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

Laboratory Exam II (Example I)

Key to Laboratory Exam II – December 10-11, 2003

1. Define:

Cultured food – Cultured foods are those containing live microbial cultures and include dairy products such as cheese, yogurt, buttermilk and sour cream.

Acetoin – Acetoin (acetylmethylcarbinol) is an organic compound formed during the fermentation of glucose by certain heterofermentative bacteria, such as members of the genus Enterobacter. The presence of acetoin is indicated by a wine-red or pink color when 1 mL of a 48 hour broth culture is mixed with 18 drops of Barrit’s reagents A and B and then shaken vigorously for 1 minute, allowed to rest for 10-15 minutes and then shaken again.

Beta-hemolysis – Bacteria capable of beta-hemolysis will cause the complete lysis of red blood cells present in blood agar. Beta-hemolysis is characterized by the presence of clear zones behind colonies growing on blood agar plates.

2. Lactobacillus and Leuconostoc/ fermentative/ lactic acid

3. Saccharomyces cerevisiae/ carbon dioxide/ ethanol

4. Unripened cheese/ secondary flora (fungus)

5. Lactococcus lactis/ curd/ lactic acid

6. Streptococcus/ Lactobacillus/ acetaldehyde

7. Colon cancer and high serum cholesterol levels.

8. Replica plating/ time and materials

9. Staphylococcus/ mannitol/ The answer is variable here and require observation of the materials provided. Organisms that can ferment mannitol and form acid will cause the medium to turn yellow, those that cannot will grow without causing a change in the medium color (it will remain red).

10. Selective/ differential/ lactose/ Answers are variable here and require observation of the 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.

11. 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 whether they are alive or dead.

12. 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.

13. 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). Tubes containing fermentative organisms will be yellow.

14. 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 capable of fermenting the carbohydrate will grow (using the peptone in the medium) but will not cause a color change, so the medium will remain red.

15. Methyl red/ acid/ Answer is variable here and requires observation of data. The tube containing red-colored broth is positive. The tube containing organisms not capable of acid production will turn yellow when the methyl red indicator is added.

16. Amino acid/ hydrogen sulfide/ iron sulfide/ Answer is variable here and requires observation of data. Tubes containing black precipitate are positive.

17. Motility/ tryptophan (an amino acid)

18. Answers are variable here. Be familiar with the pH indicators used in enzymatic test media (phenol red in carbohydrate deeps, TSI and urea, bromothymal blue in citrate and O/F tubes and bromocresol purple in the amino acid decarboxylation test).

19. Decarboxylate/ amine/ 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.

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

21. Citrate (citric acid)/ Answer is variable and requires the observation of data/ alkaline

22. Ammonia/ Answer is variable and requires the observation of data.

23. 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.

24. Oxidase/ cytochrome C

25. 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.

26. Hydrogen peroxide (3%)/ A positive catalase reaction is indicated by the formation of bubbles.

27. Define:

Genomics – Genomics is the science or study of genomes from a large number of organisms all at once, or as a single unit. This requires the acquisition of nucleotide sequence data from multiple organisms and computer technology allowing this data to be stored, accessed, compared and analyzed.

Bacteriophage – A bacteriophage is a virus that infects bacteria. Those used in the laboratory included bacteriophage X-174 and T2.

Taq polymerase – Taq polymerase is an enzyme formed by hyperthermophilic bacteria identified as Thermus aquaticus. This DNA-dependent DNA polymerase is commonly used to catalyze DNA synthesis in polymerase chain reactions because it is thermostable, i.e., able to function after multiple exposures to the high temperatures (94 degrees centigrade) used to denature DNA.

28. Polymerase chain reaction/ DNA

29. 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

30. 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’.

31. Plasmids/ ampicillin (an antibiotic)

32. Restriction endonucleases/ EcoRI/ cohesive termini/ four (4)

33. Modification (often methylases)

34. DafEII

35. Restriction fragment length polymorphism/ RFLP patterns are created by cutting segments of DNA with restriction endonucleases/ This answer is variable and requires that the student compare the unknown RFLP pattern with labeled patterns to determine which one it matches with.

36. 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.

37. Keep track of the DNA samples as they moved down the gel. Because the dye samples move at a pace similar to our DNA samples, they could be used to track the progress of the DNA as it moved.

38. 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.

39. National Center for Biotechnology Information/ To determine the identity of the organisms the nucleotide sequence came from.

40. 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.

41. Transformation/ plasmid DNA/ There is far more growth on the TSA plates than on the TSA-AMP plates because almost all of the cells plated on the TSA were able to grow. The bacteria were not damaged by centrifugation or ice cold calcium chloride treatments. Only transformed cells,i.e., those carrying plasmid DNA, were able to grow on the TSA plates and they represnt only about 10% of the original population.

42. Mid-log (middle of their exponential or logarithmic growth phase)/ competent/ calcium chloride

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

44. 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.

45. 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.

46. Prodigiosin/ Answer is variable here and requires observation of data. The plate containing bright, red-orange colored colonies was kept at room temperature.

47. Ultra violet/ The overall effect of UV radiation on bacteria is cell death. Most of the bacteria on plates exposed to UV radiation for one minute were killed. Ultra violet radiation cannot be used to control pathogens inside cardboard or plastic containers because UV radiation has poor penetrating ability.

48. Disinfectants

49. 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.

50. The MIC (minimal inhibitory concentration) would be found at the outermost edge of the zone of inhibition.

51. The presence of 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. Tubes containing clear, red-colored medium with no gas in the Durham tube indicate negative results.

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.

54. Serology

55. Ouchterlony test/ 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. Alleles/ homozygous

57. Agglutinogens/ hemagglutination

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/ antibodies

62. Answers are variable and require observation of white blood cells under magnification. Note - the five different types of white blood cells are illustrated on the web site.

63. France

64. Blood/ hepatitis