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Unformatted text preview: Chapter 6 How Cells Read the Genome:
From DNA to Protein
FROM DNA TO RNA 6
In This Chapter
FROM DNA TO RNA DEFINITIONS
6–1 General transcription factor 6–2 snRNA (small nuclear RNA) 6–3 RNA splicing 6–8 Terminator 6–9 Trans-splicing 6–10 Exon 6–11 A146 mRNA (messenger RNA) 6–7 THE RNA WORLD AND
THE ORIGINS OF LIFE rRNA (ribosomal RNA) 6–6 A132 Exosome 6–5 FROM RNA TO
PROTEIN Promoter 6–4 A121 Nuclear pore complex TRUE/FALSE
6–12 True. Errors in DNA replication have the potential to affect future generations of cells, while errors in transcription have no genetic consequence.
Errors in transcription lead to mistakes in a small fraction of RNAs, whose
functions are further monitored by downstream quality-control mechanisms. The essential feature is that errors in DNA replication change the
gene and, thereby, affect all the copies of RNA (and protein) made in the
original cell and all its progeny cells. By contrast, errors in transcription are
limited to a small number of defective RNAs (and proteins), and are not
passed on to progeny cells.
These considerations are reflected in the intrinsic error rates for RNA and
DNA polymerases: RNA polymerases typically make 1 mistake in copying
104 nucleotides, while DNA polymerases make about 1 error per 107
nucleotides. Such significant differences in error rates suggest that natural
selection is stronger against errors in replication than against errors in transcription. 6–13 False. The s subunit associates with the bacterial RNA polymerase core
enzyme to form the RNA polymerase holoenzyme only during the initiation
phase of RNA synthesis. The s subunit helps the core enzyme bind to the
promoter and stays associated with the core enzyme until RNA synthesis has
been properly initiated, and then it dissociates. A121 A122 Chapter 6: How Cells Read the Genome: From DNA to Protein 6–14 True. At its 3¢ end each eucaryotic mRNA has a string of adenine nucleotides,
the last of which has a terminal ribose with a free 3¢-OH group. At its 5¢ end
each mRNA carries a 7-methylguanosine that is linked 5¢ to 5¢ with the first
nucleotide in the mRNA. This linkage leaves a free 3¢-OH group on the ribose
of the capping nucleotide. 6–15 False. Although intron sequences are mostly dispensable, they must be
removed precisely. An error of even one nucleotide during removal would
shift the reading frame in the spliced mRNA molecule and produce an aberrant protein. 6–16 False. The 3¢ ends of most pre-mRNA transcripts produced by RNA polymerase II are defined, not by the termination point of transcription, but by
cleavage of the RNA chain 10–30 nucleotides downstream of the sequence
AAUAAA and 30 or so nucleotides upstream of a GU- or U-rich sequence. THOUGHT PROBLEMS
6–17 The answer is best given by Francis Crick himself, who coined the terms ‘the
sequence hypothesis,’ which proposes that genetic information is encoded
in the sequence of the DNA bases, and ‘the central dogma,’ which states that
DNA makes RNA makes protein, in 1957.
“I called this idea the central dogma, for two reasons, I suspect. I had
already used the obvious word hypothesis in the sequence hypothesis, and
in addition I wanted to suggest that this new assumption was more central
and more powerful. I did remark that their speculative nature was emphasized by their names.
As it turned out, the use of the word dogma caused almost more trouble
than it was worth. Many years later Jacques Monod pointed out to me that I
did not appear to understand the correct use of the word dogma, which is a
belief that cannot be doubted. I did apprehend this in a vague sort of way but
since I thought that all religious beliefs were without serious foundation, I
used the word in the way I myself thought about it, not as most of the rest of
the world does, and simply applied it to a grand hypothesis that, however
plausible, had little direct experimental support.”
Reference: Crick F (1988) What Mad Pursuit: A Personal View of Scientific
Discovery, p 109. New York: Basic Books, Inc. 6–18 Actually, the RNA polymerases are not moving at all because they have been
fixed and coated with metal to prepare the sample for viewing in the electron microscope. Before they were fixed, they were moving from left to right,
as indicated by the gradual lengthening of the RNA transcripts. The RNA
transcripts are shorter than the DNA that encodes them because they begin
to fold up (acquire a three-dimensional structure) as soon as they are synthesized, whereas the DNA is an extended double helix. 6–19 A with 7, B with 4, C with 2, D with 5, E with 8, F with 3, G with 1, and H
with 6. 6–20 If the polymerase transcribes the sequence from left to right, it will use the
bottom strand as a template to make the sequence 5¢-GUAACGGAUG (the
RNA sequence corresponding to the top strand of the DNA). If the polymerase moves from right to left, it will use the top strand as a template to
make the sequence 5¢-CAUCCGUUAC (the RNA sequence corresponding to
the bottom strand of the DNA, written 5¢ Æ 3¢). 6–21 General transcription factors play several roles in promoting transcription
by RNA polymerase II. They help position the RNA polymerase correctly at
the promoter, they aid in pulling apart the two strands of DNA to allow transcription to begin, and they release RNA polymerase from the promoter
once transcription has begun. They are called ‘general’ because they assemble on all promoters used by RNA polymerase II; they are identified by A123 FROM DNA TO RNA
(A) (B) Figure 6–49 Rotation of duplex due to
movement relative to RNA polymerase
(Answer 6–23). (A) Direction of rotation of
the magnetic bead. (B) Direction of
rotation of the DNA duplex. magnet magnetic
polymerase RNA glass slide names beginning with TFII (transcription factor for RNA polymerase II).
Labeling them ‘general’ transcription factors also serves to distinguish them
from more specialized gene regulatory proteins that enhance transcription
at selected promoters in certain cell types.
6–22 The RNA polymerase must be moving from right to left in Figure 6–2. If the
RNA polymerase does not rotate around the template as it moves, it will
overwind the DNA ahead of it, causing positive supercoils, and underwind
the DNA behind it, causing negative supercoils. If the RNA polymerase were
free to rotate about the template as it moved along the DNA, it would not
overwind or underwind the DNA, and no supercoils would be generated.
Reference: Liu LF & Wang JC (1987) Supercoiling of the DNA template during transcription. Proc. Natl Acad. Sci. U.S.A. 84, 7024–7027. 6–23 The bead would rotate clockwise from the perspective of the magnet, as
shown in Figure 6–49A. As shown in Figure 6–49B, the motion of the helix
relative to a fixed RNA polymerase causes the helix to rotate.
Reference: Harada Y, Ohara O, Takatsuki A, Itoh H, Shimamoto N & Kinosita
K (2001) Direct observation of DNA rotation during transcription by
Escherichia coli RNA polymerase. Nature 409, 113–115. 6–24 From electron micrographs such as that shown in Figure 6–1 it seems clear
that RNA does not become wrapped around the DNA as it is spun out
behind RNA polymerase. Thus, RNA polymerase doesn’t seem to revolve
around DNA as it moves. Consistent with this, other evidence suggests that
RNA polymerase does induce positive supercoiling ahead of it and negative
supercoiling behind it. But the level of supercoiling tension is not so high as
you would expect if the continued movement of RNA polymerase generated
ever-higher levels of coiling. These observations suggest that supercoiling
tension is relieved by the action of topoisomerases in the cell, so that supercoiling around the RNA polymerase is maintained at optimal levels. 6–25 Phosphorylation of the CTD is the event that permits release of RNA polymerase from the other proteins present at the start point of transcription.
Phosphorylation also allows association of a new set of proteins that are
involved in processing the nascent RNA transcript. These proteins include
components required for capping, splicing, and polyadenylation. 6–26 The tails arise because the ends of the mRNA are not complementary to the
ends of the restriction fragment. Thus, one of the single-strand tails at each
end is DNA from the restriction fragment. A single-strand tail at one end corresponds to the 5¢ end of the mRNA, which must come from an upstream
exon that is not present in the restriction fragment. A single-strand tail at the
other end corresponds to the 3¢ end of the mRNA, which may come from a
downstream exon not present in the restriction fragment or simply be the
poly-A tail itself. Without additional information you cannot identify which
single strand comes from which source. A124 Chapter 6: How Cells Read the Genome: From DNA to Protein
(A) NORMAL (B) MUTANT
mutation inactivates 3¢ splice site 173 bp exon 1 exon 2 exon 2 exon 1 exon 3 gene TRANSCRIPTION cap AAA
pre-mRNA TRANSCRIPTION cap AAA
TRANSLATION normal protein exon 3 gene cap AAA
TRANSLATION mutant protein Reference: Berget SM, Berk AJ, Harrison T & Sharp PA (1977) Spliced segments at the 5¢ termini of adenovirus-2 late mRNA: a role for heterogeneous
nuclear RNA in mammalian cells. Cold Spring Harbor Symp. Quant. Biol. 42,
A. A single nucleotide change in a gene could cause an internal deletion in the
mRNA if it altered splicing so that an exon that was usually incorporated was
B. Removal of 173 nucleotides from the protein-coding portion of the mRNA
would cause a shift in the reading frame for translation into amino acids.
Because a codon is three nucleotides, a loss of 173 nucleotides does not correspond to an integral number of codons. Thus, the Smilin encoded by the
deleted mRNA would be fine up to the missing exon, but would encode an
unrelated sequence of amino acids thereafter until a stop codon was reached.
C. The simplest explanation is that the Smilin gene contains a 173-nucleotidelong exon (exon 2 in Figure 6–50A) that is lost during the processing of the
mutant precursor mRNA. This could occur, for example, if the mutation
changed the 3¢ splice site in the preceding intron so that it was no longer recognized by the splicing machinery (a change in the conserved AG at the
intron–exon boundary could do this). Use of the next available 3¢ splice
site—adjacent to exon 3—would cause loss of exon 2 from the mutant
mRNA (Figure 6–50B). During protein synthesis, the absence of exon 2 (173
nucleotides) would throw the ribosomes out of the correct reading frame as
they moved from exon 1 to exon 3. At that junction the ribosomes would
begin synthesizing a protein sequence unrelated to that normally encoded
by exon 3.
6–28 Statement C is the only one that is necessarily true for exons 2 and 3. It is also
true for exons 7 and 8. While statements A and B could be true, they don’t
have to be. Because the protein sequence is the same in segments of the
mRNA that correspond to exons 1 and 10, neither choice of alternative exons
(2 versus 3, or 7 versus 8) can alter the reading frame. To maintain the normal reading frame—whatever that is—the alternative exons must have a
number of nucleotides that when divided by 3 (the number of nucleotides in
a codon) give the same remainder.
Since the sequence of the a-tropomyosin gene is known, it is possible to
check to see the actual state of affairs. Exons 2 and 3 both contain the same
number of nucleotides, 126, which is divisible by 3 with no remainder. Exons
7 and 8 also contain the same number of nucleotides, 76, which is divisible
by 3 with a remainder of 1. 6–29 Since introns evolve faster than exons, the introns of the different species
will be more variable than the exons. It is difficult to scan these sequences by Figure 6–50 Splicing of the Smilin
transcript (Answer 6–27). (A) Normal
transcript. (B) Mutant transcript. A125 FROM DNA TO RNA GGTGGTGAGGCCCTGGGCAG GTAGGTATCCCACTTACAAG
EXON INTRON eye and decide, with confidence, which side is the more conserved. One way
to quantify the differences is to pick one sequence, for example, the cow, and
count up how often the other sequences differ at each position, as shown in
Figure 6–51. Summing the differences on each side of the junction makes it
clear that sequences on the left are much more similar to one another than
are the sequences on the right. (Similar differences exist no matter which
sequence is chosen for comparison.) Thus, the more conserved sequences,
which are on the left in Figure 6–6, correspond to exons, and the less conserved sequences, which are on the right, correspond to introns. Figure 6–51 Sum of differences from the
cow b-globin sequence (Answer 6–29).
The b-globin sequence from the cow was
compared nucleotide by nucleotide with
the b-globin sequences from the other
six species. The number of differences at
each position is summed below each
nucleotide of the b-globin sequence. The
total number of differences on each side
of the exon–intron boundary (dashed
line) is shown at the bottom. (A) MINIGENE 1
5¢-to-3¢ scanning 6–30
A. If the splicing machinery bound to one splice site and scanned across the
intron to find its complementary splice site, it would use the first appropriate splice site it encounters. The products predicted from the intron-scanning model are shown in Figure 6–52. If the splicing machinery bound to a
5¢ splice site and scanned toward a 3¢ splice site, minigene 1 would generate
one product (Figure 6–52A) and minigene 2 would generate two products
(Figure 6–52B). By contrast, if the splicing machinery bound to a 3¢ splice site
and scanned toward a 5¢ splice site, minigene 1 would generate two products
(Figure 6–52A) and minigene 2 would generate one product (Figure 6–52B).
B. The results of this experiment do not match the expectations for either scanning model, suggesting that scanning models are incorrect. The ordering
mechanism by which cells avoid exon skipping probably depends on two
factors: first, that assembly of the spliceosome occurs as the pre-mRNA
emerges from the RNA polymerase; and second, that exons may be defined
as an independent step prior to the assembly of the spliceosome.
Reference: Kuhne T, Wieringa B, Reiser J & Weissmann C (1983) Evidence
against a scanning model for RNA splicing. EMBO J. 2, 727–733.
6–31 6–32 6–33 Group I excised introns are linear, and they carry the activated G in covalent
linkage at their 5¢ ends. Group II excised introns are lariats, with the activated A having reacted with the 5¢-most nucleotide of the intron. The mechanism of pre-mRNA splicing catalyzed by the spliceosome is more similar to
the mechanism used by the Group II self-splicing introns.
‘Export ready’ means that an mRNA is bound by the appropriate set of proteins. Proteins such as the cap-binding complex, the exon-junction complex, and the poly-A-binding protein must be present, while proteins such
as spliceosome components must be absent. RNA fragments from excised
introns do not acquire the necessary set of proteins and are thus doomed to
The structure of the nucleolus depends on the rRNA genes, which are located
in clusters at the tips of each copy of five different chromosomes in humans.
During interphase the transcribed rRNA genes associate to form the visible
nucleolus. At mitosis, the chromosomes disperse and the nucleolus breaks up.
After mitosis the tips of the chromosomes again coalesce and the nucleolus
reforms in a process that depends on transcription of the rRNA genes. CALCULATIONS
A. Since RNA polymerase is blocked by pyrimidine dimers, the sensitivity of
transcription to UV damage will depend on the distance between the promoter for a gene and the probe. It is a simple measure of the target size for UV 5¢ 3¢ 3¢ 3¢ 3¢ 3¢-to-5¢ scanning 5¢ + (B) MINIGENE 2
5¢-to-3¢ scanning 5¢ 5¢ 3¢ 5¢ 3¢ + 3¢-to-5¢ scanning 5¢ Figure 6–52 Expected products in a test
of intron scanning (Answer 6–30).
(A) Expected products for 5¢-to-3¢
scanning and 3¢-to-5¢ scanning in
minigene 1. (B) Expected products for
5¢-to-3¢ scanning and 3¢-to-5¢ scanning in
minigene 2. Open boxes indicate
complete exons; shaded boxes represent
partial exons. A126 Chapter 6: How Cells Read the Genome: From DNA to Protein
damage. If the polymerase must travel twice as far to complete a transcript,
the chances of its encountering a block to transcription are twice as great.
B. Transcription through the Vsg gene is seven times more sensitive to UV irradiation than transcription through the ribosomal transcription unit at the
site of rRNA probe 4, which is about 7 kb from its promoter (see Figure
6–9A). Thus, the beginning of the Vsg gene, which is where the probe was
located, is about 50 kb (7 ¥ 7 kb) away from its promoter. This calculation
assumes that the DNA between the Vsg promoter and the Vsg gene has
about the same sensitivity to UV light as the DNA in the ribosomal RNA transcription unit. It also assumes that multiple UV-induced pyrimidine dimers
are not common enough to skew the linear relationship between UV dose
C. If the nearby gene is 20% less sensitive to UV irradiation than the Vsg gene,
it is inactivated at 80% the rate of the Vsg gene. Therefore, its promoter is 40
kb away (0.80 ¥ 50 kb). Given that the nearby gene is 10 kb in front of the Vsg
gene, its promoter must map very near the promoter for the Vsg gene. Thus,
it is likely that the two genes are transcribed from the same promoter.
Reference: Johnson PJ, Kooter JM & Borst P (1987) Inactivation of transcription by UV irradiation of T. brucei provides evidence for a multicistronic
transcription unit including a VSG gene. Cell 51, 273–281. 6–35
A. At the plateau there are 7.2 ¥ 109 transcripts per reaction. The reaction incorporates 2.4 pmol of CMP, and each transcript is 400 nucleotides long, of
which 200 nucleotides are CMP.
transcripts 2.4 pmol CMP 6 ¥ 1023 CMP
= 7.2 ¥ 10
B. Each reaction contains 1.0 ¥ 1011 templates. At 16 mg/mL a 25-mL reaction
volume contains 0.4 mg of template (16 mg/mL ¥ 0.025 mL = 0.4 mg). Each
template is 3500 nucleotide pairs (np) long.
6 ¥ 1017 d
= 1.0 ¥ 1011 templates/reaction
C. There are about 0.07 transcripts/template, which is equivalent to about 1
RNA transcript per 14 template molecules.
transcripts 7.2 ¥ 109 transcripts
1.0 ¥ 1011 templates
= 0.07 transcripts/template
The poor efficiency of the reaction is typical of in vitro transcription. The
ratio of 1 transcript per 14 templates does not necessarily mean that 1 out of
14 templates functions in RNA synthesis. It may be that a much smaller fraction of the templates makes a large number of transcripts. For example, 1 in
140 templates might synthesize 10 transcripts each.
Reference: Sawadogo M & Roeder RG (1985) Factors involved in specific
transcription by human RNA polymerase II: analysis by a rapid and quantitative in vitro assay. Proc. Natl Acad. Sci. U.S.A. 82, 4394–4398. DATA HANDLING
A. Starting with a complex in which a C had been incorporated at position +34,
the RNA polymerase could be walked down to position +43 in two steps: (1)
incubate with a mixture of ATP and UTP, and then wash away the A127 FROM DNA TO RNA
nucleotides; and (2) incubate with a mixture of UTP and CTP, and then wash.
This protocol will allow the polymerase to incorporate all the nucleotides up
to position +43, but not the A at position +44.
B. Incorporation of the correct A nucleotide occurred 130 times faster than the
incorrect G nucleotide at position +44 (0.20/0.0015 = 133, see Table 6–1).
Incorporation of the next nucleotide—a C—after the correct A occurred
about 5 times faster than incorporation of C after the incorrect G (0.17/0.036
= 4.7). Thus, when the RNA polymerase makes a mistake, it incorporates the
next nucleotide more slowly. This pause allows a window of opportunity for
removal of the incorrect nucleotide before the next nucleotide is added.
Reference: Thomas MJ, Platas AA & Hawley DK (1998) Transcriptional
fidelity and proofreading by RNA polymerase II. Cell 93, 627–637.
6–37 The consensus sequence for this set of promoters is shown in Figure 6–53. In
this set of 13 promoters there are clear examples of common nucleotides
outside the –35 and –10 regions. Also, one of the accepted consensus
nucleotides (the terminal A in the –35 sequence) doesn’t even show up as
common. When 300 promoters recognized by s70 are compared, the consensus sequence is TTGACA (–35) and TATAAT (–10). It’s always better to
compare more sequences! 6–38
A. There are three locations where the linker-scanning mutations drastically
decreased the amount of transcript. These locations are from about –15 to
–30, –45 to –60, and –80 to –105. Linker-scanning through the promoter element at –15 to –30 seems to have the most dramatic effect, whereas scanning through the other two sensitive sites gives lesser effects that are about
equal to one another.
You may also have noticed the effect of a linker-scanning mutation that
overlapped the start site of transcription. In the absence of the usual
sequence at the start site, the transcript is initiated at a variety of positions.
B. The segment from –15 to –30 likely includes the TATA box, which is usually
located about 25 nucleotides or so upstream of the transcription start site.
Reference: McKnight SL & Kingsbury R (1982) Transcriptional control signals of a eukaryotic protein-coding gene. Science 217, 316–324.
gene promoters bacteriophage
GATACAAATCTCCGTTGTACTTTGTT..TCGCGCTTGGTATAATCG.CTGGGGGTCAAAGATGAGTG frequency (%) AA TTGA C TATAA T C 100
AA - - - - - - - T T G A C –35 T A T AA T G C - - C - C C
–10 Figure 6–53 Consensus sequence for promoters recognized by s70 factor (Answer 6–37). Nucleotides that are perfectly
conserved and the first nucleotide of the transcripts are shaded for reference. Below the sequences themselves are
indicated the consensus nucleotides and, on an expanded scale, the frequencies of the conserved nucleotides around the
–35 and –10 regions. A128 Chapter 6: How Cells Read the Genome: From DNA to Protein 6–39
A. The 400-nucleotide transcript is absent from lane 4, Figure 6–13B, because
GTP was included in the reaction mixture. GTP allows transcription to proceed beyond the end of the C-minus sequence (the synthetic sequence
lacking C nucleotides), thereby generating transcripts longer than 400
nucleotides. In the absence of GTP (see lane 2) transcription cannot proceed beyond the C-minus sequence. In the presence of GTP and RNase T1
(see lane 6) the longer transcripts are cleaved at the first G to yield the 400nucleotide transcript. In the presence of GTP, RNase T1, and 3¢ O-methyl
GTP (see lane 8) any long transcripts that escape termination by 3¢ Omethyl GTP are cleaved by RNase T1 to yield the 400-nucleotide transcript.
B. One of the difficulties in assaying promoter function in vitro is the high
background of nonspecific initiation of transcription. It is this background
that is so evident in Figure 6–13B, lane 3. Its source is not altogether clear,
but transcription may start at sequences in the rest of the plasmid that
weakly resemble true RNA polymerase II promoters.
C. A transcript of about 400 nucleotides is present in Figure 6–13B, lane 5,
because cleavage with RNase T1 liberates it from any randomly initiated
transcript that has traversed the C-minus sequence. It is actually a few
nucleotides longer than the specifically initiated transcript since its 5¢ and 3¢
ends are defined, respectively, by the G nucleotides that immediately precede and follow the C-minus sequence.
The 400-nucleotide transcript is absent from Figure 6–13B, lane 7, because
3¢ O-methyl GTP will terminate most transcripts that are initiated in front of
the C-minus sequence. The combination of 3¢ O-methyl GTP and RNase T1
eliminates virtually all the background synthesis from the control plasmid.
D. As shown in Figure 6–13B, lanes 7 and 8, specific transcription can be
assayed in the presence of G nucleotides if 3¢ O-methyl GTP and RNase T1
are included (to inhibit background transcription and to cleave any random
transcripts to small pieces).
Reference: Sawadogo M & Roeder RG (1985) Factors involved in specific
transcription by human RNA polymerase II: analysis by a rapid and quantitative in vitro assay. Proc. Natl Acad. Sci. U.S.A. 82, 4394–4398.
6–40 These experiments provided the most convincing early demonstration that
there were three forms of RNA polymerase in eucaryotic cells: one—peak
1—that was insensitive to a-amanitin (RNA polymerase I), one—peak 2—
that was inhibited by both 1 mg/mL and 10 mg/mL a-amanitin (RNA polymerase II), and one—peak 3—that was inhibited by 10 mg/mL a-amanitin,
but not by 1 mg/mL (RNA polymerase III). It would be unlikely that different
forms of the same polymerase could have such different sensitivities to the
same molecule. These results also indicate the ways the RNA polymerases
were named: by the order in which they were eluted from the column.
Reference: Roeder RG (1974) Multiple forms of deoxyribonucleic aciddependent ribonucleic acid polymerase in Xenopus laevis. Isolation and
partial characterization. J. Biol. Chem. 249, 241–248. 6–41
A. Since an equal amount of transcription from each template was observed
when the preincubation was carried out with the individual templates or a
mixture (see Figure 6–15C, lanes 1 to 3), Srb2 protein does not show a preference for either template.
B. Srb2 acts stoichiometrically. If Srb2 acted catalytically, it should have been
able to modify the second template after the two were mixed. Catalytic activity would have produced transcripts from both templates regardless of
which one was originally included in the preincubation with Srb2. When
excess Srb2 was added at the beginning of the preincubation with one template, transcription was observed from both templates after mixing. This is
consistent with a stoichiometric requirement for Srb2 and rules out the possibility that Srb2 was inactivated during the preincubation—and was for that A129 FROM DNA TO RNA
P P P P
P P P P CTD RNA polymerase II CTD kinase,
NTPs RNA CTD
phosphatase reason unable to act (catalytically) on the second template after mixing.
C. The production of transcripts solely from the template that was preincubated with Srb2 indicates that Srb2 is part of the preinitiation complex. If
Srb2 were able to act after transcription had begun, transcripts would have
been produced from both templates regardless of which one was included
in the preincubation.
D. During preincubation of the template with the extract and Srb2, a number of
proteins including Srb2 bind at the promoter to form a preinitiation complex. Evidently, the preinitiation complex, once formed, is stable and does
not readily exchange proteins with other templates that are added later.
E. Although the Srb2 gene was originally identified as a suppressor of the coldsensitive phenotype of yeast carrying an RNA polymerase II gene with a
short CTD, neither those genetic results nor the transcription assays
described here provide evidence that Srb2 binds to the CTD. (Nor do they
argue against direct interaction; they simply do not speak to the issue.) Additional experiments have shown that Srb2 is part of a complex of proteins
known as the mediator. The mediator binds to the dephosphorylated CTD,
entering and leaving initiation complexes at every round of transcription in
a process that may be coupled to C-terminal domain phosphorylation and
the release of RNA polymerase at initiation of transcription (Figure 6–54).
References: Koleske AJ, Buratowski S, Nonet M & Young RA (1992) A novel
transcription factor reveals a functional link between the RNA polymerase II
CTD and TFIID. Cell 69, 883–894.
Thompson CM, Koleske AJ, Chao DM & Young RA (1993) A multisubunit
complex associated with the RNA polymerase II CTD and TATA-binding protein in yeast. Cell 73, 1361–1375.
Svejstrup JQ, Li Yang, Fellows J, Gnatt A, Bjorklund S & Kornberg RD (1997)
Evidence for a mediator cycle at the initiation of transcription. Proc. Natl
Acad. Sci. U.S.A. 94, 6075–6078.
6–42 The pattern of reaction with DNA and protein is strikingly clear. When the U
analog was closer than 10 nucleotides to the 3¢ end of the RNA, it reacted
predominantly with its pairing partner in the template strand. By contrast,
when it was 10 nucleotides or farther from the 3¢ end, it no longer reacted
with DNA at all, and reacted strongly with protein. These patterns of reactivity indicate that the newly synthesized RNA remains paired with the DNA
template over a stretch of 8–9 nucleotides from the 3¢ end, and then separates from the template strand. It also seems that the U analog must be
closely associated with the RNA polymerase even when it is up to 24
nucleotides from the 3¢ end.
Reference: Nudler E, Mustaev A, Lukhtanov E & Goldfarb A (1997) The
RNA–DNA hybrid maintains the register of transcription by preventing
backtracking of RNA polymerase. Cell 89, 33–41. 6–43 A schematic diagram of the structure of the mRNA–DNA hybrid and the
intron–exon structure of the gene are shown in Figure 6–55. Figure 6–54 A summary of the proposed
role of the mediator complex in initiation
of transcription (Answer 6–41). A130 Chapter 6: How Cells Read the Genome: From DNA to Protein
C D Figure 6–55 Interpretation of the electron
micrograph and the intron–exon structure
of the chicken ovalbumin gene (Answer
6–43). The letters A to G identify introns
and the black boxes identify exons. F G
nucleotides A B C E D E F G 1000
nucleotides References: Dugaiczyk A, Woo SL, Lai EC, Mace Jr ML, McReynolds L &
O’Malley BW (1978) The natural ovalbumin gene contains seven intervening
sequences. Nature 274, 328–333.
Garapin AC, Cami B, Roskam W, Kourilsky P Le Pennec JP, Perrin F, Gerlinger
P Cochet M & Chambon P (1978) Electron microscopy and restriction
enzyme mapping reveal additional intervening sequences in the chicken
ovalbumin split gene. Cell 14, 629–639.
6–44 The results of these experiments argue convincingly that base-pairing
occurs between the pre-mRNA and the U1 RNA. The base-pairing between
the U1 snRNA and the pre-mRNA is not extensive, but it is better when splicing occurs successfully than when it does not. These experiments illustrate
a classic approach for testing the reality of proposed base-pairing schemes.
If a scheme is important, then a base change in one component should
interfere with the process. A compensating change in the second component (to restore base-pairing) should then reestablish the process. That is
just what was observed in these experiments. A change of a G Æ A in the premRNA inhibited splicing, whereas a compensating C Æ U change in the U1
RNA, which should restore base-pairing (GC Æ AU), reestablished splicing.
Moreover, none of the other mutations in the U1 RNA (which would not
restore base-pairing) overcame the splicing defect.
Reference: Zhuang Y & Weiner AM (1986) A compensatory base change in
U1 snRNA suppresses a 5¢ splice site mutation. Cell 46, 827–835. 6–45
A. If the RNAs are joined by splicing, then there should be a 5¢-splice-site
sequence (GU) in the leader RNA and a 3¢-splice-site sequence (AG) in the
actin RNAs. When joined according to the usual splicing rules, these splice
sites should generate the actin mRNA sequences. As indicated in Figure 6–56
for actin gene 1, splicing could join the leader RNA and the actin RNA to generate actin mRNA with the correct sequence. If the RNAs are joined by splicing, then the leader gene contributes 22 nucleotides to the actin mRNA.
B. The leader RNA sequences and the actin RNA sequences cannot be part of
the same precursor RNA because actin genes 1, 2, and 3 are not all transcribed in the same direction. (The observation that the leader RNA genes
seem to be transcribed into a discrete 100-nucleotide transcript is a weaker
5¢ XGUUUAAUUACCCAAGUUUGAG 5¢ actin RNA GUAAA.... ....UUCAG GUACAUUAAAAACUAAUCAAAAUG XGUUUAAUUACCCAAGUUUGAG GUACAUUAAAAACUAAUCAAA AUG
actin mRNA Figure 6–56 Splicing of leader RNA and
actin gene 1 RNA to generate actin
mRNA (Answer 6–45). The GU at the 5¢
splice site associated with the leader RNA
is underlined, as is the AG at the 3¢ splice
site associated with the actin RNA. Also
underlined are the leader segment
identified by sequence comparisons and
the start site for translation. FROM DNA TO RNA
argument against a precursor. Since the leader RNA genes are present in
about 100 copies, it would be difficult to rule out that one or a few of them
initiated a longer transcript.)
C. The presence of proper splice signals that could generate the correct actin
mRNA, taken together with the impossibility of a single large precursor, suggests that the leader RNA may be attached to actin RNA by splicing between
two separate RNA molecules. It turns out that trans-splicing produces about
1% of a nematode’s mRNAs; it is even more common in trypanosomes where
it is responsible for all of the mRNAs. In both organisms a common exon is
spliced onto the 5¢ ends of many different RNA transcripts.
Reference: Krause M & Hirsh D (1987) A trans-spliced leader sequence on
actin mRNA in C. elegans. Cell 49, 753–761.
A. The 5¢ ends of the RNA molecules were labeled. Only labeled fragments
show up in the autoradiograph (see Figure 6–22). Thus, if the shortest fragments (those that were at the bottom of the gel) are from the 5¢ end, the 5¢
end must have been labeled.
B. The bands corresponding to the As in the AAUAAA signal sequence are missing from the ladder of bands in polyadenylated and cleaved RNA (see Figure
6–22, lanes 3 and 4) because modification of any one of those As interfered
with cleavage and with polyadenylation. Thus, RNA molecules that carry a
single modification in the signal sequence are not recognized by the components of the extract and, as a result, do not show up in the population of
molecules that carry poly-A tails (see lane 3) or in the population of
molecules that are cleaved (see lane 4).
C. The band at the arrow in Figure 6–22 is absent from the polyadenylated RNA
but present in the cleaved RNA because modification of this A does not prevent cleavage, but it does prevent polyadenylation. Thus, RNA molecules
with this A modified are present in the cleaved molecules (see Figure 6–22,
lane 4) but are not present in the polyadenylated molecules (see lane 3).
D. The analysis of the missing bands in parts B and C indicates that the AAUAAA
signal sequence is important for the cleavage of precursor RNAs and that the
AAUAAA sequence and the single A are required for polyadenylation.
E. If the other end—the 3¢ end—of the RNA molecules were labeled, it would
have been possible to determine whether any of the As or Gs on the 3¢ side
of the cleavage site were important for polyadenylation. These experiments
have been done; they show that no single modification 3¢ of the polyadenylation site prevents polyadenylation. The sequence requirements on the 3¢
side of the cleavage site (GU- or U-rich) are not so specific as those on the 5¢
side and would not be expected to be inactivated by single changes.
Reference: Conway L & Wickens M (1987) Analysis of mRNA 3¢-end formation by modification interference: the only modifications which prevent
processing lie in AAUAAA and the poly(A) site. EMBO J. 6, 4177–4184.
A. The idea behind the oligonucleotide experiment was to try to cleave the RNA
component of the snRNP that was suspected of interacting with the conserved
sequence at the 3¢ end of the histone precursor. If the snRNP interacted by
hybridizing to the precursor RNA, then an oligonucleotide that matched the
sequence in the precursor RNA should be able to hybridize to the snRNA. Formation of a DNA–RNA hybrid would render the snRNA sensitive to cleavage
by added RNase H. Cleavage of the snRNA in this critical region of interaction
should render the extract incapable of processing the precursor. This result
was the one observed for the mouse and consensus oligonucleotides.
B. The inability of the human oligonucleotide to block processing was not
anticipated, since a human extract was being used. Examination of the
hybrids that can form between the various oligonucleotides and the U7
snRNA reveals that the mouse and consensus oligonucleotides can hybridize
perfectly for a 10-nucleotide and a 9-nucleotide stretch, respectively A131 A132 Chapter 6: How Cells Read the Genome: From DNA to Protein
3¢ GTGTCGATGAAACCA 5¢ human ||||||
5¢ m3G-NNGUGUUACAGCUCUUUUAGAAUUUGUCUAGU 3¢ human U7 snRNA
3¢ TTGTCGAGAAAGGC 5¢ mouse
3¢ TGGTCGAGAAAGAAA 5¢ consensus (Figure 6–57). By contrast, hybridization to the human oligonucleotide is split
by an unmatched nucleotide into two segments of 6 and 4 nucleotides. The
stability of pairing of two separate segments is not so great as for a continuous
pairing segment. Hence, the human oligonucleotide does not pair with sufficient stability to render the U7 snRNA sensitive to RNase H cleavage.
Reference: Mowry KL & Steitz JA (1987) Identification of human U7 snRNP
as one of several factors involved in the 3¢-end maturation of histone premessenger RNAs. Science 238, 1682–1687.
6–48 These results indicate that box elements C and D are important for the accumulation of U85 RNA. The presence of E2 RNA in Figure 6–25B, lanes 6 and
12, shows that the transfections were successful. Thus the absence of U85
RNA from those lanes is meaningful. It is unclear from these studies whether
the altered box elements prevent processing from the intronic RNA or render the processed RNA unstable.
Reference: Jady B & Kiss T (2001) A small nucleolar guide RNA functions
both in 2¢-O-ribose methylation and pseudouridylation of the U5 spliceosomal RNA. EMBO J. 20, 541–551. 6–49 For nucleotides in U5 snRNA that are true targets for U85 snoRNA-dependent modification, the expectation is that the modification will be dependent on pairing between U5 snRNA and U85 snoRNA. Thus, a bona fide
modification should be present in the transfection with U2–U5, which can
be modified by the endogenous U85 snoRNA; however, it should be absent
in the transfection with U2–U5m, which cannot pair stably with the endogenous U85 snoRNA. Most importantly, modification of U2–U5m should be
restored in the presence of U85m snoRNA, which can pair with U2–U5m.
Both pseudouridine y46 and methylated C45 behave according to these
expectations (see Table 6–2). Thus, pseudouridine y46 and 2¢-O-methylation at C45 are both dependent on U85 snoRNA.
Methylated U41 is present in all transfections, indicating that it is modified by some other component of the cell in a way that is not dependent on
the nearby sequences that were modified in U2–U5m. Pseudouridine y43 is
not modified under any conditions in these experiments, suggesting that it
cannot be modified as part of the fragment of U5 that was inserted into U2.
Thus, these experiments do not rule it out as a potential target for modification by U85 snoRNA. Nevertheless, studies with many other box H/ACA
snoRNAs indicate that the guide sequences universally position the U to be
converted to y in the location shown for y46 in Figure 6–26A and B.
Reference: Jady B & Kiss T (2001) A small nucleolar guide RNA functions
both in 2¢-O-ribose methylation and pseudouridylation of the U5 spliceosomal RNA. EMBO J. 20, 541–551. FROM RNA TO PROTEIN
6–50 Proteasome 6–51 Genetic code 6–52 Initiator tRNA 6–53 Anticodon Figure 6–57 Pairing between the three
oligonucleotides and the human U7
snRNA (Answer 6–47). A133 FROM RNA TO PROTEIN
6–54 Ribozyme FRAME 1 6–55 Nonsense-mediated mRNA decay AGU CUA GGC ACU GA-3¢
T 6–56 Reading frame 6–57 Aminoacyl-tRNA synthetase 6–58 Prion disease 6–59 Molecular chaperone FRAME 2 A GUC UAG GCA CUG A-3¢
FRAME 3 TRUE/FALSE AG UCU AGG CAC UGA-3¢
3-FRAME TRANSLATION 6–60 False. Wobble pairing occurs between the third position in the codon and
the first position in the anticodon. 6–61 False. Although the mechanism for identifying a target for digestion is a significant difference, an equally important—perhaps more important—difference is the processivity of digestion. In contrast to a simple protease, which
cleaves a substrate’s polypeptide chain just once before dissociating, the
proteasome keeps the entire substrate bound until all of it has been converted to short peptides. THOUGHT PROBLEMS
6–62 The amino acids encoded in each of the three reading frames are shown in
Figure 6–58. If this segment of RNA encoded part of a larger protein, it would
have to be translated in reading frame 1, which is the only one that does not
contain a stop codon. 6–63 The only codon assignments consistent with the observed changes, and
with the assumption that single-nucleotide changes were involved, are
GUG for valine, GCG for alanine, AUG for methionine, and ACG for threonine. It is unlikely that you would be able to isolate a valine-to-threonine
mutant in one step because that would require two nucleotide changes.
Typically, two changes would be expected to occur at a frequency equal to
the product of the frequencies for each of the single changes; hence, the
double mutant would be very rare. 6–64
A. UUUUUUUUUUUU… codes for a polymer of phenylalanine.
B. AUAUAUAUAUAU… codes for a polymer of alternating isoleucines and
tyrosines. Because the start point of the ribosome on the RNA is random, the
ribosomes will generate a mixture of polymers, some of which start with
isoleucine and some with tyrosine.
C. AUCAUCAUCAUC… codes for a mixture of three different polymers. Ribosomes start translation in each of the three reading frames: AUC–AUC–
AUC–AUC-… codes for a polymer of isoleucine; UCA–UCA–UCA–UCA-…
codes for a polymer of serine; and CAU–CAU–CAU–CAU-… codes for a polymer of histidine.
A. The genetic code can be used to convert the two amino acid sequences into
a set of potential mRNA sequences. At the sites where two or more
nucleotides are shown, different mRNA sequences are consistent with the
same sequence of encoded amino acids.
AAU AUG AAU GGU AAA
S L G T
V * A L
S R H * Figure 6–58 Amino acids encoded in the
three reading frames of an RNA (Answer
6–62). The amino acids encoded in each
reading frame are shown separately and
all together, as they are usually
represented. Asterisks identify stop codons. A134 Chapter 6: How Cells Read the Genome: From DNA to Protein
AAU AUG AUU UGG CAA AUU UGU GUU AUG AAA GAU
Comparison of the potential mRNA sequences shows that the mutant carries an extra U at the eighth position, the first one that differs between the
two potential mRNA sequences. If you compare only the codons for N (wild
type) with I (mutant) there are three possibilities for the difference: insertion of a U, deletion of an A, or a nucleotide change of A Æ U. Comparison
of the next codon in the two sequences rules out a nucleotide change,
which would leave the adjacent codon unaffected and thus would not give
rise to a frameshift mutation. It also allows one to distinguish between
insertion and deletion of a nucleotide. Note that the Gs in the first and second positions of the glycine codon (GG–) in the wild type have become the
Gs in the second and third positions in the tryptophan codon (–GG) in the
mutant. This shift can only be explained by an insertion. Thus, the
frameshift mutation arose by insertion of an AT base pair at the eighth position in this DNA sequence.
B. Removing the extra U from the mutant sequence resolves some of the ambiguity in the wild-type sequence and extends the sequence for the wild type,
which can be converted back into an amino acid sequence as shown below.
AAU AUG AAU GGC AAA UUU GUG UUA UGA AAG
The first seven codons in the extended mRNA clearly code for amino acids,
although the identities of some are ambiguous because the mRNA sequence
is not fully defined. The eighth codon could either code for an amino acid or
be a stop codon. The ninth codon is definitely a stop codon. Thus, since the
asparagine (N) at the beginning of the peptide sequence is amino acid 263,
the intact polypeptide would be either 269 or 270 amino acids in length. Two
stop codons in tandem are commonly found in the end of coding sequences. 6–66 Mutations of the type described in 2 and 4 are often the most harmful. In
both cases, the reading frame would be changed. Because these frameshifts
occur early in the coding sequence or in the middle of it, the encoded protein will contain a nonsensical and usually truncated sequence of amino
acids. In contrast, a reading frameshift that occurs toward the end of the
coding sequence, as described in 1, will result in a largely correct protein
that may be functional.
Deletion of three consecutive nucleotides, as in scenario 3, leads to the
deletion of one amino acid, but does not alter the reading frame. The deleted
amino acid may or may not be important for the folding or activity of the
protein. In many cases such mutations are silent; that is, they have insignificant consequences for the organism.
Substitution of one nucleotide for another, as in scenario 5, is often completely harmless, because it does not change the encoded amino acid. In
other cases it may change an amino acid, which may be deleterious or
benign, depending on the location and functional significance of that particular amino acid. Often, the most deleterious kind of single-nucleotide
change creates a new stop codon, which gives rise to a truncated protein. 6–67
A. A genetic code that used pairs of nucleotides would have 16 different codons
(4 possible nucleotides in the first position ¥ 4 possible nucleotides in the A135 FROM RNA TO PROTEIN
second position). Thus, it could specify a maximum of 16 different amino
B. A triplet code that depended only on codon composition would have 20 different codons (4 codons composed all of one base; 12 codons with two bases
the same and one different; and 4 codons with three different bases). Such a
code could specify a maximum of 20 different amino acids.
C. It is relatively easy to see how a doublet code could be translated by a mechanism similar to that used with the standard genetic code. It is more difficult
to see how the nucleotide composition of a stretch of three nucleotides
could be translated without regard to the order of nucleotides, because
base-pairing could no longer be used. An AUG, for example, would not basepair with the same anticodon as a UGA.
6–68 In present-day cells, there is some wobble in the matching of codons to anticodons. In several cases, the same tRNA pairs with multiple codons that
specify the same amino acid but differ in their nucleotide sequence, generally at the third position of the codon. It seems likely that in the early biological world, without highly evolved ribosomes to help in the pairing process,
the converse may also have been true: several different tRNAs, with similar
anticodons, may have been able to bind to the same codon. This would have
played havoc with the translation of the genetic message into protein, unless
the amino acids carried by these tRNAs were chemically similar. In the
absence of perfect specificity, natural selection may have operated to ensure
that tRNAs with related anticodons carried chemically similar amino acids.
Alternatively, in the early world, before modern aminoacyl-tRNA synthetases had evolved, there was probably some ambiguity in the matching of
tRNAs with appropriate amino acids. The same tRNA might have become coupled to any of a number of amino acids that were chemically similar. One can
imagine the evolution of the genetic code by refinement of a matching process
that was originally imprecise and gave only a blurred relationship between sets
of roughly similar codons and sets of roughly similar amino acids. 6–69 The rules for wobble pairing between the anticodon and the codon,
expressed both ways, are shown in Table 6–5. It is striking that A in the wobble position of the anticodon in eucaryotes does not have a pairing partner
in codons. It turns out that A is not used in the wobble position in eucaryotic
tRNAs. In many cases an A is encoded in the wobble position, but it is
changed to inosine (I) after transcription, giving rise to a mature tRNA that
recognizes U or C in the wobble position of the codon.
Reference: Lander ES et al (2001) Initial sequencing and analysis of the
human genome. Nature 409, 860–921. 6–70 In eucaryotes a minimum of 45 tRNAs would be required to recognize all 61
codons, given the rules for wobble base-pairing. Pairs of codons that end in Table 6–5 Rules for wobble base-pairing between codon and anticodon (Answer 6–69).
BASE Bacteria U
G A, G, or I
G or I
U or I
C or U Eucaryotes U
G G or I
G or I
BASE Bacteria U
I A or G
U or C
U, C, or A Eucaryotes U
U or C
U or C A136 Chapter 6: How Cells Read the Genome: From DNA to Protein
U or C always encode the same amino acid (for example, CAU and CAC
encode histidine, and GGU and GGC encode glycine). Thus a single tRNA
with an I (or a G) in the wobble position of the anticodon would be required
for each such pair of codons (see Table 6–5, Answer 6–69). The 16 pairs of
such codons (32 codons) would require 16 tRNAs. Each of the remaining 29
codons, which end in either an A or a G, would require a specific tRNA with
a corresponding U or C in the wobble position. Thus, the minimum number
of tRNAs is 16 plus 29, or 45. 6–71 The single codon for tryptophan is 5¢-UGG. The anticodon of the normal
tryptophan tRNA is 5¢-CCA, which pairs specifically with this codon. A mutation that changes the anticodon to 5¢-UCA would allow the tRNA to recognize the 5¢-UGA stop codon, which would lead to insertion of tryptophan at
UGA and prevent termination of translation. Because of wobble (see Table
6–5), the mutant anticodon would also recognize the normal 5¢-UGG codon,
so that, in principle, its ability to insert tryptophan at the normal UGG
codons would not be compromised.
Many genes use UGA codons as the natural stop sites for their encoded
proteins. These stop codons would also be affected by the mutant tRNA. In
reality there is a competition between the mutant tRNA and the termination
factors. Whenever the tRNA wins the race, the affected proteins would be
made with additional amino acids at their C-terminal ends. The additional
lengths would depend on the number of codons before the ribosomes
encounter another stop codon in the mRNA. The potential chaos that such
mutations might cause is mitigated by two factors: the efficiency of translation of stop codons by such mutant tRNAs is usually low, and many bacterial
genes are ‘protected’ by double stop codons at their ends. In reality, such
suppressors have been invaluable for genetic studies in bacteria. 6–72 This experiment beautifully demonstrates that the ribosome does not check
the amino acid that is attached to a tRNA. Once an amino acid has been coupled to a tRNA, the ribosome will ‘blindly’ incorporate that amino acid
according to the match between the codon and anticodon. We can therefore
conclude that a significant part of the correct reading of the genetic code—
namely, the matching of a codon with the correct amino acid—is performed
by the synthetase enzymes that attach amino acids to tRNAs.
Reference: Chapeville F, Lipmann F, von Ehrenstein G, Weisblum B, Ray WJ
& Benzer S (1962) On the role of soluble ribonucleic acid in coding for amino
acids. Proc. Natl Acad. Sci. U.S.A. 48, 1086–1092. 6–73 One effective way of driving a reaction to completion is to remove one of the
products. The flow of substrates to products then increases to reestablish the
equilibrium ratio—the principle of mass action. All three of the products of
this reaction are removed. The concentration of AMP is constantly reduced
by conversion to ADP and then to ATP by other reactions in the cell. Similarly, the aminoacyl-tRNAs are used in protein synthesis, constantly
decreasing their concentrations. But by far the most dramatic influence is
the removal of PPi by hydrolysis to two phosphates. That reaction yields as
much free energy as the hydrolysis of ATP to ADP, which means that essentially all of the PPi will be converted to free phosphates. As a result, the linked
reactions for charging a tRNA and hydrolyzing PPi—the reactions as they
occur in cells—have a DG° of –6.9 kcal/mole. 6–74
A. The ratio of N-terminal to total radioactivity will increase with increasing
time of exposure. Because you isolate only complete protein for your analysis, the position that will be labeled at the shortest time point will be the Cterminus. At the shortest time point the radioactivity at the N-terminus will
be nearly nonexistent, giving a very low ratio of N-terminal to total radioactivity. With increasing time more and more protein will carry label at the Nterminus so the ratio is expected to rise. By the time the N-terminus
becomes labeled, all the rest of the leucines in the protein will also be FROM RNA TO PROTEIN
labeled; thus, at late times the ratio of N-terminal to total radioactivity will
equal the ratio of N-terminal to total leucines, which is 1:5 or 0.2.
6–75 The proportion of a cell’s total energy devoted to protein synthesis is typically
determined by measuring oxygen consumption in the presence and absence
of inhibitors of protein synthesis. Because oxygen is used principally for generation of ATP via oxidative phosphorylation, and nearly all the cell’s energy
is derived from oxidative phosphorylation, oxygen consumption is a fairly
direct measure of energy usage. The fractional drop in oxygen consumption
in the absence of protein synthesis (when it’s inhibited) indicates the proportion of the cell’s energy normally devoted to protein synthesis. 6–76
B. This short mRNA encodes three different peptides because there are three
different reading frames. In the second reading frame, the first codon is the
stop codon UAG; thus, it is unlikely that the subsequent codons will be
Frame 1 V A Y P *
Frame 2 * P T H R
S L P I
The other possible mRNA from this DNA would read
Frame 1 P M G R L
L W V G Y
Y G * A
Thus, the sequence of the peptides encoded by the complementary DNA
strand would be completely different. Be careful to keep the polarity of the
strands correct; don’t fall into the trap of thinking that the complementary
sequence of the first mRNA is 5¢-CAUCGGAUGGGUAUCC-3¢, which is incorrect because it is the wrong polarity.
C. If translation begins at the 5¢ end of the RNA, the synthesized protein would
be valine-alanine-tyrosine-proline (VAYP). Only after a peptide bond has
been formed between alanine and tyrosine will tRNAAla leave the ribosome.
Thus, the next tRNA that will bind to the ribosome after tRNAAla has left is
tRNAPro. When the amino group of alanine forms a peptide bond, the ester
bond between valine and tRNAVal is broken, tRNAVal moves from the P-site
to the E-site (exit site), and tRNAAla moves from the A-site to the P-site.
6–77 When EF-Tu has positioned an aminoacyl-tRNA in the acceptor site on the
ribosome, it hydrolyzes its bound GTP and exits, leaving the aminoacyltRNA behind. The first delay comes about because the rate of GTP hydrolysis is faster for a correct codon–anticodon pair than for an incorrect one. As
a result, an incorrectly bound tRNA has more time to dissociate from the
ribosome. The second delay occurs between dissociation of EF-Tu and full
accommodation of the tRNA into the A-site. This time delay, which is also
shorter for correct than incorrect tRNAs, allows a second opportunity for the
incorrect tRNA to dissociate from the ribosome. An incorrect tRNA will, on
average, dissociate more rapidly than a correct one because its codon–anticodon match contains fewer hydrogen bonds. 6–78 In eucaryotic cells protein synthesis is normally initiated by scanning from the
5¢ end of the mRNA until the first AUG codon is found. (Sometimes the second
or third AUG codon may be used instead—a phenomenon known as leaky
scanning.) This mechanism of initiation ensures that ribosomes will all start
translating near the 5¢ end of the mRNA. When the ribosomes complete synthesis of the protein, they fall off the mRNA and must reinitiate by scanning A137 A138 Chapter 6: How Cells Read the Genome: From DNA to Protein
from the 5¢ end. By contrast, in procaryotic cells protein synthesis is initiated
by base-pairing between mRNA sequences adjacent to an initiation AUG
codon and sequences in the 16S rRNA of the small ribosomal subunit. The
procaryotic initiation strategy allows ribosomes to recognize several start
sites in the same mRNA. This key difference in mechanism underlies their
ability to make several proteins from a single polycistronic mRNA. 6–79 A broken mRNA when translated would produce a truncated protein that
could be harmful to the cell. A protein fragment can retain some of the functions of the whole protein, allowing it, for example, to bind to a target protein but trap it in an unproductive complex. Alternatively, a protein fragment can display new, aberrant binding surfaces that allow it to bind to
novel partners, interfering with their function. 6–80
A. Edeine specifically inhibits initiation of protein synthesis by preventing the
joining of the 60S ribosomal subunit to the 40S subunit/mRNA/initiator
tRNA complex. Since elongation is not blocked, ribosomes that have already
begun synthesis complete their individual chains and fall off the mRNA,
leaving attached only the small subunit and the initiator tRNA. Edeine is an
antibiotic produced by certain strains of Bacillus brevis.
B. A lag occurs before protein synthesis shuts off because edeine inhibits initiation but has no effect on elongation. Thus, a ribosome that has just started
making a new polypeptide is free to complete it. Incorporation of label continues for just the length of time it takes to complete the protein (in this case,
the globin chains of hemoglobin), which takes about a minute.
C. If cycloheximide (or any other elongation inhibitor) is added at the same
time as an initiation inhibitor, the polyribosomes are ‘frozen.’ Polyribosome
breakdown by initiation inhibitors requires ribosome movement, which is
blocked by elongation inhibitors.
Reference: Safer B, Kemper W & Jagus R (1978) Identification of a 48S preinitiation complex in reticulocyte lysate. J. Biol. Chem. 253, 3384–3386.
D. The formation of one peptide bond, but no more, eliminates all choices
except D. If farsomycin inhibited formation of the 80S initiation complex
(choice A), inhibited binding of aminoacyl-tRNAs to the A-site (choice B), or
inactivated peptidyl transferase (choice C), no peptide would have been
formed. If it interfered with chain termination and release (choice E), the
entire peptide would have been made.
6–82 In a well-folded protein, the majority of hydrophobic amino acids will be
sequestered in the interior away from water. Exposed hydrophobic patches
thus indicate that a protein is abnormal in some way. Some proteins initially
fold with exposed hydrophobic patches that are used in binding to other
proteins, ultimately burying those hydrophobic amino acids as well. As a
result, hydrophobic amino acids are usually not exposed on the surface of a
protein, and any significant patch is a good indicator that something has
gone awry. The protein may have failed to fold properly after leaving the
ribosome, it may have suffered an accident that partly unfolded it at a later
time, or it may have failed to find its normal partner subunit in a larger protein complex. 6–83 Molecular chaperones fold like any other protein. Molecules in the act of
synthesis on ribosomes are bound by Hsp70 chaperones. And incorrectly
folded molecules are helped by Hsp60-like chaperones. That they function
as chaperones when they have folded correctly makes no difference to the
way they are treated before they reach their final, functional conformation.
Of course, properly folded Hsp60-like and Hsp70 chaperones must already
be present to help fold the newly made chaperones. At cell division, each
daughter cell inherits a starter set of such chaperones from the parental
cell. A139 FROM RNA TO PROTEIN
E1 is the ubiquitin-activating enzyme. It repetitively donates activated ubiquitin monomers to the E2 component of the E2-E3 ubiquitin ligase, which
then transfers ubiquitin units repetitively to bound substrate molecules.
Particular combinations of E2 and E3, of which there are many in the cell,
provide binding specificity so that only certain species of proteins are targeted for ubiquitylation and destruction. CALCULATIONS 120
A. As shown in Figure 6–59, the rate of synthesis is linear with time. The curvature so apparent in the autoradiograph in Figure 6–29 results from the nonlinear migration of proteins in SDS polyacrylamide gels.
B. The rate of protein synthesis can be determined from the slope of the line in
Figure 6–59. This system is synthesizing roughly 52,000 daltons of protein
per 10 minutes, or 5200 daltons per minute, which corresponds to about 47
amino acids per minute [(5200 daltons/minute)/(110 daltons/amino acid)].
This rate is less than the rate in E. coli, which is about 10 times faster. The
rate is also about three times less than that of globin synthesis in the same
reticulocyte lysate. As discussed in part C, part of the reason for the low rates
may be that the mix of tRNAs in the reticulocyte lysate is not optimal for this
plant virus protein.
C. The autoradiograph contains many bands, rather than just a few, because
ribosomes keep loading onto the mRNA throughout the course of the experiment. You could obtain the theoretical result in Figure 6–30B by adding an
inhibitor of initiation after 5 minutes or, alternatively, by adding unlabeled
methionine in vast excess after 5 minutes.
The presence of discrete bands rather than a continuous background fuzz
suggests that there are specific hang-up points along the mRNA, perhaps
where ribosomes must wait for rare tRNAs. The tRNA population in the
reticulocyte is specialized for making globin, not a protein from a plant
A. Since an average protein contains about 455 amino acids [(50,000 d/protein)
¥ (amino acid/110 d)], it will take a muscle cell about 3.8 minutes to make it
[(455 amino acids) ¥ (sec/2 amino acids) ¥ (min/60 sec)]. Since titin is 60
times the size of an average protein, the muscle cell will require 3.8 hours to
make it [(3,000,000 d/titin) ¥ (amino acid/110 d) ¥ (sec/2 amino acids) ¥
B. It will take a muscle cell about 23 minutes to transcribe an average gene
and 23 hours to transcribe titin. For the average protein 455 amino acids
corresponds to 1365 nucleotides of RNA [(3 nt/codon) ¥ (455 codons)].
Given that 5% of the initial transcript is converted to mRNA, the initial
transcript is 2.7 ¥ 104 nucleotides (1365 nt ¥ 20), which would require
about 23 minutes to transcribe [(2.73 ¥ 104 nt) ¥ (sec/20 nt) ¥ (min/60
sec)]. Because titin is 60 times as big, the muscle cell will require about 23
hours to transcribe it.
6–87 140 molecular mass of
largest peptide (kd) 6–84 The energy cost of translation and transcription will be equal when 30 protein molecules have been made from one mRNA. Protein synthesis requires
four high-energy phosphate bonds per codon (per three nucleotides). Transcription consumes six high-energy phosphate bonds to make a codon, but
also consumes 19 times more energy synthesizing RNA that will be discarded
(95%). Thus, transcription consumes a grand total of 120 high-energy phosphate bonds per codon (6 + 114), compared to four per codon for translation.
The ratio of energy costs per codon (120/4 = 30) defines the number of proteins that will have been made when the energy cost of translation matches
that of transcription. Because most mRNAs are used to make hundreds to
thousands of proteins, translation consumes a much higher fraction of the
cell’s energy than does transcription. 5 10 15 20 25 time (minutes) Figure 6–59 Rate of synthesis of a TMV
protein (Answer 6–85). A140 Chapter 6: How Cells Read the Genome: From DNA to Protein 6–88
A. Since the bacteria were labeled for one generation, which represents a doubling in mass, 4 mg of the 8 mg of flagellin isolated from the gel were synthesized in the presence of 35S-cysteine. The amount of radioactivity in the
sample indicates that about 1 out of every 1670 flagellin (flgn) molecules
contains a cysteine.
Cys 300 cpm Cys
pmol Cys 4 ¥ 104 mg flgn
4 mg flgn
5 ¥ 103 cpm
= 6 ¥ 10 pmol Cys
100 pmol flgn
Cys = 6 ¥ 10–4
which is equal to 1 cysteine per 1670 flagellin molecules [1/(6 ¥ 10–4)].
B. The normal codons for cysteine are UGU and UGC. Thus, the error in anticodon–codon interaction is a mistake at the first position of the codon (third
position of the anticodon). The experiment described, as well as others, suggest that ribosomes tend to mistake U for C and C for U in the first two positions of the codon, and C and U for A in the first position.
C. Assuming that all six arginine codons are equally frequent, there should be
six sensitive (CGC and CGU) arginine codons [(2/6) ¥ 18] in a flagellin
molecule. Therefore, the actual error frequency per codon-at-risk is
1670 flagellin molecules 6 sensitive codons
error frequency = 10–4
error frequency = D. If the probability of making a mistake at each codon is 10–4, the probability
of not making a mistake at each codon is (1 – 10–4). The probability of not
making a mistake at n codons is then (1 – 10–4)n. Thus, the percentage of correctly synthesized molecules 100 amino acids in length is (1 – 10–4)100, or
99%. For a protein 1000 amino acids long, 90% are correct. For a protein
10,000 amino acids long, only 37% are correct. Given these sorts of estimates, it is perhaps not surprising that proteins more than 3000 amino acids
long are rare. If you are curious, you might calculate the fraction of titin
molecules that you would expect to be made correctly (see Problem 6–86)?
Reference: Edelman P & Gallant J (1977) Mistranslation in E. coli. Cell 10,
131–137. DATA HANDLING acceptor
A. A simple change of anticodon allows tRNAVal to be charged by IleRS at about
60% normal efficiency. Thus, the anticodon is the most important part for
charging. As the rest of the tRNAVal molecule becomes more like tRNAIle, the
efficiency steadily increases, suggesting that additional sequences that aid
in the charging with isoleucine are dispersed throughout tRNAIle.
B. Results with the chimeric tRNAs show that a tRNAVal carrying just the D-loop
and anticodon from tRNAIle, is very nearly as effective for valine editing as
normal tRNAIle. In fact, a close examination of the sequences of the D-loops
from the two tRNAs reveals just three nucleotide differences. These
nucleotides are located at the elbow in the ‘L’ shaped three-dimensional
structure of tRNAs (Figure 6–60).
C. Generally no. However, the D-loop is crucial to valine editing and it also
improves the efficiency of tRNAIle charging by IleRS.
Reference: Hale SP, Auld DS, Schmidt E & Schimmel P (1997) Discrete determinants in transfer RNA for editing and aminoacylation. Science 276,
1250–1252. anticodon Figure 6–60 Schematic structure of
tRNAIle (Answer 6–89). Arrows indicate the
locations of the nucleotide differences
between the D-loops in tRNAIle and
tRNAVal. Dotted lines indicate hydrogen
bonds between nucleotides that help to
establish the three-dimensional structure. A141 FROM RNA TO PROTEIN
(A) RIBOSOMES WITH ATTACHED PEPTIDES
8 16 N
8 16 N
8 16 nascent
N radioactivity in peptide (B) EVENLY SPACED RIBOSOMES (C) BLOCKED RIBOSOMES block amino acid number amino acid number 6–90
A. The data in Figure 6–32 indicate that the N-terminus of the protein is synthesized first. The steadily decreasing level of radioactivity from the N-terminus to the C-terminus is exactly what you would expect if synthesis began
at the N-terminus. As illustrated in Figure 6–61A, all the ribosomes carry a
labeled lysine at position 8 in their nascent chains, but the ribosome at the
5¢ end of the mRNA has not yet reached the lysine at position 16. Thus, when
digested with trypsin, all of these nascent chains will yield a labeled N-terminal peptide, but a smaller fraction will yield the second peptide. Fewer
still will contain the third peptide, and so on. Almost none of the ribosomes
will carry a nascent chain with the labeled lysine nearest the C-terminus.
B. The lines for the a and b chains in Figure 6–32 are very similar, with nearly
identical intercepts on both axes, which indicates that roughly equal numbers of each chain are being synthesized. However, there is not enough information to decide whether the numbers of a- and b-globin mRNA molecules
are equal. You would need to know how many ribosomes there were on each
mRNA—the average polyribosome size for a- and b-globin mRNAs—to
deduce their relative abundance from these kinds of data. Actually, there is
about twice as much a-globin mRNA as b-globin mRNA, but the a-globin
mRNA is less efficiently translated; that is, fewer ribosomes initiate synthesis on a-globin mRNA per unit time than on b-globin mRNA. These factors
cancel out to give a balanced production of the two chains.
C. The graph in Figure 6–32 hits zero right at the end of the coding region,
which indicates that chains are released from ribosomes as soon as they
encounter the stop codon—or at least they do so without a measurable
pause on this time scale.
D. If there were a significant roadblock to ribosome movement, the data would
resemble that in Figure 6–33A. A roadblock would result in more densely
packed ribosomes in front of the block and less densely packed ribosomes
beyond the block. The consequences of normal and inhibited ribosome
movement are illustrated schematically in Figure 6–61B and C. Figure 6–61 Relationship of ribosome
position to peptide length and labeling
pattern (Answer 6–90). (A) Lengths of
peptides associated with ribosomes at
various positions along b-globin mRNA.
Numbers refer to positions of the first two
lysines. (B) Pattern of peptide labeling for
evenly spaced ribosomes. (C) Pattern of
peptide labeling for ribosomes whose
movement is inhibited at a point about
midway down the mRNA. Peptides
associated with each ribosome are shown
on the graphs in B and C as lines. Small
circles correspond to the C-termini of the
polypeptides and are aligned immediately
below the ribosome on which the peptides
are synthesized. Dashed lines through the
small circles show the expected patterns
of peptide labeling. A142 Chapter 6: How Cells Read the Genome: From DNA to Protein 6–91
A. The DNA sequence GGG TAT CTT TGA CTA CGA CGC C should not encode
the protein sequence of RF2, since UGA is a termination codon. It appears
that this sequence must break the usual rules of the triplet code, with a
leucyl-tRNA decoding the quadruplet shown in italics below.
GGG TAT CTTT GAC TAC GAC GCC
In essence, the ribosome must shift its reading frame in the middle of the
Frameshift mutations were originally isolated by Seymour Benzer in his
work on the rII genes of bacteriophage T4 and exploited by Francis Crick in
a proof of the triplet nature of the genetic code. Later, mutant tRNA
molecules that could read four bases at a time were isolated by clever
genetic selection and shown to suppress certain frameshift mutations. It
came as a great surprise, however, to find natural examples of frameshift
suppression. The first example was found in the bacteriophage T7 gene 10.
Since then, several retroviruses and retroposons have been found to use
frameshift suppression of termination codons as a way of making minor
gene products. The mechanism of suppression in these cases is not clear.
B. The occurrence of an in-frame suppressible UGA codon (which is recognized uniquely by RF2) in the sequence of RF2 immediately suggests a novel
form of gene control. Although the mechanism of frameshifting is undefined, frameshifting and termination at the UGA codon probably compete
with one another. When the level of RF2 in the cell is high, termination
should occur more frequently at the UGA codon than when the level of RF2
is low. Thus, new RF2 would be synthesized infrequently when its levels were
already adequate, but if the levels fell, the chances of ribosomal frameshifting would increase and more RF2 would be made. Thus, this situation seems
to be a very cleverly appropriate autoregulation.
References: Craigen WJ, Cook RG, Tate WP & Caskey CT (1985) Bacterial
peptide chain release factors: conserved primary structure and possible
frameshift regulation of release factor 2. Proc. Natl Acad. Sci. U.S.A. 82,
Jacks T & Varmus H (1985) Expression of the Rous sarcoma virus pol gene by
ribosomal frameshifting. Science 230, 1237–1242.
6–92 SmpB evidently plays no role in the charging of tmRNA with alanine, since
that reaction is unaffected by the presence or absence of SmpB. SmpB is critical for the association of tmRNA with ribosomes (see Figure 6–35A), which
presumably explains why protein fragments are not tagged and degraded in
SmpB-deficient cells (see Figure 6–35B). Although these studies identify
roughly where in the process SmpB acts, they do not define its precise function.
Reference: Karzai AW, Susskind MM & Sauer RT (1999) SmpB, a unique
RNA-binding protein essential for the peptide-tagging activity of SsrA
(tmRNA). EMBO J. 18, 3793–3799. 6–93
A. The sequence data for the Tetrahymena protein are unusual because they
indicate that UAG and UAA, which are stop codons in other organisms, specify glutamine (Q) in Tetrahymena.
B. The minor protein above the full-length, 116-kd protein is produced from
the pure TMV mRNA by readthrough of the normal stop codon. Although it
is difficult to be sure exactly how such a rare event occurs, the amount of this
protein is thought to represent the frequency with which the reticulocyte
translation system mistakenly inserts an amino acid at the site of the stop
codon instead of terminating properly. It is a little surprising that a second
termination codon is not encountered for 506 codons (about 50 kd of additional protein). FROM RNA TO PROTEIN
C. Given that Tetrahymena uses UAG and UAA as codons for glutamine, the
increase in the proportion of the readthrough TMV protein is most likely due
to the presence of a tRNAGln species with an anticodon complementary to
the normal TMV stop codon (which is UAG). The addition of Tetrahymena
RNA causes a small shift in the proportions because it contains some
charged tRNAGln. The cytoplasm causes a larger shift because it also contains the appropriate aminoacyl-tRNA synthetase. (The additional shift with
the cytoplasm suggests that the tRNA synthetases in the reticulocyte lysate
cannot recharge the special Tetrahymena tRNA.) These results suggest that
at least two components from Tetrahymena—a special tRNA and its cognate
aminoacyl-tRNA synthetase—must be added to a reticulocyte lysate to allow
Tetrahymena mRNA to be translated efficiently. These components compete
effectively with the reticulocyte release factors, allowing the Tetrahymena
mRNAs to be read.
D. Although slight variations in the genetic code were originally discovered in
mitochondrial genomes, they were not as surprising as the Tetrahymena
changes. After all, mitochondrial genomes are small and encode relatively
few proteins, so it is less difficult to imagine how changes might occur. By
contrast, the Tetrahymena genome encodes thousands of proteins. It is
much more surprising that it managed to survive the presumptive transition
from the standard code to its present-day code.
References: Horowitz S & Gorovsky MA (1985) An unusual genetic code in
nuclear genes of Tetrahymena. Proc. Natl Acad. Sci. U.S.A. 82, 2452–2455.
Andreasen PH, Dreisig H & Kristiansen K (1987) Unusual ciliate-specific
codons in Tetrahymena mRNAs are translated correctly in a rabbit reticulocyte lysate supplemented with a subcellular fraction from Tetrahymena.
Biochem. J. 244, 331–335.
A. The set of control experiments argues convincingly that the association
between DnaK and the labeled proteins is meaningful; that is, it reflects
some biological function.
In the presence of SDS, which eliminates protein–protein interactions,
antibodies precipitate only DnaK (see Figure 6–37A, lane 2), suggesting that
protein–protein associations are required for precipitation of the labeled
proteins. The absence of labeled proteins from DnaK-deletion cells (see lane
3) indicates that precipitation depends on DnaK and is not the result, for
example, of nonspecific association with the antibodies. The lack of precipitation of labeled proteins from a mixture of labeled DnaK-deletion cells and
unlabeled wild-type cells (see lane 4), argues that the associations of proteins with DnaK were established in cells and not during subsequent experimental procedures.
B. ATP would be expected to interfere with precipitation of labeled proteins if
Hsp70 used it in the normal way; that is, to power the cycling of Hsp70 on
and off the protein. In the absence of ATP, DnaK will have hydrolyzed a
bound molecule of ATP—that it acquired in the cell—to ADP, altering its own
conformation and allowing it to latch onto a hydrophobic patch in a nascent
protein. If ATP is present in the extract, it will displace the ADP, reversing the
conformational change and releasing DnaK from the nascent protein. In the
more dilute conditions in the extract, the presence of ATP greatly favors the
off reaction, and as a result labeled proteins are not precipitated.
C. The pulse-chase experiment in Figure 6–37B indicates that DnaK binds the
labeled proteins only for a few minutes after 35S-methionine has been incorporated. A natural interpretation of this experiment is that some of the labeled
proteins are incorrectly folded initially, and bind DnaK as a consequence.
Antibodies against DnaK precipitate these bound proteins. With time and
with help from DnaK, the proteins correctly fold and are no longer substrates
for DnaK binding; hence, they disappear from the immunoprecipitates.
D. Nothing in these experiments shows directly that the proteins bound by
DnaK are in the process of being translated on ribosomes. The very short A143 A144 Chapter 6: How Cells Read the Genome: From DNA to Protein
duration of the labeling pulse (15 seconds) would be expected to label proteins in the process of being translated, but it would also label proteins that
were completed during the pulse and, therefore, clear of the ribosome. The
category of protein that DnaK binds is not clear from these experiments. In
additional experiments the authors show convincingly that some of the proteins that are bound by DnaK were indeed attached to ribosomes.
References: Teter SA, Houry WA, Ang D, Tradler T, Rockabrand D, Fischer G,
Blum P Georgopoulos C & Hartl FU (1999) Polypeptide flux through bacte,
rial Hsp70: DnaK cooperates with trigger factor in chaperoning nascent
chains. Cell 97, 755–765.
Deuerling E, Schulze-Specking A, Tomoyasu T, Mogk A & Bukau B (1999)
Trigger factor and DnaK cooperate in folding of newly synthesized proteins.
Nature 400, 693–696. 6–95
A. When IPTG is present, the strains that do not express TF (DTig) or DnaK (IDnaK) grow as well as the wild-type strain at all temperatures. When IPTG is
absent, the DTig strain continues to grow whereas, in contrast, the I-DnaK
strain does not do so at either 15°C or 42°C, although it grows fine at the
intermediate temperatures. Thus, DnaK appears to be the more critical
chaperone under the stressful conditions of high and low temperature.
B. Unlike the single mutants, the double mutant does not grow at any temperature in the absence of both TF and DnaK (that is, when there is no IPTG in
the plate). The lethality of the double mutant at 30°C and 37°C, where both
of the single mutants grow perfectly well, is termed synthetic lethality; it
indicates that the two gene products cooperate in some way, which need not
be direct. Because both of these genes encode Hsp70 chaperones, it seems
likely that they collaborate or cover for one another in some way in protein
folding. DnaK can fully compensate for the loss of TF throughout the temperature ranges tested in these experiments, and TF can compensate for the
loss of DnaK at intermediate, but not extreme, temperatures. When neither
chaperone is functional, misfolding of proteins even at intermediate temperatures is lethal to the cells. Additional experiments demonstrated that in
the absence of both chaperones cytosolic proteins undergo massive aggregation.
References: Deuerling E, Schulze-Specking A, Tomoyasu T, Mogk A & Bukau
B (1999) Trigger factor and DnaK cooperate in folding of newly synthesized
proteins. Nature 400, 693–696.
Teter SA, Houry WA, Ang D, Tradler T, Rockabrand D, Fischer G, Blum P,
Georgopoulos C & Hartl FU (1999) Polypeptide flux through bacterial Hsp70:
DnaK cooperates with trigger factor in chaperoning nascent chains. Cell 97,
A. The results suggest that tritium exchange occurs within one cycle. Although
there is time for multiple cycles—and they presumably occur in the presence of ATP—the results with AMPPNP indicate that a single cycle is sufficient. Because AMPPNP cannot be hydrolyzed, the chaperone will not be
able to eject the protein and repeat the cycle.
B. The accelerated exchange of tritium in the presence of the chaperone and
ATP indicates that the protein is unfolded before it is refolded. The isolationchamber model starts from the premise that aggregation limits the folding
of a protein. If the cavity of GroEL facilitated proper folding by reducing
inappropriate interactions, it would not seem essential that the protein
first be unfolded. By contrast, if a stable but incorrectly folded domain
blocked correct folding, the protein would, by necessity, have to be
unfolded first. Thus, the results with the particular protein used in these
experiments, which was the plant CO2-fixation protein, RuBisCo, support
an active unfolding model, which is powered not by ATP hydrolysis but by FROM RNA TO PROTEIN
ATP binding. More recent experiments indicate that accurate refolding of
RuBisCo occurs inside the GroEL chamber, which in some way helps the
protein to avoid misfolded intermediates by encouraging it along the productive folding pathway.
References: Shtilerman M, Lorimer GH & Englander SW (1999) Chaperonin
function: Folding by forced unfolding. Science 284, 822–825.
Brinker A, Pfeifer G, Kerner MJ, Naylor DJ, Hartl FU & Hayer-Hartl M (2001)
Dual function of protein confinement in chaperonin-assisted protein folding. Cell 107, 223–233.
A. In this experiment the plateau value for the radioactive proteins in the
absence of proteasome inhibitors was about 30% lower than it was in the
presence of inhibitors, suggesting that 30% of newly synthesized proteins
are degraded in proteasomes in lymph node cells. Similar results in other
cell types suggest that substantial degradation of newly synthesized proteins
may be common in most cells.
B. The absence of differential affects on specific proteins is surprising. One
possibility is that a constant fraction (about 30% of all proteins) misfolds
and is degraded by proteasomes; however, given the variety of proteins it
seems unlikely that they would all misfold to the same degree. Another possibility is that newly synthesized proteins are sampled randomly for degradation to serve some other biological function. (One such function is to provide an array of peptides for display on the cell surface to inform the
immune system of the proteins the cell is currently making—which is
thought to give it a head start in identifying infected cells.) Alternatively, it
may be that ribosomes make a high fraction of mistakes (30%) in the form of
peptide fragments and misincorporations that cannot fold properly and are
normally removed. These errors would contribute to a background of
radioactivity throughout the gel, accounting for an increased overall intensity in the absence of proteasomes.
Reference: Schubert U, Antón LC, Gibbs J, Norbury CC, Yewdell JW & Bennink JR (2000) Rapid degradation of a large fraction of newly synthesized
proteins by proteasomes. Nature 404, 770–774.
6–98 Overexpression of dm-N70 does not cause accumulation of any of the proteasome substrates. That is the expected result if the D-box is critical for its
interaction with APC. Because it doesn’t bind to APC, it doesn’t influence
Overexpression of K0-N70 gives the result that was initially anticipated: Dbox proteins accumulate, but non-D-box proteins do not. This result suggests that K0-N70 specifically interferes with destruction of D-box proteins,
probably by competing for binding to APC.
The most difficult result to understand is the original one: overexpression
of N70 causes accumulation of both D-box and non-D-box proteins.
Because removal of its lysines eliminates this effect, it seems likely that the
effect is caused by ubiquitylation of N70. Overexpression of N70 and its
ubiquitylation is thought to sequester a large fraction of the cellular supply
of ubiquitin; thus, interfering with ubiquitylation of other proteins by
decreasing the availability of ubiquitin.
Reference: Yamano H, Tsurumi C, Gannon J & Hunt T (1998) The role of the
destruction box and its neighbouring lysine residues in cyclin B for
anaphase ubiquitin-dependent proteolysis in fission yeast: defining the Dbox receptor. EMBO J. 17, 5670–5678. 6–99
A. Although the first codon of b-galactosidase could have been changed by
recombinant DNA techniques, it would no longer have served as a start site
for translation. All proteins, bacterial and eucaryotic, are initially translated A145 A146 Chapter 6: How Cells Read the Genome: From DNA to Protein
with methionine at their N-termini. In many cases methionine is removed
(and occasionally additional amino acids as well), leaving a new N-terminus.
The procedure described here was arrived at by chance! The investigators
were originally interested in whether ubiquitin at the N-terminus would
cause a protein to be degraded. This question led them to generate the
fusion gene. In bacteria, which do not have a ubiquitin-dependent protease,
the fusion protein was made as they anticipated; however, in yeast the same
plasmid produced only b-galactosidase, suggesting that the ubiquitin was
removed. To try to prevent this cleavage, they altered the codons at the junction. The ubiquitin was still removed, but now the resulting b-galactosidases
differed remarkably in stability. The focus of their study quickly changed,
leading to insights into the role of the N-terminus in determining the stability of proteins.
B. The half-lives of the different b-galactosidases can be estimated from the
graph in Figure 6–44B by finding the time at which half the b-galactosidase
remains. The three b-galactosidases have very different half-lives: R-b-galactosidase has a half-life of about 2 minutes; I-b-galactosidase has a half-life of
about 30 minutes; and M-b-galactosidase has a half-life that is too long to be
measured in this experiment (it was estimated to be greater than 20 hours).
Reference: Bachmair A, Finley D & Varshavsky A (1986) In vivo half-life of a
protein is a function of its amino-terminal residue. Science 234, 179–186. 6–100
A. The absence of radioactivity at the position of b-galactosidase in Figure 6–45
indicates that the labeled antibodies against ubiquitin do not react with the
protein at that position. Thus, the band that is marked b-galactosidase in
Figure 6–44A does not carry any attached ubiquitin, indicating that ubiquitin was removed from the N-terminus of the fusion protein.
B. The ubiquitin above the position of b-galactosidase in Figure 6–45 must be
attached to b-galactosidase, since the enzyme was purified by binding to
antibodies specific for b-galactosidase. Ubiquitin attached to the two unstable enzymes—but not to the stable enzyme—suggests that ubiquitin marks
the protein for degradation. The ladderlike appearance of the bands suggests that a variable number of copies of ubiquitin are attached to each protein (at lysines) before the enzyme is degraded.
Reference: Bachmair A, Finley D & Varshavsky A (1986) In vivo half-life of a
protein is a function of its amino-terminal residue. Science 234, 179–186. THE RNA WORLD AND THE ORIGINS OF LIFE
6–101 RNA world TRUE/FALSE
6–102 False. Although only a few types of reactions are represented among the
ribozymes in present-day cells, ribozymes that have been selected in the
laboratory can catalyze a wide variety of biochemical reactions, with reaction rates similar to those of proteins. In light of these results, it is unclear
why ribozymes are so underrepresented in modern cells. It seems likely that
the availability of 20 amino acids versus four bases affords proteins with a
greater number of catalytic strategies than ribozymes, as well as endowing
them with the ability to bind productively to a wider range of substrates (for
example, hydrophobic substrates, which ribozymes have difficulty with). A147 THE RNA WORLD AND THE ORIGINS OF LIFE THOUGHT PROBLEMS
6–103 RNA has the ability to store genetic information like DNA and the ability to
catalyze chemical reactions like proteins. Having both of these essential features of ‘life’ in a single type of molecule makes it easier to understand how
life might have arisen from nonliving matter. The use of RNA molecules as
catalysts in several fundamental reactions in modern-day cells supports this
idea. Nevertheless, it is not yet possible to specify a plausible pathway from
the ‘primordial’ soup to an RNA world, and many have speculated that there
may have been a precursor molecule to RNA—one that also had catalytic
and informational properties. 6–104 Although RNA is thought to have played an important role in the evolution
of life on Earth, possibly as a replicating catalyst, it is unclear that it was the
first replicating catalyst. Other less efficient molecular systems that combined informational and catalytic properties may have preceded RNA.
Regardless of its original role, it is clear that RNA now plays a larger role than
that of mere messenger in information flow; RNA provides critical functions
in replication, gene regulation, splicing, translation, peptide-bond formation, membrane transport of proteins, and telomere maintenance. 6–105 The RNA molecule will not be able to catalyze its own replication. As a single molecule with a single catalytic site, it cannot be both template and catalyst simultaneously. (To visualize the critical difficulty, try to imagine how
the active site of the RNA could copy itself.) Once a second molecule—either
template or catalyst—was generated, then replication could begin.
Reference: Bartel DP & Szostak JW (1993) Isolation of new ribozymes from a
large pool of random sequences. Science 261, 1411–1418. 6–106 The complement of this hairpin RNA could also form a similar hairpin, as
shown in Figure 6–62. The two structures would be identical in the doublestranded regions that involved standard GC and AU base pairs. They would
differ in the sequence of the single-stranded regions. Because GU base pairs
are stable in RNA, whereas CA base pairs are not, one hairpin would be predicted to contain an additional base pair, as shown. 6–107 Compartments are essential for evolution in the RNA world for two reasons.
First, a set of mutually beneficial RNA molecules would have had to remain
in proximity to have been of any use to one another. Second, selection of a
set of RNA molecules according to the quality of the self-replicating systems
they generated—the basis for natural selection and evolution—could not
have occurred efficiently until some form of compartment evolved to contain the molecules and thereby make them available only to the RNA that
had generated them.
Reference: Szostak JW, Bartel DP & Luisi L (2001) Synthesizing life. Nature
409, 387–390. 6–108 The deoxyribose sugar of DNA makes the molecule much less susceptible to
breakage. The hydroxyl group on carbon 2 of the ribose sugar is an agent for
catalysis of the adjacent 3¢-5¢ phosphodiester bond that links nucleotides
together in RNA. Its absence from DNA eliminates that mechanism of chain
breakage. In addition, the double helical structure of DNA provides two
complementary strands, which allows damage in one strand to be repaired
accurately by reference to the sequence of the second strand. Finally, the use
of T in DNA instead of U, as in RNA, builds in a protection against the effects
of deamination—a common form of damage. Deamination of T produces an
aberrant base (methyl C), whereas deamination of U generates C, a normal
base. The cell’s job of recognizing damaged bases is much easier when the
damage produces an abnormal base. 6–109
A. Ligation of the substrate oligonucleotide to the pool RNA is analogous to C-U
5¢-G-C-A C-C-G 3¢-C-G-U G-G-C U A-C 5¢-GCACUCCGUCGGCAUGC-3¢
5¢-G-C-A-U C-C-G 3¢-C-G-U-G G-G-C A A Figure 6–62 Hairpins formed by an RNA
strand and by its complement (Answer
6–106). An RNA and its complement are
shown as double-stranded RNA in the
middle. The structures formed by each
strand are shown above and below the
duplex. The nonstandard GU base pair in
the lower hair pin is highlighted with a
dashed box. A148 Chapter 6: How Cells Read the Genome: From DNA to Protein
(A) POLYMERIZATION (B) LIGATION Figure 6–63 Similarity of polymerization
and ligation (Answer 6–109). (A) RNA
polymerization. (B) RNA ligation.
Analogous parts of the reactions are
3¢ template 5¢ 3¢ PPi 3¢-5¢ bond PPi G-C
C-G primer template 3¢-5¢ bond 5¢
3¢ B. C. D. E. 5¢ G-C
3¢ chain elongation during RNA polymerization (Figure 6–63). In both cases,
the growing strand (primer) and the nucleoside triphosphate (NTP) or its
analog base-pairs to a template. In both cases, the 3¢ hydroxyl of the growing
strand attacks the a-phosphate of the 5¢ triphosphate and displaces
pyrophosphate (PPi) with concomitant formation of a 3¢-5¢ phosphodiester
bond (Figure 6–63).
It is critical to the selection and amplification scheme that the catalytic RNA
becomes attached to the tag. The tag is used to fish out specific RNA
molecules from the large pool of random molecules. If the tag were not
attached to the ribozyme that catalyzed the linkage, no selection and amplification of the relevant ribozyme (the point of the whole scheme) would
The random segment in the middle is the part of the molecule that guarantees that a very large number of different sequences—hence, conformations
and catalytic activities—will be present in the starting pool. It is your hope
that one or a few such molecules can catalyze the intended reaction.
The constant regions at the ends of each pool RNA molecule serve different purposes. The constant region at the 5¢ end of the pool RNAs serves as a
binding site for the substrate oligonucleotide, so that the ends can be juxtaposed to create the substrate for ligation. This constant region also serves as
one site required for regenerating a pool of RNA by T7 RNA polymerase transcription. This is an essential step if the cycle of selection and amplification
is to be repeated. The constant region at the 3¢ end of the pool RNAs serves
as a site for attachment to the agarose bead for ease of manipulation, for
specific amplification of linked substrate and catalytic RNAs, and for amplification to link the T7 promoter so that the DNA oligonucleotides can be
reconverted to RNA oligonucleotides for subsequent cycles.
A catalytic RNA molecule is selected by passing the pool of RNA through an
affinity column that carries oligonucleotides that are complementary to the
substrate oligonucleotide. Only in those molecules that have undergone a
ligation reaction will the catalytic RNA be attached to the substrate. The vast
majority of noncatalytic RNAs will pass through such an affinity column.
When the RNA is eluted from the column, it will contain a mixture of the
sought-after catalytic RNA and contaminating noncatalytic RNA. The catalytic RNA can be specifically amplified using PCR primers, one of which is
specific for the substrate RNA and the other for the pool RNA. Such a pair of
PCR primers will selectively amplify catalytic RNAs, because only the catalytic RNAs will be attached to the substrate RNAs and be amplified.
Even assuming that one cycle of selection and amplification is sufficient to
remove all contaminating noncatalytic RNA molecules, which is probably
not the case, there is still a critical reason for carrying out multiple cycles of
selection and amplification. In the starting pool of RNA molecules it is THE RNA WORLD AND THE ORIGINS OF LIFE
unlikely that any molecule will be represented more than once. Thus, at the
end of the first time period for ligation, the very best catalyst in the population, many much weaker catalysts, and even some noncatalytic RNAs that
are linked by an uncatalyzed mechanism, will all be attached to the substrate RNA. They will all be represented equally in the amplified pool. Purification at this stage would yield an extensive mixture of RNA molecules with
a very wide range of catalytic activities.
Subsequent rounds of selection and amplification allow the best catalysts
to win out over the weaker ones. Consider, for example, the second cycle. In
the window for ligation, most of the good catalysts will attach themselves to
the substrate, while many fewer of the weaker catalysts and essentially none
of the noncatalytic RNAs will do so. Thus, the amplification step in the second cycle will enrich considerably for the better catalysts. By decreasing the
time for ligation in subsequent cycles, better and better catalysts can be
Reference: Bartel DP & Szostak JW (1993) Isolation of new ribozymes from a
large pool of random sequences. Science 261, 1411–1418. CALCULATIONS
A. There are 6 ¥ 1015 molecules, 300 nucleotides (nt) in length, in 1 mg of RNA.
6 ¥ 1020 d
1 RNA molecule
= 6 ¥ 1015 RNA molecules number = B. If the 220 nucleotide segment were completely random, there would be four
choices of nucleotide at each of 220 positions, which is 4220 or about 3 ¥ 10132
possible different RNA molecules. Thus, in a 1-mg sample, there will be 2 ¥
10–117 [(6 ¥ 1015)/(3 ¥ 10132)] of all possible sequences represented…a trivial
fraction of the whole. (A sample large enough to have one copy of each possible RNA would outweigh the known universe by more than 30 orders of
C. If a single 50-nucleotide RNA were required to catalyze the ligation, your
chances of success would be close to nil. There are about 1018 different 50nucleotide sequences represented in a 1-mg sample of RNA. Considering
just the random 220 nucleotides, there would be about 170 different 50-mers
in each of 6 ¥ 1015 molecules (imagine sliding a 50-nucleotide window
across the 220 nucleotides one nucleotide at a time) for a total of 170 ¥ 6 ¥
1015, or 1018 different molecules. Since there are 450 or about 1030 (45 @ 103)
different 50-mers, your chances would be roughly 1 in a trillion (10–12) of
having the unique catalytic RNA in your sample. Are you feelin’ lucky?
That so many ribozymes have been successfully isolated from such pools
argues that a very large number of different sequences must be able to catalyze any given reaction, or that the catalytic RNAs must be very small. Since
the identified ribozymes are not particularly small, it must be that many different sequences are capable of catalysis.
Reference: Bartel DP & Szostak JW (1993) Isolation of new ribozymes from a
large pool of random sequences. Science 261, 1411–1418. DATA HANDLING
A. Error-prone PCR was used to introduce mutations into the pool of RNA
molecules in some rounds in order to try to generate ever more efficient catalysts of ligation. Because all possible molecules cannot be present in the
starting material (see Problem 6–110), this technique gives you a way to
increase the diversity of molecules that are closely related to those with A149 A150 Chapter 6: How Cells Read the Genome: From DNA to Protein
demonstrated catalytic activity. It is likely that better catalysts will be found
in the ‘sequence neighborhood’ of existing catalysts. You waited until round
5 to apply error-prone PCR to give time for some moderately good catalysts
B. By making ligation more and more difficult—by lowering the concentration
of Mg2+ and by decreasing the time available for ligation—you are selecting
for better and better catalysts.
C. Your scheme for selection and amplification has improved the ligation rate
about 3 million-fold from 0.000003 ligations per hour for the starting RNA
pool to 8.0 per hour after round 10. Thus, your final pool of ribozymes catalyzes ligation about 3 ¥ 106-fold faster than the uncatalyzed reaction.
D. The diversity evident in your round-10 pool of RNA molecules indicates that
many sequences can carry out efficient ligation. Since 11 of 15 of the cloned
and sequenced molecules are clearly similar, they form a single sequence
family presumably with a very similar overall conformation. The other
molecules may represent additional catalytically active conformations. You
will tell your audience that additional structural and enzymological studies
will be needed to determine the catalytic mechanism(s) represented in your
pool of ribozymes.
Reference: Bartel DP & Szostak JW (1993) Isolation of new ribozymes from a
large pool of random sequences. Science 261, 1411–1418. ...
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