Biochemistry and Molecular Biology II – Tutorial club: Questions on
transcriptional regulation (07-05-18)
Q1. a) Describe what happens at the molecular level with the lac operon when E.coli
bacteria go from a glucose‐containing medium to a lactose‐containing medium.
In absence of glucose: cAMP levels are really high CAP is recognized by CRP and
then CAP binds to RNA-polymerase stimulating transcription. CAP is a protein
with an activation domain, that binds to the alpha domain of RNA-polymerase,
stimulating transcription.
Lactose binds to the lac-repressor, which results in the lac-repressor no longer
blocking the operator sequence this allows RNA-polymerase to bind to the
operator. When RNA-polymerase binds to the operator, it recruits sigma factors
(especially sigma70) which then together transcribe b-galactosidase, which then in
turn converts lactose into glucose and galactose.
b) X‐gal is an artificial substrate of beta‐galactosidase, one of the enzymes the lac operon
encodes for. Knowing that beta‐galactosidase turns X‐gal into a bright blue compound,
design a simple experiment to identify E coli mutants defective in beta‐galactosidase
production.
1. Use an E.coli culture, add X-gal measure with UV-VIS.
2. Use X-gal lactose IPTG
3. Use a mutagen (e.g. chemical IMS / spontaneous mutations / radiation e.g. UV
or X-ray)
Put E.coli into IPTG-rich medium and X-gal. The blue ones are wildtype, the other
ones (white) are defective in their B-galactosidase activity.
An antibody in combination with a fluorophore (e.g. GFP) that specifically binds to the
‘blue compound’ can be used and detected by immunofluorescence, with the
immunofluorescent microscope. When nothing can be seen under the microscope,
you can conclude that the blue compound did not form, thus b-galactosidase did not
properly work; converting X-gal into this blue compound.
c) What kinds of proteins or control elements might be defective in these mutants?
There could be a defect in either:
1. The core promoter elements
2. Promoter proximal elements
3. Enhancer/silencers
4. The cAMP binding site of CAP is mutated, by e.g. nonsense mutation.
5. Lac-repressor unable to bind lactose/IPTG.
6. Point mutation in LacZ
7. Mutations in the Shine-Dalgarno sequence, upstream of the beta-
galactosidase ORF
Q2. The figure below describes the relation between the nucleotide sequence of the
template and non‐ template strands of DNA and the corresponding RNA product. The
, author however has made three apparent mistakes in this illustration regarding
transcriptional and translational initiation. Which are these mistakes?
Three mistakes regarding transcriptional and translational initiation in this illustration:
1. No sigma factors at the promoter site?
2. No sensor and activator domains
3. RNA polymerase is not shown
4. The pribnow box (TATA-sequence/promoter sequence) should start at ± -10, the -
35 sequence is not shown/-35 elements are not shown.
5. There is no startcodon.
Q3. A. You are asked to set up an in vitro transcription reaction from a TATA‐box
containing promoter of a protein coding gene. A) Which protein(complexe)s would this
reaction need to contain?
The pre-initiation complex (PIC) and general transcription factors (e.g. TF II A),
together with TATA-binding protein (TBP) and TBP associated factors (TAFs).
CpG-islands facilitate the recruitment of other transcription factors.
This experiment is performed on a very strong viral promoter
B) Would your reaction work if you replace RNA polymerase II with RNA polymerase
I? Explain why yes or why not.
No, because only RNA polymerase II contains the TATA-binding protein (TBP), which
recognizes the promoter. RNA polymerase I and II also contain TBP, but in these
TBP is not involved in promoter recognition.
C) What would be the consequence if you do not add TFIIE to you reaction?
TFIIE enables binding of TFIIH;
- TFIIH uses ATP to pry apart the double helix at the transcription initiation site
allowing transcription to begin.
- TFIIH phosphorylates RNA polymerase II, releasing it from general factors, so
it can begin the elongation phase.
Therefore, if TFIIE is not added, the double helix would not be unwinded, thus
transcription could not take place. Also, RNA polymerase II would not be
phosphorylated and released from general factors, inhibiting the elongation phase.
Q4. Many key transcriptional activators also enhance their own expression by binding
to an upstream regulatory sequence within their own promoter. You have identified a
novel DNA binding protein that you suspect might work in a similar fashion. Design an
experiment to prove that the novel DNA binding protein indeed binds to its own
promoter AND that this binding event is indeed relevant for the gene’s regulation.
(There are multiple correct answers!).
Control: mutant DNA
CHiP-seq assay could be performed which makes the interactions between proteins
and DNA visible;
- The protein should be cross-linked to the DNA using formaldehyde (at the
promoter sequence), then the DNA-strands of interest should be sheared (cut
into pieces by e.g. ultrasonication), protein-specific antibodies are then
transcriptional regulation (07-05-18)
Q1. a) Describe what happens at the molecular level with the lac operon when E.coli
bacteria go from a glucose‐containing medium to a lactose‐containing medium.
In absence of glucose: cAMP levels are really high CAP is recognized by CRP and
then CAP binds to RNA-polymerase stimulating transcription. CAP is a protein
with an activation domain, that binds to the alpha domain of RNA-polymerase,
stimulating transcription.
Lactose binds to the lac-repressor, which results in the lac-repressor no longer
blocking the operator sequence this allows RNA-polymerase to bind to the
operator. When RNA-polymerase binds to the operator, it recruits sigma factors
(especially sigma70) which then together transcribe b-galactosidase, which then in
turn converts lactose into glucose and galactose.
b) X‐gal is an artificial substrate of beta‐galactosidase, one of the enzymes the lac operon
encodes for. Knowing that beta‐galactosidase turns X‐gal into a bright blue compound,
design a simple experiment to identify E coli mutants defective in beta‐galactosidase
production.
1. Use an E.coli culture, add X-gal measure with UV-VIS.
2. Use X-gal lactose IPTG
3. Use a mutagen (e.g. chemical IMS / spontaneous mutations / radiation e.g. UV
or X-ray)
Put E.coli into IPTG-rich medium and X-gal. The blue ones are wildtype, the other
ones (white) are defective in their B-galactosidase activity.
An antibody in combination with a fluorophore (e.g. GFP) that specifically binds to the
‘blue compound’ can be used and detected by immunofluorescence, with the
immunofluorescent microscope. When nothing can be seen under the microscope,
you can conclude that the blue compound did not form, thus b-galactosidase did not
properly work; converting X-gal into this blue compound.
c) What kinds of proteins or control elements might be defective in these mutants?
There could be a defect in either:
1. The core promoter elements
2. Promoter proximal elements
3. Enhancer/silencers
4. The cAMP binding site of CAP is mutated, by e.g. nonsense mutation.
5. Lac-repressor unable to bind lactose/IPTG.
6. Point mutation in LacZ
7. Mutations in the Shine-Dalgarno sequence, upstream of the beta-
galactosidase ORF
Q2. The figure below describes the relation between the nucleotide sequence of the
template and non‐ template strands of DNA and the corresponding RNA product. The
, author however has made three apparent mistakes in this illustration regarding
transcriptional and translational initiation. Which are these mistakes?
Three mistakes regarding transcriptional and translational initiation in this illustration:
1. No sigma factors at the promoter site?
2. No sensor and activator domains
3. RNA polymerase is not shown
4. The pribnow box (TATA-sequence/promoter sequence) should start at ± -10, the -
35 sequence is not shown/-35 elements are not shown.
5. There is no startcodon.
Q3. A. You are asked to set up an in vitro transcription reaction from a TATA‐box
containing promoter of a protein coding gene. A) Which protein(complexe)s would this
reaction need to contain?
The pre-initiation complex (PIC) and general transcription factors (e.g. TF II A),
together with TATA-binding protein (TBP) and TBP associated factors (TAFs).
CpG-islands facilitate the recruitment of other transcription factors.
This experiment is performed on a very strong viral promoter
B) Would your reaction work if you replace RNA polymerase II with RNA polymerase
I? Explain why yes or why not.
No, because only RNA polymerase II contains the TATA-binding protein (TBP), which
recognizes the promoter. RNA polymerase I and II also contain TBP, but in these
TBP is not involved in promoter recognition.
C) What would be the consequence if you do not add TFIIE to you reaction?
TFIIE enables binding of TFIIH;
- TFIIH uses ATP to pry apart the double helix at the transcription initiation site
allowing transcription to begin.
- TFIIH phosphorylates RNA polymerase II, releasing it from general factors, so
it can begin the elongation phase.
Therefore, if TFIIE is not added, the double helix would not be unwinded, thus
transcription could not take place. Also, RNA polymerase II would not be
phosphorylated and released from general factors, inhibiting the elongation phase.
Q4. Many key transcriptional activators also enhance their own expression by binding
to an upstream regulatory sequence within their own promoter. You have identified a
novel DNA binding protein that you suspect might work in a similar fashion. Design an
experiment to prove that the novel DNA binding protein indeed binds to its own
promoter AND that this binding event is indeed relevant for the gene’s regulation.
(There are multiple correct answers!).
Control: mutant DNA
CHiP-seq assay could be performed which makes the interactions between proteins
and DNA visible;
- The protein should be cross-linked to the DNA using formaldehyde (at the
promoter sequence), then the DNA-strands of interest should be sheared (cut
into pieces by e.g. ultrasonication), protein-specific antibodies are then
