Consider an experiment in which an Hfr strain of E. coli bacteria wi.pdf
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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+..
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Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
F plasmid organisation
organization and integration of F plamid into bacterial strain through conjugation
HFR cells are formed from homologous recombination due to integratio.pdf
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.
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.
Gene and genetic mapping by recombination
Gene mapping in bacteria by recombination process
by arif PHD scholar Animal Biotechnology
Please explain your answer. Thanks! In order to determine the overal.pdf
Please explain your answer. Thanks! In order to determine the overall chromosome map for a
new wild type strain of E. coli, three Hfr strains were obtained by independent insertions of the F
factor. The start times of transfer of several genes were determined during conjugation matings
between each Hfr strain and an F recipient carrying six auxotrophic mutations. Using the times
taken to begin transfer of each marker, draw a unified chromosome map of the chromosome of
this strain of E. coli, showing its total length in minutes of transfer and the order and relative
distances between these markers. Show the position and orientation of the F factor in each Hfr
strain.
Solution
A. Hfr 1:
Ade--Bio----His--Met-----Ile
Hfr 2:
Met-His-----Bio-Ade
Hfr 3:
Bio-Ade-----------Lys--Ile
B.
Hfr 1: F factor-Ade------->
Hfr 2: F factor-Met------->
Hfr 3: F-factor-Bio------->.
Recommended
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
F plasmid organisation
organization and integration of F plamid into bacterial strain through conjugation
HFR cells are formed from homologous recombination due to integratio.pdf
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.
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.
Gene and genetic mapping by recombination
Gene mapping in bacteria by recombination process
by arif PHD scholar Animal Biotechnology
Please explain your answer. Thanks! In order to determine the overal.pdf
Please explain your answer. Thanks! In order to determine the overall chromosome map for a
new wild type strain of E. coli, three Hfr strains were obtained by independent insertions of the F
factor. The start times of transfer of several genes were determined during conjugation matings
between each Hfr strain and an F recipient carrying six auxotrophic mutations. Using the times
taken to begin transfer of each marker, draw a unified chromosome map of the chromosome of
this strain of E. coli, showing its total length in minutes of transfer and the order and relative
distances between these markers. Show the position and orientation of the F factor in each Hfr
strain.
Solution
A. Hfr 1:
Ade--Bio----His--Met-----Ile
Hfr 2:
Met-His-----Bio-Ade
Hfr 3:
Bio-Ade-----------Lys--Ile
B.
Hfr 1: F factor-Ade------->
Hfr 2: F factor-Met------->
Hfr 3: F-factor-Bio------->.
How would you solve this problem The correct answer is in red. Cons.pdf
How would you solve this problem? The correct answer is in red. Consider the cross: Hfr: neo^S
pro + met + gal-man + lys- x F^-: neo^R pro - met - gal + man - lys + where the genetic map is
shown below (the
Solution
Answer
The Hfr strains have a certain F factor incorporated into the bacterial sites in different
orientations. The DNA transfer is initiated at the origin of transfer (Ori T) with the F factor.
During conjugation, the Hfr is able to pull all the positive genes into F-. Since, strains are plated
on a specific media, they require minimal media, the antibiotics F- is resistant to but Hfr is
sensitive to and the remaining genes necessary for the growth. The recipient F- will never receive
the complete F factor sequence; they will not become F+
According to the genetic map of both Hfr and F-, the neomycin sensitive and resistance part is
located on other part from other genes which are pro, met, man, and lysine. After conjugation,
the resistance part will be present while the other genes are replaced by the Hfr genes. Thus, we
should select a minimal media containing neomycin, mannose methionine, lysine, and proline..
This is the second time I post this question. I got an incorrect ans.pdf
This is the second time I post this question. I got an incorrect answer, so I had to post it again....
Please read the question; you need to draw. 8. In order to determine the overall chromosome map
for a new wild type strain of E. coli, three Hfr strains were obtained by independent insertions of
the F factor. The start times of transfer of several genes were determined during conjugation
matings between each Hfr strain and an F recipient carrying six auxotrophic mutations Time
(min) of marker transfer 2 Hfr Strain: Gene Ade His 15 15 37 65 35 13 50 40 10 Lys Bio Met 30
20 45 A. Using the times taken to begin transfer of each marker, draw a unified chromosome
map of the chromosome of this strain of E. coli, showing its total length in minutes of transfer
and the order and relative distances between these markers B. Show the position and orientation
of the F factor in each Hfr strain.
Solution
Short time duration represents the marker is near the start codon and longest time duration
represents the codon present at end of the sequence..
Final horizontal gene transfer
Ppt is made vailable for public for scientifc use.
gene transfer in bacteria its concept and its mechanism explained by using practical examples in a simple language. data is significant for competitive examinations
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.
Genetic Analysis and Mapping in Bacteria and Bacteriophages
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1) Pen R contains penicillin resistance gene .Usually + or - is not .pdf
1) Pen R contains penicillin resistance gene .Usually + or - is not denoted for antibiotics.
therefore the answer is option e.
2) a) conjugation is the exchange of genetic material by cell-cell contact via sex-pili.
3)a) High-Frequency recombination bacteria have F plasmid integrated into its chromosome and
transfers part of the chromosome along with F+ gene ,thereby resulting in F\' cells.
4) e) an F\' lac- E.coli when mates with F- Lac cell, the result will be F+, Lac-(from F\'),
Lac+(already present in recipient cell)
5) a) true, transformation is the transfer of an exogenous genetic material(DNA).
Solution
1) Pen R contains penicillin resistance gene .Usually + or - is not denoted for antibiotics.
therefore the answer is option e.
2) a) conjugation is the exchange of genetic material by cell-cell contact via sex-pili.
3)a) High-Frequency recombination bacteria have F plasmid integrated into its chromosome and
transfers part of the chromosome along with F+ gene ,thereby resulting in F\' cells.
4) e) an F\' lac- E.coli when mates with F- Lac cell, the result will be F+, Lac-(from F\'),
Lac+(already present in recipient cell)
5) a) true, transformation is the transfer of an exogenous genetic material(DNA)..
Conjuagtion types and application
this presentation basically focusses on the types of conjugation namely Hfr and F' conjugation.
Are hydrophobic interactions weak or strongSolutionHydrophobi.pdf
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Solution
Hydrophobic interactions describe the relations between water and hydrophobes (Low-water
soluble molecule). Hydrophobes are non-polar molecules and usually have a long chain of
carbon that do not interact with water molecules.
Strength of hydrophobic bonds are relatively stronger than other weak intermolecular forces (i.e.,
Vander Waals forces and hydrogen bonds). The strength of hydrophobic interactions depends on
several factors like:
1. Temperature: As the temperature increases, the strength of hydrophobic interaction also
increases also. However, at extreme temperatures, hydrophobic interactions denature.
2. Number of carbons on hydrophobes: Molecules with greatest number of carbon atoms will
have the strongest hydrophobic interactions.
3. The shape of hydrophobes: Aliphatic organic molecules will have stronger interactions than
aromatic organic molecules..
What is another word for account based forecast What is anothe.pdf
What is another word for account based forecast?
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Solution
The other word for account based forecasting is Budgeting.
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What function can be used to return multiple numeric data such as in.pdf
What function can be used to return multiple numeric data such as integer and/or float entered at
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A) Input
B)Eval
C)Output
D)return
Solution
Answer: Input
Input function can be used to return multiple numeric data such as integer and/or float entered at
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More Related Content
Similar to Consider an experiment in which an Hfr strain of E. coli bacteria wi.pdf
How would you solve this problem The correct answer is in red. Cons.pdf
How would you solve this problem? The correct answer is in red. Consider the cross: Hfr: neo^S
pro + met + gal-man + lys- x F^-: neo^R pro - met - gal + man - lys + where the genetic map is
shown below (the
Solution
Answer
The Hfr strains have a certain F factor incorporated into the bacterial sites in different
orientations. The DNA transfer is initiated at the origin of transfer (Ori T) with the F factor.
During conjugation, the Hfr is able to pull all the positive genes into F-. Since, strains are plated
on a specific media, they require minimal media, the antibiotics F- is resistant to but Hfr is
sensitive to and the remaining genes necessary for the growth. The recipient F- will never receive
the complete F factor sequence; they will not become F+
According to the genetic map of both Hfr and F-, the neomycin sensitive and resistance part is
located on other part from other genes which are pro, met, man, and lysine. After conjugation,
the resistance part will be present while the other genes are replaced by the Hfr genes. Thus, we
should select a minimal media containing neomycin, mannose methionine, lysine, and proline..
This is the second time I post this question. I got an incorrect ans.pdf
This is the second time I post this question. I got an incorrect answer, so I had to post it again....
Please read the question; you need to draw. 8. In order to determine the overall chromosome map
for a new wild type strain of E. coli, three Hfr strains were obtained by independent insertions of
the F factor. The start times of transfer of several genes were determined during conjugation
matings between each Hfr strain and an F recipient carrying six auxotrophic mutations Time
(min) of marker transfer 2 Hfr Strain: Gene Ade His 15 15 37 65 35 13 50 40 10 Lys Bio Met 30
20 45 A. Using the times taken to begin transfer of each marker, draw a unified chromosome
map of the chromosome of this strain of E. coli, showing its total length in minutes of transfer
and the order and relative distances between these markers B. Show the position and orientation
of the F factor in each Hfr strain.
Solution
Short time duration represents the marker is near the start codon and longest time duration
represents the codon present at end of the sequence..
Final horizontal gene transfer
Ppt is made vailable for public for scientifc use.
gene transfer in bacteria its concept and its mechanism explained by using practical examples in a simple language. data is significant for competitive examinations
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.
Genetic Analysis and Mapping in Bacteria and Bacteriophages
Essentials of Genetics - Chapter 8 - Genetic Analysis and Mapping in Bacteria and Bacteriophages
1) Pen R contains penicillin resistance gene .Usually + or - is not .pdf
1) Pen R contains penicillin resistance gene .Usually + or - is not denoted for antibiotics.
therefore the answer is option e.
2) a) conjugation is the exchange of genetic material by cell-cell contact via sex-pili.
3)a) High-Frequency recombination bacteria have F plasmid integrated into its chromosome and
transfers part of the chromosome along with F+ gene ,thereby resulting in F\' cells.
4) e) an F\' lac- E.coli when mates with F- Lac cell, the result will be F+, Lac-(from F\'),
Lac+(already present in recipient cell)
5) a) true, transformation is the transfer of an exogenous genetic material(DNA).
Solution
1) Pen R contains penicillin resistance gene .Usually + or - is not denoted for antibiotics.
therefore the answer is option e.
2) a) conjugation is the exchange of genetic material by cell-cell contact via sex-pili.
3)a) High-Frequency recombination bacteria have F plasmid integrated into its chromosome and
transfers part of the chromosome along with F+ gene ,thereby resulting in F\' cells.
4) e) an F\' lac- E.coli when mates with F- Lac cell, the result will be F+, Lac-(from F\'),
Lac+(already present in recipient cell)
5) a) true, transformation is the transfer of an exogenous genetic material(DNA)..
Conjuagtion types and application
this presentation basically focusses on the types of conjugation namely Hfr and F' conjugation.
Similar to Consider an experiment in which an Hfr strain of E. coli bacteria wi.pdf (19)
How would you solve this problem The correct answer is in red. Cons.pdf
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This is the second time I post this question. I got an incorrect ans.pdf
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 mat.pdf
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carbon that do not interact with water molecules.
Strength of hydrophobic bonds are relatively stronger than other weak intermolecular forces (i.e.,
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several factors like:
1. Temperature: As the temperature increases, the strength of hydrophobic interaction also
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6 raised to the 2nd power
2 raised to the 6th power
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2 raised to the 3rd power
3 raised to the 2nd power
How many tetrads align at the metaphase plate of meiosis I?
0
12
6
3
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Ans.1 No. of possible combination during independent assortment =
2n
Where, 2 indicates ploidy of the somatic cells undergoing meiosis,
n = number of haploid chromosomes.
Given, 2n = 6
Hence, n = 3.
Thus,
No. of possible combination during independent assortment = 2n = 23
Correct option: 2 raised to 3rd power.
Ans. 2. Number of tetrad during meiosis = Number of haploid chromosomes (n)
To understand it, let’s number the three haploid chromosomes as A, B and C. A diploid cell has
two copies of each chromosome- one inherited from mother and the other from father. So,
diploid cell has following chromosomes- A1, A2, B1, B2, C1 and C3, where 1= maternal
chromosomes and 2 = fraternal chromosomes. All the chromosomes remain independent of each
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The following events occur when a diploid cell enters meiosis-
A1 and A2 are called homologues pair. During S-phase, DNA is duplicated.
The resultant-
I. A1A1 remain joined to each other. Are called sister chromatids.
The same happens to A2A2, and rest other chromosomes.
II. Homologous chromosome A1 and A2 pair together prophase I. Since each of A1 and A2 are
duplicated, they are (A1A1) and (A2A2) respectively.
III. Homologous pairing forms a tetrad – A1A1/A2A2 i.e. four chromosomes remaining together.
Note: Both the maternal and fraternal chromosomes and their duplicated copies remain together
as a tetrad. That is, the number of tetrad in half of diploid number (because A1 from mother and
A2 from father remain together as one single homologous chromosome).
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Correct option: D (3).
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it would act as capacitors in parallel.
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Non C1= A/(d/2)*K1=2C/k1
Simlarly C2= 2C/K2
Since it can be treated as capacitor is parallel so Cfinal= 2C/K1 + 2c/k2 = 2C(K1+K2/K1*K2).
The incandescent lightbulb was actually invented by Humphry Davy in 1.pdf
The incandescent lightbulb was actually invented by Humphry Davy in 1802. It was not
commercializable in that form. Joseph Swan was successful in making a commercializable unit
in the late 1870s. He may have been granted the first light bulb patent (British patent 4933,
issued in 1880). Thomas Edison was also developing electric lighting at that time. He applied for
US patents as early as 1878 So why is Edison credited as the inventor? Partly because he was the
first to design and manufacture a robust lighting system having a bulb with an effective
incandescent material, higher vacuum inside the bulb, and high resistance so power distribution
from a central source was possible. Perhaps more importantly was the development ofsupply
sources and materials for distribution networks. A modem incandescent lightbulb is shown
below. Overall efficiency, defined as light power out divided by electrical power in, is about
2.5% for a 100 W unit. 1 1. outline of glass bulb 2. Low pressure inert gas (argon, nitrogen,
krypton, xenon) 3. Tun ten filament 5) 4. Contact wire (goes out of stem) 5. Contact wire (goes
into stem) 6. Support wires (one end embedded in stem; conduct no current) 7. Stem (glass
mount) 8. Contact wire (goes out of Identify five different systems that could be used for an
energy analysis (some combination Draw arrows for each systems\' energy flow process(es)-W
Qooeduction, Qconvection, radiation.
Solution
To conduction and convection of heat must have the medium.
in vaccum only radiation is there.
given problem the heat coming out from the tungsten filament(3).
the work supplied by the electrical current through the wires.
first humphry davy invented incandescent bulb having vaccum inside so there is no possibilty of
heat transfer through conduction or convection.
next edison filled with the low pressure inert gases. beacuse of that from hot Tungsten filamet to
inert gases heat transfer occurs through convection.
next inertgasses to glass or stem heated by the convection (fluid and solid interaction)
again glass to sourroundings is convection(solid and fluid interaction)
radiation is always there beacuse it possible with medium and with out medium
contact wires are heated up through the conduction..
The fossil record provides little information about ancient mosses. D.pdf
The fossil record provides little information about ancient mosses. Do you think nonvascular
plants could have ever been as large as trees or shrubs? Provide evidence based on your
knowledge of nonvascular plant anatomy.
Solution
2). No, the nonvascular plants could have never been as large as trees or shrubs because they
cannot provide the water and nutrients to the plant parts that are away from the source of water.
The nonvascular plants cannot control their water supply internally, it is solely depends upon
their external environment that provides water to the plant. So, they most likely grown over a
lake or rocks to get the water and propagated vegetative (clonal moss)..
The DNA sequence below is a shortened version of the sequence on the .pdf
The DNA sequence below is a shortened version of the sequence on the previous page. This
DNA sequence corresponds to the mRNA sequence only; two introns are not included. The
numbers indicate the position of nucleotides within the sequence and indicates intervening
nucleotides that are not shown. The gene is 626 nucleotides long. Recall that chromosomes are
very long DNA molecules with millions of nucleotides, so this sequence is preceded and
followed by seemingly endless sequences of nucleotides. IDNA is double-stranded. Write the
sequence of the first 10 nucleotides of the complementary strand of DNA.
Solution
ans. DNA is a double stranded which consist of two strands and one strand is complemetary to
the other.DNA contains complemetary base pairs in which adenine is linke d to tymine by two
hydrogen bonds guanine to cytosine by three hydrogen bonds
The seque of complementary strand of the DNA molecule is
5\' TGTAAACGAA.
the conducting tissue found in vascular plants that functions in tra.pdf
the conducting tissue found in vascular plants that functions in transportation of materials
throughout the plant, but does not lend structure support to the stem is
A. xylem
B. phloem
C.sporopollenin
D.endosperm
e. LYCOPHYLL
Solution
Answer : B. Phloem
Phloem is responsible for the transporting sugars,proteins, and other organic molecules in plants.
xylem is responsible for the transport of water and water soluble nutrients,taken up by the roots
and also primary component of wood. Vascular plants are able to higher than other plants due to
the rigidity of xylem cells, which support the plant..
Show that f(x)=57^x+1 and f^-1(x)=7x-75 are inverses Choose the ap.pdf
Show that f(x)=5/7^x+1 and f^-1(x)=7x-7/5 are inverses Choose the appropriate equality that
will show that f(x) and f^-1(x) are inverses (f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
(f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
Solution
Replace x by f^-1(x) in f(x)
=> 5/7(7x-7)/5+1
=> (7x - 7)/7 + 1 = > x - 1 + 1 => x
=> f(f^-1(x)) = x.
Read the short article Horizontal Gene Transfer in E Coli and answer.pdf
Read the short article Horizontal Gene Transfer in E Coli and answer the following
questions.(https://www.sciencedaily.com/releases/2015/05/150519105900.htm) Remember to
use your own words; plagiarism of any kind will not be tolerated (see syllabus for further
information on what constitutes plagiarism).
(1) What is horizontal gene transfer (HGT)?
(2) In the posted article, how many gene transfer elements were found for E. coli O104:H4?
(3) What is the source of these gene transfer elements?
(4) Should we be concerned about HGT in prokaryotes? If so, why?
Solution
(1) Horizontal gene transfer (HGT) is the lateral gene movement, wherein the movement of
genetic material takes place between unicellular or multicellular organisms across species. This
gene transfer is different from the vertical gene transfer in which the transmission of DNA
occurs from parent to offspring. Often genetic material swaping occurs between bacteria of same
as well as different species, through this means by aquiring one of the processes; transformation,
transduction or conjugation. During HGT, plasmids, transposons or integrons act as vectors that
transfer genes.
(2) In the given article, the team from Madurai Kamaraj University in Madurai, Tamil Nadu,
identified 38 gene transfer elements for E. coli O104:H4.
(3) The research team identified 38 gene transfer prophage elements for E.coli O104:H4. They
showed that the these elements act as genetic weapons protecting the bacteria from antibiotics.
The source of these elements were viruses (the bacteriophages), that infect bacteria, and in the
process bacteria aquired these elements from viruses.
(4) In prokaryotes, horizontal gene transfer is a major concern. Aquisition of new genes through
HGT is ability of Bacteria and Archaea to adapt to new changing environments. Druing HGT,
genes that are responsible for antibiotic resistance in one species of bacteria can be transferred to
another species of bacteria, and is the primary reason for the spread of antibiotic resistance.
Besides antibiotic resistance, HGT also gives bacteria an ability to degrade novel compounds
(such as different insecticides and pesticides). Thus the bacterial strains after HGT are much
more dangerous as they are multi drug resistant and more virulent. These transfer elements when
transferred to a pathogenic bacteia, can become a serious health hazard as they are resistant to
antibiotics available with us and thus cannot be trated easily..
Please select the best answer and click submit. The owners of fou.pdf
Please select the best answer and click \"submit.\" The owners of four companies competing for
a contract are shown in the table below. If a report is released that advocates company A. which
of the people having funded the report should result in the most skepticism?
Solution
Option d.
Debby Smith would result in the most skepticism because his company is in the second place in
the list..
One function of the spliceosome is to (Select all that apply.)A. .pdf
One function of the spliceosome is to: (Select all that apply.)
A. recognize the beginning and end of the introns.
B. recognize the 5\' end of the primary transcript.
C recognize the beginning and end of the exons.
D recognize the 3\' end of the primary transcript.
Solution
The correct answer is :A. recognize the beginning and end of the introns.
Reason: Spliceosome consists of snRNAs and snRNPs which together make a complex.This
complex binds to the beginning and end of the introns and forms a loop which is later excised
and the ends of exons are sealed..
Match the following organelles with their key functions in the cell .pdf
Match the following organelles with their key functions in the cell nucleus rough endoplasmic
reticulum Golgi apparatus lysosomes mitochondria chloroplasts peroxisomes cytosol carbon
fixation site of protein synthesis ready for cellular distribution recycling of intracellular
materials modification and packing of proteins for secreation site of protein synthesis that
contains the majority of metabolic pathways containment of toxic reactions the site of the core
genome oxidative phosphorylation
Solution
1: Nucleus: The site of core genome;g
2: Rough ER: site of protein synthesis ready for cellular distribution; b
3: Golgi apparatus: Modification and packing of proteins for secretion; d
4: Lysosome: recylcing of intracellular material; c
5: Mitochondria: Oxidative phosphorylation;h
6: Chloroplast: Carbon fixation; a
7: Peroxisomes: Containment of toxic reaction; f
8: Cytosol: Site of protein synthesis that contains majority of metabolic pathways; e.
please choose the correct answerIn the ABO blood-type phenotype, w.pdf
please choose the correct answer
In the ABO blood-type phenotype, which gene is the most epistatic?
a.) A
b.) B
c.) O
d.)I
e.) H
Solution
Answer is e) H- Epistasis is a product of gene- gene interaction. when one genes expression is
controlled by other gene, it is called epistasis. H antigen is the precursor of A and B antigens.
mo Press Submit until all questions are colored green and you have do.pdf
mo Press Submit until all questions are colored green and you have double c Question 1 Convert
the following degree measure to radians: 210 14 21 20 12 7 d) f ONone of the above.
Solution
We know that,
Radians = Degree*(pi/180)
Given Degree measures = 210o
Using the formunla
Radian Measures = 210(pi/180) = 7(pi/6) Radians
Hence (d) is the correct answer..
matching chose the stage of the light reactions which the statement.pdf
matching: chose the stage of the light reactions which the statement occurs in.
1. _Produces ATP.
2. _Passes energized electrons down an electron transport chain to create ATP.
3. _Produces NADPH.
4. _Splits water to get electrons to replace energized electrons.
5. _Replaces energized electrons from electrons that have completed the 1st electron transport
chain.
a. water-splitting photosystem.
b. NADPH-Producing photosystem.
Solution
During water splitting photosystem ,the electrons in water are removed and passed to the
chlorophyll. The hydrogen ions extracted from the water molecule, and after two water
molecules are splitted,the oxygen atoms takes electron and form molecular oxygen (O 2 ).
In photosynthesis, water splitting donates electrons to the electron transport chain in
photosystem II.
So it splits water to get electrons to replace energized electrons.
In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons
from water. The electrons travel via the chloroplast electron transport chain to photosystem I
(PSI), which reduces NADP+ to NADPH.
So it produces NADPH.
Let E be the set of all even integers, and let O be the set of all o.pdf
Let E be the set of all even integers, and let O be the set of all odd integers. Explain why Z E O.
A. Every integer is both even and odd.
B. Some integers are both even and odd.
C. Even integers and odd integers are integers.
D. Every integer is either even or odd.
E. No integer is both even and odd.
Solution
ans : D
take any integer z, z is either odd or even so, for all z , Z is a subset of EUO.
JavaIm not sure how to implement this code can someone help me .pdf
Java
I\'m not sure how to implement this code can someone help me / guide me? thank you
(Updated with PrefixMap interface included at the bottom of the code).
import java.util.List;
import java.util.Map;
public class Storage implements PrefixMap {
private Map keys;
private Map> prefixes;
public Storage() {
// TODO Initialise your maps
}
/*
* How many keys are stored in the map
*/
@Override
public int size() {
// TODO Implement this, then remove this comment
return 0;
}
@Override
public boolean isEmpty() {
// TODO Implement this, then remove this comment
return false;
}
/*
* Get the value corresponding to the key (or null if the key is not found)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String get(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
@Override
public String put(String key, String value) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String remove(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public int countKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return 0;
}
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public List getKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8
* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
@Override
public int countPrefixes() {
// TODO Implement this, then remove this comment
return 0;
}
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
@Override
public int sumKeyLengths() {
// TODO Implement this, then remove this comment
return 0;
}
}
import java.util.List;
public interface PrefixMap {
public boole.
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