which event in this life cycle is comparable to fertilizarion .pdf
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which event in this life cycle is comparable to fertilizarion? 9. Distinguish among the various groups of fungi as outlined in the table below: Asexual Spores Fruiting Body Structure (microscopic) Giving sexual stage) Rise to Sexually Produced Spore (large; asexually produced spores from sporangium. Zygomycota Ascomycota Basidiomycota Solution Structure Fruiting body Asexual spores Zygomycetes Chitin,chitosan,polyglucornic acid No fleshy fruiting body reproduce by conjugation, sporangiospores, conidia Ascomycetes Chitin,both alpha and beta Glucans,glactomannoproteins ascus asexually produced spores from sporangium, ascospores, conidia Basisiomycetes Chitin,beta Glucans basidia contain basidium, the spore forming bodies, basidiospores,conidia.
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Foofy obtained a significant Fobt from an experiment with five level.pdf
Foofy obtained a significant Fobt from an experiment with five levels. She therefore concludes
that changing every condition of the independent variable results in a significant change in the
dependent variable. (a) Is she correct? Why or why not? (b) What must she now do?
Solution
She is not correct because changing every condition of the independent variable results in a
significant change in the dependent variable suggests that at least one independent variable is
significantly different from others and null hypothesis is rejected. She should do post-hoc test
now..
Explain how water interacts with the following types of compounds to.pdf
Explain how water interacts with the following types of compounds to influence their water
solubility.
1.Compounds that can hydrogen bond
2. Compounds that are polar but cannot hydrogen bond
3. Compounds that are non-polar
4. Compounds that dissociate into cations and anions.
Thank you so much.
Solution
Compounds that can hydrogen bond
Water tends to disassociate into H+ and OH- ions. In this disassociation, the oxygen retains the
electrons and only one of the hydrogens, becoming a negatively charged ion known as
hydroxide. Pure water has the same number (or concentration) of H+ as OH- ions. Acidic
solutions have more H+ ions than OH- ions. Basic solutions have the opposite. An acid causes an
increase in the numbers of H+ ions and a base causes an increase in the numbers of OH- ions.
Water, which not only dissolves many compounds but also dissolves more substances than any
other liquid, is considered the universal solvent. A polar molecule with partially-positive and
negative charges, it readily dissolves ions and polar molecules. Water is therefore referred to as a
solvent: a substance capable of dissolving other polar molecules and ionic compounds.
The charges associated with these molecules form hydrogen bonds with water, surrounding the
particle with water molecules. This is referred to as a sphere of hydration, or a hydration shell,
and serves to keep the particles separated or dispersed in the water.
Compounds that dissociate into cations and anions.-When ionic compounds are added to water,
individual ions interact with the polar regions of the water molecules during the dissociation
process, disrupting their ionic bonds. Dissociation occurs when atoms or groups of atoms break
off from molecules and form ions. Consider table salt (NaCl, or sodium chloride): when NaCl
crystals are added to water, the molecules of NaCl dissociate into Na+ and Cl– ions, and spheres
of hydration form around the ions. The positively-charged sodium ion is surrounded by the
partially-negative charge of the water molecule\'s oxygen; the negatively-charged chloride ion is
surrounded by the partially-positive charge of the hydrogen in the water molecule.
Compounds that are polar and non-polar-
-Since many biomolecules are either polar or charged, water readily dissolves these hydrophilic
compounds. Water is a poor solvent, however, for hydrophobic molecules such as lipids.
Nonpolar molecules experience hydrophobic interactions in water: the water changes its
hydrogen bonding patterns around the hydrophobic molecules to produce a cage-like structure
called a clathrate. This change in the hydrogen-bonding pattern of the water solvent causes the
system\'s overall entropy to greatly decrease, as the molecules become more ordered than in
liquid water. Thermodynamically, such a large decrease in entropy is not spontaneous, and the
hydrophobic molecule will not dissolve.
-Polar substances tend to dissolve well in other polar substances, but not nonpolar substances,
whil.
Explain how the financial reporting of gains and losses on assets.pdf
Explain how the financial reporting of gains and losses on assets differs from the financial
reporting of unrealized gains and losses on martketable securities.
Explain how the financial reporting of gains and losses on assets differs from the financial
reporting of unrealized gains and losses on martketable securities.
Solution
Particulars Gains and Losses on Assets unrealized gains and losses on martketable securities
Actual or Not Actual It is actual or real in nature. It is differrence between the values of
securities Income Tax It is taxable / deductible under income tax It is not considered under
income tax Cash flow There is cash inflow There is no any cash flow Realisation It is realised
gain or loss It may be realised or not. It is unrealised.
Explain two ways in which microbiologists have overcome the difficul.pdf
Explain two ways in which microbiologists have overcome the difficulty of viewing microbes in
wet preparations
Solution
First we have to find two ways in which microbiologists have overcome the difficulty of viewing
microbes in wet preparations are a bacteria which has the intrinsic ability of movement in the
surrounding medium in which it remains suspended is a motile bacteria.
A bacteria which does not have the intrinsic ability of movement in the surrounding medium in
which it remains suspended is a non-motile bacteria.
In wet mount though the shape and size of bacteria can be observed motility may be hampered as
the suspension is pressed between the slide and the cover slip.
Motility test is performed for clear observation of the motility of bacteria, besides their shape and
size.
The hanging drop is observed under a microscope using oil immersion objective its cells can be
seen to have erratic movement in the surrounding medium.
If it is non motile its cells remain static in the medium without any movement or may show
brownian movement resulting from the bombardment by the water molecules in the medium..
During the preparation of the bank reconciliation for Building Conce.pdf
During the preparation of the bank reconciliation for Building Concepts Co., Joel Kimmel, the
assistant controller, discovered that Lone Peak National Bank incorrectly recorded a $3,290
check written by Building Concepts Co. as $329. Joel has decided not to notify the bank but wait
for the bank to detect the error. Joel plans to record the $2,961 error as Other Income if the bank
fails to detect the error within the next three months.
Discuss whether Joel is behaving in a professional manner.
Solution
Joel is not behaving in a professional manner,
Joel record differece amount ($2961) as other income, this treatment is wrong
because that amount can not be include in income of Building concepts Co..
C++ CodeConsider the LinkedList class and the Node class that we s.pdf
C++ Code
Consider the LinkedList class and the Node class that we saw in lecture. The LinkedList class
basically represents a singly linked list.
Now, for this lab, consider a doubly linked list, i.e. each node has a next and previous pointers to
the next and previous node, respectively.
Modify the Node class to reflect this change and implement the following methods for
DoublyLinkedList class:
addFirst
addLast
addAtIndex
removeFirst
removeLast
removeAtIndex
Note: We saw how to implement these methods for a singly linked list and the code is added as
an attachment.
C++ Code
*****************************************************************************
******************
*****************************************************************************
******************
C++ Code
Solution
#include
#include
using namespace std;
template
class Node
{
public:
T element; // Element contained in the node
Node *next; // Pointer to the next node
Node *previous;
Node(T element) // Constructor
{
this->element = element;
next = NULL;
}
};
template
class LinkedList
{
public:
LinkedList();
void addFirst(T element);
void addLast(T element);
T removeFirst() ;
T removeLast();
void addAtIndex(int index, T element);
T removeAtIndex(int index);
void traverseList();
private:
Node *head;
Node *tail;
int size;
};
template
LinkedList::LinkedList()
{
head = NULL;
tail = NULL;
size = 0;
}
template
void LinkedList::addFirst(T element)
{
Node *newNode = new Node(element);
newNode->next = head;
if(head != NULL) {
head->previous = newNode;
}
newNode->previous = NULL;
head = newNode;
size++;
if (tail == NULL)
tail = head;
}
template
void LinkedList::addLast(T element)
{
if (tail == NULL)
{
head = tail = new Node(element);
}
else {
tail->next = new Node(element);
tail = tail->next;
}
size++;
}
template
void LinkedList::addAtIndex(int index, T element)
{
if (index == 0)
addFirst(element);
else if (index >= size)
addLast(element);
else
{
Node *current = head;
for (int i = 1; i < index; i++)
current = current->next;
Node *temp = current->next;
current->next = new Node(element);
(current->next)->next = temp;
size++;
}
}
template
T LinkedList::removeFirst()
{
if (size == 0)
exit(1);
else
{
Node *temp = head;
head = head->next;
size--;
T element = temp->element;
delete temp;
return element;
}
}
template
T LinkedList::removeLast()
{
if (size == 0)
exit(1);
else if (size == 1)
{
Node *temp = head;
head = tail = NULL;
size = 0;
T element = temp->element;
delete temp;
return element;
}
else
{
Node *current = head;
for (int i = 0; i < size - 2; i++)
current = current->next;
Node *temp = tail;
tail = current;
tail->next = NULL;
size--;
T element = temp->element;
delete temp;
return element;
}
}
template
T LinkedList::removeAtIndex(int index)
{
if (index < 0 || index >= size)
exit(1);
else if (index == 0)
return removeFirst();
else if (index == size - 1)
return removeLast();
else {
Node *previous = head;
for (int i = 1; i < index; i++)
{
previous = previous->next;
}
Node *current = previous-.
DNA Paternity Testing Simulation Post-lab QuestionsA)Why do differ.pdf
DNA Paternity Testing Simulation Post-lab Questions
A)Why do different individuals such as siblings have different restriction enzyme recognition
sites?
B) What is the function of probes in DNA praternity analysis?
C) Why is there more than one single locus probe used in an actual paternity DNA test?
D)Why do we not use probes in this DNA paternity simulation and still obtain results?
Solution
A)Why do different individuals such as siblings have different restriction enzyme recognition
sites?
The different individuals such as siblings have different restriction enzyme recognition because
the size of the gene fragments are mainly relies upon the “distance between the recognition
sites”. This is due to polymorphism i.e. single nucleotide polymorphism (SNPs) with intervening
repetitive base sequences in siblings. The paternal parents are goiung to inherit genetic
variability into the sibling due to gametes reduction division & crossing over
B) What is the function of probes in DNA praternity analysis?
The functions of DNA probes are meticulously to identify “length differences between these
intervening repetitive gene sequences via isotopic & non-isotopic labeling by the use of “single
stranded DNA finally these DNA probes are going to complement and hybridize (attach) to
single stranded DNA due to SNP during paternity analysis.
C) Why is there more than one single locus probe used in an actual paternity DNA test?
More than single locus DNA probe or Multiple- Locus probes are used to identify “multiple
intervening repetitive DNA sequences” located on many chromosomes result in generation of 20
to 30 bands so that is possible to avoid chances of same pattern due to the multi-band patterns,.
Describe how the lung cancer travels to the first site of metastasis.pdf
Describe how the lung cancer travels to the first site of metastasis.
•Given your primary tumor site, how did lung cancer cells of the primary tumor spread to the
initial secondary site?
How does lung cancer cell get access to the body during metastasis?
What pathway did your cancer cells’ take?
Solution
answer:
metastasis:
delopment of secondry malignant growth at a distance from primary site of cancer
lung cancer tumor travel process:
primary tumor willl be prolification of the clone and invesion of vessel through the metastasis
clone evolves and its transport by circulation then its leads to the embolisam finally new tumor
formation at the site of metestasis.
Recommended
Foofy obtained a significant Fobt from an experiment with five level.pdf
Foofy obtained a significant Fobt from an experiment with five levels. She therefore concludes
that changing every condition of the independent variable results in a significant change in the
dependent variable. (a) Is she correct? Why or why not? (b) What must she now do?
Solution
She is not correct because changing every condition of the independent variable results in a
significant change in the dependent variable suggests that at least one independent variable is
significantly different from others and null hypothesis is rejected. She should do post-hoc test
now..
Explain how water interacts with the following types of compounds to.pdf
Explain how water interacts with the following types of compounds to influence their water
solubility.
1.Compounds that can hydrogen bond
2. Compounds that are polar but cannot hydrogen bond
3. Compounds that are non-polar
4. Compounds that dissociate into cations and anions.
Thank you so much.
Solution
Compounds that can hydrogen bond
Water tends to disassociate into H+ and OH- ions. In this disassociation, the oxygen retains the
electrons and only one of the hydrogens, becoming a negatively charged ion known as
hydroxide. Pure water has the same number (or concentration) of H+ as OH- ions. Acidic
solutions have more H+ ions than OH- ions. Basic solutions have the opposite. An acid causes an
increase in the numbers of H+ ions and a base causes an increase in the numbers of OH- ions.
Water, which not only dissolves many compounds but also dissolves more substances than any
other liquid, is considered the universal solvent. A polar molecule with partially-positive and
negative charges, it readily dissolves ions and polar molecules. Water is therefore referred to as a
solvent: a substance capable of dissolving other polar molecules and ionic compounds.
The charges associated with these molecules form hydrogen bonds with water, surrounding the
particle with water molecules. This is referred to as a sphere of hydration, or a hydration shell,
and serves to keep the particles separated or dispersed in the water.
Compounds that dissociate into cations and anions.-When ionic compounds are added to water,
individual ions interact with the polar regions of the water molecules during the dissociation
process, disrupting their ionic bonds. Dissociation occurs when atoms or groups of atoms break
off from molecules and form ions. Consider table salt (NaCl, or sodium chloride): when NaCl
crystals are added to water, the molecules of NaCl dissociate into Na+ and Cl– ions, and spheres
of hydration form around the ions. The positively-charged sodium ion is surrounded by the
partially-negative charge of the water molecule\'s oxygen; the negatively-charged chloride ion is
surrounded by the partially-positive charge of the hydrogen in the water molecule.
Compounds that are polar and non-polar-
-Since many biomolecules are either polar or charged, water readily dissolves these hydrophilic
compounds. Water is a poor solvent, however, for hydrophobic molecules such as lipids.
Nonpolar molecules experience hydrophobic interactions in water: the water changes its
hydrogen bonding patterns around the hydrophobic molecules to produce a cage-like structure
called a clathrate. This change in the hydrogen-bonding pattern of the water solvent causes the
system\'s overall entropy to greatly decrease, as the molecules become more ordered than in
liquid water. Thermodynamically, such a large decrease in entropy is not spontaneous, and the
hydrophobic molecule will not dissolve.
-Polar substances tend to dissolve well in other polar substances, but not nonpolar substances,
whil.
Explain how the financial reporting of gains and losses on assets.pdf
Explain how the financial reporting of gains and losses on assets differs from the financial
reporting of unrealized gains and losses on martketable securities.
Explain how the financial reporting of gains and losses on assets differs from the financial
reporting of unrealized gains and losses on martketable securities.
Solution
Particulars Gains and Losses on Assets unrealized gains and losses on martketable securities
Actual or Not Actual It is actual or real in nature. It is differrence between the values of
securities Income Tax It is taxable / deductible under income tax It is not considered under
income tax Cash flow There is cash inflow There is no any cash flow Realisation It is realised
gain or loss It may be realised or not. It is unrealised.
Explain two ways in which microbiologists have overcome the difficul.pdf
Explain two ways in which microbiologists have overcome the difficulty of viewing microbes in
wet preparations
Solution
First we have to find two ways in which microbiologists have overcome the difficulty of viewing
microbes in wet preparations are a bacteria which has the intrinsic ability of movement in the
surrounding medium in which it remains suspended is a motile bacteria.
A bacteria which does not have the intrinsic ability of movement in the surrounding medium in
which it remains suspended is a non-motile bacteria.
In wet mount though the shape and size of bacteria can be observed motility may be hampered as
the suspension is pressed between the slide and the cover slip.
Motility test is performed for clear observation of the motility of bacteria, besides their shape and
size.
The hanging drop is observed under a microscope using oil immersion objective its cells can be
seen to have erratic movement in the surrounding medium.
If it is non motile its cells remain static in the medium without any movement or may show
brownian movement resulting from the bombardment by the water molecules in the medium..
During the preparation of the bank reconciliation for Building Conce.pdf
During the preparation of the bank reconciliation for Building Concepts Co., Joel Kimmel, the
assistant controller, discovered that Lone Peak National Bank incorrectly recorded a $3,290
check written by Building Concepts Co. as $329. Joel has decided not to notify the bank but wait
for the bank to detect the error. Joel plans to record the $2,961 error as Other Income if the bank
fails to detect the error within the next three months.
Discuss whether Joel is behaving in a professional manner.
Solution
Joel is not behaving in a professional manner,
Joel record differece amount ($2961) as other income, this treatment is wrong
because that amount can not be include in income of Building concepts Co..
C++ CodeConsider the LinkedList class and the Node class that we s.pdf
C++ Code
Consider the LinkedList class and the Node class that we saw in lecture. The LinkedList class
basically represents a singly linked list.
Now, for this lab, consider a doubly linked list, i.e. each node has a next and previous pointers to
the next and previous node, respectively.
Modify the Node class to reflect this change and implement the following methods for
DoublyLinkedList class:
addFirst
addLast
addAtIndex
removeFirst
removeLast
removeAtIndex
Note: We saw how to implement these methods for a singly linked list and the code is added as
an attachment.
C++ Code
*****************************************************************************
******************
*****************************************************************************
******************
C++ Code
Solution
#include
#include
using namespace std;
template
class Node
{
public:
T element; // Element contained in the node
Node *next; // Pointer to the next node
Node *previous;
Node(T element) // Constructor
{
this->element = element;
next = NULL;
}
};
template
class LinkedList
{
public:
LinkedList();
void addFirst(T element);
void addLast(T element);
T removeFirst() ;
T removeLast();
void addAtIndex(int index, T element);
T removeAtIndex(int index);
void traverseList();
private:
Node *head;
Node *tail;
int size;
};
template
LinkedList::LinkedList()
{
head = NULL;
tail = NULL;
size = 0;
}
template
void LinkedList::addFirst(T element)
{
Node *newNode = new Node(element);
newNode->next = head;
if(head != NULL) {
head->previous = newNode;
}
newNode->previous = NULL;
head = newNode;
size++;
if (tail == NULL)
tail = head;
}
template
void LinkedList::addLast(T element)
{
if (tail == NULL)
{
head = tail = new Node(element);
}
else {
tail->next = new Node(element);
tail = tail->next;
}
size++;
}
template
void LinkedList::addAtIndex(int index, T element)
{
if (index == 0)
addFirst(element);
else if (index >= size)
addLast(element);
else
{
Node *current = head;
for (int i = 1; i < index; i++)
current = current->next;
Node *temp = current->next;
current->next = new Node(element);
(current->next)->next = temp;
size++;
}
}
template
T LinkedList::removeFirst()
{
if (size == 0)
exit(1);
else
{
Node *temp = head;
head = head->next;
size--;
T element = temp->element;
delete temp;
return element;
}
}
template
T LinkedList::removeLast()
{
if (size == 0)
exit(1);
else if (size == 1)
{
Node *temp = head;
head = tail = NULL;
size = 0;
T element = temp->element;
delete temp;
return element;
}
else
{
Node *current = head;
for (int i = 0; i < size - 2; i++)
current = current->next;
Node *temp = tail;
tail = current;
tail->next = NULL;
size--;
T element = temp->element;
delete temp;
return element;
}
}
template
T LinkedList::removeAtIndex(int index)
{
if (index < 0 || index >= size)
exit(1);
else if (index == 0)
return removeFirst();
else if (index == size - 1)
return removeLast();
else {
Node *previous = head;
for (int i = 1; i < index; i++)
{
previous = previous->next;
}
Node *current = previous-.
DNA Paternity Testing Simulation Post-lab QuestionsA)Why do differ.pdf
DNA Paternity Testing Simulation Post-lab Questions
A)Why do different individuals such as siblings have different restriction enzyme recognition
sites?
B) What is the function of probes in DNA praternity analysis?
C) Why is there more than one single locus probe used in an actual paternity DNA test?
D)Why do we not use probes in this DNA paternity simulation and still obtain results?
Solution
A)Why do different individuals such as siblings have different restriction enzyme recognition
sites?
The different individuals such as siblings have different restriction enzyme recognition because
the size of the gene fragments are mainly relies upon the “distance between the recognition
sites”. This is due to polymorphism i.e. single nucleotide polymorphism (SNPs) with intervening
repetitive base sequences in siblings. The paternal parents are goiung to inherit genetic
variability into the sibling due to gametes reduction division & crossing over
B) What is the function of probes in DNA praternity analysis?
The functions of DNA probes are meticulously to identify “length differences between these
intervening repetitive gene sequences via isotopic & non-isotopic labeling by the use of “single
stranded DNA finally these DNA probes are going to complement and hybridize (attach) to
single stranded DNA due to SNP during paternity analysis.
C) Why is there more than one single locus probe used in an actual paternity DNA test?
More than single locus DNA probe or Multiple- Locus probes are used to identify “multiple
intervening repetitive DNA sequences” located on many chromosomes result in generation of 20
to 30 bands so that is possible to avoid chances of same pattern due to the multi-band patterns,.
Describe how the lung cancer travels to the first site of metastasis.pdf
Describe how the lung cancer travels to the first site of metastasis.
•Given your primary tumor site, how did lung cancer cells of the primary tumor spread to the
initial secondary site?
How does lung cancer cell get access to the body during metastasis?
What pathway did your cancer cells’ take?
Solution
answer:
metastasis:
delopment of secondry malignant growth at a distance from primary site of cancer
lung cancer tumor travel process:
primary tumor willl be prolification of the clone and invesion of vessel through the metastasis
clone evolves and its transport by circulation then its leads to the embolisam finally new tumor
formation at the site of metestasis.
Define the following engineering terms1. engineering ethics2. W.pdf
Define the following engineering terms:
1. engineering ethics
2. Whistleblowing
3. Responsibility of engineers
4. recent developments
5.general principles
6 obligation of society
7. conduct
8. case studies and key individuals
Solution
1. Engineering Ethics - Engineering Ethics is the study of moral issues and decisions confronting
individuals and organizations engaged in engineering.The field examines and sets the
obligations by engineers to society, to their clients, and to the profession.
2. Whistleblowing - The duty of an engineer to report to the concerned authority about a possible
risk to others when a client or employer fails to follow the engineer\'s directions.
3. Responsibility of engineers - The responsibility of an engineer to commit to serving society,
attending to the welfare and progress of the majority. The engineer should reject any project that
is intended to harm the general interest, thus avoiding a situation that might be hazardous or
threatening to the environment, life, health, or other rights of human beings.
4. recent developments - Recent efforts to uphold engineering ethics include addressing of
bribery and political corruption by several professional societies and business groups around the
world while getting ready to face challenges like offshoring, sustainable development, and
environmental protection.
5. general principles - General guidelines set by various engineering professional societies which
tend to include
6. obligation of society - The paramount value recognized by engineers is the safety and welfare
of the public. All the major engineering societies around the world. These societies include
Institute of Electrical and Electronics Engineers, Institution of Civil Engineers, American
Society of Mechanical Engineers, Institute of Industrial Engineers etc.
7. conduct - Several ethical issues faced by the engineers include considerations of business
conduct. The issues faced are relationships with clients, ensuring legal compliance by clients,
treatment of confidential or proprietary information, conflict of interest etc.
8. case studies and key individuals - The study of engineering failures in the last century
indicates that not all engineering failures involve ethical issues. Many failures involve the failure
of the design process or management culture. The following mentioned failures include ethical as
well as technical issues..
Data structuresUsing java language and develop a prot.pdf
Data structures:
****Using java language and develop a prototype for a demo program that reads in strings that
look like:
Using the infix to postfix code below:
A = 3; // your code should store A and 3 as a key-value pair
B = 4; // your code should store B and 4 as a key-value pair
A; // your code should now look-up A and then display its value
later we (you) will add the ability to write
Y = A * B + 3;
Y; // your code should show 16
Using this code
// infix.java
// converts infix arithmetic expressions to postfix
// to run this program: C>java InfixApp
import java.io.*; // for I/O
////////////////////////////////////////////////////////////////
class StackX {
private int maxSize;
private char[] stackArray;
private int top;
// --------------------------------------------------------------
public StackX(int s) // constructor
{
maxSize = s;
stackArray = new char[maxSize];
top = -1;
}
// --------------------------------------------------------------
public void push(char j) // put item on top of stack
{
stackArray[++top] = j;
}
// --------------------------------------------------------------
public char pop() // take item from top of stack
{
return stackArray[top--];
}
// --------------------------------------------------------------
public char peek() // peek at top of stack
{
return stackArray[top];
}
// --------------------------------------------------------------
public boolean isEmpty() // true if stack is empty
{
return (top == -1);
}
// -------------------------------------------------------------
public int size() // return size
{
return top + 1;
}
// --------------------------------------------------------------
public char peekN(int n) // return item at index n
{
return stackArray[n];
}
// --------------------------------------------------------------
public void displayStack(String s) {
System.out.print(s);
System.out.print(\"Stack (bottom-->top): \");
for (int j = 0; j < size(); j++) {
System.out.print(peekN(j));
System.out.print(\' \');
}
System.out.println(\" \");
}
// --------------------------------------------------------------
} // end class StackX
////////////////////////////////////////////////////////////////
class InToPost // infix to postfix conversion
{
private StackX theStack;
private String input;
private String output = \"\";
// --------------------------------------------------------------
public InToPost(String in) // constructor
{
input = in;
int stackSize = input.length();
theStack = new StackX(stackSize);
}
// --------------------------------------------------------------
public String doTrans() // do translation to postfix
{
for (int j = 0; j < input.length(); j++) {
char ch = input.charAt(j);
theStack.displayStack(\"For \" + ch + \" \"); // *diagnostic*
switch (ch) {
case \'+\': // it’s + or -
case \'-\':
gotOper(ch, 1); // go pop operators
break; // (precedence 1)
case \'*\': // it’s * or /
case \'/\':
gotOper(ch, 2); // go pop operators
break; // (precedence 2)
case \'(\': // it’s a left paren
theStack.push(.
You have been cut! While you wage war in the streets of Gotham for y.pdf
You have been cut! While you wage war in the streets of Gotham for your life, your cells wage
war against incoming invaders. Place the following statement concerning the fates of unwanted
invaders/immigrants in your body in order:
1.The invader is cut apart while alive and digested in the harsh environment of your new super-
max detention/recycling center. Die invader, Die!
2.Contact with one of your cells triggers phagocytosis and a signal trough transmembrane
proteins triggers the cytoskeleton.
3.The vesicle containing the bacterium is pinched off.
4.A bacterium that has stormed the outter walls of your body (the skin through the cut) is
recognized and tagged as foreign.
5.Other vesicles containing hydrolases, enzmes that depolymerize proteins, carbohydrates, lipids,
and nucleic acids, as well as other proteins that act to increase the pH of the larger vesicle that
contains the invader arrive on the cell\'s microtubule transportation system and fuse.
6.Kinesin walks its vesicle down the microtubule highways into the cytoplasm.
7.Actin polymerizes to form rods that push out the membrane to form fingers or pseudopods that
begin to envelop the bacterium.
8.The vesicle containing the bacterium is moved by kinesin.
Solution
4A bacterium that has stormed the outter walls of your body (the skin through the cut) is
recognized and tagged as foreign.
2Contact with one of your cells triggers phagocytosis and a signal trough transmembrane
proteins triggers the cytoskeleton.
7Actin polymerizes to form rods that push out the membrane to form fingers or pseudopods that
begin to envelop the bacterium.
3 The vesicle containing the bacterium is pinched off.
8 The vesicle containing the bacterium is moved by kinesin
6 Kinesin walks its vesicle down the microtubule highways into the cytoplasm
5 .Other vesicles containing hydrolases, enzmes that depolymerize proteins, carbohydrates,
lipids, and nucleic acids, as well as other proteins that act to increase the pH of the larger vesicle
that contains the invader arrive on the cell\'s microtubule transportation system and fuse
1 The invader is cut apart while alive and digested in the harsh environment of your new super-
max detention/recycling center..
Why has Mexico encountered such a significant collapse from an econom.pdf
Why has Mexico encountered such a significant collapse from an economic perspective?
Solution
From an economic perspective Mexico had a significant collapse due to two major factors:
1. The first one was a collapse in export demand
2. The second one being the susequent constraints that it had in accessing the International
Financial markets.
Due to the above factors Mexico started feeling a significant negative effect. As the negative
effects increased, the country\'s fiscal position weakened. This was mainly due to the fall in oil
revenues. Additionally, a drop in international energy prices brought about by the global
recession along with a sharp decline in domestic oil production attributed to the crisis.
A downward rigidity in public expenditure was observed which along with the above mentioned
factors raised concerns about the sustainability of fiscal policy.All these factors played a major
spoilsport and generated doubts about the country’s capacity to finance the growth in the current
account deficit for 2009.
The Mexican economy was forced to adjust to a new external environment that was characterised
by lower foreign currency revenues with a limited access to external borrowing..
Which of the following isare true of BOTH natural selection and gen.pdf
Which of the following is/are true of BOTH natural selection and genetic drift?
I. Promotes traits that increase survival or reproduction
II. Depends on genetic mutations as the source of variation
III. Can lead to evolutionary change
Question 5 options:
I and III
I only
I, II, and III
II and III
I and II
I and III
I only
I, II, and III
II and III
I and II
Solution
answer will be I,II and III, because in a population genetic drift and natural selection always play
a role together..
Which of the following information-management systems uses artificia.pdf
Which of the following information-management systems uses artificial intelligence?
A. Polyinstantiation
B. Known signature scanning
C. Application programming interface
D. Knowledge discovery in databases
Solution
D. Knowledge Discovery in Databases (KDD) is an artificial intelligence method used to identify
useful patterns in data; as such, it provides a type of automatic analysis. Answer A is incorrect
because polyinstantiation is a technique used to prevent inference violations. Answer B is
incorrect because known signature scanning is a method used to detect computer viruses. Answer
C is incorrect because the application programming interface (API) is in no way associated with
artificial intelligence..
What steps and tools are used for Pen TestingSolutionSteps fo.pdf
What steps and tools are used for Pen Testing?
Solution
Steps for Pen Testing:
1:Attack Surface analysis
2:Threat Modelling
3:Test Configuration
4:automatic and manual scans
5:validation of vulnerabilities
6:false positive elimination
7:proof of concept
8:Check for the service/version running on the particular port.
9:Search the vulnerability in the service/version.
10:Exploit the target using tools.
Tools used for Pen Testing
1:Metasploit
2:Wireshark
3:W3af
4:core Impact
5:Back track.
what is the value of nursing educationSolutionThe main value o.pdf
what is the value of nursing education
Solution
The main value of nursing education is to deliver efficient ethical based health care to the
patients suffering from various devastating disease finally to implement clinical protocol &
clinical trials on the patients’ with moral & ethical aspects to collect epidemiological clinical
data.
As a clinical nurse leader is to obtain a better clinical outcome is main value of \"nursing
education\" by respecting patent’s autonomy. The main characteristics are to develop a strategic
plan to implement a “do” and “step “policy to active organizational goals, allocation of resources
as a clinical nurse leader. The other aspects of nursing education is to implement the treatment
protocols, epidemiological studies of disease, and disorders within time- frame using
interdisciplinary teams and according ethical based informatics. The responsibilities are allocated
to each RN, LPN to obtain an efficient health out come from patient and within the community.
Nursing education is crucial to inform the fellow nurses about the current organizational
philosophy of nursing in clinical services. Finally, collecting the patient data and monitoring,
analysis & evaluation is essential using nursing informatics to develop future prospects to
prevent diseases.
Nursing education has specific impact on the role of health care workers and clinicians to
maintain efficient patient’s electronic health care records using information technology tools. In
order to automate the aspects of acquisition of patient health information and management.
What is the exact output of the following code#include stdio.h.pdf
What is the exact output of the following code?
#include
#include
main()
{
int i,j;
i=fork();
for (j=0; j<3; j++)
{
if (i == 0 && j == 0)
{
sleep(3);
printf(\"Cats\ \");
}
else if (i == 0)
{
sleep(2);
printf(\"Dogs\ \");
}
else
{
sleep(2);
printf(\"Raining\ \");
}
}
}
Solution
Answer for the above code is as follows:
Status : runtime error
Output:
Raining
Raining
Raining.
Using a Fabry-Perot interferometer, we can Observe small objects. D.pdf
Using a Fabry-Perot interferometer, we can Observe small objects. Detect fine optical
spectrum lines. Examine optical quality of lens. Measure light speed in different media.
Solution
ANSWER is :
B) Detect fine optical spectrum lines
It consists of two highly reflective and strictly parallel plates called an etalon. Because of the
high reflectivity of the plates of the etalon, the successive multiple reflections of light waves
diminish very slowly in intensity and form very narrow, sharp fringes. These may be used to
reveal hyperfine structures in line spectra..
TRUE of falseA researcher found 95 confident interval for the aver.pdf
TRUE of false
A researcher found 95 confident interval for the average age of SMU students. It does mean that
there is a 95% chance that the population average lies in the inverval
Solution
95% confidence interval implies that we are 95% confident that for random samples of large size
the population average is more likely to fall in this interval..
The major type of interactive forces between molecules of HCl are i.pdf
The major type of interactive forces between molecules of HCl are: ionic covalent hydrogen
bonding dipole-dipole attractions dispersion forces
Solution
Dipole - Dipole attraction is the major type of interactive forces between HCl molecules.
these interaction are basically the electrostatic interaction between the permanent dipoles in the
mlecules. this interaction occurs when the partially negatively charged part of one molecule (in
case of HCl it is Cl- ) interacts with partially positive charge of the other molecule (H+ in case of
HCl)..
Tetrad AnalysisYou should be able to explain and you should know e.pdf
Tetrad Analysis
You should be able to explain and you should know examples of these:
1) Incomplete dominance
2) dominant negative
3) co-dominance,
4) pleotropic effect
5) epistasis
6) suppressors
7) developmental noise
8) norm of reaction (phenotypic plasticity).
9) You should be able to explain the biochemistry of ABO blood type and epistasis between h
and I genes.
Solution
1) incompletes dominance is a form of intermediate inheritance in which one allele for a specific
trait is not completely exprssed over its paired allele. It leads to the formation of a third
phenotype in which the expressed physical trait is a combination of phenotypes of noth
alleles.Example is a flower that is pink in colour due to cross pollination between a red and a
white flower,when neither the white or red alleles are dominant.
2. dominant negative is used to describe a gene that has a dominant effect similar to that
described genetically,one copy of the gene gives a mutant phenotype effect and a negative copy
and the negative effect in that it prevents.For example a protein that functional as a dimer.A
mutation that removes the functional domain but reatins the dimerization domain would cause a
dominant negatice phenotype because some part of the dimer would be missing functional
domains.
3. Co dominance is a type of dominance in which both the alleles in a gene pair in a heterozygote
are fully expressed. The human ABO blood groups provide an example of co dominance.
4. Pleiotropric effect is to have multiple effects from the same gene.Example , marfan gene is
pleiotropic .it had many effects such as long fingers and toes,disslocation of the lens of the eye.
5.epistasis is a phenomena in which the effect of one gene is dependant on the prsesnce of one or
more modifier genes.In case of dominant epistasis there is complete dominance at both gene
pairs however when one gene is dominant it hides the phenotype of the other gene.
in recessive epistasis recessive genes at one locus mask the phenotypic expression of other gene
locus.
example, the gene for male pattern baldness is epistatic to the gene for red hair..
Security ImplicationsWhat are the implications of Shannon’s work o.pdf
Security Implications
What are the implications of Shannon’s work on security?
Shannon, C. (1948, July). A Mathematical Theory of Communication. The Bell System
Technical Journal, 27(3), 379-423.
Solution
Implications of Shannon’s work on security :-
1) Shannon\'s work on cryptography security systems.
2) Shannon proved the cryptographic one-time password is unbreakable in his classified research
that was later published in October 1949.
3) He can proved the unbreakable system must have essentially the same characteristics as the
one-time padssword system. the key must be truly random as large as the plaintext information
never reused in whole or part of the kept secret key value.
4) The one-time password system by the Soviets was partially broken by the National Security
Agency but this was because of misuses of the one-time password by Soviet cryptographic
technicians..
Protect tissues Spinal cord functions to includes pituitary gland ski.pdf
Protect tissues Spinal cord functions to includes pituitary gland skin includes Controls Nervous
Integumentary includes 16. Endocrine ncludes bones 15 functions in support Body Systems
Muscular includes functions in ovarie Urinary Digestive Circulatory posture in includes bladder
filters males functions in includes breaks down ncludes 11 from Heart stomach (2 to be Word
Bank into females, brain, kidneys, regulate temperature, skeletal, ligaments, wastes, transport,
food, thyroid, esophagus, testes, coordination, absorbed reproductive, movement, blood
Solution
1. Coordination
2. Brain
3. Regulate temperature
4. Ligaments
5. Skeletal
6. Movement
7. Kidney
8. Waste
9. Blood
10. Transport
11. Food
12. Esophagus
13. Females
14. Testes
15. Reproductive
16. Thyroid.
Present and explain each step in human perception process. To what e.pdf
Present and explain each step in human perception process. To what extent we perceive the real
world differently, how do we manage that (all types of perceptual selections) and why (such as
protecting ourselves)
Solution
Perception starts with visualization.
The main sensory component of vision involves the gathering and recording of light scattered
from objects in the surrounding scene, and the forming of a two - dimensional function on the
photoreceptors. Human eyes can perceive visible light. It involves receiving signals in the
nervous system. These signals form chemical and physical stimulation.
Perception starts with processing sensory input which transforms these low-level information to
higher-level information. The next step is processing as per a person\'s knowledge and concepts.
We perceive the real world based on our knowledge and experience. The process of perception
begins with an object in the real world, termed the distal stimulus ordistal object. Our knowledge
of the distal stimulus influences how we percieve the real world.
We manage all types of perceptual selections by looking for familiar cues when we face an
unknown distal stimulus. We do this to bind our senses and stimulate our high level information..
please help with this genetics problem You have 3 Serratia strains (.pdf
please help with this genetics problem You have 3 Serratia strains (A, B, C) that contain different
mutations in a pathway for making prodigiosin. When grown by themselves A, B, and C are all
white. You are given the information that C feeds A and B through diffusion, and A feeds B
through the air. If you streak out the strains A, C, B in this order and next to each other on one
plate, what color would you observe at day 4? A: pink. B: pink. C: pink A: red. B: red. C:
white. A: red. B: red in colonies close to C. C: white. A: red. B: white. C: white. A: red in
colonies close to C. B: red. C: white If these strains are plated together, which strain(s) will be
able to produce the product of CANDY? Strain Sweet Has a defective y and n enzyme. Strain
Sour has a defective s enzyme Neither can feed the other Sour and Sweet can feed each other
Sour strain can feed Sweet strain but Sweet cannot feed sour Sweet strain can feed Sour strain
but Sour cannot feed sweet
Solution
5)
The prodigiosin is a red pigment. In the pigment production pathway C feeds A and B through
diffusion. Therefore, C is white, B is red colored colony located near C and A is red colored
colony.
6)
Sweet strain have defective \'y\' and \'n\' enzymes and sour strain contains defective \'s\' enzyme.
In both the sweet and sour candy production starts with colony A. With the defective ;y\' and \'n\'
enzymes sweet strain cannot feed sour strain, but with defective \'s\' enzyme, sour strain can feed
sweet strain. Option C is correct..
Please briefly describe Two types of statics where you will NOT o.pdf
Please briefly describe Two types of statics where you will NOT observe a classic Medelian
9.3:3:1 ratio in a F; generation.
Solution
In dihybridism without epistasis,double heterozygous parents cross-breed and 4 phenotypical
forms appear in F2.The proportion is 9 individuals double dominant,3 individuals dominant for
the first pair and recessive for the second pair,3 individuals recessive for the first pair and
dominant for the second pair, and 1 individual double recessive so the final ratio becomes
9:3:3:1.Considering that the epistatic gene is the second pair and that the recessive genotype of
the hypostatic gene implies the lack of the characteristic,in the F2 generation of dominant
epistasis,the following phenotypic forms would emerge:13 individuals dominant for the second
pair or recessive for the first pair,meaning that,the characteristic is not manifest;3 individuals
dominant for the first pair and recessive for the second pair,meaning that the characteristic is
manifest.The phenotypical proportion would be 13:3.In recessive epistasis,the phenotypical
forms that would emerge in F2 are:9 individuals double dominant and 7 individuals recessive for
the first pair or recessive for the second pair,meaning that the characteristic is not manifest.
Therefore,the phenotypical proportion would be 9:7.
These examples show how epistasis changes phenotypical forms and proportions,from the
normal 9:3:3:1 in F2 to 13:3 in dominant epistasis or to 9:7 in recessive epistasis..
Northwest Bakery produces seven types of loaf bread and distributes .pdf
Northwest Bakery produces seven types of loaf bread and distributes the bread to local grocery
stores each day of the week. Juanita Gomez is the marketing manager for Northwest Bakery and
she is trying to figure out how to improve sales. She may want to discontinue some types of
bread or add new types. She asked the production department to record the production levels of
the seven types of bread and Bob McGeorge, the production manager, put these data in a file
named Northwest Bakery. The file is attached.
Juanita has just asked you, who has just been hired in the marketing department, to do some
analysis on the data. She would like:
1. (5 pts) A chart showing percentages of each type of bread produced.
2. (7 pts) A table showing the average amount of each type of bread produced per day and the
variability of the production of each type of bread.
3. (8 pts) She also mentioned that Bob would like to see a chart of average daily production so he
can better manage workloads of the production staff.
4. Produce the appropriate charts, graphs and tables in Excel
5. (5 pts) Compile a brief report in a Word document explaining what you did and include the
charts, graphs and/or tables in the document.Day of WeekWhiteWheatMultigrainDark
RyeCinnamon RaisinSour DoughLight
OatSunday496490403251114108228Monday786611570373165148304Tuesday5474744242651
44104256Wednesday51344338031210092180Thursday817669622375181152308Friday375390
29920612488172Saturday700678564366173136248Sunday597502457276140144312Monday53
6530428261135112356Tuesday675703605289201188356Wednesday421433336212100104224
Thursday867576541298152144304Friday506461406254135116264Saturday4703523772088492
172Sunday748643599341153148316Monday376367310199128104208Tuesday6326465863401
74160264Wednesday591504408238140120276Thursday656497415221107120212Friday81767
3644249200180348Saturday653571557316129140224
Solution
Day of Week
White
Wheat
Multigrain
Dark Rye
Cinnamon Raisin
Sour Dough
Light Oat
Sunday
496
490
403
251
114
108
228
Monday
786
611
570
373
165
148
304
Tuesday
547
474
424
265
144
104
256
Wednesday
513
443
380
312
100
92
180
Thursday
817
669
622
375
181
152
308
Friday
375
390
299
206
124
88
172
Saturday
700
678
564
366
173
136
248
Sunday
597
502
457
276
140
144
312
Monday
536
530
428
261
135
112
356
Tuesday
675
703
605
289
201
188
356
Wednesday
421
433
336
212
100
104
224
Thursday
867
576
541
298
152
144
304
Friday
506
461
406
254
135
116
264
Saturday
470
352
377
208
84
92
172
Sunday
748
643
599
341
153
148
316
Monday
376
367
310
199
128
104
208
Tuesday
632
646
586
340
174
160
264
Wednesday
591
504
408
238
140
120
276
Thursday
656
497
415
221
107
120
212
Friday
817
673
644
249
200
180
348
Saturday
653
571
557
316
129
140
224
Total
12779
11213
9931
5850
2979
2700
5532
Percentage
25.06
21.99
19.48
11.5
5.84
5.3
10.9
Day of Week
White
Wheat
Multigrain
Dark Rye
Cinnamon Raisin
Sour Dough
Light Oat
Average
Sunday
496
490
403
251
114
108
228
298.57
Monday
786
611
570
373
165
148
304
422.43
Tuesday.
Interface four common anode seven-segment LEDs to the PIC18F452 micr.pdf
Interface four common anode seven-segment LEDs to the PIC18F452 microcontroller using
PORTB and PORTC. Write instructions to display a four-digit number stored in data registers
starting from REG20 (address 20H).
Solution
instructions
STA
LDA
LHLD
SHLD
MOV A, X Binary Octal Decimal
Hexadecimal0000000000111100102220011333010044401015550110666011177710001088100
1119910101210A10111311B11001412C11011513D11101614E11111715F.
The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
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Define the following engineering terms1. engineering ethics2. W.pdf
Define the following engineering terms:
1. engineering ethics
2. Whistleblowing
3. Responsibility of engineers
4. recent developments
5.general principles
6 obligation of society
7. conduct
8. case studies and key individuals
Solution
1. Engineering Ethics - Engineering Ethics is the study of moral issues and decisions confronting
individuals and organizations engaged in engineering.The field examines and sets the
obligations by engineers to society, to their clients, and to the profession.
2. Whistleblowing - The duty of an engineer to report to the concerned authority about a possible
risk to others when a client or employer fails to follow the engineer\'s directions.
3. Responsibility of engineers - The responsibility of an engineer to commit to serving society,
attending to the welfare and progress of the majority. The engineer should reject any project that
is intended to harm the general interest, thus avoiding a situation that might be hazardous or
threatening to the environment, life, health, or other rights of human beings.
4. recent developments - Recent efforts to uphold engineering ethics include addressing of
bribery and political corruption by several professional societies and business groups around the
world while getting ready to face challenges like offshoring, sustainable development, and
environmental protection.
5. general principles - General guidelines set by various engineering professional societies which
tend to include
6. obligation of society - The paramount value recognized by engineers is the safety and welfare
of the public. All the major engineering societies around the world. These societies include
Institute of Electrical and Electronics Engineers, Institution of Civil Engineers, American
Society of Mechanical Engineers, Institute of Industrial Engineers etc.
7. conduct - Several ethical issues faced by the engineers include considerations of business
conduct. The issues faced are relationships with clients, ensuring legal compliance by clients,
treatment of confidential or proprietary information, conflict of interest etc.
8. case studies and key individuals - The study of engineering failures in the last century
indicates that not all engineering failures involve ethical issues. Many failures involve the failure
of the design process or management culture. The following mentioned failures include ethical as
well as technical issues..
Data structuresUsing java language and develop a prot.pdf
Data structures:
****Using java language and develop a prototype for a demo program that reads in strings that
look like:
Using the infix to postfix code below:
A = 3; // your code should store A and 3 as a key-value pair
B = 4; // your code should store B and 4 as a key-value pair
A; // your code should now look-up A and then display its value
later we (you) will add the ability to write
Y = A * B + 3;
Y; // your code should show 16
Using this code
// infix.java
// converts infix arithmetic expressions to postfix
// to run this program: C>java InfixApp
import java.io.*; // for I/O
////////////////////////////////////////////////////////////////
class StackX {
private int maxSize;
private char[] stackArray;
private int top;
// --------------------------------------------------------------
public StackX(int s) // constructor
{
maxSize = s;
stackArray = new char[maxSize];
top = -1;
}
// --------------------------------------------------------------
public void push(char j) // put item on top of stack
{
stackArray[++top] = j;
}
// --------------------------------------------------------------
public char pop() // take item from top of stack
{
return stackArray[top--];
}
// --------------------------------------------------------------
public char peek() // peek at top of stack
{
return stackArray[top];
}
// --------------------------------------------------------------
public boolean isEmpty() // true if stack is empty
{
return (top == -1);
}
// -------------------------------------------------------------
public int size() // return size
{
return top + 1;
}
// --------------------------------------------------------------
public char peekN(int n) // return item at index n
{
return stackArray[n];
}
// --------------------------------------------------------------
public void displayStack(String s) {
System.out.print(s);
System.out.print(\"Stack (bottom-->top): \");
for (int j = 0; j < size(); j++) {
System.out.print(peekN(j));
System.out.print(\' \');
}
System.out.println(\" \");
}
// --------------------------------------------------------------
} // end class StackX
////////////////////////////////////////////////////////////////
class InToPost // infix to postfix conversion
{
private StackX theStack;
private String input;
private String output = \"\";
// --------------------------------------------------------------
public InToPost(String in) // constructor
{
input = in;
int stackSize = input.length();
theStack = new StackX(stackSize);
}
// --------------------------------------------------------------
public String doTrans() // do translation to postfix
{
for (int j = 0; j < input.length(); j++) {
char ch = input.charAt(j);
theStack.displayStack(\"For \" + ch + \" \"); // *diagnostic*
switch (ch) {
case \'+\': // it’s + or -
case \'-\':
gotOper(ch, 1); // go pop operators
break; // (precedence 1)
case \'*\': // it’s * or /
case \'/\':
gotOper(ch, 2); // go pop operators
break; // (precedence 2)
case \'(\': // it’s a left paren
theStack.push(.
You have been cut! While you wage war in the streets of Gotham for y.pdf
You have been cut! While you wage war in the streets of Gotham for your life, your cells wage
war against incoming invaders. Place the following statement concerning the fates of unwanted
invaders/immigrants in your body in order:
1.The invader is cut apart while alive and digested in the harsh environment of your new super-
max detention/recycling center. Die invader, Die!
2.Contact with one of your cells triggers phagocytosis and a signal trough transmembrane
proteins triggers the cytoskeleton.
3.The vesicle containing the bacterium is pinched off.
4.A bacterium that has stormed the outter walls of your body (the skin through the cut) is
recognized and tagged as foreign.
5.Other vesicles containing hydrolases, enzmes that depolymerize proteins, carbohydrates, lipids,
and nucleic acids, as well as other proteins that act to increase the pH of the larger vesicle that
contains the invader arrive on the cell\'s microtubule transportation system and fuse.
6.Kinesin walks its vesicle down the microtubule highways into the cytoplasm.
7.Actin polymerizes to form rods that push out the membrane to form fingers or pseudopods that
begin to envelop the bacterium.
8.The vesicle containing the bacterium is moved by kinesin.
Solution
4A bacterium that has stormed the outter walls of your body (the skin through the cut) is
recognized and tagged as foreign.
2Contact with one of your cells triggers phagocytosis and a signal trough transmembrane
proteins triggers the cytoskeleton.
7Actin polymerizes to form rods that push out the membrane to form fingers or pseudopods that
begin to envelop the bacterium.
3 The vesicle containing the bacterium is pinched off.
8 The vesicle containing the bacterium is moved by kinesin
6 Kinesin walks its vesicle down the microtubule highways into the cytoplasm
5 .Other vesicles containing hydrolases, enzmes that depolymerize proteins, carbohydrates,
lipids, and nucleic acids, as well as other proteins that act to increase the pH of the larger vesicle
that contains the invader arrive on the cell\'s microtubule transportation system and fuse
1 The invader is cut apart while alive and digested in the harsh environment of your new super-
max detention/recycling center..
Why has Mexico encountered such a significant collapse from an econom.pdf
Why has Mexico encountered such a significant collapse from an economic perspective?
Solution
From an economic perspective Mexico had a significant collapse due to two major factors:
1. The first one was a collapse in export demand
2. The second one being the susequent constraints that it had in accessing the International
Financial markets.
Due to the above factors Mexico started feeling a significant negative effect. As the negative
effects increased, the country\'s fiscal position weakened. This was mainly due to the fall in oil
revenues. Additionally, a drop in international energy prices brought about by the global
recession along with a sharp decline in domestic oil production attributed to the crisis.
A downward rigidity in public expenditure was observed which along with the above mentioned
factors raised concerns about the sustainability of fiscal policy.All these factors played a major
spoilsport and generated doubts about the country’s capacity to finance the growth in the current
account deficit for 2009.
The Mexican economy was forced to adjust to a new external environment that was characterised
by lower foreign currency revenues with a limited access to external borrowing..
Which of the following isare true of BOTH natural selection and gen.pdf
Which of the following is/are true of BOTH natural selection and genetic drift?
I. Promotes traits that increase survival or reproduction
II. Depends on genetic mutations as the source of variation
III. Can lead to evolutionary change
Question 5 options:
I and III
I only
I, II, and III
II and III
I and II
I and III
I only
I, II, and III
II and III
I and II
Solution
answer will be I,II and III, because in a population genetic drift and natural selection always play
a role together..
Which of the following information-management systems uses artificia.pdf
Which of the following information-management systems uses artificial intelligence?
A. Polyinstantiation
B. Known signature scanning
C. Application programming interface
D. Knowledge discovery in databases
Solution
D. Knowledge Discovery in Databases (KDD) is an artificial intelligence method used to identify
useful patterns in data; as such, it provides a type of automatic analysis. Answer A is incorrect
because polyinstantiation is a technique used to prevent inference violations. Answer B is
incorrect because known signature scanning is a method used to detect computer viruses. Answer
C is incorrect because the application programming interface (API) is in no way associated with
artificial intelligence..
What steps and tools are used for Pen TestingSolutionSteps fo.pdf
What steps and tools are used for Pen Testing?
Solution
Steps for Pen Testing:
1:Attack Surface analysis
2:Threat Modelling
3:Test Configuration
4:automatic and manual scans
5:validation of vulnerabilities
6:false positive elimination
7:proof of concept
8:Check for the service/version running on the particular port.
9:Search the vulnerability in the service/version.
10:Exploit the target using tools.
Tools used for Pen Testing
1:Metasploit
2:Wireshark
3:W3af
4:core Impact
5:Back track.
what is the value of nursing educationSolutionThe main value o.pdf
what is the value of nursing education
Solution
The main value of nursing education is to deliver efficient ethical based health care to the
patients suffering from various devastating disease finally to implement clinical protocol &
clinical trials on the patients’ with moral & ethical aspects to collect epidemiological clinical
data.
As a clinical nurse leader is to obtain a better clinical outcome is main value of \"nursing
education\" by respecting patent’s autonomy. The main characteristics are to develop a strategic
plan to implement a “do” and “step “policy to active organizational goals, allocation of resources
as a clinical nurse leader. The other aspects of nursing education is to implement the treatment
protocols, epidemiological studies of disease, and disorders within time- frame using
interdisciplinary teams and according ethical based informatics. The responsibilities are allocated
to each RN, LPN to obtain an efficient health out come from patient and within the community.
Nursing education is crucial to inform the fellow nurses about the current organizational
philosophy of nursing in clinical services. Finally, collecting the patient data and monitoring,
analysis & evaluation is essential using nursing informatics to develop future prospects to
prevent diseases.
Nursing education has specific impact on the role of health care workers and clinicians to
maintain efficient patient’s electronic health care records using information technology tools. In
order to automate the aspects of acquisition of patient health information and management.
What is the exact output of the following code#include stdio.h.pdf
What is the exact output of the following code?
#include
#include
main()
{
int i,j;
i=fork();
for (j=0; j<3; j++)
{
if (i == 0 && j == 0)
{
sleep(3);
printf(\"Cats\ \");
}
else if (i == 0)
{
sleep(2);
printf(\"Dogs\ \");
}
else
{
sleep(2);
printf(\"Raining\ \");
}
}
}
Solution
Answer for the above code is as follows:
Status : runtime error
Output:
Raining
Raining
Raining.
Using a Fabry-Perot interferometer, we can Observe small objects. D.pdf
Using a Fabry-Perot interferometer, we can Observe small objects. Detect fine optical
spectrum lines. Examine optical quality of lens. Measure light speed in different media.
Solution
ANSWER is :
B) Detect fine optical spectrum lines
It consists of two highly reflective and strictly parallel plates called an etalon. Because of the
high reflectivity of the plates of the etalon, the successive multiple reflections of light waves
diminish very slowly in intensity and form very narrow, sharp fringes. These may be used to
reveal hyperfine structures in line spectra..
TRUE of falseA researcher found 95 confident interval for the aver.pdf
TRUE of false
A researcher found 95 confident interval for the average age of SMU students. It does mean that
there is a 95% chance that the population average lies in the inverval
Solution
95% confidence interval implies that we are 95% confident that for random samples of large size
the population average is more likely to fall in this interval..
The major type of interactive forces between molecules of HCl are i.pdf
The major type of interactive forces between molecules of HCl are: ionic covalent hydrogen
bonding dipole-dipole attractions dispersion forces
Solution
Dipole - Dipole attraction is the major type of interactive forces between HCl molecules.
these interaction are basically the electrostatic interaction between the permanent dipoles in the
mlecules. this interaction occurs when the partially negatively charged part of one molecule (in
case of HCl it is Cl- ) interacts with partially positive charge of the other molecule (H+ in case of
HCl)..
Tetrad AnalysisYou should be able to explain and you should know e.pdf
Tetrad Analysis
You should be able to explain and you should know examples of these:
1) Incomplete dominance
2) dominant negative
3) co-dominance,
4) pleotropic effect
5) epistasis
6) suppressors
7) developmental noise
8) norm of reaction (phenotypic plasticity).
9) You should be able to explain the biochemistry of ABO blood type and epistasis between h
and I genes.
Solution
1) incompletes dominance is a form of intermediate inheritance in which one allele for a specific
trait is not completely exprssed over its paired allele. It leads to the formation of a third
phenotype in which the expressed physical trait is a combination of phenotypes of noth
alleles.Example is a flower that is pink in colour due to cross pollination between a red and a
white flower,when neither the white or red alleles are dominant.
2. dominant negative is used to describe a gene that has a dominant effect similar to that
described genetically,one copy of the gene gives a mutant phenotype effect and a negative copy
and the negative effect in that it prevents.For example a protein that functional as a dimer.A
mutation that removes the functional domain but reatins the dimerization domain would cause a
dominant negatice phenotype because some part of the dimer would be missing functional
domains.
3. Co dominance is a type of dominance in which both the alleles in a gene pair in a heterozygote
are fully expressed. The human ABO blood groups provide an example of co dominance.
4. Pleiotropric effect is to have multiple effects from the same gene.Example , marfan gene is
pleiotropic .it had many effects such as long fingers and toes,disslocation of the lens of the eye.
5.epistasis is a phenomena in which the effect of one gene is dependant on the prsesnce of one or
more modifier genes.In case of dominant epistasis there is complete dominance at both gene
pairs however when one gene is dominant it hides the phenotype of the other gene.
in recessive epistasis recessive genes at one locus mask the phenotypic expression of other gene
locus.
example, the gene for male pattern baldness is epistatic to the gene for red hair..
Security ImplicationsWhat are the implications of Shannon’s work o.pdf
Security Implications
What are the implications of Shannon’s work on security?
Shannon, C. (1948, July). A Mathematical Theory of Communication. The Bell System
Technical Journal, 27(3), 379-423.
Solution
Implications of Shannon’s work on security :-
1) Shannon\'s work on cryptography security systems.
2) Shannon proved the cryptographic one-time password is unbreakable in his classified research
that was later published in October 1949.
3) He can proved the unbreakable system must have essentially the same characteristics as the
one-time padssword system. the key must be truly random as large as the plaintext information
never reused in whole or part of the kept secret key value.
4) The one-time password system by the Soviets was partially broken by the National Security
Agency but this was because of misuses of the one-time password by Soviet cryptographic
technicians..
Protect tissues Spinal cord functions to includes pituitary gland ski.pdf
Protect tissues Spinal cord functions to includes pituitary gland skin includes Controls Nervous
Integumentary includes 16. Endocrine ncludes bones 15 functions in support Body Systems
Muscular includes functions in ovarie Urinary Digestive Circulatory posture in includes bladder
filters males functions in includes breaks down ncludes 11 from Heart stomach (2 to be Word
Bank into females, brain, kidneys, regulate temperature, skeletal, ligaments, wastes, transport,
food, thyroid, esophagus, testes, coordination, absorbed reproductive, movement, blood
Solution
1. Coordination
2. Brain
3. Regulate temperature
4. Ligaments
5. Skeletal
6. Movement
7. Kidney
8. Waste
9. Blood
10. Transport
11. Food
12. Esophagus
13. Females
14. Testes
15. Reproductive
16. Thyroid.
Present and explain each step in human perception process. To what e.pdf
Present and explain each step in human perception process. To what extent we perceive the real
world differently, how do we manage that (all types of perceptual selections) and why (such as
protecting ourselves)
Solution
Perception starts with visualization.
The main sensory component of vision involves the gathering and recording of light scattered
from objects in the surrounding scene, and the forming of a two - dimensional function on the
photoreceptors. Human eyes can perceive visible light. It involves receiving signals in the
nervous system. These signals form chemical and physical stimulation.
Perception starts with processing sensory input which transforms these low-level information to
higher-level information. The next step is processing as per a person\'s knowledge and concepts.
We perceive the real world based on our knowledge and experience. The process of perception
begins with an object in the real world, termed the distal stimulus ordistal object. Our knowledge
of the distal stimulus influences how we percieve the real world.
We manage all types of perceptual selections by looking for familiar cues when we face an
unknown distal stimulus. We do this to bind our senses and stimulate our high level information..
please help with this genetics problem You have 3 Serratia strains (.pdf
please help with this genetics problem You have 3 Serratia strains (A, B, C) that contain different
mutations in a pathway for making prodigiosin. When grown by themselves A, B, and C are all
white. You are given the information that C feeds A and B through diffusion, and A feeds B
through the air. If you streak out the strains A, C, B in this order and next to each other on one
plate, what color would you observe at day 4? A: pink. B: pink. C: pink A: red. B: red. C:
white. A: red. B: red in colonies close to C. C: white. A: red. B: white. C: white. A: red in
colonies close to C. B: red. C: white If these strains are plated together, which strain(s) will be
able to produce the product of CANDY? Strain Sweet Has a defective y and n enzyme. Strain
Sour has a defective s enzyme Neither can feed the other Sour and Sweet can feed each other
Sour strain can feed Sweet strain but Sweet cannot feed sour Sweet strain can feed Sour strain
but Sour cannot feed sweet
Solution
5)
The prodigiosin is a red pigment. In the pigment production pathway C feeds A and B through
diffusion. Therefore, C is white, B is red colored colony located near C and A is red colored
colony.
6)
Sweet strain have defective \'y\' and \'n\' enzymes and sour strain contains defective \'s\' enzyme.
In both the sweet and sour candy production starts with colony A. With the defective ;y\' and \'n\'
enzymes sweet strain cannot feed sour strain, but with defective \'s\' enzyme, sour strain can feed
sweet strain. Option C is correct..
Please briefly describe Two types of statics where you will NOT o.pdf
Please briefly describe Two types of statics where you will NOT observe a classic Medelian
9.3:3:1 ratio in a F; generation.
Solution
In dihybridism without epistasis,double heterozygous parents cross-breed and 4 phenotypical
forms appear in F2.The proportion is 9 individuals double dominant,3 individuals dominant for
the first pair and recessive for the second pair,3 individuals recessive for the first pair and
dominant for the second pair, and 1 individual double recessive so the final ratio becomes
9:3:3:1.Considering that the epistatic gene is the second pair and that the recessive genotype of
the hypostatic gene implies the lack of the characteristic,in the F2 generation of dominant
epistasis,the following phenotypic forms would emerge:13 individuals dominant for the second
pair or recessive for the first pair,meaning that,the characteristic is not manifest;3 individuals
dominant for the first pair and recessive for the second pair,meaning that the characteristic is
manifest.The phenotypical proportion would be 13:3.In recessive epistasis,the phenotypical
forms that would emerge in F2 are:9 individuals double dominant and 7 individuals recessive for
the first pair or recessive for the second pair,meaning that the characteristic is not manifest.
Therefore,the phenotypical proportion would be 9:7.
These examples show how epistasis changes phenotypical forms and proportions,from the
normal 9:3:3:1 in F2 to 13:3 in dominant epistasis or to 9:7 in recessive epistasis..
Northwest Bakery produces seven types of loaf bread and distributes .pdf
Northwest Bakery produces seven types of loaf bread and distributes the bread to local grocery
stores each day of the week. Juanita Gomez is the marketing manager for Northwest Bakery and
she is trying to figure out how to improve sales. She may want to discontinue some types of
bread or add new types. She asked the production department to record the production levels of
the seven types of bread and Bob McGeorge, the production manager, put these data in a file
named Northwest Bakery. The file is attached.
Juanita has just asked you, who has just been hired in the marketing department, to do some
analysis on the data. She would like:
1. (5 pts) A chart showing percentages of each type of bread produced.
2. (7 pts) A table showing the average amount of each type of bread produced per day and the
variability of the production of each type of bread.
3. (8 pts) She also mentioned that Bob would like to see a chart of average daily production so he
can better manage workloads of the production staff.
4. Produce the appropriate charts, graphs and tables in Excel
5. (5 pts) Compile a brief report in a Word document explaining what you did and include the
charts, graphs and/or tables in the document.Day of WeekWhiteWheatMultigrainDark
RyeCinnamon RaisinSour DoughLight
OatSunday496490403251114108228Monday786611570373165148304Tuesday5474744242651
44104256Wednesday51344338031210092180Thursday817669622375181152308Friday375390
29920612488172Saturday700678564366173136248Sunday597502457276140144312Monday53
6530428261135112356Tuesday675703605289201188356Wednesday421433336212100104224
Thursday867576541298152144304Friday506461406254135116264Saturday4703523772088492
172Sunday748643599341153148316Monday376367310199128104208Tuesday6326465863401
74160264Wednesday591504408238140120276Thursday656497415221107120212Friday81767
3644249200180348Saturday653571557316129140224
Solution
Day of Week
White
Wheat
Multigrain
Dark Rye
Cinnamon Raisin
Sour Dough
Light Oat
Sunday
496
490
403
251
114
108
228
Monday
786
611
570
373
165
148
304
Tuesday
547
474
424
265
144
104
256
Wednesday
513
443
380
312
100
92
180
Thursday
817
669
622
375
181
152
308
Friday
375
390
299
206
124
88
172
Saturday
700
678
564
366
173
136
248
Sunday
597
502
457
276
140
144
312
Monday
536
530
428
261
135
112
356
Tuesday
675
703
605
289
201
188
356
Wednesday
421
433
336
212
100
104
224
Thursday
867
576
541
298
152
144
304
Friday
506
461
406
254
135
116
264
Saturday
470
352
377
208
84
92
172
Sunday
748
643
599
341
153
148
316
Monday
376
367
310
199
128
104
208
Tuesday
632
646
586
340
174
160
264
Wednesday
591
504
408
238
140
120
276
Thursday
656
497
415
221
107
120
212
Friday
817
673
644
249
200
180
348
Saturday
653
571
557
316
129
140
224
Total
12779
11213
9931
5850
2979
2700
5532
Percentage
25.06
21.99
19.48
11.5
5.84
5.3
10.9
Day of Week
White
Wheat
Multigrain
Dark Rye
Cinnamon Raisin
Sour Dough
Light Oat
Average
Sunday
496
490
403
251
114
108
228
298.57
Monday
786
611
570
373
165
148
304
422.43
Tuesday.
Interface four common anode seven-segment LEDs to the PIC18F452 micr.pdf
Interface four common anode seven-segment LEDs to the PIC18F452 microcontroller using
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starting from REG20 (address 20H).
Solution
instructions
STA
LDA
LHLD
SHLD
MOV A, X Binary Octal Decimal
Hexadecimal0000000000111100102220011333010044401015550110666011177710001088100
1119910101210A10111311B11001412C11011513D11101614E11111715F.
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