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Chapter 38. Microbiology Experiment

Recommended articles: 【Biology】 Biology Table of Contents, 【Biology】 Chapter 37. Biology Experiment


1. Streak Plate Method

2. Gram Staining

3. Antibiotic Susceptibility Test

4. Ames Test

5. Cross-Feeding Experiment

6. Interrupted Mating Technique


a. Microbiological Testing and Genetic Monitoring



1. Streak Plate Method (streak plate)

⑴ Definition: Method used for isolating a single colony

⑵ Procedure


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Figure. 1. Streak Plate Method Procedure


① 1st. Flame to sterilize inoculation loop.

② 2nd. Collect bacteria onto loop.

③ 3rd. Inoculate plate.

④ 4th. Inoculation procedure (flame loop between streaks)

⑤ 5th. Formation of discrete colonies



2. Gram Staining

⑴ Gram Staining: Staining technique developed by Gram

⑵ Gram Staining: Crystal violet → Safranin O

① 1st. Heat-fix the bacteria to the slide glass by passing it 2–3 times through an alcohol lamp flame.

② 2nd. Flood the smeared sample with crystal violet stain and leave for 20 seconds.

○ At this point, all cells appear purple.

○ Because crystal violet is a basic dye carrying a (+) charge, it binds strongly to the teichoic acids of Gram-positive bacteria.

③ 3rd. Briefly rinse the stain off in a tube filled with distilled water and remove any remaining moisture.

④ 4th. Flood the smear with iodine solution (I2–KI) and leave for 1 minute.

○ Iodine forms a crystal violet–iodine complex, preventing color change and decolorization (acts as a mordant agent).

⑤ 5th. Dip the slide in a tube containing 95% EtOH to wash off the iodine solution, moving it in and out.

○ Only Gram-positive bacteria remain purple.

○ In Gram-positive bacteria, alcohol shrinks the cell wall, trapping the I2–crystal violet complex within the cytoplasm.

⑥ 6th. Gently rinse the slide by dipping it in distilled water to stop decolorization.

⑦ 7th. Flood the smear with safranin and leave for 1 minute.

○ Gram-positive bacteria remain purple.

○ Gram-negative bacteria take up safranin and appear red.

⑧ 8th. Rinse carefully with water for a few seconds, then remove moisture.

⑶ Gram-positive bacteria

① Structure: protoplasm → cell membrane → thick peptidoglycan layer.

② Peptidoglycan: alternating N-acetylglucosamine (N-AG) and N-acetylmuramic acid (N-AM) linked by β 1→4 bonds.

③ Form endospores, enabling resistance to high temperature and pressure.

④ Teichoic acids protruding beyond the peptidoglycan are covalently linked and observed only in Gram-positives.

○ Teichoic acids contain glutamic acid and thus carry a negative charge.

Examples: Bacillus subtilis, Bacillus anthracis, Staphylococcus.

⑷ Gram-negative bacteria

① Structure: protoplasm → cell membrane → periplasmic space → thin peptidoglycan → periplasmic space → LPS, outer membrane.

② LPS (lipopolysaccharide): toxic and a target of the immune system.

○ The toxicity of botox is also attributed to LPS.

○ Penicillin cannot pass through LPS.

○ Causes blood coagulation and high fever.

③ Outer membrane: contains porins and is more permeable than the cell membrane.

④ Periplasmic space:

○ The compartment separating the peptidoglycan layer from the cell membrane and outer membrane.

○ A cellular space important for functions such as metabolism and transport.

Examples: Escherichia coli, Helicobacter pylori.



3. Antibiotic Susceptibility Test (AST)

⑴ Overview

① CFU (colony forming unit): Number of colonies formed by bacteria or fungal spores in a sample

○ Formula can be used if necessary: Sample’s CFU = CFU of diluted sample × dilution factor

② MIC (minimum inhibitory concentration): Minimum concentration of an antibiotic that inhibits microbial growth

③ MBC (minimum bactericidal concentration): Minimum concentration at which no microorganisms appear even after subculture on a nutrient medium

○ Essentially, the minimum concentration at which bacteria are killed

○ MBC ≤ MIC

④ Can provide information about the bacteria’s identity

⑵ Broth dilution test (broth microdilution method)

① 1st. Prepare varying antibiotic concentrations, then add microbes at the same CFU.

② 2nd. MIC determination: after overnight incubation, determine the lowest antibiotic concentration at which no visible turbidity appears.

③ 3rd. MBC determination: subculture each sample onto nutrient medium and determine the lowest antibiotic concentration that yields no colonies.

④ A limitation is that it relies on optical density.

⑶ Disc diffusion method

① Measure the diameter of the inhibition zone produced by the drug.

② Follow the EUCAST guidelines.

⑷ Epsilometer test (E-test)

① By generating a gradient of antibiotic concentrations on a single plate, the MIC can be measured directly and conveniently.

② The MBC cannot be measured.

⑸ SCMA (single-cell morphological analysis)

① Motivation: when determining the MIC from blood samples, the result strongly depends on the initial bacterial concentration.

② MIC can be determined easily using optical microscopy.



4. Ames Test

⑴ Overview

① Genetic toxicity assessment method developed by Dr. Bruce Ames’ team at the University of California, Berkeley

② Used to identify mutagenic and carcinogenic agents: Shows a strong correlation with carcinogenicity test results; quick and simple

③ Commonly used for screening before phase 1 of drug development

④ Verify if back mutation occurs using auxotrophic mutations

○ Reversion mutation: Requires the same mutation mechanism, typically a point mutation

⑤ Uses Salmonella typhimurium sensitive to mutagens

⑥ Colonies decrease when using strains with normal DNA repair functions

⑦ Commonly used test strains

○ Salmonella typhimurium TA98

○ Salmonella typhimurium TA100

○ Salmonella typhimurium TA1535

○ Salmonella typhimurium TA1537

○ Escherichia coli WP2 uvrA (pKM101)

⑵ Experimental procedure

① Prepare S. typhimurium mutants by strain.

② Prepare solid minimal-medium agar plates and spread the strains uniformly.

③ Place, at the center of each plate, a disk impregnated with one of the following:

○ Water: negative control

○ Liver enzymes: negative control

○ Chemical: experimental group

○ Water + liver enzymes: negative control

○ Chemical + liver enzymes: If the colony count is similar to that of the chemical-only group, the chemical is a direct mutagen, i.e., it does not require metabolic activation in the pathway to become mutagenic.

④ Incubate at 37 °C for 16 hours.

⑤ If colonies appear, they are revertants that have lost the auxotrophic requirement due to a reverse mutation.


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Figure 2. Example of Ames Test Results



5. Cross-Feeding Experiment

⑴ Overview: Experiment to determine the sequence of metabolic actions

⑵ Premises

① Mutant A has a mutation in gene a, B has a mutation in gene b, C has a mutation in gene c

○ For convenience, let’s denote the substance produced by gene x as X

② In mutants unable to synthesize proline, the intermediates of proline biosynthesis accumulate inside the cell.

③ The accumulated proline-biosynthesis intermediates diffuse through the agar medium.

⑶ Results

① 22 °C: A—only some strains grow well; B—all strains grow well; C—no strains grow.

② 30 °C: A—no strains grow; B and C—all strains grow well.

③ 42 °C: A—no strains grow; B—only some strains grow well; C—all strains grow well.


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Figure 3. Experiment under normal conditions


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Figure 4. Experiment for temperature dependency assessment


⑷ Interpretation

① Analysis at 22 °C

○ Substance C acts later than substance A: ?? → A → C; the A strain can survive because it has enzyme c.

○ The B strain is prototrophic (non-auxotrophic).

② At 30 °C

○ The B strain is prototrophic.

○ The C strain is prototrophic.

○ Given point (②) above, the C strain does not accumulate the substance; therefore nothing is transferred to the A strain.

③ At 42 °C

○ The B strain is auxotrophic; it is a temperature-sensitive mutant.

○ Substance A acts later than substance B: ?? → B → A; the B strain can survive because it has enzyme a.

○ The C strain is prototrophic.

⑸ Tips for interpretation

① The later an enzyme acts in the pathway, the more severe the mutation.

② From the 22 °C experiment, we infer A → C.

③ From the 42 °C experiment, we infer B → C.



**6. Interrupted Mating Experiment


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Figure 5. Interrupted Mating Experiment Process


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Figure 6. Interrupted Mating Experiment Results


⑴ 1st. Hfr strain has strs leu+ thr+ azir tonr lac+ gal+

① Conditions: Antibiotic sensitivity, nutritional auxotrophy

⑵ 2nd. F- strain has strr leu- thr- azis tons lac- gal-

① Conditions: Antibiotic resistance, nutritional requirements

⑶ 3rd. Cultured in a medium with streptomycin and without threonine and leucine: Only strains with strr leu+ thr+ survive

⑷ 4th. After 8 minutes of mating, azir survives additionally: Adjacent to strr leu+ thr+, there’s azir

⑸ 5th. After 10 minutes of mating, tons survives additionally: Adjacent to strr leu+ thr+ azir, there’s tons

⑹ 6th. After 16 minutes of mating, lac+ survives additionally: Adjacent to strr leu+ thr+ azir tons, there’s lac+

⑺ 7th. After 25 minutes of mating, gal+ survives additionally: Adjacent to strr leu+ thr+ azir tons lac+, there’s gal+

⑻ In this way, it is applied in making genetic map.



Input: 2019.03.17 16:28

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