JOURNAL OF MICROBIOLOGY & BIOLOGY EDUCATION, May 2016, p. 261-268 DOI: http://dx.doi.org/10.1128/jmbe.v17i2.981
Supplemental Materials for Battle of the Bacteria: Characterizing the Evolutionary Advantage of Stationary Phase Growth Karin E. Kram1*, Kristina M. Yim2, Aaron B. Coleman3, and Brian K. Sato4* Department of Biology, California State University, Dominguez Hills, CA 90747, 2 Department of Genetics, Yale University, New Haven, CT 06510, 3 Division of Biological Sciences, University of California, San Diego, CA 92093, 4 Department of Molecular Biology and Biochemistry, University of California, Irvine, CA 92697 1
Table of Contents (Total pages 87)
Appendix 1: Reagent/equipment list and faculty instructions Appendix 2: Student handout and protocol Appendix 3: Lecture slides Appendix 4: GASP worksheet Appendix 5: Module pre-/post-test Appendix 6: Self-assessment and exam questions Appendix 7: Pipetting test instructions Appendix 8: Example student hypotheses
*Corresponding authors. Mailing addresses: Brian Sato: 2238 McGaugh Hall MC3900, Irvine, CA 92697. Phone: 949-824-0661. Fax: 949-824-8551. E-mail:
[email protected]; Karin Kram: 1000 E. Victoria St. NSM A-137, Carson, CA 90747. Phone: 310-243-1090. Fax: 301-243-2350. E-mail:
[email protected]. ©2016 Author(s). Published by the American Society for Microbiology. This is an Open Access article distributed under the terms of the Creative Commons AttributionNoncommercial-NoDerivatives 4.0 International license (https://creativecommons.org/licenses/by-nc-nd/4.0/ and https://creativecommons.org/licenses/by-nc-nd/4.0/legalcode), which grants the public the nonexclusive right to copy, distribute, or display the published work.
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Appendix 1: Reagent/Equipment List and Faculty Instructions. -Micropipettors (P20, P200, P1000) – one set per group of four -Bench top centrifuge – one per section of 20 students -Glass tubes (20mm x 150mm) -Spectrophotometer E. coli – both strains express the bolA-lacZ fusion protein -Naladixic acid resistant – SF2054 (W3110 ∆lacU169 tna2 Nalr lambda-bolA::lacZ-Kan-r) -Streptomycin resistant – SF2055 (W3110 ∆lacU169 tna2 Strr lambda-bolA::lacZ-Kan-r) More information can be found in Farrell and Finkel, 2003. Lysogeny broth (LB) – for bacteria growth LB + Drug plates -Rifampicin - 100µg/ml (100mg/ml Rifampicin stock in DMSO) -Naladixic acid - 20µg/ml (20mg/ml Naladixic acid stock in water, add NaOH to get in solution) -Streptomycin - 25µg/ml (25mg/ml Streptomycin stock in water) Reagents per experiment (per pair) 25X Variable solutions (others can be used as well) – make in water, sterile filter before use -12.5% NaCl, 1.25M CaCl2,12.5% tryptone, 12.5% yeast extract, 5% glucose, 5% fructose 1. Growth of cells into long-term stationary phase and GASP assay Starting cultures to generate LTSP conditions -2 glass tubes, 15ml LB total (to be distributed to each glass tube for culture inoculation) Co-culture and GASP assay -3 glass tubes, 20ml LB -6 LB + Nalidixic acid plates, 6 LB + Streptomycin plates, and 6 empty petri dishes -5ml sterile water 2. Examination of RpoS activity -a few drops of 3% H2O2 solution, microscope slides 3. Cell stress measurement -Z buffer (must be made within a few days of use) – 5ml -To make 150ml - 2.4g Na2HPO4•7H2O, 0.84g NaH2PO4•H2O, 1.5ml 1M KCl, 0.15ml 1M MgSO4, 0.405ml β-mercaptoethanol adjust pH to 7.0, then bring to 150ml with water, keep at 4°C -ONPG solution (must be made within a few days of use) – 500µl -4mg/ml in water -0.1% SDS solution - 150µl -1M Na2CO3 – 1.5ml -1 LB plate, 2 empty plastic petri dishes -2ml sterile water 4. Examination of mutation frequency -1 LB plates, 2 LB + Rif plates, 1 empty plastic petri dishes -2ml sterile water
Appendix 2: Student Handout and Protocol.
Week 1 – Begin Bacteria Evolution Experiment Evolution of populations occurs in two steps. First, organisms within those populations gain mutations as part of normal biological processes. Second, conditions in the environment allow a few organisms with beneficial mutations to survive longer than their non-mutated counterparts. Those organisms with beneficial mutations are more fit, and will be selected for over time. Experiments conducted by scientists have shown that mutations arise in a population regardless of the environment. This is because there is a constant mutation rate simply due to mistakes made in DNA replication and repair. One example is the famous fluctuation test done by Luria and Delbruck (1943). The scientists demonstrated that mutations, which confer resistance to bacteriophage attack, occur randomly during growth of bacterial cultures, even when the bacteriophages are not present. This shows that mutations occur during growth with or without the selective environment. If, however, we place the bacteria into selective conditions, we can show that those mutations can be selected out of the population, which leads to evolution. These selective environments generate a number of interesting questions regarding the surviving mutant population. Are the mutations beneficial in other environments, what characteristics of the environment enhance this phenotype, etc? In this experiment, we will be growing bacteria in a selective condition, and showing that selection causes the bacteria to be better adapted to their environment after merely a few days of exposure. The selective environment will be long-term stationary phase (LTSP), during which populations of bacteria must survive harsh conditions including low nutrients, basic pH, and high waste product levels. Bacteria experience five phases of growth during their life cycle: lag phase, exponential phase, stationary phase, death phase, and LTSP.
Most often when we grow bacteria in the lab, we examine them during log phase or early stationary phase growth. However, in the natural environment, and in certain laboratory conditions, bacteria more often experience LTSP. Some strains of Escherichia coli can survive in LTSP without the addition of nutrients for years. We can take bacterial populations that have been grown for over 10 days into LTSP, and compete them with un-aged bacteria in fresh media. The aged bacteria will be better adapted to the environment, and will therefore grow faster than the un-aged cells. We call this phenomenon growth advantage in stationary phase (GASP). Further, we will age our cells in a slightly different environment, which will change the conditions to which it will adapt, and determine whether or not those adaptations also help them in the original environment. Some conditions may select mutations that give the bacteria an advantage in a wide variety of environments, while other conditions may select for mutations that only give an advantage to the bacteria in that specific environment. Using data from this experiment, we can determine what factors influence adaptation. We will also characterize the GASP phenotype displayed by your cells. Are cells exhibiting the GASP phenotype surviving due to physiological adaptations or genetic mutations? Does increased environmental stress assist
or hinder the growth advantage? Do GASP populations also experience a higher mutation frequency? With these experiments, we will illustrate that our aged strain is genetically different strain from the strain that began the experiment. We will see evolution occurring in real-time! Keep in mind that this is also novel work you will be conducting in collaboration with Dr. Steven Finkel’s lab at USC. You have a shot of identifying something that nobody has ever seen! Below is an outline of the timeline of each experiment, and the specific questions that experiment will be answering. In addition, you will be examining how cells grow in LTSP when the media is altered. You and your group can choose one of the below variations of LB: LB + 0.5% NaCl LB + 50mM CaCl or MgCl LB + 0.5% tryptone LB + 0.5% yeast extract LB + 0.2% glucose or fructose or sucrose
Assay I - Growth of cells into long-term stationary phase and GASP assay We need to select for cells that exhibit the GASP phenotype, by first growing a culture of E. coli into long-term stationary phase and then adding the survivors to a fresh bacteria culture. We will then compare survival of the aged versus the un-aged cells using different antibiotic resistance markers to distinguish each (see diagram below). In addition, you will be modifying the culture media in a specific manner (by adding Variable X) to determine how it impacts the GASP phenotype. For this aspect of the experiment, we will need to make slight modifications to the general experimental protocol outlined below. Experimental Questions: 1. 2. 3.
Does growth into long-term stationary phase select for certain mutations (existing or spontaneous) producing a culture with a growth advantage? Does the addition of Variable X to the LB alter the growth advantage for long-term stationary phase cells? Is the growth advantage observed in LB + Variable X specific for growth in LB + X or evident in regular LB as well?
Assay II - Examination of RpoS activity – Genotypic vs. Phenotypic change in extreme conditions RpoS is a transcription factor that activates genes responsible for fighting cell stress and allowing cells to enter stationary phase. Surprisingly, RpoS activity appears to be detrimental to cell survival in long-term stationary phase, and many cells will downregulate RpoS in LTSP. In this part of the experiment, you will determine whether your LTSP cells have decreased RpoS activity. If so, this may be accomplished through one of two mechanisms, cellular downregulation of RpoS (through decreased transcription or increased protein degradation, for example) or through mutation, which results in chronically decreased RpoS levels. One of the targets of RpoS is catalase, and you will measure RpoS activity by examining hydrogen peroxide breakdown by your cells. By conducting this test at various intervals throughout the experiment, you will answer the following questions: 1. 2. 3.
Does RpoS activity change in cultures that are grown to LTSP? If RpoS is being downregulated in LTSP, is this due to a phenotypic or genotypic modification of RpoS? Do the answers to these questions differ for cells grown in LB + Variable X?
There are a few potential results you may see: 1. If catalase expression is similar on day 2, 14 and the following day 2, RpoS activity is unaltered by LTSP. 2. If catalase expression decreases on day 14, but is restored when the cells are added back to fresh media, RpoS activity is altered by LTSP conditions. This signifies a physiological, but not genomic change. 3. If catalase expression decreases on day 14 and remains decreased in fresh media, this signifies a genomic change to RpoS.
Assay III - Cell stress measurement Environmental stress can be a major factor in driving the evolution of organisms. Bacteria are equipped with cellular mechanisms to handle a variety of stresses, including elevated temperatures, oxidative stress, and low nutrient environments, among others. But in extreme scenarios, it is likely that the stress will overwhelm these pathways and the cells will be unable to survive. In rare cases, it is also possible that cells contain or acquire beneficial mutations that allow them to survive to a greater extent in these environments compared to the rest of the population. We will measure cell stress using a lacZ fusion protein, driven by the bolA promoter. bolA is part of the E. coli stress response and is upregulated under carbon starvation, increased osmolarity and oxidative stress. When stressed, the cell will transcribe from the bolA promoter, turning on the lacZ gene resulting in the production of β-galactosidase protein. We will measure β-galactosidase activity with the addition of its substrate, ortho-nitrophenyl-β-galactoside (ONPG), which is broken down to produce ortho-nitrophenol, a molecule with a yellow color. We can measure the amount of this molecule using the spectrophotometer. With this reporter protein, we can ask the following questions: 1. 2.
Does the amount of stress experienced by the cells correlate with the intensity of the GASP phenotype? Does the addition of Variable X to the media alter cell stress?
Assay IV - Examination of mutation frequency Environmental stresses can promote the formation of mutations, for example through increased DNA alterations caused by base damage or decreased activity of DNA repair pathways. While most are likely detrimental or have no effect on the cell, it is also possible a cell will acquire a beneficial mutation. Normally your cells are sensitive to rifampicin, but it is possible for them to become antibiotic resistant. In this experiment, you will be measuring the mutation frequency in your bacterial culture to answer the following questions: 1. 2. 3.
What is the frequency of mutations in the bacteria population? Does this frequency change in LB + Variable X media? Does the frequency of mutation correlate with the results of the GASP experiment?
Week 1 Lab: Assay I - Growth of cells into long-term stationary phase and GASP assay Materials: 2 test tubes, variable X stock solution, StrepR E. coli culture, LB medium This week you will start the cultures you will use for the GASP experiment in two weeks. You will be setting up 2 cultures, one with streptomycin resistant E. coli in LB and a second with the same strain in LB + Variable X. When choosing your variable, be sure to think about how it might affect the experiment. Think about the role that the variable plays in the media. This experiment will be conducted in pairs, although your group (of 4) will choose a variable together. That way each pair will perform the same experiment and you can verify each other’s results. The variable you and your group have chosen: 1. 2. 3.
4.
Set up 2 tubes. On a piece of tape, label one as LB and one as LB + (your variable). Include your and your partner’s initials, the date, and your section day (TUES/WED), time (AM/PM) and room number (226/230). Place the tape on the tube, not the lid. In each tube, add 5ml of LB and then 5µl of the overnight culture (the StrepR E. coli). You now want to add your variable X. The stock variable solutions are all 25X, meaning they are 25 times as concentrated as you need. Calculate how much of the stock you will add to the LB + X culture. In the c1v1 = c2v2 equation, v2 = 5ml. In addition, to keep the volumes in each tube equal, you will add an equal amount of LB to the LB only tube. Volume you will add: Place each tube in your section’s rack for incubation at 37°C.
Week 1 Discussion: Assay II - Examination of RpoS activity Materials: 2 cultures from lab (LB only, LB + X), microscope slide, H2O2 We want to measure catalase activity (and thus RpoS activity) before the cells reach LTSP. This test will be performed similarly to the catalase test performed earlier in the quarter. 1. 2. 3. 4.
Agitate each tube if a pellet is at the bottom of the tube. Remove 5µl of each of your cultures and add them to a glass slide. Add one drop of H2O2 to each aliquot. Note whether bubbles form (large/small, many/few, form immediately/after time, etc).
Assay IV - Examination of mutation frequency Materials: 1 LB plate, 1 LB + Rif plate, 1 empty petri dish, liquid LB Normally your bacteria are rifampicin sensitive. It is possible that they acquire a mutation that renders them rifampicin resistant. You will examine this by plating your cultures on an LB + Rif plate. In addition, you will plate them on an LB plate to determine the total viable cells in the population to calculate the percentage of the culture that is rifampicin resistant. For most tests of cell viability, like the LB media plating you will perform today, you will make 1:10 serial dilutions, plating 10µl of each dilution on the plate, eight dilutions total.
Each 1:10 dilution will involve adding 90µl media and 10µl cells. These dilutions will be performed using an empty petri dish (either the lid or the dish), where you add eight spots of 90µl LB. You will then transfer 10µl of your culture to the first spot, pipette up and down a few times and mix the drop with the pipette tip. The next spot will be made with 10µl of the first spot and 90µl LB, mixing by pipetting up and down and using the pipette tip. This will be repeated for all eight spots. Your TA will demonstrate this procedure for you. To determine the frequency of rifampicin resistant cells in the population, you must identify how many cells are rifampicin resistant (by plating on LB + Rif) and comparing this to how many viable cells there are total (by plating on LB alone). We anticipate the frequency of rifampicin resistant cells to be low, so there will be no need to dilute the culture plated on LB + Rif. 1. 2.
Take 1 LB plate and 1 LB + Rif plate. Label the bottom of these, along with your identifying information. Prepare your LB plate first. In an empty petri dish, you will make eight spots with 90µl LB in each. Do not let the drops merge. You should plate one culture in its entirety before moving on to the next. 3. Take 10µl of one of your cultures (well-mixed) and add it into the first drop. Pipette up and down and stir the drop with the pipette tip to mix the dilution. 4. Using the same tip, pipette 10µl of the first drop onto your LB plate. You will use the grid below to keep track of your spots. 5. Take a new pipette tip and remove 10µl of the first drop and pipette it in the second drop. Mix the solution as above and pipette 10µl onto the LB plate (onto the #2 on the grid). 6. Repeat for drops three through eight. 7. Repeat steps 1-5 for your second culture. 8. Let the plate dry face up. Then flip upside down and incubate at room temperature. If it does not dry in time, leave face up on the room temperature shelf. 9. Now prepare the LB + Rif plate. Because the number of rifampicin resistant cells will be low, you will not do the above procedure. Instead, plate 100µl of your first culture (well-mixed) on an LB + Rif plate. 10. Spread the 100µl using the ethanol spreader (dip in ethanol, briefly flame, let cool by placing on a region on the plate where the cells are not). 11. Repeat this with your second culture using another LB + Rif plate.
Next lab period, you will count the number of surviving cells on the LB and LB + Rif plates.
Week 2 – Continue Bacteria Evolution Experiment This week you will continue incubating your cultures so they can reach long-term stationary phase by the following lab period. There are two questions you will be working to answer: Assay IV - Examination of mutation frequency This is a continuation from your work from last discussion where you will analyze the growth observed to determine the frequency of rifampicin resistance. Your mutation frequency will be based on counting colony forming units (CFU). CFU estimates the number of viable cells by assuming each viable cell is capable of cell division resulting in the formation of a colony. So CFU is determined by counting colony number on a plate. CFU/ml Rif Res Bacteria (on LB + Rif plate) CFU/ml Total Cells (on LB plate)
The equation for the mutation frequency will be:
1. To determine the CFU/ml for Rif resistant bacteria you will count the total colonies on the LB + Rif plate. Note that number here: (LB only) (LB + X media) Remember, you plated 100µl of your original cultures onto the LB + Rif plate. So rather than identifying the CFU/ml, your values above represent the CFU/0.1ml. This means you need to multiply the numbers above by a factor of . Your CFU/ml values for Rif resistance cells are:
(LB)
(LB + X)
2. To determine the CFU/ml for all viable cells (plated on the LB media), you will look at your dilution spots. The more concentrated spots are likely completely covered with growth and it is no longer possible to count individual colonies. Instead, select the spot with individual colonies. Colonies from LB only: Spot # Colonies from LB + X: Spot # Now you need to determine the CFU/ml. Imagine you had 2 colonies on spot 5. The very first spot was created by adding 10µl of cells to 90µl LB and plating 10µl (1/10th) of this dilution. Since you plated 1/10th of that original 10µl (from the culture), you actually only added 1µl of the original culture. This means the growth in the first spot signifies the CFU/0.001ml or CFU/10-3 ml. Spot number two is a 1:10 dilution of the first, so growth in this spot signifies the CFU/10-4 ml. This means the 2 colonies on spot 5 signify a value of 2 CFU/10-______ ml or
CFU/ml.
Now, based on this little tutorial, calculate your CFU/ml for all viable cells using the values you noted from your LB plates. LB only CFU/ml:
LB + X CFU/ml:
And finally, you can calculate your mutation frequency: CFU/ml Rif Resistant Bacteria (on LB + Rif plate) CFU/ml Total Cells (on LB plate) LB only:
LB + X:
Assay III - Cell stress measurement Materials: Z buffer (60mM Na2HPO4, 40mM NaH2PO4, 10mM KCl, 1mM MgSO4, 50mM BME), phosphate buffer (60mM Na2HPO4, 40mM NaH2PO4), 0.1% SDS, 4mg/ml ONPG, 1M Na2CO3, spectrophotometer Another goal in our experiment is determine how stressed the cells are in a given medium. This will allow us to see if there is any correlation between cell stress and the GASP phenotype. As mentioned previously, we will measure stress with our bolA-lacZ reporter gene. Stress will lead to bolA promoter activation, lacZ transcription, and βgalactosidase production. β-galactosidase activity can be determined by adding a substrate, ONPG, which is converted to a yellow product. The amount of this yellow product can be measured with a spectrophotometer. Spectrophotometry allows one to measure the components in a solution. Different molecules are capable of absorption of specific wavelengths of light. This absorption can be measured with a spectrophotometer. A solution is added to a cuvette that is placed into the spectrophotometer. Light of a specific wavelength is directed at the cuvette and may be absorbed by specific molecules. The amount of light, which is not absorbed and passes through the cuvette, is read resulting in the calculation of an absorbance. More of that specific molecule results in more light absorbed and a higher absorbance value, known as the optical density (O.D.). The figure below illustrates this process:
For our purposes, there will never be any need to use this equation for absorbance, but it is helpful for understanding how the spectrophotometer is able to measure this value. Since we want to know the light absorbed by a specific molecule (in this case ortho-nitrophenol), we need to eliminate any light absorbed by the buffer. To eliminate the absorbance by any components in the buffer, we first “blank” the spectrophotometer with a cuvette containing only buffer. Thus, when we add the cuvette with ONPG and β-galactosidase, the resulting absorbance will be due only to ortho-nitrophenol. Below is the protocol for the spectrophotometer based β-galactosidase assay. The purpose of the experiment is to lyse the cells to expose the β-galactosidase protein to the ONPG substrate. In addition, you will need to determine the viable cell count in the population similarly to what you did the past week with the rifampicin assay. This is necessary because higher β-galactosidase activity could merely be due to a greater number of viable cells, and not necessarily greater bolA activation in each cell.
Before you begin, familiarize yourself with the spectrophotometer.
1. 2. 3. 4.
On the screen, confirm that the wavelength is set at 420nm (not 600nm as above). Next, add your blank cuvette (Z buffer only) to the spectrophotometer and hit “blank”. After this is complete, you will add your cuvette containing your β-galactosidase assay. Hit the green read sample button. This will generate an absorbance (OD) value.
Week 2 Lab: 1.
Remove 0.5ml of each culture (1 week old LB only, LB + X cultures) to two 1.5ml eppendorf tubes. The following steps will be performed for each of the tubes. 2. Centrifuge on full speed for 1 minute to pellet the cells. Pipette off the liquid but do not disturb the pellet. This removes the media, which may contain molecules that impair β-galactosidase activity or light absorbance by the spectrophotometer. 3. Resuspend the pellet in 0.5ml chilled Z buffer. 4. In a new 2.0ml tube, add 900µl Z buffer and 100µl of cells (resuspended in Z buffer). 5. Add 50µl SDS and mix by inverting the tube a few times. SDS is a detergent, which will lyse the cells. 6. Let the tubes sit on your bench for 5 minutes at room temperature. 7. Add 200µl ONPG to the tube and mix by inverting a few times. 8. Incubate the tube in the 28°C water bath until you see color (usually around 15 minutes). 9. Once you see yellow color in both tubes, stop the reaction by adding 0.5ml Na2CO3 and mix by inverting a few times. The sodium carbonate raises the pH to 11, which inactivates the β-galactosidase protein. 10. Transfer 1ml of the solution to a new 1.5ml tube and centrifuge for 5 minutes at max speed. 11. Instructions for using the spectrophotometer are on the following page. Blank the spectrophotometer with 1ml Z buffer in a cuvette (at OD420). 12. Remove the supernatant from the centrifuged tubes to cuvettes. 13. Determine the OD of the tubes.
LB only OD420:
LB + X OD420:
Determine CFU count for cells As mentioned above, to accurately measure the β-galactosidase activity, it is necessary to determine the viable cell count in the population. This will be accomplished by performing a dilution assay using LB plates. 1. 2. 3. 4. 5. 6. 7.
Prepare your LB plate. In an empty petri dish, you will make eight spots with 90µl LB in each. Do not let the drops merge. You should plate one culture in its entirety before moving on to the next. Take 10µl of one of your cultures (well mixed) and add it into the first drop. Pipette up and down and stir the drop with the pipette tip to mix the dilution. Using the same tip, pipette 10µl of the first drop onto your LB plate. You will use the grid below to keep track of your spots. Take a new pipette tip and remove 10µl of the first drop and pipette it in the second drop. Mix the solution as above and pipette 10µl onto the LB plate (onto the #2 on the grid). Repeat for drops three through eight. Repeat steps 1-5 for your second culture. Let the plate dry face up. Then flip upside down and incubate at room temperature. If it does not dry in time, leave face up on the room temperature shelf.
Week 2 Discussion: To determine the CFU/ml for viable cells, you will look at your dilution spots. The more concentrated spots are likely completely covered with growth and it is no longer possible to count individual colonies. Instead, select the spot with individual colonies. Colonies from LB only:
Spot #
Colonies from LB + X:
Spot #
Now you need to determine the CFU/ml using the example from the previous class. Imagine you had 2 colonies on spot 5. The very first spot was created by adding 10µl of cells to 90µl LB and plating 10µl (1/10th) of this dilution. Since you plated 1/10th of that original 10µl (from the culture), you actually only added 1µl of the original culture. This means the growth in the first spot signifies the CFU/0.001ml or CFU/10-3 ml. Spot number two is a 1:10 dilution of the first, so growth in this spot signifies the CFU/10-4 ml. This means the 2 colonies on spot 5 signify a value of 2 CFU/10-______ ml or
CFU/ml.
Now, based on this little tutorial, calculate your CFU/ml for all viable cells using the values you noted from your LB plates. LB only CFU/ml:
LB + X CFU/ml:
And finally, you can calculate your cell stress: OD420 CFU/ml LB only OD420: LB only:
LB + X OD420:
(values from last class) LB + X:
Week 3 – Initiate GASP Experiment This week you will begin your competition experiment to determine whether the LTSP E. coli are better able to grow compared to a fresh culture. You will conduct two separate experiments related to this study in lab today. Lab: Assay I - Growth of cells into long-term stationary phase and GASP assay Materials: Three test tubes, 2 aged StrepR E. coli cultures (LB, LB + X), fresh NalR E. coli cultures, 2 LB + Strep plates, 2 LB + Nal plates Your StrepR E. coli have now been growing for two weeks, well into long-term stationary phase. You will now compete them against another strain of E. coli, which is resistant to nalidixic acid, but is otherwise identical to the strains you aged. It is because each strain possesses a different resistance marker that we can tell them apart. The competition experiment will start in favor of the new NalR E. coli, as they will be found at roughly 1,000 times higher concentration than the aged StrepR strain. To see whether the LTSP culture possesses the GASP phenotype, you will survey the mixed culture today, in discussion and next week in lab. What results would you expect to see if the StrepR E. coli are capable of GASP? In addition, keep in mind that you grew two cultures, one with LB only and one with LB + Variable X. To determine whether growing in Variable X generated a phenotype specific for Variable X or a more general one (that is evident in LB only as well), you will examine three different conditions. I. LTSP StrepR E. coli grown in LB ⇒ Add cells to fresh NalR bacteria in LB II. LTSP StrepR E. coli grown in LB + X ⇒ Add cells to fresh NalR bacteria in LB III. LTSP StrepR E. coli grown in LB + X ⇒ Add cells to fresh NalR bacteria in LB + X Below is the protocol for today’s part of the experiment. This will be similar to the set-up that you used to start the StrepR culture in week 1. 1.
2. 3. 4. 5.
6.
Set up 3 tubes. On a piece of tape, label one as Condition I (which will consist of LB media/StrepR E. coli grown in LB), one as Condition II (which will consist of LB media/StrepR E. coli grown in LB + X), and one as Condition III (which will consist of LB + X media/StrepR E. coli grown in LB + X). Include your and your partner’s initials, the date, and your section day (TUES/WED), time (AM/PM) and room number (226/230). In each, add 5ml of the NalR overnight culture. This is the NalR bacteria grown in LB. You now want to add your variable X to condition III. Add the same amount of the stock solution to tube III as you did in week 7. In addition, to keep the volumes in each tube equal, you will add an equal amount of LB to tubes I and II. Shake each tube to mix. Now you need to add the aged StrepR bacteria. Pay careful attention that you add the correct bacteria to the correct tubes! For each, you will add only 5µl of the LTSP culture. When done, keep the LTSP cultures for the next experiment! To tube I – add 5µl of the LB only LTSP culture To tube II – add 5µl of the LB + X LTSP culture To tube III – add 5µl of the LB + X LTSP culture At the end of the class period, place each tube in your section’s rack for incubation at 37°C. But in the meantime there are a few additional things you need to do.
Remember, the goal is to monitor how the LTSP StrepR bacteria grow compared to the fresh NalR bacteria. This will be determined by looking at growth on an LB + streptomycin plate (where only the StrepR bacteria will survive) versus growth on an LB + nalidixic acid plate.
The plating will occur similarly to the rifampicin plating, although that only had 16 spots (2 cultures), whereas this will have 24 spots (3 cultures). This means you will need to use 2 of each plate rather than 1. Plate 1 will contain conditions I and II. Plate 2 will be for Condition III only. Details are below. 1.
Mix the tubes thoroughly before removing any liquid. You should plate one strain in its entirety before moving on to the next. 2. In an empty petri dish, you will make eight spots with 90µl each. Do not let the drops merge. Eventually you will have 24 spots, which may require you to use both the lid and bottom of the empty petri dish. 3. Take 10µl of one of your cultures (start with condition I) and add it into the first drop. Pipette up and down and stir the drop with the pipette tip to mix the dilution. 4. Using the same tip, pipette 10µl of the first drop onto your LB + Strep plate. You will use the grid below to keep track of your spots. Then pipette 10µl of the same drop onto your LB + Nal plate. Use the grid from both your and your partner’s manuals so that each will be used for one of the plates. 5. Take a new pipette tip and remove 10µl of the first drop and pipette into the second drop. Mix and pipette 10µl onto both plates. 6. Repeat for drops three through eight. 8. Repeat steps 2-6 for your second culture. 9. Repeat steps 2-6 for your third culture on plate 2. 10. Let the plate dry face up. Then flip upside down and incubate at room temperature. If it does not dry in time, leave face up on the room temperature shelf.
Assay II - Examination of RpoS activity Materials: 2 cultures (LB only, LB + X), microscope slide, H2O2 We want to measure catalase activity (and thus RpoS activity) now that the cells have reached LTSP. 1. 2. 3. 4.
Shake each tube if a pellet is at the bottom of the tube. Remove 5µl of each of your cultures and add them to a glass slide. Add one drop of H2O2 to each aliquot. Note whether bubbles form (large/small, many/few, form immediately/after time, etc).
We next want to see whether the catalase activity observed in LTSP is altered when the cells are returned to fresh media. To accomplish this we need to set up new cultures in fresh media. We cannot use the StrepR/NalR mixed cultures you just set up because we are unable to distinguish between the catalase activity of the LTSP StrepR cells and the NalR cells. 1. 2. 3. 4.
Take two tubes. On a piece of tape, label one as LB and the other as LB + X. Include your and your partner’s initials, the date, and your section day (TUES/WED), time (AM/PM) and room number (226/230). Add 5ml fresh LB to each. Add the appropriate amount of Variable X to one tube and the same amount of LB to the other. Add 5µl of the LTSP culture to each. The LTSP culture in LB will be added to the fresh LB while the LTSP culture in LB + X will be added to the fresh LB + X tube. Stick both tubes in your section’s rack to be incubated at 37°C.
Week 3 Discussion: Assay I - Growth of cells into long-term stationary phase and GASP assay It has now been two days since starting your mixed culture. Today, you will be analyzing the plates you inoculated in lab (at day 0) and setting up a new set to see the ratio of StrepR and NalR bacteria today. First, examine the plates from lab. You want to determine the ratio of StrepR CFU/NalR CFU. As you did with the rifampicin experiment, find the highest dilution where single colonies are easily distinguished. LB + Strep
LB + Nal
Colonies from Cond I:
Spot #
Colonies from Cond I:
Spot #
Colonies from Cond II:
Spot #
Colonies from Cond II:
Spot #
Colonies from Cond III:
Spot #
Colonies from Cond III:
Spot #
Now we need to convert this to CFU/ml for each condition/plate as you did in Week 1 and 2 lab. Be sure to set up your new LB + Strep/Nal plates for today before performing the below calculations. Instructions for this are on the following page.
Calculate the CFU/ml for all viable cells using the values you noted from your LB plates.
LB + Strep
LB + Nal
Condition I CFU/ml:
Condition I CFU/ml:
Condition II CFU/ml:
Condition II CFU/ml:
Condition III CFU/ml:
Condition III CFU/ml:
And finally, we will now see compare levels of StrepR and NalR cells in the culture. Since these were taken immediately after creating the cultures, we expect them to be in the starting ratio of 1:1,000. StrepR CFU/NalR CFU: Condition I:
Condition II:
Condition III:
LB + Strep, LB + Nal plating Materials: 2 LB + Strep plates, 2 LB + Nal plates, 3 cultures (Condition I, II, III) 1. 2. 3. 4.
5. 6. 7. 8. 9.
Mix the tubes thoroughly before removing any liquid. You should plate one strain in its entirety before moving on to the next. In an empty petri dish, you will make eight spots with 90µl each. Do not let the drops merge. Eventually you will have 24 spots, which may require you to use both the lid and bottom of the empty petri dish. Take 10µl of one of your cultures (start with condition I) and add it into the first drop. Pipette up and down and stir the drop with the pipette tip to mix the dilution. Using the same tip, pipette 10µl of the first drop onto your LB + Strep plate. You will use the grid below to keep track of your spots. Then pipette 10µl of the same drop onto your LB + Nal plate. Use the grid from both your and your partner’s manuals so that each will be used for one of the plates. Take a new pipette tip and remove 10µl of the first drop and pipette into the second drop. Mix and pipette 10µl onto both plates. Repeat for drops three through eight. Repeat steps 2-6 for your second culture. Repeat steps 2-6 for your third culture on plate 2. Let the plate dry face up. Then flip upside down and incubate at room temperature. If it does not dry in time, leave face up on the room temperature shelf.
Assay II - Examination of RpoS activity Materials: 2 StrepR cultures (LB only, LB + X), microscope slide, H2O2 We want to measure catalase activity (and thus RpoS activity) now that the cells have reached LTSP. 1. 2. 3. 4.
Shake each tube (the tubes with only StrepR E. coli) if a pellet is at the bottom of the tube. Remove 5µl of each of your cultures and add them to a glass slide. Add one drop of H2O2 to each aliquot. Note whether bubbles form (large/small, many/few, form immediately/after time, etc).
Compare the data from Week 3 Discussion (day 2), Week 4 lab (day 14) and today (following day 2) to see whether you are observing changes in RpoS downregulation that have a genetic or physiological basis. 1. If catalase expression is similar on day 2, 14 and the following day 2, RpoS activity is unaltered by LTSP. 2. If catalase expression decreases on day 14, but is restored when the cells are added back to fresh media, RpoS activity is altered by LTSP conditions. This signifies a physiological, but not genomic change. 3. If catalase expression decreases on day 14 and remains decreased in fresh media, this signifies a genomic change to RpoS.
Week 4 – Continue GASP Experiment This week you will determine whether your LTSP E. coli were able to outcompete the fresh NalR E. coli. You will be given a worksheet that must be filled out and submitted to the EEE dropbox within 24 hours after your discussion period. Failure to do so on time will result in a loss of points. As in discussion last week, you will be analyzing the results from the previous class (examining growth 2 days into the GASP experiment) and setting up the 7 day growth plates. Week 4 Lab: I. Bacteria Growth After Stationary Phase (GASP) Assay First, examine the plates from discussion. You want to determine the ratio of StrepR CFU/NalR CFU. As you did previously, find the highest dilution where single colonies are easily distinguished. LB + Strep
LB + Nal
Colonies from Cond I:
Spot #
Colonies from Cond I:
Spot #
Colonies from Cond II:
Spot #
Colonies from Cond II:
Spot #
Colonies from Cond III:
Spot #
Colonies from Cond III:
Spot #
Now we need to convert this to CFU/ml for each condition/plate. LB + Strep
LB + Nal
Condition I CFU/ml:
Condition I CFU/ml:
Condition II CFU/ml:
Condition II CFU/ml:
Condition III CFU/ml:
Condition III CFU/ml:
And finally, we will now see compare levels of StrepR and NalR cells in the culture. StrepR CFU/NalR CFU: Condition I:
Condition II:
Condition III:
You also need to set up your final LB + Strep and LB + Nal plates to examine the StrepR/NalR ratio at day 2 of the experiment. Materials: 2 LB + Strep plates, 2 LB + Nal plates, 3 cultures (Condition I, II, III) 1. 2. 3. 4.
5. 6. 7.
Mix the tubes thoroughly before removing any liquid. You should plate one strain in its entirety before moving on to the next. In an empty petri dish, you will make eight spots with 90µl each. Do not let the drops merge. Eventually you will have 24 spots, which may require you to use both the lid and bottom of the empty petri dish. Take 10µl of one of your cultures (start with condition I) and add it into the first drop. Pipette up and down and stir the drop with the pipette tip to mix the dilution. Using the same tip, pipette 10µl of the first drop onto your LB + Strep plate. You will use the grid below to keep track of your spots. Then pipette 10µl of the same drop onto your LB + Nal plate. Use the grid from both your and your partner’s manuals so that each will be used for one of the plates. Take a new pipette tip and remove 10µl of the first drop and pipette into the second drop. Mix and pipette 10µl onto both plates. Repeat for drops three through eight. Repeat steps 2-6 for your second culture.
8. 9.
Repeat steps 2-6 for your third culture on plate 2. Let the plate dry face up. Then flip upside down and incubate at room temperature. If it does not dry in time, leave face up on the room temperature shelf.
Week 4 Discussion: As you did previously, find the highest dilution where single colonies are easily distinguished. LB + Strep
LB + Nal
Colonies from Cond I:
Spot #
Colonies from Cond I:
Spot #
Colonies from Cond II:
Spot #
Colonies from Cond II:
Spot #
Colonies from Cond III:
Spot #
Colonies from Cond III:
Spot #
Now we need to convert this to CFU/ml for each condition/plate. LB + Strep
LB + Nal
Condition I CFU/ml:
Condition I CFU/ml:
Condition II CFU/ml:
Condition II CFU/ml:
Condition III CFU/ml:
Condition III CFU/ml:
And finally, we will now see compare levels of StrepR and NalR cells in the culture. StrepR CFU/NalR CFU: Condition I:
Condition II:
Condition III:
Appendix 3: Lecture Slides.
Week 1 Lecture
Evolu/on! • Define evolu/on
Evolu/on! • Darwin thought – Organisms had no control over how they evolve – Changes during their life/me have no effect on their offspring – Organisms which happen to have varia/ons which help them to adapt to their environment will have more offspring – Those offspring will have the helpful traits – Organisms which do not have helpful traits will die off
Two Steps of Evolu/on • Muta/on – Change in DNA that leads to expression of a different RNA/protein and thus different phenotype
• Selec/on – Pressure from the environment that makes that phenotype advantageous
Muta%ons occur all the %me, but without selec%on we rarely see them appear in popula%ons
How do we know that beneficial muta/ons happen in the absence of selec/on?
Examined bacteria resistance to viruses
Luria and Delbrück, 1943
Fluctua/on Tests
Red = mutant
Fluctua/on Tests
Red = mutant
Jackpot!
Evolu/on in Bacteria • Bacteria are a good model system for examining evolu/on – Why?
Phases of Bacterial Growth in Batch Culture 10
3
Sta/onary
log10 CFU/mL
9
4
8
2
7
Log
6
5
Long-‐Term Sta/onary Phase (LTSP)
1
Lag
5 4
Death
0
6 12 Hours
2
6
14 10 Days
12
Finkel SE. Nature Reviews Microbiology. 2006 Feb;4(2):113-‐20.
24
36 48 Months
60
Growth Advantage in Sta/onary Phase (GASP) a
log 10 CFU/ml
10
Aged Culture 10-day-old
8 6 4
Un-‐Aged Culture 2 0 0
*
*
9
12
1-day-old
3
6
Day
Finkel & Kolter, PNAS (1999)
GASP is a: A. Phenotypic (non-‐genotypic) change B. Genotypic change C. It depends
Genotypic – In the absence of stress the phenotype remains
Do we see the GASP phenotype?
2 weeks E. coli
Old cells
Fresh cells
Is the GASP phenotype media specific? Grow into LTSP LB
Add to fresh culture
LB + X
• Characterize GASP phenotype in LB and LB + variable with E. coli – Variables -‐ salts, yeast extract, sugars – Work in pairs, group of 4 will test same variable
Growth Advantage in Sta/onary Phase (GASP) a
log CFU/ml
10
Aged Culture 10-day-old
8 6 4 2 0 0
Do we see this? Does this effect change in different media?
Un-‐Aged Culture
*
*
9
12
1-day-old
3
6
Day
Finkel & Kolter, PNAS (1999)
How do we determine the number of colony forming u nits? 10ul culture + 90ul LB, plate 10ul to make spot 1
Spot 1 on plate is 1ul of original culture 1/10th of 100ul (plate 10ul of 10ul culture + 90ul LB) Each addi/onal spot is a 1:10 dilu/on
So growth on spot 1 signifies CFU/0.001ml or CFU/ 10-‐3 ml
How do we determine the number of colony forming u nits? 10ul culture + 90ul LB, plate 10ul to make spot 1
Each addi/onal spot is a 1:10 dilu/on
What does spot 2 signify? A. CFU/ml B. CFU/10-‐1 ml C. CFU/10-‐2 ml D. CFU/10-‐3 ml E. CFU/10-‐4 ml
How do we determine the number of colony forming u nits? 10ul culture + 90ul LB, plate 10ul to make spot 1
Each addi/onal spot is a 1:10 dilu/on
Based on this data, what is the CFU/ml in the culture? A. 6 x 10-‐6 CFU/ml B. 6 x 106 CFU/ml C. 6 x 10 -‐7 CFU/ml D. 6 x 10 7 CFU/ml E. None of the above
Is survival in sta/onary phase due to genotypic or phenotypic changes? RpoS gene is downregulated in LTSP RpoS regulates catalase expression If RpoS is downregulated, we DO/DO NOT expect to see bubbles when adding H2O2 to our cells? A. Do B. Do not
Is RpoS regulated by genotypic or phenotypic change?
Is survival in sta/onary phase due to genotypic or phenotypic changes? • Genotypic vs. Phenotypic change? In fresh cells – expect to see catalase ac/vity In LTSP cells – expect to see low catalase ac/vity When adding LTSP cells back to fresh media? If now we see (A. High B. Low) catalase ac/vity, the LTSP result would be due to a phenotypic change. If now we see (A. High B. Low) catalase ac/vity, the LTSP result would be due to a genotypic change.
Module Timeline
W1 Start StrepR culture in LB, LB + X
Catalase, Rifampicin
W2 β-‐gal stress assay
Focus on the experimental ques/ons!
W3 Start GASP Day 0 plates Catalase
GASP Day 2 plates Catalase
W4
GASP Day 7 plates
Week 2 Lecture
Evolu/on Experiment • Can we observe the GASP phenotype? • Is survival in LTSP dependent on genotypic or phenotypic changes? • Does cell stress impact GASP?
Evolu/on Experiment 10
3
Sta/onary
log10 CFU/mL
9
4
8
2
7
Log
6
5
Long-‐Term Sta/onary Phase (LTSP)
1
Lag
5 4
Death
0
6 12 Hours
2
6
14 10 Days
12
Finkel SE. Nature Reviews Microbiology. 2006 Feb;4(2):113-‐20.
24
36 48 Months
60
Growth Advantage in Sta/onary Phase (GASP) a
log 10 CFU/ml
10
Aged Culture 10-day-old
8 6 4
Un-‐Aged Culture 2 0 0
*
*
9
12
1-day-old
3
6
Day
Finkel & Kolter, PNAS (1999)
Evolu/on Experiment • Can we observe the GASP phenotype? • Is survival in LTSP dependent on genotypic or phenotypic changes? • Does cell stress impact GASP? • Does muta%on frequency impact GASP?
Evolu/on Experiment • Can we observe the GASP phenotype? • Is survival in LTSP dependent on genotypic or phenotypic changes? • Does cell stress impact GASP?
How to measure cell stress Catalase promoter
GFP
• bolA reporter gene – What is a reporter gene? • A gene that is easy to visualize – anach to your promoter of interest to measure expression at that promoter • Ex. What if we wanted to measure expression of the catalase gene using a GFP reporter? Describe the reporter gene you would use.
How to measure cell stress • bolA reporter gene Transcrip/on
bolA promoter
β-‐Galactosidase Gene
How to measure cell stress • bolA reporter gene
β-‐gal protein
ONPG
Galactose
+ Ortho-‐nitrophenol (yellow, absorbs 420nm light)
How to measure cell stress • But what if there is more cell growth in one tube? • So need to normalize OD value to CFU/ml
Determine CFU/ml 10ul culture + 90ul LB, plate 10ul to make spot 1
What is the CFU/ml for this culture? (5x109CFU/ml)
Each addi/onal spot is a 1:10 dilu/on
How to perform the dilu/on assay • Determine CFU/ml • Determine stress levels per culture OD/(CFU/ml)
How to measure muta/on frequency? Look at an/bio/c resistance of E. coli culture
Plate culture on LB + Rifampicin – calculate RifR CFU/ml But need to normalize to CFU/ml of culture – Why?
How to measure muta/on frequency? Muta/on Frequency is RifR CFU/ml _ Viable Cells CFU/ml
Plate on LB (dilu/on spoong) Plate 100ul on LB + Rif
You plate 100ul of your culture on the LB + Rif plate and see 25 colonies grow. What is the RifR CFU/ml for your culture? A. 25 RifR CFU/ml B. 250 RifR CFU/ml C. 2,500 RifR CFU/ml
D. 25,000 RifR CFU/ml E. 250,000 RifR CFU/ml
Week 3 Lecture
Evolu/on Experiment 10
3
Sta/onary
log10 CFU/mL
9
4
8
2
7
Log
6
5
Long-‐Term Sta/onary Phase (LTSP)
1
Lag
5 4
Death
0
6 12 Hours
2
6
14 10 Days
12
24
36 48 Months
But what is the GASP phenotype?
60
Growth Advantage in Sta/onary Phase (GASP) a
log 10 CFU/ml
10
Aged Culture 10-day-old
8 6 4
Un-‐Aged Culture 2 0 0
*
*
9
12
1-day-old
3
6
Day
Finkel & Kolter, PNAS (1999)
So how do we know whether our cells have the GASP phenotype? • Add 2 week old cells to culture with new cells • Measure growth (CFU/ml) of aged and un-‐aged cells at day 0, 2, and 7. • Plot on a log graph of CFU/ml vs. /me • How do we dis/nguish old vs. new cells?
Analyze this data • Determine CFU/ml for each plate • Make plot of log CFU/ml vs. /me
Day 7
Analyze this data • What is the CFU/ml for the aged cells in LB at day 0? A. 103 B. 105 C. 10-‐3 D. 10-‐5
Analyze this data
LB Culture
LB+ X Culture
The GASP phenotype is due to genotypic changes, but what about survival in LTSP? Catalase ac/vity in fresh cultures Catalase ac/vity in LTSP cultures (this week lab) Catalase ac/vity in when LTSP cultures added back to fresh media (this week discussion)
Evolu/on Expt this week • GASP – Add aged cells to fresh cells • Condi/on I – LB LTSP culture to new cells in LB • Condi/on II – LB + X LTSP culture to new cells in LB • Condi/on III -‐ LB + X LTSP culture to new cells in LB + X
• Catalase – Add aged to fresh media only (no fresh cells) • LB LTSP culture to new LB • LB + X LTSP culture to new LB + X
So each pair will set up 5 new tubes this week
Week 4 Lecture
Evolu/on experiment • Check day 2 GASP plates from discussion • Set up day 7 GASP plates in lab
Evolu/on experiment • Why are DNA muta/ons important for evolu/on? Altered DNA leads to altered RNA/protein which can cause beneficial phenotypic changes
• Why is stress important for evolu/on? Without stress, there is no driving force to keep those beneficial muta/ons in the popula/on
Evolu/on experiment • GASP phenotype – Growth of aged, unaged cells in mixed culture at day 0, 2, and 7
• Survival in LTSP – due to genotypic or phenotype changes? – Measure RpoS/catalase ac/vity in fresh media, sta/onary phase cultures, back in fresh media
Genotypic vs. Phenotypic Changes A cell is able to grow at elevated temperatures due to ac/va/on of a signaling pathway leading to overexpression of a chaperone protein. A. Phenotypic (non-‐genotypic) change B. Genotypic change C. Neither
Genotypic vs. Phenotypic Changes A cell is able to survive in the presence of a detergent due to a promoter muta/on that leads to more pep/doglycan synthesis. A. Phenotypic (non-‐genotypic) change B. Genotypic change C. Neither
Evolu/on experiment • GASP phenotype – Growth of aged, unaged cells in mixed culture at day 0, 2, and 7
• Survival in LTSP – due to genotypic or phenotype changes? – Measure RpoS/catalase ac/vity in fresh media, sta/onary phase cultures, back in fresh media
• Cell stress – related to GASP? – Measure β-‐gal ac/vity
Cell Stress • What is a reporter gene? – 1 sentence • bolA is ac/vated when the cell is stressed – Why do we use a reporter gene and not look at bolA expression?
• Draw a picture of our bolA-‐β-‐galactosidase reporter gene bolA promoter
β-‐galactosidase
Imagine we want to measure gela/nase expression with a GFP reporter. What would this reporter gene look like? A. bolA promoter with gela/nase gene B. Gela/nase promoter with bolA gene C. GFP promoter with gela/nase gene D. bolA promoter with GFP gene E. Gela/nase promoter with GFP gene
Evolu/on expt – GASP • Remember, novel data = no right answer! Unaged Cells Aged Cells Day 0
Day 2
Day 7
What conclusion would you make from this data?
Based on this data, which of the following is the best conclusion?
A. The GASP phenotype is not observed B. CaCl2 does not impact the GASP phenotype C. CaCl2 leads to greater cell stress D. CaCl2 inhibits the long-‐term sta/onary phase growth advantage E. CaCl2 is toxic to cells
Along with this data, we find that cells grown in LB + CaCl2 had 2x as much β-‐gal ac/vity. What conclusion do you make?
A. Increased cell stress may lead to an enhanced GASP phenotype B. Decreased cell stress may lead to an enhanced GASP phenotype C. Increased cell stress may lead to greater culture growth. D. Decreased cell stress may lead to greater culture growth.
Evolu/on worksheet • One per pair • Submit to EEE dropbox within 24hr arer discussion • Evolu/on data – LB vs. LB + X-‐ cell stress, muta/on frequency, GASP – GASP data in form of graphs (Log CFU/ml vs. day)
• Thought ques/on – How should variable X impact cell growth?
Appendix 4: GASP Worksheet. Names/SID: Partner: TA: Your Variable: 1. How do we expect the addition of this variable to change the media? Include a specific hypothesis in your answer. Include at least one primary literature article as support.
2. Based on this, speculate how we expect this to impact the GASP phenotype. Explain your answer using the literature.
3. In the end, how did the variable impact the GASP phenotype? Explain in regards to the following scenarios. Provide one graph of CFU/ml vs. Day for each of the scenarios below. Each graph should have 2 lines (one for aged cells, one for fresh cells). a. Initial growth in LB, added to fresh LB
b. Initial growth in LB + X, added to fresh LB
c. Initial growth in LB + X, added to fresh LB + X
4. Based on the data from part 3, was your hypothesis correct? YES or NO. If yes, elaborate on what this means regarding the GASP phenotype. If no, provide an evidence-based (including at least one primary literature article) explanation.
5. The mutation frequency was (HIGHER/LOWER/THE SAME) in LB + X media compared to LB media. Provide the RifR CFU/ml and overall CFU/ml from each culture to support your conclusion. LB RifR CFU/ml:
LB + X CFU/ml:
LB CFU/ml:
LB + X CFU/ml:
6. The variable media created a (MORE/LESS/THE SAME) stressful environment for the cells. Provide the OD420/CFU/ml values to support your conclusion. LB OD420:
LB + X OD420:
LB CFU/ml:
LB + X CFU/ml:
7. RpoS regulation a. Did RpoS expression decrease in LTSP? YES or NO. Describe the data that allowed you to come to this conclusion.
b. If RpoS expression did drop in stationary phase, was this a GENOTYPIC or PHENOTYPIC change? Describe the data that allowed you to come to this conclusion.
c. Do the answers to a or b vary in the variable medium?
8. Overall conclusions a. Is there any relationship between mutation frequency and the GASP phenotype?
b. Is there any relationship between cell stress and the GASP phenotype?
Appendix 5: Module Pre-/Post-Test. ID:
KEY
The following quiz will be used to measure the degree to which the evolution experiment you will perform in the next few weeks is successful as a teaching tool. Please put your full effort into answering these questions. As stated at the beginning of the quarter, your answers will be analyzed but only displayed in aggregate with the remainder of the class. *For any questions you do not know the answer to, write “Do not know”. 1. Mutations occur in the (PRESENCE/ABSENCE) of environmental stress. Why is environmental stress necessary for evolution? (0.5pt for absence, 0.5pt for explanation) Without selection, there is no driving force for those with a favorable mutation to pass this on to future generations. 2. All microbial growth media contain salts. What is the purpose of this ingredient? To maintain osmotic balance with the contents in the cell. 3. Reporter genes can be used to measure gene expression. Imagine you were working with a superoxide dismutase (SOD) reporter gene. Describe the structure of this reporter gene in molecular terms. SOD promoter and reporter gene open reading frame (something like GFP, lacZ, etc). 4. If you have a culture with 1010 bacterial cells, and dilute the culture 1:10 five times, how many cells will be in the final dilution? 105 5. True or False. The terms bacterial species and bacterial strain mean the same thing. If true, describe a scenario you may use one or the other. If false, explain why. False. Something along the lines of strain refers to a specific group within a species (only +0.25 for getting false correct but not a good explanation)
6. (Bacteria Evolution question) For the following scenarios, are they examples of phenotypic changes in the cell (P), genotypic changes in the cell (G), or neither (N)? a. Lactase protein is degraded in response to a lack of lactose present in the extracellular space. P b. Overexpression of a drug pump occurs when antibiotics bind to cell membrane proteins, activating a signaling pathway that increases drug pump transcription. P c. Decreased hexokinase levels are due to an altered nucleotide that leads to an altered amino acid and destabilized protein structure. G
Jane the researcher learns that for yeast grown into stationary phase, the transcription factor Phx1 may be upregulated. Once she establishes her experimental protocol, Jane generates the following hypothesis: Gene expression is important for yeast survival in stationary phase. 7. Rewrite the statement so it is a strong hypothesis. Must be more specific, ex. Phx1 upregulation (or Phx1 in general) is important for yeast survival in stationary phase. 8. Briefly describe how Jane could test the hypothesis. Take 2 strains, one with Phx1 and one with Phx1 knocked out (or one where the Phx1 upregulation pathway is shut down). Grow strains in stationary phase and measure growth in some way (with spectrophotometer, dilution on a plate, etc). I would distribute points with half credit for set up – compare growth of 2 strains in stationary phase, and half credit for how to measure growth What volume are the following pipettes set to? Include units. 9. P1000
330ul
10. P20
1.7ul
Appendix 6: Self-Assessment and Exam Questions. Self-Assessment Questions 1. I am very capable of using pipettors. A. Strongly Agree B. Agree
C. Neutral
D. Disagree
E. Strongly Disagree
2. I can comfortably describe to my peers the molecular basis of evolution. A. Strongly Agree B. Agree C. Neutral D. Disagree
E. Strongly Disagree
Exam Questions (taken from multiple exams over two quarters) 1. In one sentence or less, why does the “jackpot” phenotype found by Luria and Delbrück illustrate that mutations occur spontaneously and not necessarily in the presence of a stress? High frequency of mutations in population (jackpot) means that the mutation was introduced early in the experiment before exposure to the stress. 2. A selective environment (IS/IS NOT) a necessary for evolution to occur. Explain why. Without the selective environment there is no benefit to obtaining a given mutation and thus those that have it will not be more likely to have offspring 3. Luria and Delbruck concluded that beneficial mutations (DO/DO NOT) occur in the absence of stress. This was evident due to the jackpot phenotype they observed in their experiment. Why did the jackpot phenotype allow them to make this conclusion? Jackpot shows that mutations occurred very early on in the experiment (before stress added) 4. Luria and Delbruck found bacteria that were resistant to bacteriophages. One resistant bacterium overexpresses Protein X, a cell wall protein. To test if this is due to a phenotypic or genotypic change, they measure Protein X levels in the (ABSENCE/PRESENCE) of the bacteriophage. If resistance is due to a phenotypic change, what result would they see in this follow up experiment? Would now see lower/normal levels of Protein X 5. We would expect that an environment that causes a greater mutation frequency in cells would lead to (MORE/LESS/CAN’T DETERMINE) cells capable of overcoming a toxic stress. Explain. More mutations could lead to more cell death but it could also lead to more beneficial mutations that result in greater life, depends on the scenario and what is required to overcome the stress 6. A yeast strain is able to grow in elevated ethanol concentrations due to the addition of a mutant transcription factor. This leads to the overexpression of a dozen genes (Genes A through L). In this case, the ability to tolerate high ethanol concentrations is a due to a (GENOTYPIC/PHENOTYPIC/NEITHER/BOTH) change. Under what condition would we grow these cells to experimentally confirm this answer and what phenotype would we expect to see? Condition: low ethanol Phenotype: regular expression of A-L 7. A Pseudomonas culture is grown in media with the antibiotic ciprofloxacin. One cell is resistant due to overexpression of the target gene, DNA gyrase. What result would we need to see when growing these cells without cipro to conclude that resistance was due to a phenotypic change? Regular/low DNA gyrase levels 8. A is for Antimycin, an antibiotic that targets cytochrome c oxidase. a. A soil isolate is resistant to antimycin and also is found to overexpress cytochrome c. This overexpression could be due to a (PHENOTYPIC/GENOTYPIC/EITHER) change in the cell.
b. To test your answer in question 1b, you would grow the bacteria in the (PRESENCE/ABSENCE) of antimycin. What cellular characteristic would you be looking for in this environment to answer the phenotypic/genotypic question? Examine whether levels of cytochrome c oxidase are high or low 9. What question are we trying to answer with the stress assay experiment? Does cell stress vary by environment? 10. Rather than look at catalase activity with H2O2, what if decide to make a β-gal reporter gene to measure catalase expression. Describe what this reporter gene would look like. Catalase promoter fused to the β-galactosidase gene. 11. What question are we trying to answer using the rifampicin experiment? What is the mutation frequency in the culture? 12. Answer the questions regarding the data to the right. a. Based on this data, the GASP phenotype (IS/IS NOT/CAN’T CONCLUDE) being observed. b. For the day 0 LB + Nal culture, calculate the CFU/ml. CFU/ml LB Strep: 7x109 13. In class, you used a β-galactosidase reporter gene. This involved ONPG, which acts as the enzyme’s (SUBSTRATE/PRODUCT). We then measure reporter gene activation using a spectrophotometer set at 420nm light. On accident, it is set at 500nm. This (WILL/WILL NOT) allow us to measure cell stress. Explain why or why not. The ONPG product does not absorb 500nm light thus any light absorbed by the solution will not be indicative of β-gal activity 14. You are testing growth of cells in LB and LB + sucrose. You determine cell stress by measuring βgalactosidase activity and get an OD420 value of 0.15 for LB and 0.14 for LB + sucrose. You then perform a dilution assay (with 1:10 dilutions), plate the cells on LB, and get the data below. a. Based on this, which media causes more cell stress? LB or LB + SUCROSE. Explain. The OD value from the stress reporter is equal despite fewer cells in the LB condition (so ratio of OD/CFU is higher) b. In the data from part a, the first spot was generated by taking 1µl of a culture and diluting it in 100µl. 10µl of this dilution was then added to create that spot. Each additional spot is 10µl of a 1:10 dilution. What is the CFU/ml of the LB + sucrose culture? 4x107 CFU/ml 15. Based on the growth on the plate (assuming it is set up the same way that you did so in lab), calculate the concentration of the original culture. Include units. This is real data, analyze to the best of your ability. 6 x 109 to 1 x 1010 CFU/ml 16. Write a detailed hypothesis for a potential GASP experiment based on the info below. E. coli expresses many different DNA polymerase genes. It is possible to make mutants that only express a single gene, for example only DNA pol II or DNA pol V. When cultures are maintained at a low density, growth between pol II-only and pol V-only expressing strains is similar. When cultures are maintained at a high density, growth of pol V-only expressing strains is much greater than pol II-only expressing strains. Ex. PolV-only expressing cells will be capable of exhibiting a much stronger GASP phenotype compared to PolII-only expressing cells.
Appendix 7: Pipetting Test Instructions. Administering the Pipetting Lab Practical Following the pipetting practice making 1:4 dilutions in the 24 well plates, each student will be subject to this pipetting exercise. Students will do this twice in the class (once during lab practical) and they only need to pass once. They will not be told if they pass/fail the first time, so they’ll want to attempt to pass on both occasions. “Passing” will be based on the accuracy of their pipetting (within 0.050 OD) as measured by the OD595. The standard value will be determined by the instructor who will perform each dilution three times and take the average value. In a corner of the room you will have the following: a. b. c. d. e. f. g. h.
Full set of micro pipettes (P20, P200, P1000) in rack Boxes of yellow and blue tips Water for dilution 1% bromophenol blue dye Plastic cuvettes, spectrophotometer (set to 595 nm) Water blank Dilution cards Scratch paper, calculator, pen
You will call the students up one at a time to perform the practical (or students can volunteer). Each student will need to perform a dilution (randomly chosen with a card) and you will control the spectrophotometer (blanking and reading their sample). The dilutions we will use are 1:30, 1:45, 1:90, 1:150. Keep track of your students on a spreadsheet with the following columns: Student Name
Dilution (1:x)
OD595
When they arrive let them get comfortable, set a timer for 3 minutes, and then follow the instructions below: 1.
Have them select a dilution card and tell them: “Prepare 1ml of a 1:X dilution of the bromophenol blue into the water in a cuvette.”
2.
Start the timer for 3 minutes.
3.
Do not answer any questions the student may ask; you cannot give them any advice on which pipette to use or how to perform the procedure.
4.
As they are working, blank the spectrophotometer with the water cuvette (arrow pointing left towards the light source, then hit blank).
5.
The student must stop and set everything down at the end of 3 minutes. Place the cuvette in the spectrophotometer (again in the correct orientation, then hit read).
6.
When the student has left, reset the station before the next student arrives, including placing pipettes back in the rack, placing lids on tip boxes, and remove any used scratch paper.
7.
When complete and all the values are in your spreadsheet, email to the instructor.
Appendix 8: Example Student Hypotheses. We hypothesize that growth in media with added NaCl will add stress and allow for the aged bacterial cells to develop mechanisms or mutations that help them survive in hypertonic solutions, which will translate to overall better survival, thus improving the GASP phenotype. Based on the fact that tryptone provides additional nutrients, including nitrogen and amino acids, we speculate that more cells are able to enter the long-term stationary phase. Therefore, more cells will be seen with the GASP phenotype. Addition of glucose means the bacteria will have more carbon source compared to the control and thus, it will have a less stressful environment. The strain that survives in the LB only conditions will be better adept at surviving and show a stronger GASP phenotype. Since not all E. coli can utilize sucrose as a nutrient source, the LTSP cells will have an advantage when sucrose is in the media. Cells that have been cultured with sucrose for a long period would most likely have adapted to survive in the presence of sucrose and will better outcompete unaged cells. MgCl2 in the environment would produce a change in osmolarity, and that stress may encourage mutation. Thus, the salt and increase in mutation frequency should support and increase the growth advantage in stationary phase phenotype.
