lOMoARcPSD|30659517
Test Bank for Essential Cell Biology 4th Edition Alberts
Nursing Care of the Child (Alabama State University)
Scan to open on Studocu
Studocu is not sponsored or endorsed by any college or university
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Chapter 1: CELLS: THE FUNDAMENTAL UNITS OF LIFE
Unity and Diversity of Cells
1-1 Living systems are incredibly diverse in size, shape, environment, and behavior. It is estimated that there
are between 10 million and 100 million different species. Despite this wide variety of organisms, it
remains difficult to define what it means to say something is alive. Which of the following can be
described as the smallest living unit?
(a) DNA
(b) cell
(c) organelle
(d) protein
1-2 Indicate whether the following statements are true or false. If the statement is false, explain why it is
false.
A. The Paramecium is a multicellular microorganism covered with hair-like cilia.
B. Cells of different types can have different chemical requirements.
C. The branchlike extensions that sprout from a single nerve cell in a mammalian brain can extend over
several hundred micrometers.
1-3 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list
below. Not all words or phrases will be used; each word or phrase should be used only once.
Cells can be very diverse: superficially, they come in various sizes, ranging from bacterial cells such as
Lactobacillus, which is a few in length, to larger cells such as a frog’s egg, which has a diameter of about
one .Despite the diversity, cells resemble each other to an astonishing degree in their chemistry. For
example, the same 20 are used to make proteins.
Similarly, the genetic information of all cells is stored in their
. Although
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
contain the same types of molecules as cells, their inability to reproduce
themselves by their own efforts means that they are not considered living matter.
amino acids micrometer(s) viruses
DNA millimeter(s) yeast
fatty acids plants
meter plasma membranes
1-4 How does cellular specialization serve multicellular organisms and how might a high
degree of specialization be detrimental?
1-5 The flow of genetic information is controlled by a series of biochemical reactions that
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
result in the production of proteins, each with its own specific order of amino acids.
Choose the correct series of biochemical reactions from the options presented here.
(a) replication, transcription, translation
(b) replication, translation, transcription
(c) translation, transcription, replication
(d) translation, replication, transcription
1-6 Proteins are important architectural and catalytic components within the cell, helping to
determine its chemistry, its shape, and its ability to respond to changes in the
environment. Remarkably, all of the different proteins in a cell are made from the same
20
. By linking them in different sequences, the cell can make protein
molecules with different conformations and surface chemistries, and therefore different
functions.
(a) nucleotides.
(b) sugars.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(c) amino
acids.
(d) fatty acids.
1-7 Which statement is NOT true about mutations?
(a) A mutation is a change in the DNA that can generate offspring less fit for survival
than their parents.
(b) A mutation can be a result of imperfect DNA duplication.
(c) A mutation is a result of sexual reproduction.
(d) A mutation is a change in the DNA that can generate offspring that are as fit for
survival as their parents are.
1-8 Changes in DNA sequence from one generation to the next may result in offspring that are altered in
fitness compared with their parents. The process of change and selection
over the course of many generations is the basis of
.
(a) mutation.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(b) evolution.
(c) heredity.
(d) reproduction.
1-9 Select the option that best finishes the following statement: “Evolution is a process
.”
(a) that can be understood based on the principles of mutation and selection.
(b) that results from repeated cycles of adaptation over billions of years.
(c) by which all present-day cells arose from 4–5 different ancestral cells.
(d) that requires hundreds of thousands of years.
1-10 Select the option that correctly finishes the following statement: “A cell’s genome
.”
(a) is defined as all the genes being used to make protein.
(b) contains all of a cell’s DNA.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(c) constantly changes, depending upon the cell’s environment.
(d) is altered during embryonic development.
Cells Under the Microscope
1-11 Which statement is NOT true about the events/conclusions from studies during the mid-
1800s surrounding the discovery of cells?
(a) Cells came to be known as the smallest universal building block of living
organisms.
(b) Scientists came to the conclusion that new cells can form spontaneously from the
remnants of ruptured cells.
(c) Light microscopy was essential in demonstrating the commonalities between
plant and animal tissues.
(d) New cells arise from the growth and division of previously existing cells.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-12 What unit of length plant or animal cell?
would you generally use to measure atypical
(a) centimeters
(b) nanometers
(c) millimeters
(d) micrometers
1-13 Match the type of microscopy on the left with the corresponding description provided
below. There is one best match for each.
A. confocal
B. transmission electron
C. fluorescence
D. phase-contrast
E. scanning electron
F. bright-field
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
uses a light microscope with an optical component to take advantage of the different
refractive indices of light passing through different regions of the cell.
employs a light microscope and requires that samples be fixed and stained in order
to reveal cellular details.
requires the use of two sets of filters. The first filter narrows the wavelength range
that reaches the specimen and the second blocks out all wavelengths that pass
back up to the eyepiece except for those emitted by the dye in the sample.
scans the specimen with a focused laser beam to obtain a series of two-dimensional
optical sections, which can be used to reconstruct an image of the specimen in
three dimensions. The laser excites a fluorescent dye molecule, and the emitted
light from each illuminated point is captured through a pinhole and recorded by a
detector.
has the ability to resolve cellular components as small as 2 nm.
requires coating the sample with a thin layer of a heavy metal to produce three-
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
dimensional images of the surface of a sample.
1-14 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
A. The nucleus of an animal cell is round, small, and difficult to distinguish using
light microscopy.
B. The presence of the plasma membrane can be inferred by the well-defined
boundary of the cell.
C. The cytosol is fairly empty, containing a limited number of organelles, which
allows room for rapid movement via diffusion.
1-15 Cell biologists employ targeted fluorescent dyes or modified fluorescent proteins in both
standard fluorescence microscopy and confocal microscopy to observe specific details in
the cell. Even though fluorescence permits better visualization, the resolving power is
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
essentially the same as that of a standard light microscope because the resolving power of
a microscope is limited by the
of light.
(a) absorption
(b) intensity
(c) filtering
(d) wavelength
1-16 What is the smallest distance two points can be separated and still resolved using light
microscopy?
(a) 20 nm
(b) 0.2 µm
(c) 2 µm
(d) 200 µm
The Prokaryotic Cell
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-17 By definition, prokaryotic cells do not possess
.
(a) a nucleus.
(b) replication machinery.
(c) ribosomes.
(d) membrane bilayers.
1-18 Although there are many distinct prokaryotic species, most have a small range of shapes,
sizes, and growth rates. Which of the following characteristics are not observed in
prokaryotes?
(a) a highly structured cytoplasm
(b) endoplasmic reticulum
(c) the ability to divide rapidly
(d) a cell wall
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-19 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
A. The terms “prokaryote” and “bacterium” are synonyms.
B. Prokaryotes can adopt several different basic shapes, including spherical, rod-
shaped, and spiral.
C. Some prokaryotes have cell walls surrounding the plasma membrane.
1-20 Prokaryotic cells are able to evolve very fast, which helps them to rapidly adapt to new
food sources and develop resistance to antibiotics. Which of the options below lists the
three main characteristics that support the rapid evolution of prokaryotic populations?
(a) microscopic, motile, anaerobic
(b) aerobic, motile, rapid growth
(c) no organelles, cell wall, can exchange DNA
(d) large population, rapid growth, can exchange DNA
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-21 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
A. Oxygen is toxic to certain prokaryotic organisms.
B. Mitochondria are thought to have evolved from anaerobic bacteria.
C.
Photosynthetic bacteria contain chloroplasts.
1-22 Some prokaryotes can live by utilizing entirely inorganic materials. Which of the
following inorganic molecules would you predict to be the predominant building block
for fats, sugars, and proteins?
(a) O2
(b) N2
(c) CO2
(d) H2
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
The Eukaryotic Cell
1-23 Use the list of structures below to label the schematic drawing of an animal cell in Figure
Q1-23.
Figure Q1-23
A. plasma membrane
B. nuclear envelope
C. cytosol
D. Golgi apparatus
E. endoplasmic reticulum
F. mitochondrion
G. transport vesicles
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-24 For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or phrase
should be used only once.
Eukaryotic cells are bigger and more elaborate than prokaryotic cells. By
definition, all eukaryotic cells have a
, usually the most
prominent organelle. Another organelle found in essentially all eukaryotic cells is
the for the cell. In
, which generates the chemical energy
contrast, the
is a type of organelle found only in the cells
of plants and algae, and performs photosynthesis. If we were to strip away the
plasma membrane from a eukaryotic cell and remove all of its membrane-
enclosed organelles, we would be left with the
, which
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
contains many long, fine filaments of protein that are responsible for cell shape
and structure and thereby form the cell’s
chloroplast cytosol
.
nucleus
chromosome endoplasmic reticulum ribosomes
cytoskeleton mitochondrion
1-25 The
is made up of two concentric membranes and is continuous
with the membrane of the endoplasmic reticulum.
(a) plasma membrane
(b) Golgi network
(c) mitochondrial membrane
(d) nuclear envelope
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-26 The nucleus, an organelle found in eukaryotic cells, confines the
, keeping
them separated from other components of the cell.
(a) lysosomes
(b) chromosomes
(c) peroxisomes
(d) ribosomes
1-27 Which of the following organelles has both an outer and an inner membrane?
(a) endoplasmic reticulum
(b) mitochondrion
(c) lysosome
(d) peroxisome
1-28 Mitochondria perform cellular respiration, a process that uses oxygen, generates carbon
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
dioxide, and produces chemical energy for the cell. Which answer below indicates a
correct pairing of material “burned” and the form of energy produced during cellular
respiration?
(a) fat, ADP
(b) sugar, fat
(c) sugar, ATP
(d) fat, protein
1-29 You fertilize egg cells from a healthy plant with pollen (which contains the male germ
cells) that has been treated with DNA-damaging agents. You find that some of the
offspring have defective chloroplasts, and that this characteristic can be passed on to
future generations. This surprises you at first because you happen to know that the male
germ cell in the the egg cell and thus
pollen grain contributes no chloroplasts to fertilized
to the offspring. What can you deduce from these results?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-30 Mitochondria contain their own genome, are able to duplicate, and actually divide on a
different time line from the rest of the cell. Nevertheless, mitochondria cannot function
for long when isolated from the cell because they are
.
(a) viruses.
(b) parasites.
(c) endosymbionts.
(d) anaerobes.
1-31 The mitochondrial proteins found in the inner membrane are involved in the conversion
of ADP to ATP, a source of energy for the cell. This process consumes which of the
following substances?
(a) oxygen
(b) nitrogen
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(c) sulfur
(d) carbon dioxide
1-32 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
A. With respect to cellular respiration, the only organelles used by animal cells are
mitochondria, while plant cells use both mitochondria and chloroplasts.
B. The number of mitochondria inside a cell remains constant over the life of the
cell.
1-33 Chloroplasts are found only in eukaryotic cells that carry out photosynthesis: plants and
algae. Plants and algae appear green as a result of the presence of chlorophyll. Where is
chlorophyll located in the chloroplast?
(a) in the first, outer membrane
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(b) in the space between the first and second membranes
(c) in the second, inner membrane
(d) in the third, innermost membrane
1-34 Photosynthesis enables plants to capture the energy from sunlight. In this essential
process, plants incorporate the carbon from CO2 into high-energy
molecules,
which the plant cell mitochondria use to produce ATP.
(a) fat
(b) sugar
(c) protein
(d) fiber
1-35 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
A. Membrane components in the cell are made in the endoplasmic reticulum.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
B. The Golgi apparatus is made up of a series of membrane-enclosed compartments
through
which materials destined for secretion must pass.
C. Lysosomes are small organelles where fatty acid synthesis occurs.
1-36 Circle the appropriate cell type in which the listed structure or molecule can be found.
Note that the structure or molecule can be found in more than one type of cell.
1-37 Which of the following choices best describes the role of the lysosome?
(a) transport of material to the Golgi
(b) clean-up, recycling, and disposal of macromolecules
(c) sorting of transport vesicles
(d) the storage of excess macromolecules
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-38 The protozoan Didinium feeds on other organisms by engulfing them. Why are bacteria,
in general, unable to feed on other cells in this way?
1-39 The cell constantly exchanges materials by bringing nutrients in from the external
environment and shuttling unwanted by-products back out. Which term describes the
process by which external materials are captured inside vesicles and brought into the
cell?
(a) degradation
(b) exocytosis
(c) phagocytosis
(d) endocytosis
1-40 Eukaryotic cells are able to trigger the release of material from secretory vesicles to the
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
extracellular space using a process called exocytosis. An example of materials
commonly released this way is
.
(a) hormones.
(b) nucleic acids.
(c) sugars.
(d) cytosolic proteins.
1-41
are fairly small organelles that provide a safe place within the cell to carry
out certain biochemical reactions that generate harmful, highly reactive oxygen species.
These chemicals are both generated and broken down in the same location.
(a) N
ucleosomes
(b) Lysosomes
(c) Peroxisomes
(d) Endosomes
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-42 The cytoskeleton provides support, structure, motility, and organization, and it forms
tracks to direct organelle and vesicle transport. Which of the cytoskeletal elements listed
below is the thickest?
(a) actin filaments
(b) microtubules
(c) intermediate filaments
(d) none of the above (all the same thickness)
1-43 Despite the differences between eukaryotic and prokaryotic cells, prokaryotes have
proteins that are distantly related to eukaryotic actin filaments and microtubules. What is
likely to be the most ancient function of the cytoskeleton?
(a) cell motility
(b) vesicle transport
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(c) membrane support
(d) cell division
1-44 Which of the following characteristics would not support the idea that the ancestral
eukaryote was a predator cell that captured and consumed other cells?
(a) dynamic cytoskeleton
(b) large cell size
(c) ability to move
(d) rigid membrane
1-45 Choose the phrase that best completes this sentence: Microtubules
required to pull duplicated chromosomes to opposite poles of dividing cells.
(a) generate contractile forces
(b) are intermediate in thickness
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
and are
lOMoARcPSD|30659517
(c) can rapidly reorganize
(d) are found in especially large numbers in muscle cells
1-46 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
A. Plants do not require a cytoskeleton because they have a cell wall that lends
structure and support to the cell.
B. The cytoskeleton is used as a transportation grid for the efficient, directional
movement of cytosolic components.
C. Thermal energy promotes random movement of proteins, vesicles, and small
molecules in the cytosol.
1-47 Which pair of values best fills in the blanks in this statement: On average, eukaryotic
cells are
times longer and have
times more volume than
prokaryotic cells.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(a) 5, 100
(b) 10, 200
(c) 10, 100
(d) 10, 1000
1-48 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
A. Primitive plant, animal, and fungal cells probably acquired mitochondria after
they diverged from a common ancestor.
B. Protozoans are single-celled eukaryotes with cell morphologies and behaviors that
can be as complex as those of some multicellular organisms.
C. The first eukaryotic cells on Earth must have been aerobic; otherwise, they would
not have been able to survive when the planet’s atmosphere became oxygen-rich.
Model Organisms
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-49 Given what you know about the differences between prokaryotic cells and eukaryotic
cells, rate the following things as “good” or “bad” processes to study in the model
organism E. coli.
A. formation of the endoplasmic reticulum
B. DNA replication
C. how the actin cytoskeleton contributes to cell shape
D. how cells decode their genetic instructions to make proteins
E. how mitochondria get distributed to cells during cell division
1-50 Scientists learned that cell death is a normal and even important part of life by studying
the development of the nematode worm C. elegans. What was the most important feature
of C. elegans for the study of programmed cell death?
(a) The nematode is smaller and simpler than the fruit fly.
Page 10 of 21
(b) 70% of C. elegans genes have homologs in humans.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(c) The developmental pathway of each cell in the adult worm was known.
(d) Its genome was partially sequenced.
1-51 Biologists cannot possibly study all living species. Instead, they try to understand cell
behavior by studying a select subset of them. Which of the following characteristics are
useful in an organism chosen for use as a model in laboratory studies?
(a) amenability to genetic manipulation
(b) ability to grow under controlled conditions
(c) rapid rate of reproduction
(d) all of the above
1-52 Many of the mechanisms that cells use for maintenance and reproduction were first
studied at the molecular level in bacteria. Which bacterial species had a central role in
advancing the field of molecular biology?
(a) E. coli
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(b) D.
melanogaster
(c) S. pombe
(d) C. elegans
1-53 Brewer’s yeast, apart from being an irreplaceable asset in the brewery and in the bakery,
is an experimental organism used to study eukaryotic cells. However, it does have some
limitations. Which of the processes below cannot be studied in yeast?
(a) DNA replication
(b) cell motility
(c) exocytosis
(d) cell division
1-54 For each process (A–D), circle the simplest model organism from the list of three that
would be best used for investigation:
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-55 A. thaliana, or Arabidopsis, is a common weed. Biologists have selected it over hundreds
of thousands of other flowering plant species to serve as an experimental model organism
because
.
(a) it can withstand extremely cold climates.
(b) it can reproduce in 8–10 weeks.
(c) it produces thousands of offspring per plant.
(d) Both (b) and (c) are true.
1-56 Drosophila melanogaster is a/an
. This type of animal is the most abundant
of all animal species, making it an appropriate choice as an experimental model.
(a) insect
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(b) bird
(c) amphibian
(d) mammal
1-57 Caenorhabditis elegans is a nematode. During its development, it produces more than
1000 cells. However, the adult worm has only 959 somatic cells. The process by which
131 cells are specifically targeted for destruction is called
.
(a) directed cell pruning.
(b) programmed cell death.
(c) autophagy.
(d) necrosis.
1-58 Zebrafish (Danio rerio) are especially useful in the study of early development because
their embryos
.
(a) are exceptionally large.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(b) develop slowly.
(c) are transparent.
(d) are pigmented.
1-59 You wish to explore how mutations in specific genes affecting sugar metabolism might
alter tooth development. Which organism is likely to provide the best model system for
your studies, and why?
(a) horses
(b) mice
(c) E. coli
(d) Arabidopsis
1-60 Indicate whether the following statements are true or false. If the statement is false,
explain why it is false.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
A. The human genome is roughly 30 times larger than the Arabidopsis genome, but
contains approximately the same number of protein-coding genes.
B. The variation in genome size among protozoans is larger than that observed
across all species of mammals, birds, and reptiles.
C. The vast majority of our genome encodes functional RNA molecules or proteins
and most of the intervening DNA is nonfunctional.
1-61 Genes that have homologs in a variety of species have been discovered through the
analysis of genome sequences. In fact, it is not uncommon to find a family of
homologous genes encoding proteins that are unmistakably similar in amino acid
sequence in organisms as diverse as budding yeast, archaea, plants, and humans. Even
more remarkably, many of these proteins can substitute functionally for their homologs in
other organisms. Explain what it is about the origins of cells that makes it possible for
proteins expressed by homologous genes to be functionally interchangeable in different
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
organisms.
1-62 Match each biological process with the model organism that is best suited or most
specifically useful for its study, based on information provided in your textbook. You
may list individual processes more than once.
A. cell division
B. development (multicellular)
C. programmed cell death
D. photosynthesis
E. immunology
A. thaliana (Arabidopsis)
M. musculus (mouse)
S. pombe
C. elegans
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
S. cerevisiae
D. rerio (zebrafish)
D. melanogaster
Testing the Concepts
1-63 Employ the principles of evolution discussed in this chapter to explain how the specific
features and predatory behaviors of some primitive eukaryotes may have given them a
selective advantage over others 1.5 billion years ago.
1-64 Evolutionary biologists have always used a broad range of modern organisms to infer the
characteristics that ancestral organisms may have possessed. Genomic sequences are now
available for an increasing number of species, and scientists studying evolutionary
processes can take advantage of this enormous amount of data to bring evolution into the
arena of molecular studies. By aligning the sequences of homologous genes and looking
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
for regions of similarity and where changes have occurred, it is possible to infer the
sequence of the ancestral gene.
A. What term is used to describe the changes in gene sequences that have occurred?
How can we use what we know about this process to construct a time line
showing when various sequence changes occurred and when they led to the
modern sequences that we know today?
B. It is possible to express an ancestral gene sequence in modern organisms and
subsequently compare the function of its product with that of the modern protein.
Why might this approach give misleading conclusions?
1-65 The antibiotic streptomycin inhibits protein synthesis in bacteria. If this antibiotic is
added to a culture of animal cells, protein synthesis in the cytosol continues normally.
However, over time, the population of mitochondria in the cell becomes depleted.
Specifically, it is observed that the protein-synthesis machinery inside the mitochondria is
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
inhibited.
A. Explain this observation based on what you know about the origins of the modern
eukaryote.
B. What do you expect to observe if, in a new experiment, animal cells are treated
with diphtheria toxin, a compound that is known to block cytosolic protein
synthesis but does not have any impact on bacterial growth?
1-66 You have been following the recent presidential elections and have heard some
candidates disparaging excessive and “unnecessary” federal government expenditures.
One particular candidate asks: “Why are we spending millions of dollars studying fruit
flies? How can that possibly help us find a cure for cancer?” Use your knowledge of
model organisms to explain why studies in D. melanogaster (the fruit fly) are actually an
excellent use of research funding.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-67 Cellular processes are often regulated by unknown mechanisms. In many cases,
biologists work in which they are
backward in an attempt to understand a process
interested. This was the case when Nurse and Hartwell were trying to understand how
cell division is controlled in yeast. Describe the process by which they “broke” the
system and then supplied the “missing parts” to get the cell cycle running again. What
further evidence did they collect to show that human cells and yeast cells regulate the cell cycle using
a similar mechanism?
1-68 Your friend has just returned from a deep-sea mission and claims to have found a new
single-celled life-form. He believes this new life-form may not have descended from the
common ancestor that all types of life on Earth share. You are convinced that he must be
wrong, and you manage to extract DNA from the cells he has discovered. He says that the
mere presence of DNA is not enough to prove the point: his cells might have adopted
DNA as a useful molecule quite independently of all other known life-forms. What could
you do to provide additional evidence to support your argument?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
ANSWERS
1-1
(b)
1-2
A. False. The Paramecium is a single-celled
organism. B. True.
C. True.
1-3 Cells can be very diverse: superficially, they come in various sizes, ranging from bacterial
cells such as Lactobacillus, which is a few micrometers in length, to larger cells such as a frog’s
egg, which has a diameter of about one millimeter. Despite the diversity, cells resemble each
other to an astonishing degree in their chemistry. For example, the same 20 amino acids are
used to make proteins. Similarly, the genetic information of all cells is stored in their DNA.
Although viruses contain the same types of molecules as cells, their inability to reproduce
themselves by their own efforts means that they are not considered living matter.
1-4 In a multicellular organism, the specialization of cells creates a division of labor and
each type of cell relies on the activities of other cell types for survival. This cooperation
between specialized cells is essential for the organism as a whole. If one of these overly
specialized cells were removed from the context of the organism, it would not have the
capabilities needed to generate offspring and would probably not even live very long.
1-5
(a)
1-6
(c)
1-7
(c)
1-8
(b)
1-9
(a)
1-10
(b)
1-11
(b)
1-12
(d)
1-13
F
D
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
C
A
B
E
1-14 A. False. The nucleus is one of the largest organelles and is the easiest organelle to discern
within a typical cell.
B. True.
C. False. The cytosol is actually brimming with individual proteins, protein fibers,
extended membrane systems, transport vesicles, and small molecules. And
although cellular components do move by diffusion, the rate of movement is
limited by the space available and the size of the component in question.
1-15
(d)
1-16
(b)
1-17
(a)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-18 (a), (b)
1-19 A. False. Archaea that are significantly make up a class of prokaryotic organisms
different from bacteria.
B. True.
C. True.
1-20
(d)
1-21
(a) True.
(b) False. Mitochondria use oxygen to generate energy and are thought to have
evolved from aerobic bacteria.
(c) False. Photosynthetic bacteria have enzyme systems similar to those found in
chloroplasts, which allow them to harvest light energy to fix carbon dioxide.
1-22
(c)
1-23
A. Plasma membrane—3
B. Nuclear envelope—5
C. Cytosol—1
D. Golgi apparatus—2
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
E. Endoplasmic reticulum—4
F. Mitochondrion—7
G. Transport vesicles—6
1-24 Eukaryotic cells are bigger and more elaborate than prokaryotic cells. By definition, all
eukaryotic cells have a nucleus, usually the most prominent organelle. Another organelle
found in essentially all eukaryotic cells is the mitochondrion, which generates the chemical
energy for the cell. In contrast, the chloroplast is a type of organelle found only in the cells
of plants and algae, and performs photosynthesis. If we were to strip away the plasma
membrane from a eukaryotic cell and remove all of its membrane-enclosed organelles, we
would be left with the cytosol, which contains many long, fine filaments
of protein that are responsible for cell shape and structure and thereby form the cell’s
cytoskeleton.
1-25 (d)
1-26 (b)
1-27 (b)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-28 (c)
1-29 Your results show that not all of the information required for making a chloroplast is
encoded in the chloroplast’s own DNA; some, at least, must be encoded in the DNA
carried in the nucleus. The reasoning is as follows. Genetic information is carried only in
DNA, so the defect in the chloroplasts must be due to a mutation in DNA. But all of the
chloroplasts in the offspring (and thus all of the chloroplast DNA) must derive from those
in the female egg other Hence, all of
cell, since chloroplasts only arise from chloroplasts.
the chloroplasts contain undamaged DNA from the female parent’s chloroplasts. In all
the cells of the offspring, however, half of the nuclear DNA will have come from the
male germ-cell nucleus, which combined with the female egg nucleus at fertilization.
Since this DNA has been treated with DNA-damaging agents, it must be the source of the
heritable chloroplast defect. Thus, some of the information required for making a
chloroplast is encoded by the nuclear DNA.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-30 (c)
1-31 (a)
1-32 A. False. In plants, only mitochondria perform cellular respiration (using oxygen to
break down organic molecules to produce carbon dioxide) just as in animal cells.
Chloroplasts perform photosynthesis in which water molecules are split to
generate oxygen and fix carbon dioxide molecules.
B. False. Mitochondria have their own division cycle and their numbers change
based on the rate of division.
1-33 (d)
1-34 (b)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-35 A. True.
B. True.
C. False. Lysosomes house enzymes that break down nutrients for use by the cell and
help recycle materials that cannot be used, which will later be excreted from the
cell.
1-36
1-37 (b)
1-38 Didinium engulfs prey by changing its shape, and for this it uses its cytoskeleton.
Bacteria have no cytoskeleton and cannot easily change their shape because they are
generally surrounded by a tough cell wall.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-39 (d)
1-40 (a)
1-41 (c)
1-42 (b)
1-43 (d)
1-44 (d)
1-45 (c) 1-46 A. False. Although plant cells do have a cell wall that lends structure and support,
they still need a cytoskeleton, which also helps with connections between cells
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
and the transport of vesicles inside the cell.
B. True.
C. True.
1-47 (d)
1-48 A. False. The mitochondria in modern plant, animal, and fungal cells are very
similar, implying that these lines diverged after the mitochondrion was acquired
by the ancestral eukaryote.
B. True.
C. False. The first eukaryotic cells likely contained a nucleus but no mitochondria.
These ancestral eukaryotes subsequently adapted to survive in a world filled with
oxygen by engulfing primitive aerobic prokaryotic cells.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-49 A. bad
B. good
C. bad
D. good
E. bad
1-50 (c) This is the best answer because it was the prior developmental studies tracing cell
lineages from the egg to the adult that allowed scientists to identify the precise time and
location of cells that were being targeted for cell death. It was observed that this cell
death was a normal and necessary part of the developmental pathway in the worm.
Programmed cell death has since become known to be an important process in all
multicellular eukaryotic organisms.
1-51 (d)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-52 (a)
1-53 (b)
AC. elegans
1-54
BArabidopsis
Cmouse
DDrosophila 1-55 (b)
1-56 (a)
1-57 (b)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1-58 (c)
1-59 (b) Mice are likely to provide the best model system. Mice have teeth and have long been
used as a model organism. Mice reproduce relatively rapidly and the extensive scientific
community that works with mice has developed techniques to facilitate genetic
manipulations. E. coli (a bacterium) and Arabidopsis (a plant) do not have teeth. Horses
like sugar and have big teeth, but they would not be a good model organism. There is not
an extensive scientific community working on the molecular and biochemical
mechanisms of cell behaviors in horses; they are expensive and have a long reproduction
time, which makes genetic studies costly and slow; and tools for genetic manipulation
(other than traditional breeding) have not been developed.
1-60 A. True.
B. True.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
C. False. It is a relatively small proportion of our DNA that encodes RNA and protein
molecules. The majority of nonencoding sequences is probably involved
in critical regulatory processes.
1-61
All living beings on Earth (and thus, all cells) are thought to be derived from a common
ancestor. Solutions to many of the essential challenges that face a cell (such as the
synthesis of proteins, lipids, and DNA) seem to have been achieved in this ancient
common ancestor. The ancestral cell therefore possessed sets of proteins to carry out
these essential functions. Many of the essential challenges facing modern-day cells are
the same as those facing the ancestral cell, and the ancient solutions are often still
effective. Thus, it is not uncommon for organisms to use proteins and biochemical
pathways inherited from their ancestors. Although these proteins usually show some
species-specific diversification, they still retain the basic biochemical characteristics of
the ancestral protein. For example, homologous proteins often retain their ability to
interact with a specific protein target, even in cells of diverse species. Because the basic
biochemical characteristics are retained, homologous proteins are often capable of
functionally
substituting for one another.
1-62
B, D A. thaliana (Arabidopsis)
B, E M. musculus (mouse)
A S. pombe
C C. elegans
A S. cerevisiae
B D. rerio (zebrafish)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
B D. melanogaster
1-63 The Earth’s atmosphere became oxygen-rich roughly 1.5 billion years ago. If some
primitive predatory eukaryotic cells were similar to modern-day protozoans, they may
have been mobile and able to engulf other cells. These characteristics would have
been advantageous in the face of a changing atmosphere, and the establishment of a
symbiotic relationship with an engulfed aerobe would have been selected for in the
eukaryotic cell populations.
1-64
A. Changes in gene sequence occur through mutation. Mutations accumulate
over time, occurring independently and at different sites in each gene lineage.
Homologous genes that diverged recently will differ only slightly; genes that
diverged long ago will differ more. Knowing the average mutation rate, you can
estimate the time that has elapsed since the different versions of the gene
diverged. By seeing how closely the various members of the family of
homologous genes resemble one another, you can draw up a family tree, showing
the sequence of lineage splits that lead from the ancestral gene to its many modern
descendants. Suppose this family tree shows that family members A and B
diverged from one another long ago, but that C diverged from B more recently;
and suppose that at a certain site in the gene, A and B have the same sequence but
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
C is different. Then, it is likely that the sequence of A and B is ancestral, while
that of C reflects a recent mutation that has occurred in the lineage of C alone.
B. Although an inferred ancestral sequence can be reconstructed and the protein expressed,
you would be placing an inferred, ancient protein in the context of a
modern cell. If there are important interacting partners for the modern protein,
there is a chance they may not recognize the ancestral protein, and therefore any
information about its function may be inaccurate.
1-65 A. If the mitochondria originated from an ancient aerobic bacterium that was engulfed by
an ancient eukaryote, as postulated, it is possible that an antibiotic
that inhibits protein synthesis in bacteria could also block that process in
mitochondria.
B. We would expect that although cytosolic protein synthesis would stop,
mitochondrial protein synthesis should still occur normally (at least for a little
while). This result would lend further support to the idea that mitochondria are
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
derived from a noneukaryotic organism. If this were not the case, these
compounds would be expected to affect protein synthesis at both locations.
1-66
Funding research o a for several reasons: (1) n
D. melanogasteris worthwhile investment
working with insect animal models is relatively inexpensive; (2) fruit flies have historically
proven useful in helping understand eukaryotic chromosome behavior; and (3) many of the
genes in Drosophila are highly similar in sequence to the homologous human genes, and thus
can be used tostudy human diseases.
1-67 Nurse and Hartwell first treated yeast cells with a chemical mutagen. The mutated
population of cells was then grown and observed. Cells that demonstrated defects in cell-cycle
regulation (characterized by cell-cycle arrest, larger-than-normal cells, and smaller-than-normal
cells) were then isolated. The use of a library of plasmids that each express a normal gene from
yeast cells allowed the scientists to identify exactly which gene could be used to “rescue” the
mutant, because when the normal gene is expressed again, the cells return to a normal cell
cycle. After this big result, the scientistswent on to show that the homologous gene from other
organisms could also rescue the mutant phenotype. The most exciting result was obtained with
the human version of the cdc2 gene, which demonstrated that there are common principles
underlying cell-cycle regulation across a large range of eukaryotic organisms.
1-68 You could use modern technology to discover the sequence of the DNA. If you are right,
you would expect to find parts of this sequence that are unmistakably similar to corresponding
sequences in other, familiar, living organisms; it would be highly improbable that such similar
sequences would have evolved independently. You could, of course, also analyze other features
of the chemistry of his cells; forexample, do they contain proteins made of the same set of 20
amino acids? This could all be supporting evidence that this newly discovered species arose
from the same common ancestral cells as all other lifeon Earth.
ESSENTIAL CELL BIOLOGY, FOURTH EDITION
CHAPTER 2: CHEMICAL COMPONENTS OF CELLS
Chemical Bonds
2-1
Select the answer that best completes the following statement: Chemical
reactionsin living systems occur in an
environment, within a narrow range
of temperatures.
(a)
optimal
(b)
organic
(c)
extracellular
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(d)
aqueous
2-2
Indicate whether the statements below are true or false. If a statement is
false,explain why it is false.
A.
The chemistry of life is carried out and coordinated primarily by the
actionof small molecules.
B.
Carbon-based compounds make up the vast majority of molecules
foundin cells.
C.
The chemical reactions in living systems are loosely regulated, allowing
for a wide range of products and more rapid evolution.
2-3
Which subatomic particles contribute to the atomic number for any
givenelement?
(a)
protons
(b)
protons and neutrons
(c)
neutrons
(d)
protons and electrons
2-4
Which subatomic particles contribute to the atomic mass for any given element?
(a)
protons
(b)
protons and neutrons
(c)
neutrons
(d)
protons and electrons
2-5
Which subatomic particles can vary between isotopes of the same
element,without changing the observed chemical properties?
(a)
electrons
(b)
protons and neutrons
(c)
neutrons
(d)
neutrons and electrons
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
2-6
Figure Q2-6 depicts the structure of carbon. Use the information in the diagram
tochoose the correct atomic number and atomic weight, respectively, for an atom
ofcarbon.
Figure Q2-6
(a)
(b)
(c)
(d)
2-7
6, 12
12, 12
6, 18
12, 6
Carbon 14 is an unstable isotope of carbon that decays very slowly. Compared
tothe common, stable carbon 12 isotope, carbon 14 has two additional
.
(a)
(b)
(c)
(d)
2-8
electrons.
neutrons.
protons.
ions.
If the isotope 32S has 16 protons and 16 neutrons, how many protons,
neutrons,and electrons will the isotope 35S have, respectively?
(a)
(b)
(c)
(d)
16, 20, 15
16, 19, 15
16, 19, 16
16, 19, 17
2-9
A.
If 0.5 mole of glucose weighs 90 g, what is the molecular mass of
glucose?
B.
What is the concentration, in grams per liter (g/L), of a 0.25 M solution
ofglucose?
C.
How many molecules are there in 1 mole of glucose?
2-10
Which of the following elements is least abundant in living organisms?
(a)
sulfur
(b)
carbon
(c)
oxygen
(d)
nitrogen
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
2-11
You explain to a friend what you have learned about Avogadro’s number. Your
friend thinks the number is so large that he doubts there is even a mole of
livingcells on the Earth. You have recently heard that there are about 50
trillion (5 × 1013) human cells in each adult human body and that each human
contains morebacterial cells (in the digestive system) than human cells, so you
bet your friend
$5 that there is more than a mole of cells on Earth. The human population is
approximately 7 billion (7 × 109). What calculation can you show your friend toconvince
him you are right?
2-12
Avogadro’s number, calculated from the atomic weight of hydrogen, tells us
how many atoms or molecules are in a mole. The resulting base for all
calculations of moles and molarity (how many molecules are present when you
weigh out a substance or measure from a stock solution) is the following:
1 g of hydrogen atoms = 6 × 1023 hydrogen atoms = 1 mole of hydrogen
Sulfur has a molecular weight of 32. How many moles and atoms are there in 120grams of
sulfur?
(a)
3.75 and 6 × 1023
(b)
32 and 6 × 1023
(c)
1.75 and 1.05 ×1024
(d)
3.75 and 2.25 × 1024
2-13
The first task you are assigned in your summer laboratory job is to prepare a
concentrated NaOH stock solution. The molecular weight of NaOH is 40. How
many grams of solid NaOH will you need to weigh out to obtain a 500 mL
solution that has a concentration of 10 M?
(a)
800 g
(b)
200 g
(c)
400 g
(d)
160 g
2-14
You have a concentrated stock solution of 10 M NaOH and want to use it to
produce a 150 mL solution of 3 M NaOH. What volume of water and stock
solutions will you measure out to make this new solution?
(a)
135 mL of water, 15 mL of NaOH stock
(b)
115 mL of water, 35 mL of NaOH stock
(c)
100 mL of water, 50 mL of NaOH stock
(d)
105 mL of water, 45 mL of NaOH stock
2-15
Indicate whether the statements below are true or false. If a statement is
false,explain why it is false.
A.
Electron shells fill discrete regions around the nucleus of the atom
andlimit the number of electrons that can occupy a specific orbit.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
B.
C.
H, C, O, and N are the most common elements in biological molecules
because they are the most stable.
Some atoms are more stable when they lose one or two electrons,
eventhough this means they will have a net positive charge.
2-16
A covalent bond between two atoms is formed as a result of the
(a)
sharing of electrons.
(b)
loss of electrons from both atoms.
(c)
loss of a proton from one atom.
(d)
transfer of electrons from one atom to the other.
2-17
An ionic bond between two atoms is formed as a result of the
(a)
sharing of electrons.
(b)
loss of electrons from both atoms.
(c)
loss of a proton from one atom.
(d)
transfer of electrons from one atom to the other.
2-18
For each of the following sentences, fill in the blanks with the best word or
phraseselected from the list below. Not all words or phrases will be used; each
word or phrase should be used only once.
Whereas ionic bonds form a(n)
between atoms form a(n)
have a characteristic bond
more rigid when two electrons are shared in a(n)
, covalent bonds
. These covalent bonds
and become stronger and
.
Equal sharing of electrons yields a(n)
covalent
bond. If one atom participating in the bond has a stronger affinity for the
electron, this produces a partial negative charge on one atom and a partial
positive charge on the other. These
covalent bonds
should not be confused with the weaker
bonds that
are critical for the three-dimensional structure of biological molecules andfor
interactions between these molecules.
charge
covalent
double bond
ionic
2-19
length
molecule
noncovalent
nonpolar
polar
salt
single bond
weight
Indicate whether the statements below are true or false. If a statement is
false,explain why it is false.
A.
Electrons are constantly moving around the nucleus of the atom, but
theycan move only in discrete regions.
B.
There is no limit to the number of electrons that can occupy the
fourthelectron shell.
C.
Atoms with unfilled outer electron shells are especially stable and
aretherefore less reactive.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
.
.
lOMoARcPSD|30659517
2-20
Table Q2-20 indicates the electrons in the first four atomic electron shells for
selected elements. On the basis of the information in the chart and what you
knowabout atomic structure, which elements are chemically inert?
Table Q2-20
(a)
(b)
(c)
(d)
2-21
carbon, sulfur
helium, neon
sodium, potassium
magnesium, calcium
Table Q2-21 indicates the electrons in the first four atomic electron shells for
selected elements. On the basis of the information in the chart and what you
knowabout atomic structure, which elements will form ions with a net charge of
+1 in solution?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Table Q2-21
(a)
(b)
(c)
(d)
2-22
carbon, sulfur
helium, neon
sodium, potassium
magnesium, calcium
Table Q2-22 indicates the electrons in the first four atomic electron shells for
selected elements. On the basis of the information in the chart and what you
knowabout atomic structure, which elements will form ions with a net charge of
+2 in solution?
Table Q2-22
(a)
(b)
(c)
(d)
2-23
carbon, sulfur
helium, neon
sodium, potassium
magnesium, calcium
Table Q2-23 indicates the electrons in the first four atomic electron shells for
selected elements. On the basis of the information in the chart and what you
knowabout atomic structure, which elements form stable but reactive diatomic
gases?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Table Q2-23
(a)
(b)
(c)
(d)
nitrogen, oxygen
helium, neon
sodium, potassium
magnesium, calcium
2-24
Indicate whether the statements below are true or false. If a statement is
false,explain why it is false.
A.
There are four elements that constitute 99% of all the atoms found in
thehuman body.
B.
Copper, zinc, and manganese are among 11 nonessential trace
elementsthat contribute less than 0.1% of all the atoms in the human
body.
C.
Approximately 0.9% of the atoms in the human body come from seven
essential elements—Na, Mg, K, Ca, P, S, and Cl—all of which form stable
ions in aqueous solution.
2-25
Which of the following factors do not influence the length of a covalent bond?
(a)
the tendency of atoms to fill the outer electron shells
(b)
the attractive forces between negatively charged electrons and
positivelycharged nuclei
(c)
the repulsive forces between the positively charged nuclei
(d)
the minimization of repulsive forces between the two nuclei by the
cloudof shared electrons
2-26
Double covalent bonds are both shorter and stronger than single covalent
bonds, but they also limit the geometry of the molecule because they
.
(a)
(b)
(c)
(d)
create a new arrangement of electron shells.
change the reactivity of the bonded atoms.
limit the rotation of the bonded atoms.
prevent additional bonds from being formed with the bonded atoms.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
B.
2-27
Polar covalent bonds are formed when the electrons in the bond are not
sharedequally between the two nuclei. Which one of these molecules
contains polar bonds?
(a)
molecular oxygen
(b)
methane
(c)
propane
(d)
water
2-28
A.
In which scientific unit is the strength of a chemical bond usually
expressed?
If 0.5 kilocalories of energy are required to break 6 × 1023 bonds of aparticular type, what is
the strength of this bond?
2-29
Oxygen, hydrogen, carbon, and nitrogen atoms are enriched in the cells and
tissues of living organisms. The covalent bond geometries for these atoms
influence their biomolecular structures. Match the elements on the left with
thebond geometries illustrated on the right. Some elements assume more
than onebond geometry.
Figure Q2-29
A.
B.
C.
2-30
oxygen
carbon
nitrogen
Which combination of answers best completes the following statement: When
atoms are held together by
, they are typically
referredto as .
(a)
hydrogen bonds, molecules.
(b)
ionic interactions, salts.
(c)
ionic interactions, molecules.
(d)
double bonds, nonpolar.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
2-31
Although covalent bonds are 10–100 times stronger than noncovalent
interactions,many biological processes depend upon the number and type of
noncovalent interactions between molecules. Which of the noncovalent
interactions below willcontribute most to the strong and specific binding of two
molecules, such as a pair of proteins?
(a)
electrostatic attractions
(b)
hydrogen bonds
(c)
hydrophobic interactions
(d)
Van der Waals attractions
2-32
Which of the following expressions accurately describes the calculation of pH?
(a)
pH = –log10[H+]
(b)
pH = log10[H+]
(c)
pH = –log2[H+]
(d)
pH = –log10[OH–]
2-33
The pH of an aqueous solution is an indication of the concentration of available
protons. However, you should not expect to find lone protons in solution; rather,
the proton is added to a water molecule to form a(n)
ion.
(a)
hydroxide
(b)
ammonium
(c)
chloride
(d)
hydronium
2-34
The relative strengths of covalent bonds and van der Waals interactions remain
thesame when tested in a vacuum or in water. However, this is not true of
hydrogen bonds or ionic bonds, whose bond strengths are lowered considerably
in the presence of water. Explain these observations.
2-35
Indicate whether the statements below are true or false. If a statement is
false,explain why it is false.
A.
Any covalently bonded H atom can participate in a hydrogen bond if it
comes in close proximity with an oxygen atom that forms part of a
watermolecule.
B.
Protons are constantly moving between water molecules, which means
there is an overall equilibrium between hydroxyl ions and hydronium
ionsin aqueous solutions.
C.
A strong base is defined as a molecule that can readily remove
protonsfrom water.
2-36
Larger molecules have hydrogen-bonding networks that contribute to specific,
high-affinity binding. Smaller molecules such as urea can also form these
networks. How many hydrogen bonds can urea (Figure Q2-36) form if dissolved
in water?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Figure Q2-36
2-37
(a)
(b)
(c)
(d)
6
5
3
4
A.
Sketch three different ways in which three water molecules could be
heldtogether by hydrogen-bonding.
On a sketch of a single water molecule, indicate the distribution
ofpositive and negative charge (using the symbols δ+ and δ–).
How many hydrogen bonds can a hydrogen atom in a water
moleculeform? How many hydrogen bonds can the oxygen atom in
a water molecule form?
B.
C.
2-38
A.
B.
C.
What is the pH of pure water?
What concentration of hydronium ions does a solution of pH 8 contain?
Complete the following reaction:
CH3COOH + H2O ↔
D.
2-39
.
Will the reaction in (C) occur more readily (be driven to the right) if the
pH of the solution is high?
Aromatic carbon compounds such as benzene are planar and very stable. Doublebond character extends around the entire ring, which is why it is often drawn as
ahexagon with a circle inside. This characteristic is caused by electron
.
(a)
(b)
(c)
(d)
2-40
resonance.
pairing.
partial charge.
stacking.
The amino acid histidine is often found in enzymes. Depending on the pH of its
environment, sometimes histidine is neutral and at other times it acquires a
protonand becomes positively charged. Consider an enzyme with a histidine side
chain that is known to have an important role in the function of the enzyme. It is
not clear whether this histidine is required in its protonated or its unprotonated
state. To answer this question, you measure enzyme activity over a range of pH,
with the results shown in Figure Q2-40. Which form of histidine is necessary for
the active enzyme?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Figure Q2-40
2-41
Silicon is an element that, like carbon, has four vacancies in its outer electron
shell and therefore has the same bonding chemistry as carbon. Silicon is not
foundto any significant degree in the molecules found in living systems,
however. Does this difference arise because elemental carbon is more abundant
than silicon?
What other explanations are there for the preferential selection of carbon over
silicon as the basis for the molecules of life?
2-42
Selenium (Se) is an element required in the human body in trace amounts.
Selenium is obtained through the diet and levels of selenium found in food
depend greatly on the soil where it is grown. Once ingested and absorbed as
selenate, it can become incorporated into a small number of polypeptides.
These selenoproteins are formed when selenium replaces an element that is
found in twoof the twenty “standard” amino acids. Using your knowledge of
atomic structure, the periodic table in Figure 2–7, and the structure of amino
acids found in Panel 2–5, deduce which two amino acids may be converted to
“seleno” amino acids and used to make selenoproteins.
Small Molecules in Cells
2-43
Indicate whether the molecules below are inorganic (I) or organic (O).
A.
glucose
B.
ethanol
C.
sodium chloride
D.
water
E.
cholesterol
F.
adenosine
G.
calcium
H.
glycine
I.
oxygen
J.
iron
K.
phospholipid
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
2-44
Which of the following monomer building blocks is necessary to assemble
selectively permeable boundaries around and inside cells?
(a)
sugars
(b)
fatty acids
(c)
amino acids
(d)
nucleotides
2-45
The variety and arrangement of chemical groups on monomer subunits
contributeto the conformation, reactivity, and surface of the macromolecule
into which they become incorporated. What type of chemical group is circled on
the nucleotide shown in Figure Q2-45?
Figure Q2-45
(a)
(b)
(c)
(d)
2-46
pyrophosphate
phosphoryl
carbonyl
carboxyl
The amino acids glutamine and glutamic acid are shown in Figure Q2-46. They
differ only in the structure of their side chains (circled). At pH 7, glutamic acid
can participate in molecular interactions that are not possible for glutamine.
Whattypes of interactions are these?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Figure Q2-46
(a)
(b)
(c)
(d)
ionic bonds
hydrogen bonds
van der Waals interactions
covalent bonds
2-47
Cells require one particular monosaccharide as a starting material to
synthesizenucleotide building blocks. Which of the monosaccharides below
fills this important role?
(a)
glucose
(b)
fructose
(c)
ribulose
(d)
ribose
2-48
Oligosaccharides are short sugar polymers that can become covalently linked to
proteins and lipids through condensation reactions. These modified proteins
and lipids are called glycoproteins and glycolipids, respectively. Within a
protein, which of the amino acids (shown in Figure Q2-48) is the most probable
target forthis type of modification?
Figure Q2-48
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(a)
(b)
(c)
(d)
serine
glycine
phenylalanine
methionine
2-49
Which of the following are examples of isomers?
(a)
glucose and galactose
(b)
alanine and glycine
(c)
adenine and guanine
(d)
glycogen and cellulose
2-50
A.
How many carbon atoms does the molecule represented in Figure Q250have?
B.
How many hydrogen atoms does it have?
C.
What type of molecule is it?
Figure Q2-50
2-51
Most types of molecules in the cell have asymmetric (chiral) carbons.
Consequently there is the potential to have two different molecules that look much the
same but are mirror images of each other and therefore not equivalent. These special
types of isomer are called stereoisomers. Which of the four carbonscircled in Figure Q251 is the asymmetric carbon that determines whether the amino acid (threonine in this
case) is a ᴅ or an ʟ stereoisomer?
Figure Q2-51
(a)
(b)
(c)
(d)
2-52
1
2
3
4
Indicate whether the statements below are true or false. If a statement is
false,explain why it is false.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
A.
B.
C.
2-53
A disaccharide consists of a sugar covalently linked to another
moleculesuch as an amino acid or a nucleotide.
The hydroxyl groups on monosaccharides are reaction hot spots and
canbe replaced by other functional groups to produce derivatives of
the original sugar.
The presence of double bonds in the hydrocarbon tail of a fatty acid
doesnot greatly influence its structure.
On the phospholipid molecule in Figure Q2-53, label each numbered line with
thecorrect term selected from the list below.
Figure Q2-53
2-54
Many types of cells have stores of lipids in their cytoplasm, usually seen as
fatdroplets. What is the lipid most commonly found in these droplets?
(a)
cholesterol
(b)
palmitic acid
(c)
isoprene
(d)
triacylglycerol
2-55
Choose the answer that best fits the following statement: Cholesterol is an
essential component of biological membranes. Although it is much smaller
thanthe typical phospholipids and glycolipids in the membrane, it is a(n)
molecule, having both hydrophilic and hydrophobic regions.
(a)
(b)
(c)
(d)
polar
oxygen-containing
hydrophobic
amphipathic
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
2-56
For each of the following sentences, fill in the blanks with the best word or
phraseselected from the list below. Not all words or phrases will be used; each
word or phrase should be used only once.
Proteins are
built from amino acids, which each have an amino
group and a
group attached to the central
.
There are twenty possible
that differ in structure and are
generally referred to as “R.” In solutions of neutral pH, amino acids are
, carrying both a positive and negative charge. When a protein is
made, amino acids are linked together through
, which are
formed by condensation reactions between the carboxyl end of the last amino acidand
the
end of the next amino acid to be added to the
growing chain.
amino
α-carbon
carbon
carboxyl
hydroxide
ionized
length
noncovalent
peptide bonds
polypeptides
protein
R group
side chains
2-57
DNA and RNA are different types of nucleic acid polymer. Which of the
following is true of DNA but not true of RNA?
(a)
It contains uracil.
(b)
It contains thymine.
(c)
It is single-stranded.
(d)
It has 5′-to-3′ directionality.
2-58
Match each term related to the structure of nucleic acids (A–I) with one of
thedescriptions provided.
A.
base
B.
glycosidic bond
C.
nucleoside
D.
nucleotide
E.
phosphoanhydride bond
F.
phosphoester bond
G.
ribose
H.
phosphodiester bond
I.
deoxyribose
the linkage between two nucleotides
the linkage between the 5′ sugar hydroxyl and a phosphate group
the nitrogen-containing aromatic ring
five-carbon sugar found in DNA
sugar unit linked to a base
linkage between the sugar and the base
linkages between phosphate groups
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
sugar linked to a base and a phosphate
five-carbon sugar found in RNA
2-59
A.
Write out the sequence of amino acids in the following peptide, using
thefull names of the amino acids.
Pro-Val-Thr-Gly-Lys-Cys-Glu
B.
C.
2-60
Write the same sequence with the single-letter code for amino acids.
According to the conventional way of writing the sequence of a peptide
or a protein, which is the C-terminal amino acid and which is the Nterminalamino acid in the above peptide?
The cell is able to harvest energy from various processes in order to generate
ATPmolecules. These ATPs represent a form of stored energy that can be used
later todrive other important processes. Explain how the cell can convert the
chemical energy stored in ATP to generate mechanical energy, for example
changing the shape of a protein.
Macromolecules in Cells
2-61
Both DNA and RNA are synthesized by covalently linking a nucleoside
triphosphate to the previous nucleotide, constantly adding to a growing chain.
Inthe case of DNA, the new strand becomes part of a stable helix. The two
strandsare complementary in sequence and antiparallel in directionality. What
is the principal force that holds these two strands together?
(a)
ionic interactions
(b)
hydrogen bonds
(c)
covalent bonds
(d)
van der Waals interactions
2-62
Each nucleotide in DNA and RNA has an aromatic base. What is the principal
force that keeps the bases in a polymer from interacting with water?
(a)
hydrophobic interactions
(b)
hydrogen bonds
(c)
covalent bonds
(d)
van der Waals interactions
2-63
Because there are four different monomer building blocks that can be used to
assemble RNA polymers, the number of possible sequence combinations that
canbe created for an RNA molecule made of 100 nucleotides is
.
(a)
(b)
(c)
(d)
1004
4100
4 × 100
100/4
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
2-64
There are 20100 different possible sequence combinations for a protein chain
with 100 amino acids. In addition to the amino acid sequence of the protein,
what otherfactors increase the potential for diversity in these macromolecules?
(a)
free rotation around single bonds during synthesis
(b)
noncovalent interactions sampled as protein folds
(c)
the directionality of amino acids being added
(d)
the planar nature of the peptide bond
2-65
Indicate whether the statements below are true or false. If a statement is
false,explain why it is false.
A.
“Nonpolar interactions” is simply another way of saying “van der Waals
attractions.”
B.
Condensation reactions occur in the synthesis of all the
macromoleculesfound in cells.
C.
All proteins and RNAs pass through many unstable conformations as
theyare folded, finally settling on one single, preferred conformation.
2-66
Macromolecules in the cell can often interact transiently as a result of
noncovalentinteractions. These weak interactions also produce stable, highly
specific interactions between molecules. Which of the factors below is the most
significantin determining whether the interaction will be transient or stable?
(a)
the size of each molecule
(b)
the concentration of each molecule
(c)
the rate of synthesis
(d)
surface complementarity between molecules
2-67
As a protein is made, the polypeptide is in an extended conformation, with every
amino acid exposed to the aqueous environment. Although both polar and
chargedside chains can mix readily with water, this is not the case for nonpolar
side chains. Explain how hydrophobic interactions may play a role in the early
stages of protein folding, and have an influence on the final protein
conformation.
2-68
You are trying to make a synthetic copy of a particular protein but accidentally
join the amino acids together in exactly the reverse order. One of your
classmatessays the two proteins must be identical, and bets you $20 that your
synthetic protein will have exactly the same biological activity as the original.
After havingread this chapter, you have no hesitation in staking your $20 that it
won’t. What particular feature of a polypeptide chain makes you sure your $20
is safe and thatthe project must be done again.
2-69
A protein chain folds into its stable and unique three-dimensional structure, or
conformation, by making many noncovalent bonds between different parts of
the chain. Such noncovalent bonds are also critical for interactions with other
proteinsand cellular molecules. From the list provided, choose the class(es) of
amino acidsthat are most important for the interactions detailed below.
A.
forming ionic bonds with negatively charged DNA
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
B.
C.
D.
E.
acidic
basic
forming hydrogen bonds to aid solubility in water
binding to another water-soluble protein
localizing an “integral membrane” protein that spans a lipid bilayer
tightly packing the hydrophobic interior core of a globular protein
nonpolar
uncharged polar
2-70
It is now a routine task to determine the exact order in which individual
subunitshave been linked together in polynucleotides (DNA) and polypeptides
(proteins).However, it remains difficult to determine the arrangement of
monomers in a polysaccharide. Explain why this is the case.
2-71
Your lab director requests that you add new growth medium to the mammalian
cell cultures before heading home from the lab on a Friday night. Unfortunately,
you need to make fresh medium because all the premixed bottles of medium
havebeen used. One of the ingredients you know you need to add is a mix of the
essential amino acids (those that cannot be made by the cells, but are needed in
proteins). On the shelf of dry chemicals you find the amino acids you need, and
you mix them into your medium, along with all the other necessary nutrients,
andreplace the old medium with your new medium. On Sunday, you come in to
the lab just to check on your cells and find that the cells have not grown. You are
sureyou made the medium correctly, but on checking you see that somebody
wrote a note on the dry mixture of amino acids you used: “Note: this mixture
contains only ᴅ-amino acids.”
A.
What is the meaning of the note and how does it explain the lack of
cellgrowth in your culture?
B.
Are there any organisms that could grow using this mixture? Justify
youranswer.
2-72
Eukaryotic cells have their DNA molecules inside their nuclei. However, to
package all the DNA into such a small volume requires the cell to use specialized
proteins called histones. Histones have amino acid sequences enriched for
lysinesand arginines.
A.
What problem might a cell face in trying to package DNA into a small
volume without histones, and how do these special packaging
proteinsalleviate the problem?
B.
Lysine side chains are substrates for enzymes called acetylases. A diagram
of an acetylated lysine side chain is shown in Figure Q2-72. How do you
think the acetylation of lysines in histone proteins will affect the ability
ofa histone to perform its role (refer to your answer in part A)?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Figure Q2-72
CHAPTER 3: DNA REPLICATION, REPAIR, AND
RECOMBINATION
6-1
the
The process
of
DNA
parental
DNA strands
used as
of
the
replication
be
a
to
opposing
(a)
catalyst
(b)
competito
requires
that
each of
produce
a
duplicate
strand.
r(c)
template
(d)
copy
6-2
DNA
replication
is
considered
semiconservative
because
of
replication,
the
original
consists
of
the
two
.
(a)
after many rounds
double helix is
still
DNA
intact.
(b)
each daughter
newstrands
parent DNA
DNA molecule
copied from
molecule.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
DNA
lOMoARcPSD|30659517
(c)
each daughter
DNA
fromthe
parent DNA
molecule
molecule
strand.
(d)
new
and
one
new
DNA
strands
must be
6-3
The classic experiments conducted
demonstrated
that
consists
of
one
strand
copied from a
DNA
template.
by
DNA
and
Meselson
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
Stahl
lOMoARcPSD|30659517
replication is
mechanism.
accomplished by
(a)
continuous
(b)
semiconservative
(c)
dispersive
(d)
conservative
6-4
Initiator
anddisrupt
proteins
employing
a
bind to
replication
hydrogen bonds between
the
two DNA strands
being copied.
below
does not contribute to
origins
Which of
the
the
relative
ease of
strand separation
by
initiator
(a)
replication
origins
are
A-T
base pairs
(b)
the
reaction
(c)
they
only
(d)
once opened,
other proteins
replicationmachinery
can
separate
rich
in
proteins?
occur at
room temperature
a
base pairs at
a
of
bind
the
few
the
to
DNA
E.
factors
time
origin
6-5
If
the
genome
requiresabout 20
itself,
of
the
minutes
bacterium
to
how
can the
genome
bereplicated in
of
only
the
3
fruit
fly
Drosophila
minutes?
is
smaller
than the
(a)
The
genome.
(b)
rate
genome
Drosophila
Eukaryotic
than
DNA
polymerase synthesizes DNA
at
coli
replicate
a
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
E.
coli
much faster
lOMoARcPSD|30659517
prokaryotic
DNA
polymerase.
Page
of
25
1
(c)
one
The nuclear
place
in
the
(d)
Drosophila DNA
replicationthan
Meselson
isotopes
produced
A.
DNA
cell,
membrane
keeps the
which increases
contains
E.
Drosophila
the
rate
of
more origins
coli
DNA
concentrated in
polymerization.
of
DNA.
and
Stahl grew cells in
media that contained
different
of
nitrogen
(15N and 14N) so
that the
DNA molecules
from these different
isotopes
could be
distinguished by
mass.
Explain
in
the
how
“light” DNA
was
separated
from
“heavy”
model
s
for
DNA
them was
ruled
dispersive
model of
Meselsonand Stahl
B.
Describe
the
three existing
experiments.
replication when these studies
were begun, and
6 definitively by
explain
the
how
one
6-
of
experiment you
described
for
part
A.
C.
What experimental result eliminated
the
out
replication?
6- Indicate
whether
7 a
statement
is
the
following
false, explain
statements are
why it
is
true or
false.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
false. If
DNA
lOMoARcPSD|30659517
A.
When DNA is
heatingoccurs,
being replicated
allowing
two
strands
to
separate.
replication
origins
B.
DNA
base pairs.
C.
Meselson
and
Stahl
forDNA replication. D.
DNA
process that is
initiated
are
inside a
the
cell,
local
typically
rich
in
G-C
ruled out the
replication is
at multiple
dispersive
model
a
bidirectional
along chromosomes
in
eukaryotic
cells.
How many replication forks are
is
opened?
(a)
1
formed
when an
origin of
locations
6- (b)
replication
2
8
(c)
3
(d)
4
Answer
A.
is
6- the
the
following
questions
On
a
DNA
growing—the 3′
strand that
end,
5′
both ends? Explain
end,
or
about DNA replication.
is
being synthesized, which end
your
answer.
9
B.
On
a
DNA
template,where is
occurring
5′,
or
relative
both?
6- How does the
10 cells compare
strand that is
being used as
the
copying
to
the
total number
of
with the
number
(a)
1
versus 100
(b)
5
versus 500
replication
origin—3′
replication origins
of
origins
in
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
a
of
the
in
bacterial
human
cells?
origin,
lOMoARcPSD|30659517
6-
(c)
10
versus 1000
(d)
1
versus 10,000
Which of
11 for
the
following
DNA replication
Page
2
of
(a)
The replication
theconditions.
(b)
on
The
the
statements
correctly
to
be
bidirectional?
fork
can
open or
explains
what it
means
25
DNA replication
template
machinery
can
close, depending
move in
on
either direction
strand.
(c)
Replication-fork
converges
on
another
(d)
The
opposite
movement
replication
can
switch directions
at
the
6-12 The chromatin structure
complicated
than that
in
eukaryotic
observe
dreason
This
is
prokaryotic cells.
DNA replication
occurs much faster in
(a)
2×
(b)
5×
fork
fork.
replication forks formed
directions.
in
tha
t
when the
prokaryotes. How
origin move in
cells
is
much more
thought
to
be
much faster is
it?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
the
lOMoARcPSD|30659517
(c)
10×
(d)
100
×
6-13
DNA polymerase catalyzes
nucleotideto
a
the
joining
growing
of
from catalyzing
the
What prevents
this
enzyme
(a)
(Pi)
hydrolysis
+
Pi
of
pyrophosphate
(b)
release
of
PPi
from the
(c)
hybridization of
the
new
strand to
the
(d)
loss
as
an
energy
source
knowledge
explain
of
of
ATP
6-14 Use
your
synthesized to
replication
words,
must occur in
what would be
consequences
6-15
of
Figure Q6-15 shows a
to
strand.
reverse
inorganic
reaction?
phosphate
nucleotide
how a
why DNA
the
the
3′–to-5′
(PPi)
a
DNA
5′-to-3′
template
new
strand of
direction.
In
DNA
is
other
strand elongation?
replication
bubble.
Figure Q6-15
On
A. (use
the
O).
figure, indicate
where the
origin of
replication
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
was
located
lOMoARcPSD|30659517
Label the
B. of
the
leading-strand
template
and
the
fork
[R]
X
and
Y,
respectively.
by
arrows
the
direction
(indicated
by
dark lines) were synthesized.
Number
D. 3
in
the
the
Okazaki
order in
right-hand
Indicate
C. strands
E.
as
which they
were synthesized.
Indicate
where the
on
the
direction
Page
of
25
of
movement
which the
of
template
newly made DNA
each strand as
most recent DNA synthesis
Indicate
F. arrows.
3
fragments
in
lagging-strand
1,
2,
and
S).
has
occurred
(use
the
replication
forks with
Use the following information about a
series of
in
vitro DNA replication
experiments to answer
questions
6-16
through
622.
You
prepare
bacterial
cell
extracts
by
lysing the
cells
andremoving insoluble debris via
centrifugation.
These extracts
provide
the
proteins
required
for
DNA replication. Your DNA
template
is
a
small, double-stranded
circular
piece of
DNA
(a
plasmid)
with a
single origin of
replication and a
single
replication termination site. The termination site is
on
the
opposite
side of
the
plasmid
from the
origin.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
6-16 In
addition
to
the
extracts
are there any
additional
materials
and
you
should
add
system?Explain
replication
6-17
to
Which of
this in
to
your
this
in
vitro
answer.
the
following
statements is
vitro replication
the
plasmid
DNA,
true
with
respect
strand and
one
lagging
system?
(a)
There will
strand produced
be
only
using
one
leading
this
template.
(b)
of
The leading
and lagging
each
newly synthesized
DNA
strand.
(c)
The
places on
DNA replication
this
strands
compose
machinery
can
assemble
molecule
will
be
other.
at
one
half
multiple
plasmid.
(d)
One daughter
shorterthan
You
protei
n
examine
6-18
decide
to
of
DNA
the
use
different
the
replication
the
role of
normalprocess of
What part of
the
mostdirectly affected
of
bacteria
extracts?
bacterial
lacking
DNA
if
initiation
of
DNA
(b)
Okazaki
fragment
(each having one
mutated)
proteins
replication
primase
(a)
strains
machinery
individual
DNA
slightly
in
order to
in
the
replication.
process
a
would be
strain
were used to
make the
synthesis
synthesis
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
cell
lOMoARcPSD|30659517
(c)
leading-strand
elongation
(d)
lagging-strand
completion
6-19 What part of
directly
affected
of
bacteria
polymerase
the
if
DNA
a
replication
strain
process
lacking
the
exonuclease activity
were used to
make
the
cell
extracts?
Page
4
of
25
(a)
initiation
of
(b)
Okazaki
(c)
leading-strand
elongation
(d)
lagging-strand
completion
What part of
affected
if
make the cell
would be
of
most
DNA
DNA synthesis
fragment
synthesis
the
DNA replication process
would be
most
directly
a
strain of
bacteria
lacking
helicase
were used to
extracts?
6- (a)
20
initiation
of
DNA
synthesis
(b)
Okazaki
fragment
(c)
leading-strand
elongation
(d)
lagging-strand
completion
synthesis
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
What part of
affected
if
bacteria
the
a
lacking
DNA replication
strain of
the
process
cell
single-strand binding
would be
protein
most
directly
were used to
make
extracts?
621 (a)
initiation
of
(b)
Okazaki
fragment
(c)
leading-strand
elongation
(d)
lagging-strand
completion
What part of
affected
if
make the cell
DNA
synthesis
synthesis
the
DNA replication process
would be
most
directly
a
strain of
bacteria
lacking
DNA ligase were used to
extracts?
6- (a)
22
initiation
of
(b)
Okazaki
fragment
(c)
leading-strand
elongation
(d)
lagging-strand
completion
6- Which of
the
23 a
human
(a)
a
DNA
synthesis
synthesis
following
statements
chromosome is
true?
about the
newly synthesized
strand of
It
was synthesized from a
single origin solely by
continuousDNA
synthesis.
(b)
It
was
synthesized from a
single origin by
mixture
of
continuous
anddiscontinuous
DNA
synthesis.
(c)
It
was
discontinuous
synthesized from multiple
DNA
origins
solely by
synthesized from multiple
origins
by
synthesis.
(d)
It
was
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
a
lOMoARcPSD|30659517
mixture
of
continuous
anddiscontinuous
DNA synthesis.
You have discovered an
“Exo–” mutant
form of
DNA
polymerase in
which the
3′-to-5′
exonuclease function
has been destroyed
but the
ability to
join nucleotides
together
is
unchanged. Which of
the
following
properties do
you expect the
mutant
polymerase to
have?
6- (a)
It
24 and
the
slowly than
(c)
the
to
will
polymerize in
3′-to-5′
direction.
the
normal
Exo+
It
will
normal
generate
of
both the
5′-to-3′
direction
(b)
It
will
polymerize
polymerase.
fall
off
the
template
Exo+ polymerase. (d)
It
mismatched base pairs.
Page
5
6-25
A
molecule
yeastcell
nucleus
more frequently than
will
be
more
likely
25
of
is
but fails
think the
to
arises?
that is
Choose
the
most probably
bacterial
DNA
imported
replicate
with the
block to
replication
protei
n
responsible for
the
failure to
anexplanation for
introduced
into
into
yeast DNA. Where
primase
(b)
helicase
a
the
do
or
protein
complex
replicate
your
bacterial
DNA. Give
answer.
(a)
more
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
you
below
lOMoARcPSD|30659517
(c)
DNA
polymerase
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(d)
initiator
6-26
Most cells in
thetelomerase
gene is
important
proteins
turned
step
in
the
body of
enzyme
an
adult human
because
off
is
therefore
not
cell
into
cancer which circumvents
and
lack
its
expressed.
An
the
cell, growth
conversion
normal
control,
why
be
over
of
a
normal
is
the
resumption of
telomerase might
necessary
again.
for
the
of
a
replication
Page
6
of
telomerase
ability of
6-27 Which diagram
accurately
strands
at
one side
a
expression. Explain
cancer cells
represents
the
to
divide over and
directionality of
DNA
fork?
25
6- Indicate
whether
28 a
statement
is
the
following
false, explain
statements are
why it
is
A.
on
Primase
both
th
e
is
neede
to
leading d
strand
and
the
lagging
strand
.
initiat
e
true or
false.
DNA
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
false. If
replication
lOMoARcPSD|30659517
B.
The sliding clamp is
strand,where it
associated
until
loaded
remains
replication
once on
is
complete.
each DNA
C.
own
Telomerase is
RNA molecule
a
to
DNA
use
polymerase that
as
a
primer
the
end
of
a
proofreading function
at
D.
Primase
there are no
the
RNA
requires
errors in
primers
used for
DNA
the
carries
its
lagging
strand.
that
ensures
replication.
Because
all
DNA polymerases synthesize DNA in
the
5′-to-3′
direction,
and the parental
strands
are antiparallel,
DNA replication
is
accomplished with the
use of
two
whether
the
629 mechanisms:
continuous
and
following
items relate to
discontinuous
replication,
discontinuous
(1)
or
replication.
continuous replication,
(3)
both modes of
Indicate
(2)
replication.
primase
single-strand binding
protein
sliding clamp
RNA
primers
leading
strand
lagging
strand
Okazaki
fragments
DNA
helicase
DNA
ligase
6- The synthesis
of
30 direction.
However,
DNA in
living systems
scientists
synthesize
short
occurs in
the
DNA sequences
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
5′-to-3′
needed
lOMoARcPSD|30659517
for
their experiments on
an
instrument
A.
The chemical
proceeds
in
the
synthesis
3′-to-5′
of
direction.
possible
diagram
toexplain
the
Draw a
and
dedicated
to
this
by
this
instrument
show how
this
is
of
is
chemically
DNA
task.
process.
B.
Although
possible,
it
3′-to-5′
synthesisdoes
not occur in
living systems.
Why
DNA
not?
DNA polymerases are processive, which means that they remain
tightly
associated with the
template
strand while moving
rapidly
and
adding
nucleotides to
the
growing
daughter
strand.
Which piece
of
the
replication
machinery accounts
for
this characteristic?
6- (a)
31
helicase
(b)
sliding clamp
(c)
single-strand binding
(d)
primase
Page
7
6-32 Use
diagram
protein
of
25
the
of
components in
the
list
a
replication fork in
below to
Figure Q6-32.
1. DNA polymerase
2. single-strand binding
protein
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
label
the
lOMoARcPSD|30659517
3. Okazaki fragment
4. primase
5. sliding clamp
6. RNA primer
7. DNA helicase
Figure
Q6-32
6-33 Researchers have isolated
a
that carries
a
temperature
sensitive
variant
of
thepermissive
mutant
the
cells grow just
nonpermissive
grown at
compared
DNA isolated
results
are
strain of
the
enzyme
temperature, the
DNA
ligase. At
as
wild-type
temperature, all
of
the
cells
2
hours. DNA from mutant
cells
is
mutant
well
as
the
in
the
culture
the
nonpermissive
with the
temperature for
from cells
the
grown at
E.
coli
cells.
At
tube die
30
permissive
minutes
temperature.
The
shown in
size
by
Figure Q6-33, where DNA
molecules
have been separated
means of
Explain
electrophoresis
the
permissive;
NP,
nonpermissive).
distinct
band with
the
sample
a
size
appearanc of
epairs
(bp)
a
in
(P,
within
of
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
200
by
base
lOMoARcPSD|30659517
collected
at
the
nonpermissive
temperature.
Figure
Page
8
of
25
6-34 Indicate
false. If
a
whether
the
statement is
explain
it
why
is
1. The repair polymerase is
synthesized
strands
to
1. There is
and
ensure
a
lays
corresponding
DNA
fragments
following
false,
statements are
the
enzyme
that
proofreads
the
accuracy
of
DNA
replication.
the
RNA primers
on
sequence
the
to
it
is
synthesizing DNA
6-35
Which of
the
following
proofreadingduring DNA
the
that
degrades
behind
it.
seal
sugar–phosphate
the
lagging
1. The repair polymerase does not
clamp, because
only
true
or
false.
single enzyme
down the
1. DNA ligase is
required
between
all
the
DNA
Q6-33
newly
backbone
strand.
require
the
over very
aid
of
short stretches.
statements about sequence
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
the
sliding
lOMoARcPSD|30659517
replication
is
false?
(a)
DNA
The
the
exonuclease activity
polymerase.
(b)
The
direction.
is
in
exonuclease activity
a
cleaves
different
domain
DNA
the
in
(c)
The DNA proofreading activity
strandelongation.
occurs concomitantly
(d)
If
an
incorrect
beforeremoval.
added, it
6-36 The
binds to
base is
sliding clamp complex
DNA
polymerase. This
of
DNA
helps the
the
polymerase synthesize
is
“unpaired”
DNA
template
must happen
each time
made on
the
lagging
a
new
Okazaki
does the
expedite
this
process?
The DNA duplex
consists
of
covalent polymers
wrapped
around
two
each
long
it
is
strand.
How
cell
many times over their entire length. The
strands
for
separation
replicatio
causes
n
the
the
replication
to
be
the
torsional
strands
and
much longer stretches
clamp only
strand,
fragment
How
the
with
the
dissociating. While the
the
leading
loading
5′-to-3′
of
without
once
o
n
6-37
encircles
of
occurs
other
of
the
“overwound” in
DNA
front of
fork.
does the
cell
DNA duplex
relieve
during
stress created
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
along
lOMoARcPSD|30659517
replication?
(a)
will
Nothing
needs to
be
separated
be
done because
the
two
strands
the
backbone
afte
r
replication
is
complete.
(b)
break the
covalent
bonds of
Topoisomerase
s
allowing
unwinding
the
local
of
DNA
(c)
Helicase
replication is
DNA
(d)
replac
e
ahead of
the
replication
unwinds
the
complete.
DNA
and
remov
e
repair
enzymes
incorrectly
fork.
rewinds
torsiona
l
it
stress as
after
they
paired bases.
6-38 Telomeres
chromosomes.
serve as
Which of
caps at
the
ends of
the
following
is
not
regarding
the
(a)
The lagging-strand
replicatedby
true
replication
telomeres
linear
of
telomeric
are not
DNA
completely
sequences?
polymerase.
(b)
Telomeres
are
made of
(c)
Additional
repeated
templatestrand.
(d)
The leading
aprimer
repeating
sequences
strand doubles
for
sequences.
are
added to
the
back
the
on
itself
lagging
to
strand.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
form
lOMoARcPSD|30659517
Page
9
of
25
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
DNA Repair
Sickle-cell
anemia
disease.
Individuals
6-39
disorder
by
with
have misshapen
change
in
is
this
an
example
red
(sickle-shaped)
the
of
an
blood cells
inherited
caused
a
sequence
of
the
thechange?
(a)
chromosome loss
(b)
base-pair
(c)
gene duplication
(d)
base-pair
6-40 Even
function,it
in
gene. What is
the
nature of
change
insertion
though
still
the
β-globin
DNA polymerase has a
introduces
errors
proofreading
what degree
newly synthesized
strand at
does the
mismatch
arising
from DNA
a
repair system
rate
of
1
per
107nucleotides.
decrease
the
error
rate
replication?
(a)
2-fold
(b)
5-fold
(c)
10-fold
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
To
lOMoARcPSD|30659517
(d)
100-fold
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
6-41
to
Which of
repair the
shown in
Figure
below represents the
correct
way
in
DNA
Figure Q6-41?
Q6-41
6-42 A
backbone.
only
of
the
choices
mismatch
mismatched base pair causes
If
this
were the a
indication
mutatio
n
higher.
of
an
would be
Explain
replication,
the
the
overall
rate
backbone
caused
by
DNA
indicate
which
why.
Page
10
6-43
a
Beside the
distortion
in
the
mismatche base pair, what
d
additiona
lstrand
of
error in
much
distortion
25
mark is
need to
s
ther
e be
on
the
template
DNA
eukaryotic
to
repaired?
(a)
a
nick
in
(b)
a
chemical
modification of
strand
the
new
strand
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(c)
a
nick
in
the
new
strand
(d)
a
sequence
gap
in
the
6-44
A
pregnant
achemical.
litter
deformed,
new
strand
mouse is
exposed
Many of
the
to
mice
but
interbred
when they
are
high
in
levels of
her
with
each
ar
e
all
other
,
their offspring
are normal.
Which two
explain
these results?
of
(a)
In
were
mice, somatic
the
deformed
mutated.
the
following
cells
statements could
but
not
germ cells
(b)
The
mouse’s
original
(c)
In
were
the
deformed
mutated
.
(d)
The
toxic
chemical
6-45
The repair of
nucleotidesin
were mutated.
mice, germ cells
but
affects development but
mismatched base pairs or
a
requires
known
a
multistep
sequence
of
events in
this
(a)
DNA
isidentified
damage
by
an
existing
new
germ cells
not
somatic
is
not
damaged
DNA
cells
mutagenic.
strand
process.
Which choice below describes
the
recognized,
the
newly synthesized strand
nick in
the
backbone,
strand is
removed
by
a
segment
by
DNA
process?
is
repair proteins,
the
gap
is
thestrand is sealed
filled
of
the
polymerase, and
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
by
DNA
ligase.
(b)
it
DNA
and
repair polymerase simultaneously
polymerizes the
correc
sequenc
moves
along tthe etemplate.
DNA
ligase seals
(c)
DNA
isidentified
existing
new
the
damage
by
an
is
the
recognized,
the
(d)
A
nick
proteinsswitch out
in
gap
is
the
DNA is
wrong
the
insert the
correct
bases ahead of
behin
d
it
repaired
strand.
the
nick in
the
backbone,
a
strand is
removed
by
an
exonuclease, and
base and
nicks in
remove
s
repaired
as
it
newly synthesized strand
segment
by
DNA
recognized,
base, and
DNA
of
the
ligase.
DNA
repair
ligase seals
the
6-46 Human
pigmentosum
beings with the
have
seriou
s
inherited
disease
problem
scancer
with
with
lesions
repeate
d
on
their skin
and
often develop
exposur
ebeing
to
sunlight.
recognized in
DNA
damage
of
these individuals?
(a)
chemical
(b)
X-ray irradiation
(c)
mismatched bases
(d)
ultraviolet
What typ
the
ecells
of
xeroderma
damage
damage
irradiation
nick.
damage
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
skin
is
not
lOMoARcPSD|30659517
6-47
You are
theenzyme
examining
phosphofructokinase
noticethat the
sequence
tha
t
in
the
Page
11
the
DNA
in
skinks and
coding
actually
enzyme
of
25
similar
in
vary quite a
the
bit.
sequences
that
Komodo
dragons.
directs
is
the
sequence
code for
of
but that the
surrounding
most likely
explanation for
(a)
Coding
sequences
are
repaired
more efficiently.
(b)
Coding
sequences
are
replicated
more accurately.
(c)
Coding
sequences
thechromosomes
to
are packaged
protect
more tightly in
damage.
(d)
Mutations
deleterious to
in
the
organism
coding sequences
than mutations
amino acids
very
two organisms
What is
the
them from DNA
You
in
are
noncoding
more likely to
sequences
this?
be
sequences
.
In
somatic
cells, if
a
base is
mismatched in
one
new
daughter
strand during DNA replication, and is
not
repaired,
what
fraction
of
the
DNA duplexes
will
have a
permanent change
in
the
DNA sequence
after the
second
round of
DNA replication?
6- (a)
1/2
4
8 (b)
1/4
(c)
1/8
(d)
1/16
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Sometimes,
replication
correct
mutation.
and not
6- mutation
4 rounds
9
(b)
TUAT
(c)
TGAT
(d)
TAAT
chemical
damage
to
DNA can occur just before
DNA
begins,
not giving the
repair system
enough
time to
the
error before the
DNA is
duplicated. This gives rise to
If
the
cytosine
in
the
sequence
TCAT is
deaminated
repaired,
which of
the
following
is
the
point
you would observe
after this segment
has
undergone two
of
DNA replication? (a)
TTAT
During
DNA replication in
a
bacterium, a
C
is
accidentally
incorporated instead
of
an
A
into one
newly synthesized DNA
strand.
Imagine
that this error
was not corrected
and that it
has no
effect on
the
ability of
the
progeny
to
grow and
reproduce. A.
After this original
bacterium
has divided
once, what
its
6- proportion of
5
would you
expect to
contain
the
mutation?
0 progeny
B.
What proportion
tocontain
the
after three more rounds
of
its
mutation
of
progeny
DNA
would you
replication
and
expect
cell
division?
6- Sometimes, chemical
damage
to
DNA can occur just before
DNA
5 replication begins,
not giving the
repair system
enough
time to
1 correct
the
error before the
DNA is
duplicated. This gives rise to
mutation.
If
the
adenosine in
the
sequence
TCAT is
depurinated
and
not repaired,
which of
the following
is
the
point
mutation
you would observe
after this segment
has
undergone two
rounds
of
DNA replication? (a)
TCGT
(b)
TAT
(c)
TCT
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
TGTT
(d)
Page
12
of
25
6-52
Which of
the
accuratestatement
following
statements is
about thymine
not
an
dimers?
(a)
to
Thymine
stall.
dimers
can
cause the
DNA replication
machinery
(b)
on
Thymine
opposite
dimers
DNA
are
covalent
links
thymidines
exposure
to
sunlight
causes thymine
recognize
the
thymine
dimers
DNA
6-53 Indicate
false. If
a
whether
the
statement is
following
false,
statements are
explain
it
is
false.
1. Ionizing radiation
and
oxidative
between
strands.
(c)
to
Prolonged
form.
(d)
Repair proteins
adistortion
in
dimers
as
backbone.
why
damage
can
true
or
cause DNA double-strand
breaks.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1. After backbone
damaged
are
DNA has
been
repaired,
nicks in
the
phosphate maintained as
a
way
to
identify
the
repaired.
is
caused
by
mechanism
that
ensures
strand that
1. Depurination of DNA is
ultraviolet irradiation.
2. Nonhomologous end
DNA double-
strand breaks
fidelity
to
was
a
rare
event that
joining
is
a
are repaired
the
original
with a
DNA
high
degree
that
of
sequence.
6-54
the
Several
same
members
kind of
when they were unusually
the
most likely
of
the
cancer
young. Which one
explanation for
this
phenomenon?
theindividuals
with
have
same family were diagnosed
It
the
of
the
following
is
possible
cancer
.
cancer-causing
an
gene that
ancestor’s
inherited
amutation
a
mutation
in
a
gene required
for
(c)
inherited
a
mismatch
repair.
mutation
in
a
gene required
for
inherited
a
thesynthesis of
mutation
purine
in
a
gene required
for
somatic
suffered
cells.
(b)
inherited
synthesis.
(d)
is
that
a
in
(a)
with
nucleotides.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
DNA
lOMoARcPSD|30659517
6-55 You
have made a
thataredefective
collection
of
in
various
mutant
aspects
fruit
of
to
each mutant
for
hypersensitivity
DNA repair. You test
three DNA-damaging
agents:
cytosine),
acid
sunlight,
nitrous
and
formic
(which
causes depurination).
summarizedin
“yes” indicates
normal
fly,
where a
than a
blanks indicate
normal
Page
25
Table
acid
13
of
its
(which
causes deamination of
The results
Table Q6-55,
that the
and
flies
mutant
are
is
more sensitive
sensitivity.
Q6-55
1. polymerase?
Which mutant
is
most likely to
theDNA repair
1. What aspect of
other mutants?
6-56
is
nucleic
caused
The
a
repair is
most likely to
deamination of
cytosine
naturally
occurring
acid
by
base, and
damage
be
so
due
defective
be
generates
does not
to
in
affected
a
represent
in
the
uracil base. This
a
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
DNA
lesion
lOMoARcPSD|30659517
chemicals
“foreign”
or
and
irradiation. Why is
why is
it
important
for
cells
andremove
to
found in
the
duplex?
DNA
this
base recognized
have a
mechanism
uracil when it
to
is
as
recognize
Homologous Recombination
6-57
Select the
option that
Nonhomologous
joining
end is
is
best
completes
process
by
which a
statement
:
to
a
similar
thecomplementary
stretch
(b)
after repairing
mismatches.
(c)
to
(d)
after filling in
maintainthe
DNA
sequence.
6-58
of
Nonhomologous
end
the
following?
any
nearest
recovery
double-stranded
DNA
.
(a)
(a)
the
strand
following
end
a
joined
the
the
of
sequence
chromosome.
on
double-stranded
DNA
nucleotides, helping
of
to
the
available
any
lost
integrity
of
(b)
the
interruption of
(c)
loss
of
(d)
translocations
entirelydifferent
joining
can
result in
lost
nucleotides on
all
a
but
which
damaged
DNA
gene expression
nucleotides at
the
site
of
repair
of
DNA fragments
chromosome
to
an
end.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
6-59 Homologous recombination
which organisms
use a
“backup”
genetic
for
is
an
copy of
the
DNA as
double-strand breaks
without
information. Which of
homologous
recombination
to
the
important
a
mechanism
template
loss
following
is
in
to
of
fix
not
necessary
occur?
(a)
3′
DNA
strand overhangs
(b)
5′
DNA
strand overhangs
(c)
a
long
stretch
Page
14
of
25
(d)
nucleases
of
sequence
similarity
6-60 In
addition
to
the
repair of
strandbreaks, homologous recombination
is
DNA
a
mechanism
segments
of
for
generating
parental
diversity
chromosomes.
During
(a)
DNA
replication
(b)
DNA
repair
(c)
meiosis
(d)
transposition
genetic
which process
double-
by
does swapping
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
swapping
occur?
lOMoARcPSD|30659517
Recombination
has occurred
chromosomesegments shown in
6-61
Q6-61. The
b,
are
genes A
and B,
used as
markers
the
maternal
alignment
and
,
and
paternal
between
and the
on
the
Figure
recessive
chromosomes,
alleles a
and
respectively. After
have
b
homologous recombination,
B,
aand
the
specific
arrangements
of
A,
changed.
Figure Q6-61
Which of
from the
the
choices
below correctly
replication products
of
the
(a)
AB
and
aB
(b)
ab
and
Ab
(c)
AB
and
Ab
(d)
aB
and
Ab
6- The
62 to
events listed below are
occur properly:
A.
Holliday
junction
B.
strand invasion
C.
DNA
synthesis
D.
DNA
ligation
all
indicates
maternal
necessary
for
cut
ligated
and
the
gene
combination
chromosome?
homologous
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
recombination
lOMoARcPSD|30659517
E.
double-strand
F.
nucleases
Page
15
Which of
of
break
create uneven
strands
25
the
following
events during
is
the
correct
homologous
order of
recombination?
(a)
E,
B,
F,
D,
C,
A
(b)
B,
E,
F,
D,
C,
A
(c)
C,
E,
F,
B,
D,
A
(d)
E,
F,
B,
C,
D,
A
6-63 Homologous recombination
strandbreaks
(DSBs) in
is
a
chromosome. DSBs arise from DNA
harmfulchemicals or
radiation
specialized
(for
cell
produce
s the
gametes
(spermcells
intentionally
cause DSBs so
at
least one
homologous
chronot
example,
division
initiated
damage
by
X-rays).
that
and
by
caused
double-
by
During
meiosis,
the
eggs) for
sexual reproduction,
as
to
stimulate
crossover
homologousrecombination. If
not
there is
occurrence
within each pair
of
crossing-over
segregate
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
of
lOMoARcPSD|30659517
mosomes
during meiosis,
those noncrossover
chromosomes
will
properly.
Figure
Q6-43
1. Consider the
copy of
Chromosome 3
your mother.
Is
it
identical
mother
to
(her
the
Chromosome 3
that
tha
t
you
received
sh
e
received
from
from her
materna
chromosome)
lshe
received
from
or
identica
l
to
the
Chromosome 3
her
chromosome),
or
neither?
Explain.
father (her
1. Starting with
paternal
the
representation
in
maternal
and paternal
mother,draw two
chromosomes
possible
chromosomes
you
1. What does this
indicate
may
Figure Q6-43 of
the
found in
your
have received
from
about your resemblance to
double-stranded
your
mother.
your grandfather and
grandmother?
6-64 Indicate
false. If
a
whether
the
statement is
explain
it
why
is
haploid,
and therefore
thechromosome to use as
template
for
statements are
true
or
false.
1. Homologous recombination
they
are
following
false,
cannot
occur in
have no
a
prokaryotic cells, because
extra copy
of
repair.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1. The first
stretch
from
step in
repair requires
of
base pairs
5′
the
1. The 3′
can
end
overhang
be
used
as
a
primer
Page
16
of
of
to
the
site
the
remove
of
homologous DNA duplex,
the
repair DNA
polymerase.
used to
repair the
broken
a
the
break.
which
25
chromosome inherited
17
nuclease
each strand at
“invades”
for
1. The DNA template
homologous
Page
of
a
from the
strand is
the
other parent.
25
ANSWERS
6-1
(c)
6-2
(b)
Choice (c)
is
the
are
false. Although
a
correct
replication
statement,
is
called
correct
answer.
choice (d)
is
Choices
(a)
and
it
reason
that
DNA
is
not
the
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
semiconservative.
6-3
(b)
6-4
(b)
6-5
Choice (d)
is the
of
and
correct
answer.
Bacteria
has
many. Choice (a)
genome
is
is
incorrect
because
genome.
Choice (b)
eukaryotic
is
Drosophila
Drosophila
bigger than the
E.
coli
incorrect,because
polymerases are
not
faster than prokaryotic
A.
The DNA samples
centrifugetubes containing
6-6
cesium
the
light
chloride.
heavy and
DNA
have one
products
collected
the
polymerases.
were placed into
After high-speed
centrifugation
were separated
by
1. The three models
origin replication,
for
2
density.
were conservative, semiconservative,
and
dispersive.
conservative model suggested
parental
a
mechanism
by
which the
strands
stayed together
after replication
and
the
duplex
was
made
entirely
of
newly synthesized DNA. The
semiconservative
model
two
produced
during
proposedthat the
replication
DNA
duplexes
days,
original
daughte
r
were hybrid
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
The
lOMoARcPSD|30659517
molecules, each having one
oneof the
of
the
newly
synthesized strands.
newDNA
The
dispersive
model predicted
each contained
all
along
segments
of
duplexes
strands
the
molecule.
The
density-gradient
parental
conservative model was
strands
and
that
the
parental
and
daughter
ruled out
by
the
experiments.
1. The dispersive
DNA
model was
ruled out
duplexes
stranded
and then comparing
DNA. If
the
dispersive
have had
model had
an
the
been
by
using heat
to
densities
of
individual
strand
s
was
the
correct
denature
the
the
single-
should
,
intermediat density.
eheavy
strands
light
strands
However,
and
were observed,
thesemiconservative
DNA
6-7
A.
replication
this
not
case; only
which convincingly supported
model for
replication.
False. The two strands
to
occur, but this is
accomplished by
the
binding
theorigin of replication.
1. False. DNA replication
pairs, which are
held together
for
C-G
by
base
origins
only
do
need to
of
initiator
are
typically
separate
proteins
rich
at
in
A-T
of
two
hydrogen
bonds (instead
pairs), making
it
easier to
atthese sites.
separate
the
strands
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
for
base
three
lOMoARcPSD|30659517
1. True.
2. True.
Page
18
6-8
(b)
6-9
to
A.
the
nucleic
of
The 3′
end. DNA
3′-OH
end of
a
acid
1. Both, as
a
replication.
6-10
25
polymerase can
add
nucleotides only
chain.
result of
the
bidirectional nature of
chromosomal
(d)
6-11 (d)
6-12
(c)
6-13
(a)
6-14 There would be
several
to-5′ strand
elongation.
detrimental consequences
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
to
3′–
lOMoARcPSD|30659517
One of
replication
those most directly
involves
synthesis
of
the
lagging
are
degraded
the
DNA
,
segments
group. The
remaining
will
incoming
nucleotid
eprovided
will
by
have
linked to
the
strand
.
After the
have 5′
ends with
3′-OH group.
release
Without of
processes
RNA
of
DNA
primers
a
single phosphat
e
the
energy
athe
PPi
from the
5′
end,
nolonger be
energetically
the
process
of
elongation
would
favorable.
6-15
See
Figure
Figure A6-15.
A6-15
You will probably
add exogenous nucleoside triphosphates
to
serve as
the
building
blocks needed
to
make
new strands
of
DNA.
Although
these monomers will
be
present
in
the
extracts,
they
6will be
present
at
lower concentrations
than are normally
found
1
inside
the
cell. They may also be
subject
to
hydrolysis,
and
6
the
nucleoside diphosphates that are the
products
of
this hydrolysis
are not usable substrates for
DNA
replication. For both of
these
reasons,
it
is
important
to
add excess nucleotides to
the
reaction
mixture
for
efficient
DNA replication to
occur.
6- (b)
Leading
and lagging
strands
are
1 from the
replication origin, and are joined
7 the
two replication forks
meet at
the
synthesized
bidirectionally
together
by
DNA ligase where
termination site. Choice (a)
is
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
not correct,
because
this answer
implies
that the
replication
fork is
not bidirectional and that replication continues
around
the
plasmid
until the
process
makes it
back to
the
origin of
replication. Choice (c)
is
Page
19
of
25
incorrect
because
the
origin is
a
specialized sequence
where
initiator
proteins
bind and open the
DNA so
that the
DNA
replication machinery can assemble.
Choice
(d)
is
incorrect
because
the
daughter
DNA
molecules
will be
same size as
the
original
plasmid
(and each other).
6- Choice (a)
is
the
best answer
because
DNA synthesis
cannot
1 begin without
the
initial primers.
Choice (b)
is
a
good answer
8 because
lagging-strand
synthesis
requires
continual
use of
RNA
primers
for
discontinuous
replication to
occur.
61
(d)
9
6-
(a)
Because
replication
0 helicase
of
at
helicase
unwinds
both strands
the
time of
the
two
depends2
upon
initiation.
DNA template
strands,
the
of
activity
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
62
(b)
1
62
(d)
2
(d)
Each newly synthesized strand in
a
daughter
duplex
was synthesized
by
a
mixture
of
continuous and
discontinuous
DNA synthesis
from multiple
origins.
Consider
a
single replication origin. The fork
6- moving
in
one direction
synthesizes a
daughter
strand continuously
2 as
part of
leading-strand
synthesis;
the
fork moving
in
the
3 opposite
direction
synthesizes a
portion
of
the
same daughter
strand
discontinuously
as
part
of
lagging-strand
synthesis.
62
(d)
4
Choice (d)
is
the
correct
answer.
DNA from all
organisms is
6- chemically identical
except for
the
sequence
of
nucleotides. The
2 proteins
listed in
choices
(a)
to
(c)
can act
on
any DNA
5 regardless of
its
sequence.
In
contrast,
the
initiator
proteins
recognize
specific
DNA sequences at
the
origins
of
replication.
These sequences differ between
bacteria
and
yeast.
6- In
the
absence
of
telomerase, the
life-span
of
a
cell and
2 its
progeny
cells is
limited.
With each
round of
DNA replication,
6 the
length of
telomeric
DNA
will shrink,
until finally all
the
telomeric
DNA has
disappeared. Without telomeres
capping
the
chromosome
ends, the
ends might be
treated
like breaks
arising
from DNA damage,
or
crucial genetic
information might
be
lost. Cells
whose DNA lacks telomeres will stop dividing
or
die. However,
if
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
telomerase
indefinitely
despite
is
provided
to
cells, they may be
because
their telomeres
will remain
repeated
rounds
of
DNA replication.
able to
divide
a
constant
length
62
(d)
7
62
A.
True.
8
Page
20
of
25
1. False. Although
the
leading
strand,
sliding clamp is
only
loaded
once on
the
lagging
strand needs to
thepolymerase
unload
reaches
the
the
RNA primer
reloadit
previous
new
segment
primer
has
from the
where a
the
clamp once
and
then
been synthesized.
1. True.
2. False. Primase
need one
does not
have a
because
the
RNA primers
the
DNA. The primers
are
are
removed,
and a
aproofreading function
DNA
polymerase that
fills
in
gaps.
the
remaining
proofreading function,
not
a
nor
permanent part
does have
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
does it
of
lOMoARcPSD|30659517
6-29
primase
3
2
single-strand binding
3
sliding clamp
3
RNA
protein
primers
1
leading
strand
2
lagging
strand
2
Okazaki
fragments
3
DNA
helicase
2
DNA
ligase
6-30
the
A.
The actual chemical
same
regardless of
whether
direction.
reaction
going in
the
5′-to-3′
The most important
or
distinction between
synthesized in
the
these two
3′-
options
to-5′ direction,
rather
tha
n
5′
3′
of
will
th
e
the
en
the
in
DNA
synthesis
in
the
3′-to-5′
is
that
elongating
if
DNA
is
is
strand,
d
end,
have a
nucleoside
1. DNA synthesis
the
last
from 3′
triphosphate.
to
nucleotide
added is
mispaired
lastnucleotide
growing
nucleotide
strand is
coming
a
5′
and
nucleoside
does not
is
on
allow proofreading.
removed,
the
the
monophosphate
and
in
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
the
If
lOMoARcPSD|30659517
only
no
has a
hydroxyl
favorable
hydrolysis
6-31
(b)
Page
21
reaction
6-32
to
of
25
See
Figure A6-32.
3′
group on
the
drive the
addition
end.
Thus, there is
of
new
nucleotides.
Figure A6-32
DNA ligase has an
important
role in
DNA replication. After
Okazaki
6- fragments
are synthesized, they must be
ligated
(covalently connected) to
3
each other so
that they finally
form one continuous strand.
At
the
3 nonpermissive
temperature this does not happen, and although theremay
be
a
range of
fragments, the
notable
band at
200 base
pairs is
the
typical size of
an
individual
Okazaki
fragment.
6- A.
False. The
3 vacant after the
4
RNA
primers
repair polymerase is
are
used to
in
the
spaces left
degraded.
B.
False. This
is
different
enzymes.
a
two-step
First, a
nuclease
polymeras
e
the
th
e
removes
fills
in
fill
RNA
process
primers.
that
Then, the
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
requires
repair
two
lOMoARcPSD|30659517
complementary
C.
True.
D.
True.
DNA
sequence.
63
(b)
5
The cell employs
an
additional
protein
in
order to
make the
6constant
reloading
of
the
sliding clamp on
the
lagging
strand much
3
more efficient.
The protein,
called the
clamp loader,
harnesses
energy
6
from ATP
hydrolysis to
lock a
sliding clamp complex
around
the
DNA for
every successive round of
DNA synthesis.
63
(b)
7
63
(d)
8
63
(b)
9
64
(d)
0
6- (a)
4
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
1
Page
22
of
25
The distortion
in
the
DNA backbone
is
insufficient
information for
the
mismatch
repair system
to
identify
which base is
incorrect
and
which was originally
part of
the
chromosome when replication began.
Without
additional
marks that identify
the
difference
between
the
6- newly
synthesized strand and the
template
strand,
the
repair
would
42 be
corrected only 50% of
the
time by
random
chance.
The
error rate (and therefore
the
mutation
rate) would still be
less than
in
a
system
that
lacked the
mismatch
repair enzymes
(1
mistake
per
107 base pairs), but greater
than the
error rate in
a
system
that accurately identifies
the
newly synthesized
strand (1
mistake
per 109 base pairs).
6-
(c)
43
Choices
(a)
these results
in
644 was
the
or
(d)
because
original
are
a
mouse’s
correct.
mutation
germ cells
already
carrying.
Neither
mutations
in
the
germ cells of
utero would have had no
effect on
theymight have
led
to
their
offspring.
6-
Choice (b)
cannot
the
fetuses
would have no
(a)
(d)
46
6-
(d)
47
6-
effect on
can choice (c),
because
the
fetuses
while in
their development, but
mutant mice
among
45
6-
account
she
(b)
48
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
for
lOMoARcPSD|30659517
6-
(a)
49
A.
will
One-half,
create two
or
50%. DNA replication
new
DNA molecules,
mismatchedC-T
one
of
base
which will
pair.
now
So
a
one daughter
cell
completely
norma
l
of
DNA
that
division
will
molecule
with
round of
DNA
molecule;
the
other cell
themutation
to
correct
a
the
original
bacterium
carry a
carry
nucleotide.
or
25%. At
cell
division,
the
carrying
the
from the
bacterium
on
one
cell
will
have the
mispaired
B.
Onequarter,
replication and
6- pass
in
the
next
mismatched C-T
will
produce
and
50
normal
the
T
DNA
and
molecule
one
mutan So
tpair.
DNA
at
molecule
with
this
stage, one
a
fully
mutant
C-G
out
the
four
of
the
original
bacteriu
m
rise
of
mutant.
containing
base
is
Subsequent cell
divisions
of
to
mutant
these mutant
bacteria,
bacteri
a
will
give
whereas
bacteria
.
other bacteria
normalThe
will
give
proportion
rise
to
containing
25%.
mutation
will
therefore
of
6-
progeny
undamaged strand
the
progeny
remainat
the
(c)
51
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
only
lOMoARcPSD|30659517
6-
(b)
52
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Page
23
of
6-53
A.
True.
1. False. It
replication
25
is
as
believed
a
that
means of
easy identification of
aresealed
by
ligase shortly
DNA
the
nicks are
the
newly synthesized
after replication
1. False. Depurination occurs constantly
hydrolysis
of
deoxyribose
the
sugar.
involves
nucleases
6-54
bond linking the
DNA
base to
DNA
a
loss of
genetic
becauseare degraded
can
be
Choice (c)
is
affectedindividuals
the
in
correct
of
in
early-onset
mismatch
repair choice (a)
is
genes. Mutations
arising
incorrect.
would
A
defect in
probabl
be
y
spontaneous
the
repair double-strand
sequence,
always
before they
history
mutations
through
cells
change
in
nonhomologousjoining
strand but
completed.
our
can
during DNA
is
in
1. False. Homologous recombination
any
generated
breaks
but
end
information
by
ligated back
without
the
ends
together.
answer.
some families
In
with
fact,
a
colon cancer have been found to
carry
in
somatic
cells
are
inherited,
DNA
synthesis
or
nucleotide
not
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
biosynthesis
so
lOMoARcPSD|30659517
lethal, so
choices
(b)
and
(d)
are
incorrect.
6-55
to
A.
be
Mr
Self-destruct is
more likely than
defective
in
the
DNA
the
repair polymerase
repair of
becaus
e
Mr
all
thre
e
thre kinds of
e
of
kinds
DNA
damage.
The
in
later steps, including
are
damage
for
requiremen
t
similar
the
DNA
the
the
Self-destruct is
other mutants
defective
repair pathways
in
for
all
a
repair polymerase.
1. The other mutants
are
damage.
Thus, the
mutations
are likely to
thefirst stage of
specific
be
recognitio
and excision
of
n
and
Mole are likely to
for
a
particular
in
genes required
repair—the
the
damaged
type
of
for
bases. Dracula
be
defective
in
thymidinedimers;
the
recognition or
excision
Faust is
likely
to
G
be
defective
in
the
mismatched base pairs;
recognition or
excision
of
and
or
Marguerite
excision
be
defective
in
recognition
it
it
recognized
as
it
be
is
likely to
of
the
U-
of
abasic sites.
6-56
Uracil is
amutational
formed
with a
an
RNA base and
lesion because,
as
from the
guanine
deamination of
in
the
is
is
cytosine,
will
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
paired
lOMoARcPSD|30659517
context
hydrogen
of
the
DNA
bonds, similar
thymine,
forms
and is
thus a
three hydrogen
bonds
with cytosine.
The
backbone,
allowing
duplex.
to
mismatch
the
Uracil pairs by
poor partner
causes a
for
distortion
repair machinery
Becauseuracil
to
recognize
pairs preferably
the
uracil as
with adenine
deamination of
(its
partner
cytosine
to
in
double-stranded
uracil is
transition
highly mutagenic. If
unrepaired, it
of
a
C-G
base
pair
to
a
T-A
6-57
(c)
6-58
(a)
6-59
(b)
of
25
Page 24
6-60
6-61
a
forming
two
guanine,
which
of
DNA
the
lesion.
can
RNA), the
result in
base pair.
(c)
(d)
6-62 (d)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
the
lOMoARcPSD|30659517
A.
Neither.
mother
is
hybrid of
The
a
copy of
Chromosome 3
you
received
from your
the
ones she
herfather.
received
from
her
mother
and
See
Figure A6-43. The
chromosome in
which a
correct
answers
include
any
chromosome
6-63 B.
portion
and
th
e
matches
the
information from
the
paternal
matches
the
information from
the
maternal
extensive
crossing-over,
you
resemble
there were no
crossing-over,
you
one
other.
remainder
chromosome
.
Figure
C.
A6-43
As
a
result of
grandmother
and
a
your grandfather. If
much
stronger
6-
resemblance to
A. False. Homologous recombination
than the
also
occurs in
prokaryotic
both your
might have
cells, and
64
typically
occurs very shortly
replicated
duplexes
after DNA replication,
are
in
when
the
newly
is
not
close proximity.
B. True.
C. True.
D. False. Although
a
process
it
is
called homologous recombination,
that depends
as
a
on
the
proximity
of
parental
this
homologs.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
When used
lOMoARcPSD|30659517
mechanism
for
DNA repair, homologous recombination
chromatids
as
a
in
an
undamaged, newly replicated
uses
(homologous)
the
sister
DNA helix
template.
Page
25
of
25
ESSENTIAL CELL BIOLOGY, FOURTH EDITION
CHAPTER 7: FROM DNA TO PROTEIN
© 2014 GARLAND SCIENCE PUBLISHING
7-1
For each of the following sentences, fill in the blanks with the best word or phrase selected from
the list below. Not all words or phrases will be used; use each word or phrase only once.
The instructions specified by the DNA will ultimately specify the sequence of proteins. This
process involves DNA, made up of
different nucleotides, which gets
into RNA, which is then
into proteins, made up of
different amino
acids. In eukaryotic cells, DNA gets made into RNA in the
,while proteins are produced from RNA in the
. The segment of DNA called a
portionthat is copied into RNA; this process is catalyzed by RNA
is the
.
4 gene proteasome
20 Golgi replisome
109 kinase sugar-phosphate
128 nuclear pore transcribed
cytoplasm nucleus transferase
exported polymerase
translated
7-2
Use the numbers in the choices below to indicate where in the schematic diagram of a
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
eukaryotic cell (Figure Q7-2) those processes take place.
Figure Q7-2
1.
transcription
2.
translation
3.
RNA splicing
Page 1 of 29
4. polyadenylation
5. RNA capping
From DNA to RNA
7-3
Consider two genes that are next to each other on a chromosome, as arranged in
FigureQ7-3.
Figure Q7-3
Which of
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
the following statements is true?
(a) The two genes must be transcribed into RNA using the same strand of DNA.
(b) If gene A is transcribed in a cell, gene B cannot be transcribed.
(c) Gene A and gene B can be transcribed at different rates, producing
differentamounts of RNA within the same cell.
(d) If gene A is transcribed in a cell, gene B must be transcribed.
7-4
RNA in cells differs from DNA in that
.
(a) it contains the base uracil, which pairs with cytosine.
(b) it is single-stranded and cannot form base pairs.
(c) it is single-stranded and can fold up into a variety of structures.
(d) the sugar ribose contains fewer oxygen atoms than does deoxyribose.
7-5
Transcription is similar to DNA replication in that
.
(a) an RNA transcript is synthesized discontinuously and the pieces are then
joinedtogether.
(b) it uses the same enzyme as that used to synthesize RNA primers during DNA
replication.
(c) the newly synthesized RNA remains paired to the template DNA.
(d) nucleotide polymerization occurs only in the 5′-to-3′ direction.
Figure Q7-6 is to be used with Questions 7-6, 7-7, and 7-8. These three
questions can be used separately or together.
Figure Q7-6
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Page 2 of 29
Figure Q7-6 shows a ribose sugar. RNA bases are added to the part of the ribose sugar pointed to
by arrow
.
(a) 3.
7-6
(b) 4.
(c) 5.
(d) 6.
Figure Q7-6 shows a ribose sugar. The part of the ribose sugar that is different from the
deoxyribose sugar used in DNA is pointed to by arrow
.
(a) 1.
7-7
(b) 4.
(c) 5.
(d) 6.
Figure Q7-6 shows a ribose sugar. The part of the ribose sugar where a new
ribonucleotide will attach in an RNA molecule is pointed to by arrow
.
(a) 1.
7-8
(b) 3.
(c) 4.
(d) 5.
7-9
For each of the following sentences, fill in the blanks with the best word or phrase selected from
the list below. Not all words or phrases will be used; use each word or phrase only once.
For a cell’s genetic material to be used, the information is first copied from the DNA into the
nucleotide sequence of RNA in a process called
. Various kinds of RNA are
produced, each with different functions.
molecules code for proteins,
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
molecules act as adaptors for protein synthesis,
molecules are integral components of the ribosome, and
important in the splicing of RNA transcripts.
molecules are
incorporation rRNA translation
mRNA snRNA transmembrane
pRNA transcription tRNA
proteins
7-10
Match the following structures with their names.
Page 3 of 29
Figure Q7-10
Which of the following statements is false?
(a) A new RNA molecule can begin to be synthesized from a gene before
theprevious RNA molecule’s synthesis is completed.
(b) If two genes are to be expressed in a cell, these two genes can be transcribed
7-11
withdifferent efficiencies.
(c) RNA polymerase is responsible for both unwinding the DNA helix and catalyzing
the formation of the phosphodiester bonds between nucleotides.
(d) Unlike DNA, RNA uses a uracil base and a deoxyribose sugar.
7-12
Unlike DNA, which typically forms a helical structure, different molecules of RNA can fold into a
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
variety of three-dimensional shapes. This is largely because
.
(a) RNA contains uracil and uses ribose as the sugar.
(b) RNA bases cannot form hydrogen bonds with each other.
(c) RNA nucleotides use a different chemical linkage between nucleotides
comparedto DNA.
(d) RNA is single-stranded.
Which of the following molecules of RNA would you predict to be the most likely to fold into a
specific structure as a result of intramolecular base-pairing?
(a) 5′-CCCUAAAAAAAAAAAAAAAAUUUUUUUUUUUUUUUUAGGG-3′
7-13
(b) 5′-UGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUG-3′
(c) 5′-AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA-3′
(d) 5′-GGAAAAGGAGAUGGGCAAGGGGAAAAGGAGAUGGGCAAGG-3′
Which one of the following is the main reason that a typical eukaryotic gene is able to respond to a
far greater variety of regulatory signals than a typical prokaryotic gene or operon?
7-14
(a) Eukaryotes have three types of RNA polymerase.
(b) Eukaryotic RNA polymerases require general transcription factors.
Page 4 of 29
(c) The transcription of a eukaryotic gene can be influenced by proteins that bind
farfrom the promoter.
(d) Prokaryotic genes are packaged into nucleosomes.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
7-15
7-16
Match the following types of RNA with the main polymerase that transcribes them.
List three
ways in which the process of eukaryotic transcription differs from the process
of bacterial transcription.
7-17
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or phrase
should be used only once.
In eukaryotic cells, general transcription factors are required for the activity of all
promoters transcribed by RNA polymerase II. The assembly of the general
transcriptionfactors begins with the binding of the factor
to DNA, causing a
marked local distortion in the DNA. This factor binds at the DNA sequence called the
box, which is typically located 25 nucleotides upstream from the
transcription start site. Once RNA polymerase II has been brought to the promoter
DNA,it must be released to begin making transcripts. This release process is facilitated
by the addition of phosphate groups to the tail of RNA polymerase by the factor
. It must be remembered that the general transcription factors
and RNA polymerase are not sufficient to initiate transcription in the cell and are
affected by proteins bound thousands of nucleotides away from the promoter. Proteins
that link the distantly bound transcription regulators to RNA polymerase and the
general transcription
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
factors include the large complex of proteins called the
. The
packing of DNA into chromatin also affects transcriptional initiation, and histone
is an enzyme that can render the DNA less accessible to
thegeneral transcription factors.
activator lac TFIIACAP
ligase TFIID
deacetylase Mediator TFIIH
enhancer TATA
7-18
You have a piece of DNA that includes the following sequence:
5′-ATAGGCATTCGATCCGGATAGCAT-3′
3′-TATCCGTAAGCTAGGCCTATCGTA-5′
Page 5 of 29
Which of the following RNA molecules could be transcribed from this piece of DNA? (a) 5′UAUCCGUAAGCUAGGCCUAUGCUA-3′
(b) 5′-AUAGGCAUUCGAUCCGGAUAGCAU-3′
(c) 5′-UACGAUAGGCCUAGCUUACGGAUA-3′
(d) none of the above
7-19
The following segment of DNA is from a transcribed region of a chromosome. You know that RNA
polymerase moves from left to right along this piece of DNA, that the promoter for this gene is to
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
the left of the DNA shown, and that this entire region of DNA is made into RNA.
5′-GGCATGGCAATATTGTAGTA-3′
3′-CCGTACCGTTATAACATCAT-5′
Given this information, a student claims that the RNA produced from this DNA is:
3′-GGCATGGCAATATTGTAGTA-5′
Give two reasons why this answer is incorrect.
7-20
You have a segment of DNA that contains the following sequence:
5′-GGACTAGACAATAGGGACCTAGAGATTCCGAAA-3′
3′-CCTGATCTGTTATCCCTGGATCTCTAAGGCTTT-5′
You know that the RNA transcribed from this segment contains the following sequence:
5′-GGACUAGACAAUAGGGACCUAGAGAUUCCGAAA–3′
Which of the following choices best describes how transcription occurs?
(a) the top strand is the template strand; RNA polymerase moves along this
strandfrom 5′ to 3′
(b) the top strand is the template strand; RNA polymerase moves along this
strandfrom 3′ to 5′
(c) the bottom strand is the template strand; RNA polymerase moves along this
strandfrom 5′ to 3′
(d) the bottom strand is the template strand; RNA polymerase moves along this
strandfrom 3′ to 5′
7-21
Imagine that an RNA polymerase is transcribing a segment of DNA that contains the following
sequence:
5′-AGTCTAGGCACTGA-3′
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
3′-TCAGATCCGTGACT-5′
Page 6 of 29
A. If the polymerase is transcribing from this segment of DNA from left to right,
which strand (top or bottom) is the template?
B. What will be the sequence of that RNA (be sure to label the 5′ and 3′ ends of
yourRNA molecule)?
The sigma subunit of bacterial RNA polymerase
.
(a) contains the catalytic activity of the polymerase.
7-22
(b) remains part of the polymerase throughout transcription.
(c) recognizes promoter sites in the DNA.
(d) recognizes transcription termination sites in the DNA.
7-23
Which of the following might decrease the transcription of only one specific gene in a bacterial
cell?
(a) a
decrease in the amount of sigma factor
(b) a decrease in the amount of RNA polymerase
(c) a mutation that introduced a stop codon into the DNA that precedes the
gene’scoding sequence
(d) a mutation that introduced extensive sequence changes into the DNA that
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
precedes the gene’s transcription start site
There are several reasons why the primase used to make the RNA primer for DNA replication is not
suitable for gene transcription. Which of the statements below is not one of those reasons?
(a) Primase initiates RNA synthesis on a single-stranded DNA template.
7-24
(b) Primase can initiate RNA synthesis without the need for a base-paired primer.
(c)Primase synthesizes only RNAs of about 5–20 nucleotides in length.
(d) The RNA synthesized by primase remains base-paired to the DNA template.
You have a bacterial strain with a mutation that removes the transcription termination signal from
the Abd operon. Which of the following statements describes the most likely effect of this
mutation on Abd transcription?
(a) The Abd RNA will not be produced in the mutant strain.
(b) The Abd RNA from the mutant strain will be longer than normal.
7-25
(c) Sigma factor will not dissociate from RNA polymerase when the Abd operon is
being transcribed in the mutant strain.
(d) RNA polymerase will move in a backward fashion at the Abd operon in the
mutant strain.
Transcription in bacteria differs from transcription in a eukaryotic cell because
.
(a) RNA polymerase (along with its sigma subunit) can initiate transcription on its
own.
7-26
(b) RNA polymerase (along with its sigma subunit) requires the general transcription
factors to assemble at the promoter before polymerase can begin transcription. (c) the
sigma subunit must associate with the appropriate type of RNA polymerase to
produce mRNAs.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Page 7 of 29
(d) RNA polymerase must be phosphorylated at its C-terminal tail for transcription to
proceed.
Which of the following does not occur before a eukaryotic mRNA is exported from the nucleus?
(a) The ribosome binds to the mRNA.
7-27
(b) The mRNA is polyadenylated at its 3′ end.
(c) 7-methylguanosine is added in a 5′-to-5′ linkage to the mRNA.
(d) RNA polymerase dissociates.
Total nucleic acids are extracted from a culture of yeast cells and are then mixed with resin beads
to which the polynucleotide 5′-TTTTTTTTTTTTTTTTTTTTTTTTT-3′ has been covalently attached.
After a short incubation, the beads are then extracted from the mixture. When to beads, which
you analyze the cellular nucleic acids that have stuck the
of the following is most abundant?
7-28
(a) DNA
(b) tRNA
(c) rRNA
(d) mRNA
7-29
7-30
Name three covalent modifications that can be made to an RNA molecule in eukaryotic cells
before the RNA molecule becomes a mature mRNA.
Which of the following statements about RNA splicing is false?
(a) Conventional introns are not found in bacterial genes.
(b) For a gene to function properly, every exon must be removed from the
primarytranscript in the same fashion on every mRNA molecule produced from
the same
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
gene.
(c) Small RNA molecules in the nucleus perform the splicing reactions necessary for
the removal of introns.
(d) Splicing occurs after the 5′ cap has been added to the end of the
primarytranscript.
7-31
7-32
The length of a particular gene in human DNA, measured from the start site for transcription to
the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the mRNA
produced from this gene is 4000 nucleotides. What is the most likely reason for this difference?
Why is the old dogma “one gene—one protein” not always true for eukaryotic genes?
Genes in eukaryotic cells often have intronic sequences coded for within the DNA. These
sequences are ultimately not translated into proteins. Why?
(a) Intronic sequences are removed from RNA molecules by the spliceosome,
7-33
whichworks in the nucleus.
(b) Introns are not transcribed by RNA polymerase.
(c) Introns are removed by catalytic RNAs in the cytoplasm.
Page 8 of 29
(d) The ribosome will skip over intron sequences when translating RNA into protein.
7-34
snRNAs
.
(a) are translated into snRNPs.
(b) are important for producing mature mRNA transcripts in bacteria.
(c) are removed by the spliceosome during RNA splicing.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(d) can bind to specific sequences at intron–exon boundaries through
complementarybase-pairing.
7-35
Is this statement true or false? Explain your answer.
“Since introns do not contain protein-coding information, they do not have to be removed
precisely (meaning, a nucleotide here and there should not matter) from the primary transcript
during RNA splicing.”
7-36
You have discovered a gene (Figure Q7-36A) that is alternatively spliced to produce several forms
of mRNA in various cell types, three of which are shown in Figure Q7-36B. The lines connecting
the exons that are included in the mRNA indicate the splicing. From your experiments, you know
that protein translation begins in exon 1. For all forms of the mRNA, the encoded protein
sequence is the same in the regions of the mRNA that correspond to exons 1 and 10. Exons 2 and
3 are alternative exons used in different mRNA, as are exons 7 and 8. Which of the following
statements about exons 2 and 3 is the most accurate? Explain your answer.
Figure Q7-36
(a)
Exons 2 and 3 must have the same number of nucleotides.
(b)
Exons 2 and 3 must contain an integral number of codons (that
is, the number of
nucleotides divided by 3 must be an integer).
(c)
Exons 2 and 3 must contain a number of nucleotides that when
divided by 3,
leaves the same remainder (that is, 0, 1, or 2).
(d)
Exons 2 and 3 must have different numbers of nucleotides.
Page 9 of 29
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
From RNA to Protein
7-37
Which of the following statements about the genetic code is correct?
(a) All codons specify more than one amino acid.
(b) The genetic code is redundant.
(c) All amino acids are specified by more than one codon.
(d) All codons specify an amino acid.
NOTE: The following codon table is to be used for Problems Q7-38 to Q7-49.
7-38
The piece of RNA below includes the region that codes for the binding site for the initiator tRNA
needed in translation.
5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′
Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome?
(a) methionine
(b) arginine
(c) cysteine
(d) valine
7-39
The following DNA sequence includes the beginning of a sequence coding for a protein. What
would be the result of a mutation that changed the C marked by an asterisk to an A?
5′-AGGCTATGAATGGACACTGCGAGCCC…
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
*
7-40
Which amino acid would you expect a tRNA with the anticodon 5′-CUU-3′ to carry? (a) lysine
Page 10 of 29
(b) glutamic acid
(d) leucine
(d) phenylalanine
Which of the following pairs of codons might you expect to be read by the same tRNA as a result
of wobble?
(a) CUU and UUU
7-41
(b) GAU and GAA
(c) CAC and CAU
(d) AAU and AGU
7-42
Below is a segment of RNA from the middle of an mRNA.
5′-UAGUCUAGGCACUGA-3′
If you were told that this segment of RNA was part of the coding region of an mRNA for a large
protein, give the amino acid sequence for the protein that is encoded by this segment of mRNA.
Write your answer using the one-letter amino acid code.
7-43
Below is the sequence from the 3′ end of an mRNA.
5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′
If you were told that this sequence contains the stop codon for the protein encoded by this mRNA,
what is the anticodon on the tRNA in the P site of the ribosome when release factor binds to the A
site?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(a) 5′-CCA-3′
(b) 5′-CCG-3′
(c) 5′-UGG-3′
(d) 5′-UUA-3′
7-44
One strand of a section of DNA isolated from the bacterium E. coli reads:
5′-GTAGCCTACCCATAGG-3′
A. Suppose that an mRNA is transcribed from this DNA using the complementary
strand as a template. What will be the sequence of the mRNA in this region (make
sure you label the 5′ and 3′ ends of the mRNA)?
B. How many different peptides could potentially be made from this sequence
ofRNA, assuming that translation initiates upstream of this sequence?
C. What are these peptides? (Give your answer using the one-letter amino
acidcode.)
7-45
A strain of yeast translates mRNA into protein inaccurately. Individual molecules of a particular
protein isolated from this yeast have variations in the first 11 amino acids compared with the
sequence of the same protein isolated from normal yeast cells, as
Page 11 of 29
listed in Figure Q7-45. What is the most likely cause of this variation in protein
sequence?
Figure Q7-45
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(a) a mutation in the DNA coding for the protein
(b) a
mutation in the anticodon of the isoleucine-tRNA (tRNAIle)
(c) a mutation in the isoleucyl-tRNA synthetase that decreases its ability
todistinguish between different amino acids
(d) a mutation in the isoleucyl-tRNA synthetase that decreases its ability
todistinguish between different tRNA molecules
A mutation in the tRNA for the amino acid lysine results in the anticodon sequence 5′-
UAU-3′ (instead of 5′-UUU-3′). Which of the following aberrations in protein
synthesis might this tRNA cause?
7-46
(a) read-through of stop codons
(b) substitution of lysine for isoleucine
(c) substitution of lysine for tyrosine
(d) substitution of lysine for phenylalanine
After treating cells with a mutagen, you isolate two mutants. One carries alanine and the
other carries methionine at a site in the protein that normally contains valine. After
treating these two mutants again with mutagen, you isolate mutants from each that
7-47
now carry threonine at the site of the original valine (see Figure Q7-47). Assuming that
all mutations caused by the mutagen are due to single nucleotide changes, deduce the
codonsthat are used for valine, alanine, methionine, and threonine at the affected site.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Figure Q7-47
What do you predict would happen if you created a tRNA with an anticodon of 5′-CAA-
7-48
3′ that is charged with methionine, and added this modified tRNA to a cell-free
translation system that has all the normal components required for translating RNAs?
Page 12 of 29
(a) methionine would be incorporated into proteins at some positions where
glutamine should be
(b) methionine would be incorporated into proteins at some positions where
leucineshould be
(c) methionine would be incorporated into proteins at some positions where
valineshould be
(d) translation would no longer be able to initiate
7-49
7-50
In a diploid organism, the DNA encoding one of the tRNAs for the amino acid tyrosine is mutated
such that the sequence of the anticodon is now 5′-CTA-3′ instead of 5′-GTA-3′. What kind of
aberration in protein synthesis will this tRNA cause? Explain your answer.
The ribosome Which of the
is important for catalyzing the formation of peptide bonds.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
following statements is true?
(a) The number of rRNA molecules that make up a ribosome greatly exceeds
thenumber of protein molecules found in the ribosome.
(b) The large subunit of the ribosome is important for binding to the mRNA. (c) The
catalytic site for peptide bond formation is formed primarily from an rRNA. (d) Once the
large and small subunits of the ribosome assemble, they will not
separate from each other until degraded by the proteasome.
Which of the following statements is true?
(a) Ribosomes are large RNA structures composed solely of rRNA.
7-51
(b) Ribosomes are synthesized entirely in the cytoplasm.
(c) rRNA contains the catalytic activity that joins amino acids together.
(d) A ribosome binds one tRNA at a time.
7-52
Figure Q7-52A shows the stage in translation when an incoming aminoacyl-tRNA has bound to the
A site on the ribosome. Using the components shown in Figure Q7-52A as a guide, show on
Figures Q7-52B and Q7-52C what happens in the next two stages to complete the addition of the
new amino acid to the growing polypeptide chain.
Page 13 of 29
Figure Q7-52
7-53
A poison added to an in vitro translation mixture containing mRNA molecules with the sequence
5′-AUGAAAAAAAAAAAAUAA-3′ has the following effect: the only product made is a Met-Lys
dipeptide that remains attached to the ribosome. What is the most likely way in which the
poison acts to inhibit protein synthesis?
(a) It inhibits peptidyl transferase activity.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(b) It inhibits movement of the small subunit relative to the large subunit. (c) It
inhibitsrelease factor.
(d) It mimics release factor.
In eukaryotes, but not in prokaryotes, ribosomes find the start site of translation by
.
7-54
(a) binding directly to a ribosome-binding site preceding the initiation codon.
(b)scanning along the mRNA from the 5′ end.
(c) recognizing an AUG codon as the start of translation.
(d) binding an initiator tRNA.
Which of the following statements about prokaryotic mRNA molecules is false? (a) A single
prokaryotic mRNA molecule can be translated into several proteins. (b) Ribosomes must bind to
the 5′ cap before initiating translation.
(c) mRNAs are not polyadenylated.
7-55
(d) Ribosomes can start translating an mRNA molecule before transcription
iscomplete.
Page 14 of 29
7-56
Figure Q7-56 shows an mRNA molecule.
Figure Q7-56
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
A. Match the labels given in the list below with the label lines in Figure Q7-56. (a)
ribosome-binding site
(b) initiator codon
(c) stop codon
(d) untranslated 3′ region
(e) untranslated 5′ region
(f) protein-coding region
B. Is the mRNA shown prokaryotic or eukaryotic? Explain your answer.
7-57
You have discovered a protein that inhibits translation. When you add this inhibitor to a mixture
capable of translating human mRNA and centrifuge the mixture to separate polyribosomes and
single ribosomes, you obtain the results shown in Figure Q7-57. Which of the following
interpretations is consistent with these observations?
Figure Q7-57
(a)
The protein binds to the small ribosomal subunit and increases the rate of initiation of
translation.
Page 15 of 29
(b)
The protein binds to sequences in the 5′ region of the mRNA and inhibits the rate
of initiation of translation.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(c)
The protein binds to the large ribosomal subunit and slows down elongation of
thepolypeptide chain.
(d)
The protein binds to sequences in the 3′ region of the mRNA and
preventstermination of translation.
The concentration of a particular protein, X, in a normal human cell rises gradually from a low
point, immediately after cell division, to a high point, just before cell division, and then drops
sharply. The level of its mRNA in the cell remains fairly constant throughout this time. Protein X is
required for cell growth and survival, but the drop in its level just before cell division is essential
for division to proceed. You have isolated a line of human cells that grow in size in culture but
cannot divide, and on analyzing these mutants, you find that following
levels of X mRNA in the mutant cells are normal. Which of the
7-58
mutations in the gene for X could explain these results?
(a) the introduction of a stop codon that truncates protein X at the fourth amino acid
(b) achange of the first ATG codon to CCA
(c) the deletion of a sequence that encodes sites at which ubiquitin can be
attached tothe protein
(d) a change at a splice site that prevents splicing of the RNA
7-59
For each of the following sentences, fill in the blanks with the best word or phrase selected from
the list below. Not all words or phrases will be used; use each word or phrase only once.
Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed
of two subunits: the
subunit, which catalyzes the formation of the peptide
bonds that link the amino acids together into a polypeptide chain, and the
subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation
process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is
bound to the
site of the ribosome. An incoming aminoacyl-tRNA carrying
the next amino acid in the chain will bind to the
site by forming base pairs
with the exposed codon in the mRNA. The
enzyme catalyzes the formation
of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid.
The end of a protein-coding message is signaled by the presence of a stop codon, which binds the
called release factor. Eventually, most proteins will be degraded by a large
complex of proteolytic enzymes called the
.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
A medium protein
central P RNA
DNA peptidyl transferase smallE
polymerase T
large proteasome ubiquitin
Page 16 of 29
7-60
Which of the following methods is not used by cells to regulate the amount of a protein
inthe cell?
(a) Genes can be transcribed into mRNA with different efficiencies.
(b) Many ribosomes can bind to a single mRNA molecule.
(c) Proteins can be tagged with ubiquitin, marking them for degradation.
(d) Nuclear pore complexes can regulate the speed at which newly
synthesizedproteins are exported from the nucleus into the cytoplasm.
7-61
Which of the following statements about the proteasome is false?
(a) Ubiquitin is a small protein that is covalently attached to proteins to mark
themfor delivery to the proteasome.
(b) Proteases reside in the central cylinder of a proteasome.
(c) Misfolded proteins are delivered to the proteasome, where they are sequestered
from the cytoplasm and can attempt to refold.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(d) The protein stoppers that surround the central cylinder of the proteasome use
theenergy from ATP hydrolysis to move proteins into the proteasome inner
chamber. RNA and the Origins of Life
7-62
Which of the following molecules is thought to have arisen first during evolution?
(a) protein
(b) DNA
(c) RNA
(d) all came to be at the same time
7-63
According to current thinking, the minimum requirement for life to have originated
onEarth was the formation of a
.
(a) molecule that could provide a template for the production of a
complementarymolecule.
(b) double-stranded DNA helix.
(c) molecule that could direct protein synthesis.
(d) molecule that could catalyze its own replication.
7-64
Ribozymes catalyze which of the following reactions?
(a) DNA synthesis
(b) transcription
(c) RNA splicing
(d) protein hydrolysis
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
7-65
You are studying a disease that is caused by a virus, but when you purify the
virusparticles and analyze them you find they contain no trace of DNA. Which of the
following molecules are likely to contain the genetic information of the virus?
(a) high-energy phosphate groups
(b) RNA
(c) lipids
(d) carbohydrates
Page 17 of 29
7-66
Give a reason why DNA makes a better material than RNA for the storage of genetic
information, and explain your answer.
How We Know: Cracking the Genetic Code
7-67
When using a repeating trinucleotide sequence (such as 5′-AAC-3′) in a cell-free
translation system, you will obtain:
(a) three different types of peptides, each made up of a single amino acid
(b) peptides made up of three different amino acids in random order
(c) peptides made up of three different amino acids, each alternating with each
otherin a repetitive fashion
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
(d) polyasparagine, as the codon for asparagine is AAC
7-68
You have discovered an alien life-form that surprisingly uses DNA as its genetic
material, makes RNA from DNA, and reads the information from RNA to make protein
using ribosomes and tRNAs, which read triplet codons. Because it is your job to
decipherthe genetic code for this alien, you synthesize some artificial RNA molecules
and examine the protein products produced from these RNA molecules in a cell-free
translation system using purified alien tRNAs and ribosomes. You obtain the results
shown in Table Q7-68.
Table Q7-68
From this information, which of the following peptides can be produced from poly UAUC?
(a) Ile-Phe-Val-Tyr
(b) Tyr-Ser-Phe-Ala
(c) Ile-Lys-His-Tyr
(d) Cys-Pro-Lys-Ala
7-69
An extraterrestrial organism (ET) is discovered whose basic cell biology seems pretty much the
same as that of terrestrial organisms except that it uses a different genetic code
Page 18 of 29
to translate RNA into protein. You set out to break the code by translation experiments using
RNAs of known sequence and cell-free extracts of ET cells to supply the necessary protein-
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
synthesizing machinery. In experiments using the RNAs below, the following results were obtained
when the 20 possible amino acids were added either singly or in different combinations of two or
three:
RNA 1: 5′-GCGCGCGCGCGCGCGCGCGCGCGCGCGC-3′
RNA 2: 5′-GCCGCCGCCGCCGCCGCCGCCGCCGCCGCC-3′
Using RNA 1, a polypeptide was produced only if alanine and valine were added to the reaction
mixture. Using RNA 2, a polypeptide was produced only if leucine and serine and cysteine were
added to the reaction mixture. Assuming that protein synthesis can start anywhere on the
template, that the ET genetic code is nonoverlapping and linear, and that the many
each codon is the same length (like terrestrial triplet code), how
nucleotides does an ET codon contain?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
7-70
NASA has discovered an alien life-form. You are called in to help NASA scientists to deduce the
genetic code for this alien. Surprisingly, this alien life-form shares many similarities with life on
Earth: this alien uses DNA as its genetic material, makes RNA from DNA, and reads the
information from RNA to make protein using ribosomes and tRNAs. Even more amazing, this alien
uses the same 20 amino acids, like the organisms found on Earth, and also codes for each amino
acid by a triplet codon. However, the scientists at NASA have found that the genetic code used by
the alien life-form differs from that used by life on Earth. NASA scientists drew this conclusion
after creating a cell-free protein-synthesis system from alien cells and adding an mRNA made
entirely of uracil (poly U). They found that poly U directs the synthesis of a peptide containing
only glycine. NASA scientists have synthesized a poly AU mRNA and observe that it codes for a
polypeptide of alternating serine and proline amino acids. From these experiments, can you
determine which codons code for serine and proline? Explain.
Bonus question. Can you propose a mechanism for how the alien’s physiology is altered so that it
uses a different genetic code from life on Earth, despite all the similarities?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Page 19 of 29
ANSWERS
The instructions specified by the DNA will ultimately specify the sequence of proteins.
This process involves DNA, made up of 4 different nucleotides, which gets transcribed
into RNA, which is then translated into proteins, made up of 20 different amino acids. In
7-1
eukaryotic cells, DNA gets made into RNA in the nucleus, while proteins are produced
from RNA in the cytoplasm. The segment of DNA called a gene is the portion that is copied
into RNA; this process is catalyzed by RNA polymerase.
7-2
See Figure A7-2.
Figure A7-2
7-3
(c)
Choice (c) is correct. Choice (a) is untrue because although RNA contains uracil, uracil pairs with
adenine, not cytosine. Choice (b) is false because RNA can form base pairs with a complementary
7
RNA or DNA sequence.
Choice (d) is false because ribose contains one more oxygen atom than
deoxyribose. -
4
7-5
Choice (d) is correct. Choice (a) is incorrect because an RNA transcript is made by a single
polymerase molecule that proceeds from the start site to the termination site without falling off.
The enzyme used to make primers during DNA synthesis is indeed an RNA polymerase, but it is a
special enzyme, primase, and not the enzyme that is used for transcription, which is why choice
(b) is incorrect. Choice (c) is false.
7-6
(a)
7-7
(c)
Page 20 of 29
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
7-8
(c)
7-9
For a cell’s genetic material to be used, the information is first copied from the DNA into the
nucleotide sequence of RNA in a process called transcription. Various kinds of RNA are produced,
each with different functions. mRNA molecules code for proteins, tRNA molecules act as adaptors
for protein synthesis, rRNA molecules are integral components of the ribosome, and snRNA
molecules are important in the splicing of RNA transcripts.
7-10
A—4; B—2; C—1; D—3
7-11
Choice (d) is false. RNA nucleotides contain the sugar ribose.
Choice (d) why RNA
is correct. Choice
7 (a) is true, but is not the main reason different molecules can form
different three-dimensional
structures (although ribose does increase potential hydrogenbonding potentials
1 compared to deoxyribose). Choices (b) and (c) are untrue.
2
Choice (a) is correct. Choices (b) and (c) do not have any opportunity for intramolecular basepairing and thus a specific structure is unlikely. Although there is some opportunity for
intramolecular base-pairing
in choice (d), choice (a) has much more intrastrand complementarity
7
and is a better choice.
1
3
7-14
(c)
7-15
A—1; B—3; C—3; D—2; E – 2
7-16
Any three of the following are acceptable.
1. Bacterial cells contain a single RNA polymerase, whereas eukaryotic cells have
three.
2. Bacterial RNA polymerase can initiate transcription without the help of additional
proteins, whereas eukaryotic RNA polymerases need general transcription factors. 3.
Ineukaryotic cells, transcription regulators can influence transcriptional initiation
thousands of nucleotides away from the promoter, whereas bacterial regulatory
sequences are very close to the promoter.
4. Eukaryotic transcription is affected by chromatin structure and nucleosomes,
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
whereas bacteria lack nucleosomes.
In eukaryotic cells, general transcription factors are required for the activity of all promoters
transcribed by RNA polymerase II. The assembly of the general transcription factors begins with
the binding of the factor TFIID to DNA, causing a marked local distortion in the DNA. This factor
binds at the DNA sequence called the TATA box, which is typically located 25 nucleotides upstream
from the transcription
start site. Once RNA polymerase II has been brought to the promoter DNA,
7
it must be released
- to begin making transcripts. This release process is facilitated by the addition
of phosphate groups
1 to the tail of RNA polymerase by the factor TFIIH. It must be remembered
that the
7
Page 21 of 29
general transcription factors and RNA polymerase are not sufficient to initiate
transcription in the cell and are affected by proteins bound thousands of nucleotides
away from the promoter. Proteins that link the distantly bound transcription regulators
to RNA polymerase and the general transcription factors include the large complex of
proteins called the Mediator. The packing of DNA into chromatin also affects
transcriptional initiation, and histone deacetylase is an enzyme that can render the DNA
less accessible tothe general transcription factors.
7-18
Choice (b) is correct. The molecules listed in choices (a) and (c) have incorrect polarity.
First, the RNA molecule should have uracil instead of thymine bases. Second, the polarity of the
molecule is incorrectly labeled. The correct RNA molecule produced, using the bottom strand of
the DNA duplex as a template, would be:
7-19
5′-GGCAUGGCAAUAUUGUAGUA-3′
7-20
(d) The bottom strand can hybridize with the RNA molecule and thus is the template strand. The
polymerase moves along the DNA in a 3′-to-5′ direction, because the RNA nucleotides are joined
in a 5′-to-3′ polarity.
A.
The
bottom
strand.
7
B. 5′-AGUCUAGGCACUGA-3′
2
1
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
7-22
(c)
(d) Such changes would probably destroy the function of the promoter, making RNA polymerase
unable to bind to it. Decreasing the amount of sigma factor or RNA polymerase [choices (a) or (b)]
would affect the transcription of most of the genes in the cell, not just one specific gene.
7 codon before the coding sequence [choice (c)] would have no effect on
Introducing a stop
- gene, because the transcription machinery does not recognize translational
transcription of the
2
stops.
3
7-24
7-25
7-26
Choice (b) is true for both primase and RNA polymerase, so it does not describe why primase
cannot be used for gene transcription.
(b) Without the termination signal, the polymerase will not halt and release from the DNA
template at the normal location when transcribing the Abd operon. Most probably, the
polymerase will continue to transcribe RNA until it reaches a sequence in the DNA that can serve
as a termination sequence, either from the next downstream operon or in the intervening
sequence between the Abd operon and the next operon. Dissociation of sigma factor occurs once
an approximately 10-nucleotide length of RNA has been synthesized by RNA polymerase and
should not be affected by the lack of a termination signal [choice (c)].
Choice (a) is correct. Eukaryotic cells, but not bacteria, require general transcription factors
[choice (b)]. There is only a single type of RNA polymerase in bacterial cells
Page 22 of 29
[choice (c)]. The general transcription factor TFIIH phosphorylates the C-terminal tail of RNA
polymerase in eukaryotic cells but not in bacteria [choice (d)].
7-27
7-28
(a) Ribosomes are in the cytosol and will bind to the mRNA once it has been exported from the
nucleus.
(d) mRNA is the only type of RNA that is polyadenylated, and its poly-A tail would be able to basepair with the strands of poly T on the beads and thus stick to them. DNA would not be found in
the sample, because the poly-A tail is not encoded in the DNA and long runs of T are rare in DNA.
1. A poly-A tail is added.
2. A 5′ cap is added.
7-29
3.
Introns can be spliced out.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
7-30
7-31
7-32
(b) The primary transcript of a gene can sometimes be spliced differently so that different exons
can be stitched together to produce distinct proteins in a process called alternative splicing.
The gene contains one or more introns.
The transcripts from some genes can be spliced in more than one way to give mRNAs containing
different sequences, thus encoding different proteins. A single eukaryotic gene may therefore
encode more than one protein.
7-33
(a)
7-34
(d) snRNAs are part of the snRNPs, which include proteins and RNA molecules. The proteins within
the snRNPs are encoded by their own genes and not translated from snRNPs, which is why choice
(a) is incorrect. Bacteria do not have introns, which is why choice (b) is incorrect.
7-35
False. Although it is true that the sequences within the introns are mostly dispensable, the introns
must still be removed precisely because an error of one or two nucleotides would shift the reading
frame of the resulting mRNA molecule and change the protein it encodes.
7-36
Choice (c) is the only answer that must be true for exons 2 and 3. Although choices (a), (b), and (d)
could be true, they do not have to be. Because the protein sequence is the same in segments of
the mRNA corresponding to exons 1 and 10, the choice of either exon 2 or exon 3 would not alter
the reading frame. To maintain the normal reading frame, whatever it is, the alternative exons
must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon)
give the same remainder.
Page 23 of 29
7-37
(b) Most amino acids can be specified by more than one codon. Each codon specifies only one
amino acid [choice (a)]. Tryptophan and methionine are encoded by only one codon [choice (c)].
Some codons specify translational stop signals [choice (d)].
7-38
Choice (b) is correct. The initiator methionine is underlined on the RNA molecule below.
5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′
The first tRNA to bind at the A site is the second codon of the protein, because the initiator tRNA is
already bound to the P site when translation begins. The codon that follows the binding site for
the initiator tRNA is CGU, which codes for arginine.
7-39
The change creates a stop codon (TGA, or UGA in the mRNA) very near the beginning of the (the of
the coding
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
protein-coding sequence and in the correct reading frame beginning
sequence is indicated by the ATG). Thus, translation would terminate after only four amino
acids had been joined together, and the complete protein would not be made.
7-40
(a) As is conventional for nucleotide sequences, the anticodon is given reading from 5′ to 3′. The
complementary base-pairing occurs between antiparallel nucleic acid sequences, and the codon
recognized by this anticodon will therefore be 5′-AAG-3′.
7-41
Choice (c) is the correct answer. These two codons differ only in the third position and also encode
the same amino acid, which is the definition of wobble. Although the codons GAU and GAA
[choice (b)] also differ only in the third position, they are unlikely in normal circumstances to be
read by the same tRNA, because they encode different amino acids.
7-42
SLGT is the correct answer. (Reading frame two is the only reading frame that does not contain a
stop codon.)
7-43
(a) The stop codon (UAA) is underlined in the mRNA sequence below; this is the only stop codon
on this piece of mRNA. The codon (UGG) preceding the stop codon will be binding to a tRNA in the
P site of the ribosome when release factor binds to the A site. The anticodon of the tRNA will bind
to the codon UGG and will be CCA.
5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′
A.
5′-GUAGCCUACCCAUAGG-3′
B. Two. (There are three potential reading frames for each RNA. In this case,
theyare
7-44
GUA GCC UAC CCA UAG …UAG
CCU ACC CAU AGG … AGC CUA
CCC AUA GG? …
The center one cannot be used in this case, because UAG is a stop codon.)
Page 24 of 29
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
C. VAYP
SLPIG
Note: PTHR will not be a peptide because it is preceded by a stop codon.
7-45
(c) A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between
amino acids would allow an assortment of amino acids to be attached to the tRNAIle. These
assorted aminoacyl-tRNAs would then base-pair with the isoleucine codon and cause a variety of
substitutions at positions normally occupied by isoleucine. A mutation in the gene encoding the
protein would cause only a single variant protein to be made [choice (a)]. A mutation in the
anticodon loop of tRNAIle
[choice (b)] or a mutation in the isoleucyl-tRNA synthetase that decreases its ability to
distinguish between different tRNA molecules [choice (d)] would cause the substitution
ofisoleucine for some other amino acid (which is the opposite of what is observed).
(b) The mutant tRNALys will be able to pair with the codon 5-AUA-3, which codes for
′′
7-46
isoleucine.
7-47
Given that there are only single nucleotide changes, the only codons consistent with the changes
are GUG for valine, GCG for alanine, AUG for methionine, and ACG for threonine.
7-48
(b) The 5′-CAA-3′ anticodon binds to the 5′-UUG-3′ codon, which normally codes for leucine.
7-49
If the DNA sequence specifying the anticodon is changed from 5′-GTA-3′ to 5′-CTA-3′, this tRNA will
now pair with the 5′-UAG-3′ codon (instead of 5′ -UAC-3′). The UAG codon normally serves as a
stop codon. Thus, this change will result in the amino acid tyrosine being incorrectly incorporated
where there is a stop codon, resulting in the addition of amino acids at the end of proteins that
normally would come to a stop because of the UAG codon in the mRNA. (Note that the tyrosine
codons will NOT cause premature termination of translation, as tyrosine should continue to be
incorporated into proteins, as there are additional tyrosine-tRNA genes in the cell that will provide
a normal supply of tyrosine-tRNAs.)
7-50
Choice (c) is correct. A ribosome is built from many more proteins than rRNA molecules, although
the ribosome is about two-thirds RNA and one-third protein by weight. Thus, (a) is incorrect. The
small subunit binds to mRNA, so (b) is incorrect. Choice (d) is incorrect, as the assembly and
disassembly of the small subunit with the large subunit occurs every time a protein is produced
from an mRNA. When release factor binds to an mRNA, the ribosome will release the mRNA and
dissociate into its two subunits, to be recycled during another round of protein synthesis.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
7-51
Choice (c) is correct. Ribosomes contain proteins as well as rRNA [choice (a)]. rRNA is synthesized
in the nucleus, and ribosomes are partly assembled in the nucleus [choice (b)]. A ribosome must
be able to bind two tRNAs at any one time [choice (d)].
7-52
See Figure A7-52.
Page 25 of 29
Figure A7-52
7-53
Choice (b) is correct. Choice (a) would prevent all peptide bond formation. Choice (c) would have
no effect on translation until the stop codon was reached. Choice (d) would be likely to result in a
mixture of polypeptides of various lengths; a poison mimicking a release factor could conceivably
cause only Met-Lys to be made, but this dipeptide would not remain bound to the ribosome.
7-54
Choice (b) is correct. Choice (a) is true only for prokaryotes. Choices (c) and (d) are true for both
prokaryotes and eukaryotes.
(b) Bacterial mRNAs do not have 5′ caps. Instead, ribosome-binding sites upstream of the start
7-55
codon tell the ribosome where to begin searching for the start of translation.
A.
(a)—3;
(b)—2;
(c)—4;
(d)—6;(e)—1;
(f)—5
B. The mRNA is prokaryotic. It contains coding regions for more than one protein,
7-56
as shown by the multiple initiation codons, each preceded by a ribosome-binding
site. It contains an unmodified 5′ end, as shown by the three phosphate groups,
and an unmodified 3′ end, as shown by the absence of a poly-A tail.
7-57
(b) The results in Figure Q7-57 show a marked decrease in the number of polyribosomes formed
relative to normal. Polyribosomes form because the initiation of translation is fairly rapid:
ribosomes can bind successively to the free 5′ end of an mRNA molecule and start translation
before the first ribosome has had a chance to finish translating the message. Therefore, inhibition
of the rate of initiation will tend to decrease the number of
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Page 26 of 29
ribosomes in the polyribosome, and in the extreme case there will be only one ribosome per
mRNA. Conversely, increasing the rate of initiation or slowing the rate of elongation would result
in an increased number of ribosomes per polyribosome (up to a maximum point), making choices
(a) and (c) false. Choice (d) is incorrect, because preventing termination would prevent release of
the ribosomes at the end of the coding sequence and would be expected to “freeze” the
assembled polyribosomes, so that the ratio of polyribosomes to ribosomes would be much as
normal.
7-58
(c) The decrease in the level of protein X in the normal cell is most probably due to protein
degradation, because levels of mRNA remain constant. The inability of the mutant cell to divide
could be due to a mutation that inhibits protein degradation. This would be achieved by the
removal of sites for attachment of ubiquitin, which targets proteins for destruction. Choices (a),
(b), and (d) would probably not produce the results
mutant cells described, because without the production of a functional protein X,the
could not grow in size.
7-59
Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is
composed of two subunits: the large subunit, which catalyzes the formation of the peptide bonds
that link the amino acids together into a polypeptide chain, and the small subunit, which matches
the tRNAs to the codons of the mRNA. During the chain elongation process of translating an
mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the P site of the
ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the
A site by forming base pairs with the exposed codon in the mRNA. The peptidyl transferase
enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain
and the newly arriving amino acid. The end of a protein-coding message is signaled by the
presence of a stop codon, which binds the protein called release factor. Eventually, most proteins
will be degraded by a large complex of proteolytic enzymes called the proteasome.
7-60
(d) Proteins are synthesized in the cytoplasm and therefore newly synthesized proteins would not
be exported from the nucleus into the cytoplasm.
7-61
(c) Once proteins are sent to the proteasome, proteases degrade them. Chaperone proteins
provide a place for misfolded proteins to attempt to refold.
7-62
(c) Because RNA is known to catalyze reactions within the cell, because the components of RNA
are thought to be more readily formed in the conditions on primitive Earth, and because RNA can
contain genetic information, it is the most likely of the three molecules to have arisen first in
evolution.
7-63
Choice (d) is correct. Choice (a) is incorrect in that, although this may have been a step in self-
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
replication, it would not by itself be sufficient. Choices (b) and (c) are incorrect, as these stages in
the evolution of the cell must have succeeded the formation of the first self-replicating molecules.
7-64
(c)
Page 27 of 29
7-65 (b)
7-66 Three possible answers are:
1. The deoxyribose sugar of DNA makes the molecule much less susceptible than
RNA to breakage, because of the lack of the hydroxyl group on carbon 2 of the
ribose sugar.
2. DNA is double-stranded and therefore the complementary strand provides a
template from which damage can be repaired accurately.
3. The use of “T” in DNA instead of “U” (as in RNA) protects against the effect of
deamination, a common form of damage. Deamination of T produces an
aberrantbase (methyl C), whereas deamination of U generates C, a normal
base. The presence of an abnormal base eases the cell’s job of recognizing the
damaged
strand.
7-67 (a) An mRNA composed of a trinucleotide repeat of AAC can be “read” in three different
frames: AAC, ACA, and CAA. Thus, this mRNA will yield polyasparagine (codon = AAC),
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
polythreonine (codon = ACA), and polyglutamine (codon = CAA).
7-68 (d) All other answers are not possible, because poly UAUC cannot code for Tyr. Tyr must
be encoded by AUA, because both poly AUA and poly UA lead to the synthesis of Tyr (see
Table A7-68).
Table A7-68
7-69
(d) An organism having codons with an even number of nucleotides (such as 2, 4, or 6) could read
5′-GCGCGCGCGC-3′ (RNA 1) in either of two ways, namely “GC GC GC GC …” or “CG CG CG CG …”
Either of the two amino acids alone could have supported protein synthesis, so you would not
need them in combination [thus eliminating choices (a), (c), and (e)]. An organism having three
bases per codon could read the sequence 5′-GCCGCCGCCGCCGCC-3′ (RNA 2) in one of three ways,
namely “GCC GCC GCC GCC …,” “CCG CCG CCG CCG …,” or “CGC CGC CGC
Page 28 of 29
CGC …,” and so again, any one of the three amino acids could have supported synthesis of a
polypeptide, and you would not need to add all three amino acids to produce a polypeptide chain,
thus eliminating choice (b). Only a five-nucleotide code gives you two different consecutive codons
for RNA 1 and three different consecutive codons for RNA 2.
7-70
No, you cannot definitively determine the codons that code for serine or proline, because it could
be either UAU or AUA.
Bonus. The alien aminoacyl-tRNA synthetases could adapt a different amino acid to each tRNA,
thus matching an amino acid with a different codon compared with the codons used by life on
Earth.
Page 29 of 29
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
INTRACELLULAR COMPARTMENTSAND TRANSPORT
Membrane-Enclosed Organelles
15.1 Name the membrane-bounded compartments in a eucaryotic cell where each of
thefunctions listed below takes place.
A. Photosynthesis
B. Transcription
C. Oxidative phosphorylation
D. Modification of secreted proteins
E. Steroid hormone synthesis
F. Degradation of worn-out organelles
G. New membrane synthesis
H. Breakdown of lipids and toxic molecules
1
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.2
Label the structures of the cell indicated by the lines on the figure below:
Figure Q15-2
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
15.3
nucleus
free ribosomes
rough endoplasmic reticulum
Golgi apparatus
cytosol
endosome
plasma membrane
lysosome
mitochondrion
peroxisome
You discover a fungus that contains a strange star-shaped organelle not found in
anyother eucaryotic cell you have seen. On further investigation you find the
following
1.
the organelle possesses a small genome in its interior.
2.
the organelle is surrounded by two membranes.
3.
vesicles do not pinch off the organelle membrane.
4.
the interior of the organelle contains proteins similar to those of many bacteria.
5.
the interior of the organelle contains
ribosomes. How might this organelle have
arisen?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Protein Sorting
15.4
For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or
phraseshould be used only once.
Plasma membrane proteins are inserted into the membrane in the
. The address information for protein sorting in a
eucaryotic cell is contained in the
of the proteins.
Proteins enter the nucleus in their
form. Proteins
that remain in the cytosol do not contain a
.
Proteins are transported into the Golgi apparatus via
. The proteins transported into the endoplasmic
reticulum by
are in their
form.
amino acid sequence
endoplasmic reticulum
folded
15.5
Golgi apparatus
plasma membrane
protein translocators
sorting signal
transport vesicles
unfolded
What would happen in each of the following cases? Assume in each case that the
proteininvolved is a soluble protein, not a membrane protein.
A.
You add a signal sequence (for the ER) to the amino-terminal end of a normally
cytosolic protein.
B.
You change the hydrophobic amino acids in an ER signal sequence into charged
amino acids.
C.
You change the hydrophobic amino acids in an ER signal sequence into other,
hydrophobic, amino acids.
D.
You move the amino-terminal ER signal sequence to the carboxyl-terminal end
ofthe protein.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.6
You are trying to identify the peroxisome-targeting sequence in the thiolase enzyme
fromyeast. The thiolase enzyme normally resides in the peroxisome and therefore must
contain amino acid sequences that are used to target the enzyme for import into the
peroxisome. To identify the targeting sequences, you create a set of hybrid genes that
encode fusion proteins containing part of the thiolase protein fused to another protein,
histidinol dehydrogenase (HDH). HDH is a cytosolic enzyme required for the synthesis
of the amino acid histidine and cannot function if it is localized in the peroxisome. You
genetically engineer a series of yeast cells to express these fusion proteins instead of
their own versions of these enzymes. If the fusion proteins are imported into the
peroxisome, the HDH portion of the protein cannot function and the yeast cells cannot
grow on media lacking histidine. You obtain the following results:
Figure Q15-6
What region of the thiolase protein contains the peroxisomal targeting sequence? Explainyour
answer.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.7
What is the role of the nuclear localization sequence in a nuclear protein?
(a)
It is bound by cytoplasmic proteins that direct the nuclear protein to
thenuclear pore.
(b)
It is a hydrophobic sequence that enables the protein to enter
thenuclear membranes.
(c)
It aids protein unfolding in order for the protein to thread through nuclear pores.
(d)
It prevents the protein diffusing out of the nucleus via nuclear pores.
15.8
A gene regulatory protein, A, contains a typical nuclear localization signal but
surprisingly is usually found in the cytosol of cells. When the cell is exposed to
hormones, protein A moves from the cytosol into the nucleus where it turns on genes
involved in cell division. When you purify protein A from cells that have not been
treated with hormones, you find that protein B is always complexed with it. To
determine the function of protein B, you engineer cells lacking the gene for protein B.
You compare normal and defective cells by using differential centrifugation to
separate the nuclear fraction from the cytoplasmic fraction and then separate the
proteins in thesefractions by gel electrophoresis. You identify the presence of protein A
and protein B bylooking for their characteristic bands on the gel. The gel you run is
shown below:
Figure Q15-8
On the basis of these results, what is the function of protein B? Explain your conclusionand
propose a mechanism for how protein B works.
15.9
Which of the following statements about import of proteins into mitochondria are TRUE?
(a)
The signal sequences on mitochondrial proteins are usually carboxyl terminal.
(b)
The first stage of import of a mitochondrial protein is across the
outermembrane into the intermembrane space.
(c)
Most mitochondrial proteins are not imported from the cytosol but
aresynthesized inside the mitochondria.
(d)
Mitochondrial proteins are translocated across the inner and
outermembranes simultaneously.
(e)
Mitochondrial proteins cross the membrane in their native, folded state.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.10 Proteins destined to enter the endoplasmic reticulum
(a)
are transported across the membrane after their synthesis is complete.
(b)
are synthesized on free ribosomes in the cytosol.
(c)
begin to cross the membrane while still being synthesized.
(d)
cross the membrane in a folded state.
(e)
all remain within the endoplasmic reticulum.
15.11 After isolating the rough endoplasmic reticulum from the rest of the cytoplasm, you
purify the RNAs attached to it. Which of the following proteins do you expect the RNA
from the rough endoplasmic reticulum to encode?
(a)
Soluble secreted proteins
(b)
ER membrane proteins
(c)
Mitochondrial membrane proteins
(d)
Plasma membrane proteins
(e)
Ribosomal proteins
15.12 Briefly describe the mechanism by which the presence of an internal stop-transfer
sequence in a protein causes the protein to become embedded in the lipid bilayer as
a transmembrane protein with a single membrane-spanning region. Assume that the
protein has an amino terminal signal sequence and just one internal hydrophobic
stop-transfer sequence.
15.13 Using genetic engineering techniques, you have created a set of proteins that contain
two (and only two) conflicting signal sequences that specify different compartments.
Predict which signal would win out for the following combinations. Explain youranswers.
A.
Signals for import into the nucleus and import into the ER.
B.
Signals for export from the nucleus and import into the mitochondria.
C.
Signals for import into mitochondria and retention in the ER.
15.14 A protein traverses the plasma membrane three times in the orientation shown
below (N = amino terminus, C = carboxyl terminus; the hydrophobic membranespanning regions are shown as open boxes). This protein is known to have a signal
sequence thatis cleaved by signal peptidase in the ER.
Figure Q15-14
Sketch the ER membrane and the arrangement of the newly synthesized protein chainafter it
has completed its entry into the ER membrane but before any action of signal
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
peptidase. Be sure to label the cytosol, the ER lumen, the signal sequence, and the aminoand
carboxyl termini of the protein in your diagram.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.15 The figure below shows the orientation of a multipass transmembrane protein after it
hascompleted its entry into the ER membrane (part A) and after it gets delivered to the
plasma membrane (part B). This protein has an amino-terminal signal sequence
(depicted as the dark grey membrane spanning box), which is cleaved off in the
endoplasmic reticulum by signal peptidase. The other membrane-spanning domains in
the protein are depicted as open boxes. Given that any hydrophobic membranespanningdomain can act as either a start-transfer or a stop-transfer region, draw the
final consequences of the actions described below on the orientation of the protein in
the plasma membrane. Be sure to indicate on your drawing the extracellular space, the
cytosolic face, and the plasma membrane, as well as the amino- and carboxyl-termini
of the protein.
Figure Q15-15
Deleting the first signal sequence.
Changing the hydrophobic amino acids in the first, cleaved, sequence
tocharged amino acids.
C.
Changing the hydrophobic residues in every other transmembrane
sequenceto charged residues, starting with the first, cleaved, signal
sequence.
15.16 Examine the multipass transmembrane protein shown in Figure Q15-16. What would
you predict would be the effect of converting the first hydrophobic transmembrane
segment to a hydrophilic segment? Sketch the arrangement of the modified protein in
theER membrane.
A.
B.
Figure Q15-16
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Vesicular Transport
15.17 For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or
phraseshould be used only once.
Proteins are transported out of a cell via the
or
pathway. Fluids and macromolecules are
transported into the cell via the
pathway. All
proteins being transported out of the cell pass through the
and the
. Transport vesicles
link organelles of the
system. The formation of
in the endoplasmic reticulum stabilizes protein
structure.
carbohydrate
disulfide bonds
endocytic
endomembrane
endoplasmic reticulum
endosome
exocytic
Golgi apparatus
hydrogen bonds
ionic bonds
lysosome
protein
secretory
15.18 Name two functions of the protein coat of vesicles that bud from
membranousorganelles used in vesicular transport.
15.19 An individual transport vesicle
(a)
contains only one type of protein in its lumen.
(b)
will fuse with only one type of membrane.
(c)
is endocytic if it is traveling toward the plasma membrane.
(d)
is enclosed by a membrane with the same lipid and protein composition
asthe membrane of the donor organelle.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.20 In class we have discussed how v-SNAREs and t-SNARES mediate the recognition ofa
vesicle with its target membrane so that a vesicle displaying a particular type of vSNARE will only fuse with a target membrane containing a complementary type of tSNARE. It is also known that in some cases, v-SNAREs and t-SNAREs may also mediate
fusion of identical membranes. In yeast cells, right before the formation of a new cell,
vesicles derived from the vacuole will come together and fuse to form a new vacuole
destined for the new cell. Unlike the situation we’ve discussed in class, the vacuolar
vesicles contain both v-SNAREs and t-SNAREs. Your friend is trying to understand the
role of these SNAREs in the formation of the new vacuole and wants toconsult with
you regarding the interpretation of his data.
Your friend has designed an ingenious assay for the fusion of vacuolar vesicles utilizing alkaline
phosphatase. The protein alkaline phosphatase is made in a “pro” form that must be cleaved
in order for the protein to be active. Your friend has designed two different strains of yeast:
strain A produces the “pro” form of alkaline phosphatase (pro-Pase), while strain B produces
the protease that can cleave pro-Pase into the active form(Pase). Neither strain has the active
form of the alkaline phosphatase, but when vacuolarvesicles from the strains A and B are
mixed, fusion of vesicles generates active alkaline phosphates, whose activity can be measured
and quantified.
Figure Q15-20A
Your friend has taken each of these yeast strains and further engineered them so that they
express only the v-SNAREs, the t-SNAREs, both (the normal situation), or neither SNARE. He
then isolates vacuolar vesicles from all strains and tests the ability of each variant form of strain
A to fuse with each variant form of strain B, using the alkaline phosphatase assay. The data are
shown in the graph depicted in Figure Q15-20B. On this graph, the SNARE present on the
vesicle of the particular yeast strain is indicated as“v” (for the presence of the v-SNARE) and “t”
(for the presence of the t-SNARE).
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Figure Q15-20 B
What does his data say about the requirements for v-SNAREs and t-SNAREs in the vacuolar
vesicles? Be sure to comment on whether it is important to have a specific typeof SNARE (that is,
v- or t-SNARE) on each vesicle.
Secretory Pathway
15.21 N-linked oligosaccharides on secreted glycoproteins are attached to
(a)
nitrogen atoms in the polypeptide backbone.
(b)
the serine or threonine in the sequence Asn-X-Ser/Thr.
(c)
the amino terminus of the protein.
(d)
the asparagine in the sequence Asn-X-Ser/Thr.
(e)
the aspartic acid in the sequence Asp-X-Ser/Thr.
15.22 Name two types of protein modification that can occur in the ER but not in the cytosol.
15.23 If you were to remove the ER-retention signal from a protein that normally resides
inthe ER lumen, where do you expect the protein will ultimately end up? Be sure to
explain your reasoning, for full credit.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.24 Match the set of labels below with the numbered label lines on Figure 15-24.
Figure Q15-24
A.
B.
C.
D.
E.
Cisterna
Golgi stack
Secretory vesicle
trans Golgi network
cis Golgi network
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.25 A plasma membrane protein carries an oligosaccharide containing mannose (Man),
galactose (Gal), sialic acid (SA), and N-acetylglucosamine (GlcNAc). These sugars are
added to the protein as it proceeds through the secretory pathway. First, a core
oligosaccharide containing Man and GlcNAc is added, followed by Gal, Man, SA, and
GlcNAc in a particular order. Each addition is catalyzed by a different transferase acting
at a different stage as the protein proceeds through the secretory pathway. You have
isolated mutants defective for each of the transferases, purified the membrane
protein from each of the mutants, and identified which sugars are present in each
mutant protein. The results are summarized in Table Q15-25.
Table Q15-25
Cell lacking:
A. Oligosaccharide
protein
transferase
Sugars present in the purified protein
SA
GlcNAc
Ma
Gal
–
–
–
n
–
B. Galactose
transferase
C. SA transferase
D. GlcNAc
transferase
+
–
–
+
+
+
+
–
–
–
+
less than in
normal cells
From these results, match each of the transferases (A, B, C, D) to its subcellular locationselected
from the list below. (Assume that each location contains only one enzyme.)
1.
2.
3.
4.
Central Golgi cisternae
cis Golgi network
ER
trans Golgi network
15.26 For each of the following sentences, choose one of the options enclosed
insquare brackets to make a correct statement.
New plasma membrane reaches the plasma membrane by the
[regulated/constitutive] exocytosis pathway. New plasma membrane proteins
reach the plasma membrane by the [regulated/constitutive] exocytosis
pathway. Insulin is secreted from pancreatic cells by the
[regulated/constitutive] exocytosis pathway. The interior of the trans Golgi
network is [acidic/alkaline]. Proteins that are constitutively secreted
[aggregate/do not aggregate] in the trans Golgi network.
15.27 In a cell capable of regulated secretion, what are the three main classes of
proteinsthat must be separated before they leave the trans Golgi network?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Endocytic Pathways
15.28 Name three possible fates for an endocytosed molecule that has reached the endosome.
15.29 For each of the following sentences, fill in the blanks with the best word or phrase
selected from the list below. Not all words or phrases will be used; each word or
phraseshould be used only once.
Eucaryotic cells are continually taking up materials from the extracellularspace
by the process of endocytosis. One type of endocytosis is
, which involves utilizing
proteins to form small vesicles containing fluids and molecules. After these
vesicles pinch off from the plasma membrane, they will fuse with the
, where the uptaken materials are sorted. A second
type of endocytosis is , which is used to takeup large vesicles that can contain
microorganisms and cellular debris.
Macrophages are especially suited for this process, as they extend
(sheetlike projections of their plasma membrane)to
surround the invading microorganisms.
chaperone
Golgi apparatus
cholesterol
clathrin
endosome
mycobacterium
phagocytosis
pinocytosis
pseudopod
s
rough ER
SNARE
transcytosis
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.30 Fibroblast cells from patients W, X, Y, and Z, who each have a different inherited defect,
all contain “inclusion bodies,” which are lysosomes filled with undigested material. You
wish to identify the cellular basis of these defects. The possibilities are:
1.
2.
3.
a defect in one of the lysosomal hydrolases.
a defect in the phosphotransferase that is required for
mannose-6-phosphate tagging of the lysosomal hydrolases.
a defect in the mannose-6-phosphate receptor, which binds mannose-6phosphate tagged lysosomal proteins in the trans Golgi network and
deliversthem to lysosomes.
You find that when some of these mutant fibroblasts are incubated in media in which normal
cells have been grown, the inclusion bodies disappear. This leads you to suspect that lysosomal
hydrolases are being secreted by the constitutive exocytic pathway in normal cells and are
being taken up by the mutant cells. (It is known that some mannose-6-phosphate receptor
molecules are found in the plasma membrane and can take up and deliver lysosomal proteins
via the endocytic pathway.) You incubate cells from each patient with media from normal cells
and media from each of the other mutant cell cultures, and get the following results.
Cell Line
Normal
W
X
Y
Z
From
normal
cells
From
cultures
of W cells
Media
From
cultures
of X cells
+
–
+
+
+
+
–
+
+
+
+
–
–
–
–
From
cultures
of Y cells
From
cultures
of Z cells
+
–
–
–
+
+
–
–
+
–
+ indicates that the cells appear normal; – indicates that the cells still have inclusionbodies.
For each patient (W, X, Y, Z) indicate which of the defects (1, 2, 3) they are most likelyto have.
15.31 How is it that the low pH of lysosomes protects the rest of the cell
fromlysosomal enzymes in case the lysosome breaks?
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
How We Know: Tracking Protein and Vesicle Transport
15.32 You have created a GFP fusion to a protein that is normally secreted from yeast cells.
Since you have learned about the use of temperature-sensitive mutations in yeast to
study
protein and vesicle transport, you obtain a collection of three mutant yeast strains, each one
defective in some aspect of the protein secretory process. Being a good scientist, youof course,
also obtain a wild-type control strain. You decide to examine the fate of your GFP fusion protein
in these various yeast strains and engineer the mutant strains to express your GFP fusion
protein. However, in your excitement to do the experiment, yourealize that you did not label
any of the mutant yeast strains and no longer know which strain is defective in what process.
You end up numbering your strains with the numbers1 through 4, and then you carry out the
experiment anyway, obtaining the following results (note that the black dots represent your
GFP fusion protein):
Figure Q15-32
Name the process defective in each of these strains. Remember that one of these strainsis
your wild-type control.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
Answers
15.1
A
B.
C.
D.
E.
F.
G.
H.
15.2
Photosynthesis = chloroplast
Transcription = nucleus
Oxidative phosphorylation = mitochondrion
Modification of secreted proteins = Golgi apparatus and
roughendoplasmic reticulum (ER)
Steroid hormone synthesis = smooth ER
Degradation of worn-out organelles = lysosome
New membrane synthesis = ER
Breakdown of lipids and toxic molecules = peroxisome
See Figure A15-2.
Figure A15-2
15.3
A genome, a double membrane, ribosomes, and proteins similar to those found
in bacteria are evidence for an organelle having evolved from an engulfed
bacterium.
15.4
Plasma membrane proteins are inserted into the membrane in the endoplasmic
reticulum. The address information for protein sorting in a eucaryotic cell is contained in
the amino acid sequence of the proteins. Proteins enter the nucleus in their folded
form. Proteins that remain in the cytosol do not contain a sorting signal. Proteins are
transported into the Golgi apparatus via transport vesicles. The proteins transported
intothe endoplasmic reticulum by protein translocators are in their unfolded form.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.5
A.
B.
C.
D.
15.6
The protein will now be transported into the ER lumen.
The altered signal sequence will not be recognized and the protein will
remainin the cytosol.
The protein will still be delivered into the ER. It is the distribution of
hydrophobic amino acids that is important, not the actual
sequence.
The protein will not enter the ER. Because the carboxyl terminus of the
proteinis the last part to be made, the ribosomes synthesizing this protein
will not be recognized by the SRP and carried to the ER.
The peroxisomal targeting sequence lies between amino acids number 100 and number
125. Any fusion protein containing this sequence can be targeted for import into the
peroxisome (because the yeast cannot grow on media lacking histidine) while the fusion
proteins lacking this region do not target the fusion protein for import into the peroxisome
(because the yeast do grow on media lacking histidine). The most importantpieces of data are
the fusion proteins containing amino acids 100–200 of the thiolase protein fused to HDH and
the fusion protein containing amino acids 1–125 of the thiolase protein fused to HDH. Both of
these fusion proteins do not allow growth on media lacking histidine and can be used to define
the minimal region necessary for targeting thiolase for import into the peroxisome.
(Note that although these experiments show that amino acids 100–125 are necessary, these
experiments do not show that this region is sufficient for peroxisomal targeting. Itis possible
that amino acids 100–125 is sufficient, or, it could be that this region collaborates with
redundant signals between amino acids 1–100 or 125–200.)
15-7
(a)
15-8
The data on the gel shows that protein A is always found in the nucleus in the absenceof
protein B. Therefore, any mechanism that is proposed must explain this result.
On possible answer is that protein B binds protein A and masks the nuclear localization signal.
In the presence of hormone, protein B interacts with the hormone, which changesits
conformation so that it can no longer bind protein A. When protein B no longer bindsto protein
A, the nuclear localization signal on protein A is now exposed and protein A can enter the
nucleus. Therefore, in the absence of protein B, the nuclear localization signal on protein A is
always exposed and protein A resides in the nucleus.
Another possible answer is that protein B binds protein A and sequesters it by keeping protein A
in some subcellular compartment, away from the nucleus. In the presence of hormone, protein B
interacts with the hormone, changing its conformation so that it can
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
no longer bind to protein A. When protein B is not present, protein A can enter thenucleus in
the presence or absence of hormone.
15-9
(d)
15-10 (c)
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.11
(a), (b), and (d) The rough ER consists of ER membranes and polyribosomes that are in
the process of translating and translocating proteins into the ER membrane
and lumen. Thus all proteins that end up in the lysosome, Golgi apparatus, or
plasma membrane, or are secreted, will be encoded by the RNAs associated
with the rough ER. Mitochondrial and ribosomal proteins are translated on
free cytosolic ribosomes.
15.12
The amino-terminal signal sequence initiates translocation and the protein chain starts
tothread through the translocation channel. When the stop-transfer sequence enters
the translocation channel, the channel discharges both the signal sequence and the
stop- transfer sequence sideways into the lipid bilayer. The signal sequence is then
cleaved, sothat the protein remains held in the membrane by the hydrophobic stoptransfer sequence.
15.13
A.
B.
C.
15.14
The protein would enter the ER. The signal for a protein to enter the ER is
recognized as the protein is being synthesized and will end up either in the
ERor on the ER membrane. Proteins destined for the nucleus get recognized
by cytosolic nuclear transport proteins once they are fully synthesized and
fully folded.
The protein would enter in the mitochondria. In order for a nuclear export
signalto work, the protein would have to end up in the nucleus first and thus
would need a nuclear import signal for the nuclear export signal to get utilized.
The protein would enter the mitochondria. In order to be retained in the ER,
theprotein needs to enter the ER. Since there is no signal for ER import, the
ER retention signal would not function.
The N-terminal signal sequence initiates translocation of the protein across the ER
membrane. The signal sequence will be cleaved off by signal peptidase, leaving the
amino-terminus of the protein in the luminal side of the ER membrane. Upon fusion
tothe plasma membrane, the amino terminus of the protein will reside in the
extracellularspace.
Figure A15-14
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.15
A.
Deleting the first signal sequence completely would convert the next
membrane- spanning domain into an internal start-transfer signal and
wouldinvert the orientation of the protein (see Figure A15-15A).
B.
Changing the hydrophobic amino acids to charged amino acids destroys the
ability of the sequence both to act as a signal sequence and to become a
membrane-spanning sequence. Therefore, the adjacent membrane spanning
domain will now become an internal start-transfer sequence and the protein
willbe inverted, as seen above in part A. The mutated signal sequence would
not getcleaved off, since it would remain on the cytoplasmic side of the
membrane andsignal peptidase is found only inside the ER (see Figure A1515B).
C.
Mutating every other membrane spanning region so that they are now
charged (and thus cannot span the membrane) would decrease the
number oftransmembrane regions and increase the size of the internal
loops between membrane-spanning regions (see Figure 15-15C).
Figure A15-15
15.16
As shown in Figure A15-16, elimination of the first transmembrane segment (by
making it hydrophilic) would be expected to reverse the orientation of the protein in
the membrane. What originally was the second transmembrane segment (a stoptransfer signal), would now be read as a start-transfer signal and would have the
opposite orientation in the membrane—as would all the remaining transmembrane
segments. Although the N-terminus would still be in the ER lumen, all the rest of
the external
parts of the protein would swap positions so that what was in the cytosol would now bein the
ER lumen, and vice versa.
Figure A15-16
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.17
Proteins are transported out of a cell via the secretory or exocytic pathway. Fluid
andmacromolecules are transported into the cell via the endocytic pathway. All
proteins being transported out of the cell pass through the endoplasmic reticulum
and the Golgi apparatus. Transport vesicles link organelles of the endomembrane
system. The formation of disulfide bonds in the endoplasmic reticulum stabilizes
protein structure.
15.18
1.
The proteins in the coat help shape the membrane into a
bud.
2.
The proteins in the coat can also select cargoes for transport.
15.19
Choice (b) is the correct answer. An individual vesicle may contain more than one type
of protein in its lumen (choice (a)), all of which will contain the same sorting signal (or
will lack specific sorting signals). Endocytic vesicles (choice (c)) generally move away
from the plasma membrane. The vesicle membrane will not necessarily contain the
samelipid and protein composition as the donor organelle, since the vesicle is formed
from a selected subset of the organelle membrane from which it budded (choice (d)).
15.20
In order to get maximal levels of vacuolar vesicle fusion, vesicles from each strain must
carry both v-SNAREs and t-SNARES. Experiment 1, which represents the normal
scenario, is the only experiment where 100% alkaline phosphatase activity is
measured. However, as long as complementary SNAREs are present on the vesicles,
some vesicle fusion does occur (see experiments 3, 4, 6, 7, 8, 9). If both vesicles are
missing either v- SNAREs (experiment 2) or t-SNAREs (experiment 5) or both SNAREs
(experiment 10 and 11), the level of fusion is very low. It does not matter whether a tor v-SNARE is onthe vesicle of a particular strain, as long as the vesicle from the other
strain contains a complementary SNARE (compare experiments 3 and 4, 6 and 7, and 8
and 9).
15-21 (d)
15.22
15.23
1.
Proteins in the ER can undergo disulfide bond formation. (This does not occur in
the cytosol because of its reducing environment.)
2.
Proteins in the ER can undergo glycosylation. (Glycosylating enzymes are notfound
in the cytosol.)
(Signal-sequence cleavage is also an acceptable answer, although not really whatthis
question is referring to.)
The protein would end up in the extracellular space. Normally, the protein would go
from the ER to the Golgi apparatus, get captured because of its ER-retrieval signal,
andreturn to the ER. However, without the ER-retrieval signal, the protein would
evade capture, ultimately leave the Golgi via the default pathway, and become
secreted into the extracellular space. The protein would not be retained anywhere
else along the secretory pathway, as it presumably has no signals to promote such
localization since itnormally resides in the ER lumen.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15-24 A—3; B—1; C—5; D—4; E—2
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.25
A—3 (oligosaccharide protein transferase = ER) B—1
(galactose transferase = central Golgi cisternae) C
—4 (SA transferase = trans Golgi network) D—2
(GlcNAc transferase = cis Golgi network
Proteins are modified in a stepwise fashion in the Golgi apparatus, with early steps takingplace
in the cis Golgi, intermediate steps taking place in the central Golgi cisternae, and late steps
occurring in the trans Golgi network. If each enzyme produces the substrate for the next step,
then a mutant lacking the enzyme that catalyzes the addition of the first sugar will be missing all
of the sugars, a mutant lacking the enzyme that catalyzes the addition of the second sugar will
contain the first sugar but will lack the other three, and so on. By this logic, mannose and
GlcNAc must be the first sugars added, additional GlcNAc the second, galactose the third, and
SA the last. Hence, the oligosaccharide protein transferase must be in the ER, the GlcNAc
transferase in the cis Golgi, the galactose transferase in the central Golgi, and the SA transferase
in the trans Golgi.
15.26
New plasma membrane reaches the plasma membrane by the constitutive
exocytosis pathway. New plasma membrane proteins reach the plasma membrane
bythe constitutive exocytosis pathway. Insulin is secreted from pancreatic cells by
the regulated exocytosis pathway. The interior of the trans Golgi network is acidic.
Proteins that are constitutively secreted do not aggregate in the trans Golgi network.
15.27
The three main classes of proteins that must be sorted before they leave the trans
Golgi network in a cell capable of regulated secretion are (1) those destined for
lysosomes, (2)those destined for secretory vesicles, and (3) those destined for
immediate delivery to the cell surface.
15.28
1.
2.
3.
15.29
Eucaryotic cells are continually taking up materials from the extracellular space by the
process of endocytosis. One type of endocytosis is pinocytosis, which involves utilizing
clathrin proteins to form small vesicles containing fluids and molecules. After these
vesicles pinch off from the plasma membrane, they will fuse with the endosome, where
the uptaken materials are sorted. A second type of endocytosis is phagocytosis, which
is used to take up large vesicles that can contain microorganisms and cellular debris.
Macrophages are especially suited for this process, as they extend pseudopods
(sheetlike
recycled to the original membrane
destroyed in the lysosome
transcytosed across the cell to a different membrane.
projections of their plasma membrane) to surround the invading microorganisms.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)
lOMoARcPSD|30659517
15.30
W—3 (defect in mannose-6-phosphate
receptor) X—2 (defect in
phosphotransferase)
Y—1; Z—1 (defect in lysosomal hydrolases); these will be defects in two differentlysosomal acid
hydrolases.
A cell that has no mannose-6-phosphate receptor will be able to make all the lysosomal
hydrolases properly, but will not be able to send them to the lysosome and will also not be able
to scavenge hydrolases from the external media. Hence, this cell line cannot be rescued by
culture media that has had lysosomal hydrolases secreted into it and thus willnot be rescued by
any of the media tested here. A cell line that has no phosphotransferasewill be able to scavenge
hydrolases from the external medium, but since all of the cell’s own hydrolases will lack the
mannose-6-phosphate tag, it will be rescued only by media from a cell line that is able to make
all of the hydrolases. Cell lines missing one hydrolase will be rescued by media from any cell
line that is able to secrete that hydrolase in a mannose-6-phosphate tagged form; in addition,
media from cultures of cells missing a hydrolase will rescue any cell line with another type of
defect.
15.31
The lysosomal enzymes are all acid hydrolases, which have optimal activity at the low
pH (about 5.0) found in the interior of lysosomes. If a lysosome were to break, the acid
hydrolases would find themselves at pH 7.2, the pH of the cytosol, and would therefore
do little damage to cellular constituents.
15.32
Strain A has protein accumulating in the ER, which means that this cell has a mutation
that blocks transport from the ER to the Golgi apparatus. Strain B has secreted protein,
and therefore is your wild-type control. Strain C has protein accumulating in the Golgi
apparatus, and thus has a mutation that blocks exit of proteins from the Golgi
apparatus.Strain D has protein accumulating in the cis-Golgi network, and thus has
a mutationthat blocks the travel of proteins through the Golgi cisternae.
Downloaded by S? Hi?p (nguyensyhoanghiep2003@gmail.com)