((필독1. site )) Answers to the Sample Quiz Questions in the Manual (Appendix X)

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Bacteriology 102: Answers to the Sample Quiz Questions in the Manual (Appendix X)

YOU ARE HERE: Bacteriology at UW-Madison > John L's Bacteriology Pages > Bacteriology 102 Homesite > Key to Old Quiz Questions

Page Last Modified: 12/17/01

Also see the "Things to Think About."

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The question often comes up regarding how we grade items which have to be numbered in a certain order. We do not grade these items blank-by-blank with the key. We take off only for those which are out of order. Consider the following question from a final exam (given Spring, 2000) with the correct answers indicated: (Note: In writing the exam, we had forgotten to mix these items up, so they eventually appeared in the correct order to begin with!)

5.  (5 points) With numbers (1-5; 1 is the first), arrange the following procedures which are involved in the isolation and identification of enterics:

  1    Inoculate source material into the selective enrichment medium.
  2    Streak plate for isolated colonies.
  3    Inoculate Kligler Iron Agar (KIA) tubes with pure cultures.
  4    Eliminate any tubes which show the obvious presence of non-enterics (e.g., Pseudomonas).
  5    Run a number of different biochemical tests on the various cultures.

If one were to indicate the correct order except to make the last item number 1 – i.e., indicating 2 3 4 5 1 – such a simple yet significant "frameshift" would have resulted in a minus 5 if the question were to have been graded in the conventional manner (i.e., blank-by-blank). As just one item was out of order, then we would only take off 1 point – although one should know better than to put the last item in this set in the number 1 position! (The phrase "source material" is a give-away that indicates the initial step of the procedure.) At its worst, indicating the order as 5 4 3 2 1 would have everything in reverse order, and we would grade it as minus 5 – even though the correct number (3) happens to be in the proper place.

1A. Petri dish cultures

 +  are always labeled on the bottom lid and incubated in an inverted position.
 O  are examined only through the bottom lid (i.e., the lid containing the medium).
 +  are always incubated upside down (medium side up).
 O  are always prepared by streaking a culture with a loop in order to obtain isolated colonies.

1B. The following terms are always plural, never singular:

 +  bacteria singular=bacterium
 O  bacillus plural=bacilli
 +  media singular=medium

1C. A colony-forming unit (CFU)

 O  and a colony are the same thing.
 +  is microscopic, as is a cell.
 +  is composed of one or more cells which will be able to utilize the medium (on which they are placed) and eventually produce one colony through the process of cellular division.
 O  is a term used only when we are quantitating bacteria, such as determining the number of CFUs per ml or per gram of a sample.

2A. The gram stain

 +  classifies bacteria according to the structure of the cell envelope.
 O  is an alternative to the acid-fast stain; a positive reaction for each stain will mean the same thing.
 O  gives the best results when an old culture (grown 2 or more days) is used.
 +  results in red or pink cells for a gram-negative organism.
 O  should be performed on an extremely thick smear, such that the gram reaction can be seen by the naked eye.
 +  may give a false reaction for encapsulated cells, as the presence of a capsule may inhibit proper decolorization (i.e., a gram-negative cell may appear as a gram-positive cell).

2B. "Gram-variability"

 O  is a term which can be used where two gram reactions are seen due to an error in the staining procedure.
 O  applies to an organism which changes its cell wall structure from the gram-positive type to the gram-negative type as the culture ages.
 +  applies to what is ultimately seen when cells in a culture of gram-positive bacteria lose the ability to retain the primary stain during the decolorization process.
 O  indicates a mixed (i.e., impure) culture.

2C. For an "acid-fast" organism, the word "acid" refers directly to

 O  acid produced from fermentation.
 O  mycolic acids associated with the cell wall.
 +  the resistance of a stained cell to an acid decolorizing agent (such as acid alcohol).

2D. Examples of differential stains include

 +  the gram stain.
 +  the acid-fast stain.
 O  the single application of crystal violet to a smear.

2E. Give two reasons why an organism which is gram-positive may give a gram-negative staining reaction.
1. Older culture. Cell envelopes may be weakened and cells may thus be easy to decolorize.
2. Error in staining procedure in that too much alcohol-acetone was used.

2F. A wet mount will become a smear if it dries out without having had a coverslip placed on it. T/F
True. A smear is merely a dried cell suspension on a slide. (It will still need to be heat-fixed.)

2G. Why is a smear heat-fixed prior to staining?
To get cells to stick to the slide.

2H. Total magnification of a microscope is calculated by adding the magnifications of the ocular and objective lenses. T/F
False. Multipy the two magnifications.

3A. Regarding aseptic technique, the following are acceptable procedures:

 O  Leaving open tubes upright in the test tube rack while transfers are made between them.
 O  Sterilizing only the very tip of the loop or needle prior to making transfers.
 O  Taking the lids off petri plates and placing them elsewhere on the bench while streaking or examining the plate.
 +  Flaming the open ends of plugged tubes after opening and before closing them.
 O  Discarding capsule stains and other wet mounts into the container meant for discarding stained smears and broken glassware. Know page viii!!

3B. When inoculating a tube of sterile medium from a colony on a plate,

 +  one should use a needle, not a loop.
 +  one would pick from only one colony, especially if the plate contains colonies of different species of bacteria.
 O  one must transfer the entire colony in order for the inoculation to be successful.

4A. You made several dilutions of a food sample and plated 1 ml of a 10^–4dilution which resultedin 250 colonies. Your neighbor performed the same procedure with the same sample, but only 13 colonies resulted.

 +  It is likely that either one of you failed to mix the dilutions well before making further dilutions and platings.
 O  If the procedure were such that you were using a selective medium and your neighbor were using an all-purpose medium such a difference in the colony count would not seem impossible. One expects a higher count with all-purpose medium.
 +  Based on your results, there were 2.5 X 10⁶CFUs per gram of the food.

4B. In the determination of the concentration of living bacteria in a sample, we determine the number of colony-forming units (CFUs) per ml (or gram) rather than the number of cells per ml (or gram), because:

 +  There may be cells which cannot grow on the medium used and/or at the conditions of incubation. And they still can be viable.
 +  Any colony counted may have arisen originally from one or more cells.
 O  The colony count always includes a certain proportion of contaminants which were not present in the sample.Don't expect contamination as a given! You're using aseptic technique!

4C. In doing the plate count procedure:

 +  One can prepare a 1/10 dilution in a variety of ways, but the proportion of the amount to be diluted to the total amount is always 1 to 10. The proportion (ratio) of amount to be diluted to the diluent is always 1 to 9.
 +  In spreading the inocula over the surface of plates with a sterile hockey stick, one can start with the most dilute inoculum and move up to the more concentrated inoculum without having to resterilize the hockey stick.
 +  The plates are always incubated in an inverted or upside-down position.

4D. The following could result in recording a lower CFU count than what one should get:

 +  Failing to mix a dilution thoroughly before using it to inoculate a plate or make another dilution.
 +  Actually plating 0.1 ml of a dilution when one believes one is plating 1 ml.
 +  Failing to cool the hockey stick after flame-sterilizing it.

4E. If one milliliter (1 ml) of a 10^–4 dilution of a sample is inoculated into a plate of an all-purpose medium and 300 colonies result,

 +  one would expect that the number of viable (living) cells in the one ml amount would be equal to or greater than 300.
 +  one would expect, theoretically, that 0.1 ml of the same dilution would have resulted in 30 colonies.
 O  one can calculate the CFU/ml of the original, undiluted sample to be 3.0 X 10^–2. We could not have a negative exponent in the answer to this problem (a too-common mistake). A negative exponent would have a tiny part of one cell per ml giving rise to lots of colonies after being diluted.

4F. In the "total aerobic plate count" of a sample, such as what we determined in the hamburger experiment, one may make several dilutions of the sample and then inoculate plates from the dilutions. Then, after incubation,

 +  plates with greater than 300 colonies or less than 30 colonies are not used.
 O  only colonies of a certain size are counted.
 O  the colonies we see are only of strictly aerobic bacteria.

5A. Fermentation

 +  results in production of acid and possibly gas from the breakdown of sugars.
 +  is associated with the type of growth of facultative anaerobes in Thioglycollate Medium where growth is less dense in the anaerobic region.
 O  is generally associated with a positive catalase reaction for an organism.

5B. Respiration

 +  by an organism implies the ability of the organism to produce catalase.
 +  explains why a facultative anaerobe grows better in the presence of air than under anaerobic conditions.
 O  which involves the breakdown of a sugar is indicated by a large amount of acid and gas.

5C. A facultative anaerobe

 +  respires in the presence of oxygen.
 +  ferments in the absence of oxygen. If it has something to ferment.
 +  and an aerotolerant anaerobe can be differentiated by the catalase test.

5D. Chemoheterotrophic organisms

 +  derive their energy from chemical reactions; light is not involved.
 O  obtain carbon from carbon dioxide in the atmosphere.
 O  always obtain energy by means of respiration.

5E. Regarding bacteriological media:

 O  An all-purpose medium is one which will support the growth of all species of bacteria.
 O  A differential medium is defined as one which supports the growth of some organisms but not others.
 O  One would not include any carbon compound in the medium if one wanted to grow a heterotroph, as these organisms can obtain carbon from the carbon dioxide in the atmosphere.
 +  One must include a siderophore in a medium in order to support the growth of a siderophore auxotroph.
 O  Anything which can be used as a nutrient is called a growth factor.

5F. A complex medium

 O  is one in which all the different elements and compounds are known quantitatively.
 +  can be a chemically-defined medium to which a known amount of a hay infusion is then added.
 O  can be a chemically-defined medium to which a known amount of purified NaCl is then added.

5G. An "all-purpose medium"

 O  supports the growth of all bacteria.
 +  is not formulated purposely to inhibit any organism.
 +  would be something like Plate Count Agar, the medium used for the "total aerobic plate count" in the hamburger experiment.
 O  includes agar as an essential source of nutrients.

5H. In Glucose O/F Medium and Glucose Fermentation Broth,

 O  an organism needs to utilize glucose for carbon and energy in order to grow at all.
 +  one which grows and shows an acidic reaction throughout the medium may be a facultative anaerobe.
 +  glucose and the pH indicator are "differential agents."

5I. Circle the correct choices: (NOTE: On this key, the correct choices are underlined.)

a. An organism would be expected to produce less siderophore in an environment containing a relatively high/low (circle one) concentration of iron compounds.

b. Circle the term which always represents the plural form: bacterium/bacteria. Circle the term which always (and only) represents the singular form: medium/media.

c. Organisms which can respire are facultative anaerobes/aerotolerant anaerobes/strict aerobes(circle two).

d. Organisms which can ferment are facultative anaerobes/aerotolerant anaerobes/strict aerobes(circle two).

5J. In Experiment 5.2, we considered a strain of Arthrobacter flavescens which was unable to grow on a complex medium. The main purpose of the experiment was to demonstrate how

 O  A. flavescens produces iron which other organisms can utilize.
 +  an organism can grow better (or can therefore grow) when supplied with a required growth factor.
 O  all siderophores are pigmented and glow under an ultraviolet light.

5K. For growth curves in general:

 O  The growth curve will always be the same, no matter what medium, incubation conditions or organism we use.
 +  Dead and living cells both contribute to the absorbance value.
 O  In the stationary phase, all cell division ceases.

5L. In determining the growth rate and generation time for a bacterial species,

 +  we consider points only on the best straight line drawn among the data points for the exponential phase of the growth curve.
 O  we can use any two CFU/ml values recorded from plates prepared during the exponential phase.
 +  we necessarily assume that each cell will divide into two cells, not three or more.

5M. (This problem can be done without the use of the formulas!) At 2 A.M., a flask of a liquid medium is inoculated with 1X10³ bacterial cells. The lag phase lasts a half hour. At 7:30 A.M., the culture enters the stationary phase with a population of about 1X10⁶ cells in the flask.

 +  Approximately 10 generations were produced by 7:30 A.M.
 +  We can estimate the generation time from the data given. Looks like 30 minutes. After the half-hour lag phase, note how the population doubles every half hour in order to get to about 1X10⁶ by 7:30.
 +  If the cells did not separate after division during the incubation period (perhaps they formed chains or clusters), we would get the same plate count for the 7:30 A.M. sample as for the 2 A.M. sample. If the cells did not separate after division, then the number of CFUs would remain the same. See Exp. 1 introduction.
 O  In the stationary phase, we do not expect any division or death of any cell in the culture.

5N. If the growth rate for a particular organism was found to be 4 doublings per hour, the generation time would be: 15 minutes.

5P. Note the following growth curve of a typical organism (such as the strain of E. coli we used in our growth curve experiment):

 O  To generate the growth rate, one can choose points on the CFU/ml graph corresponding to 1 hour and 4 hours.
 +  One possible reason for the difference (flat vs. slanted) in the two graphs between 0 and 2 hours is that the cells/ml may be increasing but the CFUs/ml are not.
 O  The death phase begins at 4.5 hours.
 +  One could determine the generation time from a graph such as this by finding the time difference between 1X10⁶ CFUs/ml and 2X10⁶ CFUs/ml. That's a doubling of the population.

6A. When observing a wet mount, true motility of a culture is indicated when one sees

 +  cells which are moving around on their own, independently of each other.
 O  Brownian motion. Dead cells can also be subject to Brownian motion.
 O  all of the cells moving together in the same direction In this case, the cells are obviously trapped in a current. Dead cells can be moved around this way as well..

6B. A negative result in Motility Medium

 +  is indicated if growth occurs only along the line where the medium was stab-inoculated.
 +  should be confirmed by a wet mount of a young culture of the same organism.
 +  may exhibit growth over the surface of the medium.
 +  may occur for strictly aerobic, motile organisms.

7A. You have been found most worthy to be the recipient of a generous gift of 1000 cultures. But here's the catch: You must identify each culture to genus and species! So, having been given free room and board at the lab, you proceed to do the following:

 O   Immediately perform a large number of differential tests (gram stain, motility, lactose fermentation, starch hydrolysis, indole, nitrate reduction, etc.) on each culture.
 +   Test each culture with a few "primary" tests, such as the those to determine gram reaction, glucose fermentation and catalase, before considering which further tests to do.
 O   Determine the generation time and growth rate of each organism under a variety of media and growth conditions.
 O   Not do the gram stain at all, as it is too tricky and of no real importance.

7B. In Experiment 7, we could determine the probable oxygen relationship of each of the twelve organisms without inoculating the standard test medium for oxygen relationship (Thioglycollate Medium).

 +   We know enough about how the cultures grow such that we can rule out their being strict anaerobes.
 +   We can use the results from the glucose fermentation and catalase tests to differentiate between strict aerobes, facultative anaerobes and aerotolerant anaerobes.
 O   We can use the results from the lactose fermentation and amylase tests to differentiate between strict aerobes, facultative anaerobes and aerotolerant anaerobes. Lactose fermentation is relatively trivial, compared to the primary tests indicated in Experiment 7.

7C. In the test to determine whether or not starch was broken down,

 O   we add iodine to a plate of Starch Agar after incubation, and we expect the iodine to react with amylase to give a noticeable reaction.
 O   growth on a plate of Starch Agar indicates an organism's ability to break down starch, and we only do the test with iodine to confirm‎ this fact chemically.
 O   we could perform the same test on Heart Infusion Agar as on Starch Agar.

7D. In the test for nitrate reduction,

 O   growth and gas indicate fermentation.
 O   a red color – which is obtained only after the additions of the reagents and the zinc – means that one can record a positive reaction for the organism.Record reaction for the organism. Then add the zinc.
 +   we may be able to determine whether or not a "strict aerobe" (as defined according to our "oxygen relationship" definitions; see Exp. 5.1 and our oxygen relationship web page.) can grow anaerobically, using nitrate in place of oxygen.

7E. The streak plate method (i.e., done with the loop) helps us to achieve the following:

 O   determination of the number of colony-forming units per ml of the culture
 +   determination of the purity of the culture
 +   isolation of the different organisms in a mixed culture
 +   determination of colonial characteristics

7F. When testing an organism to see if it can break down a particular substance,

 +   we may see growth of the organism in a medium containing the substance – whether or not the substance is being broken down.
 O   we must have a pH indicator in the medium.

7G. Aerotolerant anaerobes generally produce colonies which are smaller than those of the usual facultative anaerobes and strict aerobes.

 +   This is related to the fact that aerotolerant anaerobes do not respire and therefore are not as efficient in their metabolism as the other types.
 +   These organisms can be differentiated from the other types by the catalase test.

8A. Consider an organism which has undergone mutation such that it is resistant to an antibiotic while normal cells of the same species are sensitive to the antibiotic.

 O   The antibiotic must have caused the mutation.
 +   The target site for the antibiotic may have been altered such that the antibiotic is no longer effective at that site.
 +   The cell membrane may have been altered such that the antibiotic is restricted entry into the cell.

8B. Conjugation and recombination

 +   refer to two separate events, one following the other.
 +   result in the potential for up to half of a population of cells to undergo a genotypic change, ifthe population is made up of equal numbers of "donor" and "recipient" cells.
 +   will probably lead to no detectable recombinants, if the donor and recipient cells are otherwise genetically identical to each other.
 +   always involves two living cells.
 O   involve donor and recipient cells fusing into one cell.

8C. The genotype possessed by a cell

 O   is always identical to the genotype of all other cells in its species.
 +   can be changed by mutation.
 +   can be changed by recombination.
 O   is readily changed by altering environmental conditions (temperature, oxygen, etc.).

8D. You have a set of 10 E. coli cells. Five are Hfr cells and 5 are F^– cells. Four pairs of the cells undergo conjugation. Only one of the F^– cells incorporates new DNA into its chromosome. Therefore, the "recombination frequency" is: 1/5. Must have total number of pairs of cells (= no. of recipient cells) in the denominator. If we put total number of cells (10) in the denominator, the highest recombination frequency – if all cells conjugated and all recipient cells underwent recombination – would only be 50%.

8E. In our qualitative recombination experiment, we cross-streaked Hfr and F^– strains of E. coli on a plate of Minimal Medium. We expected growth of recombinants in these regions (referring to the diagram in the manual): A and E

9A. Phages 1, 2, 3 and 4 are spot-inoculated onto plates of various bacteria as seen on the diagram in the manual.

a. Which phage(s) is (are) "strain-specific" (within a species)?  3 (infects strain B of E. coli but not strain A)

b. Which phage(s) is (are) specific to one species?  1 and 3 only infect E. coli; 4 only infects K. planticola.

c. Which phage(s) infect(s) all organisms tested?  2

9B. Bacteriophages

 +   cannot be seen with the microscopes we use in lab.
 +   are viruses which utilize bacterial cells as their "medium" for replication.
 +   may be utilized in some bacterial identification procedures.
 +   when added to a medium in a sufficient quantity, can make the medium selective against organisms which are sensitive to the particular bacteriophage used.
 O   may reproduce in the absence of bacterial cells and form colonies on media much like bacterial colonies.
 +   are to plaque-forming units (PFUs) as bacterial cells are to colony-forming units (CFUs).

10A. In the antibiotic disc sensitivity test, we were testing the sensitivity of

 O   various antibiotics to Streptomyces.
 O   various antibiotics to a test organism.
Antibiotics, being chemicals, are not "sensitive" to anything.
 +   a test organism to various purified antibiotics.
 O   a test organism to antibiotics produced by a Streptomyces isolate growing on a paper disc.

10B. In the antibiotic disc sensitivity test,

 +   each disc contains a certain amount of a particular antibiotic which is expected to make the medium selective in the area around the disc.
 O   more than one test organism is used in each plate.
 O   any zone of inhibition indicates that the antibiotic is inhibiting the test organism at the "target site." Not necessarily. We must measure the zone of inhibition.

10C. You have a patient suffering with an infection caused by a strain of Grodybacter gnarlii, the causative agent of gnarliosis. You run a disc sensitivity test and come up with the following data:

Bacitracin	8 mm
Chloramphenicol	19 mm
Colistin	7 mm
Erythromycin	25 mm*
(* Numerous small colonies were seen throughout this "zone.")

ZONE SIZE INTERPRETATIVE TABLE

ANTIBIOTIC ON DISC	INHIBITION ZONE DIAMETER TO NEAREST mm
	RESISTANT	INTERMEDIATE	SUSCEPTIBLE
Bacitracin	8 or less	9-12	13 or more
Chloramphenicol	12 or less	13-17	18 or more
Colistin	8 or less	9-10	11 or more
Erythromycin	13 or less	14-17	18 or more

 +   A suitable choice of antibiotic to control the infection would be chloramphenicol.
 +   The use of erythromycin (i.e., in the patient) would "enrich" for erythromycin-resistant cells of Grodybacter gnarlii which would probably continue the infection.
 O   Your results tell you that bacitracin and colistin are resistant to Grodybacter gnarlii. It's the other way around.
 +   There really is no zone of inhibition for erythromycin.

10D. Dr. X had a patient suffering with an infection caused by a particular strain of Neisseria gonorrhoeae, the causative agent of gonorrhea. By a quantitative test on a broth culture of N. gonorrhoeae which was isolated from the patient, it was found that the culture was sensitive to penicillin except for one penicillin-resistant mutant for every one million cells. Dr. X went ahead and gave his patient penicillin. At this point – given these facts – one may expect

 O   a cure. Penicillin was not shown to be 100% effective. Can't always expect the body's defenses to kill the resistant organisms.
 +   eventually the probability of isolating Neisseria gonorrhoeae cells from the patient which are penicillin-resistant.
 +   the need to employ a truly effective antibiotic.

10E. The following are associated with Streptomyces:

 O   photosynthesis
 +   geosmins
 O   coccus morphology of vegetative cells
 +   extracellular enzymes
 +   antibiotics

10F. In our isolation of Streptomyces, we used a medium which includes

 +   large-molecular weight compounds which must be broken down by extracellular enzymes in order for microorganisms to grow on the medium.
 +   a selective agent which selects against molds as much as possible.
 O   antibiotics such as streptomycin.
 +   agar.

10G. In the Streptomyces isolation experiment, we looked for antibiotic production

 O   by one or more of the test organisms streaked up to the Streptomyces growth.
 +   by the way in which the test organisms grew in the vicinity of Streptomyces.
 O   which caused inhibition of the growth of Streptomyces.

10H. You have found a strain of Grodybacter gnarlii which is resistant to all of the antibiotics which can be tested! Briefly (preferably with a diagrammed flow chart) explain how you might look for a strain of Streptomyces which will produce an antibiotic against G. gnarlii.
Simply follow Experiment 10.2 procedure, obtaining lots of Streptomyces isolates and then using G. gnarlii as the test organism in "Period 5."

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If you have not done so already, start looking at the "thought questions" we have posted on a separate web page.

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11A. When inoculating plates to obtain a desired type of organism from a natural source (or from a liquid enrichment of a natural source),

 + the plates must be streaked for isolated colonies as we cannot expect the desired organism to be present as a pure culture in the inoculum.
 + one may see different kinds of colonies corresponding to different species of the desired type of organism.
 O one always expects that the plating medium used will select for the desired type of organism such that all other types are inhibited from growing.

11B. Fill in the following table concerning growth conditions and related items regarding the variable metabolism of purple non-sulfur photosynthetic bacteria:

	cultured as a
	photoautotroph	photoheterotroph	chemoheterotroph
nature of carbon source	carbon dioxide	organic	organic
nature of electron donor	inorganic (e.g., H2)	organic	organic
mode of energy generation	phototrophic	phototrophic	respiratory
light required?	yes	yes	no
aerobically or anaerobically incubated?	anaerobic	anaerobic	aerobic

11C. When we set up our broth enrichment for purple non-sulfur photosynthetic bacteria,

 + various respiring organisms, including purple non-sulfur photosynthetic bacteria, utilize the oxygen initially dissolved in the medium, thus creating an anaerobic environment.
 O we incubated the bottle under fluorescent – rather than tungsten – light, in order to provide the wavelengths of light which these organisms prefer.
 + the stopper prevents additional oxygen from getting into the medium.

11D. When looking for purple non-sulfur photosynthetic bacteria, we had to make use of certain medium and incubation requirements in order to isolate them most efficiently, as these organisms

 O do not utilize sulfur for any purpose.
 + produce easily-distinguished pigmented colonies when the plates are incubated under anaerobic conditions and in the light.
 + produce non-pigmented colonies if the plates are incubated under aerobic conditions, and the organisms would unfortunately be virtually indistinguishable from chemotrophs. This is usually the case, but some strains may produce some pigment even when not photosynthesizing.
 + may be overrun (crowded out) by respiring chemotrophs if the plates are incubated under aerobic conditions.
 + may be overrun (crowded out) by fermenting chemotrophs if incubated under anaerobic conditions on a medium containing glucose as a carbon source.

11E. When we "heat-shock" a suspension of soil, we expect

 O immediate production of endospores by vegetative cells of Bacillus and Clostridium.
 + more endospores will germinate than would otherwise (if we had not heat-shocked the soil). This is something we often do not point out during the course of Experiment 11.2.
 + killing of most or all of the reproductive spores of such soil organisms as Streptomyces and molds.
 + killing of most or all of the vegetative cells of non-endosporeforming microorganisms.
 + killing of most or all of the vegetative cells of Bacillus.

11F. A heat-shocked suspension of soil is streaked onto an all-purpose medium which includes glucose, a fermentable sugar. When we incubate this plate under aerobic conditions, we expect the growth of

 + strictly aerobic species of Bacillus.
 + facultatively anaerobic species of Bacillus.
 O Clostridium.

11G. If we were to take the plate prepared above and incubate it under anaerobic conditions, we would expect the growth of

 O strictly aerobic species of Bacillus.
 + facultatively anaerobic species of Bacillus.
 + Clostridium.

11H. Concerning bacteriophages, antibiotics and endospores:

 + Living microorganisms are required for their development.
 + The first two, if added to a bacteriological medium (such as what may be in a plate), will make that medium selective for bacteria which are resistant to them.
 + The last two can be produced by certain organisms in soil.
 + Bacteriophages and endospores each contain nucleic acids which are essential to their continuing existence.

11I. The endospores we see in stained slides made from Bacillus colonies

 O are the very same spores which were originally inoculated onto the plate. NO! There is an entire "endospore cycle" between the endospores plated and those we see when we observe the colonies microscopically. (What are the CFUs? How do the colonies develop?)
 O come from colonies where endospores are growing and dividing.

11J. Consider this: You have a colony of Bacillus cereus which has been growing for two days on a plate, constantly expanding outward ("colonizing new territory") as the vegetative cells continue to grow and divide. Might various points from the center to the edge of the colony correspond to particular points or stages on a growth curve made for a culture of B. cereus? How might this relate to differences one may see in the gram reaction and presence of endospores between the center and the edge? (Note: Can one truly say "old" cells and "young" cells? Recall the introduction to Experiment 1.)
As the colony expands, cells on the edge are making use of fresh nutrients, similar to (but not exactly like) cells at time "zero" being inoculated into a flask of a broth medium in a growth curve experiment. Actively metabolizing Bacillus cells should reliably stain gram-positive as they are expected to be "healthy" cells with intact cell envelopes. Cells in the center of the colony are basically a two-day-old culture whose nutrients have been spent, and endospores would probably have been formed as a response to decreasing nutrients. As for the gram reaction of the remaining vegetative cells (which are probably in "bad shape"), we expect such cells to decolorize readily and stain gram-negative.

11K. Nitrogen-fixing organisms

 + can be the source of nitrogenous compounds for non-nitrogen-fixing organisms when both are growing in nitrogen-free media.
 O are those which produce gaseous nitrogen from various chemical reactions involving nitrogen-containing compounds.
 O include a wide variety of different microorganisms such as fungi, algae and protozoa – not just bacteria.

12A. Regarding the lactic acid bacteria (also known as the "lactics"):

 +  They are gram-positive and may be rods or cocci.
 O  All produce slime from sucrose and acid from the fermentation of lactose.
 +  Some are utilized in the production of fermented dairy products such as yogurt and cheese.
 +  Their metabolic activities in foods and in (or on) the human body can be a deterrent against many pathogenic organisms.

12B. Efficient isolation of lactic acid bacteria can be accomplished by:

 O  heat-shocking the inoculum.
 +  the use of a plating medium containing a fermentable sugar and an abundance of growth factors.
 +  a cytochrome inhibitor in the plating medium.
 +  the use of aerobic incubation of the isolation plates in order to inhibit strictly anaerobic bacteria which may also grow on plating media used for lactic acid bacteria.

12C. The hot-loop test

 +  distinguishes between homo- and heterofermentative lactic acid bacteria.
 O  distinguishes between lactic acid bacteria and other groups of bacteria.
 O  performs the same function as the Durham tube – i.e., a gas bubble in a Durham tube and a positive hot-loop reaction indicate the same thing.

13A. In the isolation of Staphylococcus aureus by the use of Vogel-Johnson Agar,

 O  the black color seen for any colony is due to hydrogen sulfide production.
 O  we expect any black colony to be Staphylococcus aureus.
 +  black colonies which are catalase-positive and consist of gram-positive cocci in clusters are probably of the genus Staphylococcus.
 +  suspected colonies must be tested for the coagulase reaction before any identification of Staphylococcus aureus is made.

13B. The candle jar

 +  provides an elevated level of carbon dioxide in its atmosphere.
 O  provides an environment suitable for strictly anaerobic bacteria.
 O  will only support the growth of autotrophic organisms.

13C. Hemolysis

 O  is termed "alpha" when the blood cells and hemoglobin (the pigment) are completely wiped out in the area adjacent to a colony of a Streptococcus on Blood Agar.
 O  can be used to differentiate between Streptococcus and other organisms.
 O  can be observed on Heart Infusion Agar.

14A. Regarding the enteric bacteria (also known as the "enterics"):

 +  They are facultatively anaerobic.
 O  They are bacteria which are found only in the intestinal tract.
 O  They all produce hydrogen sulfide.
 O  They include such organisms as Pseudomonas and Neisseria.
 +  They include such organisms as Salmonella and the true coliforms.
 +  They may be found in the early part of a sauerkraut fermentation.

14B. Enrichment and isolation procedures for enteric bacteria

 +  are aided by media which inhibit gram-positive bacteria.
 +  may result in the isolation of Pseudomonas.
 +  may involve media which contain lactose as the only fermentable sugar even though many enterics do not ferment lactose.

14C. Kligler Iron Agar (KIA)

 +  tests for glucose fermentation.
 +  tests for lactose fermentation.
 +  tests for hydrogen sulfide production.
 +  can differentiate enterics from strict aerobes like Pseudomonas.
 +  gives the most reliable results for pH-related reactions when examined at just one day of incubation (not at two or more days).
 +  gives reliable results when used for gram-negative rods such as Pseudomonas, E. coli and other enterics, and it isn't a recommended medium for other types of organisms (e.g., gram-positive cocci).
 +  is an example of a differential medium where both alkaline and acid-producing activities by an organism may be found.
 O   – if incubated under anaerobic conditions – should show the same result for any organism as it would when incubated under aerobic conditions.

14D. When mineral oil is placed on a broth medium inoculated with an enteric,

 +  creation of anaerobic conditions in the medium is achieved by respiration of the organism rather than by any immediate action of the mineral oil.
 +  the mineral oil prevents additional oxygen from getting into the medium.
 O  mineral oil is necessary to supply trace elements to the medium.

14E. The methyl red test

 O  can be done by adding the methyl red reagent to a slant or plate of any all-purpose medium.
 +  is done after at least two days of incubation of the culture.
 O  distinguishes between fermenting and non-fermenting bacteria.

14F. A methyl red-negative reaction in MR-VP Broth

 +  is associated with enterics which produce "neutral products."
 O  means that the pH is in the alkaline range (above pH 7).
 +  would be expected for a non-fermenter such as Pseudomonas.
 +  for a particular lactose-fermenting enteric will help to explain why the slant of a KIA culture would change from yellow to red when incubated over a two-day period.

14G. The "slide agglutination test"

 +  can be used along with the API-20E tests to identify an organism as a particular kind of Salmonella. Special note: This is probably true, but identification won't be to the level of serovar. Not a wonderful question.
 O  involves the use of known antibiotics to test for certain antigens on the cells of the unknown organism.
 O  detects catalase.
 O  detects coagulase.

14H. Using numbers (1-4), put the following procedures in the correct order. These procedures deal with the isolation and identification of enterics.

 3  Inoculate preliminary biochemical test media such as Kligler Iron Agar (KIA).
 2  Streak plates of selective-differential media for isolated colonies.
 1  Obtain source material.
 4  Inoculate various biochemical test media (Lactose Fermentation Broth, MIO, Simmons Citrate Agar, etc.).

14I. You wish to exploit certain properties of the difficult-to-isolate bacterium Excalibacterium (an enteric) in order to help you detect and isolate it from samples which are highly-contaminated with other enterics. You decide to start with MacConkey Agar which you know (from Bact. 102!) contains lactose as the only fermentable sugar. Peptone is another medium ingredient which you recall; it contains a mixture of various amino acids – none in any especially high amount. Following is a table showing important genera to consider in this situation:

genus	fermentation of					decarboxylation of
	glucose	maltose	lactose	sucrose	mannitol	lysine	arginine
Edwardsiella	+	+	–	–	–	+	–
Aquamonas	+	+	–	–	–	+	+
Excalibacterium	+	–	–	+	–	+	–
other enterics	+	+	+ or –	+ or –	+ or –	+ or –	+ or –

a. On MacConkey Agar, what would you expect the net pH reaction would be for any of the three genera specifically listed on the table above? ALKALINE

b. As these 3 genera don't ferment or respire lactose, how can they grow on MacConkey Agar? They utilize other energy sources, such as whatever amino acids they can respire.

c. What would be the best choice for a sugar to add to MacConkey Agar which will assist greatly in the differentiation of Excalibacterium colonies from the other organisms on the table? MALTOSE (Then one can pick maltose-negative colonies!)

d. If lysine were to be included in the medium in a relatively large amount, what effect would this have on the pH reaction associated with Excalibacterium colonies? MORE ALKALINE

15A. Among the features of a good "indicator organism" is/are the following:

 O  the ability to cause the problem being examined.
 +  the ease in which it is detected.
 O  the ability to remain in the contaminated environment indefinitely.

Regarding coliform detection, be sure you understand Section I on page 89. Also review the flow chart on page 151. Know that the following types of organisms can give a positive result (i.e., growth and gas) in one or more of the three selective-enrichment broth media:

• Fecal Coliforms.
  
• Other Coliforms: These are inhibited in the EC Broth/44.5C enrichment.
  
• Certain strains of Bacillus and Clostridium which are able to ferment lactose rapidly to produce acid and gas: These are inhibited in the confirm‎atory test media. The reason they may show up in the presumptive test is because the medium employed (LLTB) is only moderately selective; remember that we don't want to shock already-stressed coliforms with too highly selective a medium such as what we have in the confirm‎atory test.
  

15B. In the bacteriological examination of water, one looks for coliforms, because

 +  they indicate the possibility of various pathogenic organisms also being present.
 O  they should be present in any "normal" supply of water, and if they are not there, then something is "wrong" with the water.
 +  they are organisms which are easy to enrich for and detect.

15C. In the presumptive test for coliforms, a positive result

 O  is recorded when growth is seen, whether or not gas is present.
 O  indicates that the tube contains a pure culture of the organisms being detected.
 O  confirm‎s the presence of Escherichia coli.
 +  can be the result of growth of one coliform originally inoculated into the medium from the sample or a dilution of the sample.
 +  can be produced by organisms other than true coliforms which can withstand the selective agents in the medium.

15D. The most probable number (MPN) method of counting bacteria

 +  may be used to estimate the number of bacteria per ml of sample, if an all-purpose broth medium is inoculated from various dilutions of the sample.
 +  may be used to estimate the number of indole-producing bacteria per ml of a sample, if Tryptone Broth is inoculated from various dilutions of the sample and then tested with Kovacs reagent after incubation.
 O  gives the estimated number of fecal coliforms per ml of water from the results of the Presumptive Test.

15E. Fecal coliforms

 O  are any bacteria found in the enteric (intestinal) canal.
 O  and coliforms are two separate (non-overlapping) groups of bacteria.
 O  and enterics are two separate (non-overlapping) groups of bacteria.
 +  can grow and produce gas in Brilliant Green Lactose Bile (BGLB) Broth.
 +  include Escherichia coli. Special note: Actually the way things are worded in the manual, only strains of E. coli which ferment lactose rapidly would be considered coliforms or fecal coliforms. That would mean the vast majority of strains but not all of them. (The question could be worded better.)

15F. In the enrichment, isolation and identification process for coliforms, the "completed tests"

 +  include the isolation and identification procedures.
 O  include the "total aerobic plate count."
 +  begin with the streaking of Eosin-Methylene Blue Agar from tubes of a selective enrichment.

15G. The following media are selective for organisms which are gram-negative:

 +  Brilliant Green Lactose Bile (BGLB) Broth
 O  Blood Agar
 +  MacConkey Agar
 +  EMB (Eosin-Methylene Blue) Agar
 O  Vogel-Johnson Agar

15H. When isolating coliforms from water, why is there a two-step enrichment process (presumptive and confirm‎atory tests)? That is, why not just inoculate the water into the confirm‎atory media? If one were to inoculate the water sample directly into the highly-selective confirm‎atory broth media, one could get considerable die-off of organisms which were being stressed or injured due to such things as lack of nutrients, chlorination, osmotic problems, etc.

17A. You have cultures of five organisms as listed below. However, the labels of the tubes have come off and you need to re-label the tubes correctly! First, you consider the various reactions you know for the organisms in question:

genus	gram reaction	shape	catalase reaction	glucose fermentation	lactose fermentation	phenylalanine deaminase	citrate utilization
Bacillus	+	rod	+	+ or –	?	?	?
Staphylococcus	+	coccus	+	+	?	?	?
Enterobacter	–	rod	+	+	+	–	+
Morganella	–	rod	+	+	–	+	–
Pseudomonas	–	rod	+	–	–	–	?

a. The results obtained from what one specific laboratory procedure will differentiate Bacillus and Staphylococcus from each other and also from the remaining three genera? the gram stain

b. The remaining three genera can each be distinguished by the result of one test. Indicate a test (just the test, not the reaction) which will give the distinguishing reaction for each of the organisms:
Enterobacter lactose fermentation
Morganella phenylalanine deamination
Pseudomonas glucose fermentation

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Page last modified on 12/17/01 at 8:00 PM, CST. John Lindquist, Department of Bacteriology University of Wisconsin – Madison

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