Block 9 Prokaryot Strategy In Lab Diag MCQ
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- 1.
Bacteria might produce adhesins to counteract which of the following aspects of host defense?
- A.
Ciliated epithelial cells
- B.
Altered cell-surface receptors
- C.
Macrophage engulfment
- D.
Opsonization by antibodies
- E.
Phagosomal lysosome fusion
Correct Answer
A. Ciliated epithelial cellsExplanation
Bacteria might produce adhesins to counteract the host defense mechanism of ciliated epithelial cells. Ciliated epithelial cells have hair-like projections called cilia that help in moving mucus and trapped bacteria out of the respiratory tract. By producing adhesins, bacteria can attach to these ciliated cells and avoid being swept away, allowing them to establish an infection.Rate this question:
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- 2.
What is the role of the alcohol in the Gram stain reaction?
- A.
It dissolves lipids, allowing the crystal violet to wash out of Gram negative cells
- B.
It counterstains the decoiorized cells, allowing them to be visualized
- C.
It fixes the crystal violet to the peptidoglycan of Gram positive cells
- D.
It complexes with the crystal violet, creating larger molecules.
Correct Answer
A. It dissolves lipids, allowing the crystal violet to wash out of Gram negative cellsExplanation
The role of alcohol in the Gram stain reaction is to dissolve lipids. This allows the crystal violet, which is the primary stain, to wash out of Gram-negative cells. Gram-negative cells have a thinner peptidoglycan layer in their cell wall, which is not able to retain the crystal violet stain as effectively as Gram-positive cells. By dissolving the lipids, the alcohol helps to remove the crystal violet from the Gram-negative cells, resulting in their decolorization.Rate this question:
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- 3.
How do the ribosomes of the eukaryotic cytoplasm and prokaryotes differ?
- A.
The structures of the ribonucleotides it is composed of.
- B.
The strand that is transcribed to create the rRNA component
- C.
The sizes and number of the rRNA components.
- D.
Prokaryotic ribosomes are comprised of only rRNA and no protein
Correct Answer
C. The sizes and number of the rRNA components.Explanation
The correct answer is the sizes and number of the rRNA components. Eukaryotic ribosomes are larger and more complex than prokaryotic ribosomes. Eukaryotes have a 60S and a 40S subunit, while prokaryotes have a 50S and a 30S subunit. Additionally, eukaryotic ribosomes have more rRNA molecules compared to prokaryotic ribosomes. This size and number difference allows eukaryotic ribosomes to perform more intricate and specialized functions in protein synthesis.Rate this question:
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- 4.
How would you expect Mycobacterium tuberculosis, an acid-fast organism, to react in the Gram stain?
- A.
Gram positive
- B.
Gram negative
- C.
Mixed
- D.
No staining
Correct Answer
D. No stainingExplanation
Mycobacterium tuberculosis is an acid-fast organism, which means it has a unique cell wall composition that is resistant to the staining process used in the Gram stain. Therefore, it would not retain the crystal violet stain or the counterstain, resulting in no staining.Rate this question:
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- 5.
Which of the following species is most structurally similar to the causative agent of tuberculosis?
- A.
Actinomyces israelli
- B.
Nocardia asteroides
- C.
Bacillus anthracis
- D.
Streptococcus pneumoniae
Correct Answer
B. Nocardia asteroidesExplanation
Nocardia asteroides is the most structurally similar to the causative agent of tuberculosis. Nocardia asteroides is a Gram-positive bacterium that belongs to the same order (Actinomycetales) as Mycobacterium tuberculosis, the causative agent of tuberculosis. Both Nocardia asteroides and Mycobacterium tuberculosis are acid-fast bacteria and share similar cell wall structures, including the presence of mycolic acids. This similarity in cell wall composition and staining properties makes Nocardia asteroides the most structurally similar to Mycobacterium tuberculosis among the given options.Rate this question:
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- 6.
Lysozyme is an antibacterial component, part of our innate immune system. it works by:
- A.
Targeting transpeptidases, preventing incorporation of peptidoglycan monomers into the peptidoglycan cell wall.
- B.
Targeting Ddl, which produces the D-ala-D-ala dipeptide that forms the end of the peptide tail on the peptidoglycan monomer,
- C.
Cleaving peptide cross-bridges, leading to weakening of the cell wall and lysis of the bacterial cell.
- D.
Cleaving bonds between the NAG and NAM sugars, leading to lysis of the bacterial cell.
- E.
Degrading biosynthetic enzymes required for synthesis of the peptidoglycan cell wall.
Correct Answer
D. Cleaving bonds between the NAG and NAM sugars, leading to lysis of the bacterial cell.Explanation
Lysozyme works by cleaving bonds between the NAG and NAM sugars, which are components of the peptidoglycan cell wall in bacteria. This cleavage weakens the cell wall and ultimately leads to the lysis of the bacterial cell.Rate this question:
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- 7.
Peptidoglycan and lipid A biosynthetc pathways both require the following component:
- A.
Acyl carrier protein (ACP)
- B.
Undecaprenol phosphate (Und-P)
- C.
N-acetytglucosamine
- D.
N-acetylmuramic acid
- E.
Kdo glycosyltransferases
Correct Answer
C. N-acetytglucosamineExplanation
The correct answer is N-acetytglucosamine. Both the peptidoglycan and lipid A biosynthetic pathways require N-acetytglucosamine as a component. Peptidoglycan is a major component of the bacterial cell wall and is composed of N-acetylmuramic acid and N-acetytglucosamine. Lipid A is a component of lipopolysaccharides (LPS) found in the outer membrane of Gram-negative bacteria and also contains N-acetytglucosamine. Therefore, N-acetytglucosamine is essential for the synthesis of both peptidoglycan and lipid A.Rate this question:
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- 8.
Which of the following best explains why Gram-positive bacteria stain blue/purple during the Gram Stain process?
- A.
Peptidoglycan irreversibly binds to the appkation of crystal violet after the application of safranin.
- B.
The outer membrane irreversibly binds to the application of crystal violet after the application of ethanol or acid ethanol.
- C.
Peptidoglycan irreversibly binds to the application of crystal violet after the application of the mordant (iodine.)
- D.
The inner membrane irreversibly binds to safranin following safranin application followed by rinsrig.
- E.
The very thin layer of peptidoglycan in the cell wall irreversibly binds to the application of crystal violet after the application of the mordant (iodine.)
Correct Answer
C. Peptidoglycan irreversibly binds to the application of crystal violet after the application of the mordant (iodine.)Explanation
The correct answer explains that Gram-positive bacteria stain blue/purple during the Gram Stain process because the peptidoglycan in their cell wall irreversibly binds to the application of crystal violet after the application of the mordant (iodine). This binding forms a complex that is not easily washed away, allowing the bacteria to retain the crystal violet stain and appear blue/purple under a microscope.Rate this question:
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- 9.
The human cytomegalovirus US3 glycoprotein is responsible for binding and retaining major histocompatibility complex class I molecules in the endoplasmic reticulum of infected cells. Which immune function will be affected by US3?
- A.
B cell-mediated antigen presentation to naive CD4-' cells
- B.
CD4+ mediated B cell activation
- C.
Somatic cell-mediated antigen presentation to activated CD8+ cells
- D.
Dendritic cell-mediated antigen presentation to naïve CD4+ cells
- E.
CD4+ mediated CD8+ cell activation
Correct Answer
C. Somatic cell-mediated antigen presentation to activated CD8+ cellsExplanation
The human cytomegalovirus US3 glycoprotein binds and retains major histocompatibility complex class I molecules in the endoplasmic reticulum of infected cells. This will affect somatic cell-mediated antigen presentation to activated CD8+ cells. The major histocompatibility complex class I molecules are responsible for presenting antigens to CD8+ cells, which triggers an immune response against infected cells. Therefore, by binding and retaining these molecules, the US3 glycoprotein inhibits the presentation of antigens to CD8+ cells, compromising the immune function of somatic cell-mediated antigen presentation to activated CD8+ cells.Rate this question:
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- 10.
In the vast majority of cases how many different MHC class I molecules would be expressed on a human dendritic cell?
- A.
0
- B.
3
- C.
4
- D.
6
Correct Answer
D. 6Explanation
A human dendritic cell would typically express six different MHC class I molecules. MHC class I molecules are highly diverse and play a crucial role in presenting antigens to T cells. They are encoded by a group of genes called the human leukocyte antigen (HLA) complex. Each individual has multiple variants of MHC class I genes, resulting in a diverse repertoire of MHC class I molecules. Dendritic cells are professional antigen-presenting cells and have the ability to capture and present a wide range of antigens. Thus, they express multiple MHC class I molecules to ensure efficient presentation of diverse antigens to T cells.Rate this question:
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- 11.
The major histocompatibility complex (MHC) class II molecules on antigen presenting cells differ from MHC class I molecules in which manner?
- A.
MHC class II molecules present antigen mainly to CD8+ T cells
- B.
MHC class II molecules are found mostly on B cells, dendritic cells, and macrophages
- C.
MHC class II molecules mainly present intracellular antigens that were processed by proteasomes
- D.
MHC class II molecules contain B2-microglobulin and are found on all nucleated cells
Correct Answer
B. MHC class II molecules are found mostly on B cells, dendritic cells, and macrophagesExplanation
MHC class II molecules are found mostly on B cells, dendritic cells, and macrophages. This is in contrast to MHC class I molecules, which are found on all nucleated cells. MHC class II molecules are responsible for presenting antigens to CD4+ T cells, whereas MHC class I molecules present antigens to CD8+ T cells. Therefore, the major difference between MHC class II and MHC class I molecules lies in their distribution on different cell types and their role in antigen presentation to different types of T cells.Rate this question:
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- 12.
JR was twelve years old when she was referred to an infectious disease specialist because she had developed severe bronchiectasis and a cough that produced yellow-green sputum. She had been repeatedly diagnosed with sinusitis, otitis media and pneumonia since the age of five. Investigations reveal that she has 'bare lymphocyte syndrome' due to a lack of expression of MHC Class I molecules. Which of the following is most likely to be absent from her blood?
- A.
CD4+ cells
- B.
CD8+ cells
- C.
CD16+ cells
- D.
CD19+ cells
- E.
CD25+ cells
Correct Answer
B. CD8+ cellsExplanation
CD8 T-cells are responsible for intracellular pathogens. Intracellular pathogens cause the cell to express MHC IRate this question:
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- 13.
Which of the following events is least likely to enable tumor cells to evade the host immune system?
- A.
Down regulation of the expression of MHC Class I,
- B.
Failure to express B7.
- C.
Expression of antigens that are not found on other self tissues.
- D.
Production of immunosuppressive cytokines like TGFP.
- E.
Shedding of membrane antigens
Correct Answer
C. Expression of antigens that are not found on other self tissues.Explanation
Expression of antigens that are not found on other self tissues is least likely to enable tumor cells to evade the host immune system. This is because the immune system is more likely to recognize and target cells that express antigens that are not found on self tissues. By expressing antigens that are not found on other self tissues, tumor cells are more likely to be recognized as foreign by the immune system, leading to an immune response against the tumor cells. In contrast, down regulation of MHC Class I, failure to express B7, production of immunosuppressive cytokines like TGFP, and shedding of membrane antigens can all contribute to tumor cells evading the host immune system.Rate this question:
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- 14.
After plating 0.1 ml of a 10-5 dilution of bacteria, you get 70 colones. In order to calculate the cfu/ml you'll have to multiply the count of a plate with the dilution and the plating factor. The cfu/ml results in 7.0 x 105. Which of the following sections of the picture displays the plating factor?
- A.
Section A)
- B.
Section B)
- C.
Section C)
- D.
Section D)
- E.
Section E)
Correct Answer
C. Section C)Explanation
The plating factor is calculated by multiplying the count of a plate with the dilution. In this case, the count of the plate is 70 colonies and the dilution is 10^-5. Therefore, the plating factor is 70 x 10^-5 = 7.0 x 10^-4. Since Section C is labeled as the plating factor, it is the correct answer.Rate this question:
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- 15.
A patient presents to your clinic with a 4-day history of fever and non-productive cough. You suspect the patient has an atypical pneumonia. Your differential diagnosis includes infection with Mycoplasma pneumoniae. Based on what you know about this organism's characteristics, which of the following test methodologies would best confirm that your patient is infected with this organism?
- A.
Gram stain of the patient's sputum
- B.
Acid-fast stain of the patient's sputum
- C.
Examination of a wet prep of the patient's sputum
- D.
Culture in cell lines
- E.
Serological testing for patient's antibodies against the organism
Correct Answer
E. Serological testing for patient's antibodies against the organismExplanation
Serological testing for patient's antibodies against the organism would best confirm that the patient is infected with Mycoplasma pneumoniae. Serological testing detects the presence of specific antibodies produced by the patient's immune system in response to the infection. This method is effective because it can detect past or current infections, even when the organism is no longer present in the patient's sputum. Other methods like Gram stain, acid-fast stain, examination of wet prep, and culture in cell lines may not be as reliable for confirming the presence of Mycoplasma pneumoniae.Rate this question:
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Mar 20, 2023Quiz Edited by
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May 05, 2012Quiz Created by
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