria, there will be an area around the wafer where the bacteria have not grown enough to be visible. This is called a zone of inhibition. -The size of the zone of inhibition indicates the degree of sensitivity of bacteria to a drug. In general, a bigger area of bacteria-free media surrounding an antibiotic disk means the bacteria are more sensitive to the drug the disk contains. Size of zone correlates to how sensitive (susceptible) microbe is to the drug. *Speculate on how an antimicrobic may have a large zone of inhibition but in reality is not very effective. The antimicrobic my appear to have a large zone of inhibition because is it very soluble in the medium and can spread more easily. (not sure if you did this already, Veronica) Give examples of effective and ineffective antibiotics against the organisms we used in lab. We need to give examples for bacteria that was susceptible to antibiotics and examples of ones that were not. E.coli- sensitive to chloramphenicol and Amoxicillin. Resistant to penicillin P.aeruginosa- resistant to chloramphenicol, Amoxicillin, penicillin (may be a mistake) B.cereus- resistant to penicillin and Amoxicillin. Sensitive to chloramphenicol S.epidermidis- sensitive to chloramphenicol and Amoxicillin. Resistant to penicillin Pseudamonas - resistant to everything Please give other examples Define or explain what narrow and broad spectrum refers to for an antibiotic. -Narrow spectrum: Know you are treating those specific bacteria when using these drugs -Broad spectrum: Can be used in cases where you may not be exactly sure what bacteria is causing the infection -Narrow spectrum work against a small number of species, for example, only Gram(+). -Broad spectrum work against a wide variety, i.e. both Gram(+) and Gram(-) Explain how gel electrophoresis works A technique that uses electricity to move negatively charged DNA molecules through an agarose-containing gel matrix. This technique separates molecules of DNA based on the length of time it takes for them to migrate through the agarose (a refined form of agar which produces a gel matrix of uniform size and consistency). The term electrophoresis literally means to carry with electricity, and that is exactly what electrophoresis does. Compare the speed of DNA by size of fragment. As the gel runs, shorter pieces of DNA will travel through the pores of the gel matrix faster than longer ones. After the gel has run for awhile, the shortest pieces of DNA will be close to the positive end of the gel, while the longest pieces of DNA will remain near the wells. (Top is negative, bottom is positive) Which lane matches each description below: This lane contains the longest DNA fragment? Lane 1 This lane contains the shortest DNA fragment? Lane 2 This lane contains a 1500 base pair DNA fragment? Lane 3 Explain how size of DNA piece is related to the distance traveled in agarose. -Although all DNA molecules are attracted to, and thus move toward, a positive electrode, their movement is impeded by the agarose matrix, causing larger pieces to take a longer time to move. -The end results is that during AGE, all of the DNA will separated based on the size of the molecules. This causes the different length RFLPs to be separated into distinct bands- the RFLP fingerprint. Estimate dna fragment size by comparison to DNA ladder (also check this, on pg 104) As the DNA samples move through the gel, distinct banding patterns become more and more visible as a dye in the Green Gel agarose fluorescence where its complexed with the DNA fragments of the RFLP. This allows bands from the known sample (patient) to be compared to unknown banding patterns (environmental samples) in order to determine a match for both number of bands and distance of migration on the gel. Compare RFLP( and define it) with PCR for DNA fingerprinting. RFLP: restriction fragments length polymorphism of differing lengths, allow the direct comparison of DNA of 2 different organisms at a specific site. PCR: Polymerase chain reaction; technique that allows molecular biologists to make many copies of a particular gene What is the function of the following: Agarose- A diagnostic tool to visualize the fragments TAE or TBE buffer- This is a buffer solution containing a mixture of Tris base, acetic acid and EDTA. It is used for the separation of nucleic acids such as, DNA and RNA. Loading dye- For visual tracking of DNA migration during electrophoresis (50 bp) and contains sucrose to increase the density of the solution UV light box- Used to visualize the DNA fragments and pair up with suspect