Name: Aamer al zaman
Roll no: 08
Class : BS Botany 5th semester
Topic: Bacterial gene mapping.
What is gene mapping:
 Genome mapping is the process of
describing a genome in terms of the
relative locations of genes and other
DNA sequences
Bacterial gene mapping:
 Once inside the F− cell, the linear single-stranded
DNA molecule acts as a polymerization template
and is converted into a DNA double helix. This linear
donor fragment is the exogenote, and the resident
F− chromosome is the endogenote. ...Gene transfer
and recombination provide the key
to mapping the bacterial chromosome.
Discription:
 Gene mapping describes the methods used to identify
the locus of a gene and the distances between genes.
 The essence of all genome mapping is to place a
collection of molecular markers onto their respective
positions on the genome. Molecular markers come in all
forms. Genes can be viewed as one special type of genetic
markers in the construction of genome maps, and mapped
the same way as any other markers.
Genetics mapping via physical
mapping:
 There are two distinctive types of "Maps"
used in the field of genome mapping: genetic
maps and physical maps. While both maps are a
collection of genetic markers and gene loci,
genetic maps' distances are based on the genetic
linkage information, while physical maps use actual
physical distances usually measured in number of
base pairs
Continu:
 While the physical map could be a more "accurate"
representation of the genome, genetic maps often offer
insights into the nature of different regions of the
chromosome, e.g. the genetic distance to physical distance
ratio varies greatly at different genomic regions which reflects
different recombination rates, and such rate is often
indicative of euchromatic (usually gene-rich) vs
heterochromatic (usually gene poor) regions of the genome.
Bacterial conjugation:
 Conjugation is merely the fusion of two compatible bacterial
cells. Bringing two genotypes together and allowing them to
conjugate is the equivalent of making a cross in eukaryotes.
Our discussion of conjugation will center on the gut
bacterium Escherichia coli (E. coli). Conjugation
and gene transfer in E. coli are driven by a
circular DNA plasmid called the fertility factor or sex factor
(F), which is found in some but not all cells. Hence to
understand how to make a cross in E. coli, we have to
understand the properties of F.
Recombination between Donor and
Recipient DNA:
 Conjugation allows genes from two different parental cells to come
together in the same cell and hence provides an opportunity
for recombination to occur. Hence mapping analysis is possible.
 All conjugations (“crosses”) are by definition of the type Hfr
(donor) × F− (recipient). After cell union, the Hfr chromosome replicates in
a peculiar manner that reels out a single-stranded DNA molecule, which
is then transferred linearly into the F− cell. The replication and transfer
begin at a specific point at one side of the integrated F, called the origin
(O). Genes close to the origin are transferred first.
Continu:
 Once inside the F− cell, the linear single-stranded DNA molecule acts as a
polymerization template and is converted into a DNA double helix. This linear
donor fragment is the exogenote, and the resident F−chromosome is
the endogenote. As a free molecule, the exogenote cannot replicate and will
become lost, but because exogenotes and endogenotes are
homologous, crossing-over can take place between them. A single crossover
between a linear molecule (the exogenote) and a circular one (the endogenote)
would produce a single long molecule that would be inviable.
However, two crossovers would integrate a part of the donor genome into the
recipient.
Mapping by Interrupted Conjugation:
 In mapping by interrupted conjugation, the Hfr and F− cells
are mixed, and conjugation proceeds. Then, at fixed times,
the F− cells are sampled to determine which donor alleles
have entered. This sampling is accomplished by using a
kitchen blender to separate the joined cells, resulting in
interrupted conjugation. After separation, the Hfr cells are
selectively killed, and the remaining F− cells,
the exconjugants, are tested to see which of the donor
alleles have entered and stably recombined with
the endogenote.
Continu:
 The times at which various donor alleles first appear in the
exconjugants are calculated. If a donor allele a+enters the
recipient at 5 minutes after union and allele b+ enters at 8
minutes, then the two genes are said to be 3 minutes apart
on the chromosome. The map units in this case are minutes.
Like the maps based on crossover frequencies,
these linkage maps are purely genetic constructions.
Although the amount of DNA corresponding to a minute is
now known
Continu:
 A typical cross in which the order and map position of the genes under
study are not known. In this particular cross, the genes by which the
parents differ will be azi (resistance or sensitivity to sodium
azide), gal (ability or inability to utilize galactose as an energy
source), lac (ability or inability to utilize lactose as an energy source),
and ton (resistance or sensitivity to bacteriophage T1). A streptomycin-
sensitivity allele in the Hfr and a streptomycin-resistance allele (strr) in
the recipient are used to selectively kill the Hfr cells after conjugation.
Continu:
 The relative positions of the azi, ton, lac, and gal genes were
established in our experiment. However, the chromosomal
region containing these loci might be only a small proportion
of the entire chromosome. The complete map is obtained
from many such interrupted conjugation experiments, in
which parental strains heteroallelic for different combinations
of genes are used; then the overall map is pieced together
from the complete set of data.
High-Resolution Mapping by Recombinant
Frequency:
 Interrupted-mating experiments provide a rough set
of gene locations over the entire map. As we learned, the
genes are mapped by time of entry. In such experiments,
the exogenote must integrate by a
double recombination event, but the mapping method is not
based on any measurement of recombinant frequencies.
However, to provide a higher-resolution method for
measuring the sizes of smaller map distances, recombinant
frequencies are used.

Continu:
 Suppose that we undertake an experiment to map three
genes—met, arg, and leu—by recombinant frequency. To
measure recombination between these genes, we must set
up a merozygote that is heterozygous for all three. This can
be accomplished if we can establish which gene enters last
by an interrupted conjugation analysis. The Hfr allele of the
last-entering gene is selected among the F−exconjugants.
Continu:
 The last gene to enter is leu+; therefore we select initially
for leu+ exconjugants by plating them on medium containing no leucine
but containing methionine and arginine. Now we can proceed to calculate
map distance in the standard way by using a map unit equal to
a recombinant frequency of 1 percent. In practice, this calculation is done
by measuring the proportion of the total leu+ exconjugants that also
carries arg+ or met+ or both or neither. The recombination events needed
to produce these recombinant genotypes are know that a double
crossover must have occurred to integrate leu+: one crossover is at the
left of the leu gene, but the other can be in various positions at the right.
Refernces:
 Treanor, Brian, Aspects of alterity: Levinas, Marcel, and the
contemporary debate, Fordham University Press, 2006, p.41
 Jump up^ Klein, Ernest, A comprehensive etymological
dictionary of the English language, Vol II, Elsevier publishing
company, Amsterdam, 1969, p.1317
 Jump up^ Saeed, John, Semantics, Blackwell,
p. 12, ISBN 0-631-22693-1

Bacterial gene mapping

  • 1.
    Name: Aamer alzaman Roll no: 08 Class : BS Botany 5th semester Topic: Bacterial gene mapping.
  • 2.
    What is genemapping:  Genome mapping is the process of describing a genome in terms of the relative locations of genes and other DNA sequences
  • 3.
    Bacterial gene mapping: Once inside the F− cell, the linear single-stranded DNA molecule acts as a polymerization template and is converted into a DNA double helix. This linear donor fragment is the exogenote, and the resident F− chromosome is the endogenote. ...Gene transfer and recombination provide the key to mapping the bacterial chromosome.
  • 4.
    Discription:  Gene mappingdescribes the methods used to identify the locus of a gene and the distances between genes.  The essence of all genome mapping is to place a collection of molecular markers onto their respective positions on the genome. Molecular markers come in all forms. Genes can be viewed as one special type of genetic markers in the construction of genome maps, and mapped the same way as any other markers.
  • 5.
    Genetics mapping viaphysical mapping:  There are two distinctive types of "Maps" used in the field of genome mapping: genetic maps and physical maps. While both maps are a collection of genetic markers and gene loci, genetic maps' distances are based on the genetic linkage information, while physical maps use actual physical distances usually measured in number of base pairs
  • 6.
    Continu:  While thephysical map could be a more "accurate" representation of the genome, genetic maps often offer insights into the nature of different regions of the chromosome, e.g. the genetic distance to physical distance ratio varies greatly at different genomic regions which reflects different recombination rates, and such rate is often indicative of euchromatic (usually gene-rich) vs heterochromatic (usually gene poor) regions of the genome.
  • 7.
    Bacterial conjugation:  Conjugationis merely the fusion of two compatible bacterial cells. Bringing two genotypes together and allowing them to conjugate is the equivalent of making a cross in eukaryotes. Our discussion of conjugation will center on the gut bacterium Escherichia coli (E. coli). Conjugation and gene transfer in E. coli are driven by a circular DNA plasmid called the fertility factor or sex factor (F), which is found in some but not all cells. Hence to understand how to make a cross in E. coli, we have to understand the properties of F.
  • 8.
    Recombination between Donorand Recipient DNA:  Conjugation allows genes from two different parental cells to come together in the same cell and hence provides an opportunity for recombination to occur. Hence mapping analysis is possible.  All conjugations (“crosses”) are by definition of the type Hfr (donor) × F− (recipient). After cell union, the Hfr chromosome replicates in a peculiar manner that reels out a single-stranded DNA molecule, which is then transferred linearly into the F− cell. The replication and transfer begin at a specific point at one side of the integrated F, called the origin (O). Genes close to the origin are transferred first.
  • 9.
    Continu:  Once insidethe F− cell, the linear single-stranded DNA molecule acts as a polymerization template and is converted into a DNA double helix. This linear donor fragment is the exogenote, and the resident F−chromosome is the endogenote. As a free molecule, the exogenote cannot replicate and will become lost, but because exogenotes and endogenotes are homologous, crossing-over can take place between them. A single crossover between a linear molecule (the exogenote) and a circular one (the endogenote) would produce a single long molecule that would be inviable. However, two crossovers would integrate a part of the donor genome into the recipient.
  • 10.
    Mapping by InterruptedConjugation:  In mapping by interrupted conjugation, the Hfr and F− cells are mixed, and conjugation proceeds. Then, at fixed times, the F− cells are sampled to determine which donor alleles have entered. This sampling is accomplished by using a kitchen blender to separate the joined cells, resulting in interrupted conjugation. After separation, the Hfr cells are selectively killed, and the remaining F− cells, the exconjugants, are tested to see which of the donor alleles have entered and stably recombined with the endogenote.
  • 11.
    Continu:  The timesat which various donor alleles first appear in the exconjugants are calculated. If a donor allele a+enters the recipient at 5 minutes after union and allele b+ enters at 8 minutes, then the two genes are said to be 3 minutes apart on the chromosome. The map units in this case are minutes. Like the maps based on crossover frequencies, these linkage maps are purely genetic constructions. Although the amount of DNA corresponding to a minute is now known
  • 12.
    Continu:  A typicalcross in which the order and map position of the genes under study are not known. In this particular cross, the genes by which the parents differ will be azi (resistance or sensitivity to sodium azide), gal (ability or inability to utilize galactose as an energy source), lac (ability or inability to utilize lactose as an energy source), and ton (resistance or sensitivity to bacteriophage T1). A streptomycin- sensitivity allele in the Hfr and a streptomycin-resistance allele (strr) in the recipient are used to selectively kill the Hfr cells after conjugation.
  • 13.
    Continu:  The relativepositions of the azi, ton, lac, and gal genes were established in our experiment. However, the chromosomal region containing these loci might be only a small proportion of the entire chromosome. The complete map is obtained from many such interrupted conjugation experiments, in which parental strains heteroallelic for different combinations of genes are used; then the overall map is pieced together from the complete set of data.
  • 14.
    High-Resolution Mapping byRecombinant Frequency:  Interrupted-mating experiments provide a rough set of gene locations over the entire map. As we learned, the genes are mapped by time of entry. In such experiments, the exogenote must integrate by a double recombination event, but the mapping method is not based on any measurement of recombinant frequencies. However, to provide a higher-resolution method for measuring the sizes of smaller map distances, recombinant frequencies are used. 
  • 15.
    Continu:  Suppose thatwe undertake an experiment to map three genes—met, arg, and leu—by recombinant frequency. To measure recombination between these genes, we must set up a merozygote that is heterozygous for all three. This can be accomplished if we can establish which gene enters last by an interrupted conjugation analysis. The Hfr allele of the last-entering gene is selected among the F−exconjugants.
  • 16.
    Continu:  The lastgene to enter is leu+; therefore we select initially for leu+ exconjugants by plating them on medium containing no leucine but containing methionine and arginine. Now we can proceed to calculate map distance in the standard way by using a map unit equal to a recombinant frequency of 1 percent. In practice, this calculation is done by measuring the proportion of the total leu+ exconjugants that also carries arg+ or met+ or both or neither. The recombination events needed to produce these recombinant genotypes are know that a double crossover must have occurred to integrate leu+: one crossover is at the left of the leu gene, but the other can be in various positions at the right.
  • 17.
    Refernces:  Treanor, Brian,Aspects of alterity: Levinas, Marcel, and the contemporary debate, Fordham University Press, 2006, p.41  Jump up^ Klein, Ernest, A comprehensive etymological dictionary of the English language, Vol II, Elsevier publishing company, Amsterdam, 1969, p.1317  Jump up^ Saeed, John, Semantics, Blackwell, p. 12, ISBN 0-631-22693-1