Essential Cell Biology 5th edition

Answers A:25
that its function is very tightly constrained, probably
because of extensive interactions with other proteins
and with DNA.
C. Histone H3 is clearly not in a “privileged” site in the
genome because it undergoes synonymous nucleotide
changes at about the same rate as other genes.
ANSWER 9–17
A. The data embodied in the phylogenetic tree (Figure
Q9–17) refutes the hypothesis that plant hemoglobin
genes were acquired by horizontal transfer from animals.
Looking at the more familiar parts of the tree, we see
that the hemoglobins of vertebrates (fish to human) have
approximately the same phylogenetic relationships as
do the species themselves. Plant hemoglobins also form
a distinct group that displays accepted evolutionary
relationships, with barley, a monocot, diverging before
bean, alfalfa, and lotus, which are all dicots (and
legumes). The basic hemoglobin gene, therefore, was
in place long ago in evolution. The phylogenetic tree of
Figure Q9–17 indicates that the hemoglobin genes in
modern plant and animal species were inherited from a
common ancestor.
B. Had the plant hemoglobin genes arisen by horizontal
transfer from a nematode, then the plant sequences
would have clustered with the nematode sequences in
the phylogenetic tree in Figure Q9–17.
ANSWER 9–18 In each human lineage, new mutations will
be introduced at a rate of 10 –10 alterations per nucleotide
per cell generation, and the differences between two human
lineages will accumulate at twice this rate. To accumulate
10 –3 differences per nucleotide will thus take 10 –3 /
(2 × 10 –10 ) cell generations, corresponding to (1/200) ×
10 –3 /(2 × 10 –10 ) = 25,000 human generations, or 750,000
years. In reality, we are not descended from one pair of
genetically identical ancestral humans; rather, it is likely that
we are descended from a relatively small founder population
of humans who were already genetically diverse. More
sophisticated analysis suggests that this founder population
existed about 200,000 years ago.
ANSWER 9–19 The virus that causes AIDS in humans,
HIV, is a retrovirus, and thus synthesizes DNA from an RNA
template using reverse transcriptase. This leads to frequent
mutation of the viral genome. In fact, people who are HIVpositive
often carry many different genetic variants of HIV
that are distinct from the original virus that infected them.
This posed a problem in treating the infection: drugs that
block essential viral enzymes would work only temporarily,
because new strains of the virus resistant to these drugs
arose rapidly by mutation. Today’s strategy employs
multiple drugs simultaneously, which greatly decreases the
likelihood that a fully resistant mutant virus could arise.
Like reverse transcriptases, RNA replicases (enzymes that
synthesize RNA using RNA as a template) do not proofread.
Thus, RNA viruses that replicate their RNA genomes directly
(that is, without using DNA as an intermediate) also mutate
frequently. In such a virus, this tends to produce changes
in the coat proteins that cause the mutated virus to appear
“new” to our immune systems; the virus is therefore not
suppressed by immunity that has arisen to the previous
version. This is part of the explanation for the new strains
of the influenza (flu) virus and the common cold virus that
regularly appear.
Chapter 10
ANSWER 10–1 The presence of a mutation in a gene does
not necessarily mean that the protein expressed from it is
defective. For example, the mutation could change one
codon into another that still specifies the same amino acid,
and so does not change the amino acid sequence of the
protein. Or, the mutation may cause a change from one
amino acid to another in the protein, but in a position that
is not important for the folding or function of the protein.
In assessing the likelihood that such a mutation might cause
a defective protein, information on the known β-globin
mutations that are found in humans is essential. You would
therefore want to know the precise nucleotide change
in your mutant gene, and whether this change has any
known or predictable consequences for the function of the
encoded protein. If your mate has two normal copies of the
globin gene, 50% of your children would be carriers of your
mutant gene.
ANSWER 10–2
A. Digestion with EcoRI produces two products:
׳AATTCGGGCCTTAAGCGCCGCGTCGAGGCCTTAAA-3‎ AAGAATTGCGG‏-׳‎5‎
׳GCCCGGAATTCGCGGCGCAGCTCCGGAATTT-5‎ TTCTTAACGCCTTAA‏-׳‎3‎
B. Digestion with HaeIII produces three products:
AAGAATTGCGGAATTCGGG‏-׳‎5‎ CCTTAAGCGCCGCGTCGAGG
׳CCTTAAA-3‎ TTCTTAACGCCTTAAGCCC‏-׳‎3‎ GGAATTCGCGGCGCAGCTCC
׳GGAATTT-5‎ C. The sequence lacks a HindIII cleavage site.
D. Digestion with all three enzymes therefore produces:
AAGAATTGCGG‏-׳‎5‎ AATTCGGG CCTTAAGCGCCGCGTCGAGG
׳CCTTAAA-3‎ TTCTTAACGCCTTAA‏-׳‎3‎ GCCC GGAATTCGCGGCGCAGCTCC
׳GGAATTT-5‎ ANSWER 10–3 Protein biochemistry is still very important
because it provides the link between the amino acid
sequence (which can be deduced from DNA sequences)
and the functional properties of the protein. We are still not
able to infallibly predict the folding of a polypeptide chain
from its amino acid sequence, and in most cases information
regarding the function of the protein, such as its catalytic
activity, cannot be deduced from the gene sequence alone.
Instead, such information must be obtained experimentally
by analyzing the properties of proteins biochemically.
Furthermore, the structural information that can be deduced
from DNA sequences is necessarily incomplete. We cannot,
for example, accurately predict covalent modifications of
the protein, proteolytic processing, the presence of tightly
bound small molecules, or the association of the protein
with other subunits. Moreover, we cannot accurately predict
the effects these modifications might have on the activity of
the protein.
ANSWER 10–4
A. After an additional round of amplification there will be 2
gray, 4 green, 4 red, and 22 yellow-outlined fragments;
after a second additional round there will be 2 gray, 5
green, 5 red, and 52 yellow-outlined fragments. Thus the
DNA fragments outlined in yellow increase exponentially
and will eventually overrun the other reaction products.
Their length is determined by the DNA sequence that
spans the distance between the two primers plus the
length of the primers.
B. The mass of one DNA molecule 500 nucleotide pairs
long is 5.5 × 10 –19 g [= 2 × 500 × 330 (g/mole)/6 × 10 23
(molecules/mole)]. Ignoring the complexities of the first
few steps of the amplification reaction (which produce