Download Spr01Final Exam Answer Key

Name_______________________________
Bacterial Genetics, BIO
4443/6443
Spring Semester 2001
Final Exam
Student ID#__________________________
1.) Different bacteriophage utilize different mechanisms to ensure that their own genes (and not
host’s genes) are transcribed and regulated. Describe a mechanism that each phage uses to
transcribe and regulate their own gene expression rather than those of the host. (6pts)
phage T7:
encodes its own RNA polymerase
phage T4:
encodes a sigma factor to modify the recognition of the E.coli RNA polymerase
phage λ (lambda):
uses the host RNA polymerase and encodes antitermination proteins that regulate which genes are
tanscribed.
(Several other answers would also be acceptable for each phage)
2.)
T7 has a typical lytic life cycle. Describe two gene products that you expect to be expressed
early after infection? (9pts)
RNA polymerase
Antirestriction Protein
Kinase that inactivates the E. coli RNA polymerase
RNA polymerase
Host DNA degradation genes
DNA ligase
Describe two gene products that you expect to be expressed late after infection?
DNA Polymerase
Primase/Helicase
Phage Particle Parts
Lysis genes
Hok
lysozyme
How does the early viral DNA replication differ from late viral DNA replication?
Early: Controlled bidirectional replication initiated from a unique origin
Late: Replication become highly recombinagenic utilizing a rolling circle form of replication
to amplify the genetic material
1
Bacteriophage T4 normally produce large, fuzzy edged plaques when plated on a lawn of
E.coli . However, rII mutants (named for rapid lysis, classII) produce smaller, clear edged plaques
on E.coli. Additionally, it is known that rII mutants cannot grow at all in E.coli which carry a
lysogen of the phage lambda, E.coli λ. By contrast, wild type phage, r+, grow normally in E.coli λ.
The type of plaque formed in each case is sumarized below.
Use Wildtype T4 (r+)
to infect E.coli
Use rII mutants
to infect E.coli
Use Wildtype T4 (r+)
to infect E.coli λ
Use rII mutants
No
to infect E.coli λ Growth
3.)
Seymour Benzer made use of these properties to determine several fundamental genetic
principles. He first isolated and collected several different rII mutants based upon their plaque
morphology. Suppose you were trying to isolate rII mutants. After you mutagenize your phage,
would you infect E.coli or E.coli λ to screen for your mutants? Why? (4pts)
E.coli, because the mutants you are hoping to isolate would not grow in E.coli λ.
After isolating several mutants, Benzer then tried to determine how many genes were
represented by the rII mutations. He mixed two different rII mutations together and infected
E.coliλ cells at a high MOI. Using a sample of just four of his mutants, the type of plaques
produced by each combination is shown in the table below
rII #1
rII #2
rII #3
rII #4
rII #1
No Growth
rII #2
No Growth
No Growth
rII #3
No Growth
rII #4
No Growth
4.)
What type of analysis is this? (3pts)
Complementation analysis
5.)
How many genes are represented by the four mutants shown above? (6pts)
No Growth
Two: Gene A (rII #1 and rII #2) and Gene B (rII #3 and rII #4)
6.)
Why is it necessary to infect at a high MOI for this experiment to work? What would have
occurred had he used a very low MOI instead? (2pts)
You need to make sure most cells are infected by both mutant phages. At a very low MOI,
cells would only get infected by one of the mutations and would have no chance to assay for
complementation.
2
rIIB is a gene that causes an rII phenotype when it is inactivated. Francis Crick and
collegues utilized the rIIB gene product to demonstrate that the genetic code was most likely based
on three letters (i.e three bases code for one amino acid). As part of this experiment, they
mutagenized T4 phage by growing them in the presence of acridine dyes and isolated an rIIB
mutant (remember that acridine dyes generally induce either a one base pair insertion or a one base
pair deletion).
7.)
Then they isolated several supressors of rIIB that restored wild type function (an r+
phenotype). Using E.coli λ, what would be an easy way to detect rIIB phage that have suppressor
mutations? (5pts)
Infect and plate on E.coli λ, phage that are able to grow and form plaques are the suppressors.
Once they had obtained several rIIB suppressors, they needed to know which ones were
true revertants and which ones were second site suppressors. (The second site suppressors were
going to be central to their experiment).
To do this, they
decided to cross the
rIIB suppressors with
wild type phage and
examine the type of
recombinants that are
produced. Shown to the
right, are what each type
of phage plaque looked
like before the crosses
were performed.
If he used wild type
T4 to infect E. coli,
he observed only large,
fuzzy edged plaques.
To cross the phage, they infected E.
coli with a mixture of a wild type T4 phage
and an rIIB suppressor. Then, they allowed
the co-infected E.coli to lyse and collected
the progeny phage.
To examine the recombinants that
were formed, they infected the E.coli again,
this time using only the progeny phage
from the cross, and plated the infected cells
out on a lawn of E.coli. Shown below are
the types of plaques produced on each plate.
If he used his rIIB
mutant to infect E. coli,
he observed only
small, clear edged
plaques.
The progeny phage
from some crosses
produced a mixture of
large, fuzzy edged
plaques and small clear
edged plaques on E.coli.
If he used any of his
rIIB suppressors to
infect E. coli,
he observed only large,
fuzzy edged plaques.
However, the progeny
phage from other crosses
produced only large, fuzzy
edged plaques on E. coli.
8.)
Based upon the above results, which suppressors are likely to be a true revertants and which
are most likely to be a second site suppressors? Why? (9pts)
Second site suppressors produced the mixed plate since a cross between a double mutant and wild
type phage could produce single mutant recombinants.
True revertants produced the plate with only wild type phage since a cross between two wild type
phage can produce only wild type recombinants.
3
Phage λ is able to regulate its gene expression such that it can either enter a lytic or lysogenic life
cycle upon infecting E. coli. The principle region that regulates this decision is shown below.
pI
pL
p RM
cII
O P
tR2
cro
tR1
O R3
O R2
cI
O R1
O L3
O L2
N
O L1
cIII
tL1
int xis
tL2
attP
RnaseIII
cleavage
site
t
pR
Q
p RE
9.)
For each bacteriophage below, determine whether the bacteriophage will enter a lytic or
lysogenic cycle upon infection and explain why. (6pts)
A phage λ with a mutation inactivating the cI gene
Lytic. CI is needed for repression of pR and pL (the lytic promoters)
A phage λ with a mutation inactivating the cII gene
Lytic. CII is needed to activate pRE and pI which transcribe the cI and int genes, both of
which are required for lysogeny.
A phage λ with a mutation inactivating the cro gene
Lysogenic. Cro competes with CI for the Or1 Or2 and Or3. Without Cro, CI binds and
promotes lysogeny
10.) Phage integration into the chromosome requires that only the int gene product be present,
whereas phage excision from the chromosome requires the function of both the int and xis gene
products.
How does phage λ regulate the expression of these genes such that int and xis are only
expressed at the proper time? (4pts)
Before integratation, the transcripts from pL containing int and xis are degraded because
they also contain the RNaseIII degradation signal. Only transcripts from pI, which contain only the
int gene can be expressed in a stable form.
Following integration using the attP site, the circular lambda genome is linearized removing
the RNaseIII degradation sequence from the region. This stabilizes the pL transcript which
contains both the int and xis genes.
You need to determine if two mutant phage belong to the same complementation group. In your
experiment you decide to infect at an MOI of 2.
i
-m
11.) Using the poisson expression, Pi = ( m e )/ i!, what fraction of the bacteria in your
experiment will actually be infected by exactly two bacteriophage? Show your work. (6pts)
probability of getting infected by exactly 2 phage=
P2 = ( 22 e-2 )/ 2!
P2 = (4)(.135)/ 2
P2 = .27 or 27% or 27/100 bacteria
What fraction of the bacteria will be infected by two or more phage?
probability of getting infected by more than 2 phage= 1-(P2 +P1 +P0 )
P2+ = 1-(( 22 e-2 )/ 2! + ( 2 1. e-2 )/ 1! + ( 2 0 e-2 )/ 0!
P2+ = 1-((.27) + (.27) + (.14)
P2+ = 1-(.68)
P2 = .32 or 32% or 32/100 bacteria
4
Shown below are two operons that control genes for the metabolism of different carbon sources.
The lactose operon and the L-arabinose operon.
lacZ lacY lacA
araC
p lacI
O1
I
p BAD
O2
lacI
O
p lac
araB araA araD
p araC
The lac operon is regulated by the lacI gene product, while the L-arabinose operon is regulated by
the araC gene product. Although these operons are similar both lacZlacYlacA and
araBaraAaraD are only expressed when either lactose or L-arabinose is present in the media, the
are regulated through very different mechanisms.
12.)
How does LacI regulate the lac operon? What type of regulation is this called? (2pts)
In the absence of lactose (allolactose), LacI binds to the Operator (O) and represses
transcription.
In the presence of lactose (allolactose), allolactose binds to LacI preventing it from binding
to the Operator, allowing the lac operon to be transcribed.
Negative regulation.
13.) How does AraC regulate the L-arabinose operon? What type of regulation is this called?
(2pts)
In the presence of arabinose, arabinose binds to AraC allowing it to bind to the Operator (I)
and activate transcription.
In the absence of arabinose, AraC cannot bind to the Operator and the operon cannot be
transcribed.
Positive regulation.
Suppose that you were trying to isolate mutants that constitutively express either the lactose operon
or the L-arabinose operon.
14.)
What is the most likely type of mutation would you expect to get that would produce a lac
constitutive phenotype? (2pts)
Any mutation that inactivates the lacI gene
15.) What what is the most likely type of mutation would you expect to get that would produce
an ara constitutive phenotype? (2pts)
A specific (subtle) mutation that changes the conformation of AraC so it will bind the
operator in the presence or absence of arabinose.
16.) If you were to screen equal numbers of total cells in each case, would you expect to obtain
more lac constitutive mutants or more ara constitutive mutants? Why? (2pts)
You would expect to get many more lac constitutive mutations since a mutation anywhere
within the lacI gene or promoter that inactivate it would produce the desireed phenotype, whereas
there are probably only one or two bases in the araC gene that would produce the subtle change that
you are looking for.
17.) If a wild type lac operon was introduced on an F’ into the lac constitutive mutant, would the
constitutive phenotype be dominant or recessive to the wild type operon? Why? (2pts)
Recessive. The wild type copy of LacI could still bund and repress both repressors.
18.) If a wild type ara operon was introduced on an F’ into the ara constitutive mutant, would
the constitutive phenotype be dominant or recessive to the wild type operon? Why? (2pts)
Dominant. *If we assume AraC is simply an activator. Then, the constitutive mutant would
always bind both copies of the operon and activate them. (In reality, AraC also functions as an antiactivator protein, so Recessive would also be correct as long as you explained why.)
5
You have an auxotrophic bacterial strain that is resistant to naladixic acid, his-, trp-, nalR. Because
all three markers appear to map closely, you decide to map these genes by transduction using a
three factor cross.
You use a wild type strain, his+, trp+, nalS , as your donor
and the his-, trp-, nalR as your recipient
You select 100 his+ transductant colonies for further analysis and score
them as having the following genotypes:
his+trp+nalR
his+trp+nalS
his+trp-nalR
his+trp-nalS
5 colonies
19 colonies
49 colonies
27 colonies
19.)
What is the cotransduction frequency of nal with his? (2pts)
How often does his+ also bring in nalS with it...
(19+27)/(5+19+49+27)=
47/100= 47% cotransducable
20.)
What is the cotransduction frequency of nal with trp? (2pts)
Typo. Everybody gets this one for free.
21.)
What is the map order of the three genes? Show your reasoning. (6pts)
Rarest phenotype is likely to involve the most exchanges.
his+trp+nalR 5 colonies
his+
nalS
trp+
his-
nalR
trp-
This gene order requires that 4 exchanges must occur.
The other two possible gene orders would only require 2 exchanges
A pathogenic strain has been isolated from a local hospital. They suspect that the strain may be a
mutator (ie. it has an abnormally high mutation rate) and they have asked you to determine its
6
mutation rate. Since this bacteria is sensitive to rifampicin, you decide to measure the mutation rate
for rifR (risistance to rifampicin) a commonly used and widely accepted measurement of an
organism’s overall mutation rate.
22.) How would you screen for rifR mutants? Does this screen utilize a positive or negative
selection? (6pts)
Plate cells on plates containing rifampicin. Cells that grow are rifR mutants.
This is a positive selection.
The number of rifampicin resistant colonies that you obtained from 30 individual, 1 ml
cultures of bacteria are shown in the table. By counting the bacteria from a separate culture, you
determine that there were 1 * 106 bacteria/ml in each culture.
PLATE
#1
#2
#3
#4
#5
#6
#7
#8
#9
#10
# of rifR
colonies
12
4
34
126
0
0
6
15
1
42
PLATE
#11
#12
#13
#14
#15
#16
#17
#18
#19
#20
# of rifR
colonies
22
16
33
14
0
0
1
22
1
11
PLATE
#21
#22
#23
#24
#25
#26
#27
#28
#29
#30
# of rifR
colonies
2
4
136
25
0
0
9
71
11
4
23.) Assuming that the mutation rate, a = m/N, estimate the mutation rate for rifR mutations by
using the Poisson expression where the probability of having i mutations per culture is represented
i -m
by Pi = ( m e )/ i!. Is the isolate a mutator? (10pts)
6 out of 30 cultures had 0 mutations, so the probability of having zero mutations (P0 ) is 6/30, and
i=0 mutational events per culture in this situation.
Pi
=
( mi e-m )/ i!
0 -m
(6/30) =
( m e )/ 0!
0.2
=
( 1 e-m )/ 1
0.2
=
e-m
-ln(0.2)
=
m
1.61
=
m
1.61 mutational events per culture
and each culture has 1 x 108 cells
a
=
m/N
a
=
1.61 mutational events/1 x 106 cells
a
=
1.61 x 10-6 rifR mutational events per cell division
This is a high mutation rate for a bacteria so it probably is a mutator. (No points off if you got
this part of the question wrong though)
7