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Genetics: Analysis and Principles
Robert J. Brooker
CHAPTER 6
GENETIC TRANSFER AND
MAPPING IN BACTERIA
AND BACTERIOPHAGES
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
INTRODUCTION


Bacteria and viruses account for a quarter to a third
of human deaths worldwide. Their impact on health
is a major reason for studying them.
Like eukaryotes, bacteria often possess allelic
differences that affect their cellular traits

However, these allelic differences (such as different
sensitivity to antibiotics) are between different strains of
bacteria because

HOW ??? They do this.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-2
INTRODUCTION

genetic transfer

In this process, a segment of bacterial DNA is
transferred from one bacterium to another
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6-3
6.1 GENETIC TRANSFER AND
MAPPING IN BACTERIA


Like sexual reproduction in eukaryotes, genetic
transfer in bacteria enhances genetic diversity
Transfer of genetic material from one bacterium to
another can occur in three ways:

Conjugation


Transduction


Involves viruses
Transformation


Involves direct physical contact
Involves uptake from the environment
See Table 6.1
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6-4
6-5

movie
Conjugation

Genetic transfer in bacteria was discovered in 1946
by Joshua Lederberg and Edward Tatum

They were studying strains of Escherichia coli that
had different nutritional growth requirements



Auxotrophs cannot synthesize a needed nutrient
Prototrophs make all their nutrients from basic
components
One auxotroph strain was designated bio– met– phe+ thr+



The other strain was designated bio+ met+ phe– thr–


It required one vitamin (biotin) and one amino acid (methionine)
It could produce the amino acids phenylalanine and threonine
Had the opposite requirements for growth
Their experiment is described in Figure 6.1
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-6
Figure 6.1
6-7

The genotype of the bacterial cells that grew on
the plates has to be bio+ met+ phe+ thr+

Lederberg and Tatum reasoned that some genetic
material was transferred between the two strains



Either the bio– met– phe+ thr+ strain got the ability to
synthesize biotin and methionine (bio+ met+)
Or the bio+ met+ phe– thr– strain got the ability to
synthesize phenylalanine and threonine (phe+ thr+)
The results of this experiment cannot distinguish
between the two possibilities
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-8

Bernard Davis later showed that the bacterial strains
must make physical contact for transfer to occur

He used an apparatus known as U-tube

It contains at the bottom a filter which has pores that were





Large enough to allow the passage of the genetic material
But small enough to prevent the passage of bacterial cells
Davis placed the two strains in question on opposite sides
of the filter
Application of pressure or suction promoted the movement
of liquid through the filter
Refer to Figure 6.2
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6-9
Figure 6.2
No colonies

Nutrient agar
plates lacking
biotin, methionine,
phenylalanine and
threonine
No colonies
Thus, without physical contact, the two bacterial strains did not
transfer genetic material to one another
6-10

The term conjugation now refers to the transfer of
DNA from one bacterium to another following direct
cell-to cell contact

Many, but not all, species of bacteria can conjugate
Moreover, only certain strains of a bacterium can act
as donor cells


Those strains contains a small circular piece of DNA
termed the F factor (for Fertility factor)
+
 Strains containing the F factor are designated F
–
 Those lacking it are F

Plasmid is the general term used to describe extrachromosomal DNA
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-11

Plasmids, such as F factors, which are transmitted via
conjugation are termed conjugative plasmids

These plasmids carry genes required for conjugation
These genes play a role in the transfer of DNA
They are thus designated tra and trb followed by a capital letter
Figure 6.3
6-12

The first step in conjugation is the contact between
donor and recipient cells

This is mediated by sex pili (or F pili) which are made only
by F+ strains



Once contact is made, the pili shorten




These pili act as attachment sites for the F– bacteria
Refer to Figure 6.4b
Donor and recipient cell are drawn closer together
A conjugation bridge is formed between the two cells
The successful contact stimulates the donor cells to begin
the transfer process
Refer to Figure 6.4a for the molecular details of
conjugation
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6-13
a complex of 10-15
proteins encoded by the
F factor that span both
inner and outer
membranes
Together, these form
the conjugation bridge
Protein complex
encoded by the F factor
Accessory proteins of the
relaxosome are released
One protein, relaxase,
remains bound to the end of
the T-DNA
Transferred DNA
Figure 6.4
6-14
6-15

The result of conjugation is that the recipient cell
has acquired an F factor



In some cases, the F factor may carry genes that
were once found on the bacterial chromosome


Thus, it is converted from an F– to an F+ cell
The F+ cell remains unchanged
These types of F factors are called F’ factors
F’ factors can be transferred through conjugation

This may introduce new genes into the recipient and
thereby alter its genotype
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-16
Hfr Strains

In the 1950s, Luca Cavalli-Sforza discovered a strain of E.
coli that was very efficient at transferring chromosomal genes


He designated this strain as Hfr (for High frequency of recombination)
Hfr Strains Contain an F Factor Integrated into the Bacterial
Chromosome
An episome is a
segment of DNA
that can exist as a
plasmid and
integrate into the
chromosome
6-17
Hfr Strains Can Transfer a Portion of
the Bacterial Chromosome to
Recipient Cells

William Hayes demonstrated that conjugation
between an Hfr and an F– strain involves the
transfer of a portion of the Hfr bacterial
chromosome

The origin of transfer of the integrated F factor
determines the starting point and direction of the
transfer process


The cut, or nicked site is the starting point that will enter
the F– cell
Then, a strand of bacterial DNA begins to enter in a
linear manner
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-18

It generally takes about 1.5-2 hours for the entire
Hfr chromosome to be passed into the F– cell

Most matings do not last that long



Only a portion of the Hfr chromosome gets into the F– cell
Since the nick is internal to the integrated F factor, only part of
the plasmid is transferred and the F– cells does not become F+
The F– cell does pick up chromosomal DNA



This DNA can recombine with the homologous region
on the chromosome of the recipient cell
This may provide the recipient cell with new
combination of alleles
Refer to Figure 6.5b
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-19
lac+  Ability to
metabolize lactose
lac–  Inability
pro+  Ability to
synthesize proline
pro–  Inability
Therefore, the order of
transfer is lac+ – pro+
Figure 6.5b
Transfer
of bacterial
F– cell received
short segment
of the genes by an Hfr strain
Hfr chromosome
It has become lac+ but remains pro–
F– cell received longer segment of
the Hfr chromosome
It has become lac+ AND pro+
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-20
Experiment 6A Interrupted Mating
Technique
Conjugation Experiments Can Map Genes Along
the E. coli Chromosome
 The rationale behind this mapping strategy



The time it takes genes to enter the recipient cell is directly
related to their order along the bacterial chromosome
The Hfr chromosome is transferred linearly to the F–
recipient cell


Therefore, interrupted mating at different times would lead to
various lengths being transferred
The order of genes along the chromosome can be
deduced by determining the genes transferred during short
matings vs. those transferred during long matings
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-21

Wollman and Jacob started the experiment with two
E. coli strains

The donor (Hfr) strain had the following genetic
composition








thr+ : Able to synthesize the essential amino acid threonine
leu+ : Able to synthesize the essential amino acid leucine
azis : Sensitive to killing by azide (a toxic chemical)
tons : Sensitive to infection by T1 (a bacterial virus)
lac+ : Able to metabolize lactose and use it for growth
gal+ : Able to metabolize galactose and use it for growth
strs : Sensitive to killing by streptomycin (an antibiotic)
The recipient (F–) strain had the opposite genotype


thr– leu– azir tonr lac – gal – strr
r = resistant
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6-22

Wollman and Jacob already knew that



The thr+ and leu+ genes were transferred first, in that
order
Both were transferred within 5-10 minutes of mating
Therefore their main goal was to determine the
times at which genes azis, tons, lac+, and gal+ were
transferred

The transfer of the strs was not examined


Streptomycin was used to kill the donor (Hfr) cell following
conjugation
The recipient (F– cell) is streptomycin resistant
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6-23
The Hypothesis

The chromosome of the donor strain in an Hfr
mating is transferred in a linear manner to the
recipient strain

The order of genes along the chromosome can be
deduced by determining the time various genes take
to enter the recipient cell
Testing the Hypothesis

Refer to Figure 6.6
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6-24
Figure 6.6
6-25
The Data
Minutes that
Bacterial
Cells were
Allowed to
Mate Before
Blender
Treatment
Percent of Surviving Bacterial Colonies
with the Following Genotypes
thr+ leu+
azis
tons
lac+
gal+
5
––
––
––
––
––
10
100
12
3
0
0
15
100
70
31
0
0
20
100
88
71
12
0
25
100
92
80
28
0.6
30
100
90
75
36
5
40
100
90
75
38
20
50
100
91
78
42
27
60
100
91
78
42
27
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-26
Interpreting the Data
Minutes that
Bacterial
Cells were
Allowed to
Mate Before
Blender
Treatment
After 10 minutes,
the thr+ leu+
genotype was
obtained
There were no surviving colonies
after 5 minutes of mating
Percent of Surviving Bacterial Colonies
with the Following Genotypes
thr+ leu+
azis
tons
lac+
gal+
5
––
––
––
––
––
10
100
12
3
0
0
15
100
70
31
0
0
20
100
88
71
12
0
25
100
92
80
28
0.6
30
100
90
75
36
5
40
100
90
75
38
20
50
100
91
78
42
27
60
100
91
78
42
27
The azis gene is
transferred first
It is followed by
the tons gene
The lac+ gene
enters between 15
and 20 minutes
The gal+ gene
enters between
20 and 25
minutes
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6-27

From these data, Wollman and Jacob constructed
the following genetic map:

They also identified various Hfr strains in which the
origin of transfer had been integrated at different
places in the chromosome

Comparison of the order of genes among these strains,
demonstrated that the E. coli chromosome is circular
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6-28
The E. coli Chromosome

Conjugation experiments have been used to map
more than 1,000 genes on the E. coli chromosome

The E. coli genetic map is 100 minutes long

Approximately the time it takes to transfer the complete
chromosome in an Hfr mating

Refer to Figure 6.7
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6-29
Arbitrarily assigned the starting point
Units are minutes
Refer to the relative time
it takes for genes to first
enter an F– recipient
during a conjugation
experiment
Figure 6.7
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-30

The distance between genes is determined by comparing
their times of entry during an interrupted mating experiment

The approximate time of entry is computed by extrapolating the time
back to the origin
Figure 6.8

Therefore these two genes are approximately 9 minutes
apart along the E. coli chromosome
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6-31
Plasmids

Plasmids occur naturally in many strains of
bacteria and a few eukaryotic cells such as yeast

Range in size from a few thousand to 500,000 bp
Carry from one to hundreds of genes
Different plasmids will have one to 100 copies per cell

Have their own origins of replication which are strong or weak
Plasmids can provide a growth advantage to the cell



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6-32
Plasmids

Plasmids fall into five different categories:





1. Fertility plasmids-allow bacteria to mate to each other
2. Resistance plasmids-confer resistance to antibiotics or toxins
3. Degradative plasmids-enable the digestion of unusual
substances
4. Col-plasmids-encode colicines which are proteins that kill other
bacteria
5. Virulence plasmids-turn a bacterium into a pathogenic strain
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6-33
Transduction

Transduction is the transfer of DNA from one
bacterium to another via a bacteriophage

A bacteriophage is a virus that specifically attacks
bacterial cells


It is composed of genetic material surrounded by a
protein coat
It can undergo two types of cycles



Lytic
Lysogenic
Refer to Figure 6.9
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6-34
It will switch to
the lytic cycle
Prophage can
exist in a dormant
state for a long
time
Virulent phages only
undergo a lytic cycle
Temperate phages can
follow both cycles
The lytic and lysogenic reproductive cycles of certain bacteriophages. Some bacteriophages,
such as temperate phages, can follow both cycles. Other phages, known as virulent phages, can
follow only a lytic cycle.
Transduction


Phages that can transfer bacterial DNA include

P22, which infects Salmonella typhimurium

P1, which infects Escherichia coli

Both are temperate phages
Figure 6.10 illustrates the process of transduction
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6-36
Any piece of bacterial
DNA can be incorporated
into the phage
This type of transduction is
termed generalized transduction
Figure 6.10
6-37



Transduction was discovered in 1952 by Joshua
Lederberg and Norton Zinder
They used an experimental strategy similar to that of
Figure 6.1
They used two strains of the bacterium Salmonella
typhimurium

One strain, designated LA-22, was phe– trp– met+ his+



The other strain, designated LA-2, was phe+ trp+ met– his–



Unable to synthesize phenylalanine or tryptophan
Able to synthesize methionine and histidine
Able to synthesize phenylalanine and tryptophan
Unable to synthesize methionine or histidine
Their experiment is described next
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6-38
phe– trp– met+ his+
phe+ trp+ met– his–
Nutrient agar plates lacking the four amino acids
Genotypes of surviving
bacteria must be
phe+ trp+ met+ his+
~ 1 cell in 100,000
was observed to grow
Therefore, genetic
material had been
transferred between the
two strains
However, Lederberg and Zinder obtained novel results when repeating
the experiment using the U-tube apparatus
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6-39
LA-22
LA-2
phe– trp– met+ his+
phe+ trp+ met– his–
Nutrient agar
plates lacking the
four amino acids
No colonies
Colonies
Genotypes of surviving
bacteria must be
phe+ trp+ met+ his+
6-40


Therefore, some agent was being transferred from
LA-2 to LA-22 through the filter
Norton and Zinder conducted the same experiment
with filters of different pore sizes



They found out that the filterable agent was less then
0.1mm in diameter
They correctly concluded that the filterable agent was a
bacteriophage
In this case, the LA-2 strain contained a prophage
(such as P22)

The prophage switched to the lytic cycle


Packaged a segment of DNA containing the phe+ and trp+ genes
Passed through the filter and injected the DNA into the LA-22
strain
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6-41
Transformation

Transformation is the process by which a bacterium
will take up extracellular DNA released by a dead
bacterium

It was discovered by Frederick Griffith in 1928 while
working with strains of Streptococcus pneumoniae

There are two types

Natural transformation


DNA uptake occurs without outside help
Artificial transformation

DNA uptake occurs with the help of special techniques
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6-48
Transformation

Natural transformation occurs in a wide variety of
bacteria

Bacterial cells able to take up DNA are termed
competent cells

They carry genes that encode proteins called
competence factors


These proteins facilitate the binding, uptake and subsequent
corporation of the DNA into the bacterial chromosome
The steps of bacterial transformation are presented
in Figure 6.12
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6-49
A region of mismatch caused
by sequence differences
between the two alleles
By DNA repair enzymes
Figure 6.12
6-50
Transformation

Sometimes, the DNA that enters the cell is not
homologous to any genes on the chromosome



It may be incorporated at a random site on the
chromosome
This process is termed nonhomologous or illegitimate
recombination
Some bacteria preferentially take up DNA of
bacteria from the same or related species

Directed by DNA uptake signal sequences


9 or 10 bp long
repeated 1-2,000 times throughout genome
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6-51
Horizontal Gene Transfer

Horizontal gene transfer is the transfer of genes
between two different species

Vertical gene transfer is the transfer of genes from
mother to daughter cell or from parents to offspring

A sizable fraction of bacterial genes are derived
from horizontal gene transfer

Roughly 17% of E. coli and S. typhimurium genes during
the past 100 million years
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6-52
Horizontal Gene Transfer

The types of genes acquired through horizontal
gene transfer are quite varied and include




Genes that confer the ability to cause disease
Genes that confer antibiotic resistance
Genes that give the ability to degrade toxins
Horizontal gene transfer has dramatically
contributed to the phenomenon of acquired
antibiotic resistance

Bacterial resistance to antibiotics is a serious problem
worldwide

In many countries, nearly 50% of Streptococcus pneumoniae
strains are resistant to penicillin
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6-53
6.2 INTRAGENIC MAPPING
IN BACTERIOPHAGES

Viruses are not living



They rely on a host cell for existence and replication
However, they have unique biological structures and
functions, and therefore have traits
We will focus our attention on bacteriophage T4

Its genetic material contains several dozen genes


These genes encode a variety of proteins needed for the viral
cycle
Refer to Figure 6.13 for the T4 structure
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6-54
Contains the
genetic material
Figure 6.13
Used for attachment to
the bacterial surface
6-55

In the 1950s, Seymour Benzer embarked on a ten-year study
focusing on the function of the T4 genes


He conducted a detailed type of genetic mapping known as intragenic
or fine structure mapping
The difference between intragenic and intergenic mapping is:
6-56
Plaques


A plaque is a clear
area on an otherwise
opaque bacterial lawn
on the agar surface of
a petri dish
It is caused by the
lysis of bacterial cells
as a result of the
growth and
reproduction of
phages
Figure 6.14
6-57

Some mutations in the phage’s genetic material can
alter the ability of the phage to produce plaques


Plaques are visible with the naked eye


Thus, plaques can be viewed as traits of bacteriophages
So mutations affecting them lend themselves to easier
genetic analysis
An example is a rapid-lysis mutant of bacteriophage
T4, which forms unusually large plaques


Refer to Figure 6.15
This mutant lyses bacterial cells more rapidly than do the
wild-type phages


Rapid-lysis mutant forms large, clearly defined plaques
Wild-type phages produce smaller, fuzzy-edged plaques
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-58


Benzer studied one category of T4 phage mutant, designated rII (r stands
for rapid lysis)
It behaved differently in three different strains of E. coli

In E. coli B

rII phages produced unusually large plaques that had poor yields of
bacteriophages


In E. coli K12S


rII phages produced normal plaques that gave good yields of phages
In E. coli K12(l) (has phage lambda DNA integrated into its
chromosome)


The bacterium lyses so quickly that it does not have time to produce many new
phages
rII phages were not able to produce plaques at all
As expected, the wild-type phage could infect all three strains
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-59
Complementation Tests

Benzer collected many rII mutant strains that can
form large plaques in E. coli B and none in E. coli
K12(l)

But, are the mutations in the same gene or in
different genes?

To answer this question, he conducted
complementation experiments
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
6-60

Figure 6.16 shows the possible outcomes of
complementation experiments involving coinfection
of plaque formation mutants
Figure 6.16
6-61

Benzer carefully considered the pattern of
complementation and noncomplementation


Benzer coined the term cistron to refer to the
smallest genetic unit that gives a negative
complementation test


He determined that the rII mutations occurred in two
different genes, which were termed rIIA and rIIB
So, if two mutations occur in the same cistron, they
cannot complement each other
A cistron is equivalent to a gene

However, it is not as commonly used
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6-62

At an extremely low rate, two noncomplementing strains of
viruses can produce an occasional viral plaque, if intragenic
recombination has occurred
rII mutations
Viruses cannot
form plaques in
E. coli K12(l)
rII mutations
Viruses cannot
form plaques in
E. coli K12(l)
Figure 6.17
Function of protein A will
be restored
Therefore new phages can
be made in E. coli K12(l)
Viral plaques will
now be formed
6-63

Figure 6.18 describes the general strategy for intragenic mapping of
rII phage mutations
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6-64
r103
r104
Both rII mutants
and wild-type
phages can infect
this strain
Total number
of phages
rII mutants cannot
infect this strain
Number of wild-type
phages produced by
intragenic recombination
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6-65

The data from Figure 6.18 can be used to estimate the
distance between the two mutations in the same gene

The phage preparation used to infect E. coli B was diluted
by 108 (1:100,000,000)


1 ml of this dilution was used and 66 plaques were produced

Therefore, the total number of phages in the original preparation is
66 X 108 = 6.6 X 109 or 6.6 billion phages per milliliter
The phage preparation used to infect E. coli k12(l) was
diluted by 106 (1:1,000,000)

1 ml of this dilution was used and 11 plaques were produced

Therefore, the total number of wild-type phages is
11 X 106 or 11 million phages per milliliter
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
In this experiment, the intragenic recombination produces
an equal number of recombinants


Wild-type phages and double mutant phages
However, only the wild-type phages are detected in the
infection of E. coli k12(l)

Therefore, the total number of recombinants is the number of wildtype phages multiplied by two
Frequency of recombinants =
Frequency of recombinants =
2 [wild-type plaques
obtained in E. coli k12(l)]
Total number of plaques
obtained in E. coli B
2(11 X 106)
6.6 X 109
= 3.3 X 10–3 = 0.0033
In this example, there was approximately 3.3 recombinants per 1,000 phages
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6-67

As in eukaryotic mapping, the frequency of recombinants can
provide a measure of map distance along the bacteriophage
chromosome


The frequency of intragenic recombinants is correlated with
the distance between the two mutations


In this case the map distance is between two mutations in the same
gene
The farther apart they are the higher the frequency of recombinants
Homoallelic mutations


Mutations that happen to be located at exactly the same site in a gene
They are not able to produce any wild-type recombinants

So the map distance would be zero
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Deletion Mapping

Benzer used deletion mapping to localize many rII
mutations to a fairly short region in gene A or gene B

He utilized deletion strains of phage T4

Each is missing a known segment of the rIIA and/or rIIB
genes
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6-69

Let’s suppose that the goal is to know the
approximate location of an rII mutation, such as r103

E. coli k12(l) is coinfected with r103 and a deletion strain


If the deleted region includes the same region that
contains the r103 mutation
 No intragenic wild-type recombinants are produced
 Therefore, plaques will not be formed
If the deleted region does not overlap with the r103
mutation
 Intragenic wild-type recombinants can be produced
 And plaques will be formed
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Mutation must be in region
contained in BP242 but
not in PT1. This
corresponds to A4 in the
rIIA gene
Figure 6.19
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
As described in Figure 6.19, the first step in the deletion
mapping strategy localized rII mutations to seven regions


Other strains were used to eventually localize each rII
mutation to one of 47 regions


36 in rIIA and 11 in rIIB
At this point, pairwise coinfections were made between
mutant strains that had been localized to the same region


Six in rIIA and one in rIIB
This would precisely map their location relative to each other
This resulted in a fine structure map with depicting the
locations of hundreds of different rII mutations

Refer to Figure 6.20
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Contain many mutations
at exactly the same site
within the gene
Figure 6.20
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


Intragenic mapping studies were a pivotal
achievement in our early understanding of gene
structure
Some scientists had envisioned a gene as being a
particle-like entity that could not be further
subdivided
However, intragenic mapping revealed convincingly
that this is not the case
 It showed that


Mutations can occur at different parts within a single
gene
Intragenic crossing over can recombine these
mutations, resulting in wild-type genes
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