Bio. Sci. 4 | Microbiology Study Guide "Answers"

Laboratory Exam II (Example II)

Key to Laboratory Exam II – May 19-20, 2004

1. Define:

Fermentation – Fermentation can be defined as the anaerobic decomposition of organic compounds (especially carbohydrates) that involves an organic compound (usually pyruvic acid) as the final electron acceptor.

Decarboxylation – Chemical reactions that involve the removal of carboxyl groups from organic compounds are called decarboxylation reactions. Testing the ability of bacteria to decarboxylate lysine (or ornithine) and produce an amine (and carbon dioxide) can be useful in their identification.

Catalase – The catalase test is used to detect the presence of catalase enzymes, i.e., enzymes catalyzing chemical reactions resulting in the conversion of hydrogen peroxide to water and oxygen gas. Bacteria subjected to a catalase test are exposed to 3% hydrogen peroxide, and will release bubbles of oxygen gas if they make catalase enzymes.

2. Lactobacillus and Leuconostoc/ lactic acid

3. Saccharomyces cerevisiae/ carbon dioxide/ ethanol

4. Lactococcus lactis/ lactic acid

5. Secondary flora (usually fungus)

6. Rennet/ curd/ whey

7. Streptococcus / Lactobacillus/ acetaldehyde

8. Replica plating/ transfer of a colony pattern (many colonies) from one plate to a series of plates, quickly and easily.

9. Selective/ differential/ lactose/ Answers are variable here and require observation of materials provided. Lactose fermenting colonies will be bright pink on MAC and yellow on T-7, while non-fermenting forms will be pale tan or yellowish on MAC and blue on T-7.

10. Mannitol/ phenol red

11. Serial dilution/ Answer is variable here and requires that the colonies present be counted and their number be multiplied by the dilution factor expressed as a positive number (drop the minus sign on the exponent). Remember to write your number in correct scientific notation. / Only viable cells are capable of growing into colonies that can be seen and counted. Individual bacteria (cells) cannot been seen with the naked eye, regardless of whether they are alive or dead.

12. Respiratory/ Answer is variable, but tubes containing respiratory organisms will be green under the vaspar seal and inside the Durham tube. There may be yellow color toward the top of the medium in the unsealed tube due to the production of aerobic acid (acids formed in association with glycolysis and the Krebs cycle). Fermentative organisms will cause the medium in both tubes to turn yellow.

13. Answer is variable here and requires observation of data. The tube containing fermentative organisms will be yellow in color due to acid production and will show evidence of gas production (split or lifted agar, bubbles, or cracks in the medium). Organisms not able to ferment the sugar present will grow (using peptone as a nutrient), but will not cause a color change in the medium (it will stay red).

14. Methyl red/ acid/ Answer is variable here and requires observation of data. The tube containing red-colored broth is positive for acid production while the one containing yellow-colored broth is negative.

15. Acetoin or acetylmethylcarbinol/ Answer is variable here and requires observation of data. The tube containing red-colored broth after addition of Barrit’s reagents and shaking, is positive for acetoin, while the tube containing yellow-colored medium is negative.

16. Amino acid/ hydrogen sulfide

17. Answer is variable here and requires observation of data. The bottle containing Kovac’s reagent is used to test for the presence of indole. / Indole is formed when bacteria catabolize the amino acid tryptophan. / Motility is indicated by the appearance of the stab line in the SIM medium. If the stab line is distinct and easy to see (clearly visible) the bacteria are not motile. If they are motile, they swim through the medium and the stab line will appear cloudy or vague. Bacteria that swim throughout the medium and form hydrogen sulfide will turn the medium black throughout the tube.

18. Amine/ Bromocresol purple/ Answer is variable here and requires observation of the data. If both tubes contain yellow-colored broth, there is no cadaverine present. If the control tube contains yellow-colored broth and the lysine tube contains purple-colored broth, cadaverine is present.

19. The vaspar seal is applied to create an anaerobic environment and to ensure glucose fermentation. The seal also keeps volatile amines in solution.

20. Citrate (citric acid)/ Bromothymal blue/ Answer is variable and requires the observation of data/ The blue-colored medium within the tube has become alkaline

21. Urease/ ammonia/ Answer is variable and requires the observation of data. If the medium becomes hot-pink in color, the urease test is positive. If the medium stays peach-colored or turns yellow, the urease test is negative.

22. Answer is variable here and requires the observation of data. Know how to interpret enzymatic test data and use an identification chart to determine the identity of bacteria.

23. Cytochrome C

24. Coagulase/ Answer is variable and requires observation of data. Tubes containing coagulated (solidified) plasma were inoculated with organisms capable of forming coagulase enzymes. Tubes containing liquid plasma were not.

25. Catalase

26. Define:

Polymerase Chain Reaction – The polymerase chain reaction is a process allowing specific segments of DNA to be amplified (reproduced about a million times) in vitro. The process requires a DNA template, oligonucleotide primers, thermal stable DNA polymerase enzymes and dNTPs. These materials are combined in a container and subjected to temperature modulations allowing for DNA denaturation, annealing of primers, and extension of new DNA strands in a cyclic manner repeated 35 to 40 times.

Electropherogram – An electropherogram is a chromatogram (visual record) composed of colored peaks arranged in a series. Each peak represents an accumulation of DNA fragments of a particular size that terminate with the same ddNTP (adenine, guanine, cytosine or thymine) carrying a fluorescent label (green, blue, yellow or red). The sequence of colored peaks indicates the sequence of bases in the DNA strand being studied.

Coliphage – A coliphage is a bacteriophage (virus that infects bacteria) capable of infecting Escherichia coli. In our laboratory we worked with two types of coliphage identified as X-174 and T2.

27. Taq polymerase/ The polymerase chain reaction requires that DNA samples be heated to around 94 degrees centigrade to break the hydrogen bonds holding the complimentary strands together (to denature the DNA), and this heat treatment is applied multiple times (35-40). Enzymes that were not stable at high temperatures would be denatured by this treatment. / Yellowstone National Park

28. Primers/ The primers determine which region of the template DNA strand will be amplified, because they hybridize with it (one at each end). The primers also provide the free 3’ ends necessary for DNA replication. Recall that DNA polymerase enzymes can only add nucleotides to the free or exposed 3’ ends of DNA strands, i.e., they build 5’ to 3’.

The region of DNA being amplified was the gene coding for 16S ribosomal-RNA (The gene itself was 16S ribosomal-DNA).

29. Plasmids/ cloning vectors

30. Restriction endonucleases

31. EcoRI/ AluI/ cohesive termini

32. Modification (often methylases)

33. NasTI

34. Electrophoresis/ Answer is variable here. Look at the gel to determine if or not the wells are located at the end of the chamber farthest from the anode (positive electrode). If they are, the gel is in the correct position; if they are not, then the gel is in the box backward. / If power is applied while the gel is positioned incorrectly relative to the electrodes, the DNA samples will travel toward the anode through the short end of the gel and will likely be lost in the buffer solution.

35. The dye is used to keep track of where the DNA is as it travels through the gel. The DNA is not visible, but the dye is, and most of our DNA samples will stay somewhere between the two dye bands.

36. Answers are variable here and require that you look at the banding patterns in the gel. First find the bacteriophage lambda DNA (the standard) , and then use it to determine the sizes of the plasmids present. The plasmid pUC19 is about 2000 bp and pBR322 is about 4000 bp. Bands formed by DNA from these plasmids will therfore line up with the 2000 bp and 4000 bp bands of the bacteriophage lambda standard. The PCR product DNA is about 1500 bp, so will be smaller (will move farther down the gel) than pUC19.

37. Restriction fragment length polymorphism/ This answer is variable and requires that the student compare the unknown RFLP pattern with labeled patterns to determine which one it matches with./ DNA fingerprints

38. Genomics/ National Center for Biotechnology Information/ BLAST

39. Answer is variable here and requires that you look at the data provided. / The lineage (taxonomic information) is shown below the technical name (genus and specific epithet), listed just to the right of the word “organism” and will usually include the domain, phylum, class, order, family and genus names of the organisms being identified.

40. Competent cells/ Mid-log (middle of their exponential or logarithmic growth phase)/ The calcium was expected to bind with the cells giving their surfaces a slight positive charge. This would tend to increase the attachment of negatively-charged DNA molecules to the cell surfaces.

41. Transformation/ resistance to the antibiotic ampicillin/ All or nearly all of the E. coli cells present will grow on the TSA plates, while only transformed cells can grow on the TAS-AMP plates. Since transformation efficiency is only about 10%, fewer cells grew on the TSA-AMP plates inoculated with bacteria that had been given plasmid DNA. No growth was observed on the plate containing host cells without plasmid DNA.

42. Phage typing/ Answer is variable here and requires that students observe the data presented.

43. Plaques/ Answer is variable here and requires the plaques present be counted and their number multiplied by the dilution factor expressed as a positive number (drop the minus sign on the exponent). Remember to write your answer in correct scientific notation.

44. Answers are variable here and require some calculations. Look for a sharp increase in the number of plaque–forming units (pfu) to determine the time required for the infected cells to be lysed. This indicates the end of the latent period. To calculate the burst size, divide the maximum number of pfu present by the initial number of pfu present.

45. Prodigiosin/ Answer is variable here and requires observation of data. The plate containing bright, red-orange colored colonies was kept at room temperature, the plate with white colonies was grown in the 37 degree C incubator.

46. Ultra violet/ die/ poor penetrating.

47. Antiseptics/ Gram-positive

48. Zones of inhibition/ Answers are variable and require observation of data. The zones of inhibition must be measured (diameter in millimeters) and the values obtained compared to the chart provided to determine sensitivity.

49. Minimal inhibitory concentration or MIC/ The MIC would be found at the outermost edge of the zone of inhibition.

50. The answer is variable and requires the observation of data. Broad-spectrum drugs will be effective against multiple types of bacteria (both Gram-positive and Gram-negative, while narrow-spectrum drugs will be effective against only a few.

51. Escherichia coli/ These bacteria live in the gut, are easy to grow in vitro, and are easy to test for.

52. Presumptive test/ Answer is variable and requires observation of data. Tubes containing medium that is cloudy, yellow in color and shows gas in the Durham tube are positive for lactose fermentation. These features are indicative of E. coli.

53. Confirmatory test/ Answer is variable and requires observation of data. Plates containing colonies that are dark with a metallic green sheen are indicative of E. coli and represent positive test results.

54. Serological

55. Precipitation reaction/ Answer is variable and requires observation of data. The presence of a precipitate band between the central well and any of the outer wells indicates that the serum sample being tested contains antibodies against the fungus.

56. Agglutinogens/ agglutinins/ hemagglutination

57. Alleles/ heterozygous

58. Answer is variable and requires observation of data. Remember that serum samples contain antibodies (agglutinins) that react with specific antigens (agglutinogens) . Anti-A serum reacts with A-antigens, anti-B with B-antigens and anti-D with Rh-antigens. Clumping (agglutination) indicates the antigen is present. The person could receive blood from a donor with type A, Rh-positive blood only if they were type A or AB and Rh-positive. Persons with type O or B blood and persons with Rh-negative blood could (and probably would) be carrying antibodies that would react with the antigens on the A, Rh-positive RBCs. In addition to causing agglutination, which would tend to disrupt circulation, the interaction of antibodies with the donated RBCs could trigger the complement cascade resulting in lysis of the donated RBCs.

59. Red blood cells designated as type O, Rh-negative do not have A, B or D antigens on their cell surfaces, so cannot react with anti-A, anti-B or anti-D (Rh) antibodies.

60. Enzyme-linked immunosorbent assay/ The presence of the enzyme.

61. Plasma

62. Answers are variable and require observation of white blood cells under magnification. Note that illustrated representations of five different types of WBCs are available on the web page.

63. France/ No, they called it LAV

64. Blood/ hepatitis B