Essential Cell Biology 5th edition

A:50 Answers
ANSWER 18–24 Antibodies bind tightly to the antigen
(in this case myosin) to which they were raised. When
bound, an antibody can interfere with the function of the
antigen by preventing it from interacting properly with
other cell components. (A) The movement of chromosomes
at anaphase depends on microtubules and their motor
proteins and does not depend on actin or myosin. Injection
of an anti-myosin antibody into a cell will therefore have no
effect on chromosome movement during anaphase.
(B) Cytokinesis, on the other hand, depends on the
assembly and contraction of a ring of actin and myosin
filaments, which forms the cleavage furrow that splits
the cell in two. Injection of an anti-myosin antibody will
therefore block cytokinesis.
ANSWER 18–25 The plasma membrane of the cell that died
by necrosis in Figure 18−38A is ruptured; a clear break is
visible, for example, at a position corresponding to the
12 o’clock mark on a watch. The cell’s contents, mostly
membranous and cytoskeletal debris, are seen spilling
into the surroundings through these breaks. The cytosol
stains lightly, because most soluble cell components were
lost before the cell was fixed. In contrast, the cell that
underwent apoptosis in Figure 18−38B is surrounded by
an intact membrane, and its cytosol is densely stained,
indicating a normal concentration of cell components. The
cell’s interior is remarkably different from a normal cell,
however. Particularly characteristic are the large “blobs”
that extrude from the nucleus, probably as the result of the
breakdown of the nuclear lamina. The cytosol also contains
many large, round, membrane-enclosed vesicles of unknown
origin, which are not normally seen in healthy cells. The
pictures visually confirm the notion that necrosis involves
cell lysis, whereas cells undergoing apoptosis remain
relatively intact until they are phagocytosed and digested
by another cell.
ANSWER 18–26
A. False. There is no G 1 to M phase transition. The
statement is correct, however, for the G 1 to S phase
transition, in which cells commit themselves to a division
cycle.
B. True. Apoptosis is an active process carried out by
special proteases (caspases).
C. True. This mechanism is thought to adjust the number of
neurons to the number of specific target cells to which
the neurons connect.
D. True. An amazing evolutionary conservation!
E. True. Association of a Cdk protein with a cyclin is
required for its activity (hence its name cyclin-dependent
kinase). Furthermore, dephosphorylation at specific
sites on the Cdk protein is required for the cyclin–Cdk
complex to be active.
ANSWER 18–27 Cells in an animal must behave for the
good of the organism as a whole—to a much greater
extent than people generally act for the good of society as
a whole. In the context of an organism, unsocial behavior
would lead to a loss of organization and possibly to
cancer. Many of the rules that cells have to obey would be
unacceptable in a human society. Most people, for example,
would be reluctant to kill themselves for the good of
society, yet our cells do it all the time.
ANSWER 18–28 The most likely approach to success (if that
is what the goal should be called) is plan C, which should
Figure A18–28
Courtesy of Ralph Brinster
result in an increase in cell numbers. The problem is, of
course, that cell numbers of each tissue must be increased
similarly to maintain balanced proportions in the organism,
yet different cells respond to different growth factors. As
shown in Figure A18–28, however, the approach has indeed
met with limited success. A mouse producing very large
quantities of growth hormone (left)—which acts to stimulate
the production of a secreted protein that acts as a survival
factor, growth factor, or mitogen, depending on the cell
type—grows to almost twice the size of a normal mouse
(right). To achieve this ECB5 twofold EA18.28/A18.28 change in size, however,
growth hormone was massively overproduced (about
fiftyfold). And note that the mouse did not even attain the
size of a rat, let alone a dog.
The other two approaches have conceptual problems:
A. Blocking all apoptosis would lead to defects in
development, as rat development requires the selective
death of many cells. It is unlikely that a viable animal
would be obtained.
B. Blocking p53 function would eliminate an important
mechanism in the cell cycle that detects DNA damage
and stops the cycle so that the cell can repair the
damage; removing p53 would increase mutation rates
and lead to cancer. Indeed, mice without p53 usually
develop normally but die of cancer at a young age.
ANSWER 18–29 The on-demand, limited release of PDGF
at a wound site triggers cell division of neighboring cells
for a limited amount of time, until the PDGF is degraded.
This is different from the continuous release of PDGF from
mutant cells, where PDGF is made in an uncontrolled way
at high levels. Moreover, the mutant cells that make PDGF
often express their own PDGF receptor inappropriately,
so that they can stimulate their own proliferation, thereby
promoting the development of cancer.
ANSWER 18–30 All three types of mutant cells would be
unable to divide. The cells:
A. would enter mitosis but would not be able to exit
mitosis.
B. would arrest permanently in G 1 because the cyclin–Cdk
complexes that act in G 1 would be inactivated.
C. would not be able to activate the transcription of
genes required for cell division because the required
transcription regulators would be constantly inhibited by
unphosphorylated Rb.
ANSWER 18–31 In alcoholism, liver cells proliferate because
the organ is overburdened and becomes damaged by the