Exercise for BCM-1
Week 9 (Constam)
November 10, 2014
1) True/False:
a. DNA footprinting can reveal the sequences of DNA elements that are bound by
interacting proteins
b. A DNA footprint refers to a region in a radioactively labelled fragment that is
completely degraded
c. The concentration of DNase I must be carefully adjusted to reach complete
degradation of the region of the fooprint
d. phylogenetic footprinting is a method to predict TF binding sites simply based on
their likely conservation in genomes of multiple phyla.
The point of this question is to make sure you understand the idea of partial
digestion, and that a footprint reflects protection against DNase
a: true; the immediate result of a footprinting assay is an autoradiogram with a
ladder of fragments, not a sequence. However, if the sequence of the labelled DNA
fragment, and the sizes of the fragments from the partial DNase digest are
determined, one can deduce the sequence of the binding site.
b: false; the region of a good footprint is blank on the autoradiogram because
protection of that area by associated protein prevents the production of fragments
of the corresponding size
c: false; the concentration (and time) must indeed be adjusted. However, not to
reach complete digestion, but rather to avoid it and thereby ensure that cleavage is
only partial.
d: true; Evolutionary changes in functionally important gene regulatory elements will
have deleterious effects, Similar to coding sequences, gene regulatory elements
cannot change
2) The cyclin D2 regulatory region contains several motifs that match the short
palindromic consensus sequence CANN'TG of an E-box (where E stands for
'enhancer'. E-boxes bind 'E proteins' and other heterodimeric bHLH factors
such as c-Myc/Mad.
a. How would you test which of the multiple E-boxes in Ccnd2 contribute to
its expression?
You can fuse a luciferase cDNA to a minimal promoter and add the
enhancer with wild-type or mutated E-boxes. You then compare the
expression levels of the luciferase reporter in transfected cells with or
without E-boxes to test whether expression decreases when one or
several of these E-boxes are mutated individually or in combination
b. Given the short nature of this motif, how do you explain that bHLH factors
do not control every gene in our genome?
E-boxes are so short that they can indeed be found all over the genome,
but it depends on the context of other TF binding sites and on chromatin
status whether they are functional.
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c. Like many other gene regulatory proteins, bHLH factors bind DNA as
dimers. What are the advantages of dimers compared to monomers
i) each half site contributes half of the binding energy. This increases the
affinity not just additively (X+Y), but rather synergistically (X * Y)
ii) a second advantage is increased sequence specificity (i.e. a longer
consensus binding motif is less likely than a short motif to randomly occur
in a genome)
iii) one monomeric, larger factor might still achieve i) and ii), but
compared to a smaller dimeric factor it will be less adaptable than dimers
(the latter can often form homo- or heterodimers)
3) Of 4 partial (incomplete) cDNA clones of the same bZIP protein, you want
to test which ones contain the region encoding the DNA binding domain. You
therefore translate the corresponding mRNAs in vitro into protein fragments
(using the translation machinery of whole rabbit reticulocyte extracts). You
then mix the translation products with radioactively labelled DNA containing a
TPA-responsive element (TRE) to monitor DNA binding by EMSA. On a
native polyacrylamide gel (b), you observe that the electrophoretic mobility of
the labelled DNA probe is shifted by the proteins derived from cDNA clones B,
C and D depicted in (a):
TRE shifted by
bound proteins
a. What does it mean that there is still a lot
of free (unbound) probe in each lane?
☜ free probe (TRE)
It means the labelled probe was added to the mix in large molar excess
b. Why is the shifted band in lane 3 less intense than those in other lanes?
The protein fragment analyzed in lane 3 shifts a smaller amount of the
labelled DNA probe than the proteins in the other lanes, either because it
was less efficiently translated or folded, or because it lacks part of the
DNA binding domain
c. Why is the probe shifted to different positions in lanes 3-5?
Electrophoresis was from top to bottom.The positions correspond to the
mobility of the proteins that have bound the probe. The mobility of the
complexes is determined by the size of the proteins (the DNA probe is
small by comparison). The shortest protein that binds (encoded by cDNA
clone B) retards the migration of the DNA fragment the least. The longest
protein (encoded by cDNA clone D) retards migration the most, so it
migrates less far and thus is found closer to the top of the gel.
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d. Which region of the cDNA above encodes the DNA-binding domain of
this bZIP protein?
The absence of binding by the protein A indicates that it does not include
the complete binding domain. Thus an essential portion of the binding
domain must map between the 5'ends of cDNA clones A and B. An
additional portion of the binding domain may map between the 5' ends of
clones B and C, since protein B seems to only weakly bind compared to
proteins C and D. Note: These experiments do not define how far the
binding domain extends in the 5' direction.
e. In lane 6, clones C and D were translated and loaded together. Why are
there 3 shifted bands rather than two?
The presence of a third distinct shifted band in the mixture of cDNA
clones C and D indicates that both fragments C and D can bind together
to the same DNA molecule, thus retarding migration of the probe to an
intermediate extent. This observation suggests that the protein normally
binds to its recognition sequence as a dimer (as is normal for bZIP TFs).
The new band arises because one molecule from clone C interacts with
one molecule from clone D, thereby forming a dimer with a molecular
mass intermediate between those of the homogeneous dimers.
f. What additional experiment would allow you to distinguish whether
clone D encodes the entire bZIP protein, or whether its 5' end is still
incomplete compared to a full length cDNA? Which enzymes would you
have to buy for this experiment?
Mapping of the transcription start site by a primer extension reaction on
the mRNA would tell you exactly how much farther the mRNA extends
towards 5'.
Polynucleotide kinase (to label the primer); Reverse transcriptase (for the
primer extension); DNA polymerase (for a sequencing reaction that is
then analyzed side-by-side with the extended cDNA fragment on a
polyacrylamide gel)
4) The POU domain factor OCT1 was found in a search for TFs that regulate
immunoglobulin expression.
a. To find novel OCT1 target genes in B lymphocytes in a ChIPseq
experiment, you will perform next generation sequencing of all fragments
of chromosomal DNA in immunoprecipitates of OCT1 from B-cell
extracts. True or false?
True
b. By simply comparing bound sequences from many different genes, it may
be possible to identify a consensus binding site of OCT-1 in B-cells. True
or false?
True
c. If you would use purified recombinant OCT-1 to select binding sites
among millions of synthethic oligonucleotides in vitro by the method
called SELEX, you should expect to find the same consensus motif as
the one that is bound in B-cells. True or false? Why?
False, because OCT-1 will recognize different target sequences
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depending on the conformation of its DNA-binding POUs and POUh
domains. The relative orientation of these domains is regulated by tissuespecific co-activators (e.g. BOB-1 in B-cells).
For question 2a), you may draw on the board something like:
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