Test Bank For Becker World of the Cell 9Th Ed By Jeff Hardin, Gregory Paul Bertoni

Test Bank For Becker World of the Cell 9Th Ed By Jeff Hardin, Gregory Paul Bertoni

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WITH ANSWERS

 

Beckers World of the Cell 9Th Ed By Jeff Hardin, Gregory Paul Bertoni

Test Bank

 

SAMPLE QUESTIONS

 

Beckers The World of the Cell, 9e (Hardin/Bertoni/Kleinsmith)

Chapter 1   A Preview of Cell Biology

 

1.1   Multiple-Choice Questions

 

1) Robert Hooke coined the term cell when studying thin slices of cork. These ________ were the first cells observed because ________.

1. A) dead plant cells; the thick cells walls did not require high resolution or magnification to view
2. B) dead animal cells; they were immobile and did not need to be fixed before viewing
3. C) compartments; they were actually the result of multiple cells that had merged and died to form large compartments that were easy to view
4. D) immune cells; they produce antibodies that embed in the cell membrane to make it visible
5. E) little rooms; they were 100 nm in diameter, much larger than most plant cells

Answer:  A

Chapter Section:  1.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.1

Global LO:  G1

 

2) The Latin phrase omnis cellula e cellula refers to a cellular principle. Which of the following statements is the best interpretation of this phrase?

1. A) Tissues are composed of similar cells.
2. B) Cells generally are found in clusters.
3. C) All cells arise only from preexisting cells.
4. D) Organs are composed of tissues and cells.
5. E) The cell is the basic unit of structure.

Answer:  C

Chapter Section:  1.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.1

Global LO:  G7

 

3) ________ improved the original light microscope in the late 1600s, allowing the visualization of ________.

1. A) Theodor Schwann; the internal structures of cells, such as ribosomes, nuclei, and golgi bodies
2. B) Robert Hooke; bacteria and viruses
3. C) Antonie van Leeuwenhoek; sperm cells, bacteria, algae, and other protists
4. D) Robert Brown; cell structures using fluorescent antibodies
5. E) Rudolf Virchow; collagen and muscle cells

Answer:  C

Chapter Section:  1.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

 

4) Which organelle stores most of the DNA in plant and animal cells?

1. A) Golgi complex
2. B) mitochondrion
3. C) chloroplast
4. D) nucleus
5. E) lysosome

Answer:  D

Chapter Section:  1.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.2

Global LO:  G1

 

5) Which of the following statements is false?

1. A) All organisms consist of one or more cells.
2. B) All cells arise from preexisting cells.
3. C) The cell is the basic unit of structure for all organisms.
4. D) All cells have a membrane-bound nucleus.
5. E) Cells come in a wide variety of sizes and shapes.

Answer:  D

Chapter Section:  1.2

Blooms Taxonomy:  Analysis

Learning Outcome:  1.1

Global LO:  G7

 

6) Which of the following is true of a nanometer?

1. A) A nanometer is about the size of a common bacterial cell.
2. B) A nanometer is one millionth of a meter.
3. C) A nanometer is equivalent to 10 Angstroms ().
4. D) The nanometer is the most common measurement used in measuring whole cells.
5. E) None of the above.

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.2

Global LO:  G4

 

7) Which of the following is closest to a micrometer in size?

1. A) the width of a strand of DNA
2. B) the length of a plant cell
3. C) the length of a chicken egg
4. D) a typical prokaryotic cell
5. E) the size of a ribosome

Answer:  D

Chapter Section:  1.2

Blooms Taxonomy:  Application

Learning Outcome:  1.2

Global LO:  G4

8) Cell biology emerged from which of the following fields of biology?

1. A) biochemistry
2. B) cytology
3. C) genetics
4. D) biochemistry, cytology, and genetics
5. E) cytology and biochemistry

Answer:  D

Chapter Section:  1.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.1

Global LO:  G1

 

9) Which of the following is smallest?

1. A) ribosome
2. B) virus
3. C) protein
4. D) mitochondrion
5. E) prokaryote

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Application

Learning Outcome:  1.2

Global LO:  G4

 

10) Early microscopes did not allow clear visualization of cells because they were limited by

1. A) magnification.
2. B) number of kernels.
3. C) resolution.
4. D) refraction.
5. E) both magnification and resolution.

Answer:  E

Chapter Section:  1.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.2

Global LO:  G4

 

 

11) You are working on a project that involves the direct observation of DNA molecules. The microscope that would give you the best information at this time would be the

1. A) light microscope.
2. B) phase-contrast microscope.
3. C) transmission electron microscope.
4. D) digital video microscope.
5. E) fluorescent microscope.

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Analysis

Learning Outcome:  1.2

Global LO:  G4

12) The limit of resolution can best be defined as

1. A) the distance that an object must be moved to be distinguished from its background.
2. B) the inverse of the wavelength of light; it is greatest for black light.
3. C) the distance that two objects must be apart to be distinguished as separate objects.
4. D) the solvent that must be available to remix a solution.
5. E) the magnification power of a microscope.

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

13) How does brightfield microscopy allow images to be visualized?

1. A) Specimens are illuminated with white light.
2. B) Electrons strike the specimen being examined.
3. C) Specimens are fixed and have bright fluorescent molecules attached to them.
4. D) Specimens are illuminated with blue light to visualize internal features of cells smaller than 100 nm.
5. E) Specimens are viewed under phased light to improve magnification.

Answer:  A

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G7

 

 

14) Which of the following is an application of immunofluorescence microscopy?

1. A) Visualization of the natural fluorescence of a specimen under UV light.
2. B) Identification of specific components of the immune system.
3. C) Identifying which organelle or cellular compartment contains a particular protein.
4. D) Visualization of the surface structures of a specimen.
5. E) Construction of three-dimensional images of structures smaller than 10 nm.

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Application

Learning Outcome:  1.2

Global LO:  G1

 

15) Which type of microscopy enhances and amplifies slight changes in the phase of transmitted light?

1. A) differential interference contrast microscopy
2. B) digital video microscopy
3. C) fluorescence microscopy
4. D) phase-contrast microscopy
5. E) both differential interference contrast microscopy and phase-contrast microscopy

Answer:  E

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

16) Which type of microscopy has the greatest resolving power?

1. A) electron microscopy
2. B) phase-contrast microscopy
3. C) fluorescence microscopy
4. D) digital video microscopy
5. E) confocal scanning microscopy

Answer:  A

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

 

17) Which of the following can only be viewed by electron microscopy?

1. A) frog eggs
2. B) DNA
3. C) nuclei
4. D) mitochondria
5. E) prokaryotes

Answer:  B

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

18) Which of the following types of light microscopy improves the resolution of thick specimens by illuminating one plane of the specimen at a time?

1. A) fluorescence microscopy
2. B) phase-contrast microscopy
3. C) confocal microscopy
4. D) differential interference contrast microscopy
5. E) brightfield microscopy

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

19) A scientist is examining motile protist. He wishes to determine their direction of movement. Which of the following microscopic techniques is least likely to be used to view these cells?

1. A) light microscopy
2. B) electron microscopy
3. C) differential interference contrast microscopy
4. D) fluorescence microscopy
5. E) phase-contrast microscopy

Answer:  B

Chapter Section:  1.2

Blooms Taxonomy:  Application

Learning Outcome:  1.2

Global LO:  G2

 

20) Scanning electron microscopy (SEM) is especially suited to

1. A) observing living specimens.
2. B) examining internal cellular structure.
3. C) creating a sense of depth.
4. D) both observing living specimens and creating a sense of depth.
5. E) simultaneously observing living specimens, examining internal cellular structure, and creating a sense of depth.

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Application

Learning Outcome:  1.2

Global LO:  G2

 

21) Melvin Calvin and his colleagues used which of the following to deduce the steps in the Calvin cycle?

1. A) negative staining
2. B) Drosophila melanogaster
3. C) electron microscopy
4. D) ultracentrifugation
5. E) radioisotopes

Answer:  E

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

 

22) A microtome is used to

1. A) view microscopic organisms.
2. B) slice thin sections of specimens.
3. C) focus short wavelengths of light.
4. D) manipulate tiny objects.
5. E) dissect cellular organelles.

Answer:  B

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

 

23) The classic work of Friedrich Whler (1828) that united the fields of biology and chemistry was based on the

1. A) discovery of yeast ferments.
2. B) production of urea in the laboratory.
3. C) discovery of ATP.
4. D) identification of nucleotide bases.
5. E) analysis of gene segregation.

Answer:  B

Chapter Section:  1.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.2

Global LO:  G1

24) You wish to obtain a purified sample of mitochondria from lysed cells. The best way to obtain this sample would be

1. A) centrifugation.
2. B) chromatography.
3. C) polyacrylamide gel electrophoresis.
4. D) agarose gel electrophoresis.
5. E) both centrifugation and polyacrylamide gel electrophoresis.

Answer:  A

Chapter Section:  1.2

Blooms Taxonomy:  Evaluation

Learning Outcome:  1.2

Global LO:  G2, G7

 

25) 1 mm = ________ nm

1. A) 1,000,000
2. B) 1000
3. C) 10
4. D) 1/1000
5. E) 1/1,000,000

Answer:  A

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

 

26) The outcome of the joining of cytology and biochemistry yielded a better understanding of the cell by

1. A) identification of cellular structures.
2. B) identification of cellular biochemical pathways.
3. C) creating bioinformatics.
4. D) identification of cellular structures and biochemical pathways.
5. E) identification of biochemical pathways and creating bioinformatics.

Answer:  D

Chapter Section:  1.2

Blooms Taxonomy:  Synthesis

Learning Outcome:  1.1

Global LO:  G1

 

27) Whler revolutionized biology through his demonstration that biological molecules are governed by the ordinary laws of physics and chemistry. He demonstrated this principle by

1. A) synthesizing urea in the laboratory from ammonium cyanate.
2. B) developing techniques for isolating, purifying, and analyzing subcomponents of cells.
3. C) defining the laws of heredity.
4. D) discovering active agents in cell extracts that were specific biological catalysts that have since come to be called enzymes.
5. E) inventing mass spectrometry which is commonly used to determine the size and composition of individual proteins.

Answer:  A

Chapter Section:  1.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.1

Global LO:  G1

28) Gregor Mendel was most influential in which field of biology?

1. A) genetics
2. B) chromatography
3. C) biochemistry
4. D) prokaryotic transformation
5. E) cytology

Answer:  A

Chapter Section:  1.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.1

Global LO:  G1

 

 

29) The scientific work that established DNA, rather than protein, as the molecule of heredity was performed prior to

1. A) Mendels work on heredity.
2. B) the elucidation of the double helix structure of DNA.
3. C) Antonie van Leeuwenhoeks observation of internal cell structures.
4. D) the description of the enzymatic steps of glycolysis.
5. E) the formation of the chromosome theory of heredity.

Answer:  B

Chapter Section:  1.2

Blooms Taxonomy:  Synthesis

Learning Outcome:  1.1

Global LO:  G1

 

30) Jacques Monod and Franois Jacob deduced the mechanism responsible for the regulation of prokaryotic gene expression. They are, therefore, responsible for launching the era of

1. A) the scientific method.
2. B) molecular genetics.
3. C) biochemistry.
4. D) light microscopy.
5. E) radioisotopes.

Answer:  B

Chapter Section:  1.2

Blooms Taxonomy:  Synthesis

Learning Outcome:  1.1

Global LO:  G1

 

31) Which of the following biochemical techniques uses an electrical field to separate macromolecules based on their mobility through a semisolid gel?

1. A) light microscopy.
2. B) ultracentrifugation.
3. C) chromatography.
4. D) electrophoresis.
5. E) mass spectrometry.

Answer:  D

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

 

32) To which of the following do Mendels observations relate?

1. A) thermodynamics
2. B) gravity
3. C) ideal gas laws
4. D) heredity
5. E) diffusion

Answer:  D

Chapter Section:  1.2

Blooms Taxonomy:  Synthesis

Learning Outcome:  1.1

Global LO:  G1

 

33) The steps of the scientific method, in the correct order, are

1. A) design experiments, draw conclusions, collect data, interpret results, make observations, and test the hypothesis.
2. B) make observations, formulate the hypothesis, design experiments, collect data, interpret results, and draw conclusions.
3. C) collect data, interpret results, test the hypothesis, design experiments, make observations, and draw conclusions.
4. D) collect data, interpret results, test the hypothesis, make observations, and design experiments.
5. E) none of the above.

Answer:  B

Chapter Section:  1.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.3

Global LO:  G1

 

34) Scientists use various terms to describe conclusions reached through the scientific method. Which of the following terms conveys the least degree of certainty?

1. A) theory
2. B) hypothesis
3. C) law
4. D) both hypothesis and theory
5. E) both theory and law

Answer:  B

Chapter Section:  1.3

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.3

Global LO:  G1

 

 

35) Once a scientific theory becomes a law, it

1. A) cannot be changed.
2. B) cannot be challenged.
3. C) becomes static.
4. D) is subject to modification.
5. E) is irrefutable.

Answer:  D

Chapter Section:  1.3

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.3

Global LO:  G1

36) You are studying the response of macrophages infected with the intracellular bacterium Brucella, specifically by examining which gene products are being expressed. You would be studying the macrophage ________ to obtain this information.

1. A) proteome
2. B) genome
3. C) transciptome
4. D) amplicon
5. E) metabolome

Answer:  C

Chapter Section:  1.2

Blooms Taxonomy:  Application

Learning Outcome:  1.2

Global LO:  G1

 

37) Which of the following is an important characteristic for a model organism?

1. A) marginally characterized
2. B) difficult to manipulate in the laboratory
3. C) prone to random changes that alter primary characteristics
4. D) widely studied
5. E) all of the above

Answer:  D

Chapter Section:  1.3

Blooms Taxonomy:  Comprehension

Learning Outcome:  1.3

Global LO:  G1

 

 

38) All of the following are model organisms, except

1. A) Saccharomyces cerevisiae.
2. B) Drosophila melanogaster.
3. C) Caenorhabditis elegans.
4. D) Arabidopsis thialana.
5. E) Homo sapiens.

Answer:  E

Chapter Section:  1.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.3

Global LO:  G1

 

39) In studying osteoporosis in humans, you wish to test a newly designed treatment for efficacy. Your best choice for a model organism would be

1. A) Escherichia coli.
2. B) Mus musculus.
3. C) Caenorhabditis elegans.
4. D) Arabidopsis thaliana.
5. E) Pisum sativum.

Answer:  B

Chapter Section:  1.3

Blooms Taxonomy:  Application

Learning Outcome:  1.3

Global LO:  G1

40) Which of the following is mismatched?

1. A) Escherichia coli genetics
2. B) Drosophila melanogaster embryogenesis
3. C) Mus musculus immunology
4. D) Caenorhabditis elegans photosynthesis
5. E) Arabidopsis thaliana plant gene function

Answer:  D

Chapter Section:  1.3

Blooms Taxonomy:  Analysis

Learning Outcome:  1.3

Global LO:  G1

 

1.2   Matching Questions

 

Match each scientist or group of scientists on the left with the appropriate phrase to the right.

 

1. A) fruit fly
2. B) DNA double helix
3. C) cell theory
4. D) transfer RNA
5. E) Calvin cycle
6. F) one geneone enzyme
7. G) transformation
8. H) translation
9. I) chromosome theory of heredity
10. J) embryonic bacteria
11. K) ferments of yeast
12. L) oral prokaryotes
13. M) urea
14. N) hereditary factors
15. O) pollen grain
16. P) dog saliva
17. Q) transcription

 

1) Gregor Mendel

Chapter Section: 1.2

Blooms Taxonomy: Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

2) Walter Sutton

Chapter Section: 1.2

Blooms Taxonomy: Knowledge

Learning Outcome: 1.1

Global LO:  G1

 

3) Matthias Schleiden

Chapter Section:  1.2

Blooms Taxonomy: Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

4) Oswald Avery, Colin MacLeod, and Maclyn McCarty

Chapter Section:  1.2

Blooms Taxonomy: Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

5) George Beadle and Edward Tatum

Chapter Section: 1.2

Blooms Taxonomy: Knowledge

Learning Outcome: 1.1

Global LO:  G1

 

6) James Watson and Francis Crick

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

7) Thomas Hunt Morgan

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

8) Friedrich Whler

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

9) Louis Pasteur

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

Answers: 1) N 2) I 3) C 4) G 5) F 6) B 7) A 8) M 9) K

 

Match the type of microscopy with the appropriate characteristic.

 

1. A) amplifies variations in density
2. B) light passes directly through specimen
3. C) detects electrons deflected from the surface of the specimen
4. D) shows specific molecules
5. E) uses a laser to view a single plane of a specimen
6. F) detects electrons passing through a specimen

 

10) brightfield

Chapter Section: 1.2

Blooms Taxonomy: Knowledge

Learning Outcome: 1.2

Global LO:  G4

 

11) fluorescence

Chapter Section:  1.2

Blooms Taxonomy: Knowledge

Learning Outcome:  1.2

Global LO: G4

 

12) phase-contrast

Chapter Section: 1.2

Blooms Taxonomy: Knowledge

Learning Outcome: 1.2

Global LO:  G4

 

13) confocal

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

14) transmission electron microscopy

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

15) scanning electron microscopy

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

Answers: 10) B 11) D 12) A 13) E 14) F 15) C

 

1.3   Short Answer Questions

 

1) To be useful to scientists, a hypothesis must be ________; in other words, the hypothesis must be able to be confirmed or discredited.

Answer:  testable

Chapter Section:  1.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.3

Global LO:  G1

 

2) A scientific ________ must be so thoroughly confirmed that virtually no doubt remains about its accuracy.

Answer:  law

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.3

Global LO:  G1

 

3) Glycolysis is also called the ________ pathway after the scientists who did most of the work to define it.

Answer:  Embden-Meyerhof

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

4) ________ synthesized urea in the laboratory from inorganic starting materials. Much of what is now called ________ dates from this discovery.

Answer:  Friedrich Whler; biochemistry

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

5) Melvin Calvin used ________, a specific ________, to deduce the Calvin cycle of photosynthesis.

Answer:  14C; radioisotope

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

 

6) A(n) ________ is an instrument used to separate subcellular structures and macromolecules on the basis of size, shape, and density. ________ developed this instrument in Sweden during the period 19251930.

Answer:  ultracentrifuge; Theodor Svedberg

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

7) Around 1914, ________ determined that DNA was an important component in ________ by using a staining technique that is still in use today.

Answer:  Robert Feulgen; chromosomes

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

8) Because of the low penetration power of electrons, samples for transmission electron microscopy must be extremely thin. A(n) ________ is able to cut sections as thin as 20 nm.

Answer:  ultramicrotome

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

9) In 1880, Walther Flemming identified ________, threadlike bodies seen in dividing cells.

Answer:  chromosomes

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1

 

10) The ________ was developed in the late 1920s by Theodore Svedberg. He originally used it to determine the sedimentation rate of proteins.

Answer:  ultracentrifuge

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

 

11) ________ is a biochemical technique that allows one to separate biological molecules based on size, shape, and/or affinity for specific molecules or functional groups.

Answer:  Chromatography

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

 

12) The total protein content of the cell is called the ________.

Answer:  proteome

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G1

 

13) ________ is the ability to distinguish two objects that are close together as separate. In any microscope, this ability is determined by ________.

Answer:  Resolution; wavelength

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.2

Global LO:  G4

 

1.4   Inquiry

 

Scientific discoveries have had great impact in human history. The people who make these discoveries and the circumstances that surround these discoveries are very important to our understanding of science. Can you identify the individuals as they might have described themselves?

 

1. A) Antonie van Leeuwenhoek
2. B) Melvin Calvin
3. C) Alfred Hershey and Martha Chase
4. D) Theodor Svedberg
5. E) Friedrich Whler
6. F) Robert Hooke
7. G) James Watson and Francis Crick

 

1) I am a seventeenth-century shopkeeper from Holland. My hobby involves hand-polishing glass to make lenses, some of which can magnify almost 300-fold. I was the first to observe living cells and am known as the Father of Microbiology.

Chapter Section: 1.2

Blooms Taxonomy: Knowledge

Learning Outcome: 1.1

Global LO:  G1, G7

 

2) I was the Curator of Instruments for the Royal Society of London in 1665. I developed a microscope that could magnify around 30-fold. I examined plant material and observed many small chambers that I called cells.

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1, G7

 

3) At the University of California, Berkeley, I worked with radioisotopes. In the late 1940s and early 1950s, I used 14C to identify the most common pathway for photosynthetic carbon metabolism.

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1, G7

4) We worked out the double helix model of DNA structure in 1953. We later received the Nobel Prize for this work.

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1, G7

 

 

5) I am a nineteenth-century German chemist. By synthesizing an organic molecule from inorganic components, I dispelled the idea that biological processes were exempt from the laws of chemistry.

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1, G7

 

6) My colleague and I worked with bacterial viruses. We were able to demonstrate that DNAnot proteinwas the genetic material of the cell.

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1, G7

 

7) I am a Swedish scientist. I developed the ultracentrifuge to determine sedimentation rates of proteins. The ultracentrifuge was later used to isolate subcellular fractions.

Chapter Section:  1.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  1.1

Global LO:  G1, G7

 

Answers: 1) A 2) F 3) B 4) G 5) E 6) C 7) D

 

 

8) The following paragraph describes the activities of hypothetical scientists. After reading this paragraph, list the steps of the scientific method, and list the activities that correspond to the steps of the scientific method.

 

A rancher noticed that several grazing animals had become sick after grazing in a new area. The rancher asked a team of scientists to analyze this problem. They visited the area and found that the food available to the animals was similar to the food they had been eating. The water supply in the area was adequate but limited to a single spring. Some of the scientists felt that the water might be contaminated with a pathogen. Therefore, they collected water samples from the spring in the new area and compared them with water samples taken from previous grazing sites. The scientists noticed that water from the new area was cloudier than water obtained from other areas. Culturing this water revealed that a pathogenic strain of bacteria was present. This bacterial strain was found to be identical to a strain obtained from sick animals. This strain was not present in healthy animals. They concluded that a contaminated water supply in the new area was responsible for the problem and instructed the rancher to avoid the water supply. The disease was not found in the ranchers livestock again.

Answer:  (Answers may vary.)

Observation. The rancher and the scientists made initial observations regarding the food and water that the livestock consumed.

Hypothesis. The water supply was contaminated with a pathogen.

Experimentation. Water was collected, examined, and cultured.

Collect data. The turbidity of the water was examined. The cultures were positive for a pathogenic strain of bacterium.

Interpret results. The data was compared to other water samples. The cultures were compared to those obtained from livestock.

Draw conclusion. The water was contaminated and responsible for the outbreak.

Chapter Section:  1.3

Blooms Taxonomy:  Evaluation

Learning Outcome:  1.3

Global LO:  G1

 

 

9) A number of different types of microscopy exist. Each type of microscopy has advantages and disadvantages. Can you identify the microscope that would be most advantageous for the situations below?

1. A cell biologist wishes to visualize the ribosomes of a cell.
2. A microbiologist wishes to examine the motility of a bacterium.
3. An immunologist wishes to determine if a lymphocyte possesses a certain surface protein.
4. A virologist is trying to determine the three-dimensional shape of a virus.
5. A pathologist is trying to examine the cytoplasm of a cell for changes that result from viral infection.

Answer:  (Answers may vary.)

1. Electron microscopy, preferably transmission electron microscopy, should be used.
2. Phase contrast or differential-interference-contrast would be most helpful.
3. Fluorescence microscopy is often used.
4. Scanning electron microscopy should be used.
5. Transmission electron microscopy will enable the pathologist to visualize the interior.

Chapter Section:  1.2

Blooms Taxonomy:  Evaluation

Learning Outcome:  1.2

Global LO:  G4

10) You have identified a new molecule associated with the immune system that drastically reduces cell division by tumor cells in vivo. Develop a hypothesis and design an experiment to test your hypothesis using a model organism. Include an explanation as to why it is the best model for your experiment.

Answer:  Answers will vary; however, the hypothesis would indicate the utility of the cytokine for lymphoma treatment. The obvious model organism would be the mouse model. It shares a great many similarities to humans at the cellular, anatomical, and physiological levels. It is well characterized, and the genome has been sequenced. Further, there is a mouse model of lymphoma currently available. Mice are easy to care for and require a relatively small amount of space to maintain.

Chapter Section:  1.3

Blooms Taxonomy:  Synthesis

Learning Outcome:  1.3

Global LO:  G1

 

 

11) You have been given a sample of Mimivirus, which has the largest capsid diameter of all currently known viruses (600 nm) and has the form of a 20-sided polyhedron (an icosahedron). Based on your knowledge of microscopes, what would you be able to see/determine about mimiviral structure using each of the following microscopes?

1. simple compound (light) microscope
2. fluorescent microscope using fluorescently labeled antibodies to a novel capsid protein
3. scanning electron microscope

Answer:

1. Light microscope: will be able to see basic viral shape, especially if particles are stained
2. Fluorescent microscope: should illuminate the outside of the viral particles
3. Scanning electron microscope: would allow imaging of the surface structure of the virus

Chapter Section:  1.2

Blooms Taxonomy:  Analysis

Learning Outcome:  1.2

Global LO:  G4

Beckers The World of the Cell, 9e (Hardin/Bertoni/Kleinsmith)

Chapter 3   The Macromolecules of the Cell

 

3.1   Multiple-Choice Questions

 

1) What are the three general types of amino acids?

1. A) helices, sheets, and looped segments
2. B) covalent, noncovalent, and van der Waals forces
3. C) positive, negative, and noncharged
4. D) hydrophobic, polar (noncharged), polar (charged)
5. E) acidic, basic, and neutral

Answer:  D

Chapter Section:  3.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

2) You are researching a cytoplasmic protein associated with a nerve disorder. The native form of the enzyme appears to be globular protein; however, when a sample of the purified protein is treated with a chemical that reduces disulfide bonds, the enzymatic activity decreases dramatically and multiple globular proteins can be detected in the sample. What does this tell you about the protein?

1. A) The primary structure of the protein contains multiple cysteine residues that are hydrolyzed by the chemical reductant.
2. B) The protein is most likely composed of multiple polypeptide chains that are held together by disulfide bonds.
3. C) The protein is most likely composed of helices that are held together by disulfide bonds.
4. D) The protein is most likely composed of sheets that are held together by disulfide bonds.
5. E) The primary and secondary structure of the protein depends on disulfide bonds.

Answer:  B

Chapter Section:  3.1

Blooms Taxonomy:  Evaluation

Learning Outcome:  3.1

Global LO:  G2, G5, G7

 

 

3) A peptide bond

1. A) is a covalent bond between the carboxyl carbon of one amino acid and the amino nitrogen of a second amino acid.
2. B) is a covalent bond between the functional R groups of adjacent amino acids.
3. C) is a covalent bond between the NH group of one polypeptide and the CO group of an adjacent polypeptide that holds together multimeric proteins.
4. D) is a noncovalent bond that dictates the tertiary structure of a protein.
5. E) is a covalent bond between adjacent glucose molecules in a peptide.

Answer:  A

Chapter Section:  3.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

4) Disulfide bonds are often found to stabilize which of the following levels of protein structure?

1. A) primary
2. B) secondary
3. C) tertiary
4. D) primary, secondary, and tertiary
5. E) None of these structures involve disulfide bonds.

Answer:  C

Chapter Section:  3.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G7

 

5) The primary structure of a protein

1. A) is important for determining the secondary and tertiary structure of a protein.
2. B) is simply the order of amino acids from one end of the protein to another.
3. C) is important both genetically and structurally.
4. D) is the linear sequence of amino acids that are linked together by peptide bonds.
5. E) All of these statements are true.

Answer:  E

Chapter Section:  3.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G2, G7

 

 

6) Proline is referred to as the helix breaker because

1. A) its only found in the L form, which is incompatible with helical protein structure.
2. B) it lacks the hydrogen atom needed for hydrogen bonding.
3. C) it lacks a charged functional groups for ionic bonding.
4. D) it is hydrophobic.
5. E) it has a polar functional group.

Answer:  B

Chapter Section:  3.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G2, G7

 

7) Which of the following accurately describes the structure of fibrous proteins?

1. A) Fibrous proteins usually contain a number of different domains with different structural motifs.
2. B) Fibrous proteins are composed of an equal mixture of helices and sheets with interconnecting looped segments.
3. C) Fibrous proteins have an extensive tertiary and quaternary structure that affects the strength and elasticity of each fiber.
4. D) Fibrous proteins have a simple primary structure and very little secondary structure, resulting in long, thin fibers.
5. E) Fibrous proteins are usually composed of either helices or sheets throughout the molecule, giving them a highly ordered, repetitive structure.

Answer:  E

Chapter Section:  3.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

8) Hydrogen bonding is most important in stabilizing the ________ structure of many proteins.

1. A) primary
2. B) secondary
3. C) tertiary
4. D) quaternary
5. E) primary, secondary, tertiary, and quaternary

Answer:  B

Chapter Section:  3.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G7

 

 

9) Which of the following statements is false?

1. A) There are more than 60 different kinds of amino acids present in cells.
2. B) The R group of amino acids differs from one amino acid to another.
3. C) Only around 20 amino acids are used in protein synthesis.
4. D) Equal amounts of D- and L-amino acids are found in cells.
5. E) An amino acid has an N-terminus, a C-terminus, and an R group.

Answer:  D

Chapter Section:  3.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

10) Which of the following is not a major functional class of proteins?

1. A) hereditary proteins
2. B) enzymes
3. C) motility proteins
4. D) regulatory proteins
5. E) structural proteins

Answer:  A

Chapter Section:  3.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G7

 

11) Two proteins associated with a rare neurodegenerative disorder have been sequenced. Protein A contains many polar amino acids with small regions containing nonpolar, hydrophobic amino acids. Protein B is rich in nonpolar, hydrophobic amino acids with only two small regions containing polar amino acids. What might this suggest about the two proteins?

1. A) The two proteins may have different secondary structures.
2. B) Protein A is fibrous and Protein B is globular.
3. C) Protein A may be a cytoplasmic protein and Protein B may be a membrane associate protein.
4. D) Protein A and Protein B are complementary parts of a supramolecular structure.
5. E) Protein A is most likely and enzyme and Protein B is most likely a storage protein.

Answer:  C

Chapter Section:  3.1

Blooms Taxonomy:  Evaluation

Learning Outcome:  3.1

Global LO:  G2, G5, G7

 

12) Which of the following is a possible function of a terpene?

1. A) cell surface receptor
2. B) vitamin
3. C) enzyme
4. D) motility
5. E) structure

Answer:  B

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

13) Which of the following pairs correctly matches the monomer with its polymer?

1. A) peptides; proteins
2. B) amino acids; polysaccharides
3. C) glucose; proteins
4. D) terpenes; nucleic acids
5. E) nucleotides; nucleic acids

Answer:  E

Chapter Section:  3.1, 3.2, 3.3, 3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

14) Cellulose belongs to which of the following groups of macromolecules?

1. A) lipids
2. B) carbohydrates
3. C) proteins
4. D) nucleic acids
5. E) none of these

Answer:  B

Chapter Section:  3.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

15) Which of the following has the greatest number of glycosidic bonds?

1. A) glucose
2. B) triacylglycerol
3. C) amylose
4. D) DNA
5. E) vitamin A

Answer:  C

Chapter Section:  3.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

16) To which of the following classes of sugars does glucose belong?

1. A) pentose
2. B) hexose
3. C) tetrose
4. D) heptose
5. E) triose

Answer:  B

Chapter Section:  3.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

17) You are investigating the structure of the seeds of a newly discovered tropical plant. There is storage material inside the seed. You treat the seed with peptidase (an enzyme that breaks peptide bonds), glycoside hydrolases (an enzyme that breaks glycosidic bonds), and amylase (an enzyme that breaks glycosidic bonds). Only the amylase appears to dissolve the storage material in the seed. What does this tell you about the identity of the storage material?

1. A) The seed contains fibrous proteins to store carbon and energy.
2. B) The seed contains lipids to store carbon and energy.
3. C) The seed contains starch to store carbon and energy.
4. D) The seed contains cellulose to store carbon and energy.
5. E) The seed contains globular proteins to store carbon and energy.

Answer:  C

Chapter Section:  3.3

Blooms Taxonomy:  Application

Learning Outcome:  3.1

Global LO:  G1, G2, G7

 

18) Fatty acids are ________; they function in the cell as ________.

1. A) short chains of double-bonded carbon molecules; storage lipids
2. B) short chains of double-bonded carbon molecules; vitamins and cofactors
3. C) four-ringed hydrocarbon molecules; key components of membranes
4. D) long, unbranched hydrocarbon chains with a carboxyl group at one end; building blocks for other lipids
5. E) short chains of double-bonded carbon molecules; vitamins, cofactors, and storage lipids

Answer:  D

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

19) The two strands of DNA are held together by ________; thus ________.

1. A) covalent bonds; double-stranded DNA is very stable at a range of temperatures
2. B) ionic bonds; double-stranded DNA separates into two separate strands in water
3. C) hydrogen bonds; double-stranded DNA separates into two separate strands at high temperatures
4. D) antiparallel bonds; double-stranded DNA is amphipathic
5. E) hydrophobic interactions; double-stranded DNA separates into two separate strands when dissolved in a hydrocarbon (hydrophobic) solvent

Answer:  C

Chapter Section:  3.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G7

 

20) RNA and DNA differ

1. A) in that RNA contains ribose and DNA contains deoxyribose.
2. B) in that RNA contains nucleosides and DNA contains nucleotides.
3. C) in that RNA contains uracil and DNA contains thymine.
4. D) both A and C.
5. E) All of these are correct.

Answer:  D

Chapter Section:  3.2

Blooms Taxonomy:  Analysis

Learning Outcome:  3.1

Global LO:  G7

 

21) Complementary relationships between purines and pyrimidines

1. A) allow adenine to form two hydrogen bonds with thymine (or uracil) and guanine to form three hydrogen bonds with cytosine to form double-stranded nucleic acids.
2. B) allow the interaction of the oppositely charged amino acids to form the tertiary structure of proteins.
3. C) allow adjacent bases in a nucleotide chain to stack tightly, stabilizing the DNA double helix.
4. D) provide highly ordered, repetitive bonding to form helices and sheets within proteins.
5. E) Both A and C are correct.

Answer:  A

Chapter Section:  3.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

 

22) The components of a nucleotide are

1. A) a carboxyl group, an amine group, and a variable R group.
2. B) a five-carbon sugar, a phosphate group, and a nitrogen-containing aromatic base.
3. C) a three-carbon alcohol with a hydroxyl group on each carbon and three fatty acids.
4. D) two six-carbon sugars attached with an (14) glycosidic bond.
5. E) a six-carbon sugar, an ester linkage, and a four-ringed hydrocarbon.

Answer:  B

Chapter Section:  3.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

23) The chemical nature of each amino acid is determined by which of the following groups?

1. A) amino
2. B) carboxyl
3. C) hydroxyl
4. D) R
5. E) hydrogen

Answer:  D

Chapter Section:  3.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G7

 

24) The nucleoside triphosphate molecules in DNA are linked together in the 53 by a(n) ________ bridge.

1. A) phosphate
2. B) covalent
3. C) phosphodiester
4. D) peptide
5. E) phosphatidyl

Answer:  C

Chapter Section:  3.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

 

25) The function of triglycerides is

1. A) to store energy.
2. B) to form semipermeable membranes.
3. C) to transport substances in and out of cells.
4. D) store information.
5. E) Both B and C are correct.

Answer:  A

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

26) A general trend in the structure of many biomolecules is

1. A) that the order and bonding of monomers form the basis for the secondary and tertiary structure of the polymer.
2. B) that they are all soluble in water independent of the size of the polymer.
3. C) that each class of biomolecule forms one type of secondary structure independent of the order of the monomers in the polymer.
4. D) that each class of biomolecule can form either fibrous or globular conformations depending on the chemical conditions inside the cell.
5. E) that four different monomers form the basis for the functional and structural properties of each polymer.

Answer:  A

Chapter Section:  3.1, 3.2, 3.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

27) Cholesterol is a ________, which ________.

1. A) terpene; is the basis for many animal and plant vitamins
2. B) fatty acid; functions in energy storage
3. C) steroid; is a component of eukaryotic membranes
4. D) steroid; is the basis for many animal and plant hormones
5. E) steroid; is a component of eukaryotic membranes and is the basis for many animal and plant hormones

Answer:  E

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

 

28) Which of the following is found exclusively in RNA?

1. A) thymine
2. B) guanine
3. C) uracil
4. D) adenine
5. E) cytosine

Answer:  C

Chapter Section:  3.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

29) Which of the following is true of purines?

1. A) Cytosine is a purine.
2. B) Adenines bonding to thymine is stronger than is guanines to cytosine.
3. C) Purines have a double-ringed structure.
4. D) Both adenine and thymine are purines.
5. E) Purines bind readily to deoxyribose but not to ribose.

Answer:  C

Chapter Section:  3.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1

Global LO:  G7

 

30) Which of the following contributes to the stability of the DNA double helix?

1. A) hydrophobic interactions between aromatic bases at the center of the double helix
2. B) hydrogen bonding between the phosphate and sugar groups in the backbone of the double helix
3. C) covalent bonding between complementary purine and pyrimidine bases.
4. D) ionic bonds between the negatively charged phosphate groups and the positively charged pyrimidine bases.
5. E) hydrophobic interactions between aromatic bases at the center of the double helix and ionic bonds between the negatively charged phosphate groups and the positively charged pyrimidine bases

Answer:  A

Chapter Section:  3.2

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1

Global LO:  G2, G7

 

31) Which of the following is false?

1. A) Phospholipids are important in membrane structure.
2. B) Serine is a molecule that may be part of a phosphoglyceride.
3. C) Phosphatidic acid contains two fatty acids and a phosphate group.
4. D) Sphingolipids are the predominant phospholipid in membranes.
5. E) Phospholipids are amphipathic.

Answer:  D

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

32) The term amphipathic describes the characteristic of some molecules that have

1. A) two polar regions.
2. B) only a single polar region.
3. C) both a polar and a nonpolar region.
4. D) no polar regions.
5. E) two nonpolar regions.

Answer:  C

Chapter Section:  3.1, 3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.1, 3.2

Global LO:  G7

 

33) Which of the following is not one of the six classes of lipids?

1. A) steroids
2. B) terpenes
3. C) fatty acids
4. D) triacylglycerols
5. E) pectins

Answer:  E

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

34) Which of the following statements is true?

1. A) Fatty acids are synthesized by the stepwise addition of three carbon units.
2. B) Unsaturated fatty acids are usually branched.
3. C) Fatty acids with 24 carbons are most common.
4. D) Saturated fatty acids have no double bonds between carbons.
5. E) Hormones are unsaturated fatty acids.

Answer:  D

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

35) Which of the following is true of glycolipids?

1. A) Glycolipids are usually found on the exterior surface of the plasma membrane.
2. B) Glycolipids are found in plastids and are used to store energy.
3. C) Fructose and sucrose are often part of glycolipids.
4. D) Glycolipids contain steroids.
5. E) Usually more than 10 sugar units are attached to the glycolipid.

Answer:  A

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

36) Which of the following is not a steroid?

1. A) testosterone
2. B) estradiol
3. C) cortisol
4. D) aldosterone
5. E) phenylalanine

Answer:  E

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

37) Which of the following is a terpene?

1. A) testosterone
2. B) vitamin A
3. C) glycerol
4. D) estrogen
5. E) chitin

Answer:  B

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

38) Which of the following is not a polymer of numerous monomer units?

1. A) a phospholipid
2. B) a polypeptide
3. C) an RNA molecule
4. D) cellulose
5. E) starch

Answer:  A

Chapter Section:  3.4

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.2

Global LO:  G7

39) Trans fats

1. A) are unsaturated fatty acids.
2. B) resemble saturated fatty acids in shape.
3. C) are associated with an increased risk of heart disease.
4. D) are present in small amounts in meat and dairy products.
5. E) All of these statements are true.

Answer:  E

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

40) Lipid rafts are

1. A) regions of the membrane that are high in sphingolipids, which facilitate communication with the external environment of the cell.
2. B) important regions of membrane structure comprised of phospholipids.
3. C) regions not typically associated with signal transduction.
4. D) regions where greater concentrations of sphingolipids are on the inner side of the membrane.
5. E) rafts of lipids inside of the cell that serve to store energy.

Answer:  A

Chapter Section:  3.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  3.2

Global LO:  G7

 

3.2   Matching Questions

 

Match the choice on the left with the choice on the right.

 

1. A) spatial structure
2. B) linear structure
3. C) AMP
4. D) monosaccharide
5. E) adenine
6. F) quaternary structure
7. G) starch
8. H) cytosine
9. I) collagen
10. J) amino acid
11. K) enzyme
12. L) steroids
13. M) guanine

 

1) monomer of protein

Chapter Section: 3.1, 3.2, 3.3

Blooms Taxonomy: Analysis

Learning Outcome:  3.1

Global LO:  G2, G7

 

2) polymer of glucose

Chapter Section: 3.1, 3.2, 3.3

Blooms Taxonomy: Analysis

Learning Outcome: 3.1

Global LO:  G2, G7

 

3) pyrimidine

Chapter Section:  3.1, 3.2, 3.3

Blooms Taxonomy: Analysis

Learning Outcome:  3.1

Global LO:  G2, G7

 

4) nucleotide

Chapter Section:  3.1, 3.2, 3.3

Blooms Taxonomy: Analysis

Learning Outcome:  3.1

Global LO:  G2, G7

 

5) multimeric complex

Chapter Section:  3.1, 3.2, 3.3

Blooms Taxonomy:  Analysis

Learning Outcome:  3.1

Global LO:  G2, G7

 

6) Haworth projection

Chapter Section:  3.1, 3.2, 3.3

Blooms Taxonomy:  Analysis

Learning Outcome:  3.1

Global LO:  G2, G7

 

7) a fibrous protein

Chapter Section:  3.1, 3.2, 3.3

Blooms Taxonomy:  Analysis

Learning Outcome:  3.1

Global LO:  G2, G7

 

Answers: 1) L 2) G 3) H 4) C 5) F 6) A 7) I

 

List all the functions that match with each biomolecule. Note that the functions may match with more than one biomolecule and each biomolecule may have multiple functions.

 

1. A) carbon and energy storage
2. B) enzyme cofactor
3. C) hormone
4. D) enzyme
5. E) vitamin
6. F) structural component of cells
7. G) cell receptor
8. H) informational molecule
9. I) motility

 

8) protein

Chapter Section:  3.1, 3.2, 3.3, 3.4

Blooms Taxonomy: Comprehension

Learning Outcome: 3.1, 3.2

Global LO:  G2, G7

 

9) DNA

Chapter Section:  3.1, 3.2, 3.3, 3.4

Blooms Taxonomy: Comprehension

Learning Outcome: 3.1, 3.2

Global LO:  G2, G7

 

10) lipids and steroids

Chapter Section:  3.1, 3.2, 3.3, 3.4

Blooms Taxonomy: Comprehension

Learning Outcome: 3.1, 3.2

Global LO:  G2, G7

 

11) terpenes

Chapter Section:  3.1, 3.2, 3.3, 3.4

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1, 3.2

Global LO:  G2, G7

 

12) polysaccharides

Chapter Section:  3.1, 3.2, 3.3, 3.4

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1, 3.2

Global LO:  G2, G7

 

 

13) RNA

Chapter Section:  3.1, 3.2, 3.3, 3.4

Blooms Taxonomy:  Comprehension

Learning Outcome:  3.1, 3.2

Global LO:  G2, G7

 

Answers: 8) A,

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