campbell biology 9th edition test bank

campbell biology 9th edition test bank

This is part of the test bank for CH6 of campbell biology 9th edition test bank

 

Campbell's Biology, 9e (Reece et al.)

Chapter 6   A Tour of the Cell

 

This chapter introduces the topics of microscopy and cell fractionation, followed by a review of the cell and the major organelles and structures of eukaryotic cells. A challenge with this chapter is to keep this data from simply being a list of parts. In addition to the structure and function of individual organelles, questions probe student understanding of the cell as a dynamic, interconnected system: the flow of membrane and proteins in the endomembrane system to the plasma membrane; the flow of information from the nucleus to the cytoplasm; and the connection between the cytoskeleton, the plasma membrane, and the extracellular matrix. An evolutionary perspective goes beyond structural distinctions between prokaryotes and eukaryotes to examine theories concerning the evolutionary origins of eukaryotes and key eukaryotic cell structures.

 

Multiple-Choice Questions

 

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.

Answer:  C

Topic:  Concept 6.1

Skill:  Knowledge/Comprehension

 

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.

Answer:  C

Topic:  Concept 6.1

Skill:  Knowledge/Comprehension

 

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.

Answer:  D

Topic:  Concept 6.1

Skill:  Knowledge/Comprehension

 

 

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.

Answer:  B

Topic:  Concept 6.1

Skill:  Knowledge/Comprehension

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

Answer:  E

Topic:  Concept 6.1

Skill:  Application/Analysis

 

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.

Answer:  A

Topic:  Concept 6.1

Skill:  Application/Analysis

 

7) What is the reason that a modern electron microscope (TEM) can resolve biological images to the subnanometer 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.

Answer:  C

Topic:  Concept 6.1

Skill:  Application/Analysis

 

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

Answer:  A

Topic:  Concept 6.1

Skill:  Synthesis/Evaluation

 

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.

Answer:  E

Topic:  Concept 6.2

Skill:  Knowledge/Comprehension

 

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.

Answer:  C

Topic:  Concept 6.2

Skill:  Synthesis/Evaluation

 

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.

Answer:  E

Topic:  Concept 6.2

Skill:  Application/Analysis

 

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

Answer:  D

Topic:  Concept 6.2

Skill:  Knowledge/Comprehension

 

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.

Answer:  A

Topic:  Concept 6.2

Skill:  Knowledge/Comprehension

 

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.

Answer:  A

Topic:  Concept 6.2

Skill:  Knowledge/Comprehension

 

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

Answer:  B

Topic:  Concept 6.2

Skill:  Knowledge/Comprehension

 

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

Answer:  D

Topic:  Concept 6.2

Skill:  Application/Analysis

 

17) Which organelle or structure is absent in plant cells?

A) mitochondria

B) Golgi vesicles

C) microtubules

D) centrosomes

E) peroxisomes

Answer:  D

Topic:  Concept 6.2

Skill:  Knowledge/Comprehension

 

18) Large numbers of ribosomes are present in cells that specialize in producing which of the following molecules?

A) lipids

B) glycogen

C) proteins

D) cellulose

E) nucleic acids

Answer:  C

Topic:  Concept 6.3

Skill:  Knowledge/Comprehension

 

19) The nuclear lamina is an array of filaments on the inner side of the nuclear membrane. If a method were found that could cause the lamina to fall into disarray, what would you expect to be the most likely consequence?

A) the loss of all nuclear function

B) the inability of the nucleus to divide during cell division

C) a change in the shape of the nucleus

D) failure of chromosomes to carry genetic information

E) inability of the nucleus to keep out destructive chemicals

Answer:  C

Topic:  Concept 6.3

Skill:  Synthesis/Evaluation

 

 

Full documents can be accessed at http://test-bank.us/Test-Bank-for-Campbell-Biology-9th-Edition-by-Reece