In the degradation of maltose to two glucose moleculesA. the enzy.pdf
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In the degradation of maltose to two glucose molecules: A. the enzyme, maltase, is the subtrate. B. maltose is the product. C. maltose is the subtrate. D. maltase is the product. water is the substrate. D. maltase is the product.E. water is the substrate. Solution C. maltose is the subtrate. Substrate is converted to product by the enzyme. Here maltase is the substrate and Glucose is the product.
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In a comparison of ectotherms and endotherms, which of the following.pdf
In a comparison of ectotherms and endotherms, which of the following are correct?
1. Endotherms are regulators while ectotherms are conformers
2. Endotherms have higher energetic demands compared to ectotherms.
3. Endotherms spend more energy on internal temperature regulation compared to ectotherms.
4. Endothermy is more common among animal phyla than ectothermy.
5. At the cellular level, endotherms cells always produce more ATP and more heat.
Solution
1.correct
Regulator
A regulator is also known as an endotherm. A regulator uses internal control mechanisms to
maintain the optimum temperature needed for its body to function. Mammals and birds are
examples of regulators. They are often described as warm blooded because they maintain a
constant body temperature. For example, if the climate around a regulator gets hotter, the animal
can adjust its metabolism to lessen the need for heat production.
Conformer (or ectotherm)
A conformer is also called an ectotherm. A conformer cannot regulate its own internal
temperature so it must adapt its behavior, physically moving itself to environments with
optimum temperatures. Reptiles, amphibians and insects are all conformers and are often referred
to as cold blooded. For example, a lizard may lie on a rock in the sun to raise its body
temperature, as it cannot regulate it internally.
2.not correct
Endotherms also called warm blooded animals have the ability to regulate their body temperature
by themselves.
If the endotherms are in a cold place their body increases the metabolism and generates more
heat. This will compensate for the cold outside.
In a hot region the body metabolism is tuned down (this is not as efficient as the adaptation to
cold region - as the metabolism can\'t be turned down below a certain limit - this will result in
death) a little and heat generation is somewhat reduced. The second and more efficient
mechanism is the heat dissipation by sweating and through skin (directly).
Ectotherms on the other hand, do not have efficient temperature regulation systems.
In warm regions when the temperature is high, the ectotherms seek shade and low-temperature
areas to prevent over-heating
In cool regions they seek sunlight and expose themselves to the sunlight for warmth. For the
same reason, they can also search out and stay near fire pits, campfires etc...
So when we calculate the energy demands, the ectotherms need more energy as they regulate
their own metabolisms to match the need.
3.correct
previously explaine
4.not correct,ectother
ectothermy is more common then endothermy
5.correct.
Implement Breadth-First Search with a QueueSolution Program .pdf
Implement Breadth-First Search with a Queue
Solution
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
for(i = adj[s].begin(); i != adj[s].end(); ++i)
{
if(!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}
// Driver program to test methods of graph class
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
cout << \"Following is Breadth First Traversal \"
<< \"(starting from vertex 2) \ \";
g.BFS(2);
return 0;
}
Output:
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
fo.
Immunology what are TdT and AID, and what are their roles in B c.pdf
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Solution
Answer-TdT stands for terminal deoxynucleotidel transferase.i t is a special kind of DNA
polymerase expressed immature pre B and pre T lymphocytes.Unlike other DNA polymerase
TdT has a unusual function to synthesize DNA without using template DNA.
AID-AID stands for activation induced cytidine deaminase.i T is a 24kDa enzyme in human
which is responsible for antibody diversity.d ye to its deamination activity it deaminase it causes
mutation in DNA by deamination of cytosine to uracil and thus change base pairs G-C to U-Aor
T-A due to these mutation in germinal center of B lymphocytes antibody diversity produces.
Role of TdT and AID in B cells-Bcells are immune cell that produces antibody with the binding
sites for antigen. Different types of antibodies produced by B cell which give immunity to
different type of antigens.These antibody diversity occure due to VDJ recombination .In
antibody producing pre B cells rearrangement of these variable(v), diversity (D), joining (J)gene
segments allow B cells to generate antibody with different antigen binding sites with antibody
specificity.For breaking,arranging and rejoining these gene segments district classes of enzymes
like nuclease ,ligase and polymerase work.S specific kind of polymerase used for these gene
recombination known as TdT.
RAG enzyme cut- the double standard DNA which contains genes .and the hairpin structure left
is open by Artemis complex which has endonuclease activity.once Artemis complex open
hairpin and add palindromic sequence in DNA cut end TdT start working on 3\'OH end of DNA
and add nucleotide to single stranded DNA in 5\'to3\' direction without using template.single
stranded DNA formed at both the ends of non homologous segment joining done by Watson and
Crick base pairing and unpaired bases are removed by exonuclease ,gap is filled by template
dependent polymerase and Nick is sealed by ligase.in this way due to the recombination of
antibody diversity producing gene pre B cells produces diffrent type of antibody specific to
different type of antigen.
AID produces mutation in B cells during germinal center development by deamination of
cytosine that produces antibody diversity.
CD-R are the part of variable chain in immunoglobulin generated by B cells and T cell
receptor.tdt and AID helps to crate these sites on antibody and create antibody diversity..
IDS in the Cloud Please respond to the following(a) Explain w.pdf
\"IDS in the Cloud\" Please respond to the following:
(a) Explain whether or not you believe technologies such as IDS are still relevant and useful as
there is a push toward SaaS and cloud-based solutions.
(b) Discuss from your perspective how cloud-based services change incident response, for better
or worse, and determine what you believe to be the greatest preparedness concern with cloud-
based services.
PLEASE NO COPY AND PASTING
Solution
(a)
IDS - Intrusion Detection systems
To prevent threat over once computer security firewall and encryption techniques were invented.
Although, the intruders could still hack one\'s system. To prevent this, Intrusion detection
systems were introduced which provide the following advantages:
1.Continuously monitors and analyzes the activities of the user and system.
2.Uses pattern recognition for recognizing typical kinds of attacks.
3.Tracks user policy violations.
3. Recognizes abnormal activities through pattern recognition.
SaaS: Software as a Service
1.SaaS is a model for Software licensing and delivery which is hosted on internet on subscription
basis and made available for the internet users.
2. It is one of the categories of cloud computing, where in instead of simply installing and
maintaining software you keep it over the internet which is easily maintainable from complex
software management.
Intrusion detection systems are still relevant to protect one\'s private data although Cloud
computing exists.
(b) Security and Incident response is a major subject which many vendors try to overcome.
At high level, the process defines six stages of incident response :
Preparation,identification,containment,eradication,recovery,lessons learned.
Five challenges or risks or concerns faced with cloud computing:
1.Privacy and Security.
2.Cloud transition.
3.Data access
4.Service quality
5.Data accessibility..
Gandalf the Grey started in the Forest of Mirkwood at a point with co.pdf
Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (-3, -1) and
arrived in the Iron Hills at the point with coordinates (-1,3). If he began walking in the direction
of the vector v = 5i + lj and changes direction only once, when he turns at a right angle, what are
the coordinates of the point where he makes the turn.
Solution
Geometrically, the vector 5i + 1j corresponds to a slope of 1/5.
So, at the right angle where Gandalf changes direction, the slope becomes -5.
Let\'s write out the equations.
Point (-3, -1), slope 1/5 => y + 1 = (1/5)(x + 3) or y = (1/5)(x + 3) - 1 or y = (1/5)(x - 2)
Point (-1, 3), slope - 5 => y - 3 = (-5)(x + 1) or y = - 5x - 2.
Now, we find the intersecting point of these two curves
(1/5)(x - 2) = -5x - 2
==> x = - 4/13
Substituting this into either equation yields y = - 6/13.
So, the desired point is ( - 4/13, - 6/13)..
Explain exploratory factor analysis and give real life examplesS.pdf
Explain exploratory factor analysis and give real life examples
Solution
In multivariate statistics, exploratory factor analysis (EFA) is a statistical method used to
uncover the underlying structure of a relatively large set of variables. EFA is a technique within
factor analysis whose overarching goal is to identify the underlying relationships between
measured variables.[1] It is commonly used by researchers when developing a scale (a scale is a
collection of questions used to measure a particular research topic) and serves to identify a set of
latent constructs underlying a battery of measured variables.[2] It should be used when the
researcher has no a priori hypothesis about factors or patterns of measured variables.[3]Measured
variables are any one of several attributes of people that may be observed and measured. An
example of a measured variable would be the physical height of a human being. Researchers
must carefully consider the number of measured variables to include in the analysis.[2] EFA
procedures are more accurate when each factor is represented by multiple measured variables in
the analysis.
Following is the set of EFA examples,
. • 4.1: Exploratory factor analysis with continuous factor indicators
• 4.2: Exploratory factor analysis with categorical factor indicators
• 4.3: Exploratory factor analysis with continuous, censored, categorical, and count factor
indicators*
• 4.4: Exploratory factor mixture analysis with continuous latent class indicators
• 4.5: Two-level exploratory factor analysis with continuous factor indicators
• 4.6: Two-level exploratory factor analysis with both individual- and cluster-level factor
indicators
• 4.7: Bi-factor exploratory factor analysis with continuous factor indicators
* Example uses numerical integration in the estimation of the model. This can be
computationally demanding depending on the size of the problem..
Drag the labels to the correct category to demonstrate your understan.pdf
Drag the labels to the correct category to demonstrate your understanding of cell walls in
different types of microbes. Protozoans Helminths Bacteria Fungi
Solution
Protozoans- These are classified under kingdom Protist.
Helminths-These are classified under kingdom animalia.
Fungi-These are classified under kingdom fungi.
Bacteria- Prokaryotes..
Do you agree that the Berg Letter, research moratorium, and Asilomar.pdf
Do you agree that the Berg Letter, research moratorium, and Asilomar Conference were
necessary before further recombinant DNA research was conducted? Why or why not?
Solution
Paul Berg, an American biochemist and professor emeritus at Stanford University witnessed
firsthand the history of recombinant DNA research and regulation, having been in the forefront
of both movements since he was a young man. He gained early recognition and influence when
he delineated the key steps in which DNA produces proteins.
In the mid-1970s, the National Academy of Sciences asked Berg to explore the safety of
recombinant DNA technology. He responded with the historic \"Berg letter,\" calling for a
moratorium on recombinant DNA research until safety issues could be addressed. He was one of
the key organizers of the international forum on recombinant DNA technology, the Asilomar
Conference. The ensuing dialogue resulted in the National Institutes of Health guidelines
published a year later, a milestone of responsible self-regulation in science.
The Asilomar Conference on Recombinant DNA was an influential conference organized by
Paul Berg to discuss the potential biohazards and regulation of biotechnology, held in February
1975 at a conference center at Asilomar State Beach.A group of about 140 professionals
(primarily biologists, but also including lawyers and physicians) participated in the conference to
draw up voluntary guidelines to ensure the safety of recombinant DNA technology. The
conference also placed scientific research more into the public domain, and can be seen as
applying a version of the precautionary principle.
The effects of these guidelines are still being felt through the biotechnology industry and the
participation of the general public in scientific discourse. Due to potential safety hazards,
scientists worldwide had halted experiments using recombinant DNA technology, which entailed
combining DNAs from different organisms. After the establishment of the guidelines during the
conference, scientists continued with their research, which increased fundamental knowledge
about biology and the public’s interest in biomedical research.
During the conference, the principles guiding the recommendations for how to conduct
experiments using this technology safely were established. The first principle for dealing with
potential risks was that containment should be made an essential consideration in the
experimental design. A second principle was that the effectiveness of the containment should
match the estimated risk as closely as possible.
The conference also suggested the use of biological barriers to limit the spread of recombinant
DNA. Such biological barriers included fastidious bacterial hosts that were unable to survive in
natural environments. Other barriers were nontransmissible and equally fastidious vectors
(plasmids, bacteriophages, or other viruses) that were able to grow in only specified hosts.
In addition to biological barriers, the co.
Recommended
In a comparison of ectotherms and endotherms, which of the following.pdf
In a comparison of ectotherms and endotherms, which of the following are correct?
1. Endotherms are regulators while ectotherms are conformers
2. Endotherms have higher energetic demands compared to ectotherms.
3. Endotherms spend more energy on internal temperature regulation compared to ectotherms.
4. Endothermy is more common among animal phyla than ectothermy.
5. At the cellular level, endotherms cells always produce more ATP and more heat.
Solution
1.correct
Regulator
A regulator is also known as an endotherm. A regulator uses internal control mechanisms to
maintain the optimum temperature needed for its body to function. Mammals and birds are
examples of regulators. They are often described as warm blooded because they maintain a
constant body temperature. For example, if the climate around a regulator gets hotter, the animal
can adjust its metabolism to lessen the need for heat production.
Conformer (or ectotherm)
A conformer is also called an ectotherm. A conformer cannot regulate its own internal
temperature so it must adapt its behavior, physically moving itself to environments with
optimum temperatures. Reptiles, amphibians and insects are all conformers and are often referred
to as cold blooded. For example, a lizard may lie on a rock in the sun to raise its body
temperature, as it cannot regulate it internally.
2.not correct
Endotherms also called warm blooded animals have the ability to regulate their body temperature
by themselves.
If the endotherms are in a cold place their body increases the metabolism and generates more
heat. This will compensate for the cold outside.
In a hot region the body metabolism is tuned down (this is not as efficient as the adaptation to
cold region - as the metabolism can\'t be turned down below a certain limit - this will result in
death) a little and heat generation is somewhat reduced. The second and more efficient
mechanism is the heat dissipation by sweating and through skin (directly).
Ectotherms on the other hand, do not have efficient temperature regulation systems.
In warm regions when the temperature is high, the ectotherms seek shade and low-temperature
areas to prevent over-heating
In cool regions they seek sunlight and expose themselves to the sunlight for warmth. For the
same reason, they can also search out and stay near fire pits, campfires etc...
So when we calculate the energy demands, the ectotherms need more energy as they regulate
their own metabolisms to match the need.
3.correct
previously explaine
4.not correct,ectother
ectothermy is more common then endothermy
5.correct.
Implement Breadth-First Search with a QueueSolution Program .pdf
Implement Breadth-First Search with a Queue
Solution
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
for(i = adj[s].begin(); i != adj[s].end(); ++i)
{
if(!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}
// Driver program to test methods of graph class
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
cout << \"Following is Breadth First Traversal \"
<< \"(starting from vertex 2) \ \";
g.BFS(2);
return 0;
}
Output:
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
fo.
Immunology what are TdT and AID, and what are their roles in B c.pdf
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Solution
Answer-TdT stands for terminal deoxynucleotidel transferase.i t is a special kind of DNA
polymerase expressed immature pre B and pre T lymphocytes.Unlike other DNA polymerase
TdT has a unusual function to synthesize DNA without using template DNA.
AID-AID stands for activation induced cytidine deaminase.i T is a 24kDa enzyme in human
which is responsible for antibody diversity.d ye to its deamination activity it deaminase it causes
mutation in DNA by deamination of cytosine to uracil and thus change base pairs G-C to U-Aor
T-A due to these mutation in germinal center of B lymphocytes antibody diversity produces.
Role of TdT and AID in B cells-Bcells are immune cell that produces antibody with the binding
sites for antigen. Different types of antibodies produced by B cell which give immunity to
different type of antigens.These antibody diversity occure due to VDJ recombination .In
antibody producing pre B cells rearrangement of these variable(v), diversity (D), joining (J)gene
segments allow B cells to generate antibody with different antigen binding sites with antibody
specificity.For breaking,arranging and rejoining these gene segments district classes of enzymes
like nuclease ,ligase and polymerase work.S specific kind of polymerase used for these gene
recombination known as TdT.
RAG enzyme cut- the double standard DNA which contains genes .and the hairpin structure left
is open by Artemis complex which has endonuclease activity.once Artemis complex open
hairpin and add palindromic sequence in DNA cut end TdT start working on 3\'OH end of DNA
and add nucleotide to single stranded DNA in 5\'to3\' direction without using template.single
stranded DNA formed at both the ends of non homologous segment joining done by Watson and
Crick base pairing and unpaired bases are removed by exonuclease ,gap is filled by template
dependent polymerase and Nick is sealed by ligase.in this way due to the recombination of
antibody diversity producing gene pre B cells produces diffrent type of antibody specific to
different type of antigen.
AID produces mutation in B cells during germinal center development by deamination of
cytosine that produces antibody diversity.
CD-R are the part of variable chain in immunoglobulin generated by B cells and T cell
receptor.tdt and AID helps to crate these sites on antibody and create antibody diversity..
IDS in the Cloud Please respond to the following(a) Explain w.pdf
\"IDS in the Cloud\" Please respond to the following:
(a) Explain whether or not you believe technologies such as IDS are still relevant and useful as
there is a push toward SaaS and cloud-based solutions.
(b) Discuss from your perspective how cloud-based services change incident response, for better
or worse, and determine what you believe to be the greatest preparedness concern with cloud-
based services.
PLEASE NO COPY AND PASTING
Solution
(a)
IDS - Intrusion Detection systems
To prevent threat over once computer security firewall and encryption techniques were invented.
Although, the intruders could still hack one\'s system. To prevent this, Intrusion detection
systems were introduced which provide the following advantages:
1.Continuously monitors and analyzes the activities of the user and system.
2.Uses pattern recognition for recognizing typical kinds of attacks.
3.Tracks user policy violations.
3. Recognizes abnormal activities through pattern recognition.
SaaS: Software as a Service
1.SaaS is a model for Software licensing and delivery which is hosted on internet on subscription
basis and made available for the internet users.
2. It is one of the categories of cloud computing, where in instead of simply installing and
maintaining software you keep it over the internet which is easily maintainable from complex
software management.
Intrusion detection systems are still relevant to protect one\'s private data although Cloud
computing exists.
(b) Security and Incident response is a major subject which many vendors try to overcome.
At high level, the process defines six stages of incident response :
Preparation,identification,containment,eradication,recovery,lessons learned.
Five challenges or risks or concerns faced with cloud computing:
1.Privacy and Security.
2.Cloud transition.
3.Data access
4.Service quality
5.Data accessibility..
Gandalf the Grey started in the Forest of Mirkwood at a point with co.pdf
Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (-3, -1) and
arrived in the Iron Hills at the point with coordinates (-1,3). If he began walking in the direction
of the vector v = 5i + lj and changes direction only once, when he turns at a right angle, what are
the coordinates of the point where he makes the turn.
Solution
Geometrically, the vector 5i + 1j corresponds to a slope of 1/5.
So, at the right angle where Gandalf changes direction, the slope becomes -5.
Let\'s write out the equations.
Point (-3, -1), slope 1/5 => y + 1 = (1/5)(x + 3) or y = (1/5)(x + 3) - 1 or y = (1/5)(x - 2)
Point (-1, 3), slope - 5 => y - 3 = (-5)(x + 1) or y = - 5x - 2.
Now, we find the intersecting point of these two curves
(1/5)(x - 2) = -5x - 2
==> x = - 4/13
Substituting this into either equation yields y = - 6/13.
So, the desired point is ( - 4/13, - 6/13)..
Explain exploratory factor analysis and give real life examplesS.pdf
Explain exploratory factor analysis and give real life examples
Solution
In multivariate statistics, exploratory factor analysis (EFA) is a statistical method used to
uncover the underlying structure of a relatively large set of variables. EFA is a technique within
factor analysis whose overarching goal is to identify the underlying relationships between
measured variables.[1] It is commonly used by researchers when developing a scale (a scale is a
collection of questions used to measure a particular research topic) and serves to identify a set of
latent constructs underlying a battery of measured variables.[2] It should be used when the
researcher has no a priori hypothesis about factors or patterns of measured variables.[3]Measured
variables are any one of several attributes of people that may be observed and measured. An
example of a measured variable would be the physical height of a human being. Researchers
must carefully consider the number of measured variables to include in the analysis.[2] EFA
procedures are more accurate when each factor is represented by multiple measured variables in
the analysis.
Following is the set of EFA examples,
. • 4.1: Exploratory factor analysis with continuous factor indicators
• 4.2: Exploratory factor analysis with categorical factor indicators
• 4.3: Exploratory factor analysis with continuous, censored, categorical, and count factor
indicators*
• 4.4: Exploratory factor mixture analysis with continuous latent class indicators
• 4.5: Two-level exploratory factor analysis with continuous factor indicators
• 4.6: Two-level exploratory factor analysis with both individual- and cluster-level factor
indicators
• 4.7: Bi-factor exploratory factor analysis with continuous factor indicators
* Example uses numerical integration in the estimation of the model. This can be
computationally demanding depending on the size of the problem..
Drag the labels to the correct category to demonstrate your understan.pdf
Drag the labels to the correct category to demonstrate your understanding of cell walls in
different types of microbes. Protozoans Helminths Bacteria Fungi
Solution
Protozoans- These are classified under kingdom Protist.
Helminths-These are classified under kingdom animalia.
Fungi-These are classified under kingdom fungi.
Bacteria- Prokaryotes..
Do you agree that the Berg Letter, research moratorium, and Asilomar.pdf
Do you agree that the Berg Letter, research moratorium, and Asilomar Conference were
necessary before further recombinant DNA research was conducted? Why or why not?
Solution
Paul Berg, an American biochemist and professor emeritus at Stanford University witnessed
firsthand the history of recombinant DNA research and regulation, having been in the forefront
of both movements since he was a young man. He gained early recognition and influence when
he delineated the key steps in which DNA produces proteins.
In the mid-1970s, the National Academy of Sciences asked Berg to explore the safety of
recombinant DNA technology. He responded with the historic \"Berg letter,\" calling for a
moratorium on recombinant DNA research until safety issues could be addressed. He was one of
the key organizers of the international forum on recombinant DNA technology, the Asilomar
Conference. The ensuing dialogue resulted in the National Institutes of Health guidelines
published a year later, a milestone of responsible self-regulation in science.
The Asilomar Conference on Recombinant DNA was an influential conference organized by
Paul Berg to discuss the potential biohazards and regulation of biotechnology, held in February
1975 at a conference center at Asilomar State Beach.A group of about 140 professionals
(primarily biologists, but also including lawyers and physicians) participated in the conference to
draw up voluntary guidelines to ensure the safety of recombinant DNA technology. The
conference also placed scientific research more into the public domain, and can be seen as
applying a version of the precautionary principle.
The effects of these guidelines are still being felt through the biotechnology industry and the
participation of the general public in scientific discourse. Due to potential safety hazards,
scientists worldwide had halted experiments using recombinant DNA technology, which entailed
combining DNAs from different organisms. After the establishment of the guidelines during the
conference, scientists continued with their research, which increased fundamental knowledge
about biology and the public’s interest in biomedical research.
During the conference, the principles guiding the recommendations for how to conduct
experiments using this technology safely were established. The first principle for dealing with
potential risks was that containment should be made an essential consideration in the
experimental design. A second principle was that the effectiveness of the containment should
match the estimated risk as closely as possible.
The conference also suggested the use of biological barriers to limit the spread of recombinant
DNA. Such biological barriers included fastidious bacterial hosts that were unable to survive in
natural environments. Other barriers were nontransmissible and equally fastidious vectors
(plasmids, bacteriophages, or other viruses) that were able to grow in only specified hosts.
In addition to biological barriers, the co.
Brief Concepts and Questions EXAM 2 Chapter 8 DNA RNA Protein What i.pdf
Brief Concepts and Questions EXAM 2 Chapter 8: DNA RNA Protein What is DNA? a
phosphate Structure of DNA: Building blocks are called nucleotides Each nucleotide is
composed of three br uithofenas bee. What makes DNA so special? Provide 4 reasons, below
DNA DNA (Replication): Where does DNA replication take place? When does DNA replication
take place? Explain steps involved in DNA replication: DNA RNA Protein (Gene Expression)
Involves 2 processes: 1. Transcription 2. Translation Explain the Synthesis of Proteins (Gene
Expression): o DNA RNA Protein What is RNA? What is \"codon What is \"anticodon\" What is
a protein molecule? DNA mutation; Change in nucleotide bases of DNA Duplex Point mutation
Frame shift mutation
Solution
Question
Answer
Where does DNA replication take place:
It takes place in the nucleus in case of eukaryotic cells and in the cytoplasm in case of
prokaryotic cells
When does DNA replication take place:
DNA replication occurs during the S-phase during cell cycle, so that cell can make an extra copy
of genetic material.
Explain steps involved in DNA replication:
Initiation: During initiation, the proteins will bind to the origin of replication; helicase unwinds
the DNA helix which results in the formation of two replication forks.
Elongation: A RNA primer sequence will be added to this the DNA pol III will add the
nucleotides in 5’ to 3’ direction and chain will elongate.
Termination: In case of bacteria, termination of replication occurs whenever two replication
forks meet each other from the opposite end of the parental chromosome.
Transcription
Gene expression first step is transcription, here a particular segment of DNA will be copied into
RNA with the help of the enzyme RNA polymerase
Translation
Translation is the final step of the gene expression. Here mRNA will be used to synthesize the
polypeptide chain. The information present in the mRNA in the form of codon will code for the
amino acids needed for polypeptide chain synthesis.
What is RNA?
RNA is ribonucleic acid and is found in all living cells. It acts as the messenger carrying
instructions from DNA for the synthesis of proteins.
Few viruses will have RNA as their genetic material.
What is codon?
Codon is a sequence of three nucleotides and they together form a unit of genetic code in either
DNA or RNA.
What is anticodon?
It is found on tRNA and it is a sequence of three nucleotides which forms a genetic code on
tRNA, and these anticodon is complementary to the codons found on messenger RNA.
What is a protein molecule?
During translation, when amino acids are added in a sequential manner, the condensation of
amino acids will form a peptide bond in between them and finally forms a polypeptide chain. It
is the DNA through mRNA directs the protein synthesis.
Point mutation
In point mutation, only one or very few nucleotides will be affected or mutated in a gene
sequence.
Frame shift mutation
Either insertions or deletion can result in frame shift mutation, due to th.
A population has two alleles for a particular gene, A and a, and is .pdf
A population has two alleles for a particular gene, A and a, and is in Hardy-Weinberg
equilibrium. The frequency of allele a is 0.2. What is the frequency of individuals with AA
genotype?
0.2
0.4
0.64
0.04
0.8
Solution
C). 0.64
Hardy and Weinberg mathematically proved that in a population, all dominant and recessive
alleles comprise all alleles for that gene.
This was mathematically represented as p+ q = 1.0
Where,
p = frequency of dominant alleles
q = frequency of recessive alleles.
Given that, the frequency of a = 0.2
As the population is in H-W equilibrium, the frequency of A = 1-0.2 = 0.8
The frequency of AA genotype = 0.8*0.8 = 0.64.
A 12-year-old male complained of strange odors before loss of consci.pdf
A 12-year-old male complained of strange odors before loss of consciousness and a major tonic-
clonic seizure. During the seizure, his parents rushed him to the emergency room of a local
hospital. On arrival at the hospital, he appeared to be asleep. Studies at the hospital showed
routine laboratory work within normal limits (WNL), lumbar puncture (CSF) was WNL, no
evidence of skull fracture on x-ray study was revealed, and an electroencephalograph (EEG)
showed no abnor malities. How would you interpret the episode and findings?
Solution
Charecterized by sudden loss of consciousness ,chaotic muscle movement and tone, and apnea
Tonic phase:Bilateral muscle rigidity
Clonic phase: muslce contraction and relaxation lasting 1 to 3 minutes
Signs:
Tachycardia, hyerpventilation,sweating, and intense salivation
Last 3-5 minutes
postictal state(5 to 20 mintues)follows
-gradual return to consciousness.
2.Which of the following statements about evolution is falseNonrand.pdf
2.Which of the following statements about evolution is false?Nonrandom mating can change the
genetic makeup of populations over time.Genetic drift can change the genetic makeup of
populations over time.Natural selection is one of several mechanisms of evolution.Populations,
not individuals, evolve.Evolution is goal-directed.2.Which of the following statements about
evolution is false?Nonrandom mating can change the genetic makeup of populations over
time.Genetic drift can change the genetic makeup of populations over time.Natural selection is
one of several mechanisms of evolution.Populations, not individuals, evolve.Evolution is goal-
directed.
Solution
Note:.
can someone explain in a easy way chromosome walking I know its w.pdf
can someone explain in a easy way chromosome walking? I know it\'s with overlapping
fragments and a gen nearby the gen you want to find, but I don\'t clearly understand it, I have
images in my book, but an image with steps and explanation would be nice,
Solution
Chromosome Walking:
Chromosome walking is positional cloning of gene sequences or upstream promoter sequences.
Eukaryotic DNA contains a lot of junk DNA between the genes. The technique relies upon
screening of gDNA library for overlapping fragments in a sequential manner to tap the upstream
or downstream region.
Procedure:
1. Isolation of Genomic DNA: First of all isolate genomic DNA (gDNA) of the test organism
and then partially digest it with some restriction enzyme, say Hind III that would yield unequal
random fragments.
2. Cloning of the Digested into E.coli: Clone the digested gDNA into a cloning vector (between
the Hind III sites in this case) to prepare a gDNA library of the organism. Each colony of the
library would have only one clone.
3. Preparation of marker probe: Prepare a radioactively labelled probe from the neighboring gene
(say A), from where you want to start walking.
4. Colony Hybridization: Hybridize the probe with your gDNA library. The probe would show
the signal in the colonies carrying DNA in and around your gene.
5. Sequencing: Pick the colonies showing the signal and isolate the plasmid DNA and sequence
each using the plasmid-specific primers. You would get only one sequence per bacterial colony
of your library.
The sequence, thus obtained must be analyzed. Select longer sequences towards the direction of
your walk.
6. Preparation of the probe: Prepare new probes from the region towards your target gene.
7. Repeat the steps 4-6 till you get the clone and sequence of your target gene..
9. When a bacteriophage inserts its DNA into a bacterial chromosome,.pdf
9. When a bacteriophage inserts its DNA into a bacterial chromosome, the inserted
bacteriophage DNA is called _____
(a) prophage
(b) lytogenic
(c) virulent
Solution
a.Prophage.
The genetic material of a bacteriophage ,incorporated in to the genome of bacterium or host ,now
it is ready to prouce phages if specifically activated.
This stage of Bacteriophage life cycle called as Prophage..
Write an informal negation for each of the following statements. Be .pdf
Write an informal negation for each of the following statements. Be careful to avoid negations
that are ambiguous. a. All dogs are friendly. b. All people are happy. c. Some suspicions were
substantiated. d. Some estimates are accurate.
Solution
(a)
these is a dog which is not friendly
(b)
Some people are not happy
(c)
these was a ssuspicions which was not substantiated
(d)
these is some estimates which are wrong.
Write a C++ program which generates and displays a random walk acros.pdf
Write a C++ program which generates and displays a random walk across a grid. A 2-
dimensional array of characters will be used for this grid. The cells visited by the walk will be
labeled by capital letters with A being the first cell, B the cell moved to from A, and so on,
ending either when Z is reached or when it is not possible to move to another cell.
The following libraries, constant definition, and type definition are needed in the program:
#include
#include // for srand and rand
#include // for time
using namespace std;
const int SIZE = 10;
typedef char Grid[SIZE][SIZE]
The grid must be set to have all its elements be the character ’.’ before the walk is begun.
The program must contain the following function prototypes, the main function, and the function
definitions for all the other functions:
// Function prototypes
bool can_move_up(int i, int j, Grid walk);
bool can_move_down(int i, int j, Grid walk);
bool can_move_left(int i, int j, Grid walk);
bool can_move_right(int i, int j, Grid walk);
void init_array(Grid walk);
void generate_random_walk(Grid walk);
void print_array(Grid walk);
int main(void)
{
Grid walk; // the grid in which the random walk occurs
srand((unsigned) time(NULL));
init_array(walk);
generate_random_walk(walk);
print_array(walk);
return 0;
}
The four functions can_move _* must return whether or not it is possible to move in the
particular direction in the grid. For example, a function call like if (can_move_up(i, j, walk))
could be used to determine whether or not it is possible to move up. The rule for a move being
possible is that if it doesn’t go outside of the grid and isn’t already part of the walk, then that
move is possible. The function init_array must set all characters in the grid to ’.’
The function generate_random_walk must modify the cells of grid by changing ones that are
part of the walk to letters.
The function print_array must print the resulting grid in the form shown in the sample run
below.
In the function generate_random_walk, at each step of the walk, randomly select one of the 4
directions and try to move in that direction. If that move isn’t available, try the other moves in
turn until one is possible. If all 4 directions are blocked, the walk must terminate, and the
program must continue and print out the walk. To select one of the 4 directions randomly, use
the expression: rand() % 4
The function rand returns an integer, and finding the remainder of this integer divided by 4
results in getting one of the numbers 0, 1, 2, or 3. Use these 4 numbers to represent the 4
directions up, down, left and right.
The call to the function srand from the main function seeds the pseudo-random number
generator used by rand so that different sequences of numbers will be generated by calls to rand
each time the program is run.
Solution
#include //input output functions in C++
#include //rand() and srand() functions
#include //time() function for creating seed
#include
#include // for srand and rand
#inclu.
true or false1- The grouped frequency table should ideally have.pdf
true or false???
1- The grouped frequency table should ideally have 10 class intervals
2- The width of the intervals should be half of the number of scores
3- Each score can be represented in one and only one interval
4- The intervals should end with the highest number in the sample
Solution
1.True
2.False
3.True
4.False.
Answer the questions that follow the VHDL code given below. LIBRARY .pdf
Answer the questions that follow the VHDL code given below. LIBRARY ieee; USE
ieee.std_logic_1164.ALL; ENTITY ex1_IS PORT (n, cp, n rd: IN st_logic; q: BUFFER integer
RANGE 0 TO 15); END ex1; ARCHITECTURE are OF ex1 IS BEGIN PROCESS (n_cp,
n_rd) BEGIN IF (n_rd = \'0\') THEN q
Solution
a)n_rd- used to reset the counter
n_cp-used as clock to the counter
b)4 binary bits are there in q;
Q stores 0 t0 15 number so
log2(16)=4.
The weak D phenotype is thought to arise from several mechanisms. .pdf
The weak D phenotype is thought to arise from several mechanisms. Which one of the following
statements is false?
C gene is in trans position to D gene, this results in a weakening influence on the D gene,
examples CDe/Cde and cDe/Cde.
Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in
caucasians. Also occurs in the Ro genotype in certain racial groups more commonly.
The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these pieces
is missing. The red cells are said to have a partial D.
D deletion genes gives rise to weakened expression of D antigen.
Rh null cells lack:
MNSs antigens.
Lewis antigens.
Normal oxygen-carrying capacity.
Rh antigens.
There are two closely related and closely linked genes.
Enzyme techniques, in conjunction with an IgM anti-D.
Increasing the time for the incubation of the saline phase.
Performance of an IAT using an IgG anti-D.
Don\'t need to do it in a blood bank.
C gene is in trans position to D gene, this results in a weakening influence on the D gene,
examples CDe/Cde and cDe/Cde.
Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in
caucasians. Also occurs in the Ro genotype in certain racial groups more commonly.
The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these
pieces is missing. The red cells are said to have a partial D.
D deletion genes gives rise to weakened expression of D antigen.
Solution
The weak D phenotype is thought to arise from several mechanisms. Which one of the following
statements is false?
D deletion genes gives rise to weakened expression of D antigen.
Rh null cells lack:
Rh antigens.
There are two closely related and closely linked genes.
Enzyme techniques, in conjunction with an IgM anti-D.
Increasing the time for the incubation of the saline phase.
Performance of an IAT using an IgG anti-D.
Don\'t need to do it in a blood bank.
Last one is an incomplete question
D deletion genes gives rise to weakened expression of D antigen..
The coding region of a BACTERIAL gene has the following sequence A.pdf
The coding region of a BACTERIAL gene has the following sequence:
\"AGAGCCAACATGTCGAGCATATGCTGATATACG\' Draw and label the DNA (including
ALLEABIS of the gene, sequence information) The info you need is in the sides. Label the
template/coding strands Label 5\' and 3\' ends of each DNA strand Label where RNA
polymerase binds \"Transcribe\" this gene into mRNA, as shown in Bb video. Write the mRNA
nucleotide sequence, starting at the correct nucleotide, starting at the correct nucleotide Show
the 5\' and 3\' ends of the RNA. Use the codon chart to \"translate\" the mRNA into a protein,
and write the correct amino acid sequence of the protein. Only translate the part of the gene that
a cell would translate. Label the N-terminus and C-terminus of the protein. If the sequence of
the gene were changed to: AGACCCAACATGTCGGACATATGCTGATATACG what would
happen to the properties of the protein? Explain why write out the entire process of bacterial
transcription step by step in the correct order, in detail.
Solution
4. Process of Bacterial Transcription:
Bacterial transcription is the process in which mRNA transcripts of genetic material in bacteria
are produced; it is further translated to produce proteins. Unlike in eukaryotes, bacterial
transcription and translation can occur simultaneously in the cytoplasm.
RNA polymerases are enzymes that transcribe DNA into RNA. Using a DNA template, RNA
polymerase builds a new RNA molecule through base pairing. For instance, if there is a G in the
DNA template, RNA polymerase will add a C to the new, growing RNA strand.
Three steps are involved in bacterial transcription:
Initiation: To begin transcribing a gene, RNA polymerase binds to the DNA of the gene at a
region called the promoter. Basically, the promoter tells the polymerase where to locate on the
DNA and begin transcribing.
Elongation:
Once RNA polymerase is in position at the promoter, the next step of
transcription—elongation—can begin. Basically, elongation is the stage when the RNA strand
gets longer, thanks to the addition of new nucleotides.
During elongation, RNA polymerase \"walks\" along one strand of DNA, known as the template
strand, in the 3\' to 5\' direction. For each nucleotide in the template, RNA polymerase adds a
matching (complementary) RNA nucleotide to the 3\' end of the RNA strand.
Termination:
RNA polymerase will keep transcribing until it gets signals to stop. The process of ending
transcription is called termination, and it happens once the polymerase transcribes a sequence of
DNA known as a terminator.
There are two major termination strategies found in bacteria: Rho-dependent and Rho-
independent.
In Rho-dependent termination, the RNA contains a binding site for a protein called Rho factor.
Rho factor binds to this sequence and starts \"climbing\" up the transcript towards RNA
polymerase.
When it catches up with the polymerase at the transcription bubble, Rho pulls the RNA transcript
and the template DNA strand apart, releasing .
Regarding the ABO blood group system in humans, if an individual is g.pdf
Regarding the ABO blood group system in humans, if an individual is genetically I^B i and yet
expresses the O blood type, which genotype is he or she likely to have? Hh HH hh i^A i i^B i
none of the above are correct.
Solution
Option C is the correct answer. This hh is also known as Bombay blood group. These individuals
do not express H antigen. Thus these individuals are unable to form either of A or B antigen on
their RBCs. Usual test of these individuals for blood group shows them as having O blood group..
Peptidoglycan is a unique macromolecule found in cell walls of what k.pdf
Peptidoglycan is a unique macromolecule found in cell walls of what kind of organism archea
yeast algae bacteria All of the above According to the fossil record, when did eukaryotes first
appear on Earth? 1 million years ago
Solution
1) In bacteria ,cell wall is composed of peptidoglycan
2) Eukaryotes appear 2.1 million years ago.
On this photo you see skull of Tasmanian Devil - now extinct only big.pdf
On this photo you see skull of Tasmanian Devil - now extinct only big marsupial predator from
Australia. Could you recognize bones on this skull and name them?
Solution
1. Surangular
2. Postorbital
3. Parietal
4. Frontal
5. Prefrontal
6. Nasal
7. Premaxilla
8. Margins of nares
9. Incisor
10. socket
11 Canine
12 Dentary
13 Mandible
14. Lacrimal
15 Jugal.
MonthMonthly Returns ()MonthMonthly Returns ()MonthMonth.pdf
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Dec-07
-13.241
-0.219
-0.692
Dec-06
-5.442
-5.018
1.403
Dec-05
-2.610
2.457
0.036
Nov-07
-13.794
-1.980
-4.182
Nov-06
2.544
1.767
1.899
Nov-05
8.818
8.806
3.778
Oct-07
15.861
24.632
1.590
Oct-06
4.193
18.534
3.257
Oct-05
9.249
17.595
-1.668
Sep-07
18.093
10.096
3.736
Sep-06
-12.314
6.174
2.575
Sep-05
1.561
10.650
0.810
Aug-07
-2.237
1.029
1.497
Aug-06
6.227
-2.087
2.376
Aug-05
-0.060
-0.612
-0.912
Jul-07
-14.302
-2.430
-3.097
Jul-06
-17.758
-7.805
0.616
Jul-05
-3.781
-2.172
3.717
Jun-07
-5.470
4.979
-1.660
Jun-06
4.463
12.778
0.133
Jun-05
-6.855
6.088
0.143
May-07
2.354
5.628
3.486
May-06
-3.630
-11.035
-2.875
May-05
7.826
26.032
3.179
Apr-07
-10.387
2.885
4.428
Apr-06
1.612
7.164
1.342
Apr-05
1.770
21.877
-1.896
Mar-07
1.393
1.938
1.116
Mar-06
0.624
7.551
1.244
Mar-05
5.051
-3.979
-1.769
Feb-07
9.007
-10.379
-1.950
Feb-06
-6.735
-16.188
0.271
Feb-05
-8.350
-3.900
2.103
Jan-07
10.846
8.908
1.511
Jan-06
-12.264
4.291
2.648
Jan-05
-6.555
1.468
-2.437
Question: (Use table above to solve)
Segregate the return data in the above table into bull and bear markets, using the S&P 500 index
as the proxy for the market.
A.) What was the average return for the two companies during these two types of markets? What
was the return for the S&P 500 index during those times?
B.) Using the average return measures in a), compute the beta for Yahoo and Google.
C.) According to these beta estimates, which company is riskier?
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Dec-07
-13.241
-0.219
-0.692
Dec-06
-5.442
-5.018
1.403
Dec-05
-2.610
2.457
0.036
Nov-07
-13.794
-1.980
-4.182
Nov-06
2.544
1.767
1.899
Nov-05
8.818
8.806
3.778
Oct-07
15.861
24.632
1.590
Oct-06
4.193
18.534
3.257
Oct-05
9.249
17.595
-1.668
Sep-07
18.093
10.096
3.736
Sep-06
-12.314
6.174
2.575
Sep-05
1.561
10.650
0.810
Aug-07
-2.237
1.029
1.497
Aug-06
6.227
-2.087
2.376
Aug-05
-0.060
-0.612
-0.912
Jul-07
-14.302
-2.430
-3.097
Jul-06
-17.758
-7.805
0.616
Jul-05
-3.781
-2.172
3.717
Jun-07
-5.470
4.979
-1.660
Jun-06
4.463
12.778
0.133
Jun-05
-6.855
6.088
0.143
May-07
2.354
5.628
3.486
May-06
-3.630
-11.035
-2.875
May-05
7.826
26.032
3.179
Apr-07
-10.387
2.885
4.428
Apr-06
1.612
7.164
1.342
Apr-05
1.770
21.877
-1.896
Mar-07
1.393
1.938
1.116
Mar-06
0.624
7.551
1.244
Mar-05
5.051
-3.979
-1.769
Feb-07
9.007
-10.379
-1.950
Feb-06
-6.735
-16.188
0.271
Feb-05
-8.350
-3.900
2.103
Jan-07
10.846
8.908
1.511
Jan-06
-12.264
4.291
2.648
Jan-05
-6.555
1.468
-2.437
Solution
According to Beta estimates Google is more risky than Yahoo because the beta is higher for
google( 1.255) with Yahoo(0.987)ABCMonthMonthly Returns (%)Using excel Average return
=AVERAGE(A1:A36)Using excel Average return =AVERAGE(B1:B36)Using excel Average
return =AVERAGE(C1:C36)Beta of.
Many DOD components are accomplishing more training in the simulator.pdf
Many DOD components are accomplishing more training in the simulator instead of the aircraft
due to limited financial resources and the benefit of reduced wear and tear on aging aircraft
fleets. Commercial Airlines do much of their initial training and their recurring training in the
simulator. Where do you stand on the premise of supplementing actual flight training with time
in the simulator? Be sure to identify the positive and negative aspects of replacing actual flight
time with simulator time. Please Cite Sources
Solution
Heads of the DoD Components shall establish procedures for transmission and transportation of
classified information and information-bearing material that minimize risk of compromise while
permitting use of the most cost-effective transmission or transportation means.COMSEC
information shall be transmitted and transported in accordance with National
Telecommunications and Information Systems Security Instruction 4001.NATO classified
information shall be transmitted in compliance with USSAN Instruction 1-69. Except under rules
established by the Secretary of Defense, or as provided by Section 102 of the National Security
Act, classified information originating in a department or agency other than the Department of
Defense shall not be disseminated outside the Department of Defense without the consent of the
originating department or agency.
When classified information is transmitted, it shall be enclosed in two opaque, sealed envelopes,
wrappings, or containers, durable enough to properly protect the material from accidental
exposure and facilitate detection of tampering. The following exceptions apply:
(1) If the classified material is an internal component of a packageable item of equipment, the
outside shell or body may be considered as the inner enclosure provided it does not reveal
classified information.
(2) If the classified material is an inaccessible internal component of a bulky item of equipment,
the outside or body of the item may be considered to be a sufficient enclosure provided
observation of it does not reveal classified information.
(3). If the classified material is an item or equipment that is not reasonably packageable and the
shell or body is classified, it shall be concealed with an opaque covering that will hide all
classified features.
(4) Specialized shipping containers, including closed cargo transporters, may be considered the
outer wrapping or cover when used.
(5) When classified material is hand-carried outside an activity, a locked briefcase may serve as
the outer wrapper. In such cases, the addressing requirements of paragraph 7-201a., below, do
not apply.
(6) NATO Restricted material need not be double-wrapped when transmitted within the United
States. The marking \"NATO Restricted\" shall not appear on the wrapper
Classified material shall be prepared for shipment, packaged, and sealed in ways that minimize
risk of accidental exposure or undetected deliberate compromise. Documents shoul.
Match the numbered terms to the descriptions that follow. Choose all .pdf
Match the numbered terms to the descriptions that follow. Choose all appropriate terms, but
only appropriate terms. autotroph heterotroph phototroph chemotroph a prokaryote that
obtains both energy and carbon as it decomposes dead organisms 1 only 4 only 1 and 3 2 and
4 1, 3, and 4 which of the following traits do archaea and bacteria share? composition of the
cell wall presence of plasma membrane lack of a nuclear envelope identical rRNA sequences 1
only 3 only 1 and 3 2 and 3 2 and 4 Which prokaryotes should be expected to be most
strongly resistant to plasmolysis in hypertonic environment? extreme halophiles extreme
thermophiles methanogens cyanobacteria nitrogen-fixing bacteria that live in root nodules The
group diplomonadida is characterized by a mouthlike gullet, and a hairlike surface: Paramecium
is an example. flagella and a lack of mitochondria \"Gilardia is an example. nonmotility,
parasitism and sporelike infective stages: Toxoplasma is an example. switching between
autotrphic and heterotrphic lifestyles: Euglena is an example large protein deposites and
movements by two falgella which a part of an undulating membrane: Trypanosoma is an
example Plasmodium is transmitted to humans by the bite of mosquito (Anopheles) Its lifecycle
is characterized by spores, gametes and cysts that can hide in human cells. This infective protist
belongs to the group: Apicomplexa Oomycota Ciliophora Archaeplastida Dinoflagellata
Solution
Answer:
42). D). 2 and 4
Chemotrophs are organisms that obtain energy by the oxidation of electron donors in their
environments. These molecules can be organic (chemoorganotrophs) or inorganic
(chemolithotrophs). The chemotroph designation is in contrast to phototrophs, which utilize solar
energy.
43). D). 2 and 3
Archaea and bacteria are both prokaryotes, meaning they do not have a nucleus and lack
membrane-bound organelles. They are tiny, single-cell organisms called microbes, which cannot
be seen by the naked human eye.
44). A) extreme halophiles
45). B). Flagella and a lack of mithochondria. Giardia is an example
The diplomonads are a group of flagellates, most of which are parasitic. They include Giardia
lamblia, which causes giardiasis in humans. They are placed among the metamonads, and appear
to be particularly close relatives of the retortamonads.
46). A). Apicomplexa
The Apicomplexa are a diverse group that includes organisms such as the coccidia, gregarines,
piroplasms, haemogregarines, and plasmodia. Diseases caused by Apicomplexa include:
Babesiosis (Babesia) Malaria (Plasmodium) Cryptosporidiosis (Cryptosporidium parvum).
Khaleesi Corporation leases dragonelass weapons to customers. She gai.pdf
Khaleesi Corporation leases dragonelass weapons to customers. She gains a loyal following of
customers because dragonglass is of limited supply but high demand due to its usefulness for the
upcoming winter. Further, Khaleesi tends to offer more favorable financing terms than
competitors offering substitutive products (e.g., Lannister, LLC) Khaleesi was recently
approached by the Lords of the North, Inc. (the North\") which is interested in leasing a
substantial stock of weapons over a potentially lengthy geriod.of time. The North has indicated a
willingness to pay any rate that Khaleesi Corporation demands for its dragcaglass products, but
generally receives an interest rate of 12% on all other borrowing transactions. The North\'s
management are a very noble group, so payments are reasonably assured. Further, there are no
material cost ncertainties. Khaleesi has gathered its council to discuss entering into such a
contract with the North and has invited you to provide financial council. Khaleesi\'s board has
proposed several alternative sets of lease terms (below) and would like you determine what the
North\'s annual payments will be under each scenario, if payments are made at the beginning of
the period $489,450 Fair value of weapons to be leased Lease Ternm Useful Life of leased assets
Lessee/Lessor interest rate Residual Value (guaranteed) 1. On the first tab of your Excel
workbook, prepare a schedule for Khaleesi which uses Excel to $489,450 11 years 13 years 696
So $489,450 11 years 13 years 6% $24,350 years 13 years 896 $24,350 compute the North\'s
necessary annual payment under each of the scenarios listed in the chart above. The schedule
should list for each scenario above the desired present value of lease payments, future value of
the leased asset, the lease term, and the interest rate and utilize an Excel formola to compute a
payment (Hint: -PMT). Khaleesi is a high paying client of your firm, so the schedule should be
extremely organized and professional looking (s.e, include a title, date, description in the top left-
hand corner and utilize borders, font effects, etc. in cells to make the worksheet look visually
appealing). On the second tab of the Excel workbook, prepare a lease amortization schedule
describing the pattern of payments and interest over the lease term for both Khaleesi and the
North. On the third tab of the Excel workbook, prepare the appropriate entries for both Khaleesi
and the North on 1/1/18. Numbers in journal entries should be linked to the table prepared in
number 3 above or computed using Excel formulas. On the fourth tab of the Excel workbook,
prepare the appropriate entries for both Khaleesi and the North through 1/1/21. Numbers in
journal entries should be linked to the table prepared in number 3 above or computed using Excel
formulas 2. 3. 4. 5. On the fifth tab, prepare the appropriate entries for both Khaleesi and the
North on 12/31/2028. Assume the weapons returned to Khaleesi have an actual resi.
IOError [Errno 2] No such file or directory states.xmlAny su.pdf
IOError: [Errno 2] No such file or directory: \'states.xml\'
Any suggestions on how I can fix this? By the way, it\'s in Python. Any help will be appreciated!
Solution
There can be several reasons for the given error.
1) states.xml is not present on yur system to be read by the python code
2) states.xml file is present in your system but cannot be accessed by the python code.
make sure you have give the correct path of file in the python code or place the file in same
destination as the python code i.e The file may be existing but may have a different path. Try
writing the absolute path for the file.
When the file is properly read, it will be parsed using the parse function and the contents will be
stored in tree.
Normal Labour/ Stages of Labour/ Mechanism of Labour
Normal labor is also termed spontaneous labor, defined as the natural physiological process through which the fetus, placenta, and membranes are expelled from the uterus through the birth canal at term (37 to 42 weeks
PCOS corelations and management through Ayurveda.
This presentation includes basic of PCOS their pathology and treatment and also Ayurveda correlation of PCOS and Ayurvedic line of treatment mentioned in classics.
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Brief Concepts and Questions EXAM 2 Chapter 8 DNA RNA Protein What i.pdf
Brief Concepts and Questions EXAM 2 Chapter 8: DNA RNA Protein What is DNA? a
phosphate Structure of DNA: Building blocks are called nucleotides Each nucleotide is
composed of three br uithofenas bee. What makes DNA so special? Provide 4 reasons, below
DNA DNA (Replication): Where does DNA replication take place? When does DNA replication
take place? Explain steps involved in DNA replication: DNA RNA Protein (Gene Expression)
Involves 2 processes: 1. Transcription 2. Translation Explain the Synthesis of Proteins (Gene
Expression): o DNA RNA Protein What is RNA? What is \"codon What is \"anticodon\" What is
a protein molecule? DNA mutation; Change in nucleotide bases of DNA Duplex Point mutation
Frame shift mutation
Solution
Question
Answer
Where does DNA replication take place:
It takes place in the nucleus in case of eukaryotic cells and in the cytoplasm in case of
prokaryotic cells
When does DNA replication take place:
DNA replication occurs during the S-phase during cell cycle, so that cell can make an extra copy
of genetic material.
Explain steps involved in DNA replication:
Initiation: During initiation, the proteins will bind to the origin of replication; helicase unwinds
the DNA helix which results in the formation of two replication forks.
Elongation: A RNA primer sequence will be added to this the DNA pol III will add the
nucleotides in 5’ to 3’ direction and chain will elongate.
Termination: In case of bacteria, termination of replication occurs whenever two replication
forks meet each other from the opposite end of the parental chromosome.
Transcription
Gene expression first step is transcription, here a particular segment of DNA will be copied into
RNA with the help of the enzyme RNA polymerase
Translation
Translation is the final step of the gene expression. Here mRNA will be used to synthesize the
polypeptide chain. The information present in the mRNA in the form of codon will code for the
amino acids needed for polypeptide chain synthesis.
What is RNA?
RNA is ribonucleic acid and is found in all living cells. It acts as the messenger carrying
instructions from DNA for the synthesis of proteins.
Few viruses will have RNA as their genetic material.
What is codon?
Codon is a sequence of three nucleotides and they together form a unit of genetic code in either
DNA or RNA.
What is anticodon?
It is found on tRNA and it is a sequence of three nucleotides which forms a genetic code on
tRNA, and these anticodon is complementary to the codons found on messenger RNA.
What is a protein molecule?
During translation, when amino acids are added in a sequential manner, the condensation of
amino acids will form a peptide bond in between them and finally forms a polypeptide chain. It
is the DNA through mRNA directs the protein synthesis.
Point mutation
In point mutation, only one or very few nucleotides will be affected or mutated in a gene
sequence.
Frame shift mutation
Either insertions or deletion can result in frame shift mutation, due to th.
A population has two alleles for a particular gene, A and a, and is .pdf
A population has two alleles for a particular gene, A and a, and is in Hardy-Weinberg
equilibrium. The frequency of allele a is 0.2. What is the frequency of individuals with AA
genotype?
0.2
0.4
0.64
0.04
0.8
Solution
C). 0.64
Hardy and Weinberg mathematically proved that in a population, all dominant and recessive
alleles comprise all alleles for that gene.
This was mathematically represented as p+ q = 1.0
Where,
p = frequency of dominant alleles
q = frequency of recessive alleles.
Given that, the frequency of a = 0.2
As the population is in H-W equilibrium, the frequency of A = 1-0.2 = 0.8
The frequency of AA genotype = 0.8*0.8 = 0.64.
A 12-year-old male complained of strange odors before loss of consci.pdf
A 12-year-old male complained of strange odors before loss of consciousness and a major tonic-
clonic seizure. During the seizure, his parents rushed him to the emergency room of a local
hospital. On arrival at the hospital, he appeared to be asleep. Studies at the hospital showed
routine laboratory work within normal limits (WNL), lumbar puncture (CSF) was WNL, no
evidence of skull fracture on x-ray study was revealed, and an electroencephalograph (EEG)
showed no abnor malities. How would you interpret the episode and findings?
Solution
Charecterized by sudden loss of consciousness ,chaotic muscle movement and tone, and apnea
Tonic phase:Bilateral muscle rigidity
Clonic phase: muslce contraction and relaxation lasting 1 to 3 minutes
Signs:
Tachycardia, hyerpventilation,sweating, and intense salivation
Last 3-5 minutes
postictal state(5 to 20 mintues)follows
-gradual return to consciousness.
2.Which of the following statements about evolution is falseNonrand.pdf
2.Which of the following statements about evolution is false?Nonrandom mating can change the
genetic makeup of populations over time.Genetic drift can change the genetic makeup of
populations over time.Natural selection is one of several mechanisms of evolution.Populations,
not individuals, evolve.Evolution is goal-directed.2.Which of the following statements about
evolution is false?Nonrandom mating can change the genetic makeup of populations over
time.Genetic drift can change the genetic makeup of populations over time.Natural selection is
one of several mechanisms of evolution.Populations, not individuals, evolve.Evolution is goal-
directed.
Solution
Note:.
can someone explain in a easy way chromosome walking I know its w.pdf
can someone explain in a easy way chromosome walking? I know it\'s with overlapping
fragments and a gen nearby the gen you want to find, but I don\'t clearly understand it, I have
images in my book, but an image with steps and explanation would be nice,
Solution
Chromosome Walking:
Chromosome walking is positional cloning of gene sequences or upstream promoter sequences.
Eukaryotic DNA contains a lot of junk DNA between the genes. The technique relies upon
screening of gDNA library for overlapping fragments in a sequential manner to tap the upstream
or downstream region.
Procedure:
1. Isolation of Genomic DNA: First of all isolate genomic DNA (gDNA) of the test organism
and then partially digest it with some restriction enzyme, say Hind III that would yield unequal
random fragments.
2. Cloning of the Digested into E.coli: Clone the digested gDNA into a cloning vector (between
the Hind III sites in this case) to prepare a gDNA library of the organism. Each colony of the
library would have only one clone.
3. Preparation of marker probe: Prepare a radioactively labelled probe from the neighboring gene
(say A), from where you want to start walking.
4. Colony Hybridization: Hybridize the probe with your gDNA library. The probe would show
the signal in the colonies carrying DNA in and around your gene.
5. Sequencing: Pick the colonies showing the signal and isolate the plasmid DNA and sequence
each using the plasmid-specific primers. You would get only one sequence per bacterial colony
of your library.
The sequence, thus obtained must be analyzed. Select longer sequences towards the direction of
your walk.
6. Preparation of the probe: Prepare new probes from the region towards your target gene.
7. Repeat the steps 4-6 till you get the clone and sequence of your target gene..
9. When a bacteriophage inserts its DNA into a bacterial chromosome,.pdf
9. When a bacteriophage inserts its DNA into a bacterial chromosome, the inserted
bacteriophage DNA is called _____
(a) prophage
(b) lytogenic
(c) virulent
Solution
a.Prophage.
The genetic material of a bacteriophage ,incorporated in to the genome of bacterium or host ,now
it is ready to prouce phages if specifically activated.
This stage of Bacteriophage life cycle called as Prophage..
Write an informal negation for each of the following statements. Be .pdf
Write an informal negation for each of the following statements. Be careful to avoid negations
that are ambiguous. a. All dogs are friendly. b. All people are happy. c. Some suspicions were
substantiated. d. Some estimates are accurate.
Solution
(a)
these is a dog which is not friendly
(b)
Some people are not happy
(c)
these was a ssuspicions which was not substantiated
(d)
these is some estimates which are wrong.
Write a C++ program which generates and displays a random walk acros.pdf
Write a C++ program which generates and displays a random walk across a grid. A 2-
dimensional array of characters will be used for this grid. The cells visited by the walk will be
labeled by capital letters with A being the first cell, B the cell moved to from A, and so on,
ending either when Z is reached or when it is not possible to move to another cell.
The following libraries, constant definition, and type definition are needed in the program:
#include
#include // for srand and rand
#include // for time
using namespace std;
const int SIZE = 10;
typedef char Grid[SIZE][SIZE]
The grid must be set to have all its elements be the character ’.’ before the walk is begun.
The program must contain the following function prototypes, the main function, and the function
definitions for all the other functions:
// Function prototypes
bool can_move_up(int i, int j, Grid walk);
bool can_move_down(int i, int j, Grid walk);
bool can_move_left(int i, int j, Grid walk);
bool can_move_right(int i, int j, Grid walk);
void init_array(Grid walk);
void generate_random_walk(Grid walk);
void print_array(Grid walk);
int main(void)
{
Grid walk; // the grid in which the random walk occurs
srand((unsigned) time(NULL));
init_array(walk);
generate_random_walk(walk);
print_array(walk);
return 0;
}
The four functions can_move _* must return whether or not it is possible to move in the
particular direction in the grid. For example, a function call like if (can_move_up(i, j, walk))
could be used to determine whether or not it is possible to move up. The rule for a move being
possible is that if it doesn’t go outside of the grid and isn’t already part of the walk, then that
move is possible. The function init_array must set all characters in the grid to ’.’
The function generate_random_walk must modify the cells of grid by changing ones that are
part of the walk to letters.
The function print_array must print the resulting grid in the form shown in the sample run
below.
In the function generate_random_walk, at each step of the walk, randomly select one of the 4
directions and try to move in that direction. If that move isn’t available, try the other moves in
turn until one is possible. If all 4 directions are blocked, the walk must terminate, and the
program must continue and print out the walk. To select one of the 4 directions randomly, use
the expression: rand() % 4
The function rand returns an integer, and finding the remainder of this integer divided by 4
results in getting one of the numbers 0, 1, 2, or 3. Use these 4 numbers to represent the 4
directions up, down, left and right.
The call to the function srand from the main function seeds the pseudo-random number
generator used by rand so that different sequences of numbers will be generated by calls to rand
each time the program is run.
Solution
#include //input output functions in C++
#include //rand() and srand() functions
#include //time() function for creating seed
#include
#include // for srand and rand
#inclu.
true or false1- The grouped frequency table should ideally have.pdf
true or false???
1- The grouped frequency table should ideally have 10 class intervals
2- The width of the intervals should be half of the number of scores
3- Each score can be represented in one and only one interval
4- The intervals should end with the highest number in the sample
Solution
1.True
2.False
3.True
4.False.
Answer the questions that follow the VHDL code given below. LIBRARY .pdf
Answer the questions that follow the VHDL code given below. LIBRARY ieee; USE
ieee.std_logic_1164.ALL; ENTITY ex1_IS PORT (n, cp, n rd: IN st_logic; q: BUFFER integer
RANGE 0 TO 15); END ex1; ARCHITECTURE are OF ex1 IS BEGIN PROCESS (n_cp,
n_rd) BEGIN IF (n_rd = \'0\') THEN q
Solution
a)n_rd- used to reset the counter
n_cp-used as clock to the counter
b)4 binary bits are there in q;
Q stores 0 t0 15 number so
log2(16)=4.
The weak D phenotype is thought to arise from several mechanisms. .pdf
The weak D phenotype is thought to arise from several mechanisms. Which one of the following
statements is false?
C gene is in trans position to D gene, this results in a weakening influence on the D gene,
examples CDe/Cde and cDe/Cde.
Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in
caucasians. Also occurs in the Ro genotype in certain racial groups more commonly.
The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these pieces
is missing. The red cells are said to have a partial D.
D deletion genes gives rise to weakened expression of D antigen.
Rh null cells lack:
MNSs antigens.
Lewis antigens.
Normal oxygen-carrying capacity.
Rh antigens.
There are two closely related and closely linked genes.
Enzyme techniques, in conjunction with an IgM anti-D.
Increasing the time for the incubation of the saline phase.
Performance of an IAT using an IgG anti-D.
Don\'t need to do it in a blood bank.
C gene is in trans position to D gene, this results in a weakening influence on the D gene,
examples CDe/Cde and cDe/Cde.
Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in
caucasians. Also occurs in the Ro genotype in certain racial groups more commonly.
The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these
pieces is missing. The red cells are said to have a partial D.
D deletion genes gives rise to weakened expression of D antigen.
Solution
The weak D phenotype is thought to arise from several mechanisms. Which one of the following
statements is false?
D deletion genes gives rise to weakened expression of D antigen.
Rh null cells lack:
Rh antigens.
There are two closely related and closely linked genes.
Enzyme techniques, in conjunction with an IgM anti-D.
Increasing the time for the incubation of the saline phase.
Performance of an IAT using an IgG anti-D.
Don\'t need to do it in a blood bank.
Last one is an incomplete question
D deletion genes gives rise to weakened expression of D antigen..
The coding region of a BACTERIAL gene has the following sequence A.pdf
The coding region of a BACTERIAL gene has the following sequence:
\"AGAGCCAACATGTCGAGCATATGCTGATATACG\' Draw and label the DNA (including
ALLEABIS of the gene, sequence information) The info you need is in the sides. Label the
template/coding strands Label 5\' and 3\' ends of each DNA strand Label where RNA
polymerase binds \"Transcribe\" this gene into mRNA, as shown in Bb video. Write the mRNA
nucleotide sequence, starting at the correct nucleotide, starting at the correct nucleotide Show
the 5\' and 3\' ends of the RNA. Use the codon chart to \"translate\" the mRNA into a protein,
and write the correct amino acid sequence of the protein. Only translate the part of the gene that
a cell would translate. Label the N-terminus and C-terminus of the protein. If the sequence of
the gene were changed to: AGACCCAACATGTCGGACATATGCTGATATACG what would
happen to the properties of the protein? Explain why write out the entire process of bacterial
transcription step by step in the correct order, in detail.
Solution
4. Process of Bacterial Transcription:
Bacterial transcription is the process in which mRNA transcripts of genetic material in bacteria
are produced; it is further translated to produce proteins. Unlike in eukaryotes, bacterial
transcription and translation can occur simultaneously in the cytoplasm.
RNA polymerases are enzymes that transcribe DNA into RNA. Using a DNA template, RNA
polymerase builds a new RNA molecule through base pairing. For instance, if there is a G in the
DNA template, RNA polymerase will add a C to the new, growing RNA strand.
Three steps are involved in bacterial transcription:
Initiation: To begin transcribing a gene, RNA polymerase binds to the DNA of the gene at a
region called the promoter. Basically, the promoter tells the polymerase where to locate on the
DNA and begin transcribing.
Elongation:
Once RNA polymerase is in position at the promoter, the next step of
transcription—elongation—can begin. Basically, elongation is the stage when the RNA strand
gets longer, thanks to the addition of new nucleotides.
During elongation, RNA polymerase \"walks\" along one strand of DNA, known as the template
strand, in the 3\' to 5\' direction. For each nucleotide in the template, RNA polymerase adds a
matching (complementary) RNA nucleotide to the 3\' end of the RNA strand.
Termination:
RNA polymerase will keep transcribing until it gets signals to stop. The process of ending
transcription is called termination, and it happens once the polymerase transcribes a sequence of
DNA known as a terminator.
There are two major termination strategies found in bacteria: Rho-dependent and Rho-
independent.
In Rho-dependent termination, the RNA contains a binding site for a protein called Rho factor.
Rho factor binds to this sequence and starts \"climbing\" up the transcript towards RNA
polymerase.
When it catches up with the polymerase at the transcription bubble, Rho pulls the RNA transcript
and the template DNA strand apart, releasing .
Regarding the ABO blood group system in humans, if an individual is g.pdf
Regarding the ABO blood group system in humans, if an individual is genetically I^B i and yet
expresses the O blood type, which genotype is he or she likely to have? Hh HH hh i^A i i^B i
none of the above are correct.
Solution
Option C is the correct answer. This hh is also known as Bombay blood group. These individuals
do not express H antigen. Thus these individuals are unable to form either of A or B antigen on
their RBCs. Usual test of these individuals for blood group shows them as having O blood group..
Peptidoglycan is a unique macromolecule found in cell walls of what k.pdf
Peptidoglycan is a unique macromolecule found in cell walls of what kind of organism archea
yeast algae bacteria All of the above According to the fossil record, when did eukaryotes first
appear on Earth? 1 million years ago
Solution
1) In bacteria ,cell wall is composed of peptidoglycan
2) Eukaryotes appear 2.1 million years ago.
On this photo you see skull of Tasmanian Devil - now extinct only big.pdf
On this photo you see skull of Tasmanian Devil - now extinct only big marsupial predator from
Australia. Could you recognize bones on this skull and name them?
Solution
1. Surangular
2. Postorbital
3. Parietal
4. Frontal
5. Prefrontal
6. Nasal
7. Premaxilla
8. Margins of nares
9. Incisor
10. socket
11 Canine
12 Dentary
13 Mandible
14. Lacrimal
15 Jugal.
MonthMonthly Returns ()MonthMonthly Returns ()MonthMonth.pdf
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Dec-07
-13.241
-0.219
-0.692
Dec-06
-5.442
-5.018
1.403
Dec-05
-2.610
2.457
0.036
Nov-07
-13.794
-1.980
-4.182
Nov-06
2.544
1.767
1.899
Nov-05
8.818
8.806
3.778
Oct-07
15.861
24.632
1.590
Oct-06
4.193
18.534
3.257
Oct-05
9.249
17.595
-1.668
Sep-07
18.093
10.096
3.736
Sep-06
-12.314
6.174
2.575
Sep-05
1.561
10.650
0.810
Aug-07
-2.237
1.029
1.497
Aug-06
6.227
-2.087
2.376
Aug-05
-0.060
-0.612
-0.912
Jul-07
-14.302
-2.430
-3.097
Jul-06
-17.758
-7.805
0.616
Jul-05
-3.781
-2.172
3.717
Jun-07
-5.470
4.979
-1.660
Jun-06
4.463
12.778
0.133
Jun-05
-6.855
6.088
0.143
May-07
2.354
5.628
3.486
May-06
-3.630
-11.035
-2.875
May-05
7.826
26.032
3.179
Apr-07
-10.387
2.885
4.428
Apr-06
1.612
7.164
1.342
Apr-05
1.770
21.877
-1.896
Mar-07
1.393
1.938
1.116
Mar-06
0.624
7.551
1.244
Mar-05
5.051
-3.979
-1.769
Feb-07
9.007
-10.379
-1.950
Feb-06
-6.735
-16.188
0.271
Feb-05
-8.350
-3.900
2.103
Jan-07
10.846
8.908
1.511
Jan-06
-12.264
4.291
2.648
Jan-05
-6.555
1.468
-2.437
Question: (Use table above to solve)
Segregate the return data in the above table into bull and bear markets, using the S&P 500 index
as the proxy for the market.
A.) What was the average return for the two companies during these two types of markets? What
was the return for the S&P 500 index during those times?
B.) Using the average return measures in a), compute the beta for Yahoo and Google.
C.) According to these beta estimates, which company is riskier?
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Month
Monthly Returns (%)
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Ended
Yahoo
Google
S&P500
Dec-07
-13.241
-0.219
-0.692
Dec-06
-5.442
-5.018
1.403
Dec-05
-2.610
2.457
0.036
Nov-07
-13.794
-1.980
-4.182
Nov-06
2.544
1.767
1.899
Nov-05
8.818
8.806
3.778
Oct-07
15.861
24.632
1.590
Oct-06
4.193
18.534
3.257
Oct-05
9.249
17.595
-1.668
Sep-07
18.093
10.096
3.736
Sep-06
-12.314
6.174
2.575
Sep-05
1.561
10.650
0.810
Aug-07
-2.237
1.029
1.497
Aug-06
6.227
-2.087
2.376
Aug-05
-0.060
-0.612
-0.912
Jul-07
-14.302
-2.430
-3.097
Jul-06
-17.758
-7.805
0.616
Jul-05
-3.781
-2.172
3.717
Jun-07
-5.470
4.979
-1.660
Jun-06
4.463
12.778
0.133
Jun-05
-6.855
6.088
0.143
May-07
2.354
5.628
3.486
May-06
-3.630
-11.035
-2.875
May-05
7.826
26.032
3.179
Apr-07
-10.387
2.885
4.428
Apr-06
1.612
7.164
1.342
Apr-05
1.770
21.877
-1.896
Mar-07
1.393
1.938
1.116
Mar-06
0.624
7.551
1.244
Mar-05
5.051
-3.979
-1.769
Feb-07
9.007
-10.379
-1.950
Feb-06
-6.735
-16.188
0.271
Feb-05
-8.350
-3.900
2.103
Jan-07
10.846
8.908
1.511
Jan-06
-12.264
4.291
2.648
Jan-05
-6.555
1.468
-2.437
Solution
According to Beta estimates Google is more risky than Yahoo because the beta is higher for
google( 1.255) with Yahoo(0.987)ABCMonthMonthly Returns (%)Using excel Average return
=AVERAGE(A1:A36)Using excel Average return =AVERAGE(B1:B36)Using excel Average
return =AVERAGE(C1:C36)Beta of.
Many DOD components are accomplishing more training in the simulator.pdf
Many DOD components are accomplishing more training in the simulator instead of the aircraft
due to limited financial resources and the benefit of reduced wear and tear on aging aircraft
fleets. Commercial Airlines do much of their initial training and their recurring training in the
simulator. Where do you stand on the premise of supplementing actual flight training with time
in the simulator? Be sure to identify the positive and negative aspects of replacing actual flight
time with simulator time. Please Cite Sources
Solution
Heads of the DoD Components shall establish procedures for transmission and transportation of
classified information and information-bearing material that minimize risk of compromise while
permitting use of the most cost-effective transmission or transportation means.COMSEC
information shall be transmitted and transported in accordance with National
Telecommunications and Information Systems Security Instruction 4001.NATO classified
information shall be transmitted in compliance with USSAN Instruction 1-69. Except under rules
established by the Secretary of Defense, or as provided by Section 102 of the National Security
Act, classified information originating in a department or agency other than the Department of
Defense shall not be disseminated outside the Department of Defense without the consent of the
originating department or agency.
When classified information is transmitted, it shall be enclosed in two opaque, sealed envelopes,
wrappings, or containers, durable enough to properly protect the material from accidental
exposure and facilitate detection of tampering. The following exceptions apply:
(1) If the classified material is an internal component of a packageable item of equipment, the
outside shell or body may be considered as the inner enclosure provided it does not reveal
classified information.
(2) If the classified material is an inaccessible internal component of a bulky item of equipment,
the outside or body of the item may be considered to be a sufficient enclosure provided
observation of it does not reveal classified information.
(3). If the classified material is an item or equipment that is not reasonably packageable and the
shell or body is classified, it shall be concealed with an opaque covering that will hide all
classified features.
(4) Specialized shipping containers, including closed cargo transporters, may be considered the
outer wrapping or cover when used.
(5) When classified material is hand-carried outside an activity, a locked briefcase may serve as
the outer wrapper. In such cases, the addressing requirements of paragraph 7-201a., below, do
not apply.
(6) NATO Restricted material need not be double-wrapped when transmitted within the United
States. The marking \"NATO Restricted\" shall not appear on the wrapper
Classified material shall be prepared for shipment, packaged, and sealed in ways that minimize
risk of accidental exposure or undetected deliberate compromise. Documents shoul.
Match the numbered terms to the descriptions that follow. Choose all .pdf
Match the numbered terms to the descriptions that follow. Choose all appropriate terms, but
only appropriate terms. autotroph heterotroph phototroph chemotroph a prokaryote that
obtains both energy and carbon as it decomposes dead organisms 1 only 4 only 1 and 3 2 and
4 1, 3, and 4 which of the following traits do archaea and bacteria share? composition of the
cell wall presence of plasma membrane lack of a nuclear envelope identical rRNA sequences 1
only 3 only 1 and 3 2 and 3 2 and 4 Which prokaryotes should be expected to be most
strongly resistant to plasmolysis in hypertonic environment? extreme halophiles extreme
thermophiles methanogens cyanobacteria nitrogen-fixing bacteria that live in root nodules The
group diplomonadida is characterized by a mouthlike gullet, and a hairlike surface: Paramecium
is an example. flagella and a lack of mitochondria \"Gilardia is an example. nonmotility,
parasitism and sporelike infective stages: Toxoplasma is an example. switching between
autotrphic and heterotrphic lifestyles: Euglena is an example large protein deposites and
movements by two falgella which a part of an undulating membrane: Trypanosoma is an
example Plasmodium is transmitted to humans by the bite of mosquito (Anopheles) Its lifecycle
is characterized by spores, gametes and cysts that can hide in human cells. This infective protist
belongs to the group: Apicomplexa Oomycota Ciliophora Archaeplastida Dinoflagellata
Solution
Answer:
42). D). 2 and 4
Chemotrophs are organisms that obtain energy by the oxidation of electron donors in their
environments. These molecules can be organic (chemoorganotrophs) or inorganic
(chemolithotrophs). The chemotroph designation is in contrast to phototrophs, which utilize solar
energy.
43). D). 2 and 3
Archaea and bacteria are both prokaryotes, meaning they do not have a nucleus and lack
membrane-bound organelles. They are tiny, single-cell organisms called microbes, which cannot
be seen by the naked human eye.
44). A) extreme halophiles
45). B). Flagella and a lack of mithochondria. Giardia is an example
The diplomonads are a group of flagellates, most of which are parasitic. They include Giardia
lamblia, which causes giardiasis in humans. They are placed among the metamonads, and appear
to be particularly close relatives of the retortamonads.
46). A). Apicomplexa
The Apicomplexa are a diverse group that includes organisms such as the coccidia, gregarines,
piroplasms, haemogregarines, and plasmodia. Diseases caused by Apicomplexa include:
Babesiosis (Babesia) Malaria (Plasmodium) Cryptosporidiosis (Cryptosporidium parvum).
Khaleesi Corporation leases dragonelass weapons to customers. She gai.pdf
Khaleesi Corporation leases dragonelass weapons to customers. She gains a loyal following of
customers because dragonglass is of limited supply but high demand due to its usefulness for the
upcoming winter. Further, Khaleesi tends to offer more favorable financing terms than
competitors offering substitutive products (e.g., Lannister, LLC) Khaleesi was recently
approached by the Lords of the North, Inc. (the North\") which is interested in leasing a
substantial stock of weapons over a potentially lengthy geriod.of time. The North has indicated a
willingness to pay any rate that Khaleesi Corporation demands for its dragcaglass products, but
generally receives an interest rate of 12% on all other borrowing transactions. The North\'s
management are a very noble group, so payments are reasonably assured. Further, there are no
material cost ncertainties. Khaleesi has gathered its council to discuss entering into such a
contract with the North and has invited you to provide financial council. Khaleesi\'s board has
proposed several alternative sets of lease terms (below) and would like you determine what the
North\'s annual payments will be under each scenario, if payments are made at the beginning of
the period $489,450 Fair value of weapons to be leased Lease Ternm Useful Life of leased assets
Lessee/Lessor interest rate Residual Value (guaranteed) 1. On the first tab of your Excel
workbook, prepare a schedule for Khaleesi which uses Excel to $489,450 11 years 13 years 696
So $489,450 11 years 13 years 6% $24,350 years 13 years 896 $24,350 compute the North\'s
necessary annual payment under each of the scenarios listed in the chart above. The schedule
should list for each scenario above the desired present value of lease payments, future value of
the leased asset, the lease term, and the interest rate and utilize an Excel formola to compute a
payment (Hint: -PMT). Khaleesi is a high paying client of your firm, so the schedule should be
extremely organized and professional looking (s.e, include a title, date, description in the top left-
hand corner and utilize borders, font effects, etc. in cells to make the worksheet look visually
appealing). On the second tab of the Excel workbook, prepare a lease amortization schedule
describing the pattern of payments and interest over the lease term for both Khaleesi and the
North. On the third tab of the Excel workbook, prepare the appropriate entries for both Khaleesi
and the North on 1/1/18. Numbers in journal entries should be linked to the table prepared in
number 3 above or computed using Excel formulas. On the fourth tab of the Excel workbook,
prepare the appropriate entries for both Khaleesi and the North through 1/1/21. Numbers in
journal entries should be linked to the table prepared in number 3 above or computed using Excel
formulas 2. 3. 4. 5. On the fifth tab, prepare the appropriate entries for both Khaleesi and the
North on 12/31/2028. Assume the weapons returned to Khaleesi have an actual resi.
IOError [Errno 2] No such file or directory states.xmlAny su.pdf
IOError: [Errno 2] No such file or directory: \'states.xml\'
Any suggestions on how I can fix this? By the way, it\'s in Python. Any help will be appreciated!
Solution
There can be several reasons for the given error.
1) states.xml is not present on yur system to be read by the python code
2) states.xml file is present in your system but cannot be accessed by the python code.
make sure you have give the correct path of file in the python code or place the file in same
destination as the python code i.e The file may be existing but may have a different path. Try
writing the absolute path for the file.
When the file is properly read, it will be parsed using the parse function and the contents will be
stored in tree.
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