Essential Cell Biology 5th edition

A:14 Answers
the sugar–phosphate backbone is vastly different in the
A-C pair compared with A-T, and the spacing between the
two sugar–phosphate strands is considerably increased
in the A-G pair, where two large purine rings interact.
Consequently, it is energetically unfavorable to incorporate
a wrong base in DNA, and such errors occur only very rarely.
ANSWER 5–7
A. The bases V, W, X, and Y can form a DNA-like doublehelical
molecule with virtually identical properties to
those of bona fide DNA. V would always pair with X,
and W with Y. Therefore, the macromolecule could
be derived from a living organism that uses the same
principles to replicate its genome as those used by
organisms on Earth. In principle, different bases, such
as V, W, X, and Y, could have been selected during
evolution on Earth as building blocks for DNA. (Similarly,
there are many more conceivable amino acid side chains
than the set of 20 selected in evolution that make up all
proteins.)
B. None of the bases V, W, X, or Y can replace A, T, G, or
C. To preserve the distance between the two sugar–
phosphate strands in a double helix, a pyrimidine always
has to pair with a purine (see, for example, Figure 5–4).
Thus, the eight possible combinations would be V-A,
V-G, W-A, W-G, X-C, X-T, Y-C, and Y-T. Because of the
positions of hydrogen-bond acceptors and hydrogenbond
donor groups, however, no stable base pairs would
form in any of these combinations, as shown for the
pairing of V and A in Figure A5–7, where only a single
hydrogen bond could form.
Figure A5−7
H
V
N
H
H
adenine
N
C
C
N
H
C
N
N
C
N
ANSWER 5–8 As the two strands are held together
by hydrogen bonds between the bases, the stability of a
DNA double helix is largely dependent on the number of
hydrogen bonds that can be formed. Thus two parameters
determine the stability: the number of nucleotide pairs and
the number of hydrogen bonds that each nucleotide pair
contributes. As shown in Figure 5–4, an A-T pair contributes
two hydrogen bonds, whereas a G-C pair contributes three
hydrogen bonds. Therefore, helix C (containing a total of
34 hydrogen bonds) would melt at the lowest temperature,
helix B (containing a total of 65 hydrogen bonds) would melt
C
C
C
C
C
H
O
H
N
N
H
H
next, and helix A (containing a total of 78 hydrogen bonds)
would melt last. Helix A is the most stable, largely owing
to its high GC content. Indeed, the DNA of organisms that
grow in extreme temperature environments, such as certain
prokaryotes that grow in geothermal vents, has an unusually
high GC content.
ANSWER 5–9 The DNA would be enlarged by a factor
of 2.5 × 10 6 (= 5 × 10 –3 /2 × 10 –9 m). Thus the extension
cord would be 2500 km long. This is approximately the
distance from London to Istanbul, San Francisco to Kansas
City, Tokyo to the southern tip of Taiwan, and Melbourne to
Cairns. Adjacent nucleotides would be about 0.85 mm apart
(which is only about the thickness of a stack of 12 pages of
this book). A gene that is 1000 nucleotide pairs long would
be about 85 cm in length.
ANSWER 5–10
A. It takes two bits to specify each nucleotide pair (for
example, 00, 01, 10, and 11 would be the binary codes
for the four different nucleotides, each paired with its
appropriate partner).
B. The entire human genome (3 × 10 9 nucleotide pairs)
could be stored on two CDs (3 × 10 9 × 2 bits/4.8 × 10 9
bits).
ANSWER 5–11
A. True.
B. False. Nucleosome core particles are approximately
11 nm in diameter.
ANSWER 5–12 The definitions of the terms can be found
in the Glossary. DNA assembles with specialized proteins
to form chromatin. At a first level of packing, histones form
the core of nucleosomes. A nucleosome includes the DNA
wrapped around this histone core plus a segment of linker
DNA. Between nuclear divisions—that is, in interphase—the
chromatin of the interphase chromosomes is in a relatively
extended form in the nucleus, although some regions of
it, the heterochromatin, remain densely packed and are
transcriptionally inactive. During nuclear division—that is, in
mitosis—replicated chromosomes become condensed into
mitotic chromosomes, which are transcriptionally inactive
and are designed to be readily distributed between the two
daughter cells.
ANSWER 5–13 Colonies are clumps of cells that originate
from a single founder cell and grow outward as the cells
divide again and again. In the lower colony of Figure Q5–13,
the Ade2 gene is inactivated when placed near a telomere,
but apparently it can become spontaneously activated in a
few cells, which then turn white. Once activated in a cell, the
Ade2 gene continues to be active in the descendants of that
cell, resulting in clumps of white cells (the white sectors)
in the colony. This result shows both that the inactivation
of a gene positioned close to a telomere can be reversed
and that this change is passed on to further generations.
This change in Ade2 expression probably results from a
spontaneous decondensation of the chromatin structure
around the gene.
ANSWER 5–14 In the electron micrographs, one can
detect chromatin regions of two different densities; the
densely stained regions correspond to heterochromatin,
while less condensed chromatin is more lightly stained.
The chromatin in (A) is mostly in the form of condensed,