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LS1a Problem Set #6
Due Friday 11/10 at noon in your TF’s drop box on the 2nd floor of the Science Center
all questions including the (*extra*) ones should be turned in
1. (18 points) Transcription in both eukaryotes and prokaryotes generates a
complementary strand of RNA to a gene on the template strand of DNA. However,
eukaryotic and prokaryotic genes are organized differently. One key difference can
be inferred from this electron micrograph, which shows newly formed eukaryotic
mRNA hybridized to the strand of genomic DNA that it was transcribed from:
*On the left is the actual electron micrograph. On the right is a schematic of the
micrograph where DNA is labeled red and RNA blue.
a) Explain the presence of this loop of DNA.
(4 points) This loop results from an intron being present in the DNA. It
has been removed from the mRNA so the DNA bases from this intron
region do not have a corresponding sequence in the mRNA to hybridize
with.
b) If this experiment was repeated using RNA transcribed from a prokaryotic gene
and the prokaryotic genomic DNA, would you expect the result to be the same or
different? Explain.
(4 points) The result should be different. Prokaryotic RNA transcripts do
not have introns and do not require processing. Thus the sequence of the
RNA is virtually identical to the DNA coding sequence (with the exception
of U’s replacing the T’s), and the RNA would anneal perfectly with no loops
to the template (non-coding) DNA strand.
c) List three differences between prokaryotic and eukaryotic mRNAs.
(6 points) Eukaryotic mRNAs have a 5’ cap, 3’ polyA tail and have splicedout introns. Prokaryotic RNAs do not have any of these features.
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d) Based on your understanding of how genes are organized in the HIV genome,
what result would you expect if this experiment were performed using HIV mRNA
hybridized to a strand of the integrated DNA genome? Explain.
(4 points) The result would depend upon which HIV RNA transcript is
used. If an early HIV transcript that has been spliced into the 4 kb form
was used, the result would look like the lefthand image above as the HIV
proviral DNA would have to loop out where the introns are. If a late,
unspliced HIV RNA is used, it would anneal to the proviral DNA without
any loops as all of the introns are still present in the RNA. For more
accurate pictures, see the answer to part e.
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e) (*extra*) Using the map of the HIV genome from Rob’s slides (Shown below).
Draw the hybridization of fully processed pol, rev, tat, env mRNAs to a strand of
the integrated HIV genome.
Note: The purpose of this question was to make students appreciate how
complex the HIV genome is. They do not need to know which RNAs encode
which genes.
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2. (18 points) After hearing Rob discuss how HIV enhances the transcription of its
RNA through the use of TAT and TAR, you attempt to utilize this feature to increase
the transcription of other genes as well. Below is the RNA sequence of TAR:
5’ - GGGUCUCUCUGGUUAGACCAGAUCUGAGCCUGGGAGCUCUCUGGCUAACUAGGGAACCCA - 3’
a) You have a company synthesize a DNA oligonucleotide containing the TAR
sequence (shown below). The sequences in bold have been added. The
addition of these sequences does not prevent TAR from functioning properly.
Briefly explain why the bold sequences do not prevent TAR from functioning.
5’ - GAATTCGGGTCTCTCTGGTTAGACCAGATCTGAGCCTGGGAGCTCTCTGGCTAACTAGGGAACCCAGAATTC - 3’
TAR
(4 points) These bold sequences do not disrupt the basepairing of the TAR
RNA to form a hairpin loop structure. Additionally, they can also basepair
and could contribute to the formation of the hairpin loop.
b) You connect the 3’ end of the synthesized DNA oligonucleotide (shown above) to
beads. You then add purified TAT protein and determine if the protein will bind
to the DNA oligonucleotide coated beads. You find that no TAT protein binds to
the TAR DNA sequence. Explain.
(4 points) TAT only binds to the TAR sequence in RNA. It does not bind to
the TAR sequence in DNA.
c) You insert the synthesized DNA oligonucleotide containing the TAR sequence at
the 5’-end of DNA encoding GFP. When this fusion gene is expressed in TAT
expressing cells versus cells that do not express TAT, an increased level of GFP
fluorescence is seen in the TAT expressing cells. Why is TAT able to bind and
enhance transcription in these cells? Was this expected in light of the result in
part b).
(5 points) When this fusion gene is expressed in cells that contain TAT,
RNA Pol II transcribes the TAR sequence into RNA. As soon as the TAR
sequence is transcribed and forms the hairpin loop structure, TAT can bind
to the TAR element in the RNA and increase transcription of the fusion
gene. Yes, this was expected in light of part b) since now the TAR RNA is
present.
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d) Next you insert the synthesized DNA oligonucleotide containing the TAR
sequence (in red) into another gene (in black below). When you express this
modified gene in TAT expressing and non-expressing cells, you do not detect any
difference in levels of the protein expressed in the TAT expressing cells.
Furthermore when you purify the TAR containing mRNA from these cells and try
to bind TAT protein to it, TAT does not bind. Explain.
5’ – AACCAGAGAGACCCTTGCTACGAATTCGGGTCTCTCTGGTTAGACCAGATCTGAGCCTGGGAGCTCTCTGGCTAACTAGGGAACCCAGAATTC ATG… - 3’
TAR
(5 points) When this fusion gene is transcribed into RNA the black
sequence at the 5’ end can form a hairpin loop with the 5’-end of the TAR
sequence (5’-AACCAGAGAGACCC-------------GGGTCTCTCTGGTT….3’). The formation of
this new hairpin loop prevents the formation of the hairpin loop that TAR
normally forms and therefore TAT cannot bind.
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3. (30 points) Below is the DNA sequence of EF-Tu from the bacterium Bacillus
stearothermophilus. The sigma factor binding sequences and the transcriptional
start site are highlighted in bold. The 5’ and 3’ ends of the genes are shown, but
for the sake of simplicity the middle sequence (77-1181, designated by …) has been
omitted in this diagram.
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5’-CGAATAATTGATTTCTCTTGCTAGTTCCGCTATAAATACTTATGTAAGTAAGACTTTT-3’
3’-GCTTATTAACTAAAGAGAACGATCAAGGCGATATTTATGAATACATTCATTCTGAAAA-5’
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64
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1181
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-AAGGAGGATCTTTCTCATGGCTAAAGCGAAATTTGAGCGCACGAAACCGCACGTCAACATT…
-TTCCTCCTAGAAAGAGTACCGATTTCGCTTTAAACTCGCGTGCTTTGGCGTGCAGTTGTAA…
…ACAGTTGGCGCTGGTTCCGTATCGGAAATCATCGAGTAATCAAGAAAAAGGATGTCCAATC…TGTCAACCGCGACCAAGGCATAGCCTTTAGTAGCTCATTAGTTCTTTTTCCTACAGGTTAG-
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-GTTGGACATCCTTTTTCTTTATCCTGGCT-3’
-CAACCTGTAGGAAAAAGAAATAGGACCGA-5’
a) Please write the first 18 bases of the mRNA that would be transcribed, 5’ to 3’.
(3 points) 5’-GUA AGU AAG ACU UUU AAG-3’
b) What will be the first five amino acids of the protein derived from the mRNA
(please denote N and C ends of the peptide and give your answer in both the
three letter code and single letter code)
(5 points) 5’-ATG GCT AAA GCG AAA-3’
N- Met- Ala- Lys- Ala- Lys-C
N-MAKAK-C
c) What will be the last five amino acids of the protein derived from the mRNA
(please denote N and C ends of the peptide and give your answer in both the
three letter code and single letter code)?
(5 points) 5’-TCG GAA ATC ATC GAG TAA-3’
N-Ser- Glu- Ile- Ile- Glu-C
N-SEIIE-C
d) What is the length of this protein (in amino acids)?
(1.5 points) 395 Amino Acids
e) Is the length of the protein the same as the length of the mRNA? Briefly explain
why or why not.
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(4 points) No, the start and stop sites for transcription are not the same as
the start and stop sites for translation.
f) For each of the following nucleotide changes, please determine their effects on
the codon, amino acid specified, and on the resulting protein (truncation, single
amino acid change, radically different protein, or no change).
(7.5 points total-1.5 for each row)
Mutation
Example:
T#→deletion
T37→deletion
Change in
Codon
UUU→UUG
Change in
Amino Acid
Phe →Leu
GCU→GCA
Ala →Ala
A1201→G1201
C65→A65
A59→T59
A1223→T1223
GUA→ GUG
CAC→AAC
AAA→ UAA
AGA→ UGA
Val → Val
His →Asn
Lys →stop
none
Effect on Protein
Radically different
protein
Radically different
protein
No change
Single aa change
truncation
Not in coding
region
g) EF-Tu is essential for bacterial survival. Antibiotics that act on EF-Tu are used to
combat infections caused by Bacillus stearothermophilus. One antibiotic,
kirromycin, freezes EF-Tu in the GTP position (GTP cannot be hydrolysed to
GDP). Briefly explain how this would effect translation?
(4 points) The hydrolysis of EF-Tu-GTP to EF-Tu-GDP is necessary for EFTu to dissociate from the aminoacyl-tRNA. EF-Tu needs to dissociate from
the aminoacyl-tRNA so that the aa-tRNA can fully occupy the A-site and
add its amino acid to the growing polypeptide chain.
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4. (10 points) Several HIV genes are encoded in overlapping portions of the genome.
For instance, TAT and REV sequences are partially overlapping. Below is the viral
coding strand (non-template strand) DNA sequence for TAT beginning with the start
codon and ending with the stop codon.
5’atggaaccag tagatcctaa aatagaaccc tggaatcagc caggaagtcg gcctaagact ccgtgtaccc
catgctattg taaaaagtgt tgctatcatt gcccaatatg cttcttaaac aagggcttag gcatttccta tggcaggaag
aagcggagac aacgacgaac agctcctcct ggcagtaaga accatcaaga tcctgtatca aagcaacccg
tatcccaaac ccaacgggag ccgacaggcc cagagaaaca gaagaaggag atggagagca aggcaacacc
agatcgattc gattag-3’
a) How many amino acids does Tat mRNA (306 nucleotides total length) encode for?
(2 points) 306/3 = 102 – 1 for the stop codon = 101
b) The protein sequence for Tat and Rev are provided below. Based on these
sequences, do they share the same translational start site? Explain.
Tat
MEPVDPKIEPWNQPGSRPKTPCTPCYCKKCCYHCPICFLNKGLG
ISYGRKKRRQRRTAPPGSKNHQDPVSKQPVSQTQREPTGPEKQK
KEMESKATPDRFD
Rev
MAGRSGDNDEQLLLAVRTIKILYQSNPYPKPNGSRQAQRNRRRR
WRARQHQIDSISQRILSPYLGRSTEPVPLQLPPIERLRLDCSED
CGNSGTQGVGDPQIPEEPGVLLGTGTKE
(3 points) No, they do not since the amino acids sequences they
encode are entirely different.
c) Based on the Rev sequence provided, please label the first Methionine codon for Rev
on the above sequence.
(2 points) atg (aug in RNA) labeled in red above encodes the first met for
Rev
d) Explain how this same stretch of nucleotides can encode two different polypeptides
with different sequences.
(3 points) Two different start sites for translation are used.
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5. (24 points) In order for mRNA to be translated by the ribosome it must exit the
nucleus of eukaryotic cells and enter the cytoplasm. The cell uses a complex system
of export machinery to accomplish this task.
a) What prevents RNA from diffusing out of the nucleus?
(3 points) Nuclear membrane. Transport into and out of the nucleus is
regulated.
b) What would be a potential consequence if RNA could freely diffuse out of the
nucleus?
(4 points) Unprocessed RNAs could potentially be translated. This would
result in the production of unusual proteins as introns may still be present
in some of the RNAs, and code for additional amino acids or aberrant
terminations.
c) Normally, a cell only exports mature or fully processed mRNA out of the nucleus.
However, we learned in lecture that HIV can co-opt the cellular machinery to
export intron-containing RNA into the cytoplasm. The protein Rev is essential for
this process. In order for Rev to function it must be able to: be transported into
the nucleus, bind the Rev-responsive element (RRE), and interact with exportin.
Briefly describe why these three things are necessary for Rev to function.
(6 points)
1. If Rev cannot enter the nucleus, it cannot bind to RRE of unspliced HIV
RNAs and mediate their nuclear export.
2. If Rev cannot bind RRE, it cannot distinguish the RNAs it is supposed to
bind to versus any other RNA.
3. Rev itself is incapable of exporting molecules out of the nucleus. It
needs to attach itself to exportin and Ran, which are the proteins that
can directly mediate nuclear export.
d)
(*extra*) The region of the REV protein that binds to RNA is rich in the amino
acid arginine. Do you think this stretch of arginines is sufficient to specifically
export RRE containing mRNAs? Briefly explain why or why not.
Probably not. The sidechain of arginine is positively charged at
physiological pH’s and is thus most likely to interact ionically with
negatively charged molecules, such as phosphates of the RNA backbone.
The backbone does not have any sequence specific information, and thus
is unlikely to be used to bind RNAs in a sequence-specific fashion.
e) In studying another virus, the mouse mammary tumor virus (MMTV), scientists
looked at a gene called MMTV-env. This gene is predicted to be around 2000
base pairs (bp) long. However, when the MMTV-env gene was introduced into
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uninfected cells, they found that the MMTV-env-derived mRNA in the cytoplasm
was 902 bases (lane 1, red arrow). Why is the size of the cytoplasmic mRNA
different from the size of the predicted primary transcript?
(3 points) an intron has been spliced out of the 902 bp
mRNA.
f) Interestingly, when the MMTV-env gene was introduced into cells infected with
MMTV, two different cytoplasmic mRNAs were identified. One was 902 bases
(lane 2, red arrow) and the second was 2000 bases (lane 2, blue arrow). What
might account for the appearance of this new cytoplasmic mRNA?
(4 points) The 2000 bp transcript was exported into the cytoplasm prior to
splicing. Therefore MMTV has a mechanism to transport unspliced RNAs
out of the nucleus.
g) When the 2000 base and 902 base RNA fragments were sequenced and
translated, they translated into two different proteins. One of them translated
into a protein with a stretch of arginines and a nucleolar localization signal.
Which band most likely contains these motifs? Explain.
(4 points) The 902 bp fragment probably has these motifs (which are
hallmarks of Rev). Normally all RNAs are completely processed prior to
nuclear export. Therefore upon initial MMTV infection, the 902 bp RNA is
the first one to be present in the cytoplasm. This RNA could then code for
a protein to help export incompletely spliced RNAs out of the nucleus, the
2000 bp fragment. If we assumed that the 2000 bp RNA encoded a
protein with arginines and a nucleolar localization signal – then the 902
bp RNA that is present in the cytoplasm first would not be capable of
encoding a REV-like protein to export the 2000 bp RNA out of the nucleus.
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