Micro LAB EXAM 3

24 July 2022
4.7 (114 reviews)
69 test answers

Unlock all answers in this set

Unlock answers (65)
question
Ex. 5-4 (Methyl Red/VP) Some protocols call for a shorter incubation time for the MR and VP tests. Other protocols allow for up to 10 days incubation with virtually no risk of producing a false positive. a) Which of the two tests would likely produce more false negatives wtih a shorter incubation time? Why? b) which test would likely benefit most from a longer incubation time? Why?
answer
a) A short incubation time runs the risk of a false negative on the VP test, since not enough time may have elapsed to produce an adequate amount of acetoin to give a positive result. The pH change associated with mixed acid fermentation occurs fairly quickly and is less likely to be affected by shorter incubation. b) The VP test would benefit most from a longer incubation time because it takes a while for the 2,3-butanodiol to be produced.
question
Ex. 5-4 (Methyl Red/VP) Would a false negative result for the VP test more likely be attributable to poor sensitivity or to poor specificity of the test system?
answer
A false negative on the VP test is likely due to a lack of sensitivity- there is not enough acetoin to react with the test reagents.
question
Ex. 5-4 (MR/VP) Why were you told to shake the VP tubes after the reagents were added?
answer
The indicator reaction requires oxygen to oxidize acetoin to diacetyl.
question
Ex. 5-4 (MR/VP) Why is the methyl red test read immediately after addition of methyl red reagent and the Voges-Prokauer read up to 60 minutes after addition of VP reagents A and B?
answer
The methyl red test is based on a pH changed, which can be demonstrated instantaneously. Reading the VP test requires a chemical reaction and time must be allowed for this reaction to occur
question
Ex. 5-4 (MR/VP) Some microbiologists recommend re incubating organisms producing methyl red negative results for an additional 2 to 3 days. Why do you think this is done?
answer
It is to give the organisms more time to produce enough acid to lower the pH to the point of making the methyl red indicator turn red. Longer incubation minimizes the possibility of a false negative
question
Ex. 5-9 Citrate Test Consider the uninoculated tube. a) Is it a positive or a negative control? b) What information is provided by the uninoculated control?
answer
a) It is a negative control b) It provides a baseline color (green) for "no change' in the medium. This allows detection of slight color changes in the experimental tubes that might not be apparent without comparison to the original color. It also verifies the sterility of the medium.
question
Ex. 5-9 Citrate Test Many bacteria that are able to metabolize citrate (as seen in the citric acid cycle) produce negative results in this test. Why? Be specific.
answer
Citrate (citric acid) is the first intermediate of the citric acid (Krebs) cycle where it is ultimately catabolized to CO2 and oxaloacetic acid (which reenters the cycle). However, the citrate test does not detect the ability of an organism to perform the citric acid cycle. It detects the ability of the organism to obtain citrate from the environment and use it. Thus, an organism could synthesize its own citrate in the citric acid cycle, but not be able to use citrate from the environment because it lacks the ability to transport it into the cell; it is citrate permease (-).
question
Ex. 5-9 Citrate Test explain how an organism that possesses the citrate lyase enzyme might not test positively on Simmons citrate agar. Is this a false negative result? Why or why not?
answer
The citrate test determines if an organism has the ability to transport citrate into the cell with the enzyme citrate permease and survive with citrate as the sole carbon source. Citrate lyase, which catalyzes the conversion of citrate into oxaloacetate and acetate, is associated with how citrate is used, not how it gets into the cell. So, an organism could be citrate-negative (it doesn't make citrate permease), but still be citrate lyase-positive.
question
Ex. 5-20 SIM Medium Consider the uninoculated SIM tube. a) Is it a positive or a negative control in each test? b) What purpose does it serve in the sulfur reduction test? c) What purpose does it serve in the indole test? d) What purpose doe sit serve in the motility test?
answer
a) In each test it is a negative control and minimizes concern about false positive results b) In the sulfur reduction test it demonstrates that nonenzymatic reduction of sulfur to H2S doesn't occur c) In the indole test it demonstrates the nonenzymatic hydrolysis of tryptophan to indole and pyruvase doesn't occur. It also shows that there is nothing in the medium that reacts with Kovac's reagent to produce a red color d) In the motility test it illustrates that the medium doesn't develop cloudiness without organisms present. It also illustrates what the medium lookds like without cloudiness. This assists with interpretation of positive results where the medium has become uniformly cloudy throughout and there is no clear medium remaining with which to compare the cloudiness
question
Ex. 5-20 SIM Medium The sulfur reduction test is not able to differentiate H2S produced by anaerobic respiration and H2S produced by putrefaction. Is this inability the result of poor sensitivity or poor specificity of the test system?
answer
This is due to a limitation in the test's specificity. It lacks the ability to discriminate between the 2 types of metabolism.
question
Ex. 5-20 SIM Medium What factors dictate the choice of tests included in a combination medium?
answer
The tests must be compatible. That is, a test relying on raising the medium's pH would not make a good partner for another test relying on lowering the pH. Likewise, two tests that use color as an indicator would not be compatible (unless the colors are separated, as in SIM medium - black in the agar, red in the alcohol layer). Further, it is most practical if the tests chosen are relevant to identification of the organisms likely to be inoculated into it.
question
Ex. 5-20 SIM Medium Which ingredient(s) could be eliminated if this medium were used strictly for testing indole production? Explain.
answer
Sodium thiosulfate, the final electron acceptor in anaerobic respiration, is easily removed, as would be the ferrous ammonium sulfate, which is an indicator of sulfur reduction. However, cysteine, involved in the second type of sulfur reduction, is an animo acid present in pancreatic digest of casein and peptic digest of animal tissue and it could not be removed in any practical way.
question
Ex. 5-21 TSIA Medium As mentioned in Theory, the fermentation reading with TSIA and KIA must take place between 18 and 24 hours after inoculation. a) Why is this true? b) Is timing as critical with H2S readings? Why or why not?
answer
a) protein digestion products alkaline end products, but these don't significantly affect pH because the acid end products of fermentation (if any) are more abundant. However, given enough time an organism will exhaust the sugar and stop producing acid end products, leaving only the alkaline end products and the pH goes up. This result in an alkaline reaction due to a reversion. Therefore, it is critical to read the tubes before a reversion can occur and replace a positive acid reaction. b) Timing isn't critical because there is no reversion associated with H2S production.
question
Ex. 5-21 TSIA Medium You learned in Theory that if the black precipitate obscures the color of the butt it must be acidic and scored as "A." Why do you think this is true?
answer
Thiosulfate reduction requires hydrogen ions. Because the medium starts out with a neutral pH, fermentation supplies the H+ when acid is an end product.
question
Ex. 5-21 TSIA Medium TSIA and KIA are complex media with many ingredients. What would be the consequnces of the following mistakes in preparing this medium? a) 1% glucose is added rather than the amount specified in the recipe.
answer
This tenfold increase in glucose would delay glucose exhaustion in the medium and would interfere with interpretation of lactose/sucrose fermentation results. That is, the medium would stay yellow longer and wouldn't revert to red in the case of a glucose fermenter that then catabolizes peptone.
question
Ex. 5-21 TSIA Medium TSIA and KIA are complex media with many ingredients. What would be the consequnces of the following mistakes in preparing this medium? b) Ferrous ammonium sulfate (or ferric ammonium citrate in KIA) is omitted.
answer
In TSIA, the loss of a sulfur source for sulfur reduction would be compensated for by the sodium thiosulfate. However, without the ferrous/ferric ion, sulfur reduction would not be indicated. These ions react with colorless H2S produced by sulfur-reducing bacteria to form the black precipitate ferric sulfide. Loss of the ammonium could have a minor effect on amino acid synthesis, but both media are richly supplied with protein sources (casein and animal tissue digest) that would compensate.
question
Ex. 5-21 TSIA Medium TSIA and KIA are complex media with many ingredients. What would be the consequnces of the following mistakes in preparing this medium? c) Casein and animal tissue are omitted
answer
The media would not be as nutrient rich and growth could be limited. Without casein and animal tissue, the only carbon sources would be glucose, lactose, and sucrose (TSIA only), and ammonium ion would be the only nitrogen source
question
Ex. 5-21 TSIA Medium TSIA and KIA are complex media with many ingredients. What would be the consequnces of the following mistakes in preparing this medium? d) Sodium thiosulfate is omitted.
answer
One source (In TSIA) and the only source for sulfur reduction due to anaerobic reparation would be lost. Unless the organism can use the sulfate from ferrous sulfate in TSIA, sulfur reduction by anaerobic respiration would not be detectable. There would be no change in sulfur reduction due to catabolisms of the amino acid cysteine.
question
Ex. 5-21 TSIA Medium TSIA and KIA are complex media with many ingredients. What would be the consequnces of the following mistakes in preparing this medium? e) phenol red is omitted
answer
No pretty colors would be seen. It is the phenol red that turns yellow and red and allows detection of fermentation and peptone catabolism
question
Ex. 5-21 TSIA Medium TSIA and KIA are complex media with many ingredients. What would be the consequnces of the following mistakes in preparing this medium? f) the initial pH is 8.2
answer
It would take greater acid production to produce a yellow color in the medium for the fermenters (remember that each pH unit represents a tenfold change in hydrogen ion concentration)
question
Ex. 5-21 TSIA Medium TSIA and KIA are complex media with many ingredients. What would be the consequnces of the following mistakes in preparing this medium? g) the agar butt is shallow rather than deep.
answer
The deep butt produces an anaerobic zone. Making the butt shallow results in more of it being close to the surface when the oxygen is. The "anaerobic" part of the medium may not actually be anaerobic, which could affect fermentation
question
Ex. 5-3 PR Broth Consider the controls a) were the uninoculated controls positive or negative controls, and what purpose did they serve? b) What purpose did the PR base broths serve?
answer
a) Uninoculated tubes are used in negative controls to show what the original color of the medium was. This allows detection of slight color changes in the experimental tubes that might not be apparent without comparison to the original. They also verify the sterility of the medium b) Inoculation of PR base medium (without sugar) verifies that the color changes are actually due to microbial activity on the carbohydrate and not due to metabolism of other ingredients in the medium
question
Ex. 5-3 PR Broth Early formulations of this medium used a smaller amount of carbohydrate and occasionally produced false (pink) results after 48 hours. This phenomenon is called reversion a) Why do you think this happened?
answer
With a smaller amount of carbohydrate, longer incubation resulted in a reversion reaction. That is, acid was produced until the carbohydrate was exhausted, then the organism performed oxidative deamination of peptone, which produces alkaline end products
question
Ex. 5-3 PR Broth Early formulations of this medium used a smaller amount of carbohydrate and occasionally produced false (pink) results after 48 hours. This phenomenon is called reversion b) List at least two steps, as a microbiologist, you could take to prevent the problem
answer
There are two obvious solutions to this problem: Formulate the medium with more carbohydrate so it would not be completely consumed, and shorten the incubation time so the organism doesn't have enough time to use up the carbohydrate.
question
Ex. 5-3 PR Broth Suppose you inoculate a PR broth with an organism known to be a slow-growing fermenter. After 48 hours, you are slight turbidity but score it as (-/-). a) Is this result a false positive or a false negative? b) Is this false result caused by poor spcificity or poor sensitivity of the test system?
answer
a) The organism is a known fermenter. Scoring it as -/- would be a false negative. b) The organism is growing in the medium, but hasn't produced enough acid for the pH indicator to change to yellow. This is a failure in the sensitivity of the medium.
question
Ex. 4-4 MSA In your own words, what is the application (purpose) of MSA?
answer
Most clinical isolates are incapable of growing in 7.5% salt, but Staphylococcus species can, which makes this a selective medium used for isolated Staphylococcus species. It is also differential, because most Staphylococcus species are not able to ferment mannitol, but S. aureus is.
question
Ex. 4-4 MSA Which ingredients in MSA supply(ies)? a) Carbon? b) Nitrogen?
answer
a) Beef extract, peptone, and mannitol all contain usable carbon b) Beef extract and peptone contain organic nitrogen
question
Ex. 4-4 MSA Is MSA a defined or an undefined medium? Provide the reasoning behind your choice and explain why this formulation is desirable.
answer
MSA is an undefined medium because the exact composition of beef extract and pepone are unknown. The goal of this medium is to isolate desired organisms, and once the undesired organisms have been inhibited you want the desired ones to grow well. The variety of biochemicals in beef extract supplemented by the peptone make the medium more likely to satisfy the varied nutritional needs of the organisms being selected for.
question
Ex. 4-4 MSA In your own words, what is the role of sodium chloride in MSA and how does it work?
answer
Sodium chloride is the selective agent in this medium. Its presence dehydrates most cells
question
Ex. 4-4 MSA In your own words, what are the roles of mannitol and phenol red in MSA?
answer
The ability to ferment mannitol to acid end products makes this a differential medium. The acid end products are detected by the pH indicator phenol red, which turns yellow.
question
Ex. 4-4 MSA Growth on the MSA and NA plates was recorded as "good growth," "poor growth," or "no growth." These are qualitative and, at least for the first tow, subjective terms. What did you use to establish what constituted "good growth"?
answer
The NA plate inoculated with the same organisms provided examples of what "good growth" for each organism looks like on a nonselective medium. Without these for comparison, it wouldn't be possible oto tell if sparse growth on MSA was due to inhibition or was just normal growth for that species.
question
Ex. 4-4 MSA Why wouldn't it be advisable to compare growth of the organisms on each plate to each other?
answer
On the NA plate, "good growth" for one organism may be very dense, wherease "good growth" for another might be fairly thing. In the context of this exercise you want to compare the amount of growth of each organism on the selective medium against its growth on NA to see if inhibition occurred. You wouldn't want to compare growth of the organisms on the MSA plate to each other because you wouldn't know which ones were inhibited which were normal.
question
Ex. 4-4 MSA Would removal of sodium chloride from MSA alter the medium's sensitivity or specificity?
answer
It would alter specificity because organisms that "shouldn't" grow on it would, and some of those might be able to ferment mannitol to acid end products. It wouldn't likely alter sensitivity because you would still probably be able to detect growth of organisms that "should" grow on it.
question
Ex. 4-4 MSA Suppose a mistake is made in preparing a batch of MSA and the starting pH is a 7.4 instead of 7.0-7.2. Would that affect the medium's sensitivity or specificity?
answer
If a batch of MSA were made at a higher than normal pH, it would take more acid production from mannitol fermentation to produce the yellow color of a positive. It's possible, then, that a true positive for mannitol fermentation might not change the medium's color in the allowed incubation time and would be incorrectly recorded as a negative. this would affect the mediums sensitivity.
question
Ex. 4-4 MSA With the diversity of microorganisms in the world, how can a single test such as MSA be used to "confidently" identify Staphylococcus aureus?
answer
If the sample plated is one taken from a random source, identification would not be as certain. However, in clinical situations, the source is known and S. aureus is likely to be there. Since the salt is selective for staphs, and S. aureus is the only staph to ferment mannitol, presumptive identification can be made.
question
Ex. 5-14 DNase Test In Theory on page 367 fo this exercise, the disassembly of DNA was described as a "depolymerization." What other term apply to the process?
answer
Catabolic and hyrolysis also apply to this reaction
question
Ex. 5-14 DNase Test A positive result for the DNA hydrolysis test does not ditinguish between Staphylococcus DNase and the DNase produced by Serratia. If you had the expertise to correct this weakness in the system, would you be improving the test's sensitivity or its specificity?
answer
You would improve its specificity because it would now be able to discriminate between the two types of Dnase
question
Ex. 5-14 DNase Test Suggest a reason why this test is read after only 24 hours while other tests may take a week. What would be the likely consequence of incubating DNase agar for a week?
answer
The suggested incubation time is related to the activity of the enzyme. Some enzymes produce a result faster than others. In fact, a DNase-positve organism may blear the entire plate if incubated too long. A gelatinase-positive organism can't over liquefy the nutrient gelatin, but an overly cleared DNAase plate is impossible to correctly evaluate
question
Ex. 5-14 DNase Test Why is the uninoculated control relatively unnecessary in this test?
answer
Usually there is plenty of medium that is not cleared (provided it is not incubated too long) to illustrate "no reaction" even when a specific sector of the plate is not designated as the uninoculated control.
question
Ex. 5-14 DNase Test Why is it advisable to use a positive control along with organisms that you are testing?
answer
As usual, a positive control indicates that the medium was prepared correctly and lends confidence in negative results being true negatives, not the result of the medium not functioning correctly.
question
Ex. 5-14 DNase Test In what ways can a DNase (+) organism use the products of DNA hydrolysis?
answer
The easiest use of the fragments would be to convert the nucleotides into their triphosphate form and incorporate them into the cell's DNA during its next round of replication. While not shown, deoxyribose can be converted into ribose and used to make ribonucleotides. Doexyribose, purines, and pyrimidines can also be used as energy sources in catabolic pathways, such as glycolysis and the citric acid cycle.
question
Ex. 5-25 Blood Agar The streak stab technique, used to promote streptolysin activity, is preferred over incubating the plates anaerobically. a) Why do you think this is so? b) Compare and contrast what you see as the advantages and disadvantages of each procedure.
answer
a) the primary goal of the microbiologist at this point is to isolate the organisms obtained from an aerobic environment - the pharynx. b) Observation of streptolysin S-positive organisms under anaerobic conditions is sometimes helpful, but incubating the entire plate anaerobically to promote streptolysin S activity could interfere with the fomation and recovery of isolated colonies. To avoid that, two plates would have to be streaked, with one grown aerobically and the other anaerobically. The stab is a simple solution to get streptolysin S results.
question
Ex. 5-25 Blood Agar Assuming that all of the organisms cultivated in this exercise came from the throats of healthy students, why is it important ot cover and tape the plates?
answer
Even healthy throats may carry pathogens, but their numbers are too small to create an infection. however, after incubation, their numbers are greatly increased and pose a greater threat for infection of people in the lab.
question
Ex. 5-25 Blood Agar Why is the streak plate preferred over the spot inoculations in this procedure?
answer
Spot inoculations are used when plating from a pure culture to demonstrate some reaction on/in the medium. The sample from the throat is a mixed culture with high density. In order to see the various hemolytic reactions isolated colonies must be produced by streaking the agar surface.
question
Ex. 5-6 Catalase Test Think about the advice given in the procedure to test a known catalase-positive organism along with an unknown. a) Is this a positive or a negative control? b) What information is provided by the results?
answer
a) This is a positive control b) Its main purpose is to verify that the hydrogen peroxide hasn't deteriorated into oxygen and water. To a lesser extent, it shows novice microbiologists what "bubbling" looks like.
question
Ex. 5-6 Catalase Test Consider the step in the tube test where hydrogen peroxide is added to the uninoculated tube. a) Is this a positive or a negative control? b) What information is provided by the results?
answer
a) This is a negative control b) It demonstrates that nothing in the medium reacts with hydrogen peroxide. If bubbling occurs in the uninoculated tube, then all positive results in inoculated tubes are in doubt and no negative results would be seen.
question
Ex. 5-6 Catalase Test When flavoprotein transfers electrons directly to oxygen, hydrogen peroxide is produced. What other consequences might result from electron carriers in the ETC being bypassed?
answer
An increase in heat and a decrease in ATP produced. This is analogous to releasing the energy from gasoline out of the context of an internal combustion engine. The explosion would not be "productive."
question
Ex. 5-6 Catalase Test Would a false positive from the reaction between the inoculating loop and hydrogen peroxide be caused by poor specificity or poor sensitivity of the test system?
answer
This would be due to poor specificity. The loop gives the same result as the enzyme catalase.
question
Ex. 5-7 Oxidase Test Think about the advice given in the procedure to test a known oxidase-positve organism along wtih an unknown. a) Is this a positive or a negative control? b) What information is provided by the results?
answer
a) This is a positive control b) Its primary prupose is to verify that the oxidase reagent hasn't degraded and is still usable. To a lesser extent, this also allows novice microbiologists to see what a positive result looks like
question
Ex. 5-7 Oxidase Test Think about the 20 second time limit on the oxidase test. a) What happens to the oxidase reagent after 20 seconds? b) Does this only happen after 20 seconds? If not, wy is a 20 second time limit set?
answer
a) It can spontaneously oxidize and turn blue after it becomes moist. b) It begins as soon as the reagent is applied to the organism, but it takes a while to produce a visible color change. If the organism is oxidase positive, it will produce an intense reaction more quickly and the spontaneous reaction become negligible. The spontaneous reaction is really only visible when the organism is oxidase negative.
question
Ex. 5-7 Oxidase Test Provide a possible explanation s to why this test identifies the presence of cytochrome c oxidase and not other oxidases
answer
The dye is an electron donor and electron transfer only occurs spontaneously ift he electrons lose free energy during the transfer. It is possible that the dye carries the electrons at an energy level too low to transfer them to any other ETC component other than cytochrome c oxidase.
question
Ex. 5-19 PYR Test Consider the uninoculated disk. a) Is it a positive or a negative control? b) What purpose does it serve in the PYR test?
answer
a) It is a negative control b) We are seeing if the p-dimethylcinnamaldehyde reagent reacts with anything on the PYR disk, which it shouldn't. It minimizes concern about false positive results.
question
Ex. 5-19 PYR Test How are the products of protein hydrolysis by an aminopeptidase used in metabolism?
answer
Aminopeptidases are enzymes that remove amino acids from the amino end of protein. These amino acids can then be used by the cell during its own protein synthesis, or they can enter a catabolic pathway and be used in ATP production
question
Ex. 5-19 PYR Test Deep red is the only color reaction that is considered positive for this test. Why do you think yellow or orange results are not positive (or weak positive)?
answer
The precipitate produced in this reaction is a deep red to purple color. The color does not develop in degrees as would a pH change; it is either formed or it is not. Colors other than red are from activity not tested for by this system.
question
Ex. 5-19 PYR Test Suppose you wanted to perform this test on an unknown organism but you were out of PYR reagent. Suppose further that a fellow lab employee said that you could substitute formaldehyde for PYR reagent and get good results. Would you be tempted to believe him? Why or why not?
answer
We would be tempted to try it in a purely experimental way. The test could not be used in a clinical lab or other serious lab work, but because the formaldehyde contains an aldehyde group, it might react with the B-naphthlyamine and form a visible precipitate.
question
Ex. 5-29 api 20 E Why is it important to perform the reagent tests last?
answer
Addition of reagents can cause complications in interpreting results of the spontaneous tests
question
Ex. 5-29 api 20 E In clinical applications of this test system, reagents are adde donly if the glucose test result is yellow or at least three other tests are positive. If these conditions are not met, a MacConkey agar plate is streaked and additional tests are performed confirming glocose metabolism, nitrate reduction, and motility. Why do you think this is so?
answer
api 20 E is a test designed to differentiate members of Enterobacteriaceae and a few other Gram-negative rods. If an organism does not ferment glucose, or produce positive results on at least three other tests in the panel then there is a good chance it is not an enteric and may be an organism not identified by the test system. There is, of course, the possibility that the organism is an enteric or related Gram-negative rod that just performed poorly in the test; hence, the reason for the additional tests. MacConkey agar would reveal its Gram status and its ability or inability to ferment lactose. Results from the motility and nitrate reduction tests combined with the other test results, would either identify the organism, significantly narrow the field, or reveal it as one not identified by the system.
question
Ex. 5-29 api 20 E Suppose after, 24 hours incubation, you notice no growth in the tubes containing mineral oil. Assuming that it is behaving properly under these conditions, what do you know about the organism and what predictions can you safely make about its performance in the decarboxylase tests, fermentation tests, and nitrate reduction test? Is it a member of Enterobacteriacease?
answer
The organism is a strict aerobe and should give negative results for decarboxylase, fermentation, and nitrate reduction tests. It is not a member because it does not ferment.
question
Ex. 11-4 Slide Agglutination What is the purpose of performing this experiment with both H and O antigens?
answer
The O antigen acts as a negative control. It should not agglutinate with anti-H antibodies
question
Ex. 11-4 Slide Agglutination Suppose you performed this test and got agglutination in both samples. Eliminating contamination of the antigen and antiserum samples as a possibility, provide an explanation of this hypothetical result. Is this attributable to poor sensitivity or specificity of the test system?
answer
One possibility is that the anti-H antibody cross-reacts with the O antigen. This would be due to poor specificity.
question
Ex. 11-4 Slide Agglutination Suppose you performed this test and neither sample produced agglutination even though you know the antigens and antibodies should react. Would this be attributable to poor sensitivity or specificity of the test system?
answer
Assuming the vials really contained what their labels said, this false negative would be due to poor sensitivity. Apparently, there is not enough H antigen in the sample to form a visible agglutinate with the ant-H antiserum.
question
Ex. 11-4 Slide Agglutination Higher vertebrates produce antibodies; Salmonella is a prokaryote. Given these facts, how can Salmonella antiserum be produced?
answer
Salmonella is the source of antigens for this exercise, not the source of antibodies. To get antibodies to Salmonella, we need to expose a vertebrate to its antigens, allow the vertebrate to mount an immune response, then isolate and purify the anti-Salmonella antibodies form the verebrate's serum. Salmonell does NOT produce Salmonella antibodies.
question
Ex. 10-3 Bacterial Transformation: The pGLO System What was the purpose of examining the original pGLO solution with and without UV illumination?
answer
Examination of the original pGLO solution indicates if it glows with UV radiation. (It shouldn't, since a plasmid is DNA and it's the gene product - GFP - that glows).
question
Ex. 10-3 Bacterial Transformation: The pGLO Sytem What was the purpose of transferring the + DNA and - DNA tubes from ice, to hot water, to ice again?
answer
The rapid change in temperature makes the cells more permeable to DNA, that is, more "competent." This process is called heat shock.
question
Ex. 10-3 Bacterial Transformation: The pGLO System Why were the microtubes incubated for 10 minutes in LB broth rather than transferring their contents directly to the plates?
answer
Ten minutes incubation in LB broth gives the transformed cells time to grow and express the ampicillin resistance gene. Otherwise, the ampicillin media would kill even the transformed cells. (The gene codes for beta-lactamase, an enzyme that hydrolyzes the beta-lactam ring of penicillin-like antibiotics).
question
Ex. 10-3 Bacterial Transformation: The pGLO System What information is provided by the LB/-DNA plate?
answer
This plate acts as a control to verify that the original (untransformed) E. coli is capable of growing on the LB media without ampicillin.
question
Ex. 10-3 Bacterial Transformation: The pGLO System Obviously, transformation could occur only if the pGLO plasmid was introduced into the solution. Which plates exhibit transformation?
answer
Growth on the LB/amp/+DNA and the LB/amp/ara/+DNA plates show transformation.
question
Ex. 10-3 Bacterial Tranformation: The pGLO System What information is provided by the LB/amp/-DNA plate?
answer
It is a control that verifies the inability of E. coli to grow in the presence of ampicillin
question
Ex. 10-3 Bacterial Transformation: The pGLO System Why does the LB/amp/ara/+DNA plate fluoresce when the LB/amp/+DNA plate does not?
answer
The GFP gene is only expressed in the presence of arabinose (because arabinose "turns on the operon" and allows transcription), so even though there were transformed cells on the LB/amp/+DNA plate, they did not fluoresce.