Biology latest questions and verified answers
graded A + 2025/2026 .RATED 100%
CORRECT
1) When biologists wish to study the internal ultrastructure of cells, they
can achieve the finest resolution by using
A) a phase-contrast light microscope.
B) a scanning electron microscope.
C) a transmission electronic microscope.
D) a confocal fluorescence microscope.
E) a super-resolution fluorescence microscope. - CORRECT ANS: C
2) The advantage of light microscopy over electron microscopy is that
A) light microscopy provides for higher magnification than electron
microscopy.
B) light microscopy provides for higher resolving power than electron
microscopy.
C) light microscopy allows one to view dynamic processes in living cells.
D) light microscopy provides higher contrast than electron microscopy.
E) specimen preparation for light microcopy does not produce artifacts. -
CORRECT ANS: c
3) A primary objective of cell fractionation is to
A) view the structure of cell membranes.
B) sort cells based on their size and weight.
C) determine the size of various organelles.
D) separate the major organelles so that their particular functions can be
determined.
E) separate lipid-soluble from water-soluble molecules. - CORRECT
ANS: CORRECT ANS: D
4) In the fractionation of homogenized cells using centrifugation, the
primary factor that determines whether a specific cellular component
ends up in the supernatant or the pellet is
A) the relative solubility of the component.
B) the size and weight of the component.
C) the percentage of carbohydrates in the component.
D) the presence or absence of nucleic acids in the component.
,E) the presence or absence of lipids in the component. - CORRECT ANS:
B
5) Which of the following correctly lists the order in which cellular
components will be found in the pellet when homogenized cells are
treated with increasingly rapid spins in a centrifuge?
A) ribosomes, nucleus, mitochondria
B) chloroplasts, ribosomes, vacuoles
C) nucleus, ribosomes, chloroplasts
D) vacuoles, ribosomes, nucleus
E) nucleus, mitochondria, ribosomes - CORRECT ANS: E
6) Green fluorescent protein (GFP) can be used to fluorescently label a
specific protein in cells by genetically engineering cells to synthesize the
target protein fused to GFP. What is the advantage of using GFP fusions
to visualize specific proteins, instead of staining cells with fluorescently
labeled probes that bind to the target protein?
A) GFP fusions enable one to track changes in the location of the protein
in living cells; staining usually requires preserved cells.
B) GFP fusions enable higher resolution than staining with fluorescent
probes.
C) GFP permits the position of the protein in the cell more precisely than
fluorescent probes.
D) GFP permits visualization of protein-protein interactions; fluorescent
probes do not.
E) GFP fusions are not subject to artifacts; fluorescent probes may
introduce background artifacts. - CORRECT ANS: A
7) What is the reason that a modern electron microscope (TEM) can
resolve biological images to the sub nanometer level, as opposed to tens
of nanometers achievable for the best super-resolution light microscope?
A) The focal length of the electron microscope is significantly longer.
B) Contrast is enhanced by staining with atoms of heavy metal.
C) Electron beams have much shorter wavelengths than visible light.
D) The electron microscope has a much greater ratio of image size to real
size.
E) The electron microscope cannot image whole cells at one time. -
CORRECT ANS: C
8) What technique would be most appropriate to use to observe the
movements of condensed chromosomes during cell division?
A) light microscopy
B) scanning electron microscopy
,C) transmission electron microscopy
D) confocal fluorescence microscopy
E) super-resolution fluorescence microscopy - CORRECT ANS: A
9) All of the following are part of a prokaryotic cell except
A) DNA.
B) a cell wall.
C) a plasma membrane.
D) ribosomes.
E) an endoplasmic reticulum. - CORRECT ANS: E
10) The volume enclosed by the plasma membrane of plant cells is often
much larger than the corresponding volume in animal cells. The most
reasonable explanation for this observation is that
A) plant cells are capable of having a much higher surface-to-volume
ratio than animal cells.
B) plant cells have a much more highly convoluted (folded) plasma
membrane than animal cells.
C) plant cells contain a large vacuole that reduces the volume of the
cytoplasm.
D) animal cells are more spherical, whereas plant cells are elongated.
E) plant cells can have lower surface-to-volume ratios than animal cells
because plant cells synthesize their own nutrients. - CORRECT ANS: C
11) A mycoplasma is an organism with a diameter between 0.1 and 1.0
µm. What does the organism's size tell you about how it might be
classified?
A) It must be a single-celled protist.
B) It must be a single-celled fungus.
C) It could be almost any typical bacterium.
D) It could be a typical virus.
E) It could be a very small bacterium. - CORRECT ANS: E
12) Which of the following is a major cause of the size limits for certain
types of cells?
A) limitation on the strength and integrity of the plasma membrane as cell
size increases
B) the difference in plasma membranes between prokaryotes and
eukaryotes
C) evolutionary progression in cell size; more primitive cells have smaller
sizes
D) the need for a surface area of sufficient area to support the cell's
metabolic needs
, E) rigid cell walls that limit cell size expansion - CORRECT ANS: D
13) Which of the following statements concerning bacteria and archaea
cells is correct?
A) Archaea cells contain small membrane-enclosed organelles; bacteria
do not.
B) Archaea cells contain a membrane-bound nucleus; bacteria do not.
C) DNA is present in both archaea cells and bacteria cells.
D) DNA is present in the mitochondria of both bacteria and archaea cells.
- CORRECT ANS: A
14) The evolution of eukaryotic cells most likely involved
A) endosymbiosis of an aerobic bacterium in a larger host cell-the
endosymbiont evolved into mitochondria.
B) anaerobic archaea taking up residence inside a larger bacterial host
cell to escape toxic oxygen-the anaerobic bacterium evolved into
chloroplasts.
C) an endosymbiotic fungal cell evolved into the nucleus.
D) acquisition of an endomembrane system, and subsequent evolution of
mitochondria from a portion of the Golgi. - CORRECT ANS: A
15) Prokaryotes are classified as belonging to two different domains.
What are the domains?
A) Bacteria and Eukarya
B) Bacteria and Archaea
C) Archaea and Protista
D) Bacteria and Protista
E) Bacteria and Fungi - CORRECT ANS: B
16) If radioactive deoxythymidine triphosphate (dTTP) is added to a
culture of rapidly growing bacterial cells, where in the cell would you
expect to find the greatest concentration of radioactivity?
A) nucleus
B) cytoplasm
C) endoplasmic reticulum
D) nucleoid
E) ribosomes - CORRECT ANS: D
17) Which organelle or structure is absent in plant cells?
A) mitochondria
B) Golgi vesicles
C) microtubules
D) centrosomes
graded A + 2025/2026 .RATED 100%
CORRECT
1) When biologists wish to study the internal ultrastructure of cells, they
can achieve the finest resolution by using
A) a phase-contrast light microscope.
B) a scanning electron microscope.
C) a transmission electronic microscope.
D) a confocal fluorescence microscope.
E) a super-resolution fluorescence microscope. - CORRECT ANS: C
2) The advantage of light microscopy over electron microscopy is that
A) light microscopy provides for higher magnification than electron
microscopy.
B) light microscopy provides for higher resolving power than electron
microscopy.
C) light microscopy allows one to view dynamic processes in living cells.
D) light microscopy provides higher contrast than electron microscopy.
E) specimen preparation for light microcopy does not produce artifacts. -
CORRECT ANS: c
3) A primary objective of cell fractionation is to
A) view the structure of cell membranes.
B) sort cells based on their size and weight.
C) determine the size of various organelles.
D) separate the major organelles so that their particular functions can be
determined.
E) separate lipid-soluble from water-soluble molecules. - CORRECT
ANS: CORRECT ANS: D
4) In the fractionation of homogenized cells using centrifugation, the
primary factor that determines whether a specific cellular component
ends up in the supernatant or the pellet is
A) the relative solubility of the component.
B) the size and weight of the component.
C) the percentage of carbohydrates in the component.
D) the presence or absence of nucleic acids in the component.
,E) the presence or absence of lipids in the component. - CORRECT ANS:
B
5) Which of the following correctly lists the order in which cellular
components will be found in the pellet when homogenized cells are
treated with increasingly rapid spins in a centrifuge?
A) ribosomes, nucleus, mitochondria
B) chloroplasts, ribosomes, vacuoles
C) nucleus, ribosomes, chloroplasts
D) vacuoles, ribosomes, nucleus
E) nucleus, mitochondria, ribosomes - CORRECT ANS: E
6) Green fluorescent protein (GFP) can be used to fluorescently label a
specific protein in cells by genetically engineering cells to synthesize the
target protein fused to GFP. What is the advantage of using GFP fusions
to visualize specific proteins, instead of staining cells with fluorescently
labeled probes that bind to the target protein?
A) GFP fusions enable one to track changes in the location of the protein
in living cells; staining usually requires preserved cells.
B) GFP fusions enable higher resolution than staining with fluorescent
probes.
C) GFP permits the position of the protein in the cell more precisely than
fluorescent probes.
D) GFP permits visualization of protein-protein interactions; fluorescent
probes do not.
E) GFP fusions are not subject to artifacts; fluorescent probes may
introduce background artifacts. - CORRECT ANS: A
7) What is the reason that a modern electron microscope (TEM) can
resolve biological images to the sub nanometer level, as opposed to tens
of nanometers achievable for the best super-resolution light microscope?
A) The focal length of the electron microscope is significantly longer.
B) Contrast is enhanced by staining with atoms of heavy metal.
C) Electron beams have much shorter wavelengths than visible light.
D) The electron microscope has a much greater ratio of image size to real
size.
E) The electron microscope cannot image whole cells at one time. -
CORRECT ANS: C
8) What technique would be most appropriate to use to observe the
movements of condensed chromosomes during cell division?
A) light microscopy
B) scanning electron microscopy
,C) transmission electron microscopy
D) confocal fluorescence microscopy
E) super-resolution fluorescence microscopy - CORRECT ANS: A
9) All of the following are part of a prokaryotic cell except
A) DNA.
B) a cell wall.
C) a plasma membrane.
D) ribosomes.
E) an endoplasmic reticulum. - CORRECT ANS: E
10) The volume enclosed by the plasma membrane of plant cells is often
much larger than the corresponding volume in animal cells. The most
reasonable explanation for this observation is that
A) plant cells are capable of having a much higher surface-to-volume
ratio than animal cells.
B) plant cells have a much more highly convoluted (folded) plasma
membrane than animal cells.
C) plant cells contain a large vacuole that reduces the volume of the
cytoplasm.
D) animal cells are more spherical, whereas plant cells are elongated.
E) plant cells can have lower surface-to-volume ratios than animal cells
because plant cells synthesize their own nutrients. - CORRECT ANS: C
11) A mycoplasma is an organism with a diameter between 0.1 and 1.0
µm. What does the organism's size tell you about how it might be
classified?
A) It must be a single-celled protist.
B) It must be a single-celled fungus.
C) It could be almost any typical bacterium.
D) It could be a typical virus.
E) It could be a very small bacterium. - CORRECT ANS: E
12) Which of the following is a major cause of the size limits for certain
types of cells?
A) limitation on the strength and integrity of the plasma membrane as cell
size increases
B) the difference in plasma membranes between prokaryotes and
eukaryotes
C) evolutionary progression in cell size; more primitive cells have smaller
sizes
D) the need for a surface area of sufficient area to support the cell's
metabolic needs
, E) rigid cell walls that limit cell size expansion - CORRECT ANS: D
13) Which of the following statements concerning bacteria and archaea
cells is correct?
A) Archaea cells contain small membrane-enclosed organelles; bacteria
do not.
B) Archaea cells contain a membrane-bound nucleus; bacteria do not.
C) DNA is present in both archaea cells and bacteria cells.
D) DNA is present in the mitochondria of both bacteria and archaea cells.
- CORRECT ANS: A
14) The evolution of eukaryotic cells most likely involved
A) endosymbiosis of an aerobic bacterium in a larger host cell-the
endosymbiont evolved into mitochondria.
B) anaerobic archaea taking up residence inside a larger bacterial host
cell to escape toxic oxygen-the anaerobic bacterium evolved into
chloroplasts.
C) an endosymbiotic fungal cell evolved into the nucleus.
D) acquisition of an endomembrane system, and subsequent evolution of
mitochondria from a portion of the Golgi. - CORRECT ANS: A
15) Prokaryotes are classified as belonging to two different domains.
What are the domains?
A) Bacteria and Eukarya
B) Bacteria and Archaea
C) Archaea and Protista
D) Bacteria and Protista
E) Bacteria and Fungi - CORRECT ANS: B
16) If radioactive deoxythymidine triphosphate (dTTP) is added to a
culture of rapidly growing bacterial cells, where in the cell would you
expect to find the greatest concentration of radioactivity?
A) nucleus
B) cytoplasm
C) endoplasmic reticulum
D) nucleoid
E) ribosomes - CORRECT ANS: D
17) Which organelle or structure is absent in plant cells?
A) mitochondria
B) Golgi vesicles
C) microtubules
D) centrosomes
