Transcript Brooker Chapter 6 - Volunteer State Community College

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Genetics: Analysis and Principles
Robert J. Brooker
CHAPTER 6
GENETIC TRANSFER AND
MAPPING IN BACTERIA
AND BACTERIOPHAGES
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INTRODUCTION

Bacteria are usually haploid


Bacteria reproduce asexually


This fact makes it easier to identify loss-of-function
mutations in bacteria than in eukaryotes
Therefore crosses are not used in the genetic analysis
of bacterial species
Bacterial genetics is based on genetic transfer

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

Transfer of genetic material from one
bacterium to another can occur in three
ways:

Conjugation


Transduction


Involves direct physical contact
Involves viruses
Transformation

Involves uptake from the environment
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6-4
Conjugation

Genetic transfer in bacteria was discovered in 1946
by Joshua Lederberg and Edward Tatum

Two strains of Escherichia coli with different
nutritional growth requirements:

One 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
The other strain was designated bio+ met+ phe– thr–
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6-5
~ 100
Figure 6.1
6-6

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

The “direction” of gene transfer was unclear
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6-7

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
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6-8
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
6-9

The term conjugation now refers to the transfer of
DNA from one bacterium to another following direct
cell-to cell contact

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
Conjugation is mediated by sex pili (or F pili) which
are made only by F+ strains

These pili act as attachment sites for the F– bacteria
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6-10

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
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6-11
Fig. 6.4b

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 (and donor genes) can be transferred
through conjugation
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6-15
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 are derived from F+ strains
An episome is a
segment of DNA
that can exist as a
plasmid and
integrate into the
chromosome
Figure 6.5a
6-16

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
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6-17
Fig. 6.5 (TE Art)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
F+ cell
Hfr cell
Bacterial
chromosome
pro+
lac+
Integration of
F factor into
chromosome
pro+
lac+
Origin of
transfer
Origin of
transfer
F factor
(a) When an F factor integrates into the chromosome, it creates an Hfr cell.
Short
time
Hfr cell
pro+
lac+
pro+
pro–
lac+
lac+
F– cell
pro–
lac–
Transfer
of Hfr
chromosome
pro–
pro+
lac+
Hfr cell
lac–
F– recipient cell
lac+
pro+
Origin of
transfer
(toward lac+)
Longer
time
(b) An Hfr donor cell can pass a portion of its chromosome to an F – recipient cell.
pro+
lac+
pro+
lac+

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
The F– cells does not become F+
The transferred DNA can recombine with the
homologous region on the chromosome of the
recipient cell
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6-18
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
F– cell received short segment of the
Hfr chromosome
It has become lac+ but remains pro–
F– cell received longer segment of
the Hfr chromosome
It has become lac+ AND pro+
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6-19
Interrupted Mating Technique



The time it takes genes to enter the recipient cell
is directly related to their order along the bacterial
chromosome
Interruptions of mating at different times 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
6-20

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
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6-21

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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6-22
Figure 6.6
6-24
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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6-26

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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6-27
Figure 6-9
Copyright © 2006 Pearson Prentice Hall, Inc.
The E. coli Chromosome

Conjugation experiments have been used to map
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
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6-28

The distance between genes is determined
by comparing their times of entry during an
interrupted mating experiment
Figure 6.7

Therefore these two genes are approximately 9 minutes
apart along the E. coli chromosome
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6-30
Figure 6-8
Copyright © 2006 Pearson Prentice Hall, Inc.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
pyrB thrA 0.0 proA,B
melA
lacA,Y,Z
uvrA
dnaB
polA
95 100/0 5
galE
10
90
oriC
dnaA
85
15
xylA
80
20
Fig. 6.7(TE Art)
75
25
70
argR
argG
trpA,B,C,D,E
30
35
65
60
pyrG
mutS
recA
55 50 45
purL
gyrA
40
uvrC
hipA
pabB
cheA

Transfer of F’
produces
merozygotes
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
 two types of cycles


Lytic
Lysogenic
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6-31
It will undergo
the lytic cycle
Prophage can
exist in a dormant
state for a long
time
Virulent phages only
undergo a lytic cycle
Figure 6.9
Temperate phages can
follow both cycles
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6-32
Transduction

Phages that can transfer bacterial DNA include

P22, which infects Salmonella typhimurium

P1, which infects Escherichia coli

Both are temperate phages
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6-33
Any piece of bacterial
DNA can be incorporated
into the phage
This type of transduction is
termed generalized transduction
Figure 6.10
6-34


Transduction was discovered in 1952 by Joshua
Lederberg and Norton Zinder
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–
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6-35
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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6-36
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+
6-37

The filterable agent was less then 0.1mm in
diameter


Conclusion: the filterable agent was a bacteriophage
In this case, the LA-2 strain contained a prophage
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6-38
Cotransduction Mapping

There is a maximum size to the DNA that can be
packaged by bacteriophages during transduction



P1 can pack up to 2-2.5% of the E. coli chromosome
P22 can pack up to 1% of the S. typhimurium
chromosome
Cotransduction refers to the packaging and transfer
of two closely-linked genes

It is used to determine the order and distance between
genes that lie fairly close together
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6-39
Cotransduction Mapping

Researchers select for the transduction of one gene


They then monitor whether a second gene is cotransduced
Consider for example the following two E. coli strains


The donor strain with genotype arg+ met+ strs
The recipient strain with genotype arg– met– strr
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6-40
Mix P1 lysate with recipient
cells that are arg– met– strr
Figure 6.11
6-41
These colonies
must be met+
Plate on minimal plates with
arginine and streptomycin
but without methionine
To determine whether
they are also arg+
streak onto plate that
lacks both amino acids
21/50
6-42
Cotransduction Mapping

Relationship between cotransduction frequency
and map distances obtained from conjugation
experiments

Cotransduction frequency = (1 – d/L)3

where

d = distance between two genes in minutes

L = the size of the transduced DNA (in minutes)
 For P1 transduction, this size is ~ 2% of the E. coli
chromosome, which equals about 2 minutes
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6-43

Let’s use the equation in our example,
Therefore, the distance
between the met+ and arg+
genes is approximately
0.5 minutes


Transduction experiments can provide very accurate
mapping data for genes that are fairly close together
Conjugation experiments, on the other hand, are usually
used for genes that are far apart on the chromosome
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6-44
Transformation

Transformation is the process by which a
bacterium will take up extracellular DNA
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6-45
Horizontal Gene Transfer

Horizontal gene transfer is the transfer of genes
between two different species

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
Genes that confer the ability to cause disease
Genes that confer antibiotic resistance
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6-49
6.2 INTRAGENIC MAPPING
IN BACTERIOPHAGES

Viruses are not living


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
6-51
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Contains the
genetic material
Figure 6.13
Used for attachment to
the bacterial surface
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6-52

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:
6-53
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
6-54

A rapid-lysis mutant of bacteriophage T4 forms
unusually large plaques

This mutant lyses bacterial cells more rapidly than
do the wild-type phages
6-55


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




The bacterium lyses so quickly that it does not have time to produce
many new phages
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)
 rII phages were not able to produce plaques at all
As expected, the wild-type phage could infect all three strains
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6-56
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
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6-57
Fig. 6.16a(TE Art)
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rII strain 1
rII strain 2
(gene A is defective, (gene A is defective,
gene B is normal)
gene B is normal)
gene A gene B
gene A gene B
Coinfect E. coli K12 (l)
Plate and observe if
plaques are formed.
No plaques
No complementation occurs since the coinfected cell is unable to make
the normal product of gene A. The coinfected cell will not produce viral
particles, thus no bacterial cell lysis and no plaque formation.
Noncomplementation: The phage mutations are in the same gene.
Fig. 6.16b(TE Art)
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rII strain 3
rII strain 4
(gene A is defective, (gene A is normal,
gene B is normal) gene B is defective)
gene A gene B
gene A gene B
Coinfect E. coli K12 (l)
Plate and observe if
plaques are formed.
Viral plaques
Complementation occurs since the coinfected cell is able to
make normal products of gene A and gene B. The coinfected
bacterial cell will produce viral particles that lyse the cell,
resulting in the appearance of clear plaques.
Complementation: The phage mutations are in different genes.

rII mutations occurred in two different genes,
which were termed rIIA and rIIB

Benzer coined the term cistron to refer to the
smallest genetic unit that gives a negative
complementation test

A cistron is equivalent to a gene

However, it is not as commonly used
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6-59

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
Coinfection (E. coli B)
Function of protein A will
be restored
Therefore new phages can
be made in E. coli K12(l)
Viral plaques will
now be formed
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6-60
General strategy for intragenic mapping of rII phage mutations
6-61
r103
r104
Take some of the phage
preparation, dilute it greatly
(10-8) and infect E. coli B
Take some of the phage
preparation, dilute it somewhat
(10-6) and infect E. coli K12(l)
Both rII mutants
and wild-type
phages can infect
this strain
Total number
of phages
66 plaques
rII mutants cannot
infect this strain
Number of wild-type
phages produced by
intragenic recombination
11 plaques
6-62


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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6-63

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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6-64

The frequency of intragenic recombinants is
correlated with the distance between the two
mutations


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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6-65
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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6-66
Figure 6.19
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6-68

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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6-67

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
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6-69
Contain many mutations
at exactly the same site
within the gene
Figure 6.20
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6-70



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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6-71