Download E. coli - UCSF Biochemistry

Bacteria Handout #1
Genetics 200A
January 6, 2014
Learning Objectives of the Lambda (λ) Lectures
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•
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Understand principles of a genetic approach to answer biological questions.
Learn the molecular interactions that underlie the λ life cycle, a paradigm for developmental processess in higher organisms.
Because we know many of the molecular details of its biology, λ provides an intellectual framework that serves as a useful grid to understand many basic biological processes (e.g. gene regulation, protein-pro
tein interactions, cooperativity, proteolysis, etc.)
Specifically, we will learn:
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•
•
•
•
•
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How to perform a genetic screen and analyze mutants in bacteria and phage.
How to work with essential genes in bacteria
How lambda programs a sequence of events – sequential gene activation
The mechanisms of a decision-making process in which lambda responds to environmental cues
How lambda interacts with its host cell, E. coli
The molecular mechanisms of important regulatory proteins
DNA recombination
Genetic Concepts in Lecture #1:
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•
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Gene and Allele
Essential gene/Lethal mutation
Conditional mutation
Genetic screen and selection
Mutagenesis/spontaneous mutation
Complementation
Suppression
Haploid
Recombination
Bacteria:
Diverse in size, shape, metabolism, and environmental niches
Have a high degree of genetic plasticity, can share their genetic material with one another
Likely all bacteria employ cell-cell communication and employ group behaviors
Despite lacking major membrane bound organelles, bacteria compartmentalize functions within the cell and
have cytoskeletons, nucleoid packaging, etc.
Bacterial viruses:
“Dark matter of the biological universe” - estimated 1031 phages on earth, ~1025 new infections per sec.
Development in Pr
Some Important Bacterial Experimental Systems
Bacillus/Clostridia
Caulobacter crescentus
Caulobacter crescentus
Bacillus subtilis
Myxobacteria
bobtailed squid
Magnetospirillum magneticum
Vibrio fischeri
E. coli
1.
2.
3.
4.
5.
Genome sequenced (haploid) – 4.6 Mb circular chromosome w/ approx. 4500 genes.
Genetically and biochemically tractable
Short doubling time – 20 min under optimal conditions.
Well established model for studying fundamental processes of all cells (transcription, translation, replication, adaptation to environment, intermediate metabolism)
Not just a host for plasmid and protein purification - many aspects of bacterial biology remain mysterious - e.g. host interactions, biofilms, quorum sensing.
1.
2.
3.
4.
Genome sequenced in 1983 – 48.5 Kb, 45 genes
“Simple” organism.
Extremely well studied (λ research predominated in the 1960s & 1970s, but continues).
Much of what we know about bacterial cells stems from the study of their associated viruses.
Bacteriophage λ (lambda)
Phage Life Cycle
Many bacteriophages employ a single mode of growth, once they infect their host they replicate and lyse the
bacterium. Thus, these viruses are termed “lytic” bacteriophages. An example is the well studied phage T4.
DNA (phage
genome)
Head
Tail
Tail Fibers
1
Life cycle of phage infection:
2
1)
2)
3)
4)
5)
4
3
phage absorption to surface receptors
DNA injection. (Nothing else gets injected)
phage DNA replication
morphogenesis (head/tail assembly, genome packaging into head)
release of phage by bacterial lysis
Infection of E. coli with a single bacteriophage kills the bacterium and
leads to release of approximately 100 viable phages.
5
λ
(50 kbp genome)
E. coli
+
Bacteriophage lambda decides
between two modes of growth
Lytic Response
host dies (like T4)
100 phages liberated
Temperate response
host lives
phage DNA integrates
In temperate response, the host lives with λ genome stably integrated into the host chromosome. λ is also referred to as a “temperate phage”. We’ll discuss the fate of λ DNA in a bit.
This λ lysogen “E. coli (λ)” has new properties:
•
immunity to λ re-infection
•
can be induced to produce λ (e.g. UV light)
λ senses intracellular environment, then makes a decision!
The λ lectures will be divided into three sections:
1.
2.
3.
Lytic pathway—production of 100 viral particles, cell death.
• What are the genes required and how is the sequence of events programmed?
Lysogenic pathway—production of viable lysogen (immunity and activation upon UV irradiation)
Choice—After discussing the genes and proteins responsible for this, we shall discuss how lambda makes a choice between (1) and (2).
We will see: (a) what proteins are needed for lambda to grow in these two different modes, (b) how regulatory genes were identified and gain an overview of how their proteins work, (c) how the action of multiple
regulatory proteins is coordinated.
I. LYTIC PATHWAY
Key Concept: There is a temporal order of
events once phage DNA is injected. Phage
replication, morphogenesis, and lysis must
be timed appropriately in order for functional phage to be released.
Fundamental intracellular requirements
1) Phage DNA replication
2) Head & tail protein production
3) Lysis and phage release
Heads/tails
How is this sequence of events programmed by λ?
The answer to this question was elucidated from many elegant genetic and biochemical experiments. In particular,
the genes required for λ growth were first identified from
genetic screens. How was this performed? Before we address this issue, we first will see how to work with phages
experimentally and general strategies to identify mutants.
DNA
replication
phage
release
0
10
20
Early
30
40
50
Late
60
The plaque assay is the fundamental method to count phages.
Working with phage
The plaque assay
0.1mL
0.1mL
10x serial
dilutions
Conc. phage
~109 λ/mL
1mL
2x107 λ
1mL
2x108 E. coli
Growth tube
1mL
~103 λ
0.9mL
buffer
lawn of uninfected
bacteria
~108
E. coli
12h
37o
Multiplicity of infection (MOI) = #phages/#bacteria, so at t=0 the MOI = ?
What is the concentration of phages at t=0?
What is the concentration of phages at t=1h?
plaque that arose from a
single phage
nutrient agar plate
How do we count phages?
Before we move on to the lambda screens, lets quickly review the difference between a genetic screen and
a genetic selection, and the basics of how to identify mutants. Here are a couple of examples:
Genetic Screen
Genetic Selection
Example - identification of E. coli mutants that can’t ferment lactose
On agar media with a visual pH indicator, WT E. coli forms red colonies.
E. coli
Lac+
(Red)
E. coli
Example - identification of ampicillin-resistant E. coli mutants
E. coli AmpS
Lac-
(White)
WT
Mutagenize
WT
Mutagenize
Plate
103
E. coli AmpR
Plate
103
Plate
106
Plate
106
AmpR
mutant
White
mutant
Agar media with ampicillin
In a genetic selection, conditions are selected in which the desired mutant can grow but all other non-mutants
cannot. For a screen, there is no selective growth advantage for the desired mutant and thus must be picked out
from all other strains by some observable phenotype.
Mutant isolation:
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Begin with an observable phenotype
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Mutagenize the organism (DNA damaging agents – UV, EMS, etc.) This is not always required due to
“background” spontaneous mutagenesis rate due to intrinsic errors in DNA replication and repair.
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Isolate individuals and score phenotype
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Pick mutants with relevant phenotype
Mutant characterization:
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Secondary screens
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Complementation tests to determine number of genes
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Dominance test
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Mapping/sequencing
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Clone the genes
We want to identify all lambda genes necessary for plaque formation but how can we mutate genes that are essential for the organism to grow? The strategy used to identify mutants defective for lytic pathway is a powerful one, the use of conditional lethal mutations. In this way, mutant alleles can be identified that are functional
under one condition (permissive condition) but non-functional under another conditions (restricitve or nonpermissive condition). In this way, the mutant can be grown under permissive conditions, and the effects of the
mutated gene can be studied under restrictive conditions.
Two different kinds of of conditional-lethal screens were performed to identify lytic genes: temperature sensitivity and nonsense suppression.
Lytic program of λ
Order of events is the same as for T4: 1. DNA synthesis
2. Morphogenesis
3. Lysis and phage liberation
What λ genes are required for the lytic pathway?
Problem! We are searching for mutations in essential genes.
e.g. A tail- mutant is DEAD, so how can we isolate a mutant?
Mutants were isolated that were conditionally lethal.
2 kinds of conditional mutants were isolated:
1. Temperature sensitive (ts)
λwt
λts
30o
42o
+
+
+
In wild-type E. coli, there are no
tRNAs that contain the anticodon
sequence for the three stop (nonsense)
codons. Under normal circumstances,
ribosomes and release factors recognize stop codons and terminate
translation. If a λ tail gene has a codon
mutated from sense to antisense,
translation will terminate at that codon
and will produce a truncated, likely
non-functional protein during infection of wild-type E. coli. However,
tRNA genes can also be mutated such
that the anticodon can now basepair
with a nonsense codon, allowing for
translation to read through individual
stop codons. E. coli strains containing
such a mutant tRNA gene allow for the
phenotype of nonsense mutations to be
suppressed. Hence, these strains are
called nonsense suppressors.
2nd kind of conditionally lethal screen Nonsense suppression
λ tail
DNA
RNA
a.a.
protein
MUTATION
...actCAGgtc...
...acuCAGguc...
Gln
Full-length
λ tail...actTAGgtc...
...acuUAGguc...
STOP (nonsense)
Truncated in WT E. coli
Useful bacterial mutants - Nonsense Suppressor strains. Have a tRNA gene that has a
mutation in the anti-codon that allows it to recognize one of three nonsense codons.
Gln
3’
Gln
5’
3’
GUC
...acuUAGguc...
Stop Codons
5’
AUC
...acuUAGguc...
3 amber suppressors
suI+ = tRNAser
UAG = “amber”
suII+ = tRNAgln
UAA = “ochre”
UGA = “opal”
suIII+ = tRNAtyr
The idea, then, is to identify λ amber mutants that cannot grow on wild-type E. coli but can grow on an
amber suppressor strain.
But how powerful is this approach? There are 61 sense codons, only 8 (13%) of which can mutate
into an amber stop codon with just one nucleotide change. Given the redundancy of the amino acid
code, aren’t there genes with codons that can’t give rise to stop codons in one step?
GAG - Glu (1 of 2)
AAG - Lys (1 of 2)
CAG - Gln (1 of 2)
UGG - Trp (only 1)
UUG - Leu (1 of 6)
UCG - Ser (1 of 6)
UAC - Tyr
UAU - Tyr (only 2)
Thus, two of the least redundant amino acid codons can mutate to amber with just one nucleotide
change. Any gene encoding a protein that contains Trp or Tyr can mutate to amber in one step.
If you were handed wild-type λ, wild-type E. coli, and E. coli
suII+, how would you isolate λ lytic mutants?
λ wt
λ amber
E. coli wt
E. coli suII+
Results: Alan Campbell isolated 130 mutants: they grow in an suII+ bacterial strain
(strain C600) but not in a wild-type bacterial
strain such as 594.
Do the mutations affect different functions/
genes? This can be determined by doing
pairwise co-infections with individual mutants. It is important that most of the cells
are infected with both phages. If the mutants
have defects in different genes, then the
defective gene product from one phage will
be helped by the presence of the normal gene
product from the other phage, and vice versa.
Phage production will occur when these two
mutants are present in the same cell. If the
mutants have defects in the same gene, the
gene products will not help each other and no
viable phages will be produced.
Complementation test: mixed infection of
NONPERMISSIVE HOST
The standard complementation test:
(1)
Infect nonpermissive wild-type E. coli
strain with two different lambda am mutants
(each at an moi of 5)
(2)
Allow infected cells to produce phage
(incubate 60 min). This is the test.
(3) Assay total phage produced on permissive strain - this is the assay to determine
if the phages complemented each other during infection of the wild-type strain. Compare the number of phages produced per
infected cell, with the number produced by
lambda wildtype and with the lambda am mutants grown singly (that is, not coinfected).
Low phage yield indicates failure to complement; high yield indicates complementation.
Campbell: 130 mutants in 18 complementation groups.
Parkinson: 310 mutants; five new genes
discovered.
This is exactly what Allan Campbell did way back in 1961 and isolated 130 λam mutants.
How many essential genes are there? Are mutations in the same or different genes?
COMPLEMENTATION TESTS. Co-infect same cell with 2 different λam mutants.
λam1: head+ tail-
head+
λam2: head- tail+
head-
tail-
Co-infection
tail+
head-
tail+
head+
tail-
What MOI would you use?
What E. coli strain would you use?
1.
Two methods for λam complementation tests
λam1 + λam2
1h
Count phages by
plaque assay
(many or few)
WT E. coli
(non-permissive)
2.
λam1 λam2
suII+ strain
(permissive)
λam1 + λam2
Complementation
gives rise to large
plaque
12h
lawn of WT E. coli
+
An example:
= many phages
= few phages
λam1
λam1
λam2
+
λam3
+
λam4
+
λam2
+
λam3
+
λam4
+
+
+
+
What can we conclude from
these pairwise co-infections?
+
Allan Campbell identified 18 genes from his collection of 130 amber mutants.
Sandy Parkinson found 5 more genes from a set of 310 amber mutants.
What do these 23 essential genes do?
The function of each gene product can be inferred from the phenotype of the mutant in the nonpermissive host. Here is the genetic map of λ (we’ll see how this was determined next) and the
phenotype of each amber mutant during infection of WT E. coli:
AWBCDEFZUVGTHMLKIJ
e.g. λEam
DNA Rep:
lysis:
heads/tails:
N
OP Q SR
e.g. λVam
+
+
+
+
+/+/-
+ +
+/-
+/-
A-F are head genes
Z-J are tail genes
S,R are required for lysis
N,O,P,Q mutants are pleotropic
O,P required for DNA replication
N,Q encode transcriptional activators - N activates O,P,Q transcription (Early genes), while
Q activates A-J,S,R (Late genes).
The relative position of the 23 essential genes on the chromosome were determined by genetic
mapping.
Mapping the 23 essential genes
During a single round of λ infection, there are many copies of the phage genomes and
recombination between them can occur relatively frequently, giving rise to hybrid progeny.
The frequency of recombination between two markers (in this case, two different amber
mutations) is directly proportional to the physical distance between the markers.
A-
R+
A+
R-
A-
R-
A+
R+
4 phages that can
arise from coinfection of λ A- R+
with λ A+ Rλ A- R+
λ A+ Rλ A- Rλ A+ R+
parental
recombinants
The frequency of recombination = # recombinants/total # phages
How would you measure the total # of phages?
How would you measure the # recombinants?
How would you do this cross in the first place? MOI, host strain, assay?
Mapping the 23 essential genes of lambda
Frequency of recombination between two markers is directly proportional to the physical distance between the markers.
Frequency of recombination = total recombinants/total phage
15% recombination between A and R; ~1% within a gene
The standard cross:
(1) Infect permissive host (suII+) with two different lambda am mutants (each at an moi = 5)
(2) Allow infected cells to produce phage (incubate 60 min)
(3) Assay phage produced for:
(a) recombinant phages (ability to form plaques on WT host 594)
(b) total phage produced (assayed on suII host, C600)
From the experiments of Campbell:
λam1
λA-R+W+
λam1 + λam2
suII+ E. coli
(permissive)
λam2
λA+R-W+
λam3
λA+R+W-
plate on suII+
serial dilute
plate on WT
(total # phage) (total # λA+R+W+)
cross
# pfu on suII+
# pfu on WT
# recombinants
% recombination
λam1 x λam2
104
76
152
15.2
103
71
142
1.4
104
70
140
14.0
λam1 x λam3
λam2 x λam3
Thus:
AW
R
pfu = plaque forming units