HFR cells are formed from homologous recombination due to integratio.pdf
•
0 likes•12 views
HFR cells are formed from homologous recombination due to integration of the conjugate plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another cell. Hfr cells do not possess a complete \"circular chromosome\" i.e. plasmid as F+ cell. Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its chromosomal DNA often referred as \"episome\". F+ cell possess F factor known as plasmid & it can donate to another cell through conjugation. Both Hfr & F+ cells are competent for conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell. Interrupted mating: F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F- factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This structure is going to promote high frequency recombination through interrupted mating into the recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting conjugation. This is completely based on the streptomycin-sensitivity allele (str-s) in the Hfr strains. On the otherhand, streptomycin-resistance allele (str-r) presence in the recipient is employed considerably to specifically destroy the Hfr donor cells after conjugation. Solution HFR cells are formed from homologous recombination due to integration of the conjugate plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another cell. Hfr cells do not possess a complete \"circular chromosome\" i.e. plasmid as F+ cell. Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its chromosomal DNA often referred as \"episome\". F+ cell possess F factor known as plasmid & it can donate to another cell through conjugation. Both Hfr & F+ cells are competent for conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell. Interrupted mating: F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F- factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This structure is going to promote high frequency recombination through interrupted mating into the recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting conjugation. This is completely based on the st.
Report
Share
Report
Share
Download to read offline
Recommended
1. How is the recipient cell different from before the two cells mat.pdf
1. How is the recipient cell different from before the two cells mated? Phenotypically as well as
genetically.
2. What do F+ cells do?
3. What are all properties of a recipient cell (which receives the F plasmid)?
Solution
1. Recipient cell (F- cell) is different after it mates with the F+ cell. Conjugation/mating is
equivalent to sex (however no new individuals are produced in this case) in that it combines
genetic material from two different individuals into a single genome.It occurs through F pili
which originate from cell membrane and project outward beyond the cell wall. Thus, like sexual
reproduction conjugation produces new genetic combinations. The pilus is essential for
recognition of recipient cell. The sex plasmid passes from the donor F+ to the recipient F– cell
through the conjugation tube, and seperates after the event is over and now called as
exconjugates. F- cell after matinh with F+ cell, now becomes F+ and can act as donor during
next conjugation event.
After conjugation event, recipient cell now contains the F factor. Genetic changes takes place in
the recipient cell. 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
termed as the exogenote, and the resident F chromosome is termed as the endogenote. Crossover
events take place which integrate a part of the donor genome into the recipient. It is in this way
that recombination takes place.
2. F+ cells are the bacterial cells which possess F factor as a plasmid independent of the bacterial
genome. The F cell plasmid contains F factor as the only DNA and no DNA from the bacterial
genome. Mating or conjugation of an F+ cell with an F cell, result in two F+ cells, both of which
are capable of transmitting the plasmid to other F cells by conjugation. Thus F+ bacterial cells
act as donor. The F+ cells also contain surface exclusion proteins; TraS and TraT on the bacterial
surface. These proteins have the capability to prevent secondary mating events involving
plasmids belonging to the same incompatibility group. Thus, each F+ bacterium can host only a
single plasmid type of any given incompatibility group.
3. Properties of Recipient cells (receiving the F plasmid)
Recipients are basically F bacteria that do not contain F factor (F plasmid), and are considered as
female. The F-plasmid is a class of conjugative plasmids that control sexual functions of bacteria
with a fertility inhibition (Fin) system. In this system, FinO, which is a trans-acting factor and
Fin P, an antisense RNAs, combine to repress the expression of the activator gene TraJ. TraJ is a
transcription factor that upregulates the tra operon, which contains the genes required for
conjugation and plasmid transfer. The F- plasmid can exist in two alternative states, viz. it can
either remain free in the cell or it can be integrated into the E. coli chromosome.called as
episomes. The F-plasmid also carr.
Bacterial Conjugation (Genetic recombination in Bacteria)
Introduction, discovery, f-plasmid and its properties, bacterial conjugation steps, Hfr strain.
For 12th, M.Sc, B.Sc students.
F plasmid organisation
organization and integration of F plamid into bacterial strain through conjugation
Bacterial conjugation
bacterial conjugation genetic transfer transfer of genetic material in bacteria F+ and F- factors bacterial biology genetics factors in bacterial genes genetics DNA synthesis mobilization cell to cell contact bridge formation
1. Escherichia coli have F plasmid which by replication gets transfe.pdf
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+.
Solution
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+..
Conjuagtion types and application
this presentation basically focusses on the types of conjugation namely Hfr and F' conjugation.
Recommended
1. How is the recipient cell different from before the two cells mat.pdf
1. How is the recipient cell different from before the two cells mated? Phenotypically as well as
genetically.
2. What do F+ cells do?
3. What are all properties of a recipient cell (which receives the F plasmid)?
Solution
1. Recipient cell (F- cell) is different after it mates with the F+ cell. Conjugation/mating is
equivalent to sex (however no new individuals are produced in this case) in that it combines
genetic material from two different individuals into a single genome.It occurs through F pili
which originate from cell membrane and project outward beyond the cell wall. Thus, like sexual
reproduction conjugation produces new genetic combinations. The pilus is essential for
recognition of recipient cell. The sex plasmid passes from the donor F+ to the recipient F– cell
through the conjugation tube, and seperates after the event is over and now called as
exconjugates. F- cell after matinh with F+ cell, now becomes F+ and can act as donor during
next conjugation event.
After conjugation event, recipient cell now contains the F factor. Genetic changes takes place in
the recipient cell. 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
termed as the exogenote, and the resident F chromosome is termed as the endogenote. Crossover
events take place which integrate a part of the donor genome into the recipient. It is in this way
that recombination takes place.
2. F+ cells are the bacterial cells which possess F factor as a plasmid independent of the bacterial
genome. The F cell plasmid contains F factor as the only DNA and no DNA from the bacterial
genome. Mating or conjugation of an F+ cell with an F cell, result in two F+ cells, both of which
are capable of transmitting the plasmid to other F cells by conjugation. Thus F+ bacterial cells
act as donor. The F+ cells also contain surface exclusion proteins; TraS and TraT on the bacterial
surface. These proteins have the capability to prevent secondary mating events involving
plasmids belonging to the same incompatibility group. Thus, each F+ bacterium can host only a
single plasmid type of any given incompatibility group.
3. Properties of Recipient cells (receiving the F plasmid)
Recipients are basically F bacteria that do not contain F factor (F plasmid), and are considered as
female. The F-plasmid is a class of conjugative plasmids that control sexual functions of bacteria
with a fertility inhibition (Fin) system. In this system, FinO, which is a trans-acting factor and
Fin P, an antisense RNAs, combine to repress the expression of the activator gene TraJ. TraJ is a
transcription factor that upregulates the tra operon, which contains the genes required for
conjugation and plasmid transfer. The F- plasmid can exist in two alternative states, viz. it can
either remain free in the cell or it can be integrated into the E. coli chromosome.called as
episomes. The F-plasmid also carr.
Bacterial Conjugation (Genetic recombination in Bacteria)
Introduction, discovery, f-plasmid and its properties, bacterial conjugation steps, Hfr strain.
For 12th, M.Sc, B.Sc students.
F plasmid organisation
organization and integration of F plamid into bacterial strain through conjugation
Bacterial conjugation
bacterial conjugation genetic transfer transfer of genetic material in bacteria F+ and F- factors bacterial biology genetics factors in bacterial genes genetics DNA synthesis mobilization cell to cell contact bridge formation
1. Escherichia coli have F plasmid which by replication gets transfe.pdf
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+.
Solution
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+..
Conjuagtion types and application
this presentation basically focusses on the types of conjugation namely Hfr and F' conjugation.
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic information? How are these similar
and how are they different? How can they be used to map bacterial genomes?
b) Explain how genes in phages are mapped. (This is different from the above question)
Solution
Ques-1: What are three ways that bacteria can exchange genetic information? How are these
similar and how are they different? How can they be used to map bacterial genomes?
Answer:
Horizontal gene transfer (HGT) or lateral gene transfer is the transfer of genes between two
organisms that are phylogenetically unrelated but can readily exchange genes by conjugation or
transduction or transformation. HGT commonly occurs in prokaryotes and it plays a key role in
their evolution. HGT accelerates evolution.
Horizontal gene transfer: 1. conjugation, 2. transformation; 3. transduction
Similarities & differences:
Horizontal gene transfer is the transfer of genomes by between bacteria by the routes that does
not involve parent-offspring transmission. Conjugation, transduction and transformation are the
three mechanism of horizontal gene transfer.
In the above case, it has clearly illustrated that the conditions, which can successfully prevents
transduction and abolish transfer of chromosomal DNA from one bacterial cell to another by
generalized transduction with P22 bacteriophage. Initially recipient cell- surface receptors are
important for the attachment of the bacteriophage (P22) and finally spikes of phage enable
bacterial membrane to lyse finally incorporate phage genome into the bacterial cell for
integration. Therefore, for efficient transduction, presence of recipient cell- surface receptors is
essential in the culture otherwise, bacteria exert resistant effects to the phages.
Transformation type of gene transfer occurs between bacteria. The genes released by dead
bacteria can be transferred to the live bacteria, and get integrated into its genome, which results
in genetic recombination of the host.
Transduction is also involving the genetic material transfer, but this needs bacteriophage as
media. The bacteriophage infects the bacteria in lytic cycle and releases the new virions, the
capsids of some of these virions carry the bacterial DNA fragments into the new host bacteria,
followed by genetic recombination of the host bacteria.
Conjugation: It is the transfer of genetic material from cell to cell by means of direct contact or a
bridge like structure. Bacterial conjugation needs the conjugation tube for genome transfer, both
transduction and transformation does not need this.
Mapping the genome -mechanisms
F factor is mainly pertaining to a class of conjugate plasmids, which meticulously originated to
control sexual functions in prokaryotes via fertility inhibition. The origin of F-factor is mainly
with most common functional segments such as gene traJ, Ori, OriC. The structure of this F
factor contains OriT for origin of transfer and acts as a starting point for gene transmission.
Gene transfer
It is defined simply as a technique to efficiently and stably introduce foreign genes into the genome of target cells.
The insertion of unrelated, therapeutic genetic information in the form of DNA into target cells
Gene and genetic mapping by recombination
Gene mapping in bacteria by recombination process
by arif PHD scholar Animal Biotechnology
Consider an experiment in which an Hfr strain of E. coli bacteria wi.pdf
Consider an experiment in which an Hfr strain of E. coli bacteria with the StrS, his+, met+
alleles in its genome is mixed with an F- strain of E. coli with the StrR, his-, and met- alleles in
its genome. After the cells have been cultured together for 8 hours, the mixture of cells is plated
on minimal media containing Streptomycin. Which genotype do you expect to be most common
among the colonies that grow?F+: StrR, his+ and met+
Solution
Answer D) F-: StrR, his+ and met+
Explanation - We know that Hfr strain is a high frequency recombination strain in which the F
plasmid (fertility factor) is integrated in its genome/chromosomal DNA. this way, Hfr has ability
to transfer the entire genome sequence to its recipient (F- strain), but it requires a long time of
physical contact with the recipient due to which the recipient get some gene (long time physical
contact is rarely achieved) and remains as F- strain. also, there is no back transfer of genes to the
Hfr cells from the F- cells.
hence, in this case, only F- strain will survive which has resistance against Streptomycin and able
to synthesize its amino acids on minimal media. which is a recombinant F-: StrR, his+ and met+..
Conjugation in Bacteria
Conjugation is a type of genetic exchange between two bacterial cells, explore further in this document.
Describe the role of plasmids in horizontal (lateral) gene transfer .pdf
Describe the role of plasmids in horizontal (lateral) gene transfer in bacteria.
Solution
Horizontal gene transfer (HGT) refers to lateral movement of genes or genetic material between
unicellar or multicellular organisms, having important role in evolution. In prokaryotes, this
HGT happens by transformation, transduction or by conjugation. Plasmids play an important role
in bacterial conjugation. During conjugation the donor bacteria transfers a conjugative genetic
element to the recipient. This genetic material is mostly a plasmid. It is the process by which the
transfer of DNA takes place via a plasmid from a donor cell to the recipient cell, through cell to
cell contact.
This mode of gene transfer is common in many bacterial strains. The contact between the donor
and recipient is stable and the DNA transferred to the recipient gets incorporated in its
chromosome by the process of homologous recombination or inserion. Cell-to-cell contact
occurs by F-pilus. Through this pilus, a copy of F- (fertility) plasmid transfers to a cell that does
not contain the plasmid or F-plasmid sometimes even gets integrated in chromosome of its
bacterial host (forming Hfr). This new incorporated DNA may recombine with that of its new
host resulting in new gene variants. Some plasmids encode for antibiotic-resistance genes, which
are transferred through HGT to other populations of bacteria by conjugation. Thus HGT becomes
an evolutionary force for creating diversity within bacterial species..
Transfer of genetic information-Bijesh
Gene transfer Methods-Conjugation, Transformation, Transduction
A microbial geneticist isolates a new mutation in E. coli and wishes.pdf
A microbial geneticist isolates a new mutation in E. coli and wishes to map its chromosomal
location. She uses interrupted-mating experiments with Hfr strains and generalized-transduction
experiments with phage P1. Explain why each technique, by itself, is insufficient for accurate
mapping.
Solution
Hfr cells undergoing conjugation transfer host genes in a linear fashion. The genes transferred
depend on both the Hfr strain and the length of time during which the transfer occurred.
F factors arise from improper excision of an Hfr from bacterial chromosome. They can have only
specific bacterial genes on them because the integration site is fixed for each strain.
P1 does not recognize pseudo-pac sites on the chromosome. P1 seems to begin packaging from
random double-stranded breaks on the chromosome that are generated during phage infection.
Generalized transduction is the process by which any bacterial gene may be transferred to
another bacterium via a bacteriophage, and very rarely a small number of phages carry the donor.
A restricted set of genes will not be able to transfer by this method. She wanted to transfer a
specific gene to a specific located which is not doen by generalized transduction, so accurate
gene mapping will be difficult by these methods..
DnaAORC This protein is responsible for initiation of DNA replicat.pdf
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined.
Solution
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined..
Answer(19)Mary’s parents are normal, therefore her brother must.pdf
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s
Solution
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s.
More Related Content
Similar to HFR cells are formed from homologous recombination due to integratio.pdf
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic information? How are these similar
and how are they different? How can they be used to map bacterial genomes?
b) Explain how genes in phages are mapped. (This is different from the above question)
Solution
Ques-1: What are three ways that bacteria can exchange genetic information? How are these
similar and how are they different? How can they be used to map bacterial genomes?
Answer:
Horizontal gene transfer (HGT) or lateral gene transfer is the transfer of genes between two
organisms that are phylogenetically unrelated but can readily exchange genes by conjugation or
transduction or transformation. HGT commonly occurs in prokaryotes and it plays a key role in
their evolution. HGT accelerates evolution.
Horizontal gene transfer: 1. conjugation, 2. transformation; 3. transduction
Similarities & differences:
Horizontal gene transfer is the transfer of genomes by between bacteria by the routes that does
not involve parent-offspring transmission. Conjugation, transduction and transformation are the
three mechanism of horizontal gene transfer.
In the above case, it has clearly illustrated that the conditions, which can successfully prevents
transduction and abolish transfer of chromosomal DNA from one bacterial cell to another by
generalized transduction with P22 bacteriophage. Initially recipient cell- surface receptors are
important for the attachment of the bacteriophage (P22) and finally spikes of phage enable
bacterial membrane to lyse finally incorporate phage genome into the bacterial cell for
integration. Therefore, for efficient transduction, presence of recipient cell- surface receptors is
essential in the culture otherwise, bacteria exert resistant effects to the phages.
Transformation type of gene transfer occurs between bacteria. The genes released by dead
bacteria can be transferred to the live bacteria, and get integrated into its genome, which results
in genetic recombination of the host.
Transduction is also involving the genetic material transfer, but this needs bacteriophage as
media. The bacteriophage infects the bacteria in lytic cycle and releases the new virions, the
capsids of some of these virions carry the bacterial DNA fragments into the new host bacteria,
followed by genetic recombination of the host bacteria.
Conjugation: It is the transfer of genetic material from cell to cell by means of direct contact or a
bridge like structure. Bacterial conjugation needs the conjugation tube for genome transfer, both
transduction and transformation does not need this.
Mapping the genome -mechanisms
F factor is mainly pertaining to a class of conjugate plasmids, which meticulously originated to
control sexual functions in prokaryotes via fertility inhibition. The origin of F-factor is mainly
with most common functional segments such as gene traJ, Ori, OriC. The structure of this F
factor contains OriT for origin of transfer and acts as a starting point for gene transmission.
Gene transfer
It is defined simply as a technique to efficiently and stably introduce foreign genes into the genome of target cells.
The insertion of unrelated, therapeutic genetic information in the form of DNA into target cells
Gene and genetic mapping by recombination
Gene mapping in bacteria by recombination process
by arif PHD scholar Animal Biotechnology
Consider an experiment in which an Hfr strain of E. coli bacteria wi.pdf
Consider an experiment in which an Hfr strain of E. coli bacteria with the StrS, his+, met+
alleles in its genome is mixed with an F- strain of E. coli with the StrR, his-, and met- alleles in
its genome. After the cells have been cultured together for 8 hours, the mixture of cells is plated
on minimal media containing Streptomycin. Which genotype do you expect to be most common
among the colonies that grow?F+: StrR, his+ and met+
Solution
Answer D) F-: StrR, his+ and met+
Explanation - We know that Hfr strain is a high frequency recombination strain in which the F
plasmid (fertility factor) is integrated in its genome/chromosomal DNA. this way, Hfr has ability
to transfer the entire genome sequence to its recipient (F- strain), but it requires a long time of
physical contact with the recipient due to which the recipient get some gene (long time physical
contact is rarely achieved) and remains as F- strain. also, there is no back transfer of genes to the
Hfr cells from the F- cells.
hence, in this case, only F- strain will survive which has resistance against Streptomycin and able
to synthesize its amino acids on minimal media. which is a recombinant F-: StrR, his+ and met+..
Conjugation in Bacteria
Conjugation is a type of genetic exchange between two bacterial cells, explore further in this document.
Describe the role of plasmids in horizontal (lateral) gene transfer .pdf
Describe the role of plasmids in horizontal (lateral) gene transfer in bacteria.
Solution
Horizontal gene transfer (HGT) refers to lateral movement of genes or genetic material between
unicellar or multicellular organisms, having important role in evolution. In prokaryotes, this
HGT happens by transformation, transduction or by conjugation. Plasmids play an important role
in bacterial conjugation. During conjugation the donor bacteria transfers a conjugative genetic
element to the recipient. This genetic material is mostly a plasmid. It is the process by which the
transfer of DNA takes place via a plasmid from a donor cell to the recipient cell, through cell to
cell contact.
This mode of gene transfer is common in many bacterial strains. The contact between the donor
and recipient is stable and the DNA transferred to the recipient gets incorporated in its
chromosome by the process of homologous recombination or inserion. Cell-to-cell contact
occurs by F-pilus. Through this pilus, a copy of F- (fertility) plasmid transfers to a cell that does
not contain the plasmid or F-plasmid sometimes even gets integrated in chromosome of its
bacterial host (forming Hfr). This new incorporated DNA may recombine with that of its new
host resulting in new gene variants. Some plasmids encode for antibiotic-resistance genes, which
are transferred through HGT to other populations of bacteria by conjugation. Thus HGT becomes
an evolutionary force for creating diversity within bacterial species..
Transfer of genetic information-Bijesh
Gene transfer Methods-Conjugation, Transformation, Transduction
A microbial geneticist isolates a new mutation in E. coli and wishes.pdf
A microbial geneticist isolates a new mutation in E. coli and wishes to map its chromosomal
location. She uses interrupted-mating experiments with Hfr strains and generalized-transduction
experiments with phage P1. Explain why each technique, by itself, is insufficient for accurate
mapping.
Solution
Hfr cells undergoing conjugation transfer host genes in a linear fashion. The genes transferred
depend on both the Hfr strain and the length of time during which the transfer occurred.
F factors arise from improper excision of an Hfr from bacterial chromosome. They can have only
specific bacterial genes on them because the integration site is fixed for each strain.
P1 does not recognize pseudo-pac sites on the chromosome. P1 seems to begin packaging from
random double-stranded breaks on the chromosome that are generated during phage infection.
Generalized transduction is the process by which any bacterial gene may be transferred to
another bacterium via a bacteriophage, and very rarely a small number of phages carry the donor.
A restricted set of genes will not be able to transfer by this method. She wanted to transfer a
specific gene to a specific located which is not doen by generalized transduction, so accurate
gene mapping will be difficult by these methods..
Similar to HFR cells are formed from homologous recombination due to integratio.pdf (20)
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic informati.pdf
Consider an experiment in which an Hfr strain of E. coli bacteria wi.pdf
Consider an experiment in which an Hfr strain of E. coli bacteria wi.pdf
Describe the role of plasmids in horizontal (lateral) gene transfer .pdf
Describe the role of plasmids in horizontal (lateral) gene transfer .pdf
A microbial geneticist isolates a new mutation in E. coli and wishes.pdf
A microbial geneticist isolates a new mutation in E. coli and wishes.pdf
More from aquacosmossystems
DnaAORC This protein is responsible for initiation of DNA replicat.pdf
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined.
Solution
DnaA/ORC: This protein is responsible for initiation of DNA replication. When DNA replication
is temperature sensitive, it affects the function of Dna A/ ORC. It cannot bind to the N-terminus
and thus DNA replication is immediately stopped. So with the increase in temperature, there is
quick stop of the DNA replication.
SSB/RPA: This is the protein complex which acts on the single stranded DNA complex. In the
presences on non-permissible temperatures, this protein complex will quick stop the process of
DNA replication. Thus replication fork is not formed since it requires presence of one of the
proteins from the complex. It can be said that since the process is of so high order, quick stop
mutation takes please due to non- permissible temperature.
DnaB/Mcm: this protein complex is responsible for separating the DNA strand. With the
increase in temperature, the protein cannot separate the DNA strands. Thus DNA replication
stops immediately. Thus it is a quick stop type mutanation.
DNA ligase: there 3 major functions of DNA ligase 1) It seals the repairs in the DNA 2) Joins
the recombinant fragment and 3) it joins Okazaki fragment. When there is temperature sensitive
mutants, DNA ligase will follow the process of slow stop mutants. It is because DNA replication
has taken place only sealing part is left. So DNA replication will be completed but Okazaki
fragments will not be joined..
Answer(19)Mary’s parents are normal, therefore her brother must.pdf
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s
Solution
Answer:
(19)
Mary’s parents are normal, therefore her brother must have got the disease allele from his mother
who is a carrier of the disease.
Parents
XY (Mary’s father) x XhX (Mary’s mother)
XhY (Mary’s brother)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
(a) Therefore Mary has 50% chance or ½ chance of being a carrier
(b) (c) and (d)
For a carrier female, with each pregnancy there is a one in two (50%) chance her sons will
inherit the disease allele and a one in two (50%) chance her daughters will be carriers.
Among mary’s sons, one will be a carrier and another will be hemophiliac
Among mary’s daughters, one will be a carrier and another will be normal
(e) the carrier daughter can transmit the disease to her son’s.
Answer Q1. Red mangroove is the last mangroove to colonise on the.pdf
Answer :
Q1. Red mangroove is the last mangroove to colonise on the shore line
Q2. The melon of Dolphins are used for the focusing of sound pulses.
Q4. Coral reef accounts for the largest pecentage of the oceans primary
Solution
Answer :
Q1. Red mangroove is the last mangroove to colonise on the shore line
Q2. The melon of Dolphins are used for the focusing of sound pulses.
Q4. Coral reef accounts for the largest pecentage of the oceans primary.
An Arrhenius acid produces H+ ions insolutionA Bronsted-Lowry acid.pdf
An Arrhenius acid produces H+ ions insolution
A Bronsted-Lowry acid is a protoon donor.
A Lewis base is an eelectron pair donor.
Water could behave as an Arrhenius acid, Bronsted-Lowryacid, and a Lewis base.
Water acting as an arrhenius acid: Water acts as asArrhenius acid as it is capable of producing
protons insolution.
H2O (l) H+ (aq) + OH-(aq)
Water acting as a Bronsted-Lowry acid: Water acts as aBronsted-Lowry acid as it can donate H+
to a base.
H2O (l) + NH3 (aq) NH4+ (aq) + OH- (aq)
Water acting as a Lewis base: Water acts as a Lewis baseas it can donate the lone pair of
electrons present on the Oxygenatom.
Al+3 (aq) + H2O (l) Al(H2O)6+3 (aq)
Solution
An Arrhenius acid produces H+ ions insolution
A Bronsted-Lowry acid is a protoon donor.
A Lewis base is an eelectron pair donor.
Water could behave as an Arrhenius acid, Bronsted-Lowryacid, and a Lewis base.
Water acting as an arrhenius acid: Water acts as asArrhenius acid as it is capable of producing
protons insolution.
H2O (l) H+ (aq) + OH-(aq)
Water acting as a Bronsted-Lowry acid: Water acts as aBronsted-Lowry acid as it can donate H+
to a base.
H2O (l) + NH3 (aq) NH4+ (aq) + OH- (aq)
Water acting as a Lewis base: Water acts as a Lewis baseas it can donate the lone pair of
electrons present on the Oxygenatom.
Al+3 (aq) + H2O (l) Al(H2O)6+3 (aq).
Ans. One of the main disadvantage is chromatic aberrasion. A more li.pdf
Ans. One of the main disadvantage is chromatic aberrasion. A more likely cost of trichromacy
arises when an image is sampled by receptors with differing spectral sensivities because it is not
possible to separate luminanance information.
Solution
Ans. One of the main disadvantage is chromatic aberrasion. A more likely cost of trichromacy
arises when an image is sampled by receptors with differing spectral sensivities because it is not
possible to separate luminanance information..
A coordination problem occurs in situations where there is no Nash E.pdf
A coordination problem occurs in situations where there is no Nash Equilibrium or there is more
than one Nash Equilibrium. it also occure when there is a unique but undireable Nash
Equilibrium.
Option (B), Option (D) and option (E) is correct answer.
Solution
A coordination problem occurs in situations where there is no Nash Equilibrium or there is more
than one Nash Equilibrium. it also occure when there is a unique but undireable Nash
Equilibrium.
Option (B), Option (D) and option (E) is correct answer..
1. When an integrated F plasmid leaves the chromosome it may take so.pdf
1. When an integrated F plasmid leaves the chromosome it may take some chromosomal genes
from one side of the integration site and it results in F\' cells. Integration of F plasmid into the
host chromosome results in Hfr (High frequency of recombination) cells
2. Genetic transfer in Archaea is by means of horizontal gene transfer where genes can be
transfered from one to other similar to bacteria conjugation or transduction.
3. Methanogenesis is the formation of methane by the methanogens. Methanogenesis actually
complete the decomposition process. The semi-final products of decomposition like carbon
dioxide, small organics and hydrogen are completely removed. If methanogenesis does not
happen, it leads to accumulation of carbon in anaerobic envrionment. Methanogens belong to
group archaea.
4. To resist high temperatures in the environment, bacteria convert all its unsaturated fatty acids
in the membrane to saturated fatty acids. It prevents the denaturation of enzymes and proteins in
the cell.
5. Methanogenesis takes place in Anaerobic condition because oxygen inhibits the growth of
methanogens.
Solution
1. When an integrated F plasmid leaves the chromosome it may take some chromosomal genes
from one side of the integration site and it results in F\' cells. Integration of F plasmid into the
host chromosome results in Hfr (High frequency of recombination) cells
2. Genetic transfer in Archaea is by means of horizontal gene transfer where genes can be
transfered from one to other similar to bacteria conjugation or transduction.
3. Methanogenesis is the formation of methane by the methanogens. Methanogenesis actually
complete the decomposition process. The semi-final products of decomposition like carbon
dioxide, small organics and hydrogen are completely removed. If methanogenesis does not
happen, it leads to accumulation of carbon in anaerobic envrionment. Methanogens belong to
group archaea.
4. To resist high temperatures in the environment, bacteria convert all its unsaturated fatty acids
in the membrane to saturated fatty acids. It prevents the denaturation of enzymes and proteins in
the cell.
5. Methanogenesis takes place in Anaerobic condition because oxygen inhibits the growth of
methanogens..
1. A DNA molecule contains two strands of poly deoxyribonucleotide s.pdf
1. A DNA molecule contains two strands of poly deoxyribonucleotide strands twisted together to
form a double helix. The two strands are antiparallel, i.e., one strand runs in the 5’ to 3’ direction
while the other in 3’ to 5’ direction. The sugar-phosphate backbone is found on the exterior of
this helix and the bases are branching towards the center.
RNA, in the majority cases, consists of one strand (polymer of ribonucleotides) of nucleic acids
linked together by 3\',5\'-phosphodiester bonds.
2. The mRNA molecules which are transcribed from genes of the total genome and have the base
sequence complementary to DNA, carry DNA,s genetic information\'s for the assembly of amino
acids into the polypeptide chains, to the cytoplasmic sites of protein synthesis, the ribosomes. to
which they become associated to participate in codon-anticodon interaction with tRNA, are
called informational or messenger RNAs (mRNA) .
Solution
1. A DNA molecule contains two strands of poly deoxyribonucleotide strands twisted together to
form a double helix. The two strands are antiparallel, i.e., one strand runs in the 5’ to 3’ direction
while the other in 3’ to 5’ direction. The sugar-phosphate backbone is found on the exterior of
this helix and the bases are branching towards the center.
RNA, in the majority cases, consists of one strand (polymer of ribonucleotides) of nucleic acids
linked together by 3\',5\'-phosphodiester bonds.
2. The mRNA molecules which are transcribed from genes of the total genome and have the base
sequence complementary to DNA, carry DNA,s genetic information\'s for the assembly of amino
acids into the polypeptide chains, to the cytoplasmic sites of protein synthesis, the ribosomes. to
which they become associated to participate in codon-anticodon interaction with tRNA, are
called informational or messenger RNAs (mRNA) ..
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neut.pdf
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neutral solutions, it will be
insoluble in water. However, in acidic solutions, the amino group will be protonated, formed a
charged salt species. As with normal salts, this is water-soluble. So; 1.0 M HCl: soluble 1.0 M
NaOH and Water: insoluble.
Solution
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neutral solutions, it will be
insoluble in water. However, in acidic solutions, the amino group will be protonated, formed a
charged salt species. As with normal salts, this is water-soluble. So; 1.0 M HCl: soluble 1.0 M
NaOH and Water: insoluble..
draw a sphere and use raytracing on the sphere in OpenGL glut. .pdf
/* draw a sphere and use raytracing on the sphere in OpenGL glut. */
#include
#include
#include
#include
#include
#define screen_width 750
#define screen_height 750
#define true 1
#define false 0
#define perpendicular 0
int gridXsize = 20;
int gridZsize = 20;
float plane[] = {0.0, 1.0, 0.0, -50.0,};
float sphere[] = {250.0, 270.0, -100.0, 100.0};
float eye[] = {0.0, 400.0, 550.0};
float light[] = {250.0, 550.0, -200.0};
float dot(float *u, float *v)
{
return u[0]*v[0] + u[1]*v[1] + u[2]*v[2];
}
void norm(float *u)
{
float norm = sqrt(abs(dot(u,u)));
for (int i =0; i <3; i++)
{
u[i] = u[i]/norm;
}
}
float plane_intersect(float *u, float *pO)
{
float normt[3] = {plane[0], plane[1], plane[2]};
float s;
if (dot(u,normt) == 0)
{
s = -10;
}
else
{
s = (plane[3]-(dot(pO,normt)))/(dot(u,normt));
}
return s;
}
float sphere_intersect(float *u, float *pO)
{
float deltaP[3] = {sphere[0]-pO[0],sphere[1]-pO[1],sphere[2]-pO[2]};
float deltLen = sqrt(abs(dot(deltaP,deltaP)));
float t=0;
float answer;
float det;
if ((det =(abs(dot(u,deltaP)*dot(u,deltaP))- (deltLen*deltLen)+sphere[3]*sphere[3])) < 0)
{
answer = -10;
}
else
{
t =-1*dot(u,deltaP)- sqrt(det) ;
if (t>0)
{
answer = t;
}
else
{
answer = -10;
}
}
return answer;
}
void find_reflect(float *u, float s, float *pO)
{
float n[3] = {pO[0]+s *u[0]-sphere[0],pO[1]+s *u[1]-sphere[1],pO[2]+s *u[2]- sphere[2]};
float l[3] = {s *u[0],s *u[1],s *u[2]};
u[0] =(2*dot(l,n)*n[0])-l[0];
u[1] = (2*dot(l,n)*n[1])-l[1];
u[2] = (2*dot(l,n)*n[2])-l[2];
}
float find_shade(float *u,float s, float *pO)
{
float answer;
float lightVec[3] = {light[0]-(pO[0]+s *u[0]), light[1]-(pO[1]+s *u[1]), light[2]-(pO[2]+s
*u[2])};
float n[3] = {pO[0]+s *u[0]-sphere[0],pO[1]+s *u[1]-sphere[1],pO[2]+s *u[2]-sphere[2]};
answer = -1*dot(lightVec,n)/(sqrt(abs(dot(lightVec,lightVec)))*sqrt(abs(dot(n,n))));
return answer;
}
void init()
{
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluOrtho2D(0,screen_width,0,screen_height);
}
void display()
{
glClear(GL_COLOR_BUFFER_BIT| GL_DEPTH_BUFFER_BIT);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
for (int i=0; i < screen_width; i++)
{
for (int j=0; j < screen_height; j++)
{
float ray[3] = {1*(eye[0]-i),-1*(eye[1]-j),1*eye[2]};
float point[3] = {i,j,0};
norm(ray);
int plotted = false;
while (!plotted)
{
float s_plane = plane_intersect(ray, point);
float s_sphere = sphere_intersect(ray, point);
if (s_plane <= 0 && s_sphere <=0)
{
glColor3f(0,0,0);
glBegin(GL_POINTS);
glVertex3f(i,j,0);
glEnd();
plotted = true;
}
else if (s_sphere >= 0 && (s_plane <=0 || s_sphere <= s_plane))
{
find_reflect(ray, s_sphere, point);
}
else if (s_plane >=0 && (s_sphere <=0 ||s_plane <= s_sphere))
{
float shade = find_shade(ray, s_plane, point);
float xx = s_plane*ray[0] + eye[0];
float z = s_plane*ray[2] + eye[2];
if (abs((int)xx/gridXsize)%2 == abs((int)z/gridZsize)%2)
{
glColor3f(shade,0,0);
}
else
{
glColor3f(shade,shade,shade);
}
glBegin(GL_POINTS);
glVertex3f(i,j,0);
glEnd();
plotted = true;
}
}
}
}
glFlush();
}
int main(.
SnO2 (ore) is initially washed and dressed. then .pdf
SnO2 (ore) is initially washed and dressed. then it is treated with red hot
coke/coall(acts as reducing agent) SnO2(s) + C(s)---------> Sn(molten) +CO(g) thus obtained
molten Sn is cooled to Pure solid form
Solution
SnO2 (ore) is initially washed and dressed. then it is treated with red hot
coke/coall(acts as reducing agent) SnO2(s) + C(s)---------> Sn(molten) +CO(g) thus obtained
molten Sn is cooled to Pure solid form.
d) Valine Note its side chain doesnt contain a.pdf
d) Valine Note: its side chain doesn\'t contain acidic H such as those in NH or
COOH group. Thus, pH value doesn\'t affect the charge since no protonation or deprotonation
happens.
Solution
d) Valine Note: its side chain doesn\'t contain acidic H such as those in NH or
COOH group. Thus, pH value doesn\'t affect the charge since no protonation or deprotonation
happens..
Amphiprotic. The preferred term is amphoteric. A .pdf
Amphiprotic. The preferred term is amphoteric. A compound that is amphoteric can
behave as either an acid or a base. These are most commonly explained when using the
Bronsted-Lowry theory of acids and bases. Sorry. NONE of these are amphoteric (or
amphiprotic). In order to behave as a Bronsted-Lowry acid it must donate a proton. That
narrows it down to HCl and HSO4-. In order to behave as a base it must accept a proton. A, D
and E can all accept protons. HCl is a strong acid and dissociates to form H+ and Cl-. There is
no tendency to gain a proton. HSO4- is a weak acid and can donate a proton, but it won\'t accept
a proton to make H2SO4 because H2SO4 is a strong acid and completely dissociates. Even if
one of the choices had been HSO3- we would still have a problem. It would be tempting to say
that HSO3- could gain a proton to make the weak acid H2SO3. The problem is that H2SO3 does
not exist in aqueous solution. It is a fictitious acid, much like H2CO3 which also cannot exist in
aqueous solution. The bottom line is that NONE of these are amphoteric. ========= Follow
up ========== See! HPV got suckered. Sorry, you can\'t possibly form H2SO4 and OH-.
H2SO4 is a strong acid and completely dissociates to form H+ and HSO4-. The H+ will combine
with OH- to make water, and you\'re right back where you started. No. HSO4- is NOT
amphoteric in aqueous solution. ======= More follow up ========== \"Would it matter
about the physical state?\" No. For the substance to gain a proton, it will be in aqueous solution.
As soon as HCl interacts with water it will dissociate to form H+ and Cl-.
=============================== You may see that HSO4- is listed as an amphoteric
substance, but only because whoever wrote that didn\'t think it through.
Solution
Amphiprotic. The preferred term is amphoteric. A compound that is amphoteric can
behave as either an acid or a base. These are most commonly explained when using the
Bronsted-Lowry theory of acids and bases. Sorry. NONE of these are amphoteric (or
amphiprotic). In order to behave as a Bronsted-Lowry acid it must donate a proton. That
narrows it down to HCl and HSO4-. In order to behave as a base it must accept a proton. A, D
and E can all accept protons. HCl is a strong acid and dissociates to form H+ and Cl-. There is
no tendency to gain a proton. HSO4- is a weak acid and can donate a proton, but it won\'t accept
a proton to make H2SO4 because H2SO4 is a strong acid and completely dissociates. Even if
one of the choices had been HSO3- we would still have a problem. It would be tempting to say
that HSO3- could gain a proton to make the weak acid H2SO3. The problem is that H2SO3 does
not exist in aqueous solution. It is a fictitious acid, much like H2CO3 which also cannot exist in
aqueous solution. The bottom line is that NONE of these are amphoteric. ========= Follow
up ========== See! HPV got suckered. Sorry, you can\'t possibly form H2SO4 and OH-.
H2SO4 is a strong acid and completely dissociates to form H+ a.
at equivalence point, Moles OH- added = moles H+.pdf
at equivalence point, Moles OH- added = moles H+ Molarity NaOH x vol. OH- x
1 OH- per NaOH = moles OH- = moles H+ = Molarity H2SO4 x vol H2SO4 x 2 H+ per
H2SO4 More generally, M1V1E1 = M2V2E2, where E is the equivalence factor, number of H+
per formula unit for acids and number of OH- per unit for bases
Solution
at equivalence point, Moles OH- added = moles H+ Molarity NaOH x vol. OH- x
1 OH- per NaOH = moles OH- = moles H+ = Molarity H2SO4 x vol H2SO4 x 2 H+ per
H2SO4 More generally, M1V1E1 = M2V2E2, where E is the equivalence factor, number of H+
per formula unit for acids and number of OH- per unit for bases.
We Know that Water is a Polar component as .pdf
We Know that: Water is a Polar component as the property ofthe solubility Like
dissolves Like methyl tertiary butyl etherwhich is polar compound is relatively more soluble
inwater when compared to other components in gasoline.
Solution
We Know that: Water is a Polar component as the property ofthe solubility Like
dissolves Like methyl tertiary butyl etherwhich is polar compound is relatively more soluble
inwater when compared to other components in gasoline..
molar mass of C2H5OH = 46.1 gmole moles of C2H5O.pdf
molar mass of C2H5OH = 46.1 g/mole moles of C2H5OH = mass / molar mass =
47.0 / 46.1 = 1.02 mole 1 moles of C2H5OH contains 6.02 x 1023 C2H5OH molecules, thus,
number of C2H5OH molecules = 1.02 x 6.02 x 1023 = 6.14 x 1023 C2H5OH molecules
Solution
molar mass of C2H5OH = 46.1 g/mole moles of C2H5OH = mass / molar mass =
47.0 / 46.1 = 1.02 mole 1 moles of C2H5OH contains 6.02 x 1023 C2H5OH molecules, thus,
number of C2H5OH molecules = 1.02 x 6.02 x 1023 = 6.14 x 1023 C2H5OH molecules.
What property of PN code makes them suitable for use in spread spect.pdf
What property of PN code makes them suitable for use in spread spectrum communication?
Solution
Pseudo random noise is used in some electronic musical instruments, either by itself or as an
input to subtractive synthesis, and in many white noise machines.
In spread-spectrum systems, the receiver correlates a locally generated signal with the received
signal. Such spread-spectrum systems require a set of one or more \"codes\" or \"sequences\"
such that
Like random noise, the local sequence has a very low correlation with any other sequence in the
set, or with the same sequence at a significantly different time offset, or with narrow band
interference, or with thermal noise.
Unlike random noise, it must be easy to generate exactly the same sequence at both the
transmitter and the receiver, so the receiver\'s locally generated sequence has a very high
correlation with the transmitted sequence.
In a direct-sequence spread spectrum system, each bit in the pseudorandom binary sequence is
known as a chip and the inverse of its period as chip rate; comparebit rate and symbol rate.
In a frequency-hopping spread spectrum sequence, each value in the pseudo random sequence is
known as a channel number and the inverse of its period as thehop rate. FCC Part 15 mandates at
least 50 different channels and at least a 2.5 Hz hop rate for narrow band frequency-hopping
systems.
GPS satellites broadcast data at a rate of 50 data bits per second – each satellite modulates its
data with one PN bit stream at 1.023 million chips per second and the same data with another PN
bit stream at 10.23 million chips per second. GPS receivers correlate the received PN bit stream
with a local reference to measure distance. GPS is a receive-only system that uses relative timing
measurements from several satellites (and the known positions of the satellites) to determine
receiver position.
Other range-finding applications involve two-way transmissions. A local station generates a
pseudo random bit sequence and transmits it to the remote location (using any modulation
technique). Some object at the remote location echoes this PN signal back to the location station
– either passively, as in some kinds of radar and sonar systems, or using an active transponder at
the remote location, as in the Apollo Unified S-band system.[4] By correlating a (delayed version
of) the transmitted signal with the received signal, a precise round trip time to the remote
location can be determined and thus the distance..
TLC can be used to help determine the number of components in a mixt.pdf
TLC can be used to help determine the number of components in a mixture, the identity of
compounds, and the purity of a compound. By observing the appearance of a product or the
disappearance of a reactant, it can also be used to monitor the progress of a reaction.
The sample is applied to the layer of adsorbent, near one edge, as a small spot of a solution.
After the solvent has evaporated, the adsorbent-coated sheet is propped more or less vertically in
a closed container, with the edge to which the spot was applied down. The spot on the thin layer
plate must be positioned above the level of the solvent in the container .If it is below the level of
the solvent, the spot will be washed off the plate into the developing solvent. The solvent, which
is in the bottom of the container, creeps up the layer of adsorbent, passes over the spot, and, as it
continues up, effects a separation of the materials in the spot (\"develops\" the chromatogram).
When the solvent front has nearly reached nearly the top of the adsorbent, the thin layer plate is
removed from the container.
Solution
TLC can be used to help determine the number of components in a mixture, the identity of
compounds, and the purity of a compound. By observing the appearance of a product or the
disappearance of a reactant, it can also be used to monitor the progress of a reaction.
The sample is applied to the layer of adsorbent, near one edge, as a small spot of a solution.
After the solvent has evaporated, the adsorbent-coated sheet is propped more or less vertically in
a closed container, with the edge to which the spot was applied down. The spot on the thin layer
plate must be positioned above the level of the solvent in the container .If it is below the level of
the solvent, the spot will be washed off the plate into the developing solvent. The solvent, which
is in the bottom of the container, creeps up the layer of adsorbent, passes over the spot, and, as it
continues up, effects a separation of the materials in the spot (\"develops\" the chromatogram).
When the solvent front has nearly reached nearly the top of the adsorbent, the thin layer plate is
removed from the container..
There may have some variables affect the model. So the error term wi.pdf
There may have some variables affect the model. So the error term will reflect a systematic
pattern. Or it is not a correct functional form. So there is a bias.
I would suggest that it can be remedial by using data transformation.
Solution
There may have some variables affect the model. So the error term will reflect a systematic
pattern. Or it is not a correct functional form. So there is a bias.
I would suggest that it can be remedial by using data transformation..
There are four major functional characteristics of skeletal muscle..pdf
There are four major functional characteristics of skeletal muscle.
(i) Contractility- for muscle movement
(ii) Excitability- response to chemical and/or electrical signals.
(iii) Extensibility- the ability to stretch to the normal resting length after contracting
(iv) Elasticity - the ability to return to the original resting length after a muscle has been
stretched.
Solution
There are four major functional characteristics of skeletal muscle.
(i) Contractility- for muscle movement
(ii) Excitability- response to chemical and/or electrical signals.
(iii) Extensibility- the ability to stretch to the normal resting length after contracting
(iv) Elasticity - the ability to return to the original resting length after a muscle has been
stretched..
More from aquacosmossystems (20)
DnaAORC This protein is responsible for initiation of DNA replicat.pdf
DnaAORC This protein is responsible for initiation of DNA replicat.pdf
Answer(19)Mary’s parents are normal, therefore her brother must.pdf
Answer(19)Mary’s parents are normal, therefore her brother must.pdf
Answer Q1. Red mangroove is the last mangroove to colonise on the.pdf
Answer Q1. Red mangroove is the last mangroove to colonise on the.pdf
An Arrhenius acid produces H+ ions insolutionA Bronsted-Lowry acid.pdf
An Arrhenius acid produces H+ ions insolutionA Bronsted-Lowry acid.pdf
Ans. One of the main disadvantage is chromatic aberrasion. A more li.pdf
Ans. One of the main disadvantage is chromatic aberrasion. A more li.pdf
A coordination problem occurs in situations where there is no Nash E.pdf
A coordination problem occurs in situations where there is no Nash E.pdf
1. When an integrated F plasmid leaves the chromosome it may take so.pdf
1. When an integrated F plasmid leaves the chromosome it may take so.pdf
1. A DNA molecule contains two strands of poly deoxyribonucleotide s.pdf
1. A DNA molecule contains two strands of poly deoxyribonucleotide s.pdf
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neut.pdf
Ethyl 4-aminobenzoate is a basic organic compound. In basic and neut.pdf
draw a sphere and use raytracing on the sphere in OpenGL glut. .pdf
draw a sphere and use raytracing on the sphere in OpenGL glut. .pdf
SnO2 (ore) is initially washed and dressed. then .pdf
SnO2 (ore) is initially washed and dressed. then .pdf
d) Valine Note its side chain doesnt contain a.pdf
d) Valine Note its side chain doesnt contain a.pdf
Amphiprotic. The preferred term is amphoteric. A .pdf
Amphiprotic. The preferred term is amphoteric. A .pdf
at equivalence point, Moles OH- added = moles H+.pdf
at equivalence point, Moles OH- added = moles H+.pdf
molar mass of C2H5OH = 46.1 gmole moles of C2H5O.pdf
molar mass of C2H5OH = 46.1 gmole moles of C2H5O.pdf
What property of PN code makes them suitable for use in spread spect.pdf
What property of PN code makes them suitable for use in spread spect.pdf
TLC can be used to help determine the number of components in a mixt.pdf
TLC can be used to help determine the number of components in a mixt.pdf
There may have some variables affect the model. So the error term wi.pdf
There may have some variables affect the model. So the error term wi.pdf
There are four major functional characteristics of skeletal muscle..pdf
There are four major functional characteristics of skeletal muscle..pdf
Recently uploaded
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
special B.ed 2nd year old paper_20240531.pdf
Instagram:-
https://instagram.com/special_education_needs_01?igshid=YmMyMTA2M2Y=
WhatsApp:-
https://chat.whatsapp.com/JVakNIYlSV94x7bwunO3Dc
YouTube:-
https://youtube.com/@special_education_needs
Teligram :- https://t.me/special_education_needs
Slide Shere :-
https://www.slideshare.net/shabnambano20?utm_campaign=profiletracking&utm_medium=sssite&utm_source=ssslideview
Polish students' mobility in the Czech Republic
Polish students mobility to the Czech Republic within eTwinning project "Medieval adventures with Marco Polo"
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Palestine last event orientationfvgnh .pptx
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Digital Tools and AI for Teaching Learning and Research
This Presentation in details discusses on Digital Tools and AI for Teaching Learning and Research
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
Class 11 CBSE Business Studies Project ( AIDS TO TRADE - INSURANCE)
Unit 2- Research Aptitude (UGC NET Paper I).pdf
This slide describes the research aptitude of unit 2 in the UGC NET paper I.
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is prepared for master's students (MIFB & MIBS) UUM. May it be useful to all.Recently uploaded (20)
Instructions for Submissions thorugh G- Classroom.pptx
Instructions for Submissions thorugh G- Classroom.pptx
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk - How can education support child empowerment.pptx
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI fund...
Digital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and Research
The French Revolution Class 9 Study Material pdf free download
The French Revolution Class 9 Study Material pdf free download
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
1.4 modern child centered education - mahatma gandhi-2.pptx
1.4 modern child centered education - mahatma gandhi-2.pptx
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptx
HFR cells are formed from homologous recombination due to integratio.pdf
- 1. HFR cells are formed from homologous recombination due to integration of the conjugate plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another cell. Hfr cells do not possess a complete "circular chromosome" i.e. plasmid as F+ cell. Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its chromosomal DNA often referred as "episome". F+ cell possess F factor known as plasmid & it can donate to another cell through conjugation. Both Hfr & F+ cells are competent for conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell. Interrupted mating: F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F- factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This structure is going to promote high frequency recombination through interrupted mating into the recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting conjugation. This is completely based on the streptomycin-sensitivity allele (str-s) in the Hfr strains. On the otherhand, streptomycin-resistance allele (str-r) presence in the recipient is employed considerably to specifically destroy the Hfr donor cells after conjugation. Solution HFR cells are formed from homologous recombination due to integration of the conjugate plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another cell. Hfr cells do not possess a complete "circular chromosome" i.e. plasmid as F+ cell. Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its chromosomal DNA often referred as "episome". F+ cell possess F factor known as plasmid & it can donate to another cell through conjugation. Both Hfr & F+ cells are competent for conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell. Interrupted mating: F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F- factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This structure is going to promote high frequency recombination through interrupted mating into the recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a
- 2. starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting conjugation. This is completely based on the streptomycin-sensitivity allele (str-s) in the Hfr strains. On the otherhand, streptomycin-resistance allele (str-r) presence in the recipient is employed considerably to specifically destroy the Hfr donor cells after conjugation.