Essential Cell Biology 5th edition

A:26 Answers
longer products that eventually make an insignificant
contribution to the total DNA amplified), this amount of
product approximately doubles for every amplification
step. Therefore, 100 × 10 –9 g = 2 N × 5.5 × 10 –19 g,
where N is the number of amplification steps of the
reaction. Solving this equation for N = log(1.81 × 10 11 )/
log(2) gives N = 37.4. Thus, only about 40 cycles of
PCR amplification are sufficient to amplify DNA from a
single molecule to a quantity that can be readily handled
and analyzed biochemically. This whole procedure is
automated and takes only a few hours in the laboratory.
ANSWER 10–5 If the ratio of dideoxyribonucleoside
triphosphates to deoxyribonucleoside triphosphates
is increased, DNA polymerization will be terminated
more frequently and thus shorter DNA strands will be
produced. Such conditions are favorable for determining
nucleotide sequences that are close to the DNA
primer used in the reaction. In contrast, decreasing
the ratio of dideoxyribonucleoside triphosphates to
deoxyribonucleoside triphosphates will produce longer
DNA fragments, thus allowing one to determine nucleotide
sequences more distant from the primer.
ANSWER 10–6 Although several explanations are
possible, the simplest is that the DNA probe has hybridized
predominantly with its corresponding mRNA, which is
typically present in many more copies per cell than is the
gene. The different extents of hybridization probably reflect
different levels of gene expression. Perhaps each of the
different cell types that make up the tissue expresses the
gene at a different level.
ANSWER 10–7 Like the vast majority of mammalian genes,
the attractase gene likely contains introns. Bacteria do not
have the splicing machinery required to remove introns,
and therefore the correct protein would not be expressed
from the gene. For expression of most mammalian genes in
bacterial cells, a cDNA version of the gene must be used.
ANSWER 10–8
A. False. Restriction sites are found at random throughout
the genome, within as well as between genes.
B. True. DNA bears a negative charge at each phosphate,
giving DNA an overall negative charge.
C. False. Clones isolated from cDNA libraries do not
contain promoter sequences. These sequences are not
transcribed and are therefore not part of the mRNAs
that are used as the templates to make cDNAs.
D. True. Each polymerization reaction produces doublestranded
DNA that must, at each cycle, be denatured to
allow new primers to hybridize so that the DNA strand
can be copied again.
E. False. Digestion of genomic DNA with restriction
enzymes that recognize four-nucleotide sequences
produces fragments that are on average 256 nucleotides
long. However, the actual lengths of the fragments
produced will vary considerably on both sides of the
average.
F. True. Reverse transcriptase is first needed to copy the
mRNA into single-stranded DNA, and DNA polymerase
is then required to make the second DNA strand.
G. True. Using a sufficient number of STRs, individuals can
be uniquely “fingerprinted” (see Figure 10–15).
H. True. If cells of the tissue do not transcribe the gene
of interest, it will not be represented in a cDNA
library prepared from this tissue. However, it will be
represented in a genomic library prepared from the
same tissue.
ANSWER 10–9
A. The DNA sequence, from its 5′ end to its 3′ end, is read
starting from the bottom of the gel, where the smallest
DNA fragments migrate. Each band results from the
incorporation of the appropriate dideoxyribonucleoside
triphosphate, and as expected there are no two
bands that have the same mobility. This allows one to
determine the DNA sequence by reading off the bands
in strict order, proceeding upward from the bottom of
the gel, and assigning the correct nucleotide according
to which lane the band is in.
The nucleotide sequence of the top strand (Figure
A10–9A) was obtained directly from the data of Figure
Q10–9, and the bottom strand was deduced from the
complementary base-pairing rules.
B. The DNA sequence can then be translated into an amino
acid sequence using the genetic code. However, there
are two strands of DNA that could be transcribed into
RNA and three possible reading frames for each strand.
Thus there are six amino acid sequences that can in
principle be encoded by this stretch of DNA. Of the
three reading frames possible from the top strand, only
one is not interrupted by a stop codon (underlined in the
DNA sequence and represented by yellow blocks in the
three amino acid sequences in Figure A10–9B).
From the bottom strand, two of the three reading
frames also have stop codons (not shown). The third
frame gives the following sequence:
SerAlaLeuGlySerSerGluAsnArgProArgThrProAlaArg
ThrGlyCysProValTyr
It is not possible from the information given to tell
which of the two open reading frames corresponds to
the actual protein encoded by this stretch of DNA. What
additional experiment could distinguish between these
two possibilities?
ANSWER 10–10
A. Cleavage of human genomic DNA with HaeIII would
generate about 11 × 10 6 different fragments
[= 3 × 10 9 /4 4 ] and with EcoRI about 730,000 different
fragments [= 3 × 10 9 /4 6 ]. There will also be some
(A)
5′ TATAAACTGGACAACCAGTTCGAGCTGGTGTTCGTGGTCGGTTTTCAGAAGATCCTAACGCTGACG 3′
3′ ATATTTGACCTGTTGGTCAAGCTCGACCACAAGCACCAGCCAAAAGTCTTCTAGGATTGCGACTGC 5′
Figure A10–9
(B)
top strand of DNA
5′ TATAAACTGGACAACCAGTTCGAGCTGGTGTTCGTGGTCGGTTTTCAGAAGATCCTAACGCTGACG 3′
1 TyrLysLeuAspAsnGlnPheGluLeuValPheValValGlyPheGlnLysIleLeuThrLeuThr
2 IleAsnTrpThrThrSerSerSerTrpCysSerTrpSerValPheArgArgSer Arg Ar
3 ThrGlyGlnProValArgAlaGlyValArgGlyArgPheSerGluAspProAsnAlaAsp