01. Draw a graph showing the growth of E. coli on a 1:1 mixture of glucose
and lactose and superimpose on it the predicted levels of ß-galactosidase and
of cAMP.
2. You find that Facilemelodius rouge uses glucose in preference to
galactose. You also find that cAMP levels are controlled in Facilemelodius
rouge like they are in E. coli. Given this information, which growth curve,
and cAMP curve, would you expect to see if you provided Facilemelodius rouge
with a medium containing 10 mM galactose and 30 mM glucose.
3. If you found a mutant of E. coli that did not ferment lactose, maltose or
melibiose (all different disaccharides) how would you check the following
possibilities that the phenotype was caused by:
A. Only one mutation and not 3 separate mutations?
You could do a reversion analysis. Chances that three independent mutations
would revert would be very low—on the order of one cell in 10-18. On the other
hand if the phenotypes are due to a single mutation, then all three
phenotypes should revert together fairly often, something like
one cell in 10-6.
B. A mutation in a glycolsis gene which would cause the cells to be unable
to degrade monosaccharides that arise when the disaccharides are broken
down.
Check to see if the mutant grows on glucose. A glycolysis mutant should not.
C. A mutation that causes the cells to be unable to make cAMP
Add cAMP to the cells and see if it fixes the phenotypes.
D. A muation in CAP
Probably the easiest thing to do would be to try complementing with a plasmid
that carries the wt gene for CAP. If the plasmid fixes the phenotypes, then
the mutation was most likely in the gene for CAP.
4a. If the E. coli genome is 4.6 million basepairs in size, how many times
would you expect to find the sequence "GATC"? Assume that the genome has
equal amounts of the 4 DNA bases.
The frequencty of “GATC” should be .25 x .25 x .25 x .25= 1/256. Therefore
there should be 4,600,000/256=17968 (roughly) GATC sites in the geneome.
4b. If you wanted to cut the into about 10 pieces with a restriction enzyme,
about how many bases would that enzyme need to recognize (for example the
enzyme EcoRI recognizes the six-basepair sequence 5'-GAATTC-3').
4c. How big would the recognition sequence have to be to cut bacteriophage
lambda into 10 pieces (lambda is 48,500 bp in length)
5. 2. Imagine that you were in charge of producing a protein that was to
be used in clinical trial for diabetes control. You decide to express
the protein in E. coli you try two different expression systems, one
that is based on expression from the lac promoter, and one that is
based on an ethanol-induced promoter. Interpret the following results
keeping in mind that overexpression can lead to misfolded protein that
is inactive, and that ethanol is membrane permeable:
A. What are the main difference between the expression systems and why
did they likely come about? Which system would you recommend using for
the expression of the protein, and what concentration of inducer (IPTG
or ethanol) would you recommend using?
6 You are studying the control of the expression of the lac operon in E. coli
so you isolate a lac mutant and then look for revertants of it. This lac
mutant grows poorly on lactose (tiny colonies on lactose minimal medium) and
has a mutation that maps upstream of lacZ. You select for spontaneous
suppressors of it on lactose minimal medium and choose those that give large
colonies. You map the suppressors in these revertants and find they have new
mutations in the locations listed below. For each, briefly explain how a
mutation in that location might suppress the original mutation and cause the
observed phenotype.
a. the gene encoding the RNA polymerase alpha subunit.
A mutation in the CAP binding site that resulted in poor binding of CAP,
could be suppressed by a mution in the alpha subunit, if enhanced binding to
alpha also helped CAP bind to its site on the DNA.
b. the gene encoding the CAP protein
A CAP binding site mutation could be fixed by an altered CAP that better
recognized the DNA site.
c. the lac promoter
Again, a bad CAP binding site could be fixed by a mutation that increased
promoter strength—something like the lacUV5 mutation.
7.
In the November 1999 issue of the Journal of Bacteriology, Ding et. al
describe a protein from Acetivibrio cellulolyticus which functions as a
scaffold to hold seven proteins needed to degrade cellulose. The protein is
called scaffoldin and the sequence for the promoter region and the first part
of the gene is given below.
“D”
-35
-10
“C”
5’ATGCTATTTATCCCAAGTGTTCTCTGGTAACATATTTGACTCGTACAAGTTTTACTTATTAACATGAG
“E”
AGACAATAACTAACATGTATTAGGTATTTGTTTTTTTTGGGTACCTTAAAGATCTTTAAATAGATCATATAA
“A”
“B”
AAATAAAATTATGGGAGGAATGGTTAAATGAAAAAGGTTATCAGTATCAGTATTGATTTTAGCTATCG-3’
The nontranscribed strand is shown above. (NOTE “B” etc. are shown above
their sites which are double underlined). The +1 site would be:
A.
B.
C.
D.
E.
At
At
At
At
At
the
the
the
the
the
site
site
site
site
site
marked
marked
marked
marked
marked
“A”
“B”
“C”
“D”
“E”
8 A friend of yours claims the promoter shown in the scaffoldin problem
(problem 16) is likely bound by sigma70, yet the sigma70 consensus binding
site is TTGACA-N15-TATAAT. Is your friend likely right?
A. No, she is wrong because sigma70 never binds promoters.
B. No, she is wrong because real binding sites always match their consensus
sequences
C. Yes, she is right because only sigma70 can bind promoters.
D. No, she is wrong because the promoter site does not contain any “U”s.
E. Yes she is right because real binding sites usually only approximate
their consensus sequences, perfect matches are rare.
9. Below is the lac promoter region. Underline the important parts. Also
compare O3 and O1--how similar are they? (Note, the sequene starts on the top
line and continues on the bottom line.
GGCAGTGAGCGCAACGCAATTAATGTGAGTTAGCTCACTCATTAGGCACCCCAGGCTTTACACT
TTATGCTTCCGGCTCGTATGTTGTGTGGAATTGTGAGCGGATAACAATTTCACACAGGAAACAG
10. Monod (1956) observed that if cells were exposed to lactose,
and then later exposed to glucose plus lactose, catabolite
repression of the lac operon was weaker in cells pre-exposed to
lactose. Later, Cohn and Horibata (1959) found that LacY mutants
did not show weak catabolite repression (CR) after pre-exposure
to lactose. These results are diagrammed below.
A. w.t. --> grow
B. w.t. --> grow
C. lacY --> grow
D. lacY --> grow
in
in
in
in
glucose
lactose
glucose
lactose
-->
-->
-->
-->
expose
expose
expose
expose
to
to
to
to
glucose+lactose
glucose+lactose
glucose+lactose
glucose+lactose
-->
-->
-->
-->
get
get
get
get
strong CR
weak CR
strong CR
strong CR
Briefly explain the preinduction effect. Your explanation should include the
reason why LacY is needed for the effect. (10 pts)
The answer is that LacY, if it is present in high amounts (like following
growth in lactose), can overcome inducer exclusion and will show weak CR. The
reason for this is that there is not enough EIIAgluc-unphos to block all of
the LacY transporter when there are a lot of them following growth on
lactose.
11. Below are operators for a transcription factor called YycB.
Write the consensus sequence for the operator using G, A, T, C
and N (for no consensus at a particular site). Also, write
whether the operator consensus represents direct repeats or
inverted repeats. (10 pts)
Operator
Operator
Operator
Operator
1:
2:
3:
4:
TGTAATCGCTAGATCTGTAATG
TGTTATGTAGTTCAGTGTAATC
TGTTACAAGACATCATGTTCTT
TGTTATTCTCGCGGTTGTAACA
A brief overview of Kd.
12. What is the concentration of lac operator in a typical E.
coli cell? See Lac Book 3.2 for answer.
---------13. What is the concentration of LacI tetramers?
---------14.What fraction of the time is lac operator occupied if the Kd
for LacI binding to operator is 10-8 10-9, 10-10 and 10-11?
---------15. What fraction of the time is lac operator occupied if the Kd
for LacI binding to operator is 10-9 and there are 20 tetramers,
50 tetramers and 100 tetramers per cell?
16. A unique class of mutations in the crp gene are designated crp*. The
properties of the crp* and crp+ mutations in either a cya or cya+ background
are shown in the table below. The cells were grown in minimal medium with
glycerol (as the carbon source) and IPTG.
Strain
crp+
crp+
crp*
crp*
cya+
cyacya+
cya-
B-galactosidase
activity
1000
20
1000
1000
What is the probable molecular mechanism for the phenotype of the crp*
mutants. Briefly explain your answer.
How would the crp* mutation affect expression of other catabolic operons such
as gal or ara?
If you constructed a partial diploid [F' crp*/crp+] which mutation would be
dominant and why? If you added a column to the table with this mutant, what
would the predicted B-galactosidase levels be?
--------------17. A. Consider an E. coli mutant that was deleted for ptsH. This strain
should be unable to use glucose as a carbon source. Why?
ptsH encodes Hpr. Without Hpr phosphates cannot flow through the PTS system
and end up converting glucose to glucose-6-P
B. Would it be able to use lactose as a carbon source?
C. How about a crr deletion. Would it be able to use glucose? Would it be
able to use lactose?
E. How about a ptsG deletion (encodes EIIBC)? Would it grow on glucose?
Use lactose?
No, it would not grow on glucose because glucose could not get into the cell.
EIIAgluc would accumulate as the phosphorylated version, because the
phosphate would have no where to go. This form activates Cya to make cAMP and
does not block LacY—therefore the mutant should grow on lactose.
----------------18. Explain why the lacUV5 mutation allows E. coli cells to express the lac
operon without either cAMP or CAP.
-------------------
19. MEMORIZE THE CONSENSUS SEQUENCE OF AN STANDARD E. COLI
PROMOTER: TTGACA-N17-TATAAT
-----------------20. All of the following typically occur during growth on glucose in a
glucose+lactose diauxie experiment except:
A. Enzyme II glucose is unphosphorylated.
B. cAMP concentrations are low.
C. Lactose is kept out of the cell.
D. Adenylate cyclase is not activated.
E. All of the above happen