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Microbiology Evolving Science 3rd Edition by Slonczewski Foster – Test Bank
CHAPTER 2: Observing the Microbial Cell
- Who proved that stomach ulcers are caused by a bacterium?
ANS: B DIF: Easy REF: Introduction TOP: VI
- The part of the human eye that is most involved in resolving an image is the:
ANS: D DIF: Medium REF: 2.1 TOP: I.A
- A ball-shaped microbe is referred to as a:
ANS: B DIF: Easy REF: 2.1 TOP: I.C.ii.b
- Resolution is the smallest distance by which two objects can be __________ and still be __________.
|a.||magnified; seen||d.||distinguished; separated|
|b.||separated; distinguished||e.||magnified; distinguished|
ANS: B DIF: Medium REF: 2.1 TOP: I.A.i
- 400 nm is equivalent to:
|a.||4.0 x 10-5 m||d.||4.0 x 10-8 m|
|b.||4.0 x 10-6 m||e.||4.0 x 10-9 m|
|c.||4.0 x 10-7 m|
ANS: C DIF: Medium REF: 2.1 TOP: I.A
- Which of these series arranges microbes from smallest to largest?
|a.||virus ® bacterium ® red blood cell ® paramecium|
|b.||virus ® red blood cell ® bacterium ® paramecium|
|c.||bacterium ® virus ® paramecium ® red blood cell|
|d.||bacterium ® virus ® red blood cell ® paramecium|
|e.||paramecium ® red blood cell ® bacterium ® virus|
ANS: A DIF: Difficult REF: 2.1 TOP: I.C
- The heating of water when exposed to light is primarily due to:
ANS: C DIF: Difficult REF: 2.2 TOP: II.B.i.a
- Wavelength interference results in small observed objects (like bacteria) being surrounded by:
|a.||a capsule||d.||a dark field|
|b.||a membrane||e.||a cell wall|
|c.||an Airy disk|
ANS: C DIF: Medium REF: 2.2 TOP: II.C
- What is the most important property that enables a lens to magnify an image?
ANS: D DIF: Medium REF: 2.2 TOP: II.C
- When two waves are out of phase by __________ wavelength, they produce destructive interference, canceling each other’s amplitude and resulting in contrast in the image.
|a.||one-tenth of a||d.||one-half of a|
|b.||one-eighth of a||e.||one|
|c.||one-quarter of a|
ANS: D DIF: Difficult REF: 2.2 TOP: II.D
- Increasing the refractive index of the medium between the object and the objective lens increases:
ANS: D DIF: Difficult REF: 2.2 TOP: II.D
- If a glass slide was submerged in a beaker of immersion oil, the slide would be:
|b.||brighter than its surroundings||e.||stained|
|c.||darker than its surroundings|
ANS: A DIF: Medium REF: 2.2 TOP: II.A.ii.a
- What would happen if a lens had the same refractive index as air?
|a.||Light would not pass through the lens.|
|b.||The image would be magnified more than with a glass lens.|
|c.||The image would be magnified, but the resolution would be less than with a glass lens.|
|d.||The image would be magnified, and the resolution would be greater than with a glass lens.|
|e.||The image would not be magnified.|
ANS: E DIF: Medium REF: 2.2 TOP: II.C
- The highest useful magnification for a light microscope is about:
ANS: C DIF: Medium REF: 2.2 TOP: II.D
- A(n) __________ acts to vary the diameter of the light column in a light microscope.
ANS: D DIF: Easy REF: 2.3 TOP: III.B.i
- Which of these arranges the steps of the Gram stain into the correct order?
|a.||iodine ® crystal violet ® decolorizer ® safranin|
|b.||safranin ® decolorizer ® crystal violet ® iodine|
|c.||crystal violet ® decolorizer ® iodine ® safranin|
|d.||crystal violet ® decolorizer ® safranin ® iodine|
|e.||crystal violet ® iodine ® decolorizer ® safranin|
ANS: E DIF: Medium REF: 2.3 TOP: III.E.ii.a
- When Gram stained, most eukaryotes appear:
ANS: C DIF: Medium REF: 2.3 TOP: III.E.ii.a
- Malachite green is commonly used to stain:
|a.||eukaryotic cells||d.||bacterial endospores|
|b.||Gram-negative cells||e.||acid-fast cells|
ANS: D DIF: Medium REF: 2.3 TOP: III.E.ii.c
- As lens strength increases, the light cone __________ and the lens must be __________ the object.
|a.||narrows; nearer to||d.||widens; farther from|
|b.||narrows; farther from||e.||widens; touching|
|c.||widens; nearer to|
ANS: C DIF: Medium REF: 2.3 TOP: III.A
- Staining helps to visualize bacteria by:
|a.||increasing the size of the cells||d.||increasing the magnification of the image|
|b.||increasing the motility of the cells||e.||increasing the aberration of the image|
|c.||increasing the contrast of the image|
ANS: C DIF: Medium REF: 2.3 TOP: III.D.iii
- In a proper Gram stain, positive cells are stained by:
|a.||crystal violet only|
|c.||both crystal violet and safranin|
|d.||neither crystal violet nor safranin|
|e.||not enough information has been provided to know|
ANS: C DIF: Medium REF: 2.3 TOP: III.E.ii.a
- Which two components of the Gram stain form a complex that is retained by Gram-positive cells?
|a.||crystal violet and iodine||d.||alcohol and safranin|
|b.||safranin and iodine||e.||alcohol and iodine|
|c.||crystal violet and safranin|
ANS: A DIF: Medium REF: 2.3 TOP: III.E.ii.a
- Which of the following is best visualized using a negative stain?
|a.||Gram-negative cell wall||d.||endospores|
|b.||acid-fast cell wall||e.||flagella|
ANS: C DIF: Medium REF: 2.3 TOP: III.E.ii.d
- Which of these numeric aperture and light combinations would give the best resolution?
|a.||numeric aperture = 0.8, wavelength = 600 nm|
|b.||numeric aperture = 0.8, wavelength = 500 nm|
|c.||numeric aperture = 1.0, wavelength = 700 nm|
|d.||numeric aperture = 1.0, wavelength = 600 nm|
|e.||numeric aperture = 0.8, wavelength = 400 nm|
ANS: E DIF: Difficult REF: 2.3 TOP: III.A
- What is the total magnification of a light microscope when using a 25X ocular and 40X objective lens?
ANS: D DIF: Medium REF: 2.3 TOP: III.B.i
- What is the best explanation for a Gram-positive bacterium appearing pink after performing a Gram stain?
|a.||The crystal violet was left on for too long.|
|b.||The iodine was left on for too long.|
|c.||The decolorizer was left on for too long.|
|d.||The safranin was left on for too long.|
|e.||The stain was properly performed.|
ANS: C DIF: Difficult REF: 2.3 TOP: III.E.ii.a
- What is the best explanation for a Gram-negative bacterium appearing purple after performing a Gram stain?
|a.||The safranin was not applied.|
|b.||The decolorizer was not applied.|
|c.||The iodine was not applied.|
|d.||The crystal violet was not applied.|
|e.||The stain was properly performed.|
ANS: B DIF: Difficult REF: 2.3 TOP: III.E.ii.a
- A useful application of dark-field optics is the study of bacterial:
ANS: A DIF: Medium REF: 2.4 TOP: IV.A
- Which of the following techniques are based upon wave interference?
|a.||X-ray diffraction and phase contrast microscopy|
|b.||phase contrast and dark-field microscopy|
|c.||bright-field and dark-field microscopy|
|d.||X-ray diffraction and atomic force microscopy|
|e.||scanning and transmission electron microscopy|
ANS: A DIF: Medium REF: 2.4 TOP: IV.A | IV.B
- In which type of microscopy do dust particles interfere the most?
|a.||bright-field microscopy||d.||interference microscopy|
|b.||dark-field microscopy||e.||fluorescence microscopy|
ANS: B DIF: Easy REF: 2.4 TOP: IV.A.iii
- Which of the following would be best suited to observe the motility of microbial cells?
|a.||Gram stain||d.||negative stain|
|b.||nuclear magnetic resonance||e.||phase-contrast microscopy|
|c.||scanning electron microscopy|
ANS: E DIF: Medium REF: 2.4 TOP: IV.B
- DAPI is a dye that is commonly used in _____________ microscopy.
ANS: E DIF: Medium REF: 2.5 TOP: V.B.i
- A fluorophore used in fluorescence microscopy that absorbs light at 260 nm would most likely fluoresce at:
|a.||100 nm||d.||400 nm|
|b.||200 nm||e.||800 nm|
ANS: D DIF: Difficult REF: 2.5 TOP: V.A
- The fluorophore DAPI specifically binds:
|b.||the cell wall||e.||DNA|
ANS: E DIF: Easy REF: 2.5 TOP: V.B.i
- The aromatic groups of the fluorophore DAPI associate exclusively with the:
|a.||cell wall||d.||cell membrane|
|b.||base pairs of DNA||e.||pili|
ANS: B DIF: Medium REF: 2.5 TOP: V.B.i
- Fluorescence microscopy using labeled antibodies is referred to as:
ANS: A DIF: Easy REF: 2.5 TOP: V.B.ii
- Which of the following can be used to localize proteins in a microbial cell?
|a.||DAPI and immunofluorescence|
|b.||acridine orange and green fluorescent protein (GFP) fusions|
|c.||DAPI and acridine orange|
|d.||GFP fusions and immunofluorescence|
|e.||DAPI and GFP fusions|
ANS: D DIF: Difficult REF: 2.5 TOP: V.B.ii | V.B.iii
- Which of these techniques can be used to localize the DNA sequence at the origin of replication in a bacterial cell?
|a.||fluorescence microscopy||d.||atomic force microscopy|
ANS: A DIF: Difficult REF: 2.5 TOP: V.B.iv
- Which form of microscopy is used with DNA microarrays to observe differences in gene expression?
|a.||light microscopy||d.||transmission electron microscopy|
|b.||atomic force microscopy||e.||confocal fluorescence microscopy|
|c.||scanning electron microscopy|
ANS: E DIF: Difficult REF: 2.5 TOP: V.C
- The knife used to cut embedded specimens for observation by transmission electron microscopy is called a:
ANS: B DIF: Easy REF: 2.6 TOP: VI.B.i
- Atomic force microscopy measures __________ between a probe and an object to map the three-dimensional topography of a cell.
|a.||hydrogen bonds||d.||pH changes|
|b.||covalent interactions||e.||magnetic interactions|
|c.||van der Waals forces|
ANS: C DIF: Easy REF: 2.6 TOP: VI.D.ii
- Which type of microscopy is particularly useful to study the surfaces of live bacteria?
ANS: A DIF: Easy REF: 2.6 TOP: VI.D.ii
- Transmission electron microscopy commonly has a resolution of __________ times the highest resolution possible for light microscopy.
ANS: C DIF: Difficult REF: 2.6 TOP: VI.A
- Which of the following would be most appropriate to visualize viral particles being assembled inside an infected bacterial cell?
|a.||dark-field microscopy||d.||scanning electron microscopy|
|b.||atomic force microscopy||e.||transmission electron microscopy|
ANS: E DIF: Medium REF: 2.6 TOP: VI.A | VI.B
- A microscopic structure that is interpreted incorrectly is a/an:
ANS: E DIF: Easy REF: 2.6 TOP: VI.C.i
- Unlike transmission electron microscopy, cryo-electron microscopy:
|a.||requires making thin slices of the sample to be viewed|
|b.||does not require staining with heavy metals|
|c.||may be used to view living tissues|
|d.||uses a weaker electron beam|
|e.||can provide a color image of the microbial cell|
ANS: B DIF: Medium REF: 2.6 TOP: VI.D
- The digitally combined images of cryo-EM can achieve resolution comparable to that of:
|a.||scanning electron microscopy||d.||X-ray crystallography|
|b.||transmission electron microscopy||e.||dark-field microscopy|
ANS: D DIF: Difficult REF: 2.6 TOP: VI.D.i
- Which of the following techniques can visualize bacteria without focusing electromagnetic radiation?
|a.||cryo-electron microscopy||d.||atomic force microscopy|
|b.||phase-contrast microscopy||e.||X-ray diffraction|
ANS: D DIF: Medium REF: 2.6 TOP: VI.D.ii
- The spots recorded on film during X-ray diffraction analyses are due to:
ANS: C DIF: Medium REF: 2.7 TOP: VII.A.i
- Which of these techniques would provide the best resolution of an enzyme’s structure?
|a.||scanning electron microscopy||d.||X-ray diffraction analysis|
|b.||transmission electron microscopy||e.||atomic force microscopy|
ANS: D DIF: Medium REF: 2.7 TOP: VII.A.i
- List and describe three common shapes of bacteria.
Bacilli (bacillus in the singular) are rod-shaped bacteria. Cocci (singular, coccus) are spherical-shaped bacteria. Spirochetes are tightly coiled spirals or corkscrew-shaped bacteria.
DIF: Easy REF: 2.1 TOP: I.C.ii MSC: Remembering
- Microbes were detected long before the invention of the microscope. How could this be?
Detection is the ability to observe the presence of an object, such as when we detect a group of bacteria in a culture tube or growing on a surface like a food product. Even though we can detect the group, we can’t resolve individual cells without the magnification afforded by microscopes.
DIF: Easy REF: 2.1 TOP: I.B.i MSC: Understanding
- Are all bacilli Bacillus? Explain.
No. Bacillus refers to a particular genus of organisms that are commonly found in the soil. Although they are rod-shaped, the members of this genus are not the only bacteria that have this cellular morphology. The term bacillus refers to any rod-shaped microbe, which means that not all bacilli belong to the genus Bacillus.
DIF: Easy REF: 2.1 TOP: I.C.ii MSC: Understanding
- If your eyes had photoreceptors packed as closely as an eagle’s (about eight times greater than humans), would you be able to resolve a virus (100 nm in size) using a light microscope? Why or why not?
No. Although your resolving power would be much improved, the light microscope’s power will still be limited by the wavelengths of light that you can see (roughly 400 nm for human eyes). Objects less than 400 nm cannot be resolved by light in the visible spectrum.
DIF: Medium REF: 2.1 TOP: I.A MSC: Applying
- Describe three conditions that are necessary for electromagnetic radiation to resolve an object.
There must be contrast between the object and its surroundings. The wavelength of the radiation must be equal to or smaller than the size of the object. The detector must have sufficient resolution for the given wavelength.
DIF: Medium REF: 2.2 TOP: II.A.ii MSC: Remembering
- List and briefly describe four ways that light interacts with objects.
(1) Absorption: light energy is absorbed by an object. (2) Reflection: a wavefront bounces off of an object at an angle equal to its incident angle. (3) Refraction: bending of light when it enters a substance that slows its speed. (4) A scattering wavefront interacts with an object of smaller dimensions than the wavelength.
DIF: Medium REF: 2.2 TOP: II.B.i MSC: Remembering
- Compare and contrast a simple stain (like methylene blue) with the Gram stain. What information about a microbial sample can be collected with each?
Both staining procedures colorize bacterial cells, thereby increasing the sample’s contrast and improving resolution. A simple stain will color all microbial cells uniformly. This allows one to record the relative size, shape, and arrangement of any cells present. The Gram stain is a differential stain. In addition to size, shape, and arrangement, this procedure allows one to determine if the cells have a Gram-positive (purple) or Gram-negative (pink) cell wall structure.
DIF: Medium REF: 2.3 TOP: III.E.i MSC: Understanding
- List three different differential stains used in microbiology. What can be detected with each?
The most common differential stain is the Gram stain. This procedure allows one to distinguish between cells having one membrane (Gram-positive) and two membranes (Gram-negative). Another common differential stain is the acid-fast stain. Carbolfuchsin stains the mycolic acid–containing acid-fast cells of the genus Mycobacterium. The endospore stain is a differential stain that stains endospores with malachite green. Negative staining and antibody staining are also included in the text.
DIF: Medium REF: 2.3 TOP: III.E.ii MSC: Understanding
- What color are Gram-positive and Gram-negative cells when properly Gram stained? For each step of the Gram stain procedure, predict the colors of a Gram-positive or Gram-negative cell if that step were omitted during staining. Explain your reasoning.
Properly Gram stained Gram-positive cells are purple and Gram-negative are pink.
(1) Skipping primary stain (crystal violet): Gram-positive and Gram-negative would both be pink. No crystal violet–iodide complex would be formed in the Gram-positive wall. All cells would be decolorized and take on the color of safranin.
(2) Skipping mordant (iodine): Gram-positive and Gram-negative would both be pink. No crystal violet–iodine complex would be formed in the Gram-positive wall. All cells would be decolorized and take on the color of safranin.
(3) Skipping decolorizer (alcohol): Gram-positive and Gram-negative would both be purple. The crystal violet–iodine complex would remain in all cells. Although safranin still binds, the purple color is so much more intense that the pink of the safranin cannot be seen.
(4) Skipping secondary stain (safranin): Gram-positive cells would be purple. The Gram-negative cells would be colorless. The dye complex will be removed from the Gram-negative cells, but they will be difficult to see since the counterstain was not applied.
DIF: Medium REF: 2.3 TOP: III.E.ii.a MSC: Understanding
- Why do some bacteria appear purple after being Gram stained, while others appear pink?
Gram-negative cells have a few layers of peptidoglycan cell wall and an outer lipopolysaccharide membrane. Gram-positive organisms have several layers of peptidoglycan and no outer membrane. The multiple layers of peptidoglycan retain the crystal violet–iodine complex, so appear purple. Gram-negative cells do not retain the crystal violet because there are few layers of peptidoglycan and the outer membrane is disrupted by the decolorizer.
DIF: Medium REF: 2.3 TOP: III.E.ii.a MSC: Understanding
- Compare and contrast the radiation sources, lenses, and image-capturing devices used in light microscopy and transmission electron microscopy.
The radiation source for light microscopy is a light, whereas for electron microscopy it is an electron source or tungsten filament. The lenses in the light microscope are glass, whereas magnets are used in electron microscopy. The lenses have similar functions and are arranged in the same order in both types of microscopy. Light microscopy uses a condenser lens, whereas the lens in electron microscopy is called the projection lens. The image-capturing device for light is the human eye, or sometimes a camera. The image-capturing device for electron microscopy is a fluorescent screen.
DIF: Difficult REF: 2.3 | 2.6 TOP: III.B | VI.A MSC: Understanding
- Why are stains used in microscopy? Compare and contrast the stains used in light versus electron microscopy.
Stains are used to increase the contrast between an object and its surroundings, so as to make it visible. The stains used in light microscopy are usually charged and interact with different cellular components. Positively charged dyes bind to negatively charged cell surfaces. They also are colored, so they impart color to a cell or its components. The stains used for electron microscopy are heavy metals or salts, which increase the density of certain components, again increasing contrast. In electron microscopy, the image of the microbe is always black and white.
DIF: Difficult REF: 2.3 | 2.6 TOP: III.D | VI.B MSC: Understanding
- Name two types of microscopy that are suitable for directly studying bacterial motility. What interaction of light with the microbe is most important for each of these techniques?
Either dark-field or phase-contrast microscopy could be used. In dark-field microscopy, the condenser contains an opaque disk held by three “spider legs” across an open ring. No light travels directly up through the specimen, so the only light that reaches the eye is light that is scattered by objects on the slide. This scattered light allows detection of objects that are too small to be resolved by light rays. Phase-contrast microscopy exploits differences in refractive index between cell components and transforms them into differences in intensity of transmitted light due to wave interference.
DIF: Medium REF: 2.4 TOP: IV.A | IV.B MSC: Understanding
- If you are interested in studying the localization of a protein in a bacterial cell, what techniques would provide you with the best information?
Fluorescence microscopy can be used to study protein localization. One method would be to use fluorescently tagged antibodies to detect the proteins using immunofluorescence microscopy. Another possibility would be to make green fluorescent protein fusions with the protein of interest. These hybrid proteins would fluoresce wherever they are in the cell.
DIF: Difficult REF: 2.5 TOP: V.B MSC: Understanding
- Define a fluorophore and give three examples of how it can be used to label cells.
A fluorophore is a fluorescent molecule that can be used to stain a specimen for observation with a fluorescence microscope. Some fluorophores, such as DAPI, have affinity for certain cell chemicals. Antibodies can be labeled with fluorescent dyes and reacted with specific targets in immunofluorescence. Short sequences of DNA attached to a fluorophore can be used to hybridize and label target DNA.
DIF: Difficult REF: 2.5 TOP: V.B MSC: Understanding
- Archaea and Bacteria differ in the genetic sequences of their ribosomal RNA genes. How can this difference be used to microscopically differentiate between members of these domains?
Short DNA sequences that are homologous to either the Bacterial or Archaeal sequences can be conjugated to fluorophores that emit different wavelength light. These probes will anneal to the complementary DNA in the corresponding cells in a sample. When viewed using a fluorescence microscope, the archaeal and bacterial cells will have different colors. This is referred to as fluorescence in situ hybridization, or FISH, analysis.
DIF: Difficult REF: 2.5 TOP: V.B.iv MSC: Applying
- Most electron micrographs in microbiology textbooks are in color. Is this normal for an electron micrograph? Why or why not?
Electron micrographs are not naturally colored. The original image is produced when the electrons bombard a fluorescent screen. The resultant image is processed by a computer to appear as black and white with intensities in the entire range of grays in between. These images are later colorized using computer software (like Photoshop) to improve the aesthetics and provide additional information.
DIF: Difficult REF: 2.6 TOP: VI.A MSC: Understanding
- Give a few reasons why living organisms may not be observed by transmission electron microscopy (TEM) or scanning electron microscopy (SEM).
In TEM, the specimens are fixed and embedded into a polymer for sectioning. The specimen is then stained with heavy metal to increase contrast. In SEM, the entire organism is shadowed with heavy metal prior to observation. Most importantly, however, the entire optical column of the EM must be maintained under vacuum, and a living specimen would be quickly destroyed by an electron beam.
DIF: Easy REF: 2.6 TOP: VI.A | VI.B MSC: Understanding
- Describe three methods of sample preparation for electron microscopy. Which method would cause the fewest artifacts? Why?
(1) Samples can be embedded in a polymer and cut into thin sections with a microtome, then coated with a heavy metal. (2) Samples can be sprayed onto a copper grid, then treated with a heavy metal. (3) Samples may be flash frozen for cryo-electron microscopy. Cryo-EM will cause the fewest artifacts. When using this technique, the cells are not fixed or artificially stained. Instead, the cells are flash frozen—leaving the cell components still hydrated and closest to their original state.
DIF: Medium REF: 2.6 TOP: VI.B MSC: Understanding
CHAPTER 10: Molecular Regulation
- __________ prevent transcription, whereas __________ stimulate transcription.
|a.||Activators; repressors||d.||Regulators; repressors|
|b.||Inducers; corepressors||e.||Repressors; activators|
ANS: E DIF: Easy REF: 10.1 TOP: I.A.iv | I.A.v
- A transmembrane sensor kinase protein senses an environmental condition outside __________ bacteria or in the periplasm of a(n) __________ bacteria.
|a.||Gram-positive; Gram-negative||d.||prokaryotic; archaean|
|b.||Gram-negative; Gram-positive||e.||Gram-positive; acid-fast|
ANS: A DIF: Easy REF: 10.1 TOP: I.B.ii
- Which form of control is the least reversible and most drastic?
|a.||posttranslational control||d.||translational control|
|b.||control of transcription||e.||mRNA stability|
|c.||alterations of DNA sequence|
ANS: C DIF: Easy REF: 10.1 TOP: I.C
- Sets of genes in operons are coordinately regulated by all of the following EXCEPT:
|e.||irreversible changes in the DNA sequence|
ANS: E DIF: Medium REF: 10.1 TOP: I.C
- In some, but not all, instances, __________ molecules bind RNA transcripts and help or hinder degradation.
ANS: D DIF: Easy REF: 10.1 TOP: I.C.ii
- Which form of control is the most rapid?
|a.||alterations of DNA sequence|
|b.||control of transcription via activators and repressors|
|e.||use of various sigma factors|
ANS: D DIF: Medium REF: 10.1 TOP: I.C.iii
- A feature common to all control mechanisms is:
|c.||ability to sense that something inside or around the cell has changed|
|e.||increased or decreased gene expression|
ANS: C DIF: Medium REF: 10.1 TOP: I.B
- In a two-component signal transduction system, a _________ is transferred from a sensor kinase to a _________ in response to an environmental signal.
|a.||phosphate; sensor domain||d.||magnesium; sensor domain|
|b.||phosphate; sensor phosphatase||e.||magnesium; response regulator|
|c.||phosphate; response regulator|
ANS: C DIF: Medium REF: 10.1 TOP: I.B.ii
- When __________ interacts with RNA polymerase, it increases the rate of transcription initiation of the lac operon.
|b.||cAMP receptor protein||e.||glucose|
ANS: B DIF: Medium REF: 10.2 TOP: II.B
- If a gene is always expressed, it is:
ANS: A DIF: Easy REF: 10.2 TOP: II.C.i
- Which of the following is the favored carbon source of Escherichia coli?
ANS: D DIF: Easy REF: 10.2 TOP: II.C.i
- Finding an inverted repeated sequence within 100–200 bases of a promoter suggests the presence of a __________ binding site.
ANS: B DIF: Easy REF: 10.2 TOP: II.E.i
- Symmetry, in terms of DNA binding, usually involves a(n):
|b.||major groove||e.||direct repeat|
ANS: A DIF: Medium REF: 10.2 TOP: II.E.i
- How many of the gene products of the lactose operon are required for the utilization of lactose to occur?
ANS: C DIF: Easy REF: 10.2 TOP: II.A
- What occurs when an inducer is added to a medium containing an organism with a metabolic pathway controlled by a repressor?
|a.||The inducer combines with the repressor and activates the repressor.|
|b.||The inducer combines with the repressor and inactivates the repressor.|
|c.||The inducer combines with the substrate and blocks induction.|
|d.||The inducer combines with the substrate and activates induction.|
|e.||The inducer does not combine with, but functions as a chaperone molecule for, the enzyme-substrate complex.|
ANS: B DIF: Medium REF: 10.2 TOP: II.A.iv
- When both glucose and lactose are added to the growth medium, Escherichia coli will have a growth curve that:
|a.||does not change|
|b.||grows slowly and then, halfway through, grows faster|
|c.||grows, levels off, and then grows again|
|d.||grows more slowly than with glucose alone|
|e.||grows more quickly than with glucose alone|
ANS: C DIF: Medium REF: 10.2 TOP: II.C.iii
- All of the following typically occur in the presence of high glucose and high lactose concentrations EXCEPT:
|a.||enzyme II glucose is unphosphorylated|
|b.||cAMP concentrations are low|
|c.||C-reactive protein levels in the cell are low|
|d.||inhibition of LacY activity|
|e.||lactose is kept out of the cell|
ANS: C DIF: Difficult REF: 10.2 TOP: II.D
- Binding of tryptophan to the __________ makes a holorepressor.
ANS: A DIF: Easy REF: 10.3 TOP: III.B
- The araBAD and araC genes are transcribed in the __________ direction with respect to each other.
ANS: B DIF: Easy REF: 10.3 TOP: III.B.i
- Transcriptional attenuation is a common regulatory strategy used to control many operons that code for what?
|a.||amino acid degradation||d.||carbohydrate biosynthesis|
|b.||amino acid biosynthesis||e.||lipid degradation|
ANS: B DIF: Easy REF: 10.3 TOP: III.C
- The AraC-like regulators share homology with each other at which locations?
|a.||both their C-terminal end and their N-terminal end|
|b.||neither their C-terminal end nor their N-terminal end|
|c.||their C-terminal end only|
|d.||their N-terminal end only|
ANS: C DIF: Medium REF: 10.3 TOP: III.C
- The stringent response relies on the production of:
ANS: E DIF: Difficult REF: 10.3 TOP: III.D
- The AraC regulatory protein turns on transcription in the presence of arabinose in which way?
|a.||With the addition of arabinose, AraC no longer dimerizes and will form a large loop by binding to araO and aria, and thus can interact with RNA polymerase to permit transcription.|
|b.||The dimer will form a large loop by binding to araO and aria, and thus can interact with RNA polymerase to permit transcription.|
|c.||With the addition, AraC no longer dimerizes and thus can interact with RNA polymerase to permit transcription.|
|d.||It becomes less flexible, so that a monomer can bind to araI1 and araI2 and interact with RNA polymerase to permit transcription.|
|e.||It becomes more flexible, so that the dimer can bind to araI1 and araI2 and interact with RNA polymerase to permit transcription.|
ANS: E DIF: Difficult REF: 10.3 TOP: III.B
- Bacterial systems using an AraC/XylS type regulator include __________ as regulated by __________.
|a.||a type III secretion system in Pseudomonas; NitR|
|b.||a type III secretion system in Pseudomonas; ExsA|
|c.||a type III secretion system in Pseudomonas; YbtA|
|d.||virulence genes and cholera toxin in Vibrio; TxtR|
|e.||virulence genes and cholera toxin in Vibrio; ExsA|
ANS: B DIF: Medium REF: 10.3 TOP: III.B
- The role of a leader sequence is to:
|a.||directly control transcription|
|b.||determine whether transcription can proceed through the genes in the operon|
|c.||determine whether RNA polymerase can bind|
|e.||direct the looping of the DNA that prevents transcription initiation|
ANS: B DIF: Medium REF: 10.3 TOP: III.C.iii
- The stringent response involves all EXCEPT:
|a.||downregulation of rRNA synthesis||d.||GTP|
|b.||downregulation of tRNA synthesis||e.||no change in gene expression|
ANS: E DIF: Difficult REF: 10.3 TOP: III.D
- The little amount of sigma H that is made at 30°C is __________ by the DnaK/GrpE/DnaJ chaperones.
ANS: C DIF: Medium REF: 10.4 TOP: IV.B.ii
- When the forespore is first produced, about how much of the chromosome is actually inside the forespore?
ANS: C DIF: Medium REF: 10.4 TOP: IV.C
- Which of the following is NOT true of sRNA genes?
|a.||They have a ribosomal-binding site.|
|b.||They occur between genes.|
|c.||They share homology between related species.|
|d.||They can act as antisense RNAs.|
|e.||Their gene product was “Molecule of the Year” in 2002.|
ANS: A DIF: Medium REF: 10.5 TOP: V.A
- __________ genes always occur in intergenic regions.
ANS: C DIF: Easy REF: 10.5 TOP: V.A.ii
- __________ RNAs bind to complementary sequences of target transcripts and stimulate or prevent translation.
ANS: D DIF: Easy REF: 10.5 TOP: V.B
- Which of the following explains the regulatory activity of sRNA?
|a.||It codes for a sigma factor.|
|b.||It prevents CRP interaction with RNA polymerase, thereby blocking its access to the promoter.|
|c.||It synthesizes ppGpp.|
|d.||It targets mRNAs for degradation.|
|e.||It creates an attenuator stem loop.|
ANS: D DIF: Medium REF: 10.5 TOP: V.A
- Some microbes use gene regulation to periodically change their appearance, in a process called:
ANS: E DIF: Easy REF: 10.6 TOP: VI.A
- Which of the following is NOT true of phase variation?
|a.||The orientation of the DNA sequence is inverted.|
|b.||It may help the bacteria avoid the immune system.|
|c.||It can result in a change in microbial appearance.|
|d.||It only occurs in prokaryotes.|
|e.||All are true.|
ANS: D DIF: Medium REF: 10.6 TOP: VI.A | VI.B | VI.C
- The role of the Hin recombinase is to:
|a.||recombine specific segments of the DNA of Haemophilus influenzae|
|b.||recombine specifc segments of flagellin genes in Neisseria gonorrhoeae|
|c.||recombine specific segments of MCP genes in Salmonella enterica|
|d.||recombine specific segments of flagellin genes in Salmonella enterica|
|e.||repair damage in DNA|
ANS: D DIF: Medium REF: 10.6 TOP: VI.B
- If the newly synthesized DNA strand slips back relative to the template strand, one copy of the repeated unit is __________ the growing strand.
|a.||inserted in||d.||folded under|
|b.||deleted from||e.||bound to|
ANS: A DIF: Medium REF: 10.6 TOP: VI.C
- Phase variation in Salmonella enterica involves:
|c.||outer membrane proteins|
ANS: A DIF: Easy REF: 10.6 TOP: VI.B.ii
- The mechanism behind slipped-strand mispairing is:
|a.||short repeats allow for slippage of the RNA polymerase during transcription of the affected genes|
|b.||short repeats allow for slippage of DNA polymerase during replication, which leads to an out-of-frame reading sequence|
|c.||an invertible promoter slips into different reading frames|
|d.||the genes coding for lipopolysaccharide in Gram-negative bacteria can be slipped in and out of frame|
|e.||short repeats slip and turn genes off, but they can never be turned back on|
ANS: B DIF: Medium REF: 10.6 TOP: VI.C
- GlnB allows cells to balance glutamine synthetase __________ with its rate of __________.
|a.||amounts; degradation||d.||synthesis; activity|
|b.||mRNA; translation||e.||activity; synthesis|
ANS: E DIF: Medium REF: 10.7 TOP: VII.B.iii
- Which of the following is a second messenger that helps to regulate biofilm production?
ANS: E DIF: Easy REF: 10.7 TOP: VII.C
- In the regulation of chemotaxis, proteins are modified in which of the following ways?
|d.||phosphorylation and methylation|
|e.||feedback inhibition, phosphorylation, and methylation|
ANS: D DIF: Medium REF: 10.7 TOP: VII.A
- Which of the following statements is true concerning the use of cyclic di-GMP as a second messenger?
|a.||It activates the expression of flagellin genes and thus enhances motility.|
|b.||It is found in all microbes: bacteria, archaea, and eukaryotic microbes.|
|c.||It is produced by the action of diguanylate cyclase.|
|d.||It decreases in level as biofilm is produced.|
|e.||Phosphodiesterase can convert GMP to cyclic di-GMP.|
ANS: C DIF: Medium REF: 10.7 TOP: VII.C.i
- Integrated circuits are involved in all of the following EXCEPT:
|a.||controlling chemotactic activities of a cell|
|b.||coupling the genetic and biochemical control of a metabolic pathway|
|c.||timing the events of a developmental cycle|
|d.||having an influence on nitrogen metabolism|
|e.||specifically inverting the flagellin genes|
ANS: E DIF: Medium REF: 10.7 TOP: VII
- In terms of controlling chemotaxis, the proteins involved control chemotaxis in which way?
|a.||They control which form of flagellin is expressed.|
|b.||They change the direction of the flagella from counterclockwise to clockwise and back.|
|c.||They promote longer “tumbles” when a chemoattractant is present.|
|d.||They increase the phosphorylation of CheY so that there are more counterclockwise flagella.|
|e.||They respond to demethylation of methyl-accepting chemotaxis proteins to bring about longer runs.|
ANS: B DIF: Difficult REF: 10.7 TOP: VII.A
- Which of the following is NOT a process regulated via quorum sensing?
|a.||pathogenesis of Pseudomonas infection in a cystic fibrosis patient|
|b.||antibiotic production by Streptomyces|
|c.||bioluminescence by Vibrio|
|d.||utilization of lactose by Escherichia coli|
|e.||exotoxin production by Staphylococcus|
ANS: D DIF: Easy REF: 10.8 TOP: VII
- Many bacteria produce some type of homoserine lactone for quorum sensing, yet some Gram-positive bacteria will produce ________________ for this same purpose.
|c.||furanosyl borate diester|
ANS: B DIF: Medium REF: 10.8 TOP: VIII
- A quorum-sensing gene system requires the accumulation of a secreted small molecule called a(n):
ANS: A DIF: Easy REF: 10.8 TOP: VIII.A
- All the cell’s expressed RNAs are collectively referred to as the cell’s:
ANS: B DIF: Medium REF: 10.9 TOP: IX.A.i
- Which molecule is used for hybridization to DNA bound on a DNA microchip?
|b.||RNA||e.||any of the above|
ANS: D DIF: Easy REF: 10.9 TOP: IX.A.iii
- Which of the following can be used to study the bacterial proteome?
|c.||two-dimensional gels and mass spectrometry|
|d.||green fluorescent protein|
|e.||networking with nanotubes|
ANS: C DIF: Easy REF: 10.9 TOP: IX.B | IX.C
- Explain a way that organisms sense and respond to their environment.
Answers vary but could include the following: Two-component signal transduction systems allow organisms to sense and respond to their environment. Activation of a sensor kinase protein that spans the membrane results in a conformational change and self-phosphorylation. The phosphate is then passed to a response regulator in the cytoplasm that controls gene expression.
DIF: Medium REF: 10.1 TOP: I.B MSC: Understanding
- Describe the function of prokaryotic activator proteins.
Most prokaryotic activators bind to activator sequences just upstream from their target genes. The function of an activator is to recruit RNAP to promoter regions to enhance transcription.
DIF: Medium REF: 10.1 | 10.3 TOP: I.A.vi MSC: Understanding
- What is diauxic growth and why does it occur?
This type of growth occurs when an organism has a choice of two carbon sources, such as glucose and lactose. Catabolite repression causes the organism to first use the glucose, since the enzymes for metabolizing glucose are constitutively produced and it is the preferred carbon source. The lactose operon will be repressed until the glucose has been consumed. After the glucose is consumed, the lactose operon will no longer be repressed and the lactose can be metabolized.
DIF: Medium REF: 10.2 TOP: II.C MSC: Remembering
- Describe what happens to the lac operon in the presence of both lactose and glucose.
In the presence of both lactose and glucose, catabolite repression will occur, which will lower cAMP levels, inhibit lactose transport into the cell, and—via lack of cAMP-CRP complex—minimize transcription of the lac operon.
DIF: Medium REF: 10.2 TOP: II.C | II.D MSC: Remembering
- Glucose transport into the cell brings about a phenomenon known as inducer exclusion.
Describe how this functions with respect to the lac operon.
In the presence of glucose, the PTS system will add phosphates from phosphoenolpyruvate (PEP) to glucose as it is transported across the cell membrane via a series of proteins. Thus, there is no phosphate left on enzyme II and LacY is inhibited. Without glucose present, these phosphorylations will not happen, enzyme II will be phosphorylated, and lacY will then be able to transport lactose into the cell.
DIF: Medium REF: 10.2 TOP: II.D MSC: Remembering
- How is DNA sequence analysis used to study DNA protein binding?
DNA sites that bind regulatory proteins typically exhibit sequence symmetry involving an inverted or direct repeat. Computer programs can be used to examine DNA sequences and find inverted repeats that suggest the presence of a regulatory binding site.
DIF: Medium REF: 10.2 TOP: II.E.i MSC: Remembering
- The AraC regulator both activates and represses transcription of the genes involved in arabinose catabolism. How is this system more advantageous to a cell than a simpler repressor system like that used on the lac operon?
The repressor in the lac operon comes completely off the DNA in the presence of the inducer allolactose. Consequently, it will disperse in the cell, and random diffusion will be necessary for it to regain repression. There will therefore be a delay between induction and repression. The AraC regulator is simply shuttled back and forth on the DNA between the activator and repressor binding sites. This means that the delay between induction and repression will be short.
DIF: Medium REF: 10.3 TOP: III.A MSC: Remembering
- How is repression of biosynthetic pathways different than repression of a catabolic pathway? Why does repression of these two types of pathways work differently?
Repressors that control catabolic pathways usually bind the substrate for that pathway. Binding the substrate decreases the repressor’s affinity for the operator sequence. Increased concentration of the substrate removes the repressor from the operator and derepresses expression of the genes. This works well because the cell needs to make enzymes that enable it to use the substrate. For biosynthetic pathways that are controlled by repressors, the repressor binds the end product of the pathway to become active. When the substrate binds the repressor, the repressor affinity for the operator sequence is increased. As the substrate builds up to high levels, the cell will shut down the biosynthetic pathway so energy is not wasted.
DIF: Difficult REF: 10.3 TOP: III.B MSC: Remembering
- Explain how a repressor protein works with respect to a biosynthetic operon.
When the endproduct of the biosynthetic operon is high, the cell does not need those enzymes. Thus, the endproduct will bind to the inactive aporepressor and become an active repressor protein, which will shut down the transcription of the genes. As the end product gets low, there is not enough to bind to the aporepressor; thus, there will be no repression, and transcription of the needed genes will occur.
DIF: Medium REF: 10.3 TOP: III.B MSC: Understanding
- Explain the process of attenuation in regulating transcription of the tryptophan operon.
Attenuation is a process of transcription regulation in which translation of a leader peptide affects transcription of a downstream structural gene. When levels of tryptophan are high, the ribosome will translate through the trp codons but runs into a stop codon, causing it to stall. This stall causes the formation of a termination loop and the ribosome is released. This prevents a waste of energy in making an amino acid that is already present in high levels. When tryptophan levels are low, the ribosome will stall at trp codons and an anti-attenuator stem loop forms that prevents formation of the attenuator stem loop. This causes the polymerase and ribosome to continue and translate the trp structural genes.
DIF: Difficult REF: 10.3 TOP: III.C MSC: Understanding
- The stringent response down-regulates rRNA synthesis. Does it similarly down-regulate ribosomal protein synthesis?
Ribosomal protein synthesis will also slow, but not due to ppGpp and the stringent response. Ribosomal proteins have the ability to bind to their own mRNA, and when they do, translation is inhibited. Binding to their own mRNA only happens when rRNA concentrations drop, thereby increasing cytoplasmic levels of free ribosomal proteins. These ribosomal proteins bind to their mRNAs and effectively shut down translation.
DIF: Difficult REF: 10.3 TOP: III.D MSC: Understanding
- How is sigma H degradation controlled by temperature, and why is this important to the cell?
At lower temperatures, chaperones bring sigma H to proteases for digestion, so levels stay low. At higher temperatures, chaperones are shuttled to the many heat-denatured proteins that are formed and away from sigma H, so sigma H levels are high. This is important because at higher temperatures sigma H is needed to transcribe heat-shock genes.
DIF: Medium REF: 10.4 TOP: IV.C MSC: Understanding
- During the process of endospore formation, different sigma factors are active in the mother cell and forespore. How is this possible, and why is it important?
When the septum is formed and the forespore is first produced, only 30% of the chromosome is in the forespore. The rest of the chromosome slowly moves in over time. Temporary asymmetrical distribution of the chromosome results in activation of different sigma factors in each compartment. Sigma F is produced in the forespore and involved in the early stages of spore formation. Without the compartment-specific sigma factors, spore formation could not occur.
DIF: Difficult REF: 10.4 TOP: IV.C MSC: Understanding
- Explain how small, untranslated RNAs (sRNAs) regulate iron uptake and iron storage.
Answers will vary but may include the following: When iron levels are low, RyhB sRNA binds to its target sequence in mRNA, encoding succinate dehydrogenase. This binding targets the mRNA for degradation by RNase, thereby downregulating succinate dehydrogenase production and lessening iron utilization. When iron levels are high, the iron-gathering enterochelin gene and the ryhB gene encoding RyhB sRNA are both repressed by binding of the Fur repressor protein. The sucCDAB gene is expressed, and succinate dehydrogenase is made.
DIF: Difficult REF: 10.5 TOP: V.A MSC: Understanding
- What does the gene inversion in Salmonella enterica accomplish with respect to its course of infection? Briefly describe how this mechanism works.
Changing its flagellar type enables the bacterium to evade the human immune response for a while longer. This mechanism involves an invertible element that contains a promoter sequence. The Salmonella has two flagellar types, H1 and H2. The gene encoding H2 is adjacent to a gene coding a repressor for H1 flagellin production. These genes are only transcribed when the invertible element is in the correct direction for the promoter to be situated properly and for the transcription of H2 flagellin and H1 flagellin repressor to occur. When the element flips, there is no promoter for the genes, and H2 flagellin and the H1 repressor are no longer produced. Without the H1 flagellin repressor, the cell can now make H1 flagellin. This mechanism ensures that only one type of flagellin will be present at any one time.
DIF: Difficult REF: 10.6 TOP: VI.B MSC: Understanding
- What happens when strand slippage occurs on the template or the newly synthesized DNA strand? How does Neisseria gonorrhoeae use slipped-strand mispairing to its advantage?
Strand slippage occurring on the template strand during DNA replication will cause one repeat to be deleted. If, however, the slippage occurs on the newly synthesized strand, one extra copy will be inserted into the growing strand. Neisseria gonorrhoeae slippage is possible due to a short repeat found in its genes for outer membrane proteins. A gene will be switched on or off by this slippage.
DIF: Difficult REF: 10.6 TOP: VI.C MSC: Understanding
- Explain what happens to motility when a chemoattractant is added to bacterial cells.
In the presence of a chemoattractant, the MCP for that chemical will become methylated. This methylation will then inhibit CheA kinase activity, which results in lower CheY-P (via CheZ dephosphorylation) and then results in the counterclockwise turning of the flagella to give a “run.” A longer run will aid the cell in moving toward the chemical.
DIF: Difficult REF: 10.7 TOP: VII.A MSC: Understanding
- Explain the process of bioluminescence in Allivibrio fischeri. How is this an example of a mutualistic relationship?
Allivibrio fischeri is bioluminescent at high cell densities due to the accumulation of an autoinducer. At high cell levels and high autoinducer concentration, the autoinducer reenters the cell and binds LuxR, which activates transcription of the luciferase genes, and the cells become bioluminescent. This organism colonizes the light organ of the Hawaiian squid. The bacteria obtain shelter by living in the light organ. The squid camouflages itself using counterillumination when swimming at night, so it is safer from predators.
DIF: Difficult REF: 10.8 TOP: VIII.A | VIII.B
- Quorum sensing plays a role in interspecies communications. Describe the relationship between green seaweed Enteromorpha and Vibrio anguillarum bacterial cells in biofilms, and explain how Escherichia coli was used to understand this interaction.
Enteromorpha makes motile zoospores that search out V. anguillarum biofilms. The eukaryote is able to sense the acyl homoserine lactone that the bacteria produce. To prove this was the communication process, E. coli were engineered to produce the Vibrio-specific acyl homoserine lactone. Enteromorpha zoospores were able to sense and attach to these engineered E. coli cells in a biofilm.
DIF: Difficult REF: 10.8 TOP: VIII.C MSC: Understanding
- Briefly explain the two-dimensional gels electrophoresis. What is the procedure used to determine, and why are both dimensions necessary?
The proteome of a cell includes all of the cell’s expressed proteins. The proteome will change throughout a cell’s life and in response to varying environmental conditions. Two-dimensional gel electrophoresis is used to compare fluctuations in the proteome under different growth conditions. The first dimension involves isoelectric focusing, which sorts proteins according to their charge or isoelectric point. The second dimension uses polyacrylamide gel electrophoresis to further separate proteins by their molecular weights.
DIF: Medium REF: 10.9 TOP: IX.B MSC: Understanding
CHAPTER 20: Eukaryotic Diversity
- Of the choices listed below, which domain has the LEAST diversity in terms of metabolism?
ANS: B DIF: Easy REF: Introduction TOP: I
- The term “mycology” would best be applied to which organism(s) listed below?
ANS: E DIF: Easy REF: 20.1 TOP: I.B.i.a
- Based on microscopy work of the eighteenth and nineteenth centuries, which of the examples listed was placed into the Eukaryotic classification of Algae?
ANS: C DIF: Easy REF: 20.1 TOP: I.B.ii.c
- Slime molds were first classified with fungi but have been reclassified as protozoa based on:
|a.||color and cellular size||d.||hyphae structure and colony formation|
|b.||reproduction rates||e.||chloroplast structure and general shape|
|c.||morphology and movement structures|
ANS: C DIF: Medium REF: 20.1 TOP: I.B.iii.a
- Protist is a classification term encompassing all EXCEPT for which of the following?
ANS: D DIF: Easy REF: 20.1 TOP: I.C.i
- Which of the following could be classified in the clade Opisthokonta?
|b.||land plants||e.||brown algae|
ANS: A DIF: Easy REF: 20.1 TOP: I.E.i
- Which of the following would NOT be classified in the clade Opisthokonta?
|a.||Saccharomyces cerevisiae||d.||Trypanosoma brucei|
|b.||Felis catus||e.||Canis lupus|
ANS: D DIF: Medium REF: 20.1 TOP: I.E.i
- The clade Opisthokonta was definitively determined by:
|a.||short sequence insertion studies of elongation factor 1α gene|
|b.||sequence variance of ribosomal RNA genes|
|c.||genomic sequence and homology studies of DNA polymerase|
|d.||flagellar movement and structural studies|
|e.||morphological and environmental comparison studies|
ANS: A DIF: Difficult REF: 20.1 TOP: I.E.i
- Which organisms are considered the possible “missing link” in the evolution from microbial eukaryotes to multicellular animals due to shared genes?
ANS: D DIF: Medium REF: 20.1 TOP: I.E.i.b
- The suffix “-mycota” refers to which taxa of eukaryotes?
ANS: D DIF: Easy REF: 20.1 TOP: I.E.i.c
- Which of the following could be classified in the clade Viridiplantae?
|a.||primary endosymbiont algae||d.||microsporidia|
ANS: A DIF: Easy REF: 20.1 TOP: I.E.ii.a
- All protists share which trait?
|a.||the presence of cilia on the cell membrane|
|b.||a nuclear membrane around the cell DNA|
|c.||the presence of mitochondria|
|d.||all of the above|
|e.||none of the above|
ANS: B DIF: Easy REF: 20.1 TOP: I.E.iii
- Which of the following is the taxa containing organisms with paired cilia or flagella and a complex outer covering called a cortex?
ANS: A DIF: Easy REF: 20.1 TOP: I.E.iii.b
- A newly discovered eukaryotic cell that is a parasitic flagellate, lacking mitochondria and Golgi organelles, would best fit which of the following taxa?
ANS: B DIF: Easy REF: 20.1 TOP: I.E.iii.e
- In which taxa can the food-grade mushrooms of morels and truffles be found?
ANS: E DIF: Medium REF: 20.2 TOP: I.E.i.c
- Which of the following is NOT a common trait shared by most mycelium-forming fungi?
|a.||absorptive nutrition||d.||chitinous cell walls|
|b.||hyphae production||e.||budding reproduction|
ANS: E DIF: Easy REF: 20.2
TOP: II.A.i | II.A.ii | II.A.iii | II.A.iv | II.A.v MSC: Remembering
- Yarrowia lipolytica is a dimorphic yeast; this means that it can:
|a.||be single-cellular or form a mycelia||d.||have one or two flagella|
|b.||form linear or branching hyphae||e.||form aerial or surface mycelium|
|c.||undergo mitosis or meiosis|
ANS: A DIF: Easy REF: 20.2 TOP: II.B
- In the organism Saccharomyces cerevisiae, which of the following induces alternation of generations from a haploid to a diploid cell?
|a.||a 10-degree decrease in optimum growth temperature|
|e.||a pH change toward acidic conditions|
ANS: C DIF: Medium REF: 20.2 TOP: II.B.ii.a
- The fungus Neocallomastix sp. is found in animal rumens and directly helps with:
|b.||digestion of tough plant material|
|c.||nutrient absorption through the intestinal wall|
|e.||preventing ulcer formation|
ANS: B DIF: Easy REF: 20.2 TOP: II.C.i.a
- Tinea pedis, the fungi that causes athlete’s foot, has male and female mycelium tips that can fuse and ultimately form ascospores. Which phyla do they belong in?
ANS: C DIF: Easy REF: 20.2 TOP: II.C.iii.a | II.C.iii.b
- Which of the following is the clade that is home to the deadly poison-producing mushrooms called Amanita or “destroying angel”?
ANS: A DIF: Medium REF: 20.2 TOP: II.C.iv.a
- Glomeromycota are obligate mutualists of plants:
|a.||that form fruiting bodies, such as basidiomycetes|
|b.||and are considered pathogenic to more than 90% of all land plants|
|c.||that form arbuscular mycorrhizae, in which the fungal filaments penetrate plant cell walls and the cytoplasmic membrane to establish symbiosis|
|d.||that receive sugars from plants and provide ammonium and phosphate to plants|
|e.||that form periarbuscular space within the fungi’s own cell wall|
ANS: D DIF: Easy REF: 20.2 TOP: II.C.v
- Enterocytozoon bieneusi is an organism classified as a microsporidia; this organism is of clinical importance because it:
|a.||causes severe allergic reactions in some people|
|c.||causes ringworm infections|
|d.||is an opportunistic pathogen in immunocompromised individuals|
|e.||infects catheters by producing biofilms|
ANS: D DIF: Medium REF: 20.2 TOP: II.C.vi.b
- Based on the criteria given in the table below, where should this newly discovered organism be placed?
|Results of unknown organism|
|Observation #1: Life cycle involves transitions between n, n+n, and 2n phases|
|Observation #2: Mycelia structures observed|
|Observation #3: Conidiophores observed|
ANS: A DIF: Difficult REF: 20.2 TOP: II.C.iii.b
- Phytoplankton refers to:
|a.||heterotrophic algae and bacteria present in aquatic and marine environments|
|b.||mixotrophic algae and amebas present in aquatic environments|
|c.||photosynthetic algae and bacteria present in aquatic and marine environments|
|d.||heterotrophic algae, bacteria, and protists present in aquatic, marine, and terrestrial environments|
|e.||photosynthetic algae and plants present in aquatic, marine, and terrestrial environments|
ANS: C DIF: Easy REF: 20.3 TOP: III.A
- An example of unicellular green algae would be found in which of the following genera?
ANS: E DIF: Easy REF: 20.3 TOP: III.B.i
- Which of the following algae reproduce by opposite mating filaments forming a cytoplasmic conjugate bridge to fuse cellular material?
|a.||filamentous green algae||d.||diatoms|
|b.||sheet-forming green algae||e.||lichens|
|c.||unicellular green algae|
ANS: A DIF: Medium REF: 20.3 TOP: III.B.ii.a
- Rhodophyta, or red algae, appear red due to:
ANS: D DIF: Easy REF: 20.3 TOP: III.C
- Heterokont is a grouping of organisms based on which of the following structural characteristics?
|a.||They all inhabit the same ecological niche.|
|b.||They all rely exclusively on photosynthesis for metabolism.|
|c.||They all have or have had pairs of flagella of different shapes and lengths.|
|d.||They all have a bipartite shell called a frustule.|
|e.||They all move with similar locomotive structures.|
ANS: C DIF: Easy REF: 20.3 TOP: III.E.iii
- Based on the criteria given in the table below, where should this newly discovered organism be placed?
|Results of unknown organism|
|Observation #1: Unicellular, photosynthetic, found in aquatic and marine environs|
|Observation #2: Unique bipartite shell made of silica|
|Observation #3: Daughter cells smaller than parent cells, with each generation|
ANS: C DIF: Difficult REF: 20.3 TOP: III.E.iii.a
- Which of the following statements regarding pseudopods is NOT true?
|a.||Pseudopods are also called “false feet”.|
|b.||Amoebozoa have lobe-shaped pseudopods.|
|c.||Lamellar pseudopods of amebas are similar to those generated by human leukocytes.|
|d.||Needlelike pseudopods are thin extensions reinforced by actin polymers.|
|e.||The tip of a pseudopod contains a gel of polymerized actin beneath its cell membrane.|
ANS: D DIF: Easy REF: 20.4 TOP: IV.A
- Amoeboid movement is controlled by:
|a.||cytoplasmic streaming into the extending pseudopodia|
|b.||paired flagella formations extending off of the cell membrane|
|c.||cilia structures surrounding the cell|
|d.||regulation of osmotic and gas vacuoles within the cytoplasm|
|e.||barophilic pumps within the cell membrane|
ANS: A DIF: Easy REF: 20.4 TOP: IV.A.i.a
- Fuligo septica is a type of plasmodial slime mold that can be seen as a yellow mass growing on decaying wood. Which of the following statements best describes this organism?
|a.||produces a fruiting body in which individual cells remain unicellular|
|b.||produces no fruiting bodies and no spore formations|
|c.||produces a fruiting body in which individual cells become a multinucleated single-cell|
|d.||produces fruiting bodies within a calcified shell structure|
|e.||produces no fruiting bodies and spores that move using pseudopodia|
ANS: C DIF: Easy REF: 20.4 TOP: IV.B.ii
- Which of the following is the signal for the slime mold Dictyostelium discoideum to switch from a unicellular existence to an aggregate slug formation?
|a.||bacterial cell wall components (e.g., LPS, teichoic acids)|
|c.||high glucose concentrations|
|d.||increased light and warmth|
ANS: B DIF: Medium REF: 20.4 TOP: IV.B.iii
- The slime mold Dictyostelium discoideum disperses spores from its fruiting body by which means?
|b.||interacting with passing animals||e.||ballistic force|
ANS: C DIF: Easy REF: 20.4 TOP: IV.B.iii.d
- Amebas that have filopodial movement structures can be found in which categories?
|a.||trypanosomes and metamonads||d.||cellular and plasmodial slime molds|
|b.||diatoms and algae||e.||radiolarians and foraminiferans|
|c.||dinoflagellates and apicomplexans|
ANS: E DIF: Difficult REF: 20.4 TOP: IV.C.i | IV.C.ii
- In 2007, the Centers for Disease Control received reports from public health authorities and ophthalmologists in 37 states and Puerto Rico identifying 221 people with Acanthamoeba keratinitis. Which other infectious agent is this disease most closely related to in terms of phylogeny?
|a.||Dictyostelium discoideum||d.||Legionella pneumophila|
|b.||Toxoplasma gondii||e.||Giardia lamblia|
ANS: D DIF: Medium REF: 20.4 TOP: IV.A
- The ciliate Paramecium multimicronucleatum beats its cilia in coordinated waves to:
|a.||distract and fool predators|
|b.||move forward and acquire food|
|c.||attach to immobile structures for filter feeding|
|d.||deliver toxins to enemies|
|e.||accumulate minerals for shell synthesis|
ANS: B DIF: Easy REF: 20.5 TOP: V.A.i
- A major group of marine phytoplankton that is essential to the marine food chain includes:
ANS: B DIF: Easy REF: 20.5 TOP: V.A.ii
- Alexandrium and Karenia are representative of flagellate, pigmented marine phototrophs responsible for the toxic red tide blooms. These organisms fit in the category of:
ANS: E DIF: Easy REF: 20.5 TOP: V.A.ii.a
- Plasmodium falciparum is a(n) __________ and is transmitted to the human host by the bite of a __________.
|a.||Apicomplexan; mosquito||d.||Trypanosome; mosquito|
|b.||Trypanosome; tsetse fly||e.||Apicomplexan; sand fly|
|c.||Apicomplexan; reduviid bug|
ANS: A DIF: Easy REF: 20.5 TOP: V.A.iii.b
- Schizogony refers to which life-cycle stage in Plasmodium falciparum?
|a.||mitotic reproduction and zygote development into spore-like form for transmission to next host; found in mosquito stage|
|b.||mitotic reproduction to achieve large population within host; found in liver stage and RBC stage|
|c.||meiotic reproduction to produce gametes; found in RBC stage|
|d.||mitotic reproduction and zygote development into spore-like form for transmission to next host; found in RBC stage|
|e.||meiotic reproduction to produce gametes; found in mosquito stage|
ANS: B DIF: Difficult REF: 20.5 TOP: V.A.iii.c
- New and emerging drug therapy targets for Plasmodium falciparum include:
|a.||spore-release inhibitors||d.||hemoglobin protease inhibitors|
|b.||chloroplast inhibitors||e.||cilia inhibitors|
ANS: D DIF: Medium REF: 20.5 TOP: V.A.iii.d
- The Apicomplexans have gone through extensive evolutionary reduction, losing their endosymbiotic chloroplast to:
|a.||allow only for phototrophic dark reactions|
|b.||produce low-levels of ATP as backup metabolism|
|c.||fix carbon dioxide only|
|d.||absorb blue/green light only|
|e.||complete one step of fatty acid metabolism|
ANS: E DIF: Medium REF: 20.5 TOP: V.A.iii
- Trypanosomes are a group of organisms that share which of the following characteristics?
|a.||free-living aquatic parasites, round cell, no flagellum|
|b.||intestinal parasites, shelled cell structure with pseudopodia|
|c.||opportunistic parasites, multicellular with paired flagellum and unique chloroplast|
|d.||obligate parasites, elongated cell with single flagellum and unique kinetoplast|
|e.||anaerobic parasites, single cellular structure with cilia|
ANS: D DIF: Easy REF: 20.6 TOP: VI.A
- A kinetoplast is a specialized bundle of DNA found where in the cell?
|a.||within a fruiting body||d.||in the pseudopodia cytoplasm|
|b.||attached to microtubules||e.||inside a large chloroplast|
|c.||inside large mitochondria|
ANS: C DIF: Easy REF: 20.6 TOP: VI.A
- Leishmania major causes a skin infection that can then disseminate to internal organs and is transmitted to a human host by:
|a.||drinking contaminated water|
|b.||eating undercooked meat harboring the organism|
|c.||skin contact with someone carrying the disease|
|d.||bite from infected sand flies|
ANS: D DIF: Easy REF: 20.6 TOP: VI.A.i
- Leishmania major infection is most problematic in:
|a.||Central and South America|
|b.||South America, Africa, the Middle East, and southern Europe|
|d.||Eastern Europe and Central Asia|
|e.||It is endemic throughout most of the world.|
ANS: B DIF: Medium REF: 20.6 TOP: VI.A.i
- Giardia lamblia is an infectious problem in day-care centers because it is a(n):
|a.||intestinal pathogen passed via fecal-oral transmission|
|b.||aerosol pathogen passed via sneezing and coughing|
|c.||mucosal pathogen passed via mucus contamination|
|d.||skin pathogen passed via direct skin contact|
|e.||blood-borne pathogen passed via contact with an open wound|
ANS: A DIF: Medium REF: 20.6 TOP: VI.B.i
- A patient who presents at the hospital office with symptoms of Chagas disease, i.e., an irregular heartbeat, inflamed/enlarged heart (cardiomyopathy), and congestive heart failure, could be infected with which of the following parasites?
|a.||Trypanosoma brucei||d.||Giardia lamblia|
|b.||Leishmania major||e.||Plasmodium falciparum|
ANS: C DIF: Medium REF: 20.6 TOP: VI.A.iii
- Why is the phylogeny data in the Eukarya domain overall harder to obtain than phylogeny evidence in either the Bacteria or Archaea domains? How has this affected Eukarya classification?
Eukaryotic genomes are several times larger than prokaryotes and 50%–90% of their DNA can consist of noncoding sequences. Therefore, the genomes take longer to sequence and are more challenging to annotate. Additionally, multiple events of endosymbiosis, reductive evolution, convergent evolution, and gene fusions/deletions have occurred in the Eukarya domain, adding to phylogeny challenge. Overall, these challenges have slowed the phylogenic classification of eukaryotes in comparison to the prokaryotes.
DIF: Medium REF: 20.1 TOP: I.D.i | I.D.ii MSC: Remembering
- Kelps (brown algae) are organisms that have four membranes surrounding their chloroplasts. What does this mean and where did each of the membranes come from?
Kelps are an example of a secondary endosymbiont (organisms that have evolved through engulfment of a primary endosymbiont). The chloroplast had two membranes in the primary endosymbiont—one from the chloroplast ancestor and one from the primary endosymbiont host cell endocytosis. The chloroplast gains two more membranes in the secondary endosymbiont—one from the primary symbiont cell membrane and one from the secondary endosymbiont host cell endocytosis.
DIF: Difficult REF: 20.1 TOP: I.E.ii.c MSC: Remembering
- What are nanoeukaryotes and how has their discovery changed our biological understanding?
Answers will vary. Nanoeukaryotes are newly discovered organisms that are much smaller than the eukaryotic cell formerly viewed (based on what a eukaryotic cell had to hold inside its cell membrane). The discovery of nanoeukaryotes in “extreme” environments (extreme temperatures, pH, pressure, etc.), thought to only harbor Archaea or Bacteria, will reshape our understanding of the Eukarya domain.
DIF: Medium REF: 20.1 TOP: I.E.iv.a | I.E.iv.b
- Pneumocystis jirovecii is an organism that causes pneumocystis pneumonia (PCP) in HIV/AIDS and other immunocompromised patient populations. Originally, Pneumocystis jirovecii was classified as protozoa but has been reclassified as fungi. Based on your knowledge of eukaryotic classification history, describe some characteristics that may have caused the initial classification and then the reclassification.
Historically, unicellular organisms that were motile and appeared (under microscope) animal-like were classified as protozoa. Pneumocystis jirovecii must have appeared animal-like and motile to the first classification observers. Reclassification based on molecular and morphological data has now placed this organism in the fungi (unicellular yeast) category.
DIF: Medium REF: 20.1 TOP: I.B.ii | I.B.iii MSC: Understanding
- Dinophysis acuminata, a dinoflagellate, is an example of a secondary endosymbiont. Explain what is meant by that designation.
Secondary endosymbionts are organisms that have evolved through engulfment of a primary endosymbiont (primary endosymbionts can trace their ancestral lineage back to a single endosymbiotic event). Many protists, including dinoflagellates, incorporated an alga and are examples of secondary endosymbionts. (The primary endosymbiotic event was when an alga incorporated the ancestral cyanobacteria; the secondary endosymbiontic event was when the alga was incorporated by the protist.)
DIF: Medium REF: 20.1 TOP: I.D.ii MSC: Understanding
- Why would an organism benefit from undergoing more than one endosymbiontic event throughout evolution? Give an example to support your argument.
Answers will vary. Primary endosymbiosis include the ancestral endosymbiosis event forming either a mitochondrion (from an ancestral proteobacterium) or a chloroplast (from an ancestral cyanobacteria). If an organism with a mitochondrion were then to undergo a secondary endosymbiosis event with an organism containing a chloroplast, this new organism would gain two different mechanism of energy production (phototrophy and heterotrophy). While primarily phototrophic, some diatoms can live in the absence of light using heterotrophy (assuming an appropriate carbon source is available). This benefits organisms by giving them the advantage to easily negotiate multiple environments.
DIF: Difficult REF: 20.1 TOP: I.D.ii.a | I.D.ii.b
- Why are sponges, jellyfish, worms, reptiles, humans, yeast, and microsporidia placed in the same eukaryotic clade?
Despite inherent structural and functional differences, all organisms mentioned above share two qualities that link them together evolutionarily. First, they share a unique short-sequence insertion in the translation elongation factor 1α gene based on genomic analysis. Second, they share the structural feature of being uniflagellar or having a uniflagellar stage at some point in their life cycle.
DIF: Medium REF: 20.1 TOP: I.E.i MSC: Understanding
- The Zygomycota taxon includes a group of insect parasites with nonmotile gametes that must grow toward each other to fuse and form a zygote. Explain why this taxon is classified under the larger taxa of Opisthokonta (whose classification involves flagellated reproductive cells).
The Zygomycota taxon falls under the True Fungi (Eumycota) classification (heterotrophs with either single-celled or nonmotile filaments of cells called “hyphae” formed from chitin cell walls). All true fungi are categorized under the larger taxa of Opisthokonta. Zygomycota are one example of a taxa of fungi that have undergone reductive evolution to produce nonmotile reproductive cells. Evolutionarily, they must have had flagella at some previous point.
DIF: Difficult REF: 20.1 TOP: I.E.i.c MSC: Understanding
- The Amanita, or “destroying angel” mushroom, produces one of the world’s most deadly poisons called alpha-amanitin. What is the target of this poison and how does it affect a eukaryotic host cell and a prokaryotic host cell?
Alpha-amanitin poison is an inhibitor of eukaryotic RNA polymerase II. RNA polymerase II transcribes messenger RNAs and several small nuclear RNAs, so essentially the process of cellular transcription would be shut down in eukaryotic cells. Not surprisingly Amanita’s own RNA polymerase II is not effected by its toxin. Since bacterial cells lack RNA polymerase II, this fungal toxin would not affect bacteria.
DIF: Medium REF: 20.2 TOP: II.C.iv.a MSC: Remembering
- You have isolated a novel fungus and are now growing it under laboratory conditions on two separate media. The first medium is a simple glucose broth and the second is a more complex broth containing pectin and starch. Will this fungus differ in its ability to acquire nutrients from the two media? Why or why not? Will this media difference affect the fungal growth?
The fungus in the first medium (glucose broth) will be able to directly transport the glucose into the absorption zone of its hyphae tip (for further metabolism) because the glucose molecule is small and soluble. The fungus in the second medium (pectin and starch broth) will have to secrete enzymes into the medium (e.g., pectinases and amylases) to break down the pectin and starch into smaller molecules, such as glucose, that can be transported across the cell wall into the absorption zone. There may be a longer lag time in growth in the second medium due to the need to produce and export the digestive enzymes.
DIF: Difficult REF: 20.2 TOP: II.A.i MSC: Understanding
- Compare and contrast the life cycles of the Chytridiomycota clade of fungi with the Zygomycota clade.
Compare: Both have alternative options of haploid and diploid forms.
Contrast: The life cycle mechanisms of each differ based on motility, cyst structures, and haploid/diploid mycelium structures (details below).
Life cycle of Chytridiomycota: Chytrids contain both haploid (gametophyte) and diploid (sporophyte) mycelia. Haploid gametophytes produce motile gametes that are attached to each other by sex-specific attractants to fuse and generate a motile zygote. The zygote forms a cyst surviving long periods until the correct environmental conditions are met to allow germination to form a diploid mycelium generating a zoosporangia full of zoospores. Two forms of zoosporangia exist: one produce more diploid mycelium; the other produce haploid mycelium (which produce more gametes).
Life cycle of Zygomycota: Zygomycota produce nonmotile haploid spore (sporangiospore) carried by air or water, forming haploid mycelium. A special hyphae with gamete cells forms at the tip, and the tips from two gamete hyphae must grow together to form zygospore. The zygospore then undergoes meiosis, forming the sporangium that releases the haploid sporangiospores.
DIF: Difficult REF: 20.2 TOP: II.C.i.b | II.C.ii.b
- The Rhodophyta algae Porphyra is known to colonize aquatic environments deeper than the Chlorophyta algae Spirogyra; why is this and what advantages does Porphyra have by claiming this niche?
Rhodophyta algae, or red algae, appear red due to the photosynthetic pigment phycoerythrin. This photosynthetic pigment absorbs in the light blue/green range. Blue and green wavelengths of light are able to penetrate deeper waters, so Rhodophyta algae colonize there. Green algae use chlorophyll pigments that absorb wavelengths of light that penetrate less deeply, so they colonize closer to the surface. This depth provides an environmental niche perfect for red algae phototrophy without competition from other phototrophic algae.
DIF: Medium REF: 20.3 TOP: III.C.i MSC: Remembering
- Amoeba proteus is a large freshwater ameba that moves using lobed-shaped pseudopodia. Describe the mechanism by which this pseudopodia structure moves this ameba using cytoplasmic streaming.
Movement by cytoplasmic streaming consists of a sol-gel transition between cortical cytoplasm and deeper interior cytoplasm. Actin dimers stream forward in liquid cytoplasm (sol) from the cortical cytoplasm on microtubules using ATP. Once in the cortical cytoplasm of the pseudopodia, dendritic polymerization of actin occurs, creating the gel state. The actin polymer then rolls forward (like a tank tread) extending the pseudopodia. As actin polymer is pushed back deeper into the cytoplasm, it resolubilizes and the cycle repeats.
DIF: Medium REF: 20.4 TOP: IV.A.i.a MSC: Understanding
- Describe how the life cycle of the slime mold Dictyostelium discoideum is dependent upon prokaryotic organisms.
Haploid ameba consume bacteria and divide asexually until food runs out. Cyclic AMP (cAMP) is then emitted by the starving ameba as an aggregation signal. Other amebas (same species) are attracted and aggregate. The amebas that join the aggregate start releasing their own cAMP, attracting even more amebas. The amebas pile on top of one another to form <1mm slug (consisting of thousands of amebas) that can migrate as a collective toward light and warmth (and presumably food). The aggregate slug forms fruiting body (a spherical sporangium), releasing spores (or cysts) on air currents. These spores can remain dormant for years, germinating when they detect a bacterial chemical signal (i.e., food source). All of the life-cycle steps above are asexual; reproductive alternatives exist in the ameba stage.
DIF: Medium REF: 20.4
TOP: IV.B.iii.a | IV.B.iii.b | IV.B.iii.c | IV.B.iii.d | IV.B.iii.e | IV.B.iii.f
- Compare and contrast the two major groups of shelled amebas: radiolarians and foraminiferans.
Compare: Both grow in marine and aquatic environments and their shells make up reef formations, sedimentary rock, and beach sand.
Contrast: Radiolarians have shells made of silica and their pseudopodia radiate out of holes in all different directions; foraminiferans have shells made of calcium carbonate as chambers laid down in helical succession, and their pseudopodia come out of the opening of the most recent chamber.
DIF: Medium REF: 20.4 TOP: IV.C | IV.C.i | IV.C.ii
- Give one reason why geologists could look at foraminiferans to determine potential sites for oil drilling.
Answers will vary. Foraminiferans are a highly diverse and highly abundant group of shelled amebas found in marine and aquatic environs. The production of a shell allows their remains to be left behind after the death of the organism. Analysis of the fossil remains of foraminiferans is one way geologists can determine the relative ages of rock strata and extrapolate the relative environmental conditions under which the rock was made. The oil industry can use this information to correlate rock type and age as indicators of petroleum deposits for possible drilling sites.
DIF: Difficult REF: 20.4 TOP: IV.C.ii MSC: Applying
- What is the life cycle of Plasmodium and how does that life cycle cause Plasmodium to be categorized as an apicomplexan?
Plasmodium falciparum parasites are injected into the bloodstream via an infected mosquito. The sporozoites travel to the liver and develop into merozoites that infect the red blood cells. Penetration of red blood cells occurs via the apical complex interacting with host cell surface receptors. Merozoite enzyme release, coupled with lipid and protein secretion, enable total RBC penetration. The presence of an apical complex structure that facilitates entry into the host cell is the characteristic that places Plasmodium in this apicomplexan grouping. Merozoites feed on hemoglobin, can undergo schizogony (mitotic reproduction of diploid form to achieve large population within host) or gamegony (meiotic reproduction to produce gametes to enter the mosquito) in the RBC stage. Sporogony (the mitotic reproduction and development of diploid zygote into spore-like form for transmission to the next human host) occurs within the mosquito.
DIF: Medium REF: 20.5 TOP: V.A.iii.c MSC: Remembering
- Why is the coral endosymbioant, zooxanthellae, an important indicator of ocean health during global warming?
The name zooxanthellae refers to the relationship between a member of the dinoflagellates and coral. The dinoflagellates live as endosymbionts, providing sugars from photosynthesis to the coral in exchange for a protected habitat within the coral. Zooxanthellae are vital to coral reef growth and are very temperature sensitive. Slight increases in ocean temperature due to global warming can cause coral bleaching (the expulsion of the zooxanthellae). This bleaching leads to the death of the coral and surrounding environmental niches.
DIF: Medium REF: 20.5 TOP: V.A.ii.b MSC: Understanding
- Organisms categorized as Metamonad are considered to be highly degenerative parasites. What does this mean and why did they evolve this way?
Answers will vary. Metamonads are considered a highly degenerate group of parasites because they have lost both mitochondria and Golgi apparatus through degenerative evolution. This evolutionary step has left them with anaerobic metabolism and an absolute dependence on the anaerobic intestinal environment of their host. It is unknown why they evolved this way, but it can be assumed that the loss of these structures and functions was somehow of benefit energetically or environmentally at some point during their evolution.
DIF: Medium REF: 20.6 TOP: VI.B MSC: Understanding
- The microarray analysis of Trypanosoma brucei gene expression demonstrated that two genes encoding cell-surface proteins PAD1 and PAD2 are expressed only in the differentiation-positive strain. Explain why PAD1 and PAD2 are important in Trypanosoma brucei infection.
Trypanosoma brucei differentiates among different forms. The “slender” form proliferates in the bloodstream of the human host, while the “stumpy” form is needed within the host to infect the tsetse fly. Within the fly the “stumpy” form differentiates into procyclic form (in the fly midgut), epimastigote form (in the fly salivary gland), and metacyclic form (in the fly salivary gland)—each form has biochemical properties enabling survival and growth in its environment. Control of differentiation is regulated by cell surface proteins PAD1 and PAD2 (proteins associated with differentiation). PAD1 protein is only expressed in stumpy form (therefore it may be required for differentiation into the stumpy form); PAD2 protein is expressed in procyclic form. The differential expression of PAD1 and PAD2 correlate to the change in temperature from humans (37°C) to flies (20°C), so temperature shift may be a signal for gene regulation.
DIF: Difficult REF: 20.6 TOP: VI.A.ii.a | VI.A.ii.b
CHAPTER 28: Clinical Microbiology and Epidemiology
- The cause of the infection that a patient is suffering from is best known as the:
|a.||clinical agent||d.||epidemiological agent|
|b.||etiological agent||e.||antibiotic menace|
ANS: B DIF: Easy REF: 28.1 TOP: II.A
- A person exhibiting glomerular nephritis and rheumatic fever must have had what type of previous infection?
|a.||Shigella sonnei||d.||Treponema pallidum|
|b.||Escherichia coli||e.||Mycobacterium tuberculosis|
ANS: C DIF: Medium REF: 28.1 TOP: II.B
- A person who begins to exhibit symptoms of bronchitis and bacterial pneumonia after recovering from the flu would be exhibiting:
|a.||drug resistance||d.||biological meltdown|
|b.||antiviral susceptibility||e.||disease sequelae|
ANS: E DIF: Easy REF: 28.1 TOP: II.B
- Shigella sonnei is the cause of an infection commonly seen in infant day-care facilities due to:
|a.||a high prevalence of the microbe in young children|
|b.||its ability to contaminate and grow on plastic toys|
|c.||its inherent antimicrobial nature (it is bleach and cleaning-product resistant)|
|d.||diaper changes and hands not being properly washed|
|e.||Gram-positive cell wall structure making the bacteria environmentally resistant|
ANS: D DIF: Medium REF: 28.1 TOP: II.B
- Penicillinase-producing Neisseria gonorrhoeae (PPNG) experienced a significant increase in the United States during the 1970s because of the:
|a.||start of the HIV epidemic and IV drug use|
|b.||feminist movement and increased sexual activity|
|c.||overuse of antibiotics during and after the Vietnam War|
|d.||doctors overprescribing quinolones for malaria|
|e.||concurrent swine flu epidemic in the mid-1970s|
ANS: C DIF: Easy REF: 28.1 TOP: II.B
- Of the following body sites, which could contain microbes in a healthy individual?
ANS: D DIF: Easy REF: 28.2 TOP: III.B
- Collection of specimen samples from places on or in the body that contain normal flora usually requires:
|a.||initial plating on nonselective media||d.||direct PCR identification|
|b.||direct Gram staining||e.||immunofluorescent staining|
|c.||initial plating on selective media|
ANS: C DIF: Easy REF: 28.2 TOP: III.B
- An abscess caused by the anaerobic species Clostridium perfringens would best be collected and transported by:
|a.||aspiration into an oxygen-filled tube||d.||draining with a catheter|
|b.||swabbed onto blood agar media||e.||draining into a syringe|
|c.||aspiration into a nitrogen-filled tube|
ANS: C DIF: Easy REF: 28.2 TOP: III.A
- Catheterization would be an appropriate method of microbial collection for:
|a.||Streptococcus pneumoniae—lung infection|
|b.||Escherichia coli—UTI infection|
|c.||Shigella dysenteriae—GI infection|
|d.||Propionibacterium acnes—skin infection|
|e.||Helicobacter pylori—stomach infection|
ANS: B DIF: Easy REF: 28.2 TOP: III.B
- Laboratory analysis of a patient suspected to have bacterial meningitis would be performed with which type of sample?
|a.||urine sample||d.||sputum sampling|
|b.||blood draw||e.||lumbar puncture|
ANS: E DIF: Easy REF: 28.2 TOP: III.B
- Collection of a sample from a patient suspected to have malaria would be performed by:
|a.||blood draw||d.||lumbar puncture|
|b.||sputum expectorate||e.||clean-catch sampling|
|c.||tissue needle aspirate|
ANS: A DIF: Medium REF: 28.2 TOP: III.B
- Hektoen agar will grow all of the following microorganisms EXCEPT:
|a.||Klebsiella pneumoniae||d.||Shigella sonnei|
|b.||Escherichia coli||e.||Streptococcus pneumoniae|
ANS: E DIF: Easy REF: 28.3 TOP: IV.C
- Chocolate agar is considered __________ in comparison to blood agar.
|a.||more nutrient-rich (capable of growing a greater variety of bacteria)|
|b.||less nutrient-rich (capable of growing fewer variety of bacteria)|
|c.||equally nutrient-rich (since they both contain red blood cells)|
|d.||a chemically defined media (whereas blood agar is not)|
|e.||better for determining hemolytic bacterial reactions|
ANS: A DIF: Medium REF: 28.3 TOP: IV.C
- Haemophilius influenzae requires __________ and __________ for confirmatory growth of an organism in vitro.
|a.||catalase; oxidase (N disk)|
|b.||Eosin-Methylene Blue (EMB) agar; MacConkey agar|
|c.||citrate; bile salts|
|d.||hemin (X factor); NAD (V factor)|
|e.||phenylalanine; ortho-Nitrophenyl-b-galactoside (ONPG)|
ANS: D DIF: Medium REF: 28.3 TOP: IV.D.ii
- The Lancefield groupings are for __________ bacteria and based on __________.
|a.||Streptococcus; carbohydrate peptidoglycans|
|c.||Staphylococcus; peptidoglycan amino acids|
|e.||Pseudomonas; optochin sensitivity|
ANS: A DIF: Easy REF: 28.3 TOP: IV.D.iii
- Catalase-positive organisms such as __________ produce __________ as end products when exposed to hydrogen peroxide.
|a.||Streptomyces; KCl and CO2||d.||Stachybotrys; HOCl and H2O|
|b.||Streptococcus; H2O2 and H2O||e.||Staphylococcus; H2O and O2|
|c.||Lactobacillales; CO2 and O2|
ANS: E DIF: Medium REF: 28.3 TOP: IV.D.iii
- A bacterial diagnosis of Mycobacterium avium in a sputum sample from an AIDS patient would be best determined by which method?
|a.||blood agar culture||d.||Gram staining|
|b.||acid-fast (Ziehl-Neelsen) staining||e.||bacitracin susceptibility|
|c.||MacConkey agar culture|
ANS: B DIF: Easy REF: 28.3 TOP: IV.A
- Low glucose and high protein levels found in the cerebrospinal fluid of a hospitalized patient most likely point to what kind of infection?
|a.||viral infection||d.||prion infection|
|b.||bacterial infection||e.||any of the above|
ANS: B DIF: Medium REF: 28.3 TOP: IV.D
- A Gram-negative diplococcus result on laboratory testing would rule out all of the following bacteria EXCEPT:
|a.||Neisseria gonorrhoeae||d.||Haemophilus influenzae|
|b.||Listeria monocytogenes||e.||Escherichia coli|
ANS: A DIF: Medium REF: 28.3 TOP: IV.D.ii
- Which of the following genera of bacteria would NOT change the color of N,N,N’,N’-tetramethyl-p-phenylenediamine to purple/black upon exposure?
ANS: A DIF: Difficult REF: 28.3 TOP: IV.D.ii
- A hemolysis reaction (on a blood agar plate) with an unidentified colony that results in a green zone due to oxidized iron in nonlysed red blood cells would be called:
ANS: D DIF: Easy REF: 28.3 TOP: IV.D.iii
- Lowenstein-Jensen agar is used to culture which of the following clinical specimens?
|a.||all Gram-negative bacteria||d.||all Gram-positive bacteria|
|b.||beta lactase–producing bacteria||e.||toxin-producing enteric bacteria|
|c.||mycolic acid–containing bacteria|
ANS: C DIF: Medium REF: 28.3 TOP: IV.A
- Symptoms of headache, fever, stiff neck, and confusion in combination with mosquito bites can indicate which viral infection?
ANS: B DIF: Easy REF: 28.4 TOP: V.B.i
- Humans are an incidental (dead-end) host in which clinical disease?
|a.||West Nile virus||d.||Staphylococcus aureas|
ANS: A DIF: Medium REF: 28.4 TOP: V.B.i
- A disease geographically spread by the movement of migrating birds is:
ANS: D DIF: Medium REF: 28.4 TOP: V.B.i
- PCR-based detection of a person thought to have HIV would require all of the following EXCEPT:
|a.||primers specific for HIV genes||d.||gel electrophoresis|
|b.||thermocycler||e.||HIV genome isolation/extraction|
|c.||HIV agar media|
ANS: C DIF: Easy REF: 28.4 TOP: V.B.i
- A proper course of action for a patient presenting with Ebola would be:
|a.||quarantine and laboratory determination of Ebola IgG antibodies|
|b.||vaccination and laboratory growth of Ebola in culture|
|c.||course of antibiotics and laboratory determination of Ebola antigen|
|d.||antiviral therapy followed by a course of antibiotics|
|e.||in-home supportive therapy and laboratory PCR detection|
ANS: A DIF: Medium REF: 28.4 TOP: V.C.i
- Direct fluorescent staining of infected tissue samples taken from a patient allows for rapid identification under what circumstances?
|a.||hard-to-culture organisms||d.||intracellular infections|
|b.||protist infections||e.||metabolically active organisms|
ANS: A DIF: Easy REF: 28.4 TOP: V.D
- Fluorescent antibody staining works well for clinical diagnosis on what types of antigens?
ANS: D DIF: Medium REF: 28.4 TOP: V.D
- Eukaryotic parasites, such as Giardia lamblia, are usually diagnosed by:
ANS: E DIF: Easy REF: 28.4 TOP: V.E
- Real-time PCR on a clinical sample from a suspected hantavirus patient can determine both:
|a.||antibodies to hantavirus and viral resistance|
|b.||viral resistance and viral load|
|c.||presence of hantavirus and viral load|
|d.||antibodies to hantavirus and strength of immune response|
|e.||hanta serotype and strength of immune response|
ANS: C DIF: Medium REF: 28.4 TOP: V.B.i
- In real-time quantitative PCR, light emission will result or will become visible in a person who has a hepatitis C infection when:
|a.||primers bind to viral hepatitis RNA|
|b.||probe oligonucleotide is degraded during PCR|
|c.||cDNA copies are formed|
|d.||fragments are run on a gel and radiolabeled|
|e.||Taq polymerase bind to the cDNA|
ANS: B DIF: Medium REF: 28.4 TOP: V.B.i
- A serum antibody ELISA would be looking for which of the following in a person who is presumed to have a necrotizing fasciitis infection?
|a.||anti–group A streptococcal antibodies||d.||enzyme-linked antibody|
|b.||antimeningococcal antibodies||e.||anti–group B streptococcal antibodies|
|c.||group A streptococcal-specific antigens|
ANS: A DIF: Medium REF: 28.4 TOP: V.C.i
- An antigen-capture ELISA could be looking for all of the following in a person who is presumed to have a severe acute lung infection EXCEPT:
|a.||mycobacterium antigens||d.||pneumococcal antigens|
|b.||SARS antigens||e.||anti-influenza antibodies|
|c.||Haemophilus sp. antigens|
ANS: E DIF: Medium REF: 28.4 TOP: V.C.i
- Point-of-care diagnostics are advantageous for all of the following reasons EXCEPT:
|a.||culturing is not required|
|b.||therapy can begin sooner|
|c.||patient compliance is improved|
|d.||antibiotics can be prescribed more effectively|
|e.||lower risk of spreading infection|
ANS: E DIF: Easy REF: 28.5 TOP: VI.A
- Most point-of-care diagnostics rely on:
|b.||Qt-polymerase chain reaction||e.||API platforms|
ANS: A DIF: Easy REF: 28.5 TOP: VI.A
- Which of the following organisms falls within risk group category II?
|a.||Ebola virus||d.||Escherichia coli strain K-12|
ANS: E DIF: Medium REF: 28.6 TOP: VII.A
- Bordetella pertussis is classified as a biosafety risk group II organism. What does this mean?
|a.||little to no pathogenic potential and standard sterile techniques are sufficient|
|b.||pathogenic potential but generally not via respiratory transmission|
|c.||pathogenic potential and respiratory transmission possible|
|d.||pathogenic potential and containment in pressurized labs|
|e.||extremely dangerous pathogens and rigorous containment in pressurized suits and labs|
ANS: B DIF: Easy REF: 28.6 TOP: VII.A
- A defining commonality between biosafety risk group II handling techniques and a biosafety IV facility would be the need for:
|a.||a personal positive-pressure suit|
|b.||an isolation facility|
|c.||a negative-pressurized lab space|
|d.||directional air flow in room|
|e.||the use of aseptic techniques|
ANS: E DIF: Easy REF: 28.6 TOP: VII.A
- The use of a laminar flow hood when studying the organism Francisella tularensis is an example of:
|a.||level 0 biological safety||d.||level III biological safety|
|b.||level I biological safety||e.||level IV biological safety|
|c.||level II biological safety|
ANS: D DIF: Difficult REF: 28.6 TOP: VII.A
- John Snow, the “father of epidemiology,” is known for deducing the source of which etiologic agent in the 1854 London outbreak?
|a.||Shigella dysenterie||d.||Giardia lamblia|
|b.||Vibrio cholerae||e.||Yersinia pestis|
ANS: B DIF: Easy REF: 28.7 TOP: VIII.C
- Which of the following is a reportable, biosafety level III organism that causes low-grade fever, chills, headache, and malaise followed by a dry cough, shortness of breath, and chest heaviness?
|a.||methicillin-resistant Staphylococcus aureas|
|c.||West Nile virus|
ANS: D DIF: Medium REF: 28.7 TOP: VIII.F
- Which infectious disease is on the CDC bioterrorism list due to the general population NOT being vaccinated for this disease since the 1970s?
ANS: C DIF: Medium REF: 28.7 TOP: VIII.F
- Which of the following is the correct definition of bioweapon?
|a.||any agent categorized as a biosafety level IV organism|
|b.||any infectious disease spread person to person with a low mortality rate|
|c.||any agent causing widespread psychological trauma, with few casualties|
|d.||any microbial agent or pieces of microbes that can be loaded into bombs|
|e.||any infectious agent or toxin that has a high virulence and/or mortality rate|
ANS: E DIF: Easy REF: 28.7 TOP: VIII.F.i
- Bubonic plague is endemic in many countries in Africa and in the former Soviet Union. This means that:
|a.||plague was once a huge problem and has now evolved to not cause infection|
|b.||plague has been eradicated from these areas|
|c.||large numbers of people are infected in these areas and it is spreading|
|d.||the disease is always present in the population at a low frequency|
|e.||plague is transferred by human-to-human contact|
ANS: D DIF: Easy REF: 28.7 TOP: VIII.B
- Gaëtan Dugas is frequently referred to as the individual who began the spread of HIV within the gay male community in North America in the 1970s. In epidemiology he is known as the:
|a.||first patient||d.||index case|
|b.||superspreader||e.||first data point|
ANS: D DIF: Easy REF: 28.7 TOP: VIII.B
- In using the PANTHER sensor for quick detection of pathogens, how does the PANTHER detect Francisella tularensis, the causative agent of tularemia?
|a.||It detects Francisella tularensis antibodies within a given sample.|
|b.||It detects Francisella tularensis DNA using RFLP technology.|
|c.||It detects Francisella tularensis antigen using antigen-specific B cells.|
|d.||It detects Francisella tularensis endotoxins using enzyme-conjugated antibodies.|
|e.||It detects Francisella tularensis antigen using antigen-specific T cells.|
ANS: C DIF: Difficult REF: 28.7 TOP: VIII.F.ii
- Tuberculosis has become a reemerging pathogen today in large part due to:
|a.||increased incidence of HIV infection|
|b.||airline travel reducing geographical distances|
|c.||surge in reservoir animal populations carrying the disease to humans|
|d.||strict compliance with drug regimens|
|e.||concurrent increases in hepatitis C infection|
ANS: A DIF: Medium REF: 28.8 TOP: IX.C.i
- Collaborations necessary to solve and control a human epidemic include:
|b.||ecologists||e.||all of the above|
ANS: E DIF: Easy REF: 28.8 TOP: IX.C.ii | IX.D
- Molecular screening is now being used to detect individuals with early stages of TB because:
|a.||skin testing is unreliable and needs refrigeration that many countries may not have|
|b.||vaccination with Bacille Calmette-Guérin does not protect against multidrug resistance|
|c.||many countries do not have disposable needles for skin testing|
|d.||it can detect infection earlier than chest X-ray and sputum sampling|
|e.||it can be used more safely on children with the disease|
ANS: D DIF: Easy REF: 28.8 TOP: IX.C.i
- Describe the general protocol you would use to diagnose a patient with a Salmonella Typhimurium infection (from collection to testing).
Answers will vary. A patient with a Salmonella Typhimurium infection would most likely be experiencing gastrointestinal symptoms. Sample collection would include a stool sample (either collected in a cup or by swab). Since normal flora would be present, plating onto selective media would be advised. Examples include Gram-negative, enteric media such as EMB, MacConkey, or xylose-lysine-desoxycholate (XLD) agar; bismuth sulphite agar; brilliant green agar; Hektoen enteric agar; or any selective chromogenic agar media specifically designed for the differentiation of Salmonella colonies. Confirmatory testing including various biochemical assays, staining, enzyme testing, and molecular methods could all be used.
DIF: Medium REF: 28.2 | 28.3 | 28.4 TOP: III.B | IV.C | V.B
- In determining the bacterial etiology of a patient with an infection, the first laboratory test should be what, and why? What type of testing would then follow?
After isolation of a pure bacterial colony from the patient, the first test is usually a Gram stain to determine whether the bacteria is Gram-negative or Gram-positive. (Sometimes the first test can be to plate on differential agar, based on ability to grow Gram-positive or Gram-negative organisms, then to confirm with a Gram stain.) This Gram-stain result will begin the process of bacterial determination through specific biochemical reactions using dichotomous keys, API strips, or computer analysis. Subsequent analysis will be dependent on the Gram-stain result and laboratory choice.
DIF: Medium REF: 28.3 TOP: IV.A MSC: Analyzing
- Compare and contrast the dichotomous key approach to bacterial determination with the seven-digit-number approach when using the API strip system.
The API strip system results from this series of biochemical tests can be interpreted two different ways:
(1) Using a dichotomous key: this is a step-wise interpretation beginning with a key reaction (e.g., lactose fermentation). Results are read as either a positive reaction or negative reaction, and the bacteria are identified based on the reactions for each of the various biochemical tests included on the strip. The inherent danger in using the dichotomous key method is that one anomalous result can lead to an incorrect identification.
(2) Generating a seven-digit number: the seven-digit number is obtained from given numerical values equated with each individual biochemical test within the strip and is based on whether the biochemical reaction is positive or negative. Once a number is generated, it can be compared to a computer database generating a probability score. The organism receiving the highest probability score is considered to be the determined species. This test is considered more accurate than the dichotomous key, in part because of the extensive and updated computer database and also because of the probability data generated (in comparison with the dichotomous key method, which simply generates data based on a yes/no format).
DIF: Difficult REF: 28.3 TOP: IV.D.i MSC: Analyzing
- Looking at the laboratory results presented here, what three interpretations can you make?
|Colony 1||Colony 2||Colony 3|
|Growth on blood agar||No growth||Growth||Growth|
|Growth on Hektoen agar||No growth||No growth||Growth|
|Growth of chocolate agar||Growth||Growth||Growth|
Many different interpretations are possible; some answers may include:
- Colony 1 is nutritionally fastidious and requires the most nutrient-rich agar (chocolate agar) to grow; may be a Neisseria or Haemophilus.
- Colony 2 is less nutritionally fastidious than colony 1, since it can grow on both chocolate and blood agar; colony is likely Gram-positive since its growth was inhibited by Hektoen media.
- Colony 3 is less nutritionally fastidious than colony 1, since it can grow on both chocolate and blood agar; colony is likely a Gram-negative enteric rod since it grew on Hektoen media.
DIF: Medium REF: 28.3 TOP: IV.D.ii MSC: Analyzing
- Looking at the laboratory results presented here, what three interpretations can you make?
|Colony 1||Colony 2||Colony 3|
|Growth on blood agar||Growth||Growth||Growth|
|Growth on CNA agar||Growth||No growth||Growth|
|Growth on MacConkey agar||No growth||Growth||Slight growth|
Many different interpretations are possible; some answers may include:
- Colony 1 is most likely a Gram-positive bacteria; blood agar will grow both Gram-negatives and Gram-positives. CNA agar is a differential medium selecting for the growth of Gram-positive organisms (due to the inhibitors colistin and naladixic acid).
- Colony 2 is most likely a Gram-negative bacteria; blood agar will grow both Gram-negatives and Gram-positives. MacConkey agar is a differential medium selecting for the growth of Gram-negative organisms (due to the inhibitors bile salts and crystal violet).
- Colony 3 is most likely a Gram-positive bacteria of either Staphylococcus aureas or Enterococcus sp. Blood agar will grow both Gram-negatives and Gram-positives. CNA agar is a differential medium selecting for the growth of Gram-positive organisms (due to the inhibitors colistin and naladixic acid), MacConkey agar is a differential medium selecting for the growth of Gram-negative organisms (due to inhibitors bile salts and crystal violet), BUT does allow for Staphylococcus aureas or Enterococcus sp. growth (a limitation of this differential medium).
DIF: Difficult REF: 28.3 TOP: IV.D.iii MSC: Analyzing
- The clinical microbiology laboratory just received a sputum (deep-lung) sample that so far has been very slow-growing in nonselective agar, as well as not Gram-stainable. What tests would you suggest next, and why?
Answers will vary. Based on the case history given (deep-lung sample, slow-growing, and non-Gram-stainable), a possible direction to pursue would be that this is a mycobacterium infection (e.g., Mycobacterium tuberculosis, Mycobacterium avium). Tests that may more closely define the etiologic agent as a member of the mycobacterium family could include:
- Growth on Löwenstein-Jensen agar—specific agar for mycobacterium growth.
- Ziehl-Neelsen (acid-fast) staining—mycobacterium stain pink in this procedure, all other bacteria stain blue, presumptive for diagnosis of mycobacterium.
- PCR or RFLP testing using mycobacterium-specific primers.
DIF: Medium REF: 28.3 | 28.4 TOP: IV.A | V.B.i MSC: Evaluating
- Haemophilus parasuis is a nonpathogenic bacteria frequently found in the upper respiratory tracts of pigs. Pathogenic strains of this bacteria form the etiological agent of Glässer’s disease in pigs. Describe how genetic screening could be used to determine if a pig had a pathogenic strain or a commensal strain.
One could take a sample from the upper respiratory tract of the pig and screen for the presence of pathogenic genes found in the pathogenic strain of Haemophilus parasuis and absent from the nonpathogenic strain. Using primers directed at specific pathogenic genes and the polymerase chain reaction, the pathogenic genes could be analyzed and their presence or absence determined. Additional note: the pathogenic genes present in pathogenic Haemophilus parasuis are 13 genes coding for trimeric autotransporters (AT-2) associated with virulence.
DIF: Medium REF: 28.4 TOP: V.B MSC: Analyzing
- How can reverse transcriptase PCR (RT-PCR) be used to determine the presence and viral load of HIV infection?
A blood sample would have to be taken from the HIV-infected patient and RNA extracted from the sample. The viral RNA would be converted to cDNA using a single primer and reverse transcriptase—the more viral RNA within the clinical sample, the more cDNA that will form in this first step. cDNA is amplified by PCR using two specific primers and Taq polymerase. Quantification occurs by addition of a fluorescent oligonucleotide probe to the PCR reaction (fluorescent dye at 3′ end, chemical dye at 5′ end). This probe binds to an area between primers. If probe binding is kept intact, no light is emitted from the sample. However, if probe binding is degraded by the action of successful PCR amplification, then two ends of the probe are separated and light is emitted. The more copies of cDNA to begin with, the fewer cycles of PCR needed to register a fluorescence increase over background levels.
DIF: Medium REF: 28.4 TOP: V.B.i MSC: Analyzing
- How does a multiplex PCR reaction differ from a traditional PCR detection method? In addition, describe an example of how multiplex can be used in diagnostics.
Traditional PCR uses primers that bind to unique genes in a pathogen’s genome to amplify either DNA or RNA present in a sample and then visualize the amplicon on an agarose gel. Multiplex PCR uses multiple primers to simultaneously search for strain variations in pathogenic organisms. The example from the text shows how multiplex PCR is used to determine different Clostridium botulinum toxins by amplifying four different toxin genes in one PCR reaction.
DIF: Medium REF: 28.4 TOP: V.B.i MSC: Analyzing
- How would a serum antibody ELISA work in the detection of Lyme disease?
Serum antibody ELISA first starts with antigen being adsorbed to the plastic wells of a 96-well plate. In this case the antigens adsorbed to the well must be from the causative agent Borrelia burgdorferi. Albumin or powered milk is first added to block remaining sites on plastic that could result in false positive reactions. Next patient serum (in this case, taken from the Lyme disease patient) is added to the wells and allowed to incubate. If anti-Lyme antibodies are present, binding reaction to the Borrelia burgdorferi antigens will occur. After a wash step, goat-antihuman IgG conjugated to an enzyme is then added to sandwich the patient serum antibody in the well. After an additional wash step, chromogenic substrate for the enzyme is added to induce a color change. The amount of color change is measured by an ELISA plate reader and is quantifiable to the amount of antibody present in the patient’s serum.
DIF: Medium REF: 28.4 TOP: V.C.i MSC: Analyzing
- Compare and contrast the use of the serum antibody ELISA with the antigen-capture ELISA in the detection of Ebola.
Serum antibody ELISA uses antigen adsorbed to the plastic wells followed by incubation with the patient serum. The addition of a secondary enzyme-linked antibody will allow for the detection and quantification of anti-Ebola antibodies present in the patient serum. This is an indirect test, measuring the body’s immune response to the infectious antigen via antibody detection.
Antigen-capture ELISA uses antibody adsorbed to the plastic wells followed by incubation with patient serum. If the serum contains antigen, it will bind to the adsorbed antibody and detection/quantification can be made using the addition of a secondary enzyme-linked antibody, followed by a chromogenic substrate. This would be considered more a direct test since it is directly measuring the presence of antigen.
In the case of Ebola, antibody levels are easier to detect than viral antigen (due to the very low levels of antigen present in a serum sample). However, since antibody development takes time, antigen detection may lead to a quicker diagnosis and therefore a better prognosis.
DIF: Difficult REF: 28.4 TOP: V.C.i MSC: Analyzing
- Why are clinical viral diseases more challenging to diagnose than are clinical bacterial diseases?
Answers will vary. Clinical viral diseases are more challenging to diagnose in a lab because viruses are intracellular parasites. Unlike bacteria, viruses do not have a viable existence outside of a host cell, and therefore laboratory diagnosis requires cell-culture techniques. Cell-culture techniques are more expensive and time consuming, and many clinical labs do not have access to the equipment needed. Newer methods of rapid detection using nucleic acid based identification is making identification of viruses much easier.
DIF: Medium REF: 28.4 TOP: V.A MSC: Evaluating
- Compare and contrast the use of point-of-care rapid tests as diagnostic methods in doctors’ offices.
Answers may vary.
- Advantages: culturing not required; antibiotic therapy can be initiated earlier and more effectively; infections chains determined more quickly; better patient compliance.
- Disadvantages: no data on antibiotic sensitivity; higher risk of technician infection, may miss co-infections; high false-positive and false negative rates (lower sensitivities).
DIF: Medium REF: 28.5 TOP: VI.A MSC: Analyzing
- Compare and contrast the handling of rabies (a category III organism) with Lassa fever (a category IV organism).
Category III organisms, such as rabies, have greater pathogenic potential than level I and II with a possibility of respiratory transmission and a high risk of infection without proper handling. Rigorous containment and procedures are necessary during handling, including but not limited to negative-pressure labs, restricted laboratory access, double-door air locks, and so on. A biosafety level III laboratory is necessary to work with these organisms.
Category IV organisms, such as Lassa fever, are extremely dangerous pathogens and need even more rigorous containment and procedures than category III organisms. Rigorous containment and procedures are necessary during handling; these include but are not limited to personal positive-pressure suits, negative-pressure labs in addition to restricted access, and double-door air locks. A biosafety level IV laboratory is necessary to work with these organisms.
DIF: Medium REF: 28.6 TOP: VII.A MSC: Analyzing
- Describe the chain of command that is followed when a patient sees their doctor and is diagnosed with a reportable disease such as listeriosis.
Notifiable diseases, such as listeriosis, are diseases that due to their severity and transmissibility must be reported by law. Physicians are required to report instances to a central health organization (local, county, and/or state offices). State and local offices report up through the chain of command. Highly infectious or dangerous organisms are reported within 48 hours of diagnosis to the Centers for Disease Control (CDC) for U.S. outbreaks. The CDC then reports to the World Health Organization (WHO), which tracks worldwide outbreaks. As a result, incidences of these diseases can be tracked and analyzed. Emerging diseases can also be detected via a cluster of patients with unusual symptoms or combinations of symptoms. An upsurge in cases sets off “alarms” that initiate efforts to determine the source and cause of the outbreak.
DIF: Medium REF: 28.7 TOP: VIII.B MSC: Analyzing
- Compare and contrast the epidemiological terms endemic, epidemic, and pandemic. Give an example of each to support your answer.
Answers will vary. Endemic diseases are always present in a population at low frequency (e.g., Lyme disease is endemic in the northeastern United States because it is found in a reservoir of deer and ticks; humans must have contact with this reservoir for infection). Epidemic diseases occur when large numbers of individuals in a population become infected over a short period of time, due in part to rapid and direct human-to-human transmission (e.g., food-borne illness, H1N1, mumps). Pandemics are epidemics that occur over a wide geographic area, usually the world; these may be short-lived (e.g., 1918 flu pandemic) or long-lived (e.g., bubonic plague, HIV).
DIF: Medium REF: 28.7 TOP: VIII.B MSC: Analyzing
- How were Koch’s postulates modified in the discovery of Tropheryma whipplei, the causative agent of Whipple’s disease?
Whipple’s disease was first diagnosed in 1907 by George Whipple, although the bacterial etiology was not determined until 1992 through amplified 16S rDNA sequencing of biopsied tissues. Symptoms from this disease include bouts of watery diarrhea, recurring fevers, flulike symptoms, profuse night sweats, and joint swelling. Stool samples have no blood or mucus, and blood and stool cultures come back from the lab as negative. Diagnosis of this disease comes from duodenal biopsy (showing large macrophages and bacterial rods) and is confirmed by electron microscopy and PCR.
The discovery of this disease happened because all bacteria from biopsied tissues had common and unique 16S rRNA sequences. Primers were developed to the common areas, and PCR amplification revealed fragments that could be sequenced. The sequences found were similar in other Whipple disease patients and different from any known species.
Koch’s postulates state that an organism must be identified and grown in culture. Since this organism (a Gram-positive soil-dwelling actinomycete named Tropheryma whipplei) defies normal culturing techniques, disease causalty was determined with molecular diagnostics.
DIF: Medium REF: 28.7 TOP: VIII.D MSC: Analyzing
- As a disease detective investigating an outbreak in the cafeteria of your college, what would be some of the first steps you would need to take to begin your investigation, other than laboratory testing?
Answers will vary. One of the first things a disease detective does is interview members of the population involved. This would involve talking with students who claim to have symptoms and recording case histories and progression of symptoms. Questions could include: when did you eat at the cafeteria? What did you eat specifically? When was the onset of symptoms? What were the symptoms? How long did the symptoms last? and so on. The disease detective would also have to conduct the same interviews with other students eating in the cafeteria at the same time, as a control population. Once all the information was collected, the disease detective could start to hypothesize about the disease based on knowing which organisms typically cause which symptoms and under what conditions. Confirmatory laboratory testing of sick students would be needed to determine the actual etiologic agent.
DIF: Medium REF: 28.7 TOP: VIII.A | VIII.B
- Describe what the One Health Initiative is and the role it played in the 2006 outbreak of food-borne O157:H7 in spinach.
Ecology influences epidemiology and evolution of pathogens. The knowledge that pathogens have reservoirs in animals, arthropods, and plants has given rise to the collaboration between scientists, clinicians, veterinarians, and ecologists called the One Health Initiative. The goal of the One Health Initiative is to control human disease through animal health and vice versa (e.g., vaccinating wild rodents can decrease Lyme disease in humans). This goal has been extended to plants, since plants can harbor some infections, too (e.g., enteric bacteria can live in plant vascular systems).
The 2006 outbreak of E. coli O157:H7 in 200 people and 20 states was caused by spinach. The source of the infection was a spinach farm in Salinas Valley, California. The organisms were found within the vascular system of the spinach (so washing the spinach did not help). A multidisciplinary collaboration allowed for this determination (otherwise the focus would have been solely on human morbidity and mortality).
- Epidemiologists and vets found a genetically identical organism in wild hogs in the same fields and in cattle close by. Ecologists and hydrologists found that surface and ground water was mixing due to drought, and irrigation systems were strained.
- One Health concluded that cows defecated in the fields that wild hogs ran through, which then brought the E. coli to the spinach fields, where irrigation water swept the pathogen into the plants’ vascular system.
- In conclusion, a holistic approach is needed to understand, promote, and protect health.
DIF: Medium REF: 28.8 TOP: IX.D MSC: Analyzing
- How have changes within our human culture caused new diseases to emerge? Explain your answer using a specific example that supports your statements.
Answers will vary. Many new diseases have emerged over the last 30 years (Lyme, MRSA, SARS, Ebola, HIV, E.coli O157:H7, etc.). Advances in technology help infectious diseases to emerge and spread (e.g., travel, blood banks, suburban sprawl).
Case studies of progress causing infections include:
- Lyme disease caused by Borrelia burgdorferi lives in deer and white-footed mice and is passed between these hosts by the deer tick. Suburban sprawl encroached on woodland habitat in Connecticut, resulting in decreased predator populations of mice and closer proximity of humans to the mice/tick/deer cycle.
- Mad cow disease—modern farming practices of feeding livestock-rendered animal protein helped to spread transmissible spongiform encephalopathies similar to Creutzfeldt-Jakob prion disease.
- Hepatitis C—spread via increasing transfusions and transplants.
- Influenza—spread via live poultry markets in Asia.
- Enterohemorrhagic E. coli—modern meat processing allows fecal matter to contaminate ground beef.
Changes in the natural environment can also cause infections. For example, in hantavirus, higher-than-usual rainfall increases plant and animal numbers, which increases the deer-mice population, which grows closer to and interacts more with human communities.
DIF: Medium REF: 28.8 TOP: IX.B MSC: Evaluating