Advanced
Molecular Biology-Dr. Schleif
October 9,
2008
90 minute
exam
Each
question is worth 10 points.
1. If proteins were
stuffed as densely as physically possible into a cell, estimate about
how many
more times concentrated they would be than normal.
10
x (from drawings
in reading or from the text that said cell interior is ~ 100 mg/ml in
protein,
~1 gm/ml would be solid protein)
2. When asked why he
made RNA polymerase transcribe so very much more slowly than he could
have
(consider elongation rate of DNA polymerase), God replied “I couldn’t
make
proteins fold any faster.” Explain. (This is not a question about why
protein folding rates are limited.)
If
protein spewed out of the ribosome much faster than it
can fold, it would aggregate. Therefore, it makes sense for ribosomes
not to go
too fast. Thus, folding rates likely
determine translation rates. There is no point in having transcription
go
faster than translation, hence the slow transcription rate.
3. Sequencing
proteins is hard, but sequencing DNA is easy. Since the pathway is DNA
to RNA
to protein, why not add purified protein to a cell free extract and
sequence
the DNA that would be produced from driving the reactions backwards?
The
vast consumption of energy in synthesizing mRNA and in
translating protein drives the reaction so far towards protein that no
DNA
could possibly be made by the reverse reactions.
4. What commonly
known enzyme might more properly be named RNA-DNA nick translatase?
DNA
polI
5. It is difficult
to build or evolve rotating machinery at the nanoscale. Not
surprisingly then,
topoisomerase II which performs the equivalent of cutting both strands,
twisting once, and rejoining, does not use rotating elements. How can
it
perform the twist operation then?
Wrap
the DNA one turn around the protein as shown in the top
diagram. At the point where the DNA
crosses, cut one duplex, pass the other duplex through the cross, and
rejoin as
shown in the second diagram.
Positive
supercoiling would help the double stranded DNA
resist melting.
7. From phage to
bacteria to eukaryotic cells, the more complicated the organism, the
more
complicated the RNA polymerase. Comment
A
larger polymerase provides more target binding sites for
proteins that must regulate gene activity. Accordingly, the larger the
number
of genes to be regulated, the larger the polymerease.
8. In eukaryotes a
nonsense codon anywhere but in the last exon of a gene leads to rapid
degradation of the gene’s mRNA. How does this lead to the prediction
that
splicing leaves a protein at a splice junction?
The
phenomenon can be restated as: a nonsense
codon upstream from a splice site leads to rapid
degradation of the mRNA. Since a
ribosome is released from the mRNA at a nonsense codon, it must mean
that a
ribosome knocks something off from a splice site, and this removal
eliminates
the rapid degradation. A protein put on
in the process of splicing is the best candidate.
9.
The only significant contacts between promoters and RNA
polymerase are made by sigma. Why then
does free sigma not bind to DNA?
Because
it is too floppy.
When bound to the core polymerase, its DNA contacting regions
are held
in the proper positions and orientations with respect to each other.
10. The following is
a protocol to accomplish what?
Radioactively label the protein on its N-terminus. While
incubating the
labeled protein in presence and absence of presumed interaction
partner,
lightly digest with protease. Separate products from the mix according
to size,
but look only at the distributions of radioactive products.
Protein
footprinting to identify the interaction region with
the partner protein.
Advanced
Molecular Biology-Dr. Schleif
November
11, 2008
90 minute
exam
Each
question is worth 10 points.
1. Gene X codes for
a single polypeptide chain, and mutation A in X, denoted XA,
leaves
the protein X inactive. The same for
mutation B. When cells contain two
mutant copies of gene X, XA
and XB, X activity is within a factor of two of being normal. What can you say about protein(s) X?
X
protein must be oligomeric, and hetero-oligomers are
active because structurally, the mutations A and B compensate.
2. What kind of
mutation might have as a suppressor, a tRNA with four nucleotides in
the
anticodon?
An
insertion of a single base.
3. In seeking fruit
fly mutations on the X chromosome, if females are mutagenized, what is
the earliest subsequent generation in
which
recessive mutations could show a mutant phenotype?
The
next generation.
4. The following
statements all refer to the haploid state. A mutation in gene X
generates no
observable phenotype. The same for a
mutation in Y. Mutations X and Y
together in the organism do display a clear phenotype.
What is an explanation for this phenomenon?
5. Describe a
selection (note, selections mean operations performed on large numbers
in
contrast to scoring, in which one tests candidates one-by-one) for a
temperature sensitive RNA polymerease in bacteria.
If you wish, you may assume that the mutant polymerase does not
properly assemble.
Select
for rifamycin resistance at high temperature,
rifamycin sensitivity at low temperature in a rifr/rifs
diploid.
6. Describe an
efficient scheme for the in vitro synthesis of a large gene keeping in
mind
that it is not possible to synthesize DNA oligonucleotides much larger
than 150
nucleotides.
Synthesize
oligos of about 100 nucleotides long, overlapping
by about 20 nucleotides and two oligonucleotide primers such that on
PCR
amplification, the full-length gene is synthesized.
7. It makes sense
that a gene coding for a restriction enzyme be closely linked to a gene
coding
for the associated modification activity. Sometimes enzymes of a
metabolic
pathway serve double roles and also act as regulators of transcription
of the
pathway. Suppose that happens in the
case of a restriction-modification pair of proteins. Why is it more
likely that
the modification protein also acts as an transcription activator of the
gene
coding for the restriction activity than the restriction protein acting
as a
transcriptional activator of the gene coding for the modification
activity?
Consider
genetic transfer of the gene cluster coding for the
restriction-modification genes. Cells receiving the cluster would
killed if the
restriction activity were expressed before the modification activity.
8. What operation is
facilitated by the following plasmid construct? The
plasmid vector contains the lac operator. It also contains
the lacI gene under control of a promoter that is always active. The
lacI gene
containts unique restriction sites located near the end of lacI such
that lacI
is fused to whatever peptide-coding or protein-coding sequence is
inserted into
them. (Hint, think phage display.)
Same
as phage display.
This is a trick to connect a protein to the DNA coding for it. In this case, in the cells, the
repressor-fusion protein product binds to the lac operator on the
plasmid. Cells are opened and the
protein-plasmid
complexes are separated from the free protein.
The protein-plasmid complexes are used in the selection of the
protein
activity.
9. How many
different maximal alignments are possible between the following two
sequences? The Needleman-Wunsch method
is a good way to start.
CBCDD
and DCCCD.
Eight. The
Needleman-Wunsch matrix for the sequences is as follows:
In this case, a
maximal alignment is a path from 3 to 2 to 1 subject to the constraint
that
after an alignment, the next alignment must be in a later column and
later
row. This matrix contains six
different paths from a 3 to a 2 to a 1 satisfying the condition.
Find
families where two or more siblings are very old. Map
the longevity genes using snp. (It is probably better to use families
rather than just a general population of old people because the trait
running in families makes the longevity more likely to be genetically
based rather than the product of good living (which however, itself,
may be genetically based).
Advanced
Molecular Biology-Dr. Schleif
December
4, 2008
90 minute
exam
Each
question is worth 10 points.
1. Estimate the
number of nucleotide changes necessary to convert a cow into a pig
(phenotypically).
If
humans and chimpanzees differ in nucleotide sequence
about 1/1000, assume the same for the difference between cows and pigs. Assume about 20,000 genes are expressed and
relevant to the differences between the animals. If the average protein
size is
about 300 amino acids, the gene plus regulatory sequences might be
about 1500
nucleotides. Thus, 30,000 nucleotides
is a reasonable first guess. Small RNAs
are unlikely to add much to this number.
2. AraC stimulates
initiation of transcription and presumably interacts with RNA
polymerase to do
so. How would you explain the failure
to detect interactions between purified AraC and purified RNA
polymerase in
ultracentrifugation experiments where the two proteins are cosedimented?
The
interaction is too weak to detect at the concentrations
that can be used in the centrifuge. The
interaction occurs in vivo because the proteins are held in the
immediate
vicinity of one another because they are bound to DNA alongside one
another. This
is just another example of the chelate effect in action.
3. Describe the
relevant genetic structure of a genetically engineered baker’s yeast
whose
haploid mating type can be controlled by the presence or absence of
leucine.
1. Nothing in MAT or
a type mating information in MAT.
2. Alpha 1 and alpha
2 are under control of a leucine-controlled promoter.
When
the alpha genes are not expressed, the cells are mating
type a, and when they are expressed, the mating type is alpha.
4. Identify two
important unanswered questions concerning the genome rearrangement and
genome
modification observed in the mammalian immune system.
Why
does J(D)JC rearrangement stop once one rearrangement is
completed?
What
is the mechanism of generation of somatic mutations?
5. Estimate the
smallest reasonable size for an insertion sequence and briefly explain
and
justify the points of your reasoning.
The
main determinant of the minimum size of an insertion
sequence would be the size of its transposase.
A few DNA binding domains of about 70 amino acids are known, so
250
nucleotides is about as small an insertion sequence as might be found.
6. What is the
possibility that in some cases, the root cause of a cancer is an
epigenetic
change (not contained in the nucleotide sequence of DNA) rather than a
nucleotide change in the genome?
There
seems to be no reason why not. We do know
that mutations in DNA sequence
are almost always involved, but there
seems to be no reason why the first step or any later step in the
development
of a cancer could not be epigenetic.
7. Suppose it were
found that the presence of IPTG in bacterial growth medium interferes
with
expression of the arabinose operon. What is a plausible explanation and
what is
an experiment that would prove or nearly prove your hypothesis?
IPTG
could bind in the arabinose binding pocket of AraC, but
not lead to folding of the N-terminal arm over the bound IPTG. Thus, the IPTG would interfere with
induction. The first experiment you
might do is check whether IPTG binds to AraC. You might expect IPTG
resistant
mutants to lie in or near the arabinose binding pocket.
Definitive evidence would be crystallization
of IPTG-AraC or IPTG-dimerization domain of AraC.
8. Tn10 hops. How
then does it manage to replicate faster
(on the whole) than the genome into which it is inserted?
Consider
a copy of Tn10 that hops. It only does so just
after a replication fork has passed by.
At the location this copy of Tn10 relocates to the Tn10 will, on
average, be replicated sooner than if had not hopped.
Thus, Tn10 replicates faster, on average than the chromosome.
9. How does the
existence of prions call into question the common assumption that the
conformation of proteins as normally found in cells is the lowest
energy state
of that polypeptide chain?
It
looks very much like susceptible proteins spontaneously
shift from their normally folded state to the prion state.
This implies that the prion state is lower
energy than the normally folded state.
