Essential Cell Biology 5th edition

Answers A:47
ANSWER 18–3 For multicellular organisms, the control of
cell division is extremely important. Individual cells must not
proliferate unless it is to the benefit of the whole organism.
The G 0 state offers protection from aberrant activation of
cell division because the cell-cycle control system is largely
dismantled. If, on the other hand, a cell just paused in G 1 ,
it would still contain all of the cell-cycle control system and
could readily be induced to divide. The cell would also have
to remake the “decision” not to divide almost continuously.
To re-enter the cell cycle from G 0 , a cell has to resynthesize
all of the components that have disappeared.
ANSWER 18–4 The cell would replicate its damaged
DNA and therefore would introduce mutations to the two
daughter cells when the cell divides. Such mutations could
increase the chances that the progeny of the affected
daughter cells would eventually become cancer cells.
ANSWER 18–5 Before injection, the frog oocytes must
contain inactive M-Cdk. Upon injection of the M-phase
cytoplasm, the small amount of the active M-Cdk in
the injected cytoplasm activates the inactive M-Cdk by
switching on the activating phosphatase (Cdc25), which
removes the inhibitory phosphate groups from the inactive
M-Cdk (see Figure 18−17). An extract of the second oocyte,
now in M phase itself, will therefore contain as much active
M-Cdk as the original cytoplasmic extract, and so on.
ANSWER 18–6 The experiment shows that kinetochores
are not preassigned to one or other spindle pole;
microtubules attach to the kinetochores that they are able
to reach. For the chromosome to remain attached to a
microtubule, however, tension has to be exerted. Tension
is normally achieved by the opposing pulling forces from
opposite spindle poles. The requirement for such tension
ensures that if two sister kinetochores ever become
attached to the same spindle pole, so that tension is not
generated, one or both of the connections would be lost,
and microtubules from the opposing spindle pole would
have another chance to attach properly.
ANSWER 18–7 Recall from Figure 18−30 that the new
nuclear envelope reassembles on the surface of the
chromosomes. The close apposition of the envelope to
the chromosomes prevents cytosolic proteins from being
trapped between the chromosomes and the envelope.
Nuclear proteins are then selectively imported through
the nuclear pores, causing the nucleus to expand while
maintaining its characteristic protein composition.
ANSWER 18–8 The membranes of the Golgi vesicles fuse
to form part of the plasma membranes of the two daughter
cells. The interiors of the vesicles, which are filled with cell
wall material, become the new cell wall matrix separating
the two daughter cells. Proteins in the membranes of the
Golgi vesicles thus become plasma membrane proteins.
Those parts of the proteins that were exposed to the lumen
of the Golgi vesicle will end up exposed to the new cell wall
(Figure A18–8).
ANSWER 18–9 In a eukaryotic organism, the genetic
information that the organism needs to survive and
reproduce is distributed between multiple chromosomes. It
is therefore crucial that each daughter cell receives a copy
of each chromosome when a cell divides; if a daughter cell
receives too few or too many chromosomes, the effects
are usually deleterious or even lethal. Only two copies of
each chromosome are produced by chromosome replication
in mitosis. If the cell were to randomly distribute the
chromosomes when it divided, it would be very unlikely
that each daughter cell would receive precisely one copy
of each chromosome. In contrast, the Golgi apparatus
fragments into tiny vesicles that are all alike, and by random
distribution it is very likely that each daughter cell will
receive an approximately equal number of them.
ANSWER 18–10 As apoptosis occurs on a large scale in
both developing and adult tissues, it must not trigger
alarm reactions that are normally associated with cell injury.
Tissue injury, for example, leads to the release of signal
molecules that stimulate the proliferation of surrounding
cells so that the wound heals. It also causes the release of
signals that can cause a destructive inflammatory reaction.
Moreover, the release of intracellular contents could elicit an
immune response against molecules that are normally not
encountered by the immune system. Such reactions would
be self-defeating if they occurred in response to the massive
cell death that occurs in normal development.
ANSWER 18–11 Because the cell population is increasing
exponentially, doubling its weight at every cell division, the
weight of the cell cluster after N cell divisions is
2 N × 10 –9 g. Therefore, 70 kg (70 × 10 3 g) = 2 N × 10 –9 g,
or 2 N = 7 × 10 13 . Taking the logarithm of both sides
allows you to solve the equation for N. Therefore,
N = ln (7 × 10 13 ) / ln 2 = 46; that is it would take only
46 days if cells proliferated exponentially. Cell division in
animals is tightly controlled, however, and most cells in
the human body stop dividing when they become highly
specialized. The example demonstrates that exponential cell
proliferation occurs only for very brief periods, even during
embryonic development.
ANSWER 18–12 The egg cells of many animals are big
and contain stores of enough cell components to last for
many cell divisions. The daughter cells that form during
the first cell divisions after fertilization are progressively
smaller in size and thus can be formed without a need for
new protein or RNA synthesis. Whereas normally dividing
cells would grow continuously in G 1 , G 2 , and S phases, until
their size doubled, there is no cell growth in these early
cleavage divisions, and both G 1 and G 2 are virtually absent.
As G 1 is usually longer than G 2 and S phase, G 1 is the most
drastically reduced phase in these divisions.
plasma membrane
cell wall
vesicle–vesicle
fusion
protein
vesicle–plasma
membrane fusion
Figure A18–8
daughter cell 1
daughter cell 2