Document Text Contents
Page 1
8
RNA: Transcription and Processing
WORKING WITH THE FIGURES
1. In Figure 8-3, why are the arrows for genes 1 and 2 pointing in opposite
directions?
Answer: The arrows for genes 1 and 2 indicate the direction of transcription,
which is always 5 to 3. The two genes are transcribed from opposite DNA
strands, which are antiparallel, so the genes must be transcribed in opposite
directions to maintain the 5 to 3 direction of transcription.
2. In Figure 8-5, draw the “one gene” at much higher resolution with the following
components: DNA, RNA polymerase(s), RNA(s).
Answer: At the higher resolution, the feathery structures become RNA
transcripts, with the longer transcripts occurring nearer the termination of the
gene. The RNA in this drawing has been straightened out to illustrate the
progressively longer transcripts.
3. In Figure 8-6, describe where the gene promoter is located.
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Chapter Eight 271
Answer: The promoter is located to the left (upstream) of the 3 end of the
template strand. From this sequence it cannot be determined how far the
promoter would be from the 5 end of the mRNA.
4. In Figure 8-9b, write a sequence that could form the hairpin loop structure.
Answer: Any sequence that contains inverted complementary regions separated
by a noncomplementary one would form a hairpin. One sequence would be:
ACGCAAGCUUAC CGAUUAUUGUAAGCUUGAAG
The two bold-faced sequences would pair and form a hairpin. The intervening
non-bold sequence would be the loop.
5. How do you know that the events in Figure 8-13 are occurring in the nucleus?
Answer: The figure shows a double-stranded DNA molecule from which RNA
is being transcribed. This process only occurs in the nucleus.
6. In Figure 8-15, what do you think would be the effect of a G to A mutation in
the first G residue of the intron?
Answer: A mutation of G to A would alter the U1 SNP binding site and prevent
formation of the spliceosome. This would prevent splicing of the intron.
7. In Figure 8-23, show how the double-stranded RNA is able to silence the
transgene. What would have to happen for the transgene to also silence the
flanking cellular gene (in yellow)?
Answer:
a. The double-stranded RNA formed from the sense and antisense transgene
transcripts would be processed by Dicer, then bind to RISC. RISC would
separate the dsRNA to produce an antisense RNA/RISC complex. The
RISC/RNA complex would bind to the transgene mRNA and deactivate it.
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Chapter Eight 275
c. Processing takes place in the cytoplasm.
d. Termination is accomplished by the use of a hairpin loop or the use of the
rho factor.
e. Many RNAs can be transcribed simultaneously from one DNA template.
Answer:
a. False. Sigma factor is required in prokaryotes, not eukaryotes.
b. True. Processing begins at the 5´ end, while the 3´ end is still being
synthesized.
c. False. Processing occurs in the nucleus and only mature RNA is transported
out to the cytoplasm.
d. False. Hairpin loops or rho factor (in conjunction with the rut site) is used to
terminate transcription in prokaryotes. In eukaryotes the conserved
sequences AAUAAA or AUUAAA, near the 3´ end of the transcript, are
recognized by an enzyme that cuts off the end of the RNA approximately
20 bases downstream.
e. True. Multiple RNA polymerases may transcribe the same template
simultaneously.
17. A researcher was mutating prokaryotic cells by inserting segments of DNA. In
this way, she made the following mutation:
Original TTGACAT 15 to 17 bp TATAAT
Mutant TATAAT 15 to 17 bp TTGACAT
a. What does this sequence represent?
b. What do you predict will be the effect of such a mutation? Explain.
Answer:
a. The original sequence represents the –35 and –10 consensus sequences
(with the correct number of intervening spaces) of a bacterial promoter.
Sigma factor, as part of the RNA polymerase holoenzyme, recognizes and
binds to these sequences.
b. The mutated (transposed) sequences would not be a binding site for sigma
factor. The two regions are not in the correct orientation to each other and
therefore would not be recognized as a promoter.
18. You will learn more about genetic engineering in Chapter 10, but for now, put
on your genetic engineer’s cap and try to solve this problem. E. coli is widely
used in laboratories to produce proteins from other organisms.
a. You have isolated a yeast gene that encodes a metabolic enzyme and want
to produce this enzyme in E. coli. You suspect that the yeast promoter will
not work in E. coli. Why?
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Chapter Eight 276
b. After replacing the yeast promoter with an E. coli promoter, you are pleased
to detect RNA from the yeast gene but are confused because the RNA is
almost twice the length of the mRNA from this gene isolated from yeast.
Explain why this result might have occurred.
Answer:
a. The promoters of eukaryotes and prokaryotes do not have the same
conserved sequences. In yeast, the promoter would have the required TATA
box located about –30, whereas bacteria would have conserved sequences at
–35 and –10 that would interact with sigma factor as part of the RNA
polymerase holoenzyme.
b. There are two possible reasons that the mRNA is longer than expected.
First, many eukaryotic genes contain introns, and bacteria would not have
the splicing machinery necessary for their removal. Second, termination of
transcription is not the same in bacteria and yeast; the sequences necessary
for correct termination in E. coli would not be expected in the yeast gene.
19. Draw a prokaryotic gene and its RNA product. Be sure to include the promoter,
transcription start site, transcription termination site, untranslated regions, and
labeled 5′ and 3′ ends.
Answer:
transcription start
promoter
5´ UT
3´ UT
translation start
translation stop termination site
( )
( )
5´
3´
mRNA
DNA
20. Draw a two-intron eukaryotic gene and its pre-mRNA and mRNA products. Be
sure to include all the features of the prokaryotic gene included in your answer
to Problem 19, plus the processing events required to produce the mRNA.
Answer:
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Chapter Eight 280
and is thus equivalent to “knocking out” the gene. To test whether a specific
mRNA encodes an essential embryonic protein, eggs or very early embryos
should be injected with the double-stranded RNA produced from your mRNA,
thus activating the RNAi pathway. The effects of knocking out the specific gene
product can then be followed by observing what happens in these, versus
control, embryos. If the encoded protein is essential, embryonic development
should be perturbed when your gene is silenced.
27. Glyphosate is an herbicide used to kill weeds. It is the main component of a
product made by the Monsanto Company called Roundup. Glyphosate kills
plants by inhibiting an enzyme in the shikimate pathway called EPSPS. This
herbicide is considered safe because animals do not have the shikimate
pathway. To sell even more of their herbicide, Monsanto commissioned its plant
geneticists to engineer several crop plants, including corn, to be resistant to
glyphosate. To do so, the scientists had to introduce an EPSPS enzyme that was
resistant to inhibition by glyphosate into crop plants and then test the
transformed plants for resistance to the herbicide.
Imagine that you are one of these scientists and that you have managed to
successfully introduce the resistant EPSPS gene into the corn chromosomes.
You find that some of the transgenic plants are resistant to the herbicide,
whereas others are not. Your supervisor is very upset and demands an
explanation of why some of the plants are not resistant even though they have
the transgene in their chromosomes. Draw a picture to help him understand.
Answer: Transgene silencing is a common phenomenon in plants. Silencing
may occur at the transcriptional or post-transcriptional level. Since you cannot
control where transgenes insert, some may insert into transcriptionally inactive
parts of the genome. Post-transcriptional silencing may be the result of
activation of the RNAi pathway due to the misexpression of both strands of
your transgene. See Figure 8-22 in the companion text as one example of how
this can happen. In these cases, the RNAi pathway will be activated and the
EPSPS gene product will be silenced.
28. Many human cancers result when a normal gene mutates and leads to
uncontrolled growth (a tumor). Genes that cause cancer when they mutate are
called oncogenes. Chemotherapy is effective against many tumors because it
targets rapidly dividing cells and kills them. Unfortunately, chemotherapy has
many side effects, such as hair loss or nausea, because it also kills many of our
normal cells that are rapidly dividing, such as those in the hair follicles or
stomach lining.
Many scientists and large pharmaceutical companies are excited about the
prospects of exploiting the RNAi pathway to selectively inhibit oncogenes in
life-threatening tumors. Explain in very general terms how gene-silencing
Page 12
Chapter Eight 281
therapy might work to treat cancer and why this type of therapy would have
fewer side effects than chemotherapy.
Answer: RNAi has the potential to selectively prevent protein production from
any targeted gene. Oncogenes are mutant versions of “normal” genes (called
proto-oncogenes) and their altered gene products are partly or wholly
responsible for causing cancer. In theory, it may be possible to design
appropriate siRNA molecules (small interfering RNAs) that specifically silence
the mutant oncogene product but do not silence the closely related proto-
oncogene product. The latter is necessary to prevent serious side effects as the
products of proto-oncogenes are essential for normal cellular function.
8
RNA: Transcription and Processing
WORKING WITH THE FIGURES
1. In Figure 8-3, why are the arrows for genes 1 and 2 pointing in opposite
directions?
Answer: The arrows for genes 1 and 2 indicate the direction of transcription,
which is always 5 to 3. The two genes are transcribed from opposite DNA
strands, which are antiparallel, so the genes must be transcribed in opposite
directions to maintain the 5 to 3 direction of transcription.
2. In Figure 8-5, draw the “one gene” at much higher resolution with the following
components: DNA, RNA polymerase(s), RNA(s).
Answer: At the higher resolution, the feathery structures become RNA
transcripts, with the longer transcripts occurring nearer the termination of the
gene. The RNA in this drawing has been straightened out to illustrate the
progressively longer transcripts.
3. In Figure 8-6, describe where the gene promoter is located.
Page 2
Chapter Eight 271
Answer: The promoter is located to the left (upstream) of the 3 end of the
template strand. From this sequence it cannot be determined how far the
promoter would be from the 5 end of the mRNA.
4. In Figure 8-9b, write a sequence that could form the hairpin loop structure.
Answer: Any sequence that contains inverted complementary regions separated
by a noncomplementary one would form a hairpin. One sequence would be:
ACGCAAGCUUAC CGAUUAUUGUAAGCUUGAAG
The two bold-faced sequences would pair and form a hairpin. The intervening
non-bold sequence would be the loop.
5. How do you know that the events in Figure 8-13 are occurring in the nucleus?
Answer: The figure shows a double-stranded DNA molecule from which RNA
is being transcribed. This process only occurs in the nucleus.
6. In Figure 8-15, what do you think would be the effect of a G to A mutation in
the first G residue of the intron?
Answer: A mutation of G to A would alter the U1 SNP binding site and prevent
formation of the spliceosome. This would prevent splicing of the intron.
7. In Figure 8-23, show how the double-stranded RNA is able to silence the
transgene. What would have to happen for the transgene to also silence the
flanking cellular gene (in yellow)?
Answer:
a. The double-stranded RNA formed from the sense and antisense transgene
transcripts would be processed by Dicer, then bind to RISC. RISC would
separate the dsRNA to produce an antisense RNA/RISC complex. The
RISC/RNA complex would bind to the transgene mRNA and deactivate it.
Page 6
Chapter Eight 275
c. Processing takes place in the cytoplasm.
d. Termination is accomplished by the use of a hairpin loop or the use of the
rho factor.
e. Many RNAs can be transcribed simultaneously from one DNA template.
Answer:
a. False. Sigma factor is required in prokaryotes, not eukaryotes.
b. True. Processing begins at the 5´ end, while the 3´ end is still being
synthesized.
c. False. Processing occurs in the nucleus and only mature RNA is transported
out to the cytoplasm.
d. False. Hairpin loops or rho factor (in conjunction with the rut site) is used to
terminate transcription in prokaryotes. In eukaryotes the conserved
sequences AAUAAA or AUUAAA, near the 3´ end of the transcript, are
recognized by an enzyme that cuts off the end of the RNA approximately
20 bases downstream.
e. True. Multiple RNA polymerases may transcribe the same template
simultaneously.
17. A researcher was mutating prokaryotic cells by inserting segments of DNA. In
this way, she made the following mutation:
Original TTGACAT 15 to 17 bp TATAAT
Mutant TATAAT 15 to 17 bp TTGACAT
a. What does this sequence represent?
b. What do you predict will be the effect of such a mutation? Explain.
Answer:
a. The original sequence represents the –35 and –10 consensus sequences
(with the correct number of intervening spaces) of a bacterial promoter.
Sigma factor, as part of the RNA polymerase holoenzyme, recognizes and
binds to these sequences.
b. The mutated (transposed) sequences would not be a binding site for sigma
factor. The two regions are not in the correct orientation to each other and
therefore would not be recognized as a promoter.
18. You will learn more about genetic engineering in Chapter 10, but for now, put
on your genetic engineer’s cap and try to solve this problem. E. coli is widely
used in laboratories to produce proteins from other organisms.
a. You have isolated a yeast gene that encodes a metabolic enzyme and want
to produce this enzyme in E. coli. You suspect that the yeast promoter will
not work in E. coli. Why?
Page 7
Chapter Eight 276
b. After replacing the yeast promoter with an E. coli promoter, you are pleased
to detect RNA from the yeast gene but are confused because the RNA is
almost twice the length of the mRNA from this gene isolated from yeast.
Explain why this result might have occurred.
Answer:
a. The promoters of eukaryotes and prokaryotes do not have the same
conserved sequences. In yeast, the promoter would have the required TATA
box located about –30, whereas bacteria would have conserved sequences at
–35 and –10 that would interact with sigma factor as part of the RNA
polymerase holoenzyme.
b. There are two possible reasons that the mRNA is longer than expected.
First, many eukaryotic genes contain introns, and bacteria would not have
the splicing machinery necessary for their removal. Second, termination of
transcription is not the same in bacteria and yeast; the sequences necessary
for correct termination in E. coli would not be expected in the yeast gene.
19. Draw a prokaryotic gene and its RNA product. Be sure to include the promoter,
transcription start site, transcription termination site, untranslated regions, and
labeled 5′ and 3′ ends.
Answer:
transcription start
promoter
5´ UT
3´ UT
translation start
translation stop termination site
( )
( )
5´
3´
mRNA
DNA
20. Draw a two-intron eukaryotic gene and its pre-mRNA and mRNA products. Be
sure to include all the features of the prokaryotic gene included in your answer
to Problem 19, plus the processing events required to produce the mRNA.
Answer:
Page 11
Chapter Eight 280
and is thus equivalent to “knocking out” the gene. To test whether a specific
mRNA encodes an essential embryonic protein, eggs or very early embryos
should be injected with the double-stranded RNA produced from your mRNA,
thus activating the RNAi pathway. The effects of knocking out the specific gene
product can then be followed by observing what happens in these, versus
control, embryos. If the encoded protein is essential, embryonic development
should be perturbed when your gene is silenced.
27. Glyphosate is an herbicide used to kill weeds. It is the main component of a
product made by the Monsanto Company called Roundup. Glyphosate kills
plants by inhibiting an enzyme in the shikimate pathway called EPSPS. This
herbicide is considered safe because animals do not have the shikimate
pathway. To sell even more of their herbicide, Monsanto commissioned its plant
geneticists to engineer several crop plants, including corn, to be resistant to
glyphosate. To do so, the scientists had to introduce an EPSPS enzyme that was
resistant to inhibition by glyphosate into crop plants and then test the
transformed plants for resistance to the herbicide.
Imagine that you are one of these scientists and that you have managed to
successfully introduce the resistant EPSPS gene into the corn chromosomes.
You find that some of the transgenic plants are resistant to the herbicide,
whereas others are not. Your supervisor is very upset and demands an
explanation of why some of the plants are not resistant even though they have
the transgene in their chromosomes. Draw a picture to help him understand.
Answer: Transgene silencing is a common phenomenon in plants. Silencing
may occur at the transcriptional or post-transcriptional level. Since you cannot
control where transgenes insert, some may insert into transcriptionally inactive
parts of the genome. Post-transcriptional silencing may be the result of
activation of the RNAi pathway due to the misexpression of both strands of
your transgene. See Figure 8-22 in the companion text as one example of how
this can happen. In these cases, the RNAi pathway will be activated and the
EPSPS gene product will be silenced.
28. Many human cancers result when a normal gene mutates and leads to
uncontrolled growth (a tumor). Genes that cause cancer when they mutate are
called oncogenes. Chemotherapy is effective against many tumors because it
targets rapidly dividing cells and kills them. Unfortunately, chemotherapy has
many side effects, such as hair loss or nausea, because it also kills many of our
normal cells that are rapidly dividing, such as those in the hair follicles or
stomach lining.
Many scientists and large pharmaceutical companies are excited about the
prospects of exploiting the RNAi pathway to selectively inhibit oncogenes in
life-threatening tumors. Explain in very general terms how gene-silencing
Page 12
Chapter Eight 281
therapy might work to treat cancer and why this type of therapy would have
fewer side effects than chemotherapy.
Answer: RNAi has the potential to selectively prevent protein production from
any targeted gene. Oncogenes are mutant versions of “normal” genes (called
proto-oncogenes) and their altered gene products are partly or wholly
responsible for causing cancer. In theory, it may be possible to design
appropriate siRNA molecules (small interfering RNAs) that specifically silence
the mutant oncogene product but do not silence the closely related proto-
oncogene product. The latter is necessary to prevent serious side effects as the
products of proto-oncogenes are essential for normal cellular function.
