8.Place PCR tube in thermal cycler, close lid, and start.
9.After 15 cycles, pause thermal cycler and transfer 7 àL to tube 15.
10. Place tube 15 on ice.
11. Return PCR tube to the thermal cycler and continue program.
13. Place tube 30 on ice.
12. After completing the last cycle, transfer 7 àL to tube 30.
✦ Theory
Ultraviolet radiation is part of the electromagnetic spec- trum, but with shorter, higher energy wavelengths than visible light. Prolonged exposure can be lethal to cells because when DNA absorbs UV radiation at 254 nm, the energy is used to form new covalent bonds between adjacent pyrimidines: cytosine-cytosine, cytosine-thymine, or thymine-thymine. Collectively, these are known as pyrimidine dimers, with thymine dimers being the most common. These dimers distort the DNA molecule and interfere with DNA replication and transcription (Figure 10-11).
Many bacteria have mechanisms to repair such DNA damage. Escherichia coli performs light repair or photo - reactivation, in which the repair enzyme, DNA photoly - ase, is activated by visible light (340–400 nm) and simply monomerizes the dimer by reversing the original reaction.
A second E. coli repair mechanism, excision repairor dark repair,involves a number of enzymes (Figure 10-12).
The thymine dimer distorts the sugar-phosphate back- bone of the strand. This is detected by an endo nuclease (UvrABC) that breaks two bonds—eight nucleo tides in the 5' direction from the dimer, and the other four nucleotides in the 3' direction. A helicase(UvrD) re moves the 13-nucleotide fragment (including the dimer), leaving single-stranded DNA. DNA polymerase I inserts the
appropriate complementary nucleotides in a 5' to 3' direction to make the molecule double-stranded again.
Finally, DNA ligasecloses the gap between the last nucleo - tide of the new segment and the first nucleotide of the
10-5 Ultraviolet Radiation Damage and Repair
P P
P S
T T
UV
S
P P
P S
T T
S
10-11 A THYMINEDIMER INONESTRAND OFDNA✦Thymine
dimers form when DNA absorbs UV radiation with a wavelength of 254 nm. The energy is used to form two new covalent bonds between the thymines, resulting in distor tion of the DNA strand.
The enzyme DNA photolyase can break this bond to return the DNA to its normal shape and function. If it doesn’t, and exci- sion repair fails, the distor tion inter feres with DNA replication and transcription.
10-12 EXCISION ORDARKREPAIR INE. COLI✦In the repair process, four enzymes are used: (1) An endonuclease (UvrABC) to break two covalent bonds in the sugar-phosphate backbone of the damaged strand, (2) a helicase (UvrD) to remove the nucleotides in the damaged segment, (3) DNA polymerase I to synthesize a new strand, and (4) DNA ligase to form a covalent bond between the new and the original strands.
old DNA, and the repair is complete. Both mechanisms are capable of repairing a small amount of damage, but long and/or intense exposures to UV produce more dam- age than the cell can repair, making UV radiation lethal.
✦ Application
Because ultraviolet radiation has a lethal effect on bac - terial cells, it can be used in decontamination. Its use is limited, however, because it penetrates materials such as glass and plastic poorly. In addition, bacterial cells have mechanisms to repair UV damage.
✦ In This Exercise
The lethal effects of ultraviolet radiation, its ability to penetrate various objects, and the cells’ ability to repair UV damage will all be demonstrated. Do not be misled by the apparent simplicity of the experimental design—
there’s a lot going on!
✦ Materials
Per Student Group
✦seven Nutrient Agar plates
✦disinfectant
✦posterboard masks with 1" to 2" cutouts (Figure 10-13)
✦UV light (shielded for eye protection)
✦sterile cotton applicators
✦24-hour trypticase soy broth culture of Serratia marcescens
Procedure
Lab One
Refer to the Procedural Diagram in Figure 10-14 as you read and follow this procedure:1
1 Using a sterile cotton applicator, streak a Nutrient Agar plate to form a bacterial lawn over the entire surface by using a tight pattern of streaks. Rotate the plate one-third of a turn and repeat, then rotate it another one-third of a turn and streak one last time. Repeat this process for all remaining plates.
2 Number the plates 1, 2, 3, 4, 5, 6, and 7.
3 Remove the lid from plate 1 and set it on a disinfectant-soaked towel. Place the plate under the UV light and cover it with a mask.
4 Turn the UV lamp on, but do not look at it. After 30 seconds, turn off the UV lamp, remove the mask, and replace its lid. If space permits, you may com- bine this step with Step 5.
5 Repeat the process for plate 2. If space permits, you may combine this step with Step 4.
6 Repeat the process for plate 3, but leave the UV lamp on for 3 minutes. If space permits, you may combine this step with Steps 7 and 8.
7 Irradiate plate 4 for 3 minutes, but leave the lid on and cover with the mask.
8 Repeat Step 7 for plate 5. If space permits, you may combine this step with Step 7.
9 Do not irradiate plates 6 and 7.
10 Incubate plates 1, 3, 4, and 6 for 24 to 48 hours at room temperature in an inverted position where they can receive natural light (e.g.,a windowsill) for 24–48 hours. Do not stack them.
11 Wrap plates 2, 5, and 7 in aluminum foil, invert them, and place with the others for 24–48 hours.
Lab Two
1 Remove the plates and examine the growth patterns.
2 Record your results on the Data Sheet.
References
Moat, Albert G., John W. Foster, and Michael P. Spector. 2002. Chapter 3 in Microbial Physiology, 4th ed. Wiley-Liss, New York.
Nelson, David L., and Michael M. Cox. 2005. Chapter 25 in Lehninger’s Principles of Biochemistry.W. H. Freeman and Co., New York.
White, David. 2000. Chapter 19 in The Physiology and Bio chemistry of Prokaryotes. Oxford University Press, Inc. New York.
1Thanks to Roberta Pettriess of Wichita State University for her helpful suggestions to improve this exercise.
Plate
Mask Agar
through opening
10-13 POSTERBOARDMASK✦This is an example of a mask placed over a Petri dish. The cutout may be any shape but should leave the outer 25% of the plate masked.
Place mask on plates with lids off
Plates 1 & 2 Plates 6 & 7
Place mask on plate with lid on
Plates 4 & 5 Place
mask on plate with lid off
Plate 3 Inoculate all seven plates
with Serratia marcescens to produce confluent growth
Expose with UV for 30 seconds
Remove the masks, replace the lids and incubate for 24–48 hours:
Plate 1 in sunlight, Plate 2 wrapped in aluminum foil.
Expose with UV for 3 minutes
Remove the mask, replace the lid and incubate in sunlight for 24–48 hours.
Expose with UV for 3 minutes
Remove the masks and incubate for 24–48 hours:
Plate 4 in sunlight, Plate 5 wrapped in aluminum foil.
Do not expose with UV
Incubate for 24–48 hours.
Plate 6 in sunlight, Plate 7 wrapped in aluminum foil.
Serratia marcescens
10-14 PROCEDURALDIAGRAM✦Inoculate and expose the plates as directed. Be sure to shield the UV light source adequately.
Do not look at the light.
✦ Theory
Bacteria that are able to enzymatically synthesize a par- ticular necessary biochemical (such as an amino acid) are called prototrophsfor that biochemical. If the bacterial strain can make the amino acid histidine, for instance, it is classified as a histidine prototroph. Bacteria that must be supplied with that biochemical (e.g.,histidine) are called (histidine) auxotrophs.Auxo trophs typically are created from prototrophs when a mutation occurs in the gene coding for an enzyme used in the pathway of the biochemical’s synthesis.
The Ames Test employs mutant strains of Salmonella typhimuriumthat have lost their ability to synthesize histidine. One strain of histidine auxotrophs possesses a frame shift mutation; they are missing one or have an extra nucleotide in the DNA sequence that otherwise would code for an enzyme necessary for histidine pro- duction. Other strains are substitution mutants, in which one nucleotide in the histidine gene has been replaced, resulting in a faulty gene product.
The Ames Test determines the ability of chemical agents to cause a reversal—aback mutation—of these auxotrophs to the original prototrophic state. In this test, histidine auxotrophs are spread ontominimal agar plates that contain all nutrients for growth but only a trace of histidine.
When a filter paper disk saturated with a suspected mutagen is placed in the middle of the minimal agar plate, the substance will diffuse outward into the medium.
If it is mutagenic, it will cause back mutation in some cells (converting them to histidine prototrophs) that freely grow into full-size colonies. (Note:Histidine ini- tially is included in the medium to allow the auxotrophs to grow for several generations and expose them to the effects of the mutagen. The unmutated auxotrophs typically exhibit faint growth but do not develop into full-size colonies because of the rapid exhaustion of the histidine.) A sample plate is shown in Figure 10-15.
Several variations of the Ames test are possible. This exercise uses two minimal agar plates and two complete agar plates. Complete agar contains all of the nutrients necessary for growth of Salmonella. All four plates are inoculated with histidine auxotrophs. One minimal agar plate and one complete agar plate each receive a filter paper disk saturated with the test substance. The second minimal agar plate and complete agar plate each receive
a filter paper disk saturated with Dimethyl Sulfoxide, DMSO (a substance known to be nonmutagenic).
The minimal agar plate containing the test substance determines mutagenicity of the test substance. Only prototrophs will grow on minimal agar, so recovery of any colonies on the minimal agar inoculated with auxo - trophs is indicative of back mutation. Depending on the strain used, the type of mutation (either frameshift or base substitution) also can be determined. The minimal agar plate with DMSO serves as a control for the mini- mal agar/test substance plate by measuring spontaneous back mutations(natural mutations that occur without the presence of a mutagen).
The purpose of the complete agar plate containing the test substance is a control to evaluate toxicity of the test substance. Creation of a zone of inhibition around the disk indicates toxicity (Figure 10-16). The more toxic the substance is, the larger the zone will be. If the substance is toxic, there may be no indication of mutagenicity because the cells are killed before they can back-mutate. Finally, the complete agar plate containing DMSO serves as a con- trol for comparison to the growth and zone of inhibition on the complete agar/test substance plate.
10-6 Ames Test
10-15 TESTING FORMUTAGENICITY✦In this example, a Salmonella histidine auxotroph was grown on histidine minimal agar while exposed to a suspected mutagen. The colonies were derived from cells that under went back mutations. It is not known from this plate alone, however, whether they are a result of the ef fects of the test substance, so it must be compared to a control minimal agar plate in which a nonmutagenic substance (DMSO) is substituted for the test substance.
✦ Application
Many substances that are mutagenic to bacteria are also carcinogenic to higher animals. The Ames Test is a rapid, inexpensive means of using specific bacteria to evaluate the mutagenic properties of potential carcinogens. Many variations of the basic Ames Test are used in specific applications.
✦ In This Exercise
You will test a household substance for its mutagenic properties. If you want the best chance of a positive result, check below your sink and in the garage for products that have many organic chemicals in the ingredient list. When handling these materials, be sure to wear gloves. Use caution in the laboratory if the material is flammable.
✦ Materials
Per Group
✦four Minimal Medium (MM) plates
✦four Complete Medium (CM) plates
(Note:This exercise can be run with two plates of each medium if replicates are not desired.)
✦centrifuge
✦two sterile centrifuge tubes
small beaker containing alcohol and forceps
✦bottle of 1⳯Vogel-Bonner salts (To make 1⳯Vogel- Bonner salts, add 1.0 mL 50⳯Vogel-Bonner solution to 49 mL water.) (Appendix A)
✦eight sterile filter disks made with a paper punch
✦two sterile Petri dishes (for soaking filter paper disks)
✦two sterile transfer pipettes
✦100 àL–1000 àL digital pipettors and tips
✦container for disposal of supernatant (to be autoclaved)
✦DMSO
✦test substance (any substance that has possible muta- genic properties and does not contain histidine or protein)
✦hand tally counter
✦broth culture1of Salmonella typhimurium TA 1535
✦broth culture1of Salmonella typhimurium TA 1538
Procedure
Day One
Follow the procedural diagram in Figure 10-17.
1 Soak four filter paper disks in DMSO and four filter paper disks in the test substance.
2 Pipette 10.0 mL TA 1535 into a sterile centrifuge tube. Do the same with TA 1538, then label the tubes.
3 Centrifuge the tubes on high speed for 10 minutes.
Be sure the centrifuge is balanced.
4 Being careful not to disturb the cell pellet at the bot- tom, decant the supernatant from each centrifuge tube using a different sterile transfer pipette.
5 Resuspend the cell pellets by adding 1.0 mL sterile 1⳯Vogel-Bonner salts to each tube and mixing well.
6 Using the spread plate technique (see Exercise 1-5), inoculate two MM plates and two CM plates with 100 àL of the resuspended TA 1535. Do the same with TA 1538.
10-16 TESTING FORTOXICITY✦Shown is a Salmonellahis- tidine auxotroph grown on complete agar (containing histidine) and exposed to the test substance. The zone of inhibition around the paper disk indicates toxicity of the substance. (Note that a zone of inhibition is visible on the plate shown in Figure 10-15, but it is not as well defined as on complete agar be- cause the growth is not as dense.)
1The cultures used for this exercise must be prepared as follows:
1. 24 hours before the test, inoculate two 10.0 mL broth tubes with TA 1535 and TA 1538. Incubate at 35Ⳳ2°C together with two sterile 90.0 mL broths (in a 100 mL diluent bottle).
2. 51⁄2hours before the test, pour the TA 1538 culture into one of the sterile 90.0 mL broths and return it to the incubator until time for the exercise.
3. 4 hours before the test, pour the TA 1535 culture into the other 90.0 mL sterile broth and return it to the incubator until time for the exercise.
10-17 PROCEDURALDIAGRAM✦Be sure to dispose of all pipettes, broth, and tubes properly as you per form the experiment.
After 48 hours of incubation, use a metric ruler to measure the diameter of the cleared zone on all CM plates. Use a colony counter to count the back-mutant colonies on all the MM plates. (Note: These will be fairly large. Do not count the tiny “hazy” growth over the plate’s sur face.) Mark each colony with a toothpick (to avoid counting it more than once) as you keep track of the number using a hand tally counter.
Salmonella typhimurium TA1535 or TA1538
Small dish with disks soaking
in DMSO
Small dish with disks soaking in suspected mutagen Sterile
centrifuge tube with 10 mL Salmonella
Pellet in centrifuge tube
after spinning
Pellet in centrifuge tube after decanting supernatant
to disposal container
TA1535/TA1538 in Vogel-Bonner salts solution Mix broth, then
pipette 10 mL to sterile centrifuge tube
Centrifuge 10 minutes at high speed
Using a sterile transfer pipette, remove broth (leave pellet, dispose of broth properly)
Add 1 mL of sterile V-B salts and mix
Dispense 100 mL to each plate
Spread inoculum with flamed
glass rod
Spread inoculum with flamed
glass rod MM
MM CM
CM
Place disks on agar surface, with alcohol flamed forceps, replace lid, then incubate for
48 hours at 3552°C
7 Flame the forceps by passing them through the Bunsen burner flame, and allowing the alcohol to burn off.
8 Using the flamed forceps, place the DMSO and test disks in the centers of the plates. Gently tap down the disks with the forceps to prevent them falling off when the plates are inverted.
9 Incubate the plates aerobically at 35Ⳳ2°C for 48 hours.
Day Two
1 Measure the diameters (in millimeters) of the zones of inhibition on the CM plates.
2 Count the colonies on the MM plates and compare.
Count only the large colonies, not the “hazy” back- ground growth. (These are the colonies produced by auxotrophs that did not back-mutate to prototrophs and only grew until the histidine in the minimal medium was exhausted.)
3 Record your observations for the Ames Test plates on the Data Sheet.
Eisenstadt, Bruce, C. Carlton, and Barbara J. Brown. 1994. Page 311 in Methods for General and Molecular Bacteriology, edited by Philipp Gerhardt, R. G. E. Murray, Willis A. Wood, and Noel R. Krieg, American Society for Microbiology, Washington, DC.
Maron, D. M. and B. N. Ames. 1983. Mutation Research, 113:173–215.
Nelson, David L., and Michael M. Cox. 2005. Chapter 25 in Lehninger’s Principles of Biochemistry.W. H. Freeman and Co., New York.
✦ Theory
Identification of organisms is a primary function of microbiologists. The best identification methods involve utilizing genetically based differences to distinguish between organisms at every level: Domain through Species. In microbiology classical identification methods include cell morphology, staining properties, biochemical testing and serological reactions, all of which are still being used. Newer methods include comparisons of nucleic acids, such as hybridization and sequencing. Phage typingallows microbiologists to differentiate between strains of bacterial species based on susceptibility to infection by different viruses.
Viruses that infect bacteria are called bacteriophages or simply just phages. The infection process (see Figure 6-12) begins with attachment by the phage to a receptor on the host cell. The phage then injects its nucleic acid into the host cytoplasm and overtakes its metabolism;
the host is now under viral control and becomes a factory for making a hundred or more copies of the virus. When completed, the host cell bursts and releases the viral progeny, which now infect other cells to repeat the replicative cycle. Depending on the virus, one round of replication can take as little as 25 minutes at 37°C!
The ability of a virus to infect a host has to do with its ability to attach to specifichost receptors. Because receptor structure is genetically determined, different genetic strains of the bacterial species can be identified by their susceptibility to a particular virus. Challenging the same host with different viruses (which use different receptors) produces a composite picture of viral suscepti- bility. Different composite pictures indicate different strains of the hostand provide a means of characterizing host strains based on their viral susceptibilities. This characterization can be used as a basis for identifying fresh isolates down to the level of “strain.”
One way of determining host susceptibility to a particular phage is to inoculate a plate with the host so it produces confluent growth (called a “lawn”). Phage can be introduced at the same time and the plate incubated for 24–48 hours. Clear zones in the lawn are evidence that the virus has gone through multiple replicative cycles and destroyed all the host cells in that region. If no clearings are seen, then the bacterial strain is not a suitable host for that virus (Figure 10-18).
✦ Application
Phage typing is used to characterize and identify strains within a bacterial species.
✦ In This Exercise
Today you will be phage typing three strains of Escheri - chia coli(wild type, B, and C) by inoculating agar plates with each strain and with two E. colibacteriophages (T2
and 174).
✦ Materials
Per Student Group
✦overnight broth cultures of
⽧E. coli (wild type)
⽧E. coli B
⽧E. coli C
✦three Nutrient Agar or Trypticase Agar Plates
✦vials of
⽧T2Phage
⽧Phage 174
✦twelve sterile cotton swabs
✦marking pen
✦one vial of sterile distilled water
10-7 Phage Typing of E. coli Strains
10-18 ZONE OFCLEARING ANDNOCLEARING✦This plate was inoculated to produce a lawn of host bacteria. Two dif - ferent viruses were spotted onto the plate, which was then incubated for 48 hours. Notice the clearing on the right and absence of clearing on the left. (For convenience, the places where virus was applied are circled.