Essential Cell Biology 5th edition

Answers A:27
additional fragments generated because the maternal
and paternal chromosomes are very similar but not
identical in DNA sequence.
B. A set of overlapping DNA fragments will be generated.
Libraries constructed from sets of overlapping fragments
are valuable because they can be used to order cloned
sequences in relation to their original order in the
genome and thus obtain the DNA sequence of a long
stretch of DNA (see Figure 10−20).
ANSWER 10–11 By comparison with the positions of the
size markers, we find that EcoRI treatment gives two
fragments of 4 kb and 6 kb; HindIII treatment gives one
fragment of 10 kb; and treatment with EcoRI + HindIII gives
three fragments of 6 kb, 3 kb, and 1 kb. This gives a total
length of 10 kb calculated as the sum of the fragments in
each lane. Thus the original DNA molecule must be
10 kb (10,000 nucleotide pairs) long. Because treatment
with HindIII gives a fragment 10 kb long it could be that
the original DNA is a linear molecule with no cutting site for
HindIII. But we can rule that out by the results of the EcoRI
+ HindIII digestion. We know that EcoRI cleavage alone
produces two fragments of 6 kb and 4 kb, and in the double
digest this 4-kb fragment is further cleaved by HindIII into
a 3-kb and a 1-kb fragment. The DNA therefore contains a
single HindIII cleavage site, and thus it must be circular, as
a single fragment of 10 kb is produced when it is cut with
HindIII alone. Arranging the cutting sites on a circular DNA
to give the appropriate sizes of fragments produces the
map illustrated in Figure A10–11.
HindIII
EcoR I
1 kb
3 kb
EcoR I
Figure A10–11
6 kb
ANSWER 10–12
A. Infants 2 and 8 have identical STR patterns and therefore
must be identical twins. ECB5 Infants EA10.11/A10.11
3 and 6 also have
identical STR patterns and must also be identical twins.
The other two sets of twins must be fraternal twins
because their STR patterns are not identical. Fraternal
twins, like any pair of siblings born to the same parents,
will have roughly half their genome in common. Thus,
roughly half the STR polymorphisms in fraternal twins
will be identical. Using this criterion, you can identify
infants 1 and 7 as fraternal twins and infants 4 and 5 as
fraternal twins.
B. You can match infants to their parents by using the
same sort of analysis of STR polymorphisms. Every
band present in the analysis of an infant should have
a matching band in one or the other of the parents,
and, on average, each infant will share half of its
polymorphisms with each parent. Thus, the degree
of match between each child and each parent will be
approximately the same as that between fraternal twins.
ANSWER 10–13 Mutant bacteria that do not produce
ice-protein have probably arisen many times in nature.
However, bacteria that produce ice-protein have a slight
growth advantage over bacteria that do not, so it would be
difficult to find such mutants in the wild. Recombinant DNA
technology makes these mutants much easier to obtain.
In this case, the consequences, both advantageous and
disadvantageous, of using a genetically modified organism
are therefore nearly indistinguishable from those of a natural
mutant. Indeed, bacterial and yeast strains have been
selected for centuries for desirable genetic traits that make
them suitable for industrial-scale applications such as cheese
and wine production. The possibilities of recombinant
DNA technology are endless, however, and as with any
technology, there is a risk of unforeseen consequences.
Recombinant DNA experimentation, therefore, is regulated,
and the risks of individual projects are carefully assessed
by review panels before permissions are granted. The
state of our knowledge is sufficiently advanced that the
consequences of some changes, such as the disruption of a
bacterial gene in the example above, can be predicted with
reasonable certainty. Other applications, such as germline
gene therapy to correct human disease, may have far
more complex outcomes, and it will take many more years
of research and ethical debate to determine whether such
treatments will eventually be used.
Chapter 11
ANSWER 11–1 Water is a liquid, and the hydrogen
bonds that form between water molecules are not static;
they are continually broken and remade again by thermal
motion. When a water molecule happens to be next to a
hydrophobic molecule, it is more restricted in motion and
has fewer neighbors with which it can interact, because it
cannot form any hydrogen bonds in the direction of the
hydrophobic molecule. It will therefore form hydrogen
bonds to the more limited number of water molecules in
its proximity. Bonding to fewer partners results in a more
ordered water structure, which represents the cagelike
structure in Figure 11–9. This structure has been likened
to ice, although it is a more transient, less organized, and
less extensive network than even a tiny ice crystal. The
formation of any ordered structure decreases the entropy of
the system and is thus energetically unfavorable (discussed
in Chapter 3).
ANSWER 11–2 (B) is the correct analogy for lipid bilayer
assembly because exclusion from water rather than
attractive forces between the lipid molecules is involved.
If the lipid molecules formed bonds with one another, the
bilayer would be less fluid, and might even become rigid,
depending on the strength of the interaction.
ANSWER 11–3 The fluidity of the bilayer is strictly confined
to one plane: lipid molecules can diffuse laterally in their
own monolayer but do not readily flip from one monolayer
to the other. Lipid molecules inserted into one monolayer
therefore remain in it unless they are actively transferred to
the other monolayer by a transporter such as a scramblase
or a flippase.
ANSWER 11–4 In both an α helix and a β barrel, the
polar peptide bonds of the polypeptide backbone can be
completely shielded from the hydrophobic environment of
the lipid bilayer by the hydrophobic amino acid side chains.