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7.28 Spring ’01
Problem Set #3 key
Question 1. You are studying the regulation of tyrosine synthesis in a unicellular eukaryote.
It is thought that the enzyme Tyr1, which adds the hydroxyl group to the aromatic ring, may
be a key regulated component of this biosynthetic pathway. Significant Tyr1 enzymatic
activity is only seen in cells grown in media without added tyrosine. After identifying the
Tyr1 gene promoter, you fuse it to a reporter construct to begin studying Tyr1 regulation.
However, you are surprised by the following data:
1a. What does this tell you about the regulation of Tyr1 activity? Describe a simple
experiment you could use to confirm your hypothesis.
Since there is no change in reporter gene product made under both conditions, the
regulation of Tyr1 gene activity does not occur at the transcriptional level.
One way to further address this would be with a Northern blot. Using a labeled
probe to the Tyr1 gene, if you saw similar levels of Tyr1 mRNA in cells grown
under both conditions, you would conclude that regulation was translational or
post-translational. (If you did see a change, then differences in mRNA stability
under different conditions would be a likely implication, since the promoter activity
appears unchanged)
1b. You suspect that RNA structure may have a role in the regulation of Tyr1. A
friendly post-doc helps you out with a series of nuclease protection assays, and you
find a region in the mature mRNA purified from these cells which seems to contain
significant secondary structure. This region is located approximately 200 bp
upstream of the start codon.
How might this region of secondary structure affect the expression of Tyr1 protein?
How would you test whether this structural element was involved in the regulation of
Tyr1?
Given the position of this secondary structural region upstream of the start codon, this
region may affect translation initiation. eIF4 cannot facilitate ribosome loading to the
mRNA adjacent to the 5’cap if the secondary structure of the RNA is not eliminated.
Although eIF4a and eIF4b are capable of “flattening out” these structures in many
mRNAs, in the case of the Tyr1 mRNA a factor may bind to and stabilize this
structure in the presence of tyrosine and block this step in translation initiation.
7.28 Spring ’01
Problem Set #3 key
You could test the role of this region in two complementary ways:
i. Delete this region from the gene and see if tyrosine-dependent inhibition of Tyr1
activity is diminished or eliminated. (Tests if region is necessary for regulation.)
ii. Insert this region into a reporter gene at a similar position upstream of its start
codon and see whether the reporter gene product’s activity is now regulated by
tyrosine. (Tests if region is sufficient for regulation.)
1c. You now generate mutations that change the activity levels of Tyr1. You identify a series
of mutant strains with point mutations that map to within the Tyr1 gene. The phenotypes of
three classes of strains are shown below:
The locations of the specific mutations within the Tyr1 gene are shown below (note: all of
the coding sequence mutations are missense):
A recently published paper has identified several domains in the Tyr1 protein. None of your
mutations fall into the catalytic domain of the protein. Also, mutant Tyr1 proteins purified
from all mutant strains show normal enzymatic activity in vitro. The mutations in type A2
strains are all in a region of the protein that is unnecessary for enzymatic function in vitro and
has no known function. The mutations in type B strains are found in a region shown to be
involved in tyrosine binding. Using all this data, suggest an explanation for the defects in
each of these mutant strains and the normal function in wild type cells for the regions of the
Tyr1 gene where these mutations lie.
Type A1: These mutants have either altered or eliminated the RNA structure found in
region 1. Normally, this structure serves to block translation of the Tyr1 mRNA in the
presence of tyrosine (by the mechanism described in part a). In these mutants, the
absence of the wildtype RNA structure allows translation to proceed under both
conditions.
Type A2: These mutants have altered or eliminated a region of the Tyr1 protein
involved in regulating translation of its own mRNA. This part of the protein might
7.28 Spring ’01
Problem Set #3 key
directly interact with the mRNA, or it might indirectly affect the mRNA through
interactions with other proteins. In either case, the net effect should be the
stabilization in the presence of tyrosine of the RNA structure found in region 1 and the
prevention of Tyr1 translation. The absence of such stabilization in the mutant strains
leads to unregulated expression of Tyr1 protein.
TypeB: The Tyr1 protein contains an allosteric regulatory binding site for tyrosine,
and it is altered in these mutants in one of two ways:
i. Tyrosine is bound with much higher affinity than in the wild type Tyr1 protein, and
thus even in the absence of externally added tyrosine the site remains occupied.
ii. The domain’s conformation is altered to stay permanently in the “tyrosine-bound”
form, even in the absence of tyrosine.
This allosteric site normally functions to sense tyrosine levels in the cell: when
tyrosine is abundant, this site is occupied and Tyr1 is capable of suppressing its own
translation (through the actions of region A2); when tyrosine is scarce, this site is
empty and the Tyr1 protein now adopts a conformation incapable of suppressing its
own translation. In the mutants, this site and the Tyr1 protein are always in the
tyrosine-bound, translation suppressing conformation.
1d. You set out to identify suppressors for one of your type A1 mutants. You find that point
mutations located in two areas can restore normally regulated Tyr1 activity to these cells: the
area defined by the type A1 mutants, as well as the area defined by the type A2 mutants.
How might these two different kinds of suppressors work? What does this imply about the
regulation of Tyr1 activity? Give as complete a model as you can.
The suppressors located within the A1 region probably restore the wildtype RNA
secondary structure identified in part (b) by restoring base pair interactions with
mutations complementary to the original A1 mutations. (See problem set #3, question
7 for more information on this type of suppression)
The existence of suppressors in region A2 of the Tyr1 protein implies that this protein
domain physically contacts the RNA structure in the 5’region of the Tyr1 mRNA. The
suppressors contain mutations that alter the protein and allow it to bind to the mutant
RNA structure. In this case, the mutant Tyr1 protein and the mutant RNA structural
element interact in manner similar to the wt Tyr1 protein and the wt RNA structural
element.
MODEL: When tyrosine levels are high in the cell, the Tyr1 protein binds a tyrosine
molecule in its allosteric site (identified by the B mutants). When tyrosine is bound by
Tyr1, the protein can use a domain (defined by the A2 mutants) to bind to and
stabilize the RNA structural element upstream of the start codon. This now prevents
eIF4 from recruiting the ribosome to the mRNA, and no new Tyr1 translation will
take place. Thus, Tyr1 protein levels will remain low in the cell.
When tyrosine levels are low, the Tyr1 protein will not have a tyrosine molecule
bound and will be unable to associate with the RNA structural element. Now eIF4 will
be able to recruit the ribosome to the mRNA, and it will be able to initiate Tyr1
message translation and boostTyr1 protein levels.
7.28 Spring ’01
Problem Set #3 key
1e. Describe one experiment you could use to test your model in d). Include controls that use
your mutant forms of the Tyr1 gene.
A gel shift experiment using the Tyr1 mRNA and the Tyr1 protein would be useful.
Tyr1 protein should only bind to the Tyr1 mRNA if tyrosine is present. In the absence
of tyrosine, Tyr1 protein should not give a gel shift with the Tyr1 mRNA. Also, there
should be (i) no binding between the wt Tyr1 protein and the A1 mutant mRNAs and
(ii) no binding between the A2 mutant Tyr1 protein and the wt mRNAs
Question 2 (12 points). You have identified a series of mutations in the Alanine tRNA that
are defective in translation. You are interested in determining what functions each mutation
is defective in. Describe briefly how you could detect mutations that were defective in the
following activities. You have access to any purified proteins, amino acids, or RNAs that you
need.
2A (3 Points). Coupling to Alanine
Several different experimental designs could be used to detect whether the Alanine
tRNA is defective in translation because it cannot be "charged" with alanine. The
simplest would be to combine your tRNA, the alanine tRNA synthetase, ATP, and
radioactively labeled alanine. Run the products of this reaction out on a gel and stain
with EtBr to ID the location of the tRNA. Then dry the gel and expose to film. If the
mutant tRNA can be charged, you would expect the radioactive signal to co-migrate
with the tRNA. However, if the tRNA is defective in charging than the radioactive
amino acids will run at the bottom of the gel.
Other possibilities include a similar in vitro "charging" reaction followed by filter
binding where the filter has been treated to bind RNA but not proteins or free amino
acids. If you see a radioactive signal on the filter, then you know the tRNA was
charged.
2B (3 Points). Interaction with EF-Tu.
Any experiment that could determine whether Ef-Tu can bind tRNA would work here.
One possibility would be a filter binding assay in which the filter was treated to bind
only RNA. You would combine the tRNA, radiolabeled Ef-Tu and GTP, pass the
mixture through the filter, wash and look for radioactivity that was retained on the
filter. If you can detect a radioactive signal than you know that Ef-Tu can bind the
tRNA. Other possibilities include: gel shift assay, IP (after crosslinking), affinity
column, etc. A modification interference assay would also work, if you assume that
tRNA/EF-Tu interaction is needed for protection of the A site. Credit was not given
for sucrose gradient centrifugation here. The resolution of a sucrose gradient is not
good enough to resolve EF-Tu that is bound to tRNA from free tRNA. Mutants or
wild type would give the same result on a sucrose gradient separation, and you may
get a false negative result.
7.28 Spring ’01
Problem Set #3 key
2C (3 Points). Interaction with the Ribosome.
Since the ribosome is a rather cumbersome complex of RNA and proteins, it is
probably easiest to use sucrose density gradients to study whether the mutant tRNA is
able to bind to the ribosome. You would want to combine the radiolabeled charged
mutant tRNA, Ef-Tu, GTP (or, better, GMP-PCP), and the ribosome together and then
add this mixture to the top of a sucrose density gradient. After centrifugation remove
the fractions and
do Northern blots (probing with a DNA fragment that is complementary to the rRNA
sequence) to detect the fractions containing the ribosome. Use scintillation counting to
ID the location of the charged tRNA. If the radioactive signal co-elutes in the same
fractions as the ribosome, you know that the tRNA is able to interact with the
ribosome. Another common answer was to use a modification interference assay,
looking for protection of the A site (due to EF-Tu and tRNA interactions).
2D (3 Points). For each of the three potential defects, which would you expect to be regions
of the Ala-tRNA that are conserved in all or most tRNAs? Which would you expect to be in
regions of the Ala-tRNA that are unique?
The region of the alanine tRNA that is needed to recognize alanine is probably unique
to this tRNA. Each tRNA will have a unique region that is involved in binding the
specific amino acid that corresponds to the anticodon of the tRNA. The tRNA
domains that interact with Ef-Tu and the ribosome are probably common to all tRNAs
(conserved) since all tRNAs-regardless of the amino acid that they are carrying--need to maintain these interactions
to be functional. (1 point each)
A lot of people wrote all the unique and conserved regions of the tRNA, which is not
what the question asked. Some partial credit was given here, depending on if you
connected these regions to defects described in 3A-3C.
7.28 Spring ’01
Problem Set #3 key
Question 3 (30 points). You are studying the regulation of the Phenylalanine operon in E.
coli. This operon encodes an mRNA for 3 proteins necessary for cells to produce Phe.
As a first step in characterizing the regulation of the operon, you fuse the coding sequence for
LacZ to the Phenylalanine operon promoter to examine regulation in the presence and
absence of Phe. You find the following results:
3A (3 Points). What can you conclude about the regulation of the operon from
these results? Pay particular attention to the stage of biosynthesis that is regulated
as well as any assumptions that you make in arriving at your conclusion.
In order to make conclusions from the LacZ expression levels we must assume that
LacZ is stable in these cells and that the activity or stability of LacZ is not affected by
the presence or absence of phenylalanine (1 point). Given these assumptions, the data
suggests that the Phe operon expression is transcriptionally regulated (1 point) and
that the regulation is dependent on the phenylalanine levels in the cell. The operon is
induced when there is low phenylalanine and repressed when phenylalanine is
abundant (1 point).
You are surprised to find only a five-fold change in the presence of Phe as studies of other
operons involved in amino acid metabolism usually show a 200-1000 fold repression in the
presence of the amino acid they synthesize. To look for additional regulation you fuse the
LacZ coding region to the C-terminus of either the PheA or the PheB coding regions. In each
case you find that the fusion protein is active for β-Galactosidase activity.
7.28 Spring ’01
Problem Set #3 key
You test for LacZ activity in strains containing one or the other construct in the presence or
absence of Phe. You obtain the following results:
3B (3 Points). What can you conclude about the regulation of the PheA and PheB
proteins based on these findings?
By making gene fusions of LacZ to both PheA and PheB we are able to look at the
translational regulation (1 point) of both of these genes. You still need to make the
same general assumptions about LacZ activity that were stated in 5a. These results
indicate that PheA is not regulated at the translational level since the LacZ fused to
PheA has the same activity as when it is was only controlled by the Phe promoter (1
point). The PheB gene is not only transcriptionally regulated but also translationally
regulated since the LacZ activity in the absence of phenylalanine is even lower in the
protein fusion. Thus, the translation of PheB is negative regulated by the presence of
Phenylalanine (1 point).
Intrigued by the LacZ replacement experiments, you want to identify mutations in the Phe
operon that alter the regulation observed. To this end, you mutagenize E. coli cells containing
the Fusion B construct. You look for colonies that are Blue (= high LacZ expression) in the
presence of Phe or that are white (=low or no LacZ expression) in the absence of Phe.
Using these screens you identify three classes of mutants that you call Control of
Phenylalanine Response (Cpr) mutants. Cpr1 and Cpr2 are constitutively White and Cpr3 is
constitutively Blue.
7.28 Spring ’01
Problem Set #3 key
3C (5 points). Describe how you would determine whether each Cpr mutant acts in cis- or in
trans- to the Phe operon.
To determine whether each of the Cpr mutants is cis or trans-acting you want to add a
second reporter gene construct to the cells. You could use a plasmid containing the
Phe promoter and a PheB-luciferase fusion (i.e. an extra Fusion B construct with
luciferase instead of LacZ). Note : since these mutants were isolated in the Fusion B
construct background, they are mutations that affect transcriptional AND translational
regulation
of PheB. A promoter fusion (as in part 5A) will NOT allow you to assay Cis- Trans,
because it does not allow observation of translational regulation. (2 points for correct
construct).
This construct will allow you to determine whether the Cpr mutants affect either the
transcriptional or translational regulation of PheB in cis or trans. If the mutation is
trans-acting, you will expect the luciferase activity to show the same phenotype as the
LacZ activity (i.e. the mutant phenotype). If the mutation is cis-acting than the
luciferase activity should exhibit wild-type regulation (low in phe+, high in phe-). (3
points for analysis of results).
Using the cis-trans test you find the following:
Cpr1 mutants acts in cis
Cpr2 mutants acts in trans
Cpr3 mutants acts in cis.
To determine where the cis-acting mutations are located, you sequence the Phe operon from
Cpr1 and Cpr3 mutant strains. You find mutations in the following locations near the PheB
AUG:
3D (4 points). Based on your findings, how do you think that the Cpr3 mutations are
influencing the regulation of the Phe operon?
The Cpr3 mutations are cis-acting and lead to the constitutive expression of PheB. The
symmetrical nature of the Cpr3 mutations suggests that the mutations may be
disrupting a hairpin loop. The position of the mutations right after the S/D sequence
suggests that a hairpin loop may be forming to inhibit the translation of PheB. We
know from the translational regulation data that translation of PheB is strongly
repressed in the presence of phenylalanine therefore the formation of this loop must be
dependent on high phenylalanine levels (2 points for hairpin, 2 points for recognizing
its dependence on phe levels and that it affects TRANSLATION). Another answer
that was given was that phenylalanine, or a phe+ dependent regulator, bound to Cpr3
to block translation of PheB. This is unlikely, because of the symmetrical nature of the
mutations. However, partial credit was given for this answer (2 points).
7.28 Spring ’01
Problem Set #3 key
3E (4 points). Briefly describe how you could test your hypothesis?
You could test your hypothesis using either a genetic or biochemical assay. We have
shown that suppressors of mutations that disrupt hairpin loops restore proper base
pairing. For example if you had a A to G mutation then the suppressor would be
nearby and you would see a change from a T to a C. This type of suppressor analysis
could give you additional data that the Cpr3 mutation do indeed disrupt a hairpin loop.
To test this biochemically you could use S1 RNase and SV Rnase in a protection
assay of wild-type and mutant RNA. You would expect to see the characteristic
double stranded stem of the loop in wild-type RNA but not in the Cpr3 mutants.
If you used a “repressor binding model” in part 5D, and you gave a good assay for
it in this section, you were given credit. However, phenylalanine binding (the aa) will
not give a DNAse or Rnase protection pattern—it’s just too small. You’d need to
hypothesize a protein binding to use that assay.
The Cpr1 and Cpr2 mutations remain mysterious to you. You initially suspect an attenuator
similar to the Trp operon but you find no start codons between the PheA gene stop codon
and the PheB gene ATG and the only terminator in the mRNA is after the PheC gene.
Nevertheless, you look at transcription of the Phe operon by Northern blot and obtain the
following results.
After once again assuring yourself that there is no terminator in the area 2000 bp from the
transcription start site, you notice that this region is near the site of the Cpr1 and Cpr3
cis-acting mutations.
Because the Cpr2 mutants also do not form the 2000 bp transcript in low Phe conditions, you
wonder if Cpr2 binds to the RNA or DNA at the Cpr1 site.
3F (5 points). Describe how you would you determine whether this is the case using purified
Cpr2 protein provided to you by a generous lab mate? You can make any RNA or DNA
probe that you want.
7.28 Spring ’01
Problem Set #3 key
To test whether Cpr2 binds either DNA or RNA you would want to perform a gel shift assay
using either purified RNA transcript from the Phe operon or a DNA fragment containing the
Phe operon. Both of these fragments will be labeled on their 5' ends to facilitate visualization.
You would load the fragments with an without purified Cpr2 and ask whether
the band is shifted up due to the binding of Cpr2.
Another good answer was to use a filter binding assay with unlabeled Cpr2 protein and labeled
RNA or dsDNA containing the Cpr1 site. The filter should retain protein, but let RNA and
dsDNA flow through. Radioactivity would only remain associated with the filter if the protein
bound the DNA or RNA.
You find that Cpr2 protein binds the dsDNA site identified by the Cpr1
mutations but that it does not bind the RNA form of the Cpr1 site. Interestingly,
you only see binding in the absence of Phe.
3G (6 Points). Propose a model that explains how Cpr2 can influence translation of
the PheB gene by binding to dsDNA.
Cpr2 could influence the translation of PheB by binding to dsDNA if it was involved in
pausing the transcription of the PheB gene. The crucial step in the translational regulation of
PheB is whether or not the stem loop that blocks the S/D sequence is able to form. Because
transcription and translation are coupled in prokaryotes, the formation of the hairpin loop
could be controlled by the timing of transcription and in particular if transcription is paused
so that the RNA necessary to form the loop has not been released
Cpr2 could influence the translation of PheB by binding to dsDNA if it was involved in
pausing the transcription of the PheB gene. The crucial step in the translational regulation of
PheB is whether or not the stem loop that blocks the S/D sequence is able to form. Because
transcription and translation are coupled in prokaryotes, the formation of the hairpin loop
could be controlled by the timing of transcription and in particular if transcription is paused
so that the RNA necessary to form the loop has not been released from the transcription
machinery. The Cpr1 and Cpr2 mutations are critical in establishing the proper timing to
prevent the hairpin loop from forming. Cpr2 probably binds to the DNA site coded for by the
Cpr1 mutations and physically blocks transcription in the absence of Phe. The RNA necessary
to form the hairpin loop is not released right away and translation has a chance to catch up.
When the pause is released, the newly transcribed RNA is immediately translated leading to
the production of PheB. In the presence of Phe, Cpr2 cannot bind and block transcription, the
hairpin loop forms, and PheB is not made. The 2000bp product seen by Northern analysis is
the RNA transcript that is made when the RNA polymerase is paused by Cpr1. You don't see
it in Cpr1 or 2 mutants since these mutations abolish pausing.
Full credit required a detailed understanding of the processes that would occur + and – phe
(why the RNAP would stall, loop formation or not, transcriptional and translational coupling,
etc.). Partial credit was given for answers that showed some understanding of the data, but
which did not provide a complete model.
Problem 4. You’re a UROP working on a new telomerase gene in a little-studied ciliate. You
clone the gene, but can’t decipher from the DNA sequence where the open reading frame
starts and stops. At wits end, you consult your 7.28 notes and recall that ciliates often have
strange genetic codes. You decide that deciphering the genetic code of the ciliate would be a
much more exciting project than working on telomeres.
7.28 Spring ’01
Problem Set #3 key
4a) You aren’t sure that your ciliate even uses a triplet code, but given that it has 20 amino
acids, why do you think that the triplet code is the most likely? (What’s wrong with a doublet
or quadruplet code?)
A doublet code would only code for 4*4= 16 different codons. A quadruplet code
would code for 4*4*4*4=256 codons, many more than necessary.
4b) What is the minimum number of tRNA’s a cell needs to be able to recognize all 64
possible triplet codons?
The first base of the anticodon can wobble pair with the last base of the codon, so
having either G or U in the first position would recognize, respectively, codons with U
or C, and codons with A or G, in the last position. Therefore, instead of 4*4*4=64
possible codons, there need only be 2*4*4= 32 anticodons. Also, since the stop codon
is not recognized by a tRNA, but by a release factor, you can subtract one anticodon
for each stop codon.
4c) Following in Ghobind Khorana’s footsteps, you make a ciliate cell extract and program it
with various RNA polymers in order to see what proteins will be made. What are the simplest
four codons to decipher, and how?
Make polymers of each ribonucleotide, and check what amino acid is synthesized.
This will tell you what UUU, AAA, GGG, and CCC code for.
4d) You then decide to make random copolymers of G and U by mixing together the
ribonucleotides, and retrieve polypeptides which have eight different amino acids. How can
you narrow down which of the eight possible codons code for which of the eight amino acids
you retrieved? Which codons would be indistinguishable by your method?
Change the ratios of Gs to Us. That should change the ratios of the different codons
present in your RNAs. You can then figure out what percentage of codons should
have 3 Us, 2 Us and 1 U, and from there match that up to the percentage of each
amino acids recovered. However, you cannot distinguish between the three different
codons with 2 Us (UUG, UGU, and GUU), and the three different codons with 1 U
(UGG, GUG, and GGU). In order to sort these out, you need to make repeating
copolymers of specific sequences, such as (UUG)n.
4e) You do the same with Gs and As, but retrieve fewer than the expected 8 amino acids.
What is the maximum number of amino acids that the 8 possible G and A codons could code
for? Why?
4! Wobble! There is only one possible tRNA anticodon that recognizes GGA (UCC),
and that anticodon will also recognize the codon GGG. Similarly, the anticodon UUC
will recognize both the GAA and GAG, the anticodon UUU will recognize both AAA
and AAG, and the anticodon UCU will recognize both AGA and AGG.
4f) You observe even fewer than the maximum number of amino acids you calculated as
possible in part e. What is an explanation for this and how could you test it?
7.28 Spring ’01
Problem Set #3 key
One of the codons codes for a stop. Run the synthesized polypeptides out on an SDS
polyacrylamide gel and see if they are shorter than those synthesized from random
polymers of Gs and Us. (If there is one stop codon, the polypeptides should on
average be only 8 amino acids long.)
Problem 5. You are interested in studying the initiation of translation with fMettRNA in the
brewer’s yeast Saccharomyces cerevisiae. In particular, you would like to investigate the
specific components of the system that direct initiation at the first AUG. Specifically, you are
intrigued by the system’s ability to distinguish between internal methionine coding AUGs
from the first AUG codon that initiates with formyl- Methionine.
To initially investigate how the system differentiates the different AUG codons, you mutate
the tRNA that binds fMethionine. You identify three bases in the sequence of the tRNA that
are responsible for its ability to uniquely recognize the first AUG as the start codon. You then
mutate these three bases such that the tRNA no longer retains its ability to differentiate the
start codon from internal AUGs, but is otherwise normal in its ability to be charged with
fMethionine and basepair with the sequence AUG. Finally, you replace the wild type allele of
the tRNA in your strain with your mutant version.
5a) What phenotype would you expect from your mutant strain? Provide an
explanation for the cause of this phenotype.
You would not expect initiation of translation. One possible method fMettRNA uses
to distinguish the first AUG from internal AUGs is its association with eIF-2. eIF-2
may recognize a sequence on the fMet-tRNA. It can then bring the tRNA to the partial
p-site of the 40s subunit of the ribosome for initiation. If the three bases mutated are
responsible for the interaction between eIF-2 and fMet-tRNA, you would fail to load
fMet-tRNA into the partial p-site and fail to initiate.
Realizing your mistake, you construct a strain that now expresses both the wild type
allele of the fMethionine binding tRNA and your mutant allele.
5b) What phenotype do you now expect? Explain.
Since you are expressing the wild type allele of the fMet binding tRNA, you may
initiate translation normally. However, since you are also expressing the mutant
tRNA, it is possible that a formyl-Methionine will be inserted at an internal AUG
rather than a normal Methionine. This can occur if the fMettRNA associates with
EF-Tu rather than eIF-2 and is loaded into the A site.
If an fMet is inserted into a polypeptide chain rather than a normal Methionine,
aberrant termination of translation will ensue. If you visualized the polypeptides in
this strain (by SDS-PAGE), you would see polypeptides of varying length, but most
would be shorter than a polypeptide in a wild type strain.
7.28 Spring ’01
Problem Set #3 key
5c) You suspect that the three bases you mutated on the fMethionine binding tRNA
are responsible for binding to initiation factors such as eIF2, eIF3, and eIF4.
Describe an assay you can use to test this hypothesis. Assume that you have access
to purified tRNAs and initiation factors.
You can perform a filter binding assay to test if the tRNA of interest binds to any of
the initiation factors. For this assay, you would use a filter that will retain proteins, but
allow nucleic acid to pass through. First, radiolabel your tRNA. Then, incubate with
the purified initiation factors. Passage your protein/tRNA mixture through your filter.
Finally, measure the radioactivity that remains on the filter. The radioactivity
represents tRNA retained on the filter because of its binding with initiation factors. In
order to test if the three mutated bases are responsible for binding to your initiation
factors, you need to test your mutant and a wild type copy of the tRNA and compare
the radioactivity from the two alleles.
5d) You decide to further test the codon specificity of the fMet-tRNA. In doing so, you
mutate the anticodon loop so that the fMet- tRNA now recognizes the codon GUG for
translation initiation. You then mutate the start codon for the HIS gene from AUG to GUG.
You monitor translation of the HIS gene and are very surprised to see a polypeptide of
inappropriate length. Confused by this, you make the same anticodon and codon mutants, but
this time in the LEU gene. Once again, you get a polypeptide of inappropriate length. Provide
an explanation for this observation (hint: normally, there is strong selection against the
presence of the AUG start codon in the region between the 5’-CAP and the first start codon).
The expectation is that normal initiation can occur when mutating the start codon to
GUG and the fMet-tRNA anticodon to recognize GUG. However, this is not the case.
It is very likely that the fMet-tRNA mutant will recognize a GUG upstream of the
correct start codon and initiate polypeptide synthesis from there. This will result in a
polypeptide of inappropriate length that may also be out of frame. Under normal
circumstances, the negative selection against AUGs between the 5’-CAP and the
correct start codon prevent this situation. However, there is no such selection against
GUG codons, so they have the opportunity to occur in this region.
Question 6) For a summer UROP project you decide to analyze gene expression in a highly
proliferative fungus isolated from your dorm shower. Since the organism appears to grow so
efficiently, you reason that it must have extremely efficient mitochondria, and focus your
initial studies on mitochondrial DNA encoded genes. You are excited to find a novel gene
encoded by the mitochondrial genome; you name this gene Fuz1.
The Fuz1 transcript is very long, much longer than you expected for coding the rather small
protein that you think is the product of Fuz1. Therefore, you investigate whether the
transcript is processed. To do this, you isolate the long Fuz1 RNA, 5’end label it with 32P,
and incubated it in vitro under reaction conditions favorable for self-splicing by both group I
and group II introns: Reactions include Mn2+, Mg2+, NaCl, Tris-Cl Buffer at pH6.5, and
32
PGTP. Below is a autoradiogram of the time course of this in vitro splicing reaction.
7.28 Spring ’01
Problem Set #3 key
6a) Based on this splicing pattern, do you think that the Fuz1 RNA contains a group I or
group II intron? Briefly justify your answer.
Fuz1 RNA contains a Group I intron. The key to figuring this out is to look carefully
at the gel. The Fuz1 RNA was labeled only on the 5’end. However the autoradiogram
shows that the initial splicing reaction generates two labeled products. If you look
back at the ingredients included in the reaction, you will notice that labeled GTP was
included. The only way to detect both the spliced out exon1 (containing the 5’end
label) and the intron+exon 3’piece would be if the labeled guanine became covalently
bound to the 3’ piece as it does in the Group I introns. The intron then disappears
because the Group 1 intron is circularized and the 15bp piece that is cut off contains
the labeled Guanine and this piece is too small to detect on an agarose gel.
6b) Draw a diagram of a possible organization for the Fuz1 transcript prior to processing,
labeling the length of any introns and exons.
Exon 1 is approximately 200 bp, the intron is approximately 3000bp, and Exon 2 is
approximately 100 bp. Providing the complex structure of group I introns is also
acceptable.
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6c) To determine if the Fuz1 gene is responsible for the unusually fast growth of the shower
fungus, you mutagenize the gene with a mutagen that gives rise to base substitutions. Once
again, your suspicion was correct; you isolate many slow growing Fuz1 mutants. You start to
characterize these mutant strains, and now you are disappointed. These mutations will not
give you much insight into the biochemical function of the Fuz1 protein, because nearly all
of them (at least 90%) are completely defective in expressing the protein. Upon seeing these
results, you kick yourself for using a poor experimental design. Why are so many of the
mutant strains completely defective in Fuz1 protein expression?
The intron in the Fuz1 gene is ten times longer than the exons. If you did a random
mutagenisis, the majority of the mutations that you would generate would fall within
the non-coding region of the intron and would not be particularly informative about
the function of the Fuz1 protein. Mutations in the intron would disrupt the complex
secondary structures that are needed to form the catalytic center of the intron and as a
result the gene would not be able to self-splice. Without proper splicing, a functional
Fuz1 gene product would not be made
6d)In an attempt to salvage some useful mechanistic data from the mutants you isolated in
part C, you decide to do a suppressor analysis with the slow growing mutant fungus strains.
This time you mutagenize each of the slow growing strains, and select for mutants that
restore the fast-growth phenotype. What type of information would you hope this experiment
would reveal about Fuz1 expression? Explain how you would analyze the mutant strains to
gain the desired information.
If your hypothesis is correct, the suppressor mutations that you isolate should restore
the secondary structure of the group one intron. As discussed in class, you would
expect to see compensatory mutations. For example, if your original mutation was a T
to C change than its suppressor would be nearby and would be a A-G change which
would re-establish proper base pairing between these residues and restore proper
secondary structure. These types of mutational analysis would allow you to define the
secondary structure of the intron as well as identify those secondary structures that are
crucial for the formation of the catalytic center of the intron.
Question 7a) You are studying the pathogenic fungus C. albicans, and when you sequence
one of its mitochondrial genes you find that it has a novel intron. Analysis of its sequence
suggests that it may be related to the group II self-splicing introns. Your first step in studying
the intron’s structure is to mutagenize it. 7a) You identify a series of mutants that can no
longer splice. Shown below is the sequence of the region of the intron (NOT the whole intron)
containing these mutations.
What is a possible explanation for the failure to splice in these mutant strains?
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The mutations in the mutant strains may disrupt the secondary structure elements
necessary for proper folding of the intron into an active catalytic structure. Inspection
of the sequence above reveals several potential stemloop structures in which the
mutations are located.
7b) You decide to further analyze these mutant introns by treating fragments containing this
region with a set of RNA nucleases. You use an RNA fragment corresponding to the intron
sequence shown above, labeled at its 5’end, as a substrate. You visualize the digested
products by autoradiography. The smallest piece you can distinguish on the gel is 5 bases in
length, based on control size standards.
You find that Mutants B and C give similar patterns to Mutant A, while Mutants E and F give
similar patterns to Mutant D. From these data, draw the following:
i. The secondary structure of the wild type intron
ii. The secondary structure of the Mut. A-like mutant introns
iii. The secondary structure of the Mut. D-like mutant introns
Indicate on each drawing regions I-VIII. Also indicate on the wild type structure where the
individual mutations A-F map.
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7c) You decide to generate new mutations that suppress the mutant phenotypes and restore
normal splicing. You isolate two different suppressors, identified as Sup1 and Sup2. Sup1
specifically suppresses only Mut A, and Sup2 specifically suppresses only Mut E. Each
suppressor turns out to map to a location within the intron, but neither is a revertant. Why are
these suppressing mutations so specific? Can you guess what the specific mutations might be,
and where they might be located?
The suppressing mutations are specific because they are individual base changes
complementary to the original mutations and restore Watson-Crick base pairing
between the two bases in question. This allows reformation of the original secondary
structure, although with a different sequence. This is known as “covariation”.
MutA has a U instead of a G at the second position in region VIII (in the first stem
structure). Sup1 must therefore have an A rather than a C at the second-to-last position
in region VI. Mut E changes a C to an A at the second position of region II (in the
second
stem). Sup2 is therefore likely a change from a G to a U at the second-tolast position
in region IV.
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7d) If you repeated the nuclease protection experiments with both S1 nuclease and snake
venom shown above using similarly labeled intron fragments from the two suppressed strains
(one of which carries the Sup1 and Mut A mutations, while the other carries the Sup2 and
Mut E mutations), what would you predict the digestion patterns to be? Explain your
reasoning.
The two suppressed strains are likely to contain introns that generate digestion
patterns with S1 and snake venom nucleases that are identical to the wild-type
digestion patterns, since wt intron splicing activity has also been restored.
Question 8) You are interested in studying eye development in mice and your lab has
recently discovered a gene, BLG1, that causes mice to develop abnormally large eyes. You
have a suspicion that BLG1 might be involved in tissue proliferation in other developmental
pathways so you decide to purify BLG1 from a variety of different developing mouse tissues.
You separate these proteins on a SDS polyacrylamide gel and then do a western blot with an
primary antibody to BLG1. You use a radioactive secondary antibody (this will recognize the
primary antibody) and expose your gel to film. You see the following results on film:
8a) In what tissues is BLG1 most likely to be playing an important developmental role?
Why?
BLG1 seems to be upregulated in eye, brain and muscle tissues suggesting that the
BLG1 protein may be important in these tissues.
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8b) Give two possible explanations that could account for the difference in the mobility of
the BLG1 protein purified from brain tissues. (Hint: Don’t worry about degradation
products.)
MODEL 1:
You would expect to see a faster migrating BLG1 band if the protein was
proteolytically processed to give a smaller protein species.
MODEL 2:
BLG1 mRNA could be spliced in alternative ways in different tissues. In brain tissues
BLG1 may be spliced in such a way that not all of the exons are included in the
processed mRNA. This would lead to a smaller protein product.
8c) You decide to test your hypotheses by comparing the cDNA prepared from brain tissue
and eye tissue. You have been given cell lysates from both tissues that contain all of the
DNA, RNA, and protein these two tissues. Describe in detail how you would isolate the
BLG1 cDNA from these two lysates.
1) Isolate the mRNA from the cell lysates. (One possibility would be to use a oligo dT
affinity column i.e. little beads with (TTTT)n attached to them. You would pass your
cell lysate over the column and mRNAs would stick to the beads due to the base
pairing of the poly-A tail to the oligo dT. You would then wash the column and elute
the mRNA.)
2) Anneal an oligo dT primer to the mRNA to provide a 3’-OH to begin reverse
transcription. Add dNTPs and reverse transcriptase.
3) Treat the RNA/DNA hybrid with base. The base will destabilize the RNA/DNA
duplex and degrade the RNA.
4) Reverse transcription of the original mRNA will create a weird little hairpin that
can serve as a primer for synthesizing the second cDNA strand. Use DNA Polymerase
I to make a double stranded cDNA.
5) Treat the double stranded cDNA with SI nuclease to digest the hairpin and create a
double stranded cDNA molecule.
Question 9) You are studying the mechanism of pre-mRNA splicing in a rare marsupial
found on an island near Australia. To initiate your analysis, you isolate a cDNA clone from
an abundant mRNA. Using the cDNA clone you isolate the genomic DNA sequence that
encodes this RNA; you name the gene kanga1. As a result of comparing the cDNA and the
genomic sequences, you conclude that kanga1 has 4 exons, each about 200 nucleotides long.
9a) Draw a schematic of the structure of the kanga1 gene, the cDNA, the pre-mRNA and the
mRNA. Label each intron and exon and label each nucleic acid at its 3' and 5' ends. Include
any elements important for further processing or expression.
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9b) By sequence comparison with eukaryotic genes, you find good matches for the consensus
5' and 3' splice sites. However, you cannot immediately find a consensus branch site. By
inspection of the sequences shown below, circle the sequence that is most likely to be used
as the branch site for splicing of each of the introns. Based on these data, write down a
consensus branch site for this marsupial.
Intron sequences are in normal type, the 3' splice sites are in bold and the exon sequences are
underlined. Where there are "..." the sequence continues. (Hint: consensus site has the same
length and separation from the 3' splice site as the yeast sequence and the nucleotide that
participates chemically in branch formation is the same in marsupials and mammals.)
9c) In order to understand the mechanism of splicing in this marsupial you set up an in vitro
splicing reaction. First you attempt to use purified splicing factors from mammalian cells.
You find that the kanga1 pre-mRNA is not spliced by these factors. Therefore, you add
components from a nuclear extract made from marsupial cells and find that splicing activity
is restored. You fractionate this extract to identify which component(s) need to be provided
by the marsupial extract. Based on your knowledge of the mechanism of mammalian splicing,
and the sequence of the kanga1 pre-mRNA, judge each of the following combinations of
factors as: unlikely, possible or very likely to support kanga1 pre-mRNA splicing. The
snRNPs are simply called U1, U2 etc. in the following table. Assume that no additional
factors are required for splicing in this experiment. Briefly justify your answers.
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U2 base pairs with the branch point and the marsupial branch site consensus is
different from the mammalian. U2 and U6 interact to form the catalytic center, it is
best if they are from the same organism. In part (b) it states that the marsupial has 5'
and 3' splice sites that match the mammalian consensus, so mammalian U1 and U2AF
are likely to function.
9d) Which two of the factor combinations in the table in part (c) would you expect to be
most likely to promote splicing of mammalian pre-mRNAs? Write down the number of the
combinations from the table, and briefly justify each answer.
1- Mammalian U2 should bind the mammalian pre-mRNA branch site. 3- Of the
remaining options this one preserves the U2-U6 interaction from the same species.
The mammalian U2 may fail to recognize the marsupial branch site since this
sequence differs substantially from the mammalian consensus.