Microbiology Chapter 2 Quiz Questions
Microbiology Chapter 2
Questions and Answers
- 1.
The __________ is the point at which a lens focuses parallel beams of light.
Correct Answer
Focal pointExplanation
The focal point is the point at which a lens focuses parallel beams of light. When parallel rays of light pass through a lens, they converge at a single point called the focal point. This is where the light rays come together after passing through the lens and is an important concept in understanding how lenses work.Rate this question:
- 2.
The __________ is the distance between the center of a lens and the point at which it focuses parallel beams of light.
Correct Answer
Focal lengthExplanation
The focal length is the distance between the center of a lens and the point at which it focuses parallel beams of light.Rate this question:
- 3.
Light rays are refracted (bent) when they cross the interface between materials with different refractive indices.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Light rays are refracted (bent) when they cross the interface between materials with different refractive indices. This is a well-known phenomenon in physics, where light changes direction as it passes from one medium to another. The amount of bending depends on the refractive indices of the materials involved. Therefore, it is true that light rays are refracted when they cross the interface between materials with different refractive indices.Rate this question:
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- 4.
Which of these microscopes can be used to create high-resolution three-dimensional images of cells?
- A.
Differential interference contrast
- B.
Dark field
- C.
Phase-contrast
- D.
Confocal
Correct Answer
D. ConfocalExplanation
Confocal microscopes can be used to create high-resolution three-dimensional images of cells. Unlike other microscopes, confocal microscopes use a pinhole aperture to eliminate out-of-focus light, resulting in sharper and more detailed images. This technique allows for the visualization of cellular structures and processes in three dimensions, providing valuable information about the organization and function of cells.Rate this question:
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- 5.
Confocal microscopes exhibit improved contrast and resolution by
- A.
Illumination of a large area of the specimen.
- B.
Blocking out stray light with an aperture located above the objective lens.
- C.
Use of light at longer wavelengths.
- D.
Use of ultraviolet light to illuminate the specimen.
Correct Answer
B. Blocking out stray light with an aperture located above the objective lens.Explanation
Confocal microscopes exhibit improved contrast and resolution by blocking out stray light with an aperture located above the objective lens. This aperture acts as a barrier, preventing unwanted light from entering the microscope and reducing background noise. By blocking out stray light, the microscope is able to capture clearer and more focused images of the specimen, resulting in improved contrast and resolution.Rate this question:
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- 6.
A 30 objective and a 20 ocular produce a total magnification of
- A.
230
- B.
320
- C.
50
- D.
600
Correct Answer
D. 600Explanation
The total magnification is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens. In this case, the objective lens has a magnification of 30 and the ocular lens has a magnification of 20. Therefore, the total magnification is 30 x 20 = 600.Rate this question:
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- 7.
A 45 objective and a 10 ocular produce a total magnification of
- A.
900
- B.
55
- C.
450
- D.
145
Correct Answer
C. 450Explanation
The total magnification is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens. In this case, the objective lens has a magnification of 45 and the ocular lens has a magnification of 10. Therefore, the total magnification is 45 x 10 = 450.Rate this question:
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- 8.
A microscope that exposes specimens to ultraviolet, violet, or blue light and forms an image with the light emitted at a different wavelength is called a __________ microscope.
- A.
Phase-contrast
- B.
Dark field
- C.
Scanning electron
- D.
Fluorescence
Correct Answer
D. FluorescenceExplanation
A microscope that exposes specimens to ultraviolet, violet, or blue light and forms an image with the light emitted at a different wavelength is called a fluorescence microscope. This type of microscope uses fluorescent dyes or proteins to label specific structures or molecules within the specimen, which then emit light of a different color when excited by the specific wavelength of light used. This allows for the visualization and study of specific cellular components or processes.Rate this question:
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- 9.
Immersion oil can be used to increase the resolution achieved with some microscope lenses because it increases the __________ between the specimen and the objective lens.
- A.
Optical density
- B.
Refractive index
- C.
Optical density and refractive index
- D.
Neither optical density nor refractive index
Correct Answer
B. Refractive indexExplanation
Immersion oil is used to increase the resolution achieved with some microscope lenses because it increases the refractive index between the specimen and the objective lens. By placing a drop of immersion oil between the lens and the specimen, the refractive index of the oil matches that of the glass lens, reducing the amount of light that is scattered or lost at the interface. This allows for better light transmission and improved resolution, allowing finer details of the specimen to be observed.Rate this question:
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- 10.
A substage condenser is used to focus light onto the specimen, which increases the resolution of a light microscope.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
A substage condenser is a component of a light microscope that is used to focus light onto the specimen. By focusing the light, the condenser increases the resolution of the microscope. This means that the microscope is able to produce a clearer and more detailed image of the specimen. Therefore, the statement "A substage condenser is used to focus light onto the specimen, which increases the resolution of a light microscope" is true.Rate this question:
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- 11.
The __________ is the distance between the specimen and the objective lens when the specimen is in focus.
Correct Answer
working distanceExplanation
The working distance refers to the distance between the specimen and the objective lens when the specimen is in focus. This distance is important because it determines the amount of space available for manipulating the specimen or inserting additional tools or equipment. A longer working distance allows for more flexibility and ease of use, while a shorter working distance may require more precision and careful handling of the specimen.Rate this question:
- 12.
The useful magnification of a light microscope is limited by the ___________ of the light source being utilized.
Correct Answer
wavelengthExplanation
The useful magnification of a light microscope is limited by the wavelength of the light source being utilized. This is because the resolution of a microscope is determined by the wavelength of the light used. As the wavelength decreases, the resolution increases, allowing for more detailed observation of the specimen. Therefore, a shorter wavelength light source will provide higher magnification and better resolution in a light microscope.Rate this question:
- 13.
The special dyes used in fluorescence microscopy that absorb light at one wavelength and emit light at a different wavelength are called __________.
Correct Answer
fluorochromesExplanation
Fluorochromes are special dyes used in fluorescence microscopy that absorb light at one wavelength and emit light at a different wavelength. These dyes are designed to specifically interact with certain molecules or structures of interest, allowing them to be visualized under a fluorescence microscope. By using fluorochromes, researchers can label and track specific molecules or structures within cells or tissues, providing valuable information about their location, behavior, and interactions.Rate this question:
- 14.
In order to view a specimen with a total magnification of 400, a __________ objective must be used if the ocular is 10.
Correct Answer
40Explanation
To view a specimen with a total magnification of 400, the ocular magnification of 10 needs to be multiplied by the objective magnification. Therefore, the objective magnification must be 40 in order to achieve a total magnification of 400.Rate this question:
- 15.
Confocal microscopes, in combination with specialized computer software, can be used to create three-dimensional images of cell structures.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Confocal microscopes are indeed capable of creating three-dimensional images of cell structures when used in combination with specialized computer software. This is achieved by capturing multiple images of the sample at different focal planes and then reconstructing them into a three-dimensional representation. This technology has greatly advanced the field of cell biology by allowing researchers to visualize and study cellular structures in greater detail.Rate this question:
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- 16.
A light microscope with an objective lens numerical aperture of 0.65 is capable of allowing two objects 400 nm apart to be distinguished when using light with a wavelength of 420 nm.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The statement is true because the resolution of a light microscope depends on the numerical aperture (NA) of the objective lens and the wavelength of light used. The formula for resolution is given by the Rayleigh criterion, which states that the minimum resolvable distance is approximately equal to half the wavelength of light divided by the numerical aperture. In this case, the numerical aperture is 0.65 and the wavelength of light is 420 nm, so the resolution is approximately 322 nm. Since the objects are 400 nm apart, they can be distinguished by the microscope.Rate this question:
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- 17.
Resolution decreases when the wavelength of the illuminating light decreases.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The resolution actually increases when the wavelength of the illuminating light decreases. This is because resolution is determined by the ability of a microscope or imaging system to distinguish between two closely spaced objects. According to the Rayleigh criterion, the resolution is inversely proportional to the wavelength of the light used. Therefore, shorter wavelengths of light (such as blue or ultraviolet light) provide better resolution compared to longer wavelengths (such as red or infrared light).Rate this question:
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- 18.
Immersion oil is used to prevent a specimen from drying out.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Immersion oil is actually used to increase the resolution and clarity of a specimen under a microscope. It has a higher refractive index than air, which reduces the amount of light that is scattered and allows more light to enter the objective lens. This improves the sharpness and detail of the image. It does not have any effect on preventing the specimen from drying out.Rate this question:
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- 19.
It is possible to build a light microscope capable of 10,000 magnification, but the image would not be sharp because resolution is independent of magnification.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The explanation for the given correct answer is that magnification refers to the ability of a microscope to enlarge an image, while resolution refers to the ability to distinguish between two separate points in an image. Even if a microscope has a high magnification power of 10,000x, it does not guarantee a sharp image if the resolution is poor. Resolution depends on factors such as the quality of the lenses and the wavelength of light used. Therefore, it is possible to have high magnification but low resolution, resulting in a blurry image.Rate this question:
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- 20.
Immersion oil increases the amount of light passing through a specimen and entering the objective lens.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Immersion oil is used in microscopy to improve the resolution and clarity of the image. It has a similar refractive index to glass, which reduces the amount of light that is lost or scattered as it passes through the specimen. This allows more light to pass through the objective lens, resulting in a brighter and clearer image. Therefore, it is true that immersion oil increases the amount of light passing through a specimen and entering the objective lens.Rate this question:
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- 21.
If the objective lenses of a microscope can be changed without losing focus on the specimen, they are said to be
- A.
Equifocal
- B.
Totifocal
- C.
Parfocal
- D.
Optifocal
Correct Answer
C. ParfocalExplanation
If the objective lenses of a microscope can be changed without losing focus on the specimen, they are said to be parfocal. This means that when switching between different objective lenses, the specimen remains in focus and there is no need for significant adjustments to be made to the focus knob. This feature is important in microscopy as it allows for efficient and convenient examination of different magnifications without losing focus on the specimen.Rate this question:
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- 22.
An instrument that magnifies slight differences in the refractive index of cell structures is called a (n) __________ microscope.
- A.
Phase-contrast
- B.
Electron
- C.
Fluorescence
- D.
Densitometric
Correct Answer
A. Phase-contrastExplanation
A phase-contrast microscope is an instrument that is used to magnify slight differences in the refractive index of cell structures. It allows for the visualization of transparent or unstained samples by enhancing the contrast between the different parts of the cell. This technique is particularly useful for studying living cells or observing dynamic processes within cells.Rate this question:
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- 23.
The instrument that produces a bright image of the specimen against a dark background is called a (n) __________ microscope.
- A.
Phase-contrast
- B.
Electron
- C.
Bright field
- D.
Dark field
Correct Answer
D. Dark fieldExplanation
A dark field microscope is an instrument that produces a bright image of the specimen against a dark background. This is achieved by blocking the direct light that would normally pass through the specimen, and instead only capturing the light that is scattered or refracted by the specimen. This technique enhances the contrast between the specimen and its surroundings, making it easier to observe and analyze.Rate this question:
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- 24.
As the magnification of a series of objective lenses increases, the working distance
- A.
Increases
- B.
Decreases
- C.
Stays the same
- D.
Cannot be predicted
Correct Answer
B. DecreasesExplanation
As the magnification of a series of objective lenses increases, the working distance decreases. This is because as the magnification increases, the lenses need to be closer to the object being observed in order to achieve a higher level of detail. Therefore, the working distance, which is the distance between the lens and the object, decreases as the magnification increases.Rate this question:
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- 25.
Prior to staining, smears of microorganisms are heat-fixed in order to
- A.
Allow eventual visualization of internal structures.
- B.
Ensure removal of dust particles from the slide surface.
- C.
Attach it firmly to the slide.
- D.
Create small pores in cells that facilitates binding of stain to cell structures.
Correct Answer
C. Attach it firmly to the slide.Explanation
Heat-fixing is a process in which smears of microorganisms are subjected to heat in order to attach them firmly to the slide. This helps in preventing the smears from being washed off during subsequent staining and washing steps. Heat-fixing also helps in preserving the overall morphology of the microorganisms, making it easier to visualize their internal structures under a microscope. Therefore, the correct answer is "attach it firmly to the slide."Rate this question:
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- 26.
Acid-fast organisms such as Mycobacterium tuberculosis contain __________ constructed from mycolic acids in their cell walls.
- A.
Proteins
- B.
Carbohydrates
- C.
Lipids
- D.
Peptidoglycan
Correct Answer
C. LipidsExplanation
Acid-fast organisms like Mycobacterium tuberculosis have cell walls composed of lipids called mycolic acids. These lipids provide a protective barrier for the organism and contribute to its acid-fast properties. Lipids are a type of biomolecule that are insoluble in water and play important roles in cell structure and function. In the case of acid-fast organisms, the presence of mycolic acids in their cell walls helps to differentiate them from other bacteria.Rate this question:
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- 27.
In the Gram-staining procedure, the primary stain is
- A.
Iodine
- B.
Safranin
- C.
Crystal violet
- D.
Alcohol
Correct Answer
C. Crystal violetExplanation
The Gram-staining procedure is a method used to differentiate bacteria into two groups: Gram-positive and Gram-negative. In this procedure, the primary stain is crystal violet. Crystal violet is a purple dye that is applied to the bacteria, allowing them to retain the color. After the crystal violet stain, a series of steps including iodine, alcohol, and safranin are used to further differentiate and stain the bacteria. However, crystal violet is the initial and primary stain used in the Gram-staining procedure.Rate this question:
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- 28.
In the Gram-staining procedure, the decolorizer is
- A.
Iodine
- B.
Safranin
- C.
Crystal violet
- D.
Ethanol or acetone
Correct Answer
D. Ethanol or acetoneExplanation
In the Gram-staining procedure, the decolorizer is ethanol or acetone. This step is crucial in differentiating between Gram-positive and Gram-negative bacteria. The decolorizer removes the crystal violet stain from the Gram-negative bacteria, making them colorless, while the Gram-positive bacteria retain the stain. This allows for the differentiation and classification of bacteria based on their cell wall composition.Rate this question:
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- 29.
In the Gram-staining procedure, the counterstain is
- A.
Iodine
- B.
Safranin
- C.
Crystal violet
- D.
Alcohol
Correct Answer
B. SafraninExplanation
The counterstain in the Gram-staining procedure is safranin. Gram staining is a technique used to differentiate bacteria into two groups: Gram-positive and Gram-negative. Crystal violet is the primary stain used to stain all bacteria, and iodine is used as a mordant to enhance the staining. After applying alcohol as a decolorizer, Gram-positive bacteria retain the crystal violet stain while Gram-negative bacteria do not. Finally, safranin is used as a counterstain to stain the Gram-negative bacteria pink or red, allowing for easy differentiation between the two groups.Rate this question:
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- 30.
In the Gram-staining procedure, the mordant is
- A.
Iodine
- B.
Safranin
- C.
Crystal violet
- D.
Alcohol
Correct Answer
A. IodineExplanation
In the Gram-staining procedure, the mordant is iodine. A mordant is a substance that helps to fix or attach a stain to a specimen. In Gram staining, iodine is used as a mordant to form a complex with crystal violet dye, which helps to stabilize the dye and prevent it from being easily washed away. This complex is then trapped within the thick peptidoglycan layer of Gram-positive bacteria, resulting in the retention of the purple color. In Gram-negative bacteria, the iodine complex is washed away during the decolorization step, leading to the uptake of the counterstain (safranin) and the pink coloration.Rate this question:
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- 31.
After the primary stain has been added but before the decolorizer has been used, gram positive organisms are stained __________ and gram-negative organisms are stained __________.
- A.
Purple; purple
- B.
Purple; colorless
- C.
Purple; pink
- D.
Pink; pink
Correct Answer
A. Purple; purpleExplanation
Before the decolorizer is used, gram-positive organisms retain the primary stain (crystal violet) and appear purple. Gram-negative organisms also retain the primary stain and appear purple. Therefore, both gram-positive and gram-negative organisms are stained purple at this stage.Rate this question:
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- 32.
After the decolorizer has been added, gram-positive organisms are stained __________ and gram-negative organisms are stained __________.
- A.
Purple; purple
- B.
Purple; colorless
- C.
Purple; pink
- D.
Pink; pink
Correct Answer
B. Purple; colorlessExplanation
After the decolorizer has been added, gram-positive organisms retain the purple stain because their cell walls have a thick layer of peptidoglycan which traps the crystal violet dye. On the other hand, gram-negative organisms lose the purple stain because their cell walls have a thinner layer of peptidoglycan and are easily decolorized by the alcohol in the decolorizer. As a result, gram-negative organisms appear colorless after the decolorizer has been added.Rate this question:
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- 33.
After the secondary stain has been added, gram-positive organisms are stained __________ and gram-negative organisms are stained __________.
- A.
Purple; purple
- B.
Purple; colorless
- C.
Purple; pink
- D.
Pink; pink
Correct Answer
C. Purple; pinkExplanation
After the secondary stain has been added, gram-positive organisms retain the purple color of the primary stain and appear purple. On the other hand, gram-negative organisms lose the purple color and take on the pink color of the secondary stain, appearing pink.Rate this question:
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- 34.
If the decolorizer is left on too long in the Gram-staining procedure, gram-positive organisms will be stained __________ and gram-negative organisms will be stained __________.
- A.
Purple; blue
- B.
Purple; colorless
- C.
Purple; pink
- D.
Pink; pink
Correct Answer
D. Pink; pinkExplanation
If the decolorizer is left on too long in the Gram-staining procedure, gram-positive organisms will be stained pink and gram-negative organisms will also be stained pink.Rate this question:
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- 35.
If the decolorizer is not left on long enough in the Gram-staining procedure, gram-positive organisms will be stained __________ and gram-negative organisms will be stained __________.
- A.
Purple; purple
- B.
Purple; colorless
- C.
Purple; pink
- D.
Pink; pink
Correct Answer
A. Purple; purpleExplanation
If the decolorizer is not left on long enough in the Gram-staining procedure, gram-positive organisms will retain the crystal violet stain and appear purple, while gram-negative organisms will lose the crystal violet stain and appear purple as well.Rate this question:
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- 36.
Which of the following is considered to be a differential staining procedure?
- A.
Gram stain
- B.
Acid-fast stain
- C.
Both Gram stain and Acid-fast stain
- D.
Leifson's flagella stain
Correct Answer
C. Both Gram stain and Acid-fast stainExplanation
Both the Gram stain and Acid-fast stain are considered to be differential staining procedures. Differential staining is a technique used to distinguish between different types of microorganisms or different parts of the same microorganism based on their staining properties. The Gram stain differentiates bacteria into Gram-positive and Gram-negative based on the differences in their cell wall composition. The Acid-fast stain is used to identify bacteria that have a waxy cell wall, such as Mycobacterium tuberculosis, which causes tuberculosis.Rate this question:
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- 37.
Basic dyes such as methylene blue bind to cellular molecules that are
- A.
Hydrophobic
- B.
Negatively charged
- C.
Positively charged
- D.
Aromatic
Correct Answer
B. Negatively chargedExplanation
Basic dyes such as methylene blue bind to cellular molecules that are negatively charged. This is because basic dyes have a positive charge and are attracted to molecules with a negative charge. The binding occurs through electrostatic interactions between the positive and negative charges, allowing the dye to attach to the cellular molecules.Rate this question:
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- 38.
The Schaeffer-Fulton procedure is used to stain
- A.
Flagella
- B.
Fat deposits
- C.
Endospores
- D.
DNA of chromosomes
Correct Answer
C. EndosporesExplanation
The Schaeffer-Fulton procedure is a staining technique commonly used to visualize endospores. Endospores are dormant, highly resistant structures formed by certain bacteria as a way to survive harsh conditions. These structures are not easily stained by conventional staining methods, hence the need for specialized techniques like the Schaeffer-Fulton procedure. This procedure involves the use of heat and multiple dyes to selectively stain the endospores, allowing them to be easily observed and distinguished from other cellular components.Rate this question:
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- 39.
Gram staining divides bacterial species into roughly two equal groups.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Gram staining is a common technique used to classify bacteria into two major groups based on their cell wall composition. These groups are referred to as Gram-positive and Gram-negative bacteria. Gram-positive bacteria have a thick layer of peptidoglycan in their cell walls, which retains the crystal violet stain used in the Gram staining process and appears purple under a microscope. On the other hand, Gram-negative bacteria have a thinner peptidoglycan layer and an outer membrane, which does not retain the stain and appears pink or red. Therefore, it can be said that Gram staining divides bacterial species into roughly two equal groups.Rate this question:
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- 40.
Negative staining facilitates the visualization of bacterial capsules which are intensely stained by the procedure.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Negative staining actually does not facilitate the visualization of bacterial capsules. Negative staining is a technique used to visualize the background or the external surface of bacteria, while bacterial capsules are visualized using positive staining techniques. Therefore, the statement is false.Rate this question:
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- 41.
Negative staining with India ink can be used to reveal the presence of capsules that surround bacterial cells.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
India ink is commonly used in microbiology to perform negative staining, a technique that allows the visualization of capsules surrounding bacterial cells. The ink particles are repelled by the capsule material, creating a dark background against which the capsules appear as clear halos. This technique is particularly useful for identifying encapsulated bacteria such as Streptococcus pneumoniae, which causes pneumonia and other infections. Therefore, the statement that negative staining with India ink can be used to reveal the presence of capsules surrounding bacterial cells is true.Rate this question:
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- 42.
Mordants increase the binding between a stain and specimen.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Mordants are substances that are used in staining techniques to enhance the binding between a stain and the specimen. They achieve this by forming a complex with the stain and the specimen, creating a stronger and more durable bond. This increased binding allows for better retention of the stain on the specimen, resulting in clearer and more visible staining. Therefore, the statement that mordants increase the binding between a stain and specimen is true.Rate this question:
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- 43.
In order to stain flagella so that they may be readily observed by light microscopy, it is usually necessary to increase their thickness.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Flagella are thin, whip-like structures that are difficult to observe under light microscopy due to their small size. In order to make them more visible, it is necessary to increase their thickness. This can be achieved by staining the flagella, which involves adding a dye or a specific chemical to the sample. The stain increases the contrast between the flagella and the surrounding environment, making them easier to observe and study under a light microscope. Therefore, the statement is true.Rate this question:
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- 44.
The procedure in which a single stain is used to visualize microorganisms is called __________ staining.
Correct Answer
simpleExplanation
The correct answer is "simple" because the question is asking for the term used to describe the procedure of using a single stain to visualize microorganisms. The term "simple" accurately describes this procedure, as it involves using only one stain instead of multiple stains.Rate this question:
- 45.
__________ is the process by which internal and external structures of cells and organisms are preserved and maintained in position.
Correct Answer
FixationExplanation
Fixation is the process by which internal and external structures of cells and organisms are preserved and maintained in position. It involves treating the cells or tissues with a fixative, such as formaldehyde, which helps to prevent decay and maintain the structure of the cells. Fixation is an important step in various biological techniques, including histology and electron microscopy, as it allows for the preservation of cellular structures and the ability to study them under a microscope. It helps to prevent the loss of cellular components and maintains the integrity of the sample for further analysis.Rate this question:
- 46.
Thin films of bacteria that have been air-dried onto a glass microscope slide are called __________.
Correct Answer
smearsExplanation
Thin films of bacteria that have been air-dried onto a glass microscope slide are called smears. This term is used to describe the process of spreading a liquid sample containing bacteria onto a slide and allowing it to dry. Smears are commonly used in microbiology to prepare samples for microscopic examination, staining, and identification of bacteria. The dried bacteria adhere to the slide, allowing for easier visualization and analysis under the microscope.Rate this question:
- 47.
A procedure that divides organisms into two or more groups depending on their individual reactions to the same staining procedure is referred to as __________ staining.
Correct Answer
differentialExplanation
A staining procedure that divides organisms into different groups based on their individual reactions is referred to as differential staining. This technique allows for the differentiation and identification of different types of organisms or structures within organisms based on their staining characteristics. By using different stains or dyes, specific structures or components can be highlighted or differentiated, providing valuable information for classification and identification purposes.Rate this question:
- 48.
The Gram-staining procedure is an example of:
- A.
Simple staining
- B.
Negative staining
- C.
Differential staining
- D.
Fluorescent staining
Correct Answer
C. Differential stainingExplanation
The Gram-staining procedure is an example of differential staining. This technique is used to differentiate bacteria into two major groups, Gram-positive and Gram-negative, based on their cell wall composition. The staining process involves multiple steps, including the application of a crystal violet dye, iodine treatment, decolorization with alcohol or acetone, and counterstaining with safranin. Gram-positive bacteria retain the crystal violet stain and appear purple, while Gram-negative bacteria lose the stain and appear pink or red after counterstaining. This staining technique is crucial for identifying and classifying bacteria in microbiology.Rate this question:
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- 49.
The Gram-staining procedure is widely used because it allows rapid identification of a microorganism with little additional testing.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The Gram-staining procedure is not widely used because it allows rapid identification of a microorganism with little additional testing. In fact, the Gram-staining procedure is used to differentiate bacteria into two major groups - Gram-positive and Gram-negative - based on their cell wall composition. This procedure does not provide rapid identification of a microorganism, but rather helps in the initial characterization of bacteria.Rate this question:
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- 50.
Regions of a specimen with higher electron density scatter ___________ electrons and, therefore, appear __________ in the image projected onto the screen of a transmission electron microscope.
- A.
More; lighter
- B.
More; darker
- C.
Fewer; darker
- D.
Fewer; lighter
Correct Answer
B. More; darkerExplanation
Regions of a specimen with higher electron density scatter more electrons, and therefore, appear darker in the image projected onto the screen of a transmission electron microscope. This is because the higher electron density causes more electrons to be deflected and scattered, resulting in a reduced intensity of the electrons reaching the screen. As a result, these regions appear darker in the image.Rate this question:
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