Determine which protist causes each of the following diseases in huma.pdf
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Determine which protist causes each of the following diseases in humans. leishmaniasis Trypanosoma cruzi malaria Plasmodium spp. Leishmania s beaver fever Toxoplasma gondii toxoplasmosis Giardia lamblia Chagas disease Solution Additional Information: 1) Leishmaniasis: This is caused from the bites of sandflies infected with the leishmania parasite.The cutaneous form presents with skin ulcers, while the mucocutaneous form presents with ulcers of the skin, mouth, and nose 2) Malaria: It is caused from the bite of infected female anopheles mosquito. The plasmodium parasite infect RBC\'s which leads to low haemoglobin count and recurrent fever. 3) Beaver fever: Also known as giardiasis, it causes diarrhea, abdominal pain, and weight loss. Vomiting, blood in the stool, and fever are less common. 4) Toxoplasmosis: The parasite attacks the lymph nodes, causing weakening of the immune system and seizures. 5) Chagas disease: Also known as sleeping sickness, the parasite attacks almost all the organs. If left untreated, the disease can prove fatal.Name of diseaseCausative protistLeishmaniasisLeishmania spp.MalariaPlasmodium sppBeaver feverGiardia lambliaToxoplasmosisToxoplasma gondiiChagas diseaseTrypanosoma cruzi.
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Need this in JAVA We have N numbers as an array, you need to find a .pdf
Need this in JAVA We have N numbers as an array, you need to find a prefix array and a suffix
array, which we can get the maximum x or value with all elements in them. Notice that for
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Solution
// C++ program for a Trie based O(n) solution to find max
// subarray XOR
#include
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
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TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
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{
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else if (temp->arr[val] != NULL)
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}
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int maxSubarrayXOR(int arr[], int n)
{
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insert(root, 0);
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int result = INT_MIN, pre_xor =0;
// Traverse all input array element
for (int i=0; i
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[1-val]!=NULL)
temp = temp-.
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middle of the smiley face it changes color. It will alternate between the yellow and red smiley
faces.
two images have been given to you , redSmiley.gif and happyFace.gif to get started.
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return (new Dimension(500, 500));
}
@Override
public void paintComponent(Graphics g) {
super.paintComponent(g);
g.clearRect(0, 0, getWidth(), getHeight());
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private cl.
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hfffdzxdtychfrdfdftfhdszexrytfudfhgjyfdftgyjcfxdgyfdxgytfdgftyudfrtydrfxtyudfxgyfcdxcgyfcyutdfxytudfxtyufyugfygufcugycfhiugcfugycfhgiughcf
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Need this in JAVA We have N numbers as an array, you need to find a .pdf
Need this in JAVA We have N numbers as an array, you need to find a prefix array and a suffix
array, which we can get the maximum x or value with all elements in them. Notice that for
prefix[0, 1] and suffix[r, n - 1], do not intersect (I
Solution
// C++ program for a Trie based O(n) solution to find max
// subarray XOR
#include
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[1-val]!=NULL)
temp = temp->arr[1-val];
// If there is no prefix with opposite
// bit, then look for same bit.
else if (temp->arr[val] != NULL)
temp = temp->arr[val];
}
return pre_xor^(temp->value);
}
// Returns maximum XOR value of a subarray in arr[0..n-1]
int maxSubarrayXOR(int arr[], int n)
{
// Create a Trie and insert 0 into it
TrieNode *root = newNode();
insert(root, 0);
// Initialize answer and xor of current prefix
int result = INT_MIN, pre_xor =0;
// Traverse all input array element
for (int i=0; i
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[1-val]!=NULL)
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Modify the following source code so that when the mouse is clicked w.pdf
Modify the following source code so that when the mouse is clicked within 30 pixels of the
middle of the smiley face it changes color. It will alternate between the yellow and red smiley
faces.
two images have been given to you , redSmiley.gif and happyFace.gif to get started.
Just for fun:
Create more than 2 colors of smiley faces and rotate through the images changing each time the
mouse is clicked within the threshold.
Use different images other than smiley faces, change images each time the mouse is clicked
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Add a sound each time the image changes.
Have the image slow down, if it has been more than x number of seconds since the image has
changed.
Solution
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import java.awt.event.*;
import java.awt.image.BufferedImage;
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public void run() {
new ChangeImageOnClick().showContentsImage();
}
});
}
}
class CustomPanel extends JComponent {
private final int SIZE = 50;
private int imageX = 100;
private int imageY = 100;
private int imageIndex;
private ImageIcon image;
private ImageIcon firstImage;
private ImageIcon secondImage;
private java.net.URL url;
private Rectangle boundsForMouse;
public CustomPanel() {
image = new ImageIcon();
try {
url = new java.net.URL(\"please mention url location of imaage\");
firstImage = new ImageIcon(url);
url = new java.net.URL(\"please mention url location of image\");
secondImage = new ImageIcon(url);
} catch(Exception e) {
e.printStackTrace();
}
imageIndex = 1;
image.setImage(firstImage.getImage());
boundsForMouse = new Rectangle(imageX - 30,
imageY - 30,
firstImage.getIconWidth() + 60,
firstImage.getIconHeight() + 60);
setOpaque(true);
addMouseListener(new MousePosition());
}
private int setImage(int counter) {
System.out.println(\"Image Index : \" + counter);
if (counter == 1) {
image = new ImageIcon();
image.setImage(secondImage.getImage());
boundsForMouse = new Rectangle(imageX - 30,
imageY - 30,
secondImage.getIconWidth() + 60,
secondImage.getIconHeight() + 60);
repaint();
counter++;
return (counter);
} else if (counter == 2) {
image = new ImageIcon();
image.setImage(firstImage.getImage());
boundsForMouse = new Rectangle(imageX - 30,
imageY - 30,
firstImage.getIconWidth() + 60,
firstImage.getIconHeight() + 60);
repaint();
return (--counter);
}
return 1;
}
@Override
public Dimension getPreferredSize() {
return (new Dimension(500, 500));
}
@Override
public void paintComponent(Graphics g) {
super.paintComponent(g);
g.clearRect(0, 0, getWidth(), getHeight());
g.drawImage(image.getImage(), imageX, imageY, null);
}
private cl.
looking for help with this question, have another question that.pdf
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Draw the Ewald sphere of reflection (in 2-dimensions) for a single-crystal sample of coppr aw
the Ewald sphere of reflection (in 2-dimensions) for a single-crystal sample of copper assuming:
- The diffracting plane is (002) ” The [010] direction lies in the plane of the paper - Include all
reciprocal lattice points in the unit cell that lie in this plane. Use = 1.542, and a = 3.615A and a
scale of 0.3A-1-1 cm. Using the drawing, verify the Bragg angle for diffraction from the (002)
planes.
Solution
The Ewald sphere, or sphere of reflection, is a sphere of radius 1/ passing through the origin O of
the reciprocal lattice. The incident direction is along a radius of the sphere, IO (Figure 1). A
reflected direction, of unit vector sh, will satisfy the diffraction condition if the diffraction vector
OH = IH – IO = sh/ – so/ (so unit vector in the direction IO) is a reciprocal lattice vector, namely
if H is a node of the reciprocal lattice (see Diffraction condition in reciprocal space) . If other
reciprocal lattice nodes, such as G, lie also on the sphere, there will be reflected beams along IG,
etc. This construction is known as the Ewald construction. When the wavelength is large, there
are seldom more than two nodes, O and H, of the reciprocal lattice simultaneously on the Ewald
sphere. When there are three or more, one speaks of multiple diffraction, multiple scattering or n-
beam diffraction. This situation becomes increasingly frequent as the wavelength decreases and
is practically routine for very short wavelengths such as those of -rays and electrons. The
curvature of Ewald sphere then becomes negligible and it can often be approximated by a plane.
Many reflections must then be taken into account at the same time.
When the wavelength changes, the radius of the Ewald sphere changes and Figure 2 illustrates
the case of the reciprocal (110) lattice plane of a silicon crystal and two Ewald spheres
corresponding to AgK (1 = 0.709 Å) and CuK (2 = 1.54 Å) radiation, respectively. If the incident
beam is a white beam, with a wavelength range minmax, there will be a nest of Ewald spheres of
radii 1/max 1/ 1/min.
A crystal can be described as a lattice of points of equal symmetry. The requirement for
constructive interference in a diffraction experiment means that in momentum or reciprocal
space the values of momentum transfer where constructive interference occurs also form a lattice
(the reciprocal lattice). For example, the reciprocal lattice of a simple cubic real-space lattice is
also a simple cubic structure. The aim of the Ewald sphere is to determine which lattice planes
(represented by the grid points on the reciprocal lattice) will result in a diffracted signal for a
given wavelength, , of incident radiation. The incident plane wave falling on the crystal has a
wave vector Ki whose length is 2p . The diffracted plane wave has a wave vector K.
In a population of Canadian Eskimos the autosomal recessive gene for.pdf
In a population of Canadian Eskimos the autosomal recessive gene for albinism exists at an allele
frequency of 0.04. What proportion of these individuals are expected to heterozygous? a. 0.02 b.
0.0285 c. 0.0384 d. 0.0768 e. 0.0112
Solution
p + q = 1
q = 0.04
p = 1- q = 1 - 0.04 = 0.96
The proportion of heterozygous individual is = 2 X 0.96 X 0.04 = 0.0768.
How is that terrorists are able to use media to their advantage an.pdf
How is that terrorists are able to use media to their advantage and why is it important for them to
capitalize the media?
How is that terrorists are able to use media to their advantage and why is it important for them to
capitalize the media?
Solution
Due to the convenience, affordability, and broad reach of social media platforms such as
YouTube, Facebook and Twitter, terrorist groups have increasingly used social media to further
their goals and spread their message.
Terror groups take to social media because social media tools are cheap and accessible, facilitate
quick, broad dissemination of messages, and allow for unfettered communication with an
audience without the filter or \"selectivity\" of mainstream news outlets.
Identify the three ventral body cavities and the two dorsal body cavi.pdf
Identify the three ventral body cavities and the two dorsal body cavities in the following
diagram. Then, name one organ found in each cavity.
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A. Cranial cavity, Brain.
B. Spinal canal, spinal cord
C. Thoracic cavity, Lungs.
D. Abdominal cavity, Liver, stomach.
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I have a Student.java class constructor like thisAnd I need to re.pdf
I have a Student.java class constructor like this:
And I need to read a text file that has information like this:
How can I read data from this file to an array (no arraylist) using loops? My code doesn\'t work.
Here\'s what I have so far:
Thanks in advance!
Solution
Correct Code:
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.lang.reflect.Array;
public class Student {
private String lastName;
private String FirstName;
private String idNo;
private Course courseArray[];
private String grade;
private double gpa;
static int filecouses=0;
public static Course courses[] = new Course[100];
public Student(String lastName,String FirstName,String idNo,Course[] courses,String
grade,double gpa)
{
this.setLastName(lastName);
this.setFirstName(FirstName);
this.setIdNo(idNo);
this.setcourseArray(courses);
this.setGrade(grade);
this.setGpa(gpa);
}
public Student(){
}
public static void main(String[] args) throws IOException {
FileReader file=new FileReader(\"student.txt\");
BufferedReader br=new BufferedReader(file);
String inputLine=br.readLine();
String nines=\"-999\";
while(inputLine!=null)
{
if(!(inputLine.equals(nines))){
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String[] arr=inputLine.split(\",\");
s.setLastName(arr[0]);
System.out.println(arr[0]);
s.setFirstName(arr[1]);
s.setIdNo(arr[2]);
inputLine=br.readLine();}
while((inputLine!=null)&&!(inputLine.equals(nines))){
String[] ar=inputLine.split(\",\");
System.out.println(ar[0]);
String course=(ar[0]);
String credits=(ar[1]);
String grade=(ar[2]);
Course c=new Course(course, credits, grade);
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System.out.println(c);
System.out.println(inputLine);*/
inputLine=br.readLine();
courses[filecouses++]=c;
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if(inputLine.equals(nines)){
inputLine=br.readLine();
String[] a=inputLine.split(\",\");
Student st=new Student();
st.setGpa(Double.parseDouble(a[1]));
st.setGrade(a[0]);
inputLine=br.readLine();
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}
}
String getLastName() {
return lastName;
}
void setLastName(String lastName) {
this.lastName = lastName;
}
String getFirstName() {
return FirstName;
}
void setFirstName(String firstName) {
FirstName = firstName;
}
String getIdNo() {
return idNo;
}
void setIdNo(String idNo) {
this.idNo = idNo;
}
Course[] getCourses() {
return courseArray;
}
void setcourseArray(Course[] courseArray) {
this.courseArray = courseArray;
}
String getGrade() {
return grade;
}
void setGrade(String grade) {
this.grade = grade;
}
double getGpa() {
return gpa;
}
void setGpa(double gpa) {
this.gpa = gpa;
}
}
class Course {
String course;
String credits;
String grade;
public Course(String course,String credits,String grade){
this.course=course;
this.credits=credits;
this.grade=grade;
}
}.
How do incomplete dominance and codominance differ There is no diff.pdf
How do incomplete dominance and codominance differ? There is no difference between
incomplete dominance and codominance. Incomplete dominance occurs when the degree of
expression of a phenotype is influenced by environmental factors; codominance when the
expression of a phenotype is influenced by the genetic background. Incomplete dominance
means the heterozygote expresses an intermediate phenotype between the homozygotes;
codominance the heterozygote expresses both homozygous phenotypes simultaneously.
Incomplete dominance means the heterozygote expresses both homozygous phenotypes
simultaneously; codominance heterozygote has an intermediate phenotype between the
homozygotes. Codominance arises for loci with three or more alleles; incomplete dominance
occurs for loci with only two alleles. True or False? Progeny from the same cross often
displaying phenotypic variability is a characteristic exhibited by a cytoplasmically inherited trait.
True False True or False? Ma
Solution
Incomplete dominance results in a offspring with phenotype which is in middle of the parental
phenotypes. Whereas in co-dominance, the offspring shows the dominant and recessive
phenotypes of parents in a single organism. The correct option is.............incomplete dominance
means the heterozygote expresses an intermediate phenotype between the homozygotes; co-
dominance get expressed when the heterozygote expresses both the homozygous phenotypes
simultaneously..
Help please, I have attached LinkedList.cpp and LinkedList.hPlease.pdf
Help please, I have attached LinkedList.cpp and LinkedList.h
Please help with the implementation of the Stack.h and Stack.cpp as well
LinkedList.cpp
~~~~~~~~~~~~~~
#include \"LinkedList.h\"
LinkedList::LinkedList()
{
first = last = NULL;
}
LinkedList::~LinkedList()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
void LinkedList::insertAtBack(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
temp->next = NULL;
if (last == NULL)
first = last = temp;
else
{
last->next = temp;
last = temp;
}
}
bool LinkedList::removeFromBack()
{
if (first == NULL)
return false;
else if (first == last)
{
free(first);
first = last = NULL;
}
else
{
Node *temp;
temp = first;
while (temp->next != last)
temp = temp->next;
last = temp;
temp = temp->next;
free(temp);
last->next = NULL;
}
return true;
}
void LinkedList::print()
{
Node *temp = first;
while (temp != last)
{
cout << temp->val << \" \";
temp = temp->next;
}
if (temp != NULL)
cout << temp->val;
}
bool LinkedList::isEmpty()
{
if (first == last)
return true;
return false;
}
int LinkedList::size()
{
int count = 0;
Node *temp = first;
if (temp == NULL)
return count;
while (temp != last)
{
count++;
temp = temp->next;
}
return count + 1;
}
void LinkedList::clear()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
/*Exercise2 */
void LinkedList::insertAtFront(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
if (last == NULL)
first = last = temp;
else
{
temp->next = first;
first = temp;
}
}
bool LinkedList::removeFromFront()
{
if (first == NULL)
return false;
Node *temp = first;
first = first->next;
free(temp);
if (first == NULL)
last = NULL;
return true;
}
LinkedList.h
~~~~~~~~~~~
#include \"LinkedList.h\"
LinkedList::LinkedList()
{
first = last = NULL;
}
LinkedList::~LinkedList()
{
while (first != NULL)
{
Node *temp;
temp = first;
first = first->next;
free(temp);
}
last = NULL;
}
void LinkedList::insertAtBack(int valueToInsert)
{
Node *temp = (Node *)malloc(sizeof(Node));
temp->val = valueToInsert;
temp->next = NULL;
if (last == NULL)
first = last = temp;
else
{
last->next = temp;
last = temp;
}
}
bool LinkedList::removeFromBack()
{
if (first == NULL)
return false;
else if (first == last)
{
free(first);
first = last = NULL;
}
else
{
Node *temp;
temp = first;
while (temp->next != last)
temp = temp->next;
last = temp;
temp = temp->next;
free(temp);
last->next = NULL;
}
return true;
}
void LinkedList::print()
{
Node *temp = first;
while (temp != last)
{
cout << temp->val << \" \";
temp = temp->next;
}
if (temp != NULL)
cout << temp->val;
}
bool LinkedList::isEmpty()
{
if (first == last)
return true;
return false;
}
int LinkedList::size()
{
int count = 0;
Node *temp = first;
if (temp == NULL)
return count;
while (temp != last)
{
count++;
temp = temp->next;
}
return count + 1;
}
void LinkedList::clear()
{
while (first != NULL)
{
Node *.
Fruits develop from microsporangia receptacles fertilized eggs ovarie.pdf
Fruits develop from microsporangia receptacles fertilized eggs ovaries ovules
Solution
Answer : The correct answer is 4 (Ovaries). Pollination in the flowering plants results in the
formation of seeds from Ovules where as the fruits are formed from Ovaries.
Supplementary information : A true fruit consists of two main parts. Seeds which are developed
from ovules and contains endosperm, embryo and seed coat
and
The pericarp developed from the walls of the ovary. Structure and thickness varies greatly in
different kinds of fruits..
Find all equilibrium solutions of the equation dydx = y^2 - 1. Use o.pdf
Find all equilibrium solutions of the equation dy/dx = y^2 - 1. Use only qualitative analysis to
sketch the graphs of solutions having various representative initial values y(0).
Solution
Equilibrium solutions are given by dy/dt=0
so y2-1=0
(y+1)(y-1)=0
y=-1,y=1
tharefore y=-1,y=1 are equilibrium solutions.
Describe the transmission of the meningococcus, and discuss the path.pdf
Describe the transmission of the meningococcus, and discuss the pathogenesis of meningococcal
disease in the body including associated signs and symptoms.
Solution
Transmission:
Meningococcal disease is caused by the bacterium Neisseria meningitidis.Meningococcal disease
is spread from person to person.The bacteria are spread by exchanging respiratory and throat
secretions such as saliva or spit,during close or lengthy contact,especially if living in the same
household.The bacteria are not spread by casual contact or by simply breathing the air where a
person with meningococcal disease has been.Sometimes bacteria spread to people who have had
close or lengthy contact with a patient with meningococcal disease.People in the same
household,roommates,or anyone with direct contact with a patient\'s oral secretions would be
considered at increased risk of getting the infection.
Pathogenesis:
Meningococcal disease causes life-threatening meningitis and sepsis conditions.In the case of
meningitis,bacteria attack the lining between the brain and skull called the meninges.The
meninges begin to swell and affects the central nervous system.The sepsis type of infection is
more deadly,and results in a severe blood poisoning called meningococcal sepsis which affects
the entire body.In this,bacterial toxins rupture blood vessels and can rapidly shut down vital
organs.Within hours,patient\'s health can change from seemingly good to mortality.
This bacterium is surrounded by a slimy outer coat that contains disease-causing endotoxin. The
levels of endotoxin produced by meningococcal bacteria are 100 to 1,000 times greater than
normal.As the bacteria multiply and move through the bloodstream,it sheds concentrated
amounts of toxin.The endotoxin directly affects the heart,reducing its ability to circulate
blood,and also build pressure on blood vessels throughout the body such that some blood vessels
start to hemorrhage,major organs like the lungs and kidneys gets damaged.Patients suffering
from meningococcal disease are treated with a large dose of antibiotic.The systemic antibiotic
flowing through the bloodstream rapidly kills the bacteria but,as the bacteria are killed,even
more toxin is released.It takes many days for the toxin to get neutralised from the body by using
treatment.
Signs and symptoms: The most common symptoms are a stiff neck,high fever,sensitivity of light
,confusion, headaches and vomitings.Bacterial meningitis may result in brain damage,hearing
loss or a learning disability.Meningococcal septicaemia is characterized by a haemorrhagic rash
and rapid circulatory collapse..
dass Defining Species art C Gene flow and the biological species conc.pdf
dass Defining Species art C Gene flow and the biological species concept Recall that the
biological species conoept defines a species as a group of whose members have the in nature and
produce viable, fertle ofspring This means that al members of these populations (called
conspecifics) can contribute to a common gene pool. Conspecifics are organisms or populations
that belong to the same species. Genes are transferred between oonspecific populations through
both direct and indrect fow. Biologists studying disease susceptibility in mice used genetic
techniques to look for gene fow in 12 wild populations of mioe. The populations (designated
Athrough L) were located along a 210km ransect The found evidenoe of gene flow between
some, but not al, pairs of populations In the grid below, a*** indcates a pair of populations in
which dieetgene flow was deteeted and a wandoares a that showed no evidence of direct gene
fiow. For help reading this grid, see Hint 2.
Solution
Look at the population A: \"+\" at K whic means Direct gene flow.
Now look at K: Straight down you get \"+\" for both D and F which means direct gene flow.
Also, H and I has \"+\" signs for D which shows indirect gene flow.
Look at population B: \"+\" at L which means direct gene flow.
Now look at L, straight down you get \"+\" for both E which shows indirect gene flow.
Look at population C: \"+\" at G and J which means direct gene flow.
So, in a nutshell :
Population A: K,D,H,I,F
Population B: L,E
Population C: G, J.
Describe one way in which you could use butterfly rearing in the fut.pdf
Describe one way in which you could use butterfly rearing in the future. If you would use it for
education, what would you want your students to learn? What grade would you utilize it in and
what standard(s) would it meet?
min 250 words
Solution
First students would learn observing the Life Cycle of a butterfly, students obserrve one
organism over time and compare its early development (caterpillar) to its later development
(butterfly). A fundamental observaion skill in science is comparing and contrasting. Students will
also compare actual characteristics of a butterfly with a functional representation of a butterfly.
There are four stages in the metamorphosis of butterflies and moths : egg, larva, pupa, and adult.
Egg: Eggs are laid on plants by the adult female butterfly. These plants will then become the
food for the hatching caterpillars. Eggs can be laid from spring, summer, or fall.This depends on
the species of butterfly. Females lay a lot of eggs at once so that at least some of them survive.
Butterfly eggs can be very small
Caterpillar: The Feeding Stage : The next stage is the larva. This is also called a caterpillarr if the
insect is a butterfly or a moth.The job of the caterpillar is to eat. As the caterpillar grows it splits
its skin and sheds it about 4 or 5 times. Food eaten at this time is stored and used later as an
adult. Caterpillars can grow 100 times size during this stage. For example, a monarch butterfly
egg is the size of a pinhead and the caterpillar that hatches from this tiny eggs isn\'t much bigger.
But it will up to 2 inches long in several weeks.
Puoa: The Transition Stage : When the caterpillar is full grown and stops eating, it becomes a
pupa. The pupa of butterflies is also called a chrysalis.Depending on the species, the pupa may
suspended under a branch, hidden in leaves or buried underground. The pupa of many moths is
protected inside a coccoon of silk.This stage can last from a few weeks, a month or even longer.
Some species have a pupal stage that lasts for two years. It may look like nothing is going on but
big changes are happening inside. Special cells that were present in the larva are now growing
rapidly. They will become the legs, wings, eyes, and other parts of the adult butterfly. Many of
the original larva cells will provide energy for these growing adult cells.
Adult: The Reproductive Stage : The adult stage is very different from the larva. The caterpillar
has a few tiny eyes, stubby legs, and very short antennae. The adult have long legs , long
antennae, and compound eyes. The can also fly by using their large and colorful wings. But they
can\'t do is grow. The adult\'s work is to mate and lay eggs. Some species of adult butterflies get
energy by feeding on nectar from flowers but many species don\'t feed at all.Flying comes in
handy but female can fly easily fly from place to place to find the right plant for its eggs. This is
important because caterpillars can\'t fly far. Most adult butterf.
Critical Thinking Course...PhilosophyIn Chapter 8, we learned the .pdf
Critical Thinking Course...Philosophy
In Chapter 8, we learned the basics of truth-functional logic. It is important to remember that the
purpose of truth-functional logic is to determine which arguments are valid and which are
invalid, and common invalid forms are known as truth-functional fallacies. Consider the
following question. Which of the following is a well-known truth-functional fallacy? (Points : 4)
a. Modus tollens
b. Modus ponens
c. Valid chain
d. Denying the consequent
Solution.
Can I get a detailed answer on the following question. Bacterial ind.pdf
Can I get a detailed answer on the following question. Bacterial indicatior to monitor faecal
contamination in water. (a) bacterial indicator (b) rationale for using indicator organism
Solution
a). Indicator bacteria are the different types of bacteria that are expected to present in the
contaminated water samples. These bacteria are not harmful to humans but indicate the presence
of pathogenic bacteria. Thus, they are used as a measure of indicating contaminated water.
b). The microbial limit of human feces is approximately 100 billion/gram. They include different
bacteria species, some of them can be pathogenic such as Campylobacter and Salmonella.
Certain viruses, protozoa and parasites may also present in the feces. The commonly used
indicator bacteria are, totalcoli formsorfecalcoli forms..
Another algebraic way to express a constraint is E_1 = E_2, where bot.pdf
Another algebraic way to express a constraint is E_1 = E_2, where both E_1 and E_2 are
relational-algebra expressions. Can this form of constraint express more than two forms ?
Solution
The other algebraic way to express a constraint is E1 = E2. Here E1and E2are the relational-
algebra expressions. Now we are expressing them in the two constraints:
E1 – E2= ø
E1 E2 AND E2 E1.
An infinitely long sheet of charge of width L lies in the xy -plane .pdf
An infinitely long sheet of charge of width L lies in the xy -plane between x=L/2 and x=L/2 .
The surface charge density is . Derive an expression for the electric field E at height z above the
centerline of the sheet. (Assume that z0 .)
Express your answer in terms of the variables , L , z , unit vector k^ , and appropriate constants.
Solution
The sheet is kept on the xy plane with a width of L along the x axis.
We would consider a section of the sheet at a distance of x from the origin of thickness dx. Since
it extends infinitely along the y axis. We have an infinite rod with at distance of x from origin
with linear charge density of dx
Now, the electic field at z due to this infinitely long strip would be dx / 2o * sqrt(z^2 + x^2)
However, by symmetry we can say that the net electric field will point along the z axis as the
components in the xy plane would eventually cancel out.
Hence dE = z dx / 2o * (z^2 + x^2)
We integrate the above to obtain:
E = ( / 2o)[atan(x/z)](from -L/2 to + L/2)
or, E = ( / 2o)[atan(L/2z) - atan(-L/2z)] = ( /o)[atan(L/2z)]
NOTE: There are two ways of approaching the problem, either you can assume a small section
of thickness dy at distance y and then proceed, or you can assume an infinitely long strip at some
distance x and then go ahead.
However, it needs to be mentioned that the first way gives a little tricky integral whereas the
second one, as above, is fairly straightforward. The integral above is of the form dx / (a^2 + x^2)
= (1/a) atan (x/a)
Further, while subtracting the max and min value for the integral, the negative. you would need
the relation tanx = tan-x
Hence the difference of atan of negative x and postiive x ends up getting added..
A study tracked incidence rates of a blood vessel disorder over time.pdf
A study tracked incidence rates of a blood vessel disorder over time for individuals 50 years and
older (see figure below).
What is the name for the type of the time trend pattern demonstrated in this chart? (Choose the
one best answer)
Secular
Cyclical
Calendar
d Influenza
Although it looks like incidence of the disease has increased over the past 40+ years, the authors
dispel this hypothesis. What might be one explanation for what appears to be higher incidence
rates of this disease over time if the incidence rates, in fact, did not increase?
The disease is being spread more rapidly due to travel.
People are living longer with the disease.
cThe population over 50 years has grown in comparison to the under-50 population.
d People who contract the disease are now more likely to be diagnosed due to more widespread
knowledge of the disease among providers.
a Women were more likely to contract the disease than men.
b The rates of disease dropped substantially in 1980.
c The cyclical nature of the disease indicates that it is infectious.
d The prevalence appears to have increased over time.a
Secularb
Cyclicalc
Calendar
d Influenza
Although it looks like incidence of the disease has increased over the past 40+ years, the authors
dispel this hypothesis. What might be one explanation for what appears to be higher incidence
rates of this disease over time if the incidence rates, in fact, did not increase?a
The disease is being spread more rapidly due to travel.b
People are living longer with the disease.
cThe population over 50 years has grown in comparison to the under-50 population.
d People who contract the disease are now more likely to be diagnosed due to more widespread
knowledge of the disease among providers. The authors of this study concluded from the data
shown in this figure that this disease is probably infectious. Why did they make this
conclusion?
a Women were more likely to contract the disease than men.
b The rates of disease dropped substantially in 1980.
c The cyclical nature of the disease indicates that it is infectious.
d The prevalence appears to have increased over time.
Solution
can you please put the figure so that the questions can be answered..
A distant galaxy was found to have a redshift of 1.8. How far away i.pdf
A distant galaxy was found to have a redshift of 1.8. How far away is the galaxy? Express you
answer in Mpc units. You can assume that the galaxy has zero peculiar velocity and that the
Hubble constant is 70 km/s/Mpc.
Solution
the distance is calculated as follows:
z = HD
D = z/H
= 1.8 km/s / 70 km/s/Mpc.
= 0.0257 Mpc.
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This slide is prepared for master's students (MIFB & MIBS) UUM. May it be useful to all.Polish students' mobility in the Czech Republic
Polish students mobility to the Czech Republic within eTwinning project "Medieval adventures with Marco Polo"
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