Initiation of DNA replication in prokaryotes involves loading and ac.pdf
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Initiation of DNA replication in prokaryotes involves loading and activation of DnaB helicase. DnaB helicase directs other necessary enzymes (particularly DnaG Primase and DNA pol III HE) to the template and help in DNA elongation. These aspects will be discussed below: DnaB protein DnaB is the major helicase, it unwinds the template in 5\'-3\' direction during oriC replication in vitro, at a rate of 700bp/sec. The direction of unwinding requires DnaB to be placed on lagging strand, ahead of the polymerase on the leading strand. It also activates primer synthesis by DnaG primase. DnaG primase The primase (DnaG) synthesizes a unique oligonucleotide primer on SSB coated single stranded phage G4 DNA. The sequence recognised by primase is over 100 nucleotides long containing a secondary structure. Since primer synthesis requires activation by DnaB, the latter is oftern described as a \'mobile promoter\' of primer synthesis. DNA polymerase III holoenzyme DNA polymerase III HE is the large multiprotein complex responsible for elongation of DNA replication in E. coli. It has atleast 10 subunits, each with specific function. Other proteins involved in DNA elongation are DNA gyrase (topoisomerase), which removes the topological strain associated with DNA unwinding SSB (single stranded DNA binding) protein, that binds tightly to single stranded DNA thus stabilizing the unwound DNA. RNase H removes RNA primers DNA Polymerase I is used for filling the gap created due to the removal of RNA primers DNA ligase converts the okasaki fragments into a continuous strand. Solution Initiation of DNA replication in prokaryotes involves loading and activation of DnaB helicase. DnaB helicase directs other necessary enzymes (particularly DnaG Primase and DNA pol III HE) to the template and help in DNA elongation. These aspects will be discussed below: DnaB protein DnaB is the major helicase, it unwinds the template in 5\'-3\' direction during oriC replication in vitro, at a rate of 700bp/sec. The direction of unwinding requires DnaB to be placed on lagging strand, ahead of the polymerase on the leading strand. It also activates primer synthesis by DnaG primase. DnaG primase The primase (DnaG) synthesizes a unique oligonucleotide primer on SSB coated single stranded phage G4 DNA. The sequence recognised by primase is over 100 nucleotides long containing a secondary structure. Since primer synthesis requires activation by DnaB, the latter is oftern described as a \'mobile promoter\' of primer synthesis. DNA polymerase III holoenzyme DNA polymerase III HE is the large multiprotein complex responsible for elongation of DNA replication in E. coli. It has atleast 10 subunits, each with specific function. Other proteins involved in DNA elongation are DNA gyrase (topoisomerase), which removes the topological strain associated with DNA unwinding SSB (single stranded DNA binding) protein, that binds tightly to single stranded DNA thus stabilizing the unwound DNA. RNase H remove.
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DNA replication in prokaryotes
This document provides an outline and summary of a presentation on DNA replication in prokaryotes. The presentation was assigned by Mam Fatima and involved several participants who discussed:
1) DNA replication in prokaryotes begins at the origin of replication site (oriC) where DnaA protein binds and unwinds the DNA. Bidirectional replication forks are formed.
2) Enzymes such as helicase, primase, DNA polymerase III and ligase are involved in semi-conservative DNA replication. The leading strand replicates continuously while the lagging strand replicates discontinuously in Okazaki fragments.
3) Elongation occurs as DNA polymerases add nucleotides to the growing DNA strands in
dna_replication.pptx
The document discusses DNA replication in prokaryotes and eukaryotes. It explains that replication involves initiation at an origin of replication, followed by unwinding of the DNA double helix by helicase. RNA primers are synthesized by primase and DNA polymerase adds nucleotides to the primers to elongate DNA strands. In prokaryotes, leading and lagging strands are synthesized continuously and discontinuously respectively to form Okazaki fragments. Enzymes like DNA polymerase, ligase, and topoisomerase ensure high fidelity and processivity of replication. Telomerase maintains telomere integrity in eukaryotes during DNA replication.
Replication
DNA replication is semi-conservative, as proved by the Meselson-Stahl experiment. It occurs during interphase and involves unwinding of the DNA double helix by helicase, followed by synthesis of new strands by DNA polymerases with RNA primers provided by primase. Replication proceeds bidirectionally from an origin of replication and results in two identical copies of the original DNA molecule.
DNA Replication in Prokaryotes and Eukaryotes .pptx
DNA replication is the process where a cell makes an exact copy of its DNA before cell division. It occurs during the S phase of the cell cycle and involves unwinding the DNA double helix, making RNA primers, and synthesizing new DNA strands in both directions from the origin of replication using the old strands as templates. This results in two identical copies of the DNA. The key enzymes involved are DNA helicase, DNA polymerase, DNA ligase, topoisomerase, and single-stranded binding proteins. Replication proceeds through the semi-conservative model and results in the production of leading and lagging strands made of long and short Okazaki fragments respectively.
Enzymes and proteins in dna replication
Multiple proteins are required for DNA replication, including DNA polymerases, helicases, primase, topoisomerases, ligase and single-stranded DNA binding proteins. Helicases unwind DNA at replication forks using ATP while primase synthesizes RNA primers and DNA polymerases use the primers to replicate DNA in the 5' to 3' direction. DNA gyrase introduces negative supercoils to relieve positive supercoiling formed during unwinding while ligase seals nicks between Okazaki fragments to complete replication.
Prokaryotic DNA replication
DNA replication in prokaryotes begins with the unwinding of DNA at the origin of replication by enzymes like DnaA and DnaB helicase. This produces two replication forks that move in opposite directions. The leading strand is replicated continuously while the lagging strand is replicated discontinuously in short segments called Okazaki fragments. DNA polymerase III is the main enzyme that synthesizes new DNA. Replication terminates at the terminus region when the DnaB helicase is stopped by protein Tus bound to Ter sequences.
Dna replication lgis
DNA replication involves the synthesis of daughter DNA strands using parental DNA as a template. It occurs through a semi-conservative process whereby each new DNA molecule contains one original and one newly synthesized strand. Replication is bidirectional, with replication forks moving in opposite directions from an origin of replication. The leading strand is synthesized continuously while the lagging strand involves the discontinuous synthesis of Okazaki fragments. DNA polymerases and other proteins such as helicases, primases, ligases and topoisomerases work together to facilitate the replication process.
DNA Replication
DNA replication occurs through a semi-conservative process in both prokaryotes and eukaryotes. In prokaryotes, replication is bidirectional from a single origin and the replication forks meet, while in eukaryotes there are multiple origins of replication activated sequentially. Key enzymes involved include DNA polymerase, helicase, primase and ligase. Replication ensures the accurate transfer of genetic information between generations.
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DNA replication in prokaryotes
This document provides an outline and summary of a presentation on DNA replication in prokaryotes. The presentation was assigned by Mam Fatima and involved several participants who discussed:
1) DNA replication in prokaryotes begins at the origin of replication site (oriC) where DnaA protein binds and unwinds the DNA. Bidirectional replication forks are formed.
2) Enzymes such as helicase, primase, DNA polymerase III and ligase are involved in semi-conservative DNA replication. The leading strand replicates continuously while the lagging strand replicates discontinuously in Okazaki fragments.
3) Elongation occurs as DNA polymerases add nucleotides to the growing DNA strands in
dna_replication.pptx
The document discusses DNA replication in prokaryotes and eukaryotes. It explains that replication involves initiation at an origin of replication, followed by unwinding of the DNA double helix by helicase. RNA primers are synthesized by primase and DNA polymerase adds nucleotides to the primers to elongate DNA strands. In prokaryotes, leading and lagging strands are synthesized continuously and discontinuously respectively to form Okazaki fragments. Enzymes like DNA polymerase, ligase, and topoisomerase ensure high fidelity and processivity of replication. Telomerase maintains telomere integrity in eukaryotes during DNA replication.
Replication
DNA replication is semi-conservative, as proved by the Meselson-Stahl experiment. It occurs during interphase and involves unwinding of the DNA double helix by helicase, followed by synthesis of new strands by DNA polymerases with RNA primers provided by primase. Replication proceeds bidirectionally from an origin of replication and results in two identical copies of the original DNA molecule.
DNA Replication in Prokaryotes and Eukaryotes .pptx
DNA replication is the process where a cell makes an exact copy of its DNA before cell division. It occurs during the S phase of the cell cycle and involves unwinding the DNA double helix, making RNA primers, and synthesizing new DNA strands in both directions from the origin of replication using the old strands as templates. This results in two identical copies of the DNA. The key enzymes involved are DNA helicase, DNA polymerase, DNA ligase, topoisomerase, and single-stranded binding proteins. Replication proceeds through the semi-conservative model and results in the production of leading and lagging strands made of long and short Okazaki fragments respectively.
Enzymes and proteins in dna replication
Multiple proteins are required for DNA replication, including DNA polymerases, helicases, primase, topoisomerases, ligase and single-stranded DNA binding proteins. Helicases unwind DNA at replication forks using ATP while primase synthesizes RNA primers and DNA polymerases use the primers to replicate DNA in the 5' to 3' direction. DNA gyrase introduces negative supercoils to relieve positive supercoiling formed during unwinding while ligase seals nicks between Okazaki fragments to complete replication.
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DNA replication in prokaryotes begins with the unwinding of DNA at the origin of replication by enzymes like DnaA and DnaB helicase. This produces two replication forks that move in opposite directions. The leading strand is replicated continuously while the lagging strand is replicated discontinuously in short segments called Okazaki fragments. DNA polymerase III is the main enzyme that synthesizes new DNA. Replication terminates at the terminus region when the DnaB helicase is stopped by protein Tus bound to Ter sequences.
Dna replication lgis
DNA replication involves the synthesis of daughter DNA strands using parental DNA as a template. It occurs through a semi-conservative process whereby each new DNA molecule contains one original and one newly synthesized strand. Replication is bidirectional, with replication forks moving in opposite directions from an origin of replication. The leading strand is synthesized continuously while the lagging strand involves the discontinuous synthesis of Okazaki fragments. DNA polymerases and other proteins such as helicases, primases, ligases and topoisomerases work together to facilitate the replication process.
DNA Replication
DNA replication occurs through a semi-conservative process in both prokaryotes and eukaryotes. In prokaryotes, replication is bidirectional from a single origin and the replication forks meet, while in eukaryotes there are multiple origins of replication activated sequentially. Key enzymes involved include DNA polymerase, helicase, primase and ligase. Replication ensures the accurate transfer of genetic information between generations.
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Dna replication in prokaryotes
DNA replication in prokaryotes occurs through a semi-conservative process whereby the parental DNA strands serve as templates for the production of new daughter strands. It begins with the unwinding of DNA at the origin of replication by helicase to form a replication fork. Primase then synthesizes an RNA primer to initiate leading and lagging strand DNA synthesis by DNA polymerase. While the leading strand is synthesized continuously, the lagging strand involves the formation of Okazaki fragments that are later joined by ligase. Polymerases also proofread and correct any errors to maintain high-fidelity replication. Replication terminates when the replication forks from opposite directions meet.
Replication
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Define Replication, its criteria, Requirements for replication, related terminology, Steps and importance of replication
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DNA replication in prokaryotes involves initiation, elongation, and termination phases. Initiation begins with the binding of initiator proteins to the origin of replication, unwinding the DNA helix to form replication forks. Elongation synthesizes the leading and lagging strands bidirectionally away from the origin using DNA polymerases. Termination occurs when the replication forks meet, completing duplication of the chromosome.
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A reaction in which daughter DNAs are synthesized using the parental DNAs as the template.
Transferring the genetic information to the descendant generation with a high fidelity
Semi-conservative replication
Bidirectional replication
Semi-continuous replication
High fidelity
Replication starts from unwinding the dsDNA at a particular point (called origin), followed by the synthesis on each strand.
The parental dsDNA and two newly formed dsDNA form a Y-shape structure called replication fork.
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DNA replication occurs through a semi-conservative process in both prokaryotes and eukaryotes. In prokaryotes:
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2. DNA polymerase III elongates the leading and lagging strands bidirectionally from the replication fork.
3. Replication terminates when the replication forks from opposite directions meet.
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Dna replication MBBS
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- The key enzymes involved are DNA helicase, topoisomerases, primase, DNA polymerase, and DNA ligase. DNA polymerase synthesizes new strands in the 5' to 3' direction using existing strands as templates.
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Prokaryotic replication
Prokaryotic replicationMarudhar Kesari Jain College for Women Vaniyambadi - 635 751, Tamil Nadu, INDIA.
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DNA replication.pptx
DNA replication in prokaryotes occurs through three main steps: initiation, elongation, and termination. Initiation begins at the origin of replication when DnaA and other proteins help separate the DNA strands. Elongation then uses DNA polymerases and other enzymes to bidirectionally copy the DNA in a semiconservative manner, producing both a leading and lagging strand. Termination occurs when the replication forks meet at the terminus region, utilizing proteins like Tus to stop replication.
Answer Every investor expects dividend from his investments.Dividen.pdf
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situation dividend is subject to taxes (either distribution tax or income tax) in the hands
of investors. So, in this case investors are not interested in dividend but they are more interested
in capital gain(which may be tax exempt). So, In this situation retained earnings fulfill investor\'s
objective.
Cons:
As explained earlier all the investors have some expectation of dividend, in case of higher
retention of earnings it may become difficult to fulfill investor\'s expectations.
Retention of earnings may affect investor\'s earning capacity when investors have available
opportunities, which gives higher return than organization earns by retaining amount. In this case
investor looses earning opportunities.
In case of dividend management\'s decision is significantly influenced by media and institutional
investors. Media is responsible for organization\'s image in society, such image significantly
affects organization\'s financial performance. This is why management always considers media\'s
point of view before making any dividend decision.
Institutional investors are usually invests their money and takes part in decision making process
of organization. Institutional investor\'s weight in decision making process is generally very
high, so it is obvious that they influence management\'s dividend decision also.
Solution
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situ.
Ans. Gene is defined as the segment of DNA that gives a functional p.pdf
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs).
Solution
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs)..
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Similar to Initiation of DNA replication in prokaryotes involves loading and ac.pdf
Replication (reviewed, 2018)
Replication is the fundamental process by which a cell copies its DNA in order to transfer genetic information to daughter cells. It involves DNA-directed DNA synthesis where the base sequence of the newly synthesized daughter DNA is identical to the parent DNA template. Replication is semi-conservative and proceeds bidirectionally from an origin of replication. It requires various enzymes and is a highly regulated process involving initiation, elongation, and termination steps to accurately duplicate the entire genome. Precise replication is essential for preventing genetic abnormalities and maintaining inheritance of traits from parent to daughter cells.
Dna replication in prokaryotes
DNA replication in prokaryotes occurs through a semi-conservative process whereby the parental DNA strands serve as templates for the production of new daughter strands. It begins with the unwinding of DNA at the origin of replication by helicase to form a replication fork. Primase then synthesizes an RNA primer to initiate leading and lagging strand DNA synthesis by DNA polymerase. While the leading strand is synthesized continuously, the lagging strand involves the formation of Okazaki fragments that are later joined by ligase. Polymerases also proofread and correct any errors to maintain high-fidelity replication. Replication terminates when the replication forks from opposite directions meet.
Replication
Updated :
Define Replication, its criteria, Requirements for replication, related terminology, Steps and importance of replication
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DNA replication in prokaryotes involves initiation, elongation, and termination phases. Initiation begins with the binding of initiator proteins to the origin of replication, unwinding the DNA helix to form replication forks. Elongation synthesizes the leading and lagging strands bidirectionally away from the origin using DNA polymerases. Termination occurs when the replication forks meet, completing duplication of the chromosome.
DNA Replication- General Concepts of DNA Replication.ppt
A reaction in which daughter DNAs are synthesized using the parental DNAs as the template.
Transferring the genetic information to the descendant generation with a high fidelity
Semi-conservative replication
Bidirectional replication
Semi-continuous replication
High fidelity
Replication starts from unwinding the dsDNA at a particular point (called origin), followed by the synthesis on each strand.
The parental dsDNA and two newly formed dsDNA form a Y-shape structure called replication fork.
Dna replication in prokaryotes
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Proposed models of replication,
Semi-conservative method by meselson and sthal
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DNA replication is the process by which DNA duplicates itself for cell division. It involves unwinding of the DNA double helix, formation of a replication fork, and synthesis of new complementary strands. Each original DNA strand acts as a template for new strand synthesis by DNA polymerase in the 5' to 3' direction. RNA primers are required for initiation, and Okazaki fragments are produced on the lagging strand. Replication proceeds bidirectionally from an origin of replication site and involves multiple enzymes until the replication fork meets, at which point termination occurs. Differences between prokaryotic and eukaryotic replication include RNA primer length, number of DNA polymerases, and presence of histones and telomeres in eukaryotes
dnareplication-151218084840 (1).pdf
DNA replication occurs through a semi-conservative process in both prokaryotes and eukaryotes. In prokaryotes:
1. Replication initiates at the origin of replication with unwinding by helicase and primer synthesis by primase.
2. DNA polymerase III elongates the leading and lagging strands bidirectionally from the replication fork.
3. Replication terminates when the replication forks from opposite directions meet.
Enzymes of DNA replication
Enzymes of DNA replicationMarudhar Kesari Jain College for Women Vaniyambadi - 635 751, Tamil Nadu, INDIA.
DNA replication is the process where one original DNA molecule is copied to produce two identical DNA molecules. It occurs through semiconservative replication where each strand of the original DNA serves as a template for the production of the complementary strand. The replication fork forms during DNA replication through the unwinding of the DNA double helix by helicases. It has two branching prongs that serve as templates for the leading and lagging strands which are synthesized by DNA polymerase.Dna replication in eukaryotes
DNA replication in eukaryotes occurs in the nucleus and proceeds in the 5' to 3' direction. It involves several enzymes including helicase, which opens the DNA helix, and DNA polymerase epsilon and delta, which synthesize the leading and lagging strands respectively. Replication occurs through initiation, elongation, and termination steps. At the origin of replication, a pre-replication complex forms and the replication fork bubbles outwards. RNA primers are laid down and DNA synthesis occurs continuously on the leading strand but discontinuously on the lagging strand. The primers are later removed and the gaps filled in.
Dna replication MBBS
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- The key enzymes involved are DNA helicase, topoisomerases, primase, DNA polymerase, and DNA ligase. DNA polymerase synthesizes new strands in the 5' to 3' direction using existing strands as templates.
- In eukaryotes, replication of the lagging strand occurs through discontinuous Okazaki fragments that are later ligated together. DNA polymerase also has a proofreading function.
- DNA damage can occur from chemicals, radiation, and replication errors but cells have multiple repair mechanisms like base excision repair, nucleotide excision repair, and
Prokaryotic replication
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Replication in prokaryotes occurs through three main stages - initiation, elongation, and termination. Initiation begins at a specific origin of replication site called OriC, where DnaA protein binds and unwinds the DNA, forming an open complex. Elongation then proceeds bidirectionally as DNA polymerase synthesizes leading and lagging strands. Termination occurs when the replication forks from opposite directions meet. The entire replication process in E. coli takes around 40 minutes to complete replication of its genome.Replication
DNA replication is a highly regulated process that exactly duplicates the genome during cell division. It involves unwinding of the DNA double helix at the origin of replication by helicase. Each single-stranded template is then used to synthesize a new complementary strand in the 5'-to-3' direction by DNA polymerase. The leading strand is replicated continuously while the lagging strand is replicated discontinuously in short segments called Okazaki fragments that are later joined by ligase. Several proteins and enzymes work together in a coordinated manner to ensure the genome is accurately duplicated and inherited by daughter cells.
Replication of DNA
DNA replication is the process where DNA copies itself to produce identical daughter molecules. It occurs during the S phase of the cell cycle and involves unwinding of the DNA double helix, synthesis of new strands complementarity to each existing strand, and production of two identical copies of the original DNA molecule. DNA polymerase synthesizes new DNA strands in the 5' to 3' direction by adding nucleotides that are complementary to the template strand, requiring primers, nucleotides, and several enzymes. Errors can occur which are corrected by DNA repair mechanisms like base excision repair.
Rolling circle mechanism ppt
Rolling circle replication is a process that can rapidly synthesize multiple copies of circular DNA or RNA molecules. It involves the unidirectional replication of circular nucleic acids. The process begins with an initiator protein nicking one strand of the circular DNA. DNA polymerase then uses the 3' end of the nicked strand to initiate replication, displacing the 5' end. Replication continues around the circle to produce a long concatemer of copies. The concatemer is then cleaved and ligated to form multiple double-stranded circular DNA molecules. Rolling circle replication is used by some viruses and plasmids to replicate their genomes and can be harnessed for applications like signal amplification in biosensing.
Prof._Moushira_DNA replication.pdf
This document summarizes DNA replication. It begins by listing the objectives of describing the process. DNA replication duplicates the entire genome before cell division so each daughter cell receives a complete copy. It proceeds in a semi-conservative, bidirectional manner starting from the origin. Several enzymes are involved including helicase, DNA polymerase, primase, ligase and topoisomerases. DNA polymerase builds new strands using the old strands as templates while primase lays down RNA primers. The leading strand is synthesized continuously but the lagging strand requires discontinuous Okazaki fragments joined by ligase. Proofreading ensures high fidelity through exonuclease activity. Eukaryotes differ in having multiple origins and using RNAse H to remove primers.
DNA polymerases (DNA manupliation Enzymes).pdf
the Secrets of DNA Manipulation: A Comprehensive Exploration of DNA Polymerase and Enzymes
In this PDF presentation entitled "Enzymes that Manipulate DNA, Specially DNA Polymerase," we delve deep into the mechanisms and functions of these remarkable enzymes that play a pivotal role in the realm of molecular biology.
🧬 Key Highlights:
Introduction to DNA Polymerase:
Uncover the fundamental aspects of DNA polymerase, a key player in DNA replication and repair. Explore its structure, functions, and the indispensable role it plays in maintaining the genetic integrity of living organisms.
Types of DNA Polymerases:
Delve into the diverse landscape of DNA polymerases, ranging from prokaryotic to eukaryotic systems. Understand how different types of DNA polymerases contribute to the precision and efficiency of DNA synthesis.
Examples of polymerases:
•DNA polymerase 1
•klenow fragment
•sequenase
•Taq polymerase
•Reverse Transcriptase
DNA Replication
Take a closer look at the intricate dance of enzymes during DNA replication. Follow the step-by-step process, and gain insights into how DNA polymerase ensures the accurate transmission of genetic information from one generation to the next.
Technological Applications:
Unleash the potential of DNA polymerase in various biotechnological applications. From PCR (Polymerase Chain Reaction) to DNA sequencing, discover how these enzymes have revolutionized molecular biology and genetic research.
Emerging Trends and Future Prospects:
Stay ahead of the curve by exploring the latest advancements and emerging trends in DNA manipulation. Witness the ongoing research that promises to unlock new possibilities in the field.
🎓 Who Should Explore This Presentation?
Students and researchers in molecular biology and genetics
Biotechnologists and professionals in the field of genetic engineering
Enthusiasts curious about the molecular machinery behind DNA manipulation
DNA REPLICATION DAMAGE AND REPAIR
This document provides lecture notes on DNA replication, damage, and repair. It discusses the process of DNA replication including the three main steps of initiation, elongation, and termination. Details are given on replication in prokaryotes and eukaryotes. The document also covers types of DNA damage including physical, chemical, and biological mutagens. Finally, various DNA repair mechanisms are summarized such as photoreactivation, excision repair, and SOS repair.
Replication In Eukaryotes and Prokaryotes
Replication Introduction , DNA replicating Models , Meselson and Stahl Experiments , Circuler Model of DNA replication , Replication in Prokaryotes , Replication In Eukaryotes , Comparison Between Prokaryotes and Eukaryotes Replicaton and PCR (Polymerease Chain Reaction)
DNA replication.pptx
DNA replication in prokaryotes occurs through three main steps: initiation, elongation, and termination. Initiation begins at the origin of replication when DnaA and other proteins help separate the DNA strands. Elongation then uses DNA polymerases and other enzymes to bidirectionally copy the DNA in a semiconservative manner, producing both a leading and lagging strand. Termination occurs when the replication forks meet at the terminus region, utilizing proteins like Tus to stop replication.
Similar to Initiation of DNA replication in prokaryotes involves loading and ac.pdf (20)
DNA Replication- General Concepts of DNA Replication.ppt
DNA Replication- General Concepts of DNA Replication.ppt
More from aquazac
Answer Every investor expects dividend from his investments.Dividen.pdf
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situation dividend is subject to taxes (either distribution tax or income tax) in the hands
of investors. So, in this case investors are not interested in dividend but they are more interested
in capital gain(which may be tax exempt). So, In this situation retained earnings fulfill investor\'s
objective.
Cons:
As explained earlier all the investors have some expectation of dividend, in case of higher
retention of earnings it may become difficult to fulfill investor\'s expectations.
Retention of earnings may affect investor\'s earning capacity when investors have available
opportunities, which gives higher return than organization earns by retaining amount. In this case
investor looses earning opportunities.
In case of dividend management\'s decision is significantly influenced by media and institutional
investors. Media is responsible for organization\'s image in society, such image significantly
affects organization\'s financial performance. This is why management always considers media\'s
point of view before making any dividend decision.
Institutional investors are usually invests their money and takes part in decision making process
of organization. Institutional investor\'s weight in decision making process is generally very
high, so it is obvious that they influence management\'s dividend decision also.
Solution
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situ.
Ans. Gene is defined as the segment of DNA that gives a functional p.pdf
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs).
Solution
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs)..
additional optmization techniques for underlying IP network must1.pdf
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch.
Solution
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch..
According to the given equation, aqueous carbon dioxide reacts with .pdf
According to the given equation, aqueous carbon dioxide reacts with water to form carbonic
acid, Carbonic acid is not stable in normal body temperatures, therefore, it ionizes into
bicarbonate ion (HCO3-) and a hydrogen ion (H+) is liberated. The liberation of hydrogen ions
decreases the pH.
During breathing, oxygen is inhaled which is carried by the blood to the different tissues of the
body. Oxygen facilitates the various metabolic reactions occurring inside our cells and helps in
respiration, or the generation of energy. Carbon dioxide, formed as a by-product of these energy-
generating metabolic reactions, diffuses out of the tissues and into the blood stream, which
carries CO2 to the lungs, where it is exhaled out.
a) In case of hypoventilation :
The exchange of gases that occurs in the lungs while breathing is impaired. Therefore, the
requisite amount of carbon dioxide is not being exhaled out. This leads to a build-up of carbon-
dioxide in the blood plasma. As a result, the forward reaction is facilitated and more H+ ions are
generated. The higher concentration of H+ ions decreases the pH (acidic pH). This can lead to
acidosis.
The build-up of carbon dioxide in the blood facilitates the forward reaction and therefore the
equilibrium shifts to the right.
The hydronium ion or H3O+ is the product of a hydrogen ion (H+) and water. Since a hydrogen
ion is too small to exist freely in water, it therefore combines with a water molecule to form a
hydronium ion which is more stable. In case of hypoventilation, the concentration of H+ ions
increase. Therefore, the concentration of hydronium ions increase during hypoventilation.
The stress concentration or partial pressure of CO2 during hypoventilation is significantly higher
than 40mmHg (around 60-65 mmHg).
b) In case of hyperventilation :
The breathing rate is more than normal. Therefore, as a result of heavier and faster breathing, the
CO2 concentration in the blood rapidly decreases, that is, CO2 easily leaves the blood plasma.
Decrease in the CO2 concentration slows down the rate of the forward reaction, reducing the
levels at which bicarbonate ion and H+ ion concentration were being generated. The bicarbonate
and hydrogen ions recombine to yield CO2 and water. This lowering in hydrogen ion
concentration consequently lowers the pH (alkaline pH). This can lead to alkalosis.
As a result of the lowering of CO2 , the reverse reaction takes place. The carbonic acid changes
back to CO2 and H2O, shifting the equilibrium to the left.
The concentration of H3O+ decreases during hyperventilation, since the H+ ion concentration is
lowered.
The stress levels or partial pressure of CO2 during hyperventilation is significantly lower than 40
mmHg (around 25-30 mmHg).
c) If sodium carbonate or Na2CO3 is added instead of sodium bicarbonate, then sodium
carbonate would react with water and it would hydrolyse into Na+ and CO32- ions. The CO32-
combine with the hydrogen (H) present in water to form H2CO3 (carbon.
a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
a) mean = 1.4*3
Thus distribution is Poisson(4.2)
P(X = 4) = 4.2^4 * e^{-4.2} / 4! = 0.1944
b)
average number of ghosts in 90min (1.5hrs) interval = 1.4*1.5 = 2.1
c)
Probability that 4th ghost will appear before 10 am
= Probability that number of ghosts appearing from 8 to 10 am >= 4
As it is a 2 hr period, the distribution of number of ghosts is Poisson(1.4*2) i.e Poisson(2.8)
P(X >= 4)
= 1 - P(X < 4)
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
= 1 - [(2.8^0 * e^{-2.8} / 0!) + (2.8^1 * e^{-2.8} / 1!) + (2.8^2 * e^{-2.8} / 2!) + (2.8^3 * e^{-
2.8} / 3!)]
= 1 - [0.0608 + 0.1703 + 0.2384 + 0.2225]
= 0.692
d)
When arrival distribution is Poisson(lambda), inter-arrival time distribution is
Exponential(lambda)
Thus, interarrival time distribution is Exponential(1.4)
Expected value of time between 5th and 6th ghost = 1/1.4 = 0.7143hr
e)
On average time taken for 10th ghost to arrive = 10*(1/1.4) = 7.143hr
Thus on an average, 10th ghost will arrive at 3:09pm
f)
Expected value of time between ghost appearances = 1/1.4 = 0.7143hr (see part d)
g)
Probability that 3rd ghost appears after 9:45 am
= Probablity that number of ghosts appearing till 9:45 am < 3
Duration from 8am to 9:45 am = 1.75hr
Mean number of ghosts = 1.4*1.75 = 2.45
Thus X is Poisson(2.45)
P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2)
= (2.45^0 * e^{-2.45} / 0!) + (2.45^1 * e^{-2.45} / 1!) + (2.45^2 * e^{-2.45} / 2!)
= 0.0863 + 0.2114 + 0.2590
= 0.5567
h)
As inter-arrival time is Exponential(1.4), inter-arrival time is memoryless
Thus, time for 7th ghost will be calculated from 1:00 pm and not from 12:35pm due to
memorylessness property.
Time duration from 1:00pm to 1:15pm = 0.25hrs
Let Y be random variable denoting inter-arrival time
probability that the 7th ghost will appear before 1:15 p.m.
= P(Y < 0.25)
= 1 - e^{-1.4 * 0.25)
= 0.2953
i)
Expected value of time for 7th ghost to appear after 1:00pm = 1/1.4 = 0.7143hr = 42.86 min
Time from 12:35pm to 1:00pm = 25min
Thus expected interarrival time = 25 + 42.86 = 67.86 min = 1hr 7.86min
j)
Expected duration after 1:00pm for 9th ghost to appear = 3 * (1/1.4) = 2.143hr = 2hr 8.58min
Thus expected time at which 9th ghost appears = 3:09pm
k)
As the inter-arrival time follows memorylessness property, the future probabilities do not change
with the observed value at past.
Thus, 4 ghosts appearing between 7:00pm to 10:00pm does not affect ghost appearance from
11:00pm to 11:30pm
Duration from 11:00pm to 11:30pm = 0.5hrs
mean number of ghosts = 1.4*0.5 = 0.7
Thus it follows Poisson(0.7) distribution
P(X > 2)
= 1 - P(X <= 2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 - [(0.7^0 * e^{-0.7} / 0!) + (0.7^1 * e^{-0.7} / 1!) + (0.7^2 * e^{-0.7} / 2!)]
= 1 - [0.4966 + 0.3476 + 0.1217]
= 0.0341
l)
Probability that 8 ghosts appear between 6pm and midnight
= Probability that number of ghosts appear from 6pm to 7pm and 10pm to midnight = 4 as it is
known that 4 ghosts appeared from 7pm to 10pm
Let A denote number of ghost from 6pm to 7pm
B denote number of ghosts .
2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf
2.
a. Wired Media Type and Explination:
Twisted-Pair Cable
Twisted-pair cable is a type of cabling that is used for telephone communications and most
modern Ethernet networks. A pair of wires forms a circuit that can transmit data. The pairs are
twisted to provide protection against crosstalk, the noise generated by adjacent pairs. When
electrical current flows through a wire, it creates a small, circular magnetic field around the wire.
When two wires in an electrical circuit are placed close together, their magnetic fields are the
exact opposite of each other. Thus, the two magnetic fields cancel each other out. They also
cancel out any outside magnetic fields. Twisting the wires can enhance this cancellation effect.
Using cancellation together with twisting the wires, cable designers can effectively provide self-
shielding for wire pairs within the network media.
Two basic types of twisted-pair cable exist: unshielded twisted pair (UTP) and shielded twisted
pair (STP). The following sections discuss UTP and STP cable in more detail.
UTP Cable
UTP cable is a medium that is composed of pairs of wires . UTP cable is used in a variety of
networks. Each of the eight individual copper wires in UTP cable \\is covered by an insulating
material. In addition, the wires in each pair are twisted around each other.
UTP cable relies solely on the cancellation effect produced by the twisted wire pairs to limit
signal degradation caused by electromagnetic interference (EMI) and radio frequency
interference (RFI). To further reduce crosstalk between the pairs in UTP cable, the number of
twists in the wire pairs varies. UTP cable must follow precise specifications governing how
many twists or braids are permitted per meter (3.28 feet) of cable.
UTP cable often is installed using a Registered Jack 45 (RJ-45) connector The RJ-45 is an eight-
wire connector used commonly to connect computers onto a local-area network (LAN),
especially Ethernets.
When used as a networking medium, UTP cable has four pairs of either 22- or 24-gauge copper
wire. UTP used as a networking medium has an impedance of 100 ohms; this differentiates it
from other types of twisted-pair wiring such as that used for telephone wiring, which has
impedance of 600 ohms.
UTP cable offers many advantages. Because UTP has an external diameter of approximately
0.43 cm (0.17 inches), its small size can be advantageous during installation. Because it has such
a small external diameter, UTP does not fill up wiring ducts as rapidly as other types of cable.
This can be an extremely important factor to consider, particularly when installing a network in
an older building. UTP cable is easy to install and is less expensive than other types of
networking media. In fact, UTP costs less per meter than any other type of LAN cabling. And
because UTP can be used with most of the major networking architectures, it continues to grow
in popularity.
Disadvantages also are involved in using twisted-pair cabling, howe.
A person may not choose to participate in the labour force due to La.pdf
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all.
Solution
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all..
Well.. 1) Ionic bonds are almost always metal to .pdf
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!
Solution
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!.
1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance.
Solution
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance..
clear clc close all Use polyfit to solve for the phase l.pdf
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point12=[0,300];
line1=polyfit(point11,point12,1);
m1=line1(1);
b1=line1(2);
xpf1=linspace(min(point11),max(point11),100);
ypf1=m1*xpf1+b1;
point21=[0,15];
point22=[700,300];
line2=polyfit(point21,point22,1);
m2=line2(1);
b2=line2(2);
xpf2=linspace(min(point21),max(point21),100);
ypf2=m2*xpf2+b2;
point31=[0,50];
point32=[700,300];
line3=polyfit(point31,point32,1);
m3=line3(1);
b3=line3(2);
xpf3=linspace(min(point31),max(point31),100);
ypf3=m3*xpf3+b3;
point41=[50,100];
point42=[300,800];
line4=polyfit(point41,point42,1);
m4=line4(1);
b4=line4(2);
xpf4=linspace(min(point41),max(point41),100);
ypf4=m4*xpf4+b4;
point51=[100,85];
point52=[800,300];
line5=polyfit(point51,point52,1);
m5=line5(1);
b5=line5(2);
xpf5=linspace(min(point51),max(point51),100);
ypf5=m5*xpf5+b5;
point61=[85,100];
point62=[300,0];
line6=polyfit(point61,point62,1);
m6=line6(1);
b6=line6(2);
xpf6=linspace(min(point61),max(point61),100);
ypf6=m6*xpf6+b6;
point71=[15,85];
point72=[300,300];
%Allow the user to set the percent of B
b=input(\'Enter the mass percent of B: \');
%Return an error if the percent of B is not a valid number
if b<0 || b>100
error(\'Please enter a value between 0 and 100.\')
end
%Allow the user to set the temperature
t=input(\'Enter the temperature [deg C]: \');
%Solve for percent composition of A
a=100-b;
%Determine phase based on temperature and percent composition of B using
%the phase lines
if t<300
if b<15 && t>m1*b+b1
phase=\'alpha\';
elseif b<15 && t=15 && b<=85
phase=\'alpha + beta\';
elseif b>85 && t>m6*b+b6
phase=\'beta\';
else phase=\'alpha + beta\';
end
elseif b<15
if t=15 && b<=50
if t50 && b<=85
if t=15 && b<=85 && t==300
fprintf(\'The provided conditions fall on the eutectic line.\ \')
end
if b==50 && t==300
fprintf(\'The provided conditions fall on the eutectic point.\ \')
end
%Output a formatted statement to display the phase
fprintf(\'For the composition of %0.2f%% A, %0.2f%% B and a temperature of %0.0f degrees
Celsius, the phase is %s.\ \',a,b,t,phase)
%Create a plot of the phase lines and the user\'s set point
figure(\'color\',\'white\')
%Set axes
axis([0 100 0 1000])
%Plot the phase lines, eutectic line, eutectic point, and the user\'s point
plot(xpf1,ypf1,\'-b\')
hold on
plot(xpf2,ypf2,\'-b\')
plot(xpf3,ypf3,\'-b\')
plot(xpf4,ypf4,\'-b\')
plot(xpf5,ypf5,\'-b\')
plot(xpf6,ypf6,\'-b\')
plot(point71,point72,\'-k\')
plot(b,t,\'ro\',\'MarkerFaceColor\',\'r\')
%Assign a title and axes labels
title(\'Phase Diagram of Elements A and B\')
xlabel(\'Percent Composition of B\')
ylabel(\'Temperature (T) [deg C]\')
%Set scaling for the axes
set(gca,\'XTick\',0:10:100,\'YTick\',0:100:1000)
%Place the phase names on the graph
text(40,200,\'alpha+beta\')
text(45,600,\'Liquid\')
text(90,300,\'beta\')
text(3,300,\'alpha\')
text(15,375,\'alpha + Liquid\')
text(62,375,\'beta + Liquid\')
Solution
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point.
The oxygen appears in both step reactions. But, i.pdf
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O)
Solution
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O).
PART A The element Si belongs to IVA group. Therefore, four electro.pdf
PART A: The element Si belongs to IVA group. Therefore, four electrons present in the
outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The
s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the
number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs
to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D:
All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons.
PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present
after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is
[Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons
present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The
element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell.
Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital
and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal.
PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital.
Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s-
electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal.
PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration
of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl.
Therefore, the number of unpaired electrons is 1
Solution
PART A: The element Si belongs to IVA group. Therefore, four electrons present in the
outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The
s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the
number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs
to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D:
All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons.
PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present
after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is
[Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons
present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The
element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell.
Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital
and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal.
PART I: The element Ba belongs to IIA group. Th.
The two contributions to the cohesive energy of t.pdf
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction.
Solution
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction..
If you are talking about an extraction design, th.pdf
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic.
Solution
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic..
Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens.
Solution
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens..
When something boils, it changes states of matter. It would go from .pdf
When something boils, it changes states of matter. It would go from a liquid to a soild when it
boils, so if more heat is added, it will just change the state of matter faster.
Solution
When something boils, it changes states of matter. It would go from a liquid to a soild when it
boils, so if more heat is added, it will just change the state of matter faster..
We need to discuss why there is an importance of adding residents to.pdf
We need to discuss why there is an importance of adding residents to the budget planning, we
can go through below explaintion and can derived various conclusions from the same.
In explaining why they did or did not seek input from citizens, respondents in- dicated that the
process was guided by the desires of the leadership. Of those who said that they did seek citizen
input, the desire (as opposed to the obligation through policy) of the staff or the govern- ing
board or staff to seek public opinion seemed to provide some motivation. Almost three-quarters
of responding cities cited staff desire, and almost two-thirds, governing board desire.
The pattern for counties was similar. The most often cited reason for extra effort to involve the
public was board interest. Almost half of the counties also cited staff interest.
Although citizen involvement is a hot topic in management literature these days, local
governments in North Caro- lina are generally not making extensive fforts to involve residents in
budgeting. Barely half of the cities responding to the survey and less than a third of the coun- ties
go beyond the single, mandated hearing to get citizen input In general, cities tend to seek input
more than counties do. This result might be explained by the different types of services offered
by the two forms of gov- ernment. Counties provide funding for, among other things, social
services, pub- lic health services, mental health ser- vices, and schools, and the boards over-
seeing these services may seek citizen input when they are preparing their own budget request
for the county.
Although there is informal interest in hearing from citizens, boards clearly want to remain
flexible in when and how they obtain input. Only 13 percent of the city respondents had a formal
pol- icy or requirement for citizen involve- ment beyond the single, state-mandated hearing.
Medium to large cities were more likely to have such requirements than small cities. Only one
county, a larger one, reported having such a policy or requirement.
Although citizen involvement is a hot topic in management literature these days, local
governments in North Caro- lina are generally not making extensive
efforts to involve residents in budgeting. Barely half of the cities responding to the survey and
less than a third of the coun- ties go beyond the single, mandated hearing to get citizen input
Of those that did not seek extra citizen involvement, the most common reason cited, on both the
city and the county level, was lack of governing board inter- est. The second most common
reason was poor response in the past, when at- tempts to get citizens involved did not seem very
effective. At the county level, lack of board interest was a far more important factor than
anything else, including poor past response: 71 percent of respondents cited lack of board
interest, whereas only 39 percent cited poor past response Although staff interest seemed to be a
big motivator for involving ci.
What is the largest decimal integer that can be represented with the.pdf
What is the largest decimal integer that can be represented with the following number of bits?
1. 3 bits
2. 8 bits
3. 32 bits
4. 64 bits
How do you find out the largest decimal integer?
Solution
In general, with k bits we can represent 2k values. Their range will depend on the system:
Unsigned: 0 to 2k-1
1. 3 bits
=> 2^3 - 1 => 7 => 111
2. 8 bits
=> 2^8 - 1 => 255
3. 32 bits
=> 2^32 - 1
4. 64 bits
=> 2^64 - 1.
Throwing.javaimport java.util.InputMismatchException; import jav.pdf
The document contains code for a Throwing class that gets integer input from the user. It uses try/catch blocks to handle exceptions. In the first code block, it catches InputMismatchException and throws an IllegalArgumentException. In the second code block, it changes the catch block to catch IOException instead of IllegalArgumentException and changes the thrown exception accordingly. Both code blocks prompt the user for an integer, catch invalid input, and print an error message.
This electron transport is accompanied by the protons transfer into .pdf
This electron transport is accompanied by the protons transfer into intermembrane space.
Protons release potential which drives the phosphorylation of ADP to ATP.
Protons return to the matrix through a proton specific channel
Solution
This electron transport is accompanied by the protons transfer into intermembrane space.
Protons release potential which drives the phosphorylation of ADP to ATP.
Protons return to the matrix through a proton specific channel.
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Answer Every investor expects dividend from his investments.Dividen.pdf
Answer Every investor expects dividend from his investments.Dividen.pdf
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additional optmization techniques for underlying IP network must1.pdf
additional optmization techniques for underlying IP network must1.pdf
According to the given equation, aqueous carbon dioxide reacts with .pdf
According to the given equation, aqueous carbon dioxide reacts with .pdf
a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf
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A person may not choose to participate in the labour force due to La.pdf
A person may not choose to participate in the labour force due to La.pdf
Well.. 1) Ionic bonds are almost always metal to .pdf
Well.. 1) Ionic bonds are almost always metal to .pdf
1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf
1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf
clear clc close all Use polyfit to solve for the phase l.pdf
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The oxygen appears in both step reactions. But, i.pdf
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PART A The element Si belongs to IVA group. Therefore, four electro.pdf
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The two contributions to the cohesive energy of t.pdf
The two contributions to the cohesive energy of t.pdf
If you are talking about an extraction design, th.pdf
If you are talking about an extraction design, th.pdf
Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf
Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf
When something boils, it changes states of matter. It would go from .pdf
When something boils, it changes states of matter. It would go from .pdf
We need to discuss why there is an importance of adding residents to.pdf
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What is the largest decimal integer that can be represented with the.pdf
What is the largest decimal integer that can be represented with the.pdf
Throwing.javaimport java.util.InputMismatchException; import jav.pdf
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- 1. Initiation of DNA replication in prokaryotes involves loading and activation of DnaB helicase. DnaB helicase directs other necessary enzymes (particularly DnaG Primase and DNA pol III HE) to the template and help in DNA elongation. These aspects will be discussed below: DnaB protein DnaB is the major helicase, it unwinds the template in 5'-3' direction during oriC replication in vitro, at a rate of 700bp/sec. The direction of unwinding requires DnaB to be placed on lagging strand, ahead of the polymerase on the leading strand. It also activates primer synthesis by DnaG primase. DnaG primase The primase (DnaG) synthesizes a unique oligonucleotide primer on SSB coated single stranded phage G4 DNA. The sequence recognised by primase is over 100 nucleotides long containing a secondary structure. Since primer synthesis requires activation by DnaB, the latter is oftern described as a 'mobile promoter' of primer synthesis. DNA polymerase III holoenzyme DNA polymerase III HE is the large multiprotein complex responsible for elongation of DNA replication in E. coli. It has atleast 10 subunits, each with specific function. Other proteins involved in DNA elongation are DNA gyrase (topoisomerase), which removes the topological strain associated with DNA unwinding SSB (single stranded DNA binding) protein, that binds tightly to single stranded DNA thus stabilizing the unwound DNA. RNase H removes RNA primers DNA Polymerase I is used for filling the gap created due to the removal of RNA primers DNA ligase converts the okasaki fragments into a continuous strand. Solution Initiation of DNA replication in prokaryotes involves loading and activation of DnaB helicase. DnaB helicase directs other necessary enzymes (particularly DnaG Primase and DNA pol III HE) to the template and help in DNA elongation. These aspects will be discussed below: DnaB protein DnaB is the major helicase, it unwinds the template in 5'-3' direction during oriC replication in vitro, at a rate of 700bp/sec. The direction of unwinding requires DnaB to be placed on lagging strand, ahead of the polymerase on the leading strand. It also activates primer synthesis by DnaG primase.
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