Practical Issues of Microbiology - Coursework Example

? Microbiology Practical Describe the full theory behind the catalase test. Why must plain agar be used? Catalase is an enzyme that converts hydrogen peroxide to water and oxygen. Aerobic and facultative organisms have developed various protective mechanisms against the toxic forms of oxygen (especially superoxide radicals as they can potentially inactivate vital cell components). Enzyme known as superoxide dismutase, eliminates superoxide radicals by enhancing the rate of reaction. During the process toxic substance hydrogen peroxide (H2O2) and hydroxyl radical (OH-) are produced, which in turn is dissipated by catalase and peroxide enzymes respectively. Catalase test is used to identify organisms that are capable of producing catalase enzyme. The enzyme converts toxic H2O2 into water and oxygen and thereby prevent the formation of highly dangerous hydroxyl radicals. 2H2O2 Catalase 2H2O + O2 To test catalase activity, the culture must be of 18- 24 hour as old culture lose catalase activity and may display false negative test. The emerging bubbles in catalase positive test is due to the formation of oxygen gas. Aerobic bacteria in general display catalase positive test while most of the anaerobes are deficient in catalase enzyme indicating the sensitivity of anaerobic bacteria towards oxygen. E.g. Catalase positive organisms are Micrococcus spp. and Staphylococcus spp. Catalase negative organisms are Enterococcus spp. and Streptococcus spp. Bacto Agar or plain agar is used for the study as it contains calcium and magnesium. On the contrary, blood agar are unsuitable to carryout catalase test as blood contains catalase (Web: Catalase test, n.d.; Pelczar, 1993). 2. Explain the principle of Oxidase test. Oxidase test is a test for the presence of cytochrome oxidase (playing vital role in electron transport chain or ETC) which catalyses oxidation of reduced cytochrome by oxygen. Cytochrome oxidase transports electrons from the ETC to oxygen, the terminal electron acceptor. The principle is utilized to test the organisms for the production of cytochrome oxidase. To test this ability of microorganisms, artificial electron donors and acceptors are provided. As soon as the electron donor (1.0-1.5 percent solution of tetramethyl p-phenylene diamine hydrochloride) is poured over the colonies, it is oxidized by cytochrome oxidase. Oxidase positive colonies become maroon, purple and black in 10-30 minutes (Pelczar MJ. 1993). E.g. Members of Pseudomonadaceae and Enterobacteriaceae are oxidase positive (Pelczar, 1993). 3. Explain how Oxoid Chromogenic UTI media (Clear) works. Oxoid Chromogenic UTI media is a nonselective medium to isolate, differentiate and enumerate the pathogens prevalent in urinary tract and are responsible for urinary tract infection. The medium facilitates the differentiation and identification of E. coli and Enterococcus without undergoing confirmatory tests (Web. Dehydrated Culture Media, n.d.). The principle utilized is based on the difference in susceptibility prototypes of the microorganisms. For the efficacy of antimicrobial therapy, the organism must be identified to the species level. However, the prevalent species Enterococcus spp. and Escherichia spp. produce enzymes for the metabolism of glucosidase or lactose or both and aids identification. On the contrary other microorganisms may not produce enzyme for such substrate fermentation (Web. Dehydrated Culture Media, n.d.). The chromogen mixture present in the Oxoid Chromogenic UTI media (Clear) comprises artificial substrates or chromogens which are responsible for releasing different colored compounds when specific microbial enzymes released by Enterococcus spp. and Escherichia spp act on substrate molecules, thereby a differentiation process works for the detection of certain species or a group of microorganism (Isenberg and Garcia, 2004). For instance the chromogen X-glucoside is a substrate for ?-glucosidase enzyme of enterococci which form blue colonies while chromogen red-galactoside is utilized by enzyme ?-galactoside generated by E. coli detected by the formation of pink colonies. On the other hand some coliforms are responsible for utilizing both the substrates and form purple colonies. The medium is incorporated with tryptophan which tests the tryptophan deaminase activity (TDA) as a result halos are observed around the colonies of Morganella, Proteus and Providencia spp (Web. Dehydrated Culture Media, n.d.). 4. What are ‘germ tubes’? What other methods can be used to differentiate Candida species? C. albicans displays the ability to proliferate either as a yeast form or a mycelial form. A germ tube can be induced from the yeast cell in vitro and proliferated into the mycelial form. Such morphological transformation is directly influenced by the environmental factors encompassing appropriate pH (Pollack, 1987), temperature (Odds, 1985) and availability of nutrition (Casanova et al., 1997). The germ tube structure could be explained from the structure of elongated daughter cells which originate from the round mother cells devoid of any constriction in the origin. Such tubular arrangement is called germ tube while the constricted hyphae around the mother cell or the generating cell is called pseudohyphae (Kim et al., 2002). With the prevailing Candida infections, identification of Candida isolates to species level require rapid pace. The most frequent fungal infection in humans is caused by C. albicans. It is responsible for causing morbidity and mortality in post-operative or immunocompromised individuals (Jarvis, 1995). However, Candida can be isolated with classical methods such as germ tube and chlamydospore formation together with sugar assimilation tests (Binkandi et al., 1998). Formation of germ tube is induced when isolation of yeast is done in serum at 37? C for the period of 2-3 hour (Lee et al., 1999). Studies reveal that only C. albicans is able to produce germ tubes at 39? C in serum free YEPD medium facilitating easy identification of the C. albicans from other species of Candida (Kim et al., 2002). 5. Explain how the BSAC method of antimicrobial susceptibility testing works. What are the advantages and disadvantages of this method? In 2001, British Society for Antimicrobial Chemotherapy (BSAC) published standardized disc susceptibility testing method. However, numerous changes have been recommended including the coordination of MIC breakpoints in Europe. The work has been carried out by European Committee on Antimicrobial Susceptibility Testing (EUCAST) with the help and alliance of national committees (Andrews, 2005). Advantages 1. Acceptable inoculum density has been defined from lightest acceptable to ideal to heaviest acceptable. 2. Measurement of the diameter of the zones of inhibition to the nearest millimeter. 3. Ignore the tiny colonies at the edge of zone, as in the Proteus due to swarming. 4. If minute colonies are present within the zone of inhibition they are required to be sub- cultured and test should be repeated for confirmation. 5. Zones of inhibition must be measured in good light. 6. Zone of inhibitions should be in acceptable range. 7. Template must be used for interpreting zone diameters (BSAC Methods for Antimicrobial Susceptibility Testing, n.d). Disadvantage The zone of inhibition should be measured using template (BSAC Methods for Antimicrobial Susceptibility Testing, n.d.). 6. Explain the function of chlamydospores. Chlamydospores are the fungal spores formed during adverse environmental conditions. They are separated by septa with an opening called pore through which their cytoplasm remains connected. Their characteristic feature helps organism to withstand adverse climatic condition and is acts as a tool in the determination of fungal pathogen C. albicans and Canditda dubliniensis. Formation of chlamydospores is controlled by environmental signals and thereby helps in developing and understanding towards the regulation of morphogenetic program and developmental process (Staib & Morschhauser, 2007). 7 Describe how the Etest MIC method works. What are the advantages and disadvantages over the BSAC method? Minimum inhibitory concentration (MIC), is regarded as the lowest or minimum concentration of any antimicrobial that can potentially inhibit the noticeable growth of an organism when inoculated for a substantial duration (overnight) (Andrews, 2001). Care is desired to procure accurate information such as size of the inoculum, growth phase of the organism, agar or broth used together with the condition in which the antimicrobial agents are stored. On the other hand, E-test is a new method of MIC used for direct antimicrobial susceptibility quantification involving agar diffusion method. It involves a plastic strip of 50X5 mm impregnated with gradient of antibiotic to be tested. The strip is applied to the surface of an agar plate inoculated with test organism followed by incubation. An elliptical zone of inhibition is formed. the minimum inhibitory concentration is read in the region where the zone of inhibition interrupts the strip. The method has an advantage of swiftness of getting information about sensitivity of the single/ fastidious species and therefore gaining prevalence for critical cases encompassing endocarditis, septicemia, aerobic infection and meningitis (Web: 208 Microbiology, n.d.). 8. Explain the principle behind the RapID NF Plus test. Gram negative bacteria other than Enterobacteriaceae could be identified on the basis of their glucose fermenting and non-fermenting ability. It is based on the enzyme technology. The method is a single step which involves inoculation of reactants in a dehydrated form into the wells of tray. On the basis of the enzyme produce by the organisms their identification is marked. E.g. anaerobic microorganisms, enterics, coryneform gram positive rods, streptococci, nonfermenters and yeast in less than 4 hours (Web: RapID™ System, n.d.). 9. Explain the theory behind the tube coagulase test. Coagulase is an enzyme produced by strains of S. aureus. The enzyme acts within host tissues to convert fibrinogen to fibrin. The fibrin meshwork which is formed by this conversion surrounds the bacterial cells or the affected cells or tissues, thereby preventing the microbe from nonspecific host immune responses encompassing phagocytosis and other action of normal serum. In the coagulase tube test free coagulase is measured. The coagulase secreted by the strain of S. aureus act with coagulase reacting factor (CRF) present in the plasma to form a complex called thrombin. A suspension of the test organism in citrated plasma is prepared and the inoculate plasma is then periodically examined for coagulation or fibrin formation. Formation of clot within 4 hours duration is inferred as positive result and also the indicative of a virulent S. aureus organism on the other hand absence of coagulation after 24 hours of incubation is an indication of negative test and avirulent strain such as S. epidermidis (Pelczar, 1993; Web: Coagulase Test Protocol, n.d.) 10. With reference to Appendix 1, complete tables 1 and 2 and answer the following question: List the antibiotics that could be used to treat an infection caused by each organism. For each organism which one antibiotic would you chose and why? NB organism A (S.aureus) was isolated from a patient with an abscess and fever, organism D (E.coli) from a patient with a urinary tract infection. Table 1 Organism Penicillin Zone size S/R Gentamicin Zone Size S/R Vancomycin Zone Size S/R Fusidic Acid Zone Size S/R Clarithromycin Zone Size S/R Mupirocin Zone Size S/R A 10 R 21 S 15 S 22 R 24 S 27 S For the patient with an abscess and fever the organism A was reported to be S. aureus, which is sensitive to Gentamicin, Vancomycin, Clarithromycin and Mupirocin. The organism is found to be resistant to Penicillin and Fusidic Acid. One antibiotic which should be selected is Clarithromycin (CLR2) as the zone of inhibition obtained for this antibiotic is 24 mm at the small concentration of 2 µg, indicating that even at low concentration the drug is efficacious in restricting the proliferation of S. aureus. Table 2 Organism Trimethoprim Zone size S/R Nitrofurantoin Zone Size S/R Ciprofloxacin Zone Size S/R Augmentin Zone Size S/R Ampicillin Zone Size S/R D 15 R 26 S 27 S 16 S 6 R For the patient with urinary tract infection (UTI) organism D was reported to be E. coli, which is sensitive to Nitrofurantoin, Ciprofloxacin and Augmentin. The organism is found to be resistant to Trimethoprim and Ampicillin. One antibiotic which should be selected is Ciprofloxacin (C1) as the zone of inhibition obtained for this antibiotic is 27 mm at the small concentration of 1 µg, indicating that even at low concentration the drug is efficacious in restricting the proliferation of E.coli. 11. Describe what the staph latex test measures and how does this differ to the tube coagulase test? A rapid identification and differentiation of S. aureus from other strains of staphylococci is highly imperative as coagulase negative staphylococci could also be virulent. There are many false results found to be associated with the coagulase test which hinder in the identification of the pathogenic staphylococci organisms. Extracellular staphylocoagulase is detected by means of tube coagulase test while bound coagulase or the clumping factor on the cell surface is determined by the slide coagulase test (Web: Laboratory Procedure, n.d.). Latex test is a latex agglutination test utilized to differentiate staphylococci possessing clumping factor alone or together with Protein A of possess Protein A alone. S. aureus possesses staphylocoagulase enzyme either in the bound form of in extracellular matrix. Almost 95 percent of the human stains of S aureus possesses Protein A which potentially bind to the Fc part of the immunoglobulin G molecule (IgG). The property of the microorganism is utilized for its detection. Latex particles are coated with the IgG as well as human fibrinogen. A cross-linking reaction is observed when the latex reagent is mixed with the colonies of staphylococci with Protein A or the clumping factor or both resulting in latex particle agglutination. However, no agglutination is observed when neither clumping factor nor Protein A is present and the test is considered as negative. E.g. Latex agglutination is positive for S. aureus while negative for S. epidermidis (Web: Laboratory Procedure, n.d.). 12.How does the Streptococcal Lancefield Grouping test work? Streptococcal Lancefield grouping works on the principle of agglutination. A six different combinations of antigen and antibody are used in this test. Every species possesses specific antigen, a carbohydrate molecule on the cell wall of the streptococci which is recognized by one of the antibodies present in the six combinations. The test involves an analysis of unknown species of Streptococcus as soon as it is subjected to Lancefield grouping test. Latex agglutination immunological assay involves- application of a set quantity of enzyme-extracted Streptococcus on the six rings of the plastic card followed by coating of a specific Lancefield group antibody on latex beads into each of the six wells. As antigen-antibody reaction is highly specific, the carbohydrate antigen present on the cell wall of the unknown Streptococcus would react with only one antibody present in one of the rings. The cross-linking reaction results in agglutination as soon as the latex beads layered with antibody particularly combine with the antigen. Agglutination or clumping of the latex beads occurs only in one well. By establishing the co-relation between Lancefield group and unknown species of Streptococcus, the unknown strain is assigned a particular group. The ease, specificity, sensitivity and convenience has made agglutination test prevalent for the identification of diverse pathogens (Web: Bacterial agglutination, n.d.). References 1. Andrews, J.M. 2001. Determination of minimum inhibitory concentrations. J Antimicrob Chemother. 48:5-16. 2. Andrews, J.M. 2005. BSAC standardized disc susceptibility testing method. J. of Antimicrobial Chemotherapy. 56: 60-76. 3. Bacterial agglutination. n.d. [online] Available at: . [Accessed 5 September 2013]. 4. BSAC Methods for Antimicrobial Susceptibility Testing, n.d. [online]. Available at: . [Accessed 5 September 2013]. 5. Bikandi, J., San Millan, R., Moragues, M.D. et al. 1998. Rapid identification of Candida dubliniensis by indirect immunofluorescence based on differential localization of antigens on C. dubliniensis blastospores and Candida albicans germ tubes. J Clin Microbiol. 36:2428–2433. 6. Casanova, M., Cervera, A.M., Gozalbo, D., Martinez, J.P. 1997. Hemin induces germ tube formation. Candida albicans. Infect Immun. 65:4360–4364. 7. Catalase Test. [online] Available at: [Accessed 6 September 2013]. 8. Coagulase Test Protocol, n.d. [online] Available at: . [Accessed 5 September 2013]. 9. Dehydrated Culture Media. n.d. [online] Available at: . [Accessed 6 September 2013]. 10. Isenberg and Garcia (ed.). 2004 (update, 2007). Clinical microbiology procedures handbook, 2nd ed. American Society for Microbiology, Washington, DC. 11. Jarvis, W.R. 1995. Epidemiology of nosocomial fungal infections with emphasis on Candida species. Clin Infect Dis. 20:1526–1530. 12. Kim, D., Shin, W.S., Lee, K.H., Kim, K., Park, J.Y., Koh, C.M. 2002. Rapid differentiation of Candida albicans from other Candida species using its unique germ tube formation at 39? C. Yeast. 19: 957-962. 13. Laboratory Procedure. n.d. [online] Available at: . [Accessed 5 September 2013]. 14. Lee, K.H., Shin, W.S., Kim, D., Koh, C.M. 1999. The presumptive identification of Candida albicans with germ tube induced by high temperature. Yon Med J. 40:420–424. 15. Odds, F.C. 1985. Morphogenesis in Candida albicans. Crit Rev Microbiol. 12: 45–93. 16. Pelczar, M.J. 1993. Microbiology: Concepts and Applications. Study Gd ed. Mcgraw-Hill College. 17. Pollack, J.H., Hashimoto, T. 1987. The role of glucose in the pH regulation of germ-tube formation. Candida albicans. J Gen Microbiol. 133: 415–424. 18. Staib, B., Morschhauser, J. 2007. Chlamydospore formation in Candida albicans and Candida dubliniensis- an enigmatic developmental programme. Mycoses, 50(1):1-12. 19. 208 Microbiology section 2. n.d. [online] Available at: . [Accessed 6th September 2013]. 20. RapID™ System, n.d. [online] Available at: . [Accessed 5th September 2013]. Read More