Download 06MicrobialGenetExamIAnswers

Name____KEY________________________
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Bacterial Genetics, BIO
4443/6443
Spring Semester 2001
Exam I
1.) Define the following. (5pts)
auxotrophic mutant: cannot make or utilize a nutrient that the wild-type organism uses for growth.
transversion mutations: mutation that changes a purine to a pyrimidine or visa versa.
missense mutations: base pair mutation in a coding region that changes an amino acid.
nonsense mutations: base pair mutation in a coding region that produces a stop codon (terminates translation)
nonsense suppressors: a mutation that allows an amino acid to be inserted at a nonsense codon during
translation (often the mutation occurs in a tRNA gene)
2.) What is unique about fmet? When or where is it found? (5pts)
a transformylase adds a formal group onto methionine. It is the first amino acid of every polypeptide
synthesized in most bacteria and is used during translation initiation
You add an antibiotic that inhibits translation to a bacterial culture.
3.) How would you expect this to affect factor independent transcriptional termination? Why?
(4pts)
It should not affect the transcription of these genes early on. Termination by this mechanism
relies upon a hairpin and stretch of UUUUs in the RNA transcript that interacts with the RNA
polymerase on its own to terminate transcription.
4.) How would you expect this to affect factor dependent transcriptional termination? Why?
(4pts)
(Either increase or decrease was acceptable with justification)
It should increase the frequency that transcription terminates since, in the absence of translation,
the termination factor, such as rho, would always have access to its binding site on the RNA
transcript (in the case of rho, it binds to the rut sequence). When translation is actively
occurring, the ribosomes may cover or hide the binding site preventing the termination factor
from functioning. (Or you could say it will decrease frequency that transcription terminates if
you thought that translation was needed to make the termination factor (such as Rho protein)
5.) A strain of Samonella enterica has approximately 6*106 base pairs in its genome, and it can
replicate and divide every 30 minutes. Assuming that it has one bidirectional origin of
replication (just like E. coli), what’s the minimum rate that the replication fork must be moving
in bases per minute? (Show your work) (6pts)
(6*106 bp/genome) / 1 (genome/30minutes) / 2 replication forks =
200,000 bp/minute / 2 replication forks =
100,000 bp/min/replication fork
6.) You have isolated two mutants that do not contain any ß-galactosidase (the mutants do not
grow on plates that have only lactose as a carbon source). They are interesting because when you
sequence the gene for ß-galactosidase in these mutants, neither mutant contains a mutation in the
coding region. You decide to sequence upstream of the start codon and find that mutant #1 has a
deletion within the TATA box. Mutant #2 contains a point mutation in the Shine Delgarno
sequence.
How can mutations in the TATA box or Shine Delgarno sequence result in a failure to
express any ß-galactosidase enzyme? (6pts)
An intact TATA box is essential for transcription initiation to occur. Without it no ßgalactosidase transcript can be made.
The Shine Delgarno sequence is important for the initiation of translation. Without it no
ß-galactosidase protein can be translated from the mRNA.
Ten years down the road….
The first round-trip space mission to Mars has just returned with large samples of soil and
ice from the polar regions of Mars. The big news in all of the journals is that scientists found an
organism that resembles bacteria on this planet and have cultured it out from one of the soil
samples! The bacteria-like organisms are able to grow under culturing conditions very similar to
the E. coli but seem to divide almost twice as fast.
At this point, scientists have examined the chemical make up the organism and found that
its DNA is identical in structure and composition to that which is found on earth. However,
nothing yet is known about how they replicate or divide.
NASA scientists have recently sent you a sample of the bacteria so that you can determine
if the DNA is replicated by a similar mechanism to that which occurs on earth.
You remember the Meselson-Stahl experiment and decide that repeating their experiment
on these new “bacteria” may be the best way to begin an analysis of replication mechanisms.
You grow three separate cultures of the bacteria.
A.) For the first culture, you grow the bacteria in media containing heavy isotopes of
nitrogen (14N) and carbon (13C) to density label the DNA during growth.
B.) For the second culture, you grow the bacteria in normal media.
C.) For the third culture, you begin by growing the culture in the media containing heavy
isotopes. Then, after several generations of growth in this media, you transfer the bacteria into
normal media and takes samples at various times ( I. immediately before transfer, II. after one
doubling time in the normal media, and III. after two doubling times in the normal media).
You lyse the cells, isolate the DNA and centrifuge each sample to equilibrium in neutral
CsCl gradients. Following centrifugation, you examine where the DNA ran in each gradient an
observe the following results shown below:
7.) NASA needs to know. Do these bacteria replicate conservatively, semiconservatively, or
distributively? (5pts)
Conservatively
8.) What would you expect the results look like if the bacteria replicated by one of the
alternative mechanisms? Circle one of the following replication mechanisms and then draw in
where you would expect the bands to run. Explain your answer below. (5pts)
Conservative. Semiconservative. Distributive.
Explain: After on generation in normal media, all DNA would contain one heavy and one light strand. After a
second generation, replication of the light strand of the hybrid creates a completely light molecule, while replication
of the heavy strand would remain a hybrid density.
Various size pieces of DNA that replicated either conservatively or semiconservatively would create a
distribution of DNA that ranges in composition but gets gradually becomes less dense on average as it replicates in
normal media.
In the course of your studies, you notice that the bacteria are able to grow on plates that
contain toluene as the sole carbon source. Realizing the potential commercial value of this
ability, you decide to try and learn more about the genes involved in this pathway by screening
for spontaneous mutants that cannot grow on toluene.
9.) Briefly describe how you would conduct your screen. (6pts)
(May wish to mutagenize the bacteria first to enrich for mutants)
1st Plate the bacteria on normal (nonselective or glucose) media: Everybody grows.
2nd Replica plate the bacterial colonies onto minimal media plates that contain only toluene as a carbon source: tol
mutants will not grow.
3rd determine which colonies didn’t grow on the toluene plates and isolate these from your original nonselective
plates: These are your mutants.
10.) Does this screen involve a positive selection or a negative selection? (4pts)
Negative selection
Through your screen, you isolated 26 mutants out of the 100,000 cells which you plated that were
could not grow on toluene. Using, the twenty-sixth mutant that you obtained (tol26), you screen
to isolate repressor mutants.
11.) Briefly describe how you would conduct your screen for revertants. (6pts)
(May wish to mutagenize the bacteria first to enrich for mutants)
1st Plate many of the tol26 bacteria onto minimal media plates that contain only toluene as a carbon source. (Also
plate an equal amount of cells on nonselective media so that you know how many cells were plated).
2nd Any supressor mutants or revertants should be able to grow on toluene plates again so these are your mutants.
12.) Does the revertant screen involve a positive selection or a negative selection? (4pts)
Positive selection
You plate more than 10 million tol26 cells and obtain only three revertants that are able to grow
once again on toluene plates.
13.) Based upon the observed reversion rate, which of the following mutation types are not likely
to have produced the original tol26 mutation? Why? (6pts)
Basepair change
Frameshift
Deletion
Deletions almost never produce revertions because large amounts of genetic information (multiple
bases) have been lost entirely.
14.) Why would you expect the reversion mutation rate (tol26 changing to wild type) to be so
much lower than the forward mutation rate (wild type changing to tol-)? Give one possible
reason. (4pts)
Toluene metabolism is likely to be controlled by many genes, so a mutation anywhere in this large target of
genetic information could produce an inability to metabolize toluene.
The tol26 mutation probably occured at a unique site in one toluene metabolism gene. So the chance that
this mutation, or a compensating one, will occur again is much, much smaller since the target nuber of bases that
could mutate to give you supression is so small.
Health concerns abound about the possibility that a foreign and potentially contagious bacteria
could accidentally be released on earth. Data involving its susceptibility to various antibiotics
and its mutation rates are important for the safety of the general public.
You find that the Mars bacteria is killed by treatment with rifampicin, a promising result. Not
surprisingly however, you notice that if you plate enough bacteria, rifampicin resistant bacteria
can be isolated. You decide to perform a fluctuation test to determine the mutation rate to
rifampicin resistance.
You grow 30 individual, 1 ml cultures of bacteria and spread each on a plate containing
rifampicin. By counting the bacteria from a separate culture, you determine that there were 1 *
108 bacteria/ml at the time you plated them on rifampicin.
After spreading each 1ml culture on separate plate, you let them grow overnight. The next day,
you count the following number of rifampicin resistant colonies that grew on each plate:
PLATE
#1
#2
#3
#4
#5
#6
#7
#8
#9
#10
# of rifR
colonies
1
4
0
0
0
0
6
6
1
4
PLATE
#11
#12
#13
#14
#15
#16
#17
#18
#19
#20
# of rifR
colonies
22
16
0
0
0
0
1
2
1
1
PLATE
#21
#22
#23
#24
#25
#26
#27
#28
#29
#30
# of rifR
colonies
2
4
0
0
0
0
9
1
1
4
Assuming that the mutations occur at random and that the mutation rate, a
= m/N.
15.) Estimate the mutation rate for rifR mutations occurring in the Martian bacteria by using the
Poisson expression where the probability of having i mutations per culture is represented by
Pi = ( mi e-m )/ i!. Show your work. (10pts)
12 out of 30 cultures had 0 mutations.
So the probability of having zero mutations (Pi ) is 12/30,
and i= 0 mutational events per culture in this situation.
Pi
(12/30)
0.4
0.4
-ln(0.4)
=
=
=
=
=
( mi e-m )/ i!
( m0 e-m )/ 0!
( 1 e-m )/ 1
e-m
m
.916
=
m
.916 mutational events per culture
and each culture has 1 x 108 cells
a
=
m/N
a
a
=
=
.916 mutational events/1 x 108 cells
9.16 x 10-9 rifR mutational events per cell division
Take Home Portion
1) (10pts):
Construct two primers (with the 5’ and 3’ ends labelled) that could be used to replace
one of the following genes with the genes for either chloramphenicol or kanamycin
resistence.
priA, priB, priC, rep, or hsdR
You may bring this anwser on a separate sheet of paper to be handed in with the exam.
Forward PriA-KanF primer
a acc cac cga att tca agt cag gat gat gct atg ccc
gttTATGGACAGCAAGCGAACCG
Reverse PriA-KanR primer
c gat ccg cct cgc atc gtg agc ggt tta acc ctc aat
cggTCAGAAGAACTCGTCAAGAAG
Forward PriB-CatF primer
t gaa gct ggg gat tct gaa gag taa ttt ctg atg acc
aacATGAGACGTTGATCGGCAC
Reverse PriB-CatR primer
acg acg gaa ata acg tgc cat atg gct a gtc tcc aga
atcCTTTCGAATTTCTGCCATTC
Forward PriC-CatF primer
t aac aat tat cat ttt cat tga ggt ctt atc gtg aaa acc
ATGAGACGTTGATCGGCAC
Reverse PriC-CatR primer
t tcc agt gac ata ttc tct cca ttg cta gcg ggt taa acg
CTTTCGAATTTCTGCCATTC
Forward Rep-KanF
t ccc ccc gtt cga aga ttg agc aat aca cct atg cgt cta
TATGGACAGCAAGCGAACCG
Reverse Rep-KanR
gct gac gca tct ttt ccg gcc ttg att att tcc ctc gtt t
TCAGAAGAACTCGTCAAGAAG
2) UV-induced mutagenesis questions for thought (10pts)
1. Why did we plate out the unirradiated cultures on rifampin-selective plates? What do
the results from this part of the experiment tell us? This is a control that tells us the
basal mutation rate for the cultures when they are not exposed to the mutagen. By
comparing it to the cultures that were exposed to UV, it allows us to determine whether
the treatment increased the mutation rate.
2. Given that UV-induced mutagenesis depends on the induction of the SOS response
following DNA damage, would you expect the number of rif-resistant mutants to be
higher before or after overnight incubation in the irradiated cultures? Why? The
frequency of rifR mutants should be higher in the cultures after overnight incubation.
Before mutations arise, the DNA damage must be replicated (inaccurately) and fixed
into the genome before it is expressed. Immediately after UV treatment, there has not
been enough time for this to occur.
3. Would you expect the mutation frequency in unirradiated cultures to change after
overnight incubation? Why? The mutation frequency after overnight incubation should
be very close to the mutation frequency before overnight incubation, since the mutation
rate in the cultures not expose to mutagens is constant and very low.
4. Which strain gave the highest number of rif-resistant mutants? The lowest? What
can you infer about the roles of the genes we tested in the cellular response to DNA
damage (mutations)? The mutS mutant had the highest mutation rate, suggesting it
plays a role in maintaining (or correcting) mistakes during replication. The umuC mutant
had the lowest mutation rate and did not exhibit any UV-induced mutagenesis,
suggesting that this gene somehow plays a role in generating the mistakes tat are
made during replication. The frequency of refR mutants in the uvrA strain was difficult
to deterimine from this expt because the cells all died at the dose of UV we utilized. The
observation indicates that uvrA is important for survival after UV-induced damage.
5. Bonus question: What might be the selective advantage for retaining an error-prone,
i.e. mutagenic, system for coping with DNA damage? This refers to the umuC gene.
One speculative idea is that the gene allows the cell to survive in a situation that would
otherwise kill the cell, although this occurs at the expense of tolerating mutations in the
genome. We’ll revisit this topic again.