making and testing a prediction suppose that after 22 months, gupp.pdf
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making and testing a prediction suppose that after 22 months, guppies from the transplanted population were returned to the source pool.what would most likely happen to those guppies. Solution Answer: 1) Due to habitat change the feeding frequency affect 2) Predation problem: Guppies have many predators, such as larger fish and birds, in their natural habitats. Some of their common predators in the wild are Crenicichla alta, Anablepsoides hartii, and Aequidens pulcher. Guppies\' small bodies and the bright coloration of males make them easy prey, and like many fish, they often school together to avoid predation..
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Please help! Computer Science C++ programming code needed.Thank yo.pdf
Please help! Computer Science C++ programming code needed.
Thank you!
Solution
interface MyListADT { void add(T var); void add(T var,int pos); void display(); T
remove(int pos); void clear(); boolean contains(Object o); } interface MyListADT {
void add(T var); void add(T var,int pos); void display(); T remove(int pos); void
clear(); boolean contains(Object o); }interface MyListADT { void add(T var); void
add(T var,int pos); void display(); T remove(int pos); void clear(); boolean
contains(Object o); }.
Please Answer, will rate! Exercise 1 - Classifying financial stateme.pdf
Please Answer, will rate! Exercise 1 - Classifying financial statement items: Below is a list of
financial statement items. For each item, mark whether is part of assets (A), liabilities (L),
shareholders’ cquity (SE), Revenue (R), or expense (E). (A) / (L)/ (SE) / (R)/ (E) Item: (1)
Accounts payable (2) Accounts receivable (3) Cash and cash equivalents (4) Common Stock (5)
Cost of goods sold (6) Dividends Payable (7) Income taxes (8) Interest expense (9) Inventories
(10) Land (11) Net sales (12) Notes payable (13) Notes receivable (14) Property, plant, and
equipment (15) Research and development expense (16) Retained earnings (17) Taxes Payable
Solution
Answer:
1
Accounts Payable
(L)Liability
2
Accounts receivable
(A) Assets
3
Cash and cash equivalents
(A) Assets
4
Common stock
(SE) Stock Holder\'s Equity
5
Cost of Goods sold
(E)Expanses
6
Dividend payable
(L)Liability
7
Income tax
(E)Expanses
8
Interest Expanses
(E)Expanses
9
Inventories
(A) Assets
10
Land
(A) Assets
11
Net sales
(,R)revenue
12
Notes payable
(L)Liability
13
Notes Receivable
(A) Assets
14
Property Plant equipment
(A) Assets
15
Research and development Expanses
(E)Expanses
16
Retained earning
(SE) Stock Holder\'s Equity
17
Taxes payable
(L)Liability
1
Accounts Payable
(L)Liability
2
Accounts receivable
(A) Assets
3
Cash and cash equivalents
(A) Assets
4
Common stock
(SE) Stock Holder\'s Equity
5
Cost of Goods sold
(E)Expanses
6
Dividend payable
(L)Liability
7
Income tax
(E)Expanses
8
Interest Expanses
(E)Expanses
9
Inventories
(A) Assets
10
Land
(A) Assets
11
Net sales
(,R)revenue
12
Notes payable
(L)Liability
13
Notes Receivable
(A) Assets
14
Property Plant equipment
(A) Assets
15
Research and development Expanses
(E)Expanses
16
Retained earning
(SE) Stock Holder\'s Equity
17
Taxes payable
(L)Liability.
Need 2 -3 paragraphConsidering the aesthetics of the late 19th cen.pdf
Need 2 -3 paragraph
Considering the aesthetics of the late 19th century, how, in your opinion, did 19th century artists
influence Modernism in the 20th century? Use at least one example of 19th century art in your
response as a basis for your analysis.
Solution
Modernism : it is a philosophical development that, alongside social patterns and changes,
emerged from wide-scale and expansive changes in Western culture amid the late nineteenth and
mid twentieth hundreds of years. Among the elements that formed innovation were the
improvement of present day modern social orders and the quick development of urban areas,
took after then by responses of ghastliness to World War I. Innovation additionally dismisses the
conviction of Enlightenment considering, and numerous pioneers rejected religious conviction.
An eminent normal for innovation is reluctance and incongruity concerning scholarly and social
conventions, which regularly prompted to explores different avenues regarding structure,
alongside the utilization of methods that attracted consideration regarding the procedures and
materials utilized as a part of making a canvas, ballad, building, and so on.
modernism unequivocally dismisses the philosophy of authenticity and makes utilization of the
works of the past by the work of repeat, joining, modifying, reiteration, update and farce. Present
day specialists explored different avenues regarding better approaches for seeing and with crisp
thoughts regarding the way of materials and elements of workmanship. An inclination far from
the account, which was trademark for the customary expressions, toward deliberation is normal
for much present day craftsmanship. Later imaginative generation is frequently called
contemporary craftsmanship or postmodern workmanship..
Match each statement to the most appropriate choices. More than one .pdf
Match each statement to the most appropriate choices. More than one choice may be appropriate.
Choices:
A – System Call
B – Library Function
C – User Program
Statements:
Enters the OS kernel via a trap instruction ______
Code is stored in user space ______
Code is stored in kernel space ______
Switches the processor to kernel (supervisor) mode ______
Encapsulates (wraps) other functions ______
Always starts at a fixed physical memory address ______
Returns control to user space once finished ______
Pushes library function arguments onto program stack ______
Calling convention is highly machine dependent ______
Implements the POSIX standard ______
Solution
A) System call - Switches the processor to the kernel (supervisor) mode
B)Library function - Returns control to user space once finished
C)User program - code is stored in kernel space.
List three sources of variability in material properties. If that va.pdf
List three sources of variability in material properties. If that variability can be reduced, describe
how.
Solution
Homogeneous materials: A material of uniform composition throughout that cannot be
mechanically separated into different materials. Examples of homogeneous materials are certain
types of plastics, ceramics, glass, metals, alloys, paper, board and resins.
Isotropic materials: Isotropic material means a material having identical values of a property in
all directions. Glass and metals are examples of isotropic materials.
Anisotropy materials: It is a term used in various scientific disciplines to indicate that certain
properties of matter (such as a material or radiation) vary with the direction from which they are
measured. For instance, if the refractive index or density of a material is different when
measured along different axes, that property is said to be anisotropic.Anisotropy is the opposite
of isotropic, a term used when properties are the same when measured from any direction.
Orthotropic materrials: In materials science and solid mechanics, orthotropicmaterials have
material properties that differ along three mutually orthogonal twofold axes of rotational
symmetry. They are a subset of anisotropic materials, because their properties change when
measured from different directions.
A design is expected to have a good concept about mechanical properties of materials. Different
manufacturing methods and heat treatment can change the materials properties.
Engineering materials are divided as follows:-
1. Metal and Metal Alloys; Example: Iron, Copper, Nickel etc.
2. Non- Metals; Example: Ceramics, rubber, glass etc.
Important Mechanical Properties of Materials are Described Below
a. Strength: This describes the ability to resist any external force. Without Yielding (Yield point
is that point at which plastic deformation starts) breaks.
b. Stiffness: This is measured by modulus of elasticity. It is ability of the material to prevent
deformation when stress is applied.
There are many mechanical properties are present .
The mechanical properties of a materialare those which effect the mechanical strength and ability
of material to be molded in suitable shape. Some of the typical mechanical properties of a
material are listed below-
Strength
Toughness
Hardness
Hardenability
Brittleness
Malleability
Ductility
Creep and Slip
Resilience
Fatigue.
Need the fill in the blank questions answeredSolution1) CamboB.pdf
Need the fill in the blank questions answered
Solution
1) CamboBox
2)CheckedListBox
3)Loop
4)infiniteloop
5)counter
6)Do While loop
7)iteration
8)Exit Do
9)Nested
10)Random
11)Next method
12)nextDouble
13)ppmt function //confirm with code functionality
14)pwt function //should be confirmed with code functionality
15)Ipmt //should be confirmed with code functionality.
Life is wholly dependent on water. List and discuss two reasons why .pdf
Life is wholly dependent on water. List and discuss two reasons why this is?!!
Solution
Life is wholly dependent on water because,
1). Plants get the carbon dioxide necessary for photosynthesis in exchange with water
2). Animals use water in several ways such as, to transport the molecules in the body, to maintain
body temperature, to excrete the waste, etc.
As plants are the primary producers, their survival is the key for survival of the remaining
organisms. The biological cycle will be terminated if there is no water because the platn\'s
survival needs water..
John Rawls argued that societys goal should be to lessen the welfar.pdf
John Rawls argued that society\'s goal should be to lessen the welfare of the most well off
property rights should be equally distributed among all citizens.
Solution
Rawls refers to “the two principles of Justice as Fairness,” but the second has two parts. These
principles address two different aspects of the basic structure of society: the “First Principle”
addresses the essentials of the constitutional structure. It holds that society must assure each
citizen “an equal claim to a fully adequate scheme of equal basic rights and liberties, which
scheme is compatible with the same scheme for all.”
So answer here will be B..
Recommended
Please help! Computer Science C++ programming code needed.Thank yo.pdf
Please help! Computer Science C++ programming code needed.
Thank you!
Solution
interface MyListADT { void add(T var); void add(T var,int pos); void display(); T
remove(int pos); void clear(); boolean contains(Object o); } interface MyListADT {
void add(T var); void add(T var,int pos); void display(); T remove(int pos); void
clear(); boolean contains(Object o); }interface MyListADT { void add(T var); void
add(T var,int pos); void display(); T remove(int pos); void clear(); boolean
contains(Object o); }.
Please Answer, will rate! Exercise 1 - Classifying financial stateme.pdf
Please Answer, will rate! Exercise 1 - Classifying financial statement items: Below is a list of
financial statement items. For each item, mark whether is part of assets (A), liabilities (L),
shareholders’ cquity (SE), Revenue (R), or expense (E). (A) / (L)/ (SE) / (R)/ (E) Item: (1)
Accounts payable (2) Accounts receivable (3) Cash and cash equivalents (4) Common Stock (5)
Cost of goods sold (6) Dividends Payable (7) Income taxes (8) Interest expense (9) Inventories
(10) Land (11) Net sales (12) Notes payable (13) Notes receivable (14) Property, plant, and
equipment (15) Research and development expense (16) Retained earnings (17) Taxes Payable
Solution
Answer:
1
Accounts Payable
(L)Liability
2
Accounts receivable
(A) Assets
3
Cash and cash equivalents
(A) Assets
4
Common stock
(SE) Stock Holder\'s Equity
5
Cost of Goods sold
(E)Expanses
6
Dividend payable
(L)Liability
7
Income tax
(E)Expanses
8
Interest Expanses
(E)Expanses
9
Inventories
(A) Assets
10
Land
(A) Assets
11
Net sales
(,R)revenue
12
Notes payable
(L)Liability
13
Notes Receivable
(A) Assets
14
Property Plant equipment
(A) Assets
15
Research and development Expanses
(E)Expanses
16
Retained earning
(SE) Stock Holder\'s Equity
17
Taxes payable
(L)Liability
1
Accounts Payable
(L)Liability
2
Accounts receivable
(A) Assets
3
Cash and cash equivalents
(A) Assets
4
Common stock
(SE) Stock Holder\'s Equity
5
Cost of Goods sold
(E)Expanses
6
Dividend payable
(L)Liability
7
Income tax
(E)Expanses
8
Interest Expanses
(E)Expanses
9
Inventories
(A) Assets
10
Land
(A) Assets
11
Net sales
(,R)revenue
12
Notes payable
(L)Liability
13
Notes Receivable
(A) Assets
14
Property Plant equipment
(A) Assets
15
Research and development Expanses
(E)Expanses
16
Retained earning
(SE) Stock Holder\'s Equity
17
Taxes payable
(L)Liability.
Need 2 -3 paragraphConsidering the aesthetics of the late 19th cen.pdf
Need 2 -3 paragraph
Considering the aesthetics of the late 19th century, how, in your opinion, did 19th century artists
influence Modernism in the 20th century? Use at least one example of 19th century art in your
response as a basis for your analysis.
Solution
Modernism : it is a philosophical development that, alongside social patterns and changes,
emerged from wide-scale and expansive changes in Western culture amid the late nineteenth and
mid twentieth hundreds of years. Among the elements that formed innovation were the
improvement of present day modern social orders and the quick development of urban areas,
took after then by responses of ghastliness to World War I. Innovation additionally dismisses the
conviction of Enlightenment considering, and numerous pioneers rejected religious conviction.
An eminent normal for innovation is reluctance and incongruity concerning scholarly and social
conventions, which regularly prompted to explores different avenues regarding structure,
alongside the utilization of methods that attracted consideration regarding the procedures and
materials utilized as a part of making a canvas, ballad, building, and so on.
modernism unequivocally dismisses the philosophy of authenticity and makes utilization of the
works of the past by the work of repeat, joining, modifying, reiteration, update and farce. Present
day specialists explored different avenues regarding better approaches for seeing and with crisp
thoughts regarding the way of materials and elements of workmanship. An inclination far from
the account, which was trademark for the customary expressions, toward deliberation is normal
for much present day craftsmanship. Later imaginative generation is frequently called
contemporary craftsmanship or postmodern workmanship..
Match each statement to the most appropriate choices. More than one .pdf
Match each statement to the most appropriate choices. More than one choice may be appropriate.
Choices:
A – System Call
B – Library Function
C – User Program
Statements:
Enters the OS kernel via a trap instruction ______
Code is stored in user space ______
Code is stored in kernel space ______
Switches the processor to kernel (supervisor) mode ______
Encapsulates (wraps) other functions ______
Always starts at a fixed physical memory address ______
Returns control to user space once finished ______
Pushes library function arguments onto program stack ______
Calling convention is highly machine dependent ______
Implements the POSIX standard ______
Solution
A) System call - Switches the processor to the kernel (supervisor) mode
B)Library function - Returns control to user space once finished
C)User program - code is stored in kernel space.
List three sources of variability in material properties. If that va.pdf
List three sources of variability in material properties. If that variability can be reduced, describe
how.
Solution
Homogeneous materials: A material of uniform composition throughout that cannot be
mechanically separated into different materials. Examples of homogeneous materials are certain
types of plastics, ceramics, glass, metals, alloys, paper, board and resins.
Isotropic materials: Isotropic material means a material having identical values of a property in
all directions. Glass and metals are examples of isotropic materials.
Anisotropy materials: It is a term used in various scientific disciplines to indicate that certain
properties of matter (such as a material or radiation) vary with the direction from which they are
measured. For instance, if the refractive index or density of a material is different when
measured along different axes, that property is said to be anisotropic.Anisotropy is the opposite
of isotropic, a term used when properties are the same when measured from any direction.
Orthotropic materrials: In materials science and solid mechanics, orthotropicmaterials have
material properties that differ along three mutually orthogonal twofold axes of rotational
symmetry. They are a subset of anisotropic materials, because their properties change when
measured from different directions.
A design is expected to have a good concept about mechanical properties of materials. Different
manufacturing methods and heat treatment can change the materials properties.
Engineering materials are divided as follows:-
1. Metal and Metal Alloys; Example: Iron, Copper, Nickel etc.
2. Non- Metals; Example: Ceramics, rubber, glass etc.
Important Mechanical Properties of Materials are Described Below
a. Strength: This describes the ability to resist any external force. Without Yielding (Yield point
is that point at which plastic deformation starts) breaks.
b. Stiffness: This is measured by modulus of elasticity. It is ability of the material to prevent
deformation when stress is applied.
There are many mechanical properties are present .
The mechanical properties of a materialare those which effect the mechanical strength and ability
of material to be molded in suitable shape. Some of the typical mechanical properties of a
material are listed below-
Strength
Toughness
Hardness
Hardenability
Brittleness
Malleability
Ductility
Creep and Slip
Resilience
Fatigue.
Need the fill in the blank questions answeredSolution1) CamboB.pdf
Need the fill in the blank questions answered
Solution
1) CamboBox
2)CheckedListBox
3)Loop
4)infiniteloop
5)counter
6)Do While loop
7)iteration
8)Exit Do
9)Nested
10)Random
11)Next method
12)nextDouble
13)ppmt function //confirm with code functionality
14)pwt function //should be confirmed with code functionality
15)Ipmt //should be confirmed with code functionality.
Life is wholly dependent on water. List and discuss two reasons why .pdf
Life is wholly dependent on water. List and discuss two reasons why this is?!!
Solution
Life is wholly dependent on water because,
1). Plants get the carbon dioxide necessary for photosynthesis in exchange with water
2). Animals use water in several ways such as, to transport the molecules in the body, to maintain
body temperature, to excrete the waste, etc.
As plants are the primary producers, their survival is the key for survival of the remaining
organisms. The biological cycle will be terminated if there is no water because the platn\'s
survival needs water..
John Rawls argued that societys goal should be to lessen the welfar.pdf
John Rawls argued that society\'s goal should be to lessen the welfare of the most well off
property rights should be equally distributed among all citizens.
Solution
Rawls refers to “the two principles of Justice as Fairness,” but the second has two parts. These
principles address two different aspects of the basic structure of society: the “First Principle”
addresses the essentials of the constitutional structure. It holds that society must assure each
citizen “an equal claim to a fully adequate scheme of equal basic rights and liberties, which
scheme is compatible with the same scheme for all.”
So answer here will be B..
In Exercises 19-24, find the volume of the solid generated by revolvi.pdf
In Exercises 19-24, find the volume of the solid generated by revolving the region bounded by
the graphs of the equations about the x-axis.
Solution
Area = Integral(ydx) from x= 0 to 3 = int[1/sqrt(1+x) dx] from x= 0 to 3 = 2
sqrt(1+x) from x = 0 to 3 = 2 [sqrt(4) - sqrt(1)] = 2 Volume when rotated along x axxis = Area *
2 *pi = 2*2*pi = 4pi.
In which step of meiosis, the members of tetra separate Solutio.pdf
In which step of meiosis, the members of tetra separate?
Solution
Answer: The tetrads separate during anaphase I of meiosis.
Explanation: Tetrads are formed towards the end of prophase I of meiosis.Separation of
homologous chromosomes (with four sister chromatids) takes place during anaphase I of
meiosis..
Identify the distinguishing characteristics of member of Phylum Cnid.pdf
Identify the distinguishing characteristics of member of Phylum Cnidaria, including symmetry,
number of embryonic tissue layers, unique structures, skeleton and polymorphism (polyp and
medusa), place Cnidaria on a cladogram
Compare and contrast how a general polyp and a medusa feed
Describe why a nerve net was such a major evolutionary adaptation
Describe the characteristics of the four classes of Cnidarians
Label the following on a drawing: tentacles, mouth, anus, gastrovascular cavity, cnidocytes,
nerve net
Solution
Answer1) Symmetry- Radially symmetrical body
Number of embryonic tissue layers- two (endoderm and ectoderm)
Unique structures- Stinging cells (Cnidocytes)
Skeleton- Hydrostatic skeleton
Polymorphism (polyp and medusa)- polyp is a flower like solitary form and medusa is a free-
floating miniature of cnidarians.
Answer 2) a tubular shaped body form is called as polyp. They are sessile and are found as
attached bottom (basal plate) to a solid surface with the opened mouth at the top. while medusa is
an umbrella or bell shape form, having their mouth facing downwards. And they are free
floating. Being a sessile, polyp filter their food from the water by their tiny-beating cilia. While
medusas are free floating and motile so they can trap their feedings, and need not to filter the
water all the time though these structures were more related to the sexual reproduction.
Answer 3) Nerve net is a diffused framework of neurons in the body which can transmit
signals/impulse in different directions from a single point of stimulus.
This became a major evolutionary adaptation because it allowed the organism to sense the
environmental stimulus from any region of its body and respond accordingly. This allowed the
readily development and motion of different body parts such as limb like structures.
Answer 4) These are the four classes of cnidarians.
Hydrozoa are diploblastic and multicellular animals. Which has only mouth opening. They have
a fixed polyp stage and the reproductive medusa stage. They have the gonads as well.
Scyphozoan are present in medusoid forms. They have tentaculocysts as Sense organs. They
have Velarium in endodermal canal (Acraspedote). They endodermal gonads.
Anthozoan are present only in polyp forms. Their medusa stage is absent. They have Mesenteries
which bear nematocyst. They also possess endodermal Gonads.
Cubozoan looks like the jelly fish and called as jelly boxes. They possess box shaped medusa
and they share typical characteristics of scyphozoans..
If most evolutionary change occurs during and immediately after spec.pdf
If most evolutionary change occurs during and immediately after speciation, could speciation
rates and extinction rates explain major patterns in macroevolution? In turn, what factors may
influence speciation and extinction rates (are they all biological, or can geological /
meteorological / astronomical factors also play a role)?
Solution
In the course of evolution, some mutations keep on occurring with every replicative step. These
mutations lead to differences in alleles, which further result in different genotypes. Some
genotypes have higher fitness while others have lower fitness. Natural selection selects the
phenotypes which have higher fitness, and slowly discards the others during the course of
evolution. So, extinction rate depends on fitness and natural selection.
As new phenotypes are selected, they result in formation of new species; the process is known as
speciation. Speciation further results in evolution. Higher the speciation rate, higher will be the
rate of evolution.
So, you can say that most evolutionary changes occur during and after speciation.
Overall the biological factors affecting speciation and extinction rates are:
It is important to note that that apart from the above two factors, geological/astronomical factors
also affect speciation, extinction and evolution. Changes in the earth since times immemorial
ultimately resulted in changes like:.
Imagine that someone picking up roses in a garden gets a scratch fro.pdf
Imagine that someone picking up roses in a garden gets a scratch from a thorn. Some bacteria
(Staphyloccocus aureus and Streptococcus pyogenes) take advantage of the wound to gain access
to the deeper layers of the skin and attempt to cause an infection. Describe the different stages of
the host response (in particular the types of cells and receptors involved) and how the bacteria
will be recognized and eliminated.
Solution
The superficial staphylococcal and streptococcal infections are very common in skin. The
physiological condition of the skin is much suitable to these bacteria. Both these bacteria are
potent toxin and enzyme producing strains.
S.auresus infection on the skin immediately stimulates the inflammatory response and
macrophages and neutrophils are proliferated at the site of infection. The organism can enter the
body through openings in the skin barrier like hair follicles and can cause infections like
folliculitis, pimples, furuncle which is type of abscess. The furuncles also damage the
neighboring tissue by invading deep in to the tissue. This is known as carbuncle. Staphylococci
cause a highly contagious skin infection in children known as impetigo. The impetigo is two
types-nonbullous impetigo is caused by both S.aureus and S.pyogenes; bullous impetigo is
caused by staphylococcal toxin. It is also known as staphylococcal scalded skin syndrome. At
later stages of toxic shock syndrome the scalded skin syndrome is the characteristic feature.
Streptococci also secrete toxin and enzymes. The toxins are called hemolysins that lyse the red
blood cells. Various hemolysins like alpha, beta and gamma are produced by various
streptococcus strains. S.pyogenes is an important beta hemolytic streptococcus. Streptococcus
skin infections are generally superficial but some times may reach the deeper tissues also.
S.pyogenes infects the dermis of the skin and also causes a disease known as erysipelas in which
the skin erupts in to reddish patches.
The skin scraping is cultured and stained to observe the bacteria under microscope.
Staphylococci and streptococci both are gram positive and round shaped bacteria. The
staphylococci occur as grape-like clusters (Staphylococci name indicates bunch of grapes in
Greek). The streptococci are arranged in chains or pairs. Topical antibiotics like Mupirocin and
oral antibiotics like first generation cephalosporins and erythromycin are used to eliminate the
bacteria..
I need help working out this question . woman has normal vision al.pdf
I need help working out this question .
woman has normal vision although her maternal grandfather (her mother’s father) had red-
green colorblindness, a sex-linked recessive trait. Her maternal grandmother and the woman’s
own father are assumed to not possess a copy of the mutant allele. The woman marries a man
with normal vision although his father was colorblind. What is the probability that the first child
of this couple will be colorblind?
Solution
As it is x linked inheritance, the genotype of women is either xx = Normal or xxc =Carrier for
disease. If she is carrier (xxc) and married normal man, then the probability of 1st chaild getting
colour blindness is 1/4 = 0.25
Color blind ness is x linked and it inherit criss cross manner. That mean, male transfer color
blindness to her female children and. female transfer her male children.
XXc x XY ----------------> XX, XY, XXc & XcY
XcY progeny is only colour blind. So probability is 1\\4.
I have a Programming is Binary Tree Search (BTS) for strings with po.pdf
I have a Programming is Binary Tree Search (BTS) for strings with pointers, but I do not know
how to change BTS for strings using arrays.
Can you help me to change the method, please?
This is programming:
#include
#include
#include
using namespace std;
struct myNode
{
string data;
struct myNode *lNode;
struct myNode *rNode;
} *roNode;
class myBTS
{
public:
void seek(string, myNode **, myNode **);
void add(myNode *, myNode*);
void rem(string);
void caseOne(myNode *, myNode *);
void caseTwo(myNode *, myNode *);
void caseThree(myNode *, myNode *);
void print(myNode *, char);
myBTS()
{
roNode = NULL;
}
};
// Main method
int main ()
{
int myOptions;
string rData;
myBTS mybst;
myNode *tmp;
while (1)
{
cout << \"-----------------------------------------\" << endl;
cout << \"Welcome to Binary Search Tree Programming\" << endl;
cout << \"-----------------------------------------\" << endl;
cout << \"1. Add Data \" << endl;
cout << \"2. Remove Data \" << endl;
cout << \"3. Print Data \" << endl;
cout << \"4. Exit \" << endl;
cout << \"Enter Option : \";
cin >> myOptions;
switch(myOptions)
{
case 1:
tmp = new myNode;
cout << \"Enter data to add : \";
cin >> tmp->data;
mybst.add(roNode, tmp);
break;
case 2:
if (roNode == NULL)
{
cout << \"Tree empty!!!\" << endl;
continue;
}
else
{
cout << \"Enter data to remove: \";
cin >> rData;
mybst.rem(rData);
}
break;
case 3:
cout << \"Print Tree:\" << endl;
mybst.print(roNode,1);
cout << endl;
break;
case 4:
exit (1);
default:
cout << \"Wrong Option!!!\" << endl;
}
}
}
// Data in BTS
void myBTS::seek(string sData, myNode **myPar, myNode **myLoc)
{
myNode *ptr1, *ptr2;
if (roNode == NULL)
{
*myLoc = NULL;
*myPar = NULL;
return;
}
if (sData == roNode-> data)
{
*myLoc = roNode;
*myPar = NULL;
return;
}
if (sData < roNode-> data)
ptr1 = roNode->lNode;
else
ptr1 = roNode-> rNode;
ptr2 = roNode;
while (ptr1 != NULL)
{
if (sData == ptr1->data)
{
*myLoc = ptr1;
*myPar = ptr2;
return;
}
ptr2 = ptr1;
if (sData < ptr1 -> data)
ptr1 = ptr1 ->lNode;
else
ptr1 = ptr1 -> rNode;
}
*myLoc = NULL;
*myPar = ptr2;
}
// Add Data into BTS
void myBTS::add(myNode *myTree, myNode *myNewNode)
{
if (roNode == NULL)
{
roNode = new myNode;
roNode -> data = myNewNode ->data;
roNode -> lNode = NULL;
roNode -> rNode = NULL;
cout << \"Root data is added successfully\" << endl;
return;
}
if (myTree -> data == myNewNode -> data)
{
cout << \"Data already in BTS\" << endl;
return;
}
if (myTree -> data > myNewNode -> data)
{
if (myTree ->lNode != NULL)
{
add (myTree -> lNode, myNewNode);
}
else
{
myTree -> lNode = myNewNode;
(myTree -> lNode) -> lNode = NULL;
(myTree -> lNode) -> rNode = NULL;
cout << \"Data added to the Left successfully\" << endl;
return;
}
}
if (myTree -> rNode != NULL)
{
add(myTree -> rNode, myNewNode);
}
else
{
myTree -> rNode = myNewNode;
(myTree -> rNode) -> lNode = NULL;
(myTree -> rNode) -> rNode = NULL;
cout << \"Data added to the Right successfully\" << endl;
return;
}
}
//Remove data from BTS
void myBTS:: rem(string sData).
How is ATP produced in mammalian cells under aerobic conditions1..pdf
How is ATP produced in mammalian cells under aerobic conditions?
1. photophosphorylation
2. oxidative phosphorylation
3. substrate-level phosphorylation
4. phosphorolysis
a. 1
b. 2,3 and 4
c. 2
d. 1 and 2
e. 2 and 3
Solution
Hi,
4.Phosphorolysis: in this process cleavage of a compound where attacking group is phosphate.
Hence by analysing above explanation it is obvious that only option c. 2 (oxidative
phosporylation) is correct.
I hope my answer is satisfactory.
Thanks!!.
How con osmosis work for and against homeostasis How do osmosis .pdf
How con osmosis work for and against homeostasis? How do osmosis and diffusion relate to
dialysis? (Short paragraph o fine)
Solution
Multi cellular organisms should maintain a balance between internal and external environments
to carry out proper metabolic activities. The maintenance of relatively stable environment
internally is known as homeostasis.
All the body functions carried out by various systems of the body contribute to the homeostasis
to maintain a stable environment in the body for its survival and function. Homeostasis is much
required and the vital force for the each cell and through performing specialized activity, each
cells maintain a stable internal environment.
The net diffusion of water down the concentration gradient through the selectively permeable
membrane is called osmosis..
Enzymes and KE. Whats the relationship in terms of substrate mo.pdf
Enzymes and KE. What\'s the relationship in terms of substrate molecules
Enzymes and KE. What\'s the relationship in terms of substrate molecules
Solution
ANS:
Enzyme KE(kinetic energy)and substrate molecules relationship:
When the temperature raises the kinetic energy of reactants is also raised, increasing in
temperature results in higher activation energy molecules and more molecular (enzyme and
substrate) collision and interaction for the reaction to proceed faster.
The enzymes present in the human body usually need 37.5 o C (optimal temperature). The
optimum temperature for most of the enzymes between 40-45o C.
In this condition rate is increased. Increasing substrate concentration means an increasing rate at
which the enzyme and substrate molecules encounter one another
In general, the enzymes are exposed to a temperature above 50 o C, denaturing leading to
derangement in native structure. Majority of enzymes became inactive at higher temperature
above 70 o C
In enzymes at higher temperatures intra and intermolecular bonds are broken down, so the
enzyme molecules gain even more kinetic energy.
After reaching the maximum point, the rate of the reaction suddenly fall down due to the enzyme
degradation..
Explain membranes. Include the definition and name the four types an.pdf
Explain membranes. Include the definition and name the four types and which tissues each type
contains.
Solution
Answer : Membranes are the flat sheets of pliable tissue that cover or line a part of the body. The
combination of an epithelial layer and an underlying connective tissue layer constitutes an
epithelial membranes. These includes,
Mucous membranes : These membranes lines the body cavity that opens directlyto the exterior.
Mucous membranes lines the entire digestive, respiratory and reproductive tracts and much of
the urinary tract.
Serous Membranes : A serous membrane or Serosa lines a body cavity that does not open
directly to the exterior and it covers the organs that lie within the cavity.serous membranes
consist of areolar connective tissue covered by mesothelium and also consists of two layers such
as Parietal and visceral layer.
Cutaneous Membrane : These membarnes covers the surface of the body and consists of a
superficial portion called the epidermis and a deeper portion called the dermis.
Synovial Membranes : Like serous membranes, synovial membranes line structures that do not
open to the exterior. They lack in epithelium and are composed of discontinuous layer of cells
called synoviocytes, which secretes synovial fluid. Synovial fluid lubricates and nourishes the
cartilage covering the bones at movable joints and contains macrophages that moves microbes
and debris from the joint cavity..
Define1. Reciprocal cross –2. True or pure-breeding –Fill in t.pdf
Define
1. Reciprocal cross –
2. True or pure-breeding –
Fill in the Blank
3. When trying to calculate the probability that two independent events will occur together you
should _____________________ the individual probability of each event.
4. When crossing dihybrid peas, Mendel noticed that there were four phenotypic types amongst
the resulting F2 progeny; two had the phenotype of the parental generation and the other two
were ___________________.
True or False
5. The dominant trait is that which is masked in the F1 generation resulting from a cross between
true breeding parents.
6. Mendel’s law of segregation says that during gamete formation different pairs of alleles
segregate independently of each other.
Short Answer
7. Calculate the probability of A.) drawing a 4 from a deck of playing cards, B.) Drawing a black
4 and a red 8 from a deck of playing cards, C.) drawing a king, queen, or jack and the ace of
spades from a deck of playing cards. (Show your work.)
8.
A. is this an autosomal, X linked or Y linked trait?
B. Is the transmission of this trait dominant or recessive?
C. what is the genotype of individual II-2? 2 2 3 4 2- 1 ! 1 1
Solution
1. Reciprocal cross: a pair of crosses between a male of one strain and a female of another, and
vice versa.
2. True or pure-breeding:True breeding means that the parents will also pass down a specific
phenotypic trait to their offspring. Remember that a phenotype is the outward appearance of
something. True bred organisms will have a pure genotype (genetic expression of a trait) and will
only produce a certain phenotype.
3. When trying to calculate the probability that two independent events will occur together you
should specify size of the individual probability of each event.
4. When crossing dihybrid peas, Mendel noticed that there were four phenotypic types amongst
the resulting F2 progeny; two had the phenotype of the parental generation and the other two
were genotype.
5. The dominant trait is that which is masked in the F1 generation resulting from a cross between
true breeding parents is True.
6. Mendel’s law of segregation says that during gamete formation different pairs of alleles
segregate independently of each other is True..
Clinical immunology questionMr. R.M., a 60-year-old Caucasian man,.pdf
Clinical immunology question
Mr. R.M., a 60-year-old Caucasian man, was stung by a bee while gardening. He had been stung
once before earlier in the summer. Within a few seconds, his hand began to itch and he began to
experience abdominal cramping. He subsequently began experiencing difficulty breathing.
Fortunately, he was able to reach a first aid kit in his garage. Inside of the kit was an EpiPen
(injectable epinephrine) that his wife kept for herself because she was allergic to bee venom. He
immediately had his wife drive him to the hospital.
He was asymptomatic on arrival at the hospital. R.M. had no history of adverse reactions to bee
venom or antibiotics. Because of the nature of the incident, a diagnosis of anaphylactic shock
due to bee venom sensitivity was made. An IgE level was ordered. The results indicated a level
more than twice the (normal) reference range value. In addition, a follow-up skin test was
performed. The patient was extremely positive for bee venom.
Questions:
1.Which of the 4 types of hypersensitivity reactions does this case represent?
2.What is the immunologic mechanism involved in anaphylaxis?
3.What types of agents can induce anaphylactic shock?
4.Describe 2 different types of testing to diagnosis this patient.
5.What is the risk in performing a skin test?
Solution
Answer:
1) There are four types of hypersensitivity reactions, they include:
1) Type - I ; Allergy , 2) Type- II ; Cytotoxic, antibody-dependent, 3) Type III; Immune
complex disease
4) Type IV; Delayed-type hypersensitivity.
2) Anaphylaxis is an Type-I reaction ; allergic reaction. During allergic reaction, IgE react with
antigen, and it activates basophils and mast cells, due this activation, histamines are released.
3) Anaphylactic shock could be due to the food agents such as milk, eggs and shellfish. In few
cases, insect stings can cause shock.
4) Diagnosis includes, 1) Allergy testing (skin allergy test), 2)Differential diagnosis.
5) Skin test is available only to test effect of penicillin drug allergy.so, it is risk for allergies..
As shown in the pedigree below, a pair of first cousins (I and K) ma.pdf
As shown in the pedigree below, a pair of first cousins (I and K) marry an unrelated pair of first
cousins (J and L). Their children M and N produce child O.
1. Calculate the inbreeding coefficient (f) of O (express answers to 1 and 3 to three decimal
places). Assume that f= 0 for all ancestors. (Hint: To help you out, I have shaded individuals
who cannot contribute to autozygosity.) To show your work, use the letter names to indicate each
path/loop that contributes to autozygosity. For example: path 1 = M-I-E…
2. Interpret your result. What does this value of f indicate?
3. Recalculate f for child O when f=1/2 for great-great grandmother D (f=0 for all other
ancestors). Hint: After you have calculated all the loops above, you’ll adjust just the loop that
includes ancester D to account for her ancestral inbreeding using the formula presented in class
(and text). As shown in the pedigree below, a pair of first cousins (I and K marry a unrelated pair
of first cousin s and L). Their children M and N produce child O. 1. Calculate the inbreeding
coefficient (f) of O (express answers to 1 and 3 to three decimal places). Assume that f 0 for all
ancestors. (Hint: To help you out, I have shaded individuals who cannot contribute to
autozyBosity.) o show your work, use the letter names to indicate each path/loop that contributes
to autoz osit For example: path 1 M-1-E 2. Interpret your result. What does this value of
findicate? 3. Recalculate f for child o when f-1/2 for great-great grandmother D (f 0 for all other
ancestors). Hint: After you have calculated all the loops above, you\'ll adjust just the loop that
includes ancester D to account for her ancestral breeding using the formula presented in class
(and text)
Solution
1. Path 1:
So, the probability that a copy of gene in A makes the way down to O via E and a copy of gene
in B makes the way down to O via F =( ½)n = (½)7=0.00781. Here n= node=7
Path 2:
Similarly like the first one the total probability in path 2 will be same that is 0.00781.
Thus inbreeding co-efficient = 0.00781+0.00781=0.01562.
2. Generally, the higher the level of inbreeding closer the coefficient of relationship approaches a
value of 1 and approaches a value of 0 for individuals with remote common ancestors.
As here the value of inbreeding co-efficient is aproximately 0 so we can conclude that
3. When a common ancestor is iself inbreed the inbreeding coefficient is = (1/2)n (1+ inbreeding
co-efficient of common ancestor) ,
The probability that a copy of gene in D makes the way to O = (1/2)7 = 0.00781
Therefore inbreeding co-efficient is = 0.00781 * (1 + 1/2 )=0.01172.
An individual with genes A, H, Se and lele has which of the followin.pdf
An individual with genes A, H, Se and lele has which of the following phenotypes?
AH, Le(a-b-)
ABH, Le(a+b-)
ABH, Le(a-b-)
AH, Le(a+b-)
The Rh phenotype results for a patient are: anti-C = 4+, anti-c = 3+, anti-D = NEG, anti-E =
NEG, anti-e = 4+. The patients most probable genotype using the Wiener nomenclature is?
r\'r
R1r
r\"r
rr
With respect to the Lewis blood group system, which statement is false.
Lewis antibodies occur frequently in the sera of pregnant women.
Lewis blood group antigens are made by tissue cells, secreted into body fluids and plasma, and
adsorbed onto the red cell membrane.
The Le gene codes for L-fucosyltransferase which adds L-fucose to the type 1 chain.
Lewis antigens are well developed at birth.
Lewis antibodies are generally IgM, and are therefore not a cause of HDN.
The major immunoglobulin class of anti-B in a group A individual is:
IgM.
None of the alternatives
IgA.
IgG.
Antibody screening cells may not detect antibodies against low frequency antigens. Which of the
following is most likely to go undetected?
K
Kpa
M
k
e
Rh antigens are inherited as?
Codominant alleles.
X linked.
Sex-linked genes.
Autosomal recessive alleles.
AH, Le(a-b-)
ABH, Le(a+b-)
ABH, Le(a-b-)
AH, Le(a+b-)
Solution
An individual with genes A, H, Se and lele has which of the following phenotypes?
answer: AH, Le(a-b-) . if B antigen is not found.
if B antigen is present then ABH, Le(a-b-)
The Rh phenotype results for a patient are: anti-C = 4+, anti-c = 3+, anti-D = NEG, anti-E =
NEG, anti-e = 4+. The patients most probable genotype using the Wiener nomenclature is?
answer is R1r
positive sign shows that antigens are present. so we can see here that D,d, C,c,e are present on
RBC. D and C are homo and heterozygous whereas e is only homozygous.
combinations are DCe/dce, Dce/dCe, Dce/DCe. when they are converted into fischer race chart.
among these phenotypes,R1r is the most probable one.
With respect to the Lewis blood group system, which statement is false.
answer: Lewis antigens are well developed at birth. they cannot be detected until 2 years of birth.
The major immunoglobulin class of anti-B in a group A individual is
answer: IgM. Major immunoglobulin class for anti-b in group A individual or anti-a in group B
individual is IgM.
Antibody screening cells may not detect antibodies against low frequency antigens. Which of the
following is most likely to go undetected?
answer: Kp-a, low frequency antigens are K,Kp-a, Le-a and Js-a
Rh antigens are inherited as?
answer: codominant alleles which are present on autosomes.
8.8 Program Playlist (Java)You will be building a linked list. Ma.pdf
8.8 Program: Playlist (Java)
You will be building a linked list. Make sure to keep track of both the head and tail nodes.
(1) Create two files to submit.
SongEntry.java - Class declaration
Playlist.java - Contains main() method
Build the SongEntry class per the following specifications. Note: Some methods can initially be
method stubs (empty methods), to be completed in later steps.
Private fields
String uniqueID - Initialized to \"none\" in default constructor
string songName - Initialized to \"none\" in default constructor
string artistName - Initialized to \"none\" in default constructor
int songLength - Initialized to 0 in default constructor
SongEntry nextNode - Initialized to null in default constructor
Default constructor (1 pt)
Parameterized constructor (1 pt)
Public member methods
insertAfter() (1 pt)
setNext() - Mutator (1 pt)
getID() - Accessor
getSongName() - Accessor
getArtistName() - Accessor
getSongLength() - Accessor
getNext() - Accessor
printPlaylistSongs()
Ex. of printPlaylistSongs output:
(2) In main(), prompt the user for the title of the playlist. (1 pt)
Ex:
(3) Implement the printMenu() method. printMenu() takes the playlist title as a parameter and
outputs a menu of options to manipulate the playlist. Each option is represented by a single
character. Build and output the menu within the method.
If an invalid character is entered, continue to prompt for a valid choice. Hint: Implement Quit
before implementing other options. Call printMenu() in the main() method. Continue to execute
the menu until the user enters q to Quit. (3 pts)
Ex:
(4) Implement \"Output full playlist\" menu option. If the list is empty, output: Playlist is empty
(3 pts)
Ex:
(5) Implement the \"Add song\" menu item. New additions are added to the end of the list. (2
pts)
Ex:
(6) Implement the \"Remove song\" method. Prompt the user for the unique ID of the song to be
removed.(4 pts)
Ex:
(7) Implement the \"Change position of song\" menu option. Prompt the user for the current
position of the song and the desired new position. Valid new positions are 1 - n (the number of
nodes). If the user enters a new position that is less than 1, move the node to the position 1 (the
head). If the user enters a new position greater than n, move the node to position n (the tail). 6
cases will be tested:
Moving the head node (1 pt)
Moving the tail node (1 pt)
Moving a node to the head (1 pt)
Moving a node to the tail (1 pt)
Moving a node up the list (1 pt)
Moving a node down the list (1 pt)
Ex:
(8) Implement the \"Output songs by specific artist\" menu option. Prompt the user for the
artist\'s name, and output the node\'s information, starting with the node\'s current position. (2 pt)
Ex:
(9) Implement the \"Output total time of playlist\" menu option. Output the sum of the time of
the playlist\'s songs (in seconds). (2 pts)
Ex:
Solution
SongEntry.java
package playlist;
public class SongEntry {
private String uniqueID;
private String songName;
private String artistName.
5. Cells with an intracellular osmolarity of 300mosmL are suspended.pdf
5. Cells with an intracellular osmolarity of 300mosm/L are suspended in the following solutions.
Describe the sequence of events that take place.
(a)0.3M urea (membrane permeable)
(b)0.3M glucose (membrane impermeable)
(c)0.6M urea
(d)0.6M glucose
Solution
Osmolarity of a solution refers to the concentration of the solution in terms of how strongly it
attracts water molecules across a semipermeable membrane. It is given the intracellular
osmolarity of the cell is 300 mosm/L. It can also be expressed as 0.3 osm/L or 0.3 M.
(a) 0.3 M Urea (membrane permeable)
This solution is isosmotic to the intracellular osmotic pressure. But, the cell membrane being
selectively permeable, allows the entry of urea until both the solutions become isotonic. This
leads to bursting ofcell as along with urea water enters the cell.
(b) 0.3 M Glucose (membrane impermeable)
This solution is also isosmotic to the intracellular osmotic pressure. But given that the membrane
is impermeable, glucose cannot enter to cell and an isotonic condition cannot be reached. Hence
the cell would be unstable and would distort due to the uneven tonicity.
(c) 0.6 M urea
This solution is hypertonic and hyperosmotic as compared to the cell\'s intracellular fluid. Thus,
water will move out through the pores of the cell membrane causing the cell to shrink or become
flaccid.
(d) 0.6 M glucose
This solution is hypertonic and hyperosmotic as compared to the cell\'s intracellular fluid. Thus,
water will move out through the pores of the cell membrane causing the cell to shrink or become
flaccid..
Which symptom of inflammation is NOT paired with its sourceA) swell.pdf
Which symptom of inflammation is NOT paired with its source?A) swelling: accumulation of
fluid in connective tissueB) heat: due to increased blood flowC) redness: blood leaking into
surrounding tissuesD) pain: pressure of fluid on nerve endingsE) pain: release of inflammatory
chemicals affect nerve endings directlyF) All of the listed responses are correctly paired.
Solution
Option C is wrong. Because both redness and heat is mainly due to increased blood flow to the
inflamed area..
Which is true of the data shown in the histogram I. The distribution.pdf
Which is true of the data shown in the histogram? I. The distribution is approximately
symmetric. II. The mean and median are approximately equal. III. The median and IQR
summarize the data better than the mean and standard deviation. Click the icon to view the
histogram. I and III I, II and III I only I and II III only
Solution
(I) The data shows that the shape of this histogram is bell-shaped. Also,if a line is drawn down
the middle of the graph, the two sides will mirror each other. So,the distribution is approximately
symmetric.
(II) In the given histogram where the values of data occur at regular frequencies. So, the mean,
median and mode occur at the same point.
(III) The IQR is the difference between the third and the first quartile. IQR makes the values by
estimate of outliers where mean and standard deviation are not effected against outliers. So, the
median and IQR is not summarize the data well.
Answer is (I) and (II).
Which of the following is a member of the domain ArchaeaA. The ce.pdf
Which of the following is a member of the domain Archaea?
A. The cell is eukaryotic
B. The cell has a nucleus
C. The cell is prokaryotic
D. The cell lack phospholipid
Solution
C. The cell is prokaryotic
Archea are the unicellular prokaryotes that can inhabit an extreme range of environments due to
their cellular adaptations. Their cell wall does not contain peptidoglycans but contains, S-layers..
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
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In Exercises 19-24, find the volume of the solid generated by revolvi.pdf
In Exercises 19-24, find the volume of the solid generated by revolving the region bounded by
the graphs of the equations about the x-axis.
Solution
Area = Integral(ydx) from x= 0 to 3 = int[1/sqrt(1+x) dx] from x= 0 to 3 = 2
sqrt(1+x) from x = 0 to 3 = 2 [sqrt(4) - sqrt(1)] = 2 Volume when rotated along x axxis = Area *
2 *pi = 2*2*pi = 4pi.
In which step of meiosis, the members of tetra separate Solutio.pdf
In which step of meiosis, the members of tetra separate?
Solution
Answer: The tetrads separate during anaphase I of meiosis.
Explanation: Tetrads are formed towards the end of prophase I of meiosis.Separation of
homologous chromosomes (with four sister chromatids) takes place during anaphase I of
meiosis..
Identify the distinguishing characteristics of member of Phylum Cnid.pdf
Identify the distinguishing characteristics of member of Phylum Cnidaria, including symmetry,
number of embryonic tissue layers, unique structures, skeleton and polymorphism (polyp and
medusa), place Cnidaria on a cladogram
Compare and contrast how a general polyp and a medusa feed
Describe why a nerve net was such a major evolutionary adaptation
Describe the characteristics of the four classes of Cnidarians
Label the following on a drawing: tentacles, mouth, anus, gastrovascular cavity, cnidocytes,
nerve net
Solution
Answer1) Symmetry- Radially symmetrical body
Number of embryonic tissue layers- two (endoderm and ectoderm)
Unique structures- Stinging cells (Cnidocytes)
Skeleton- Hydrostatic skeleton
Polymorphism (polyp and medusa)- polyp is a flower like solitary form and medusa is a free-
floating miniature of cnidarians.
Answer 2) a tubular shaped body form is called as polyp. They are sessile and are found as
attached bottom (basal plate) to a solid surface with the opened mouth at the top. while medusa is
an umbrella or bell shape form, having their mouth facing downwards. And they are free
floating. Being a sessile, polyp filter their food from the water by their tiny-beating cilia. While
medusas are free floating and motile so they can trap their feedings, and need not to filter the
water all the time though these structures were more related to the sexual reproduction.
Answer 3) Nerve net is a diffused framework of neurons in the body which can transmit
signals/impulse in different directions from a single point of stimulus.
This became a major evolutionary adaptation because it allowed the organism to sense the
environmental stimulus from any region of its body and respond accordingly. This allowed the
readily development and motion of different body parts such as limb like structures.
Answer 4) These are the four classes of cnidarians.
Hydrozoa are diploblastic and multicellular animals. Which has only mouth opening. They have
a fixed polyp stage and the reproductive medusa stage. They have the gonads as well.
Scyphozoan are present in medusoid forms. They have tentaculocysts as Sense organs. They
have Velarium in endodermal canal (Acraspedote). They endodermal gonads.
Anthozoan are present only in polyp forms. Their medusa stage is absent. They have Mesenteries
which bear nematocyst. They also possess endodermal Gonads.
Cubozoan looks like the jelly fish and called as jelly boxes. They possess box shaped medusa
and they share typical characteristics of scyphozoans..
If most evolutionary change occurs during and immediately after spec.pdf
If most evolutionary change occurs during and immediately after speciation, could speciation
rates and extinction rates explain major patterns in macroevolution? In turn, what factors may
influence speciation and extinction rates (are they all biological, or can geological /
meteorological / astronomical factors also play a role)?
Solution
In the course of evolution, some mutations keep on occurring with every replicative step. These
mutations lead to differences in alleles, which further result in different genotypes. Some
genotypes have higher fitness while others have lower fitness. Natural selection selects the
phenotypes which have higher fitness, and slowly discards the others during the course of
evolution. So, extinction rate depends on fitness and natural selection.
As new phenotypes are selected, they result in formation of new species; the process is known as
speciation. Speciation further results in evolution. Higher the speciation rate, higher will be the
rate of evolution.
So, you can say that most evolutionary changes occur during and after speciation.
Overall the biological factors affecting speciation and extinction rates are:
It is important to note that that apart from the above two factors, geological/astronomical factors
also affect speciation, extinction and evolution. Changes in the earth since times immemorial
ultimately resulted in changes like:.
Imagine that someone picking up roses in a garden gets a scratch fro.pdf
Imagine that someone picking up roses in a garden gets a scratch from a thorn. Some bacteria
(Staphyloccocus aureus and Streptococcus pyogenes) take advantage of the wound to gain access
to the deeper layers of the skin and attempt to cause an infection. Describe the different stages of
the host response (in particular the types of cells and receptors involved) and how the bacteria
will be recognized and eliminated.
Solution
The superficial staphylococcal and streptococcal infections are very common in skin. The
physiological condition of the skin is much suitable to these bacteria. Both these bacteria are
potent toxin and enzyme producing strains.
S.auresus infection on the skin immediately stimulates the inflammatory response and
macrophages and neutrophils are proliferated at the site of infection. The organism can enter the
body through openings in the skin barrier like hair follicles and can cause infections like
folliculitis, pimples, furuncle which is type of abscess. The furuncles also damage the
neighboring tissue by invading deep in to the tissue. This is known as carbuncle. Staphylococci
cause a highly contagious skin infection in children known as impetigo. The impetigo is two
types-nonbullous impetigo is caused by both S.aureus and S.pyogenes; bullous impetigo is
caused by staphylococcal toxin. It is also known as staphylococcal scalded skin syndrome. At
later stages of toxic shock syndrome the scalded skin syndrome is the characteristic feature.
Streptococci also secrete toxin and enzymes. The toxins are called hemolysins that lyse the red
blood cells. Various hemolysins like alpha, beta and gamma are produced by various
streptococcus strains. S.pyogenes is an important beta hemolytic streptococcus. Streptococcus
skin infections are generally superficial but some times may reach the deeper tissues also.
S.pyogenes infects the dermis of the skin and also causes a disease known as erysipelas in which
the skin erupts in to reddish patches.
The skin scraping is cultured and stained to observe the bacteria under microscope.
Staphylococci and streptococci both are gram positive and round shaped bacteria. The
staphylococci occur as grape-like clusters (Staphylococci name indicates bunch of grapes in
Greek). The streptococci are arranged in chains or pairs. Topical antibiotics like Mupirocin and
oral antibiotics like first generation cephalosporins and erythromycin are used to eliminate the
bacteria..
I need help working out this question . woman has normal vision al.pdf
I need help working out this question .
woman has normal vision although her maternal grandfather (her mother’s father) had red-
green colorblindness, a sex-linked recessive trait. Her maternal grandmother and the woman’s
own father are assumed to not possess a copy of the mutant allele. The woman marries a man
with normal vision although his father was colorblind. What is the probability that the first child
of this couple will be colorblind?
Solution
As it is x linked inheritance, the genotype of women is either xx = Normal or xxc =Carrier for
disease. If she is carrier (xxc) and married normal man, then the probability of 1st chaild getting
colour blindness is 1/4 = 0.25
Color blind ness is x linked and it inherit criss cross manner. That mean, male transfer color
blindness to her female children and. female transfer her male children.
XXc x XY ----------------> XX, XY, XXc & XcY
XcY progeny is only colour blind. So probability is 1\\4.
I have a Programming is Binary Tree Search (BTS) for strings with po.pdf
I have a Programming is Binary Tree Search (BTS) for strings with pointers, but I do not know
how to change BTS for strings using arrays.
Can you help me to change the method, please?
This is programming:
#include
#include
#include
using namespace std;
struct myNode
{
string data;
struct myNode *lNode;
struct myNode *rNode;
} *roNode;
class myBTS
{
public:
void seek(string, myNode **, myNode **);
void add(myNode *, myNode*);
void rem(string);
void caseOne(myNode *, myNode *);
void caseTwo(myNode *, myNode *);
void caseThree(myNode *, myNode *);
void print(myNode *, char);
myBTS()
{
roNode = NULL;
}
};
// Main method
int main ()
{
int myOptions;
string rData;
myBTS mybst;
myNode *tmp;
while (1)
{
cout << \"-----------------------------------------\" << endl;
cout << \"Welcome to Binary Search Tree Programming\" << endl;
cout << \"-----------------------------------------\" << endl;
cout << \"1. Add Data \" << endl;
cout << \"2. Remove Data \" << endl;
cout << \"3. Print Data \" << endl;
cout << \"4. Exit \" << endl;
cout << \"Enter Option : \";
cin >> myOptions;
switch(myOptions)
{
case 1:
tmp = new myNode;
cout << \"Enter data to add : \";
cin >> tmp->data;
mybst.add(roNode, tmp);
break;
case 2:
if (roNode == NULL)
{
cout << \"Tree empty!!!\" << endl;
continue;
}
else
{
cout << \"Enter data to remove: \";
cin >> rData;
mybst.rem(rData);
}
break;
case 3:
cout << \"Print Tree:\" << endl;
mybst.print(roNode,1);
cout << endl;
break;
case 4:
exit (1);
default:
cout << \"Wrong Option!!!\" << endl;
}
}
}
// Data in BTS
void myBTS::seek(string sData, myNode **myPar, myNode **myLoc)
{
myNode *ptr1, *ptr2;
if (roNode == NULL)
{
*myLoc = NULL;
*myPar = NULL;
return;
}
if (sData == roNode-> data)
{
*myLoc = roNode;
*myPar = NULL;
return;
}
if (sData < roNode-> data)
ptr1 = roNode->lNode;
else
ptr1 = roNode-> rNode;
ptr2 = roNode;
while (ptr1 != NULL)
{
if (sData == ptr1->data)
{
*myLoc = ptr1;
*myPar = ptr2;
return;
}
ptr2 = ptr1;
if (sData < ptr1 -> data)
ptr1 = ptr1 ->lNode;
else
ptr1 = ptr1 -> rNode;
}
*myLoc = NULL;
*myPar = ptr2;
}
// Add Data into BTS
void myBTS::add(myNode *myTree, myNode *myNewNode)
{
if (roNode == NULL)
{
roNode = new myNode;
roNode -> data = myNewNode ->data;
roNode -> lNode = NULL;
roNode -> rNode = NULL;
cout << \"Root data is added successfully\" << endl;
return;
}
if (myTree -> data == myNewNode -> data)
{
cout << \"Data already in BTS\" << endl;
return;
}
if (myTree -> data > myNewNode -> data)
{
if (myTree ->lNode != NULL)
{
add (myTree -> lNode, myNewNode);
}
else
{
myTree -> lNode = myNewNode;
(myTree -> lNode) -> lNode = NULL;
(myTree -> lNode) -> rNode = NULL;
cout << \"Data added to the Left successfully\" << endl;
return;
}
}
if (myTree -> rNode != NULL)
{
add(myTree -> rNode, myNewNode);
}
else
{
myTree -> rNode = myNewNode;
(myTree -> rNode) -> lNode = NULL;
(myTree -> rNode) -> rNode = NULL;
cout << \"Data added to the Right successfully\" << endl;
return;
}
}
//Remove data from BTS
void myBTS:: rem(string sData).
How is ATP produced in mammalian cells under aerobic conditions1..pdf
How is ATP produced in mammalian cells under aerobic conditions?
1. photophosphorylation
2. oxidative phosphorylation
3. substrate-level phosphorylation
4. phosphorolysis
a. 1
b. 2,3 and 4
c. 2
d. 1 and 2
e. 2 and 3
Solution
Hi,
4.Phosphorolysis: in this process cleavage of a compound where attacking group is phosphate.
Hence by analysing above explanation it is obvious that only option c. 2 (oxidative
phosporylation) is correct.
I hope my answer is satisfactory.
Thanks!!.
How con osmosis work for and against homeostasis How do osmosis .pdf
How con osmosis work for and against homeostasis? How do osmosis and diffusion relate to
dialysis? (Short paragraph o fine)
Solution
Multi cellular organisms should maintain a balance between internal and external environments
to carry out proper metabolic activities. The maintenance of relatively stable environment
internally is known as homeostasis.
All the body functions carried out by various systems of the body contribute to the homeostasis
to maintain a stable environment in the body for its survival and function. Homeostasis is much
required and the vital force for the each cell and through performing specialized activity, each
cells maintain a stable internal environment.
The net diffusion of water down the concentration gradient through the selectively permeable
membrane is called osmosis..
Enzymes and KE. Whats the relationship in terms of substrate mo.pdf
Enzymes and KE. What\'s the relationship in terms of substrate molecules
Enzymes and KE. What\'s the relationship in terms of substrate molecules
Solution
ANS:
Enzyme KE(kinetic energy)and substrate molecules relationship:
When the temperature raises the kinetic energy of reactants is also raised, increasing in
temperature results in higher activation energy molecules and more molecular (enzyme and
substrate) collision and interaction for the reaction to proceed faster.
The enzymes present in the human body usually need 37.5 o C (optimal temperature). The
optimum temperature for most of the enzymes between 40-45o C.
In this condition rate is increased. Increasing substrate concentration means an increasing rate at
which the enzyme and substrate molecules encounter one another
In general, the enzymes are exposed to a temperature above 50 o C, denaturing leading to
derangement in native structure. Majority of enzymes became inactive at higher temperature
above 70 o C
In enzymes at higher temperatures intra and intermolecular bonds are broken down, so the
enzyme molecules gain even more kinetic energy.
After reaching the maximum point, the rate of the reaction suddenly fall down due to the enzyme
degradation..
Explain membranes. Include the definition and name the four types an.pdf
Explain membranes. Include the definition and name the four types and which tissues each type
contains.
Solution
Answer : Membranes are the flat sheets of pliable tissue that cover or line a part of the body. The
combination of an epithelial layer and an underlying connective tissue layer constitutes an
epithelial membranes. These includes,
Mucous membranes : These membranes lines the body cavity that opens directlyto the exterior.
Mucous membranes lines the entire digestive, respiratory and reproductive tracts and much of
the urinary tract.
Serous Membranes : A serous membrane or Serosa lines a body cavity that does not open
directly to the exterior and it covers the organs that lie within the cavity.serous membranes
consist of areolar connective tissue covered by mesothelium and also consists of two layers such
as Parietal and visceral layer.
Cutaneous Membrane : These membarnes covers the surface of the body and consists of a
superficial portion called the epidermis and a deeper portion called the dermis.
Synovial Membranes : Like serous membranes, synovial membranes line structures that do not
open to the exterior. They lack in epithelium and are composed of discontinuous layer of cells
called synoviocytes, which secretes synovial fluid. Synovial fluid lubricates and nourishes the
cartilage covering the bones at movable joints and contains macrophages that moves microbes
and debris from the joint cavity..
Define1. Reciprocal cross –2. True or pure-breeding –Fill in t.pdf
Define
1. Reciprocal cross –
2. True or pure-breeding –
Fill in the Blank
3. When trying to calculate the probability that two independent events will occur together you
should _____________________ the individual probability of each event.
4. When crossing dihybrid peas, Mendel noticed that there were four phenotypic types amongst
the resulting F2 progeny; two had the phenotype of the parental generation and the other two
were ___________________.
True or False
5. The dominant trait is that which is masked in the F1 generation resulting from a cross between
true breeding parents.
6. Mendel’s law of segregation says that during gamete formation different pairs of alleles
segregate independently of each other.
Short Answer
7. Calculate the probability of A.) drawing a 4 from a deck of playing cards, B.) Drawing a black
4 and a red 8 from a deck of playing cards, C.) drawing a king, queen, or jack and the ace of
spades from a deck of playing cards. (Show your work.)
8.
A. is this an autosomal, X linked or Y linked trait?
B. Is the transmission of this trait dominant or recessive?
C. what is the genotype of individual II-2? 2 2 3 4 2- 1 ! 1 1
Solution
1. Reciprocal cross: a pair of crosses between a male of one strain and a female of another, and
vice versa.
2. True or pure-breeding:True breeding means that the parents will also pass down a specific
phenotypic trait to their offspring. Remember that a phenotype is the outward appearance of
something. True bred organisms will have a pure genotype (genetic expression of a trait) and will
only produce a certain phenotype.
3. When trying to calculate the probability that two independent events will occur together you
should specify size of the individual probability of each event.
4. When crossing dihybrid peas, Mendel noticed that there were four phenotypic types amongst
the resulting F2 progeny; two had the phenotype of the parental generation and the other two
were genotype.
5. The dominant trait is that which is masked in the F1 generation resulting from a cross between
true breeding parents is True.
6. Mendel’s law of segregation says that during gamete formation different pairs of alleles
segregate independently of each other is True..
Clinical immunology questionMr. R.M., a 60-year-old Caucasian man,.pdf
Clinical immunology question
Mr. R.M., a 60-year-old Caucasian man, was stung by a bee while gardening. He had been stung
once before earlier in the summer. Within a few seconds, his hand began to itch and he began to
experience abdominal cramping. He subsequently began experiencing difficulty breathing.
Fortunately, he was able to reach a first aid kit in his garage. Inside of the kit was an EpiPen
(injectable epinephrine) that his wife kept for herself because she was allergic to bee venom. He
immediately had his wife drive him to the hospital.
He was asymptomatic on arrival at the hospital. R.M. had no history of adverse reactions to bee
venom or antibiotics. Because of the nature of the incident, a diagnosis of anaphylactic shock
due to bee venom sensitivity was made. An IgE level was ordered. The results indicated a level
more than twice the (normal) reference range value. In addition, a follow-up skin test was
performed. The patient was extremely positive for bee venom.
Questions:
1.Which of the 4 types of hypersensitivity reactions does this case represent?
2.What is the immunologic mechanism involved in anaphylaxis?
3.What types of agents can induce anaphylactic shock?
4.Describe 2 different types of testing to diagnosis this patient.
5.What is the risk in performing a skin test?
Solution
Answer:
1) There are four types of hypersensitivity reactions, they include:
1) Type - I ; Allergy , 2) Type- II ; Cytotoxic, antibody-dependent, 3) Type III; Immune
complex disease
4) Type IV; Delayed-type hypersensitivity.
2) Anaphylaxis is an Type-I reaction ; allergic reaction. During allergic reaction, IgE react with
antigen, and it activates basophils and mast cells, due this activation, histamines are released.
3) Anaphylactic shock could be due to the food agents such as milk, eggs and shellfish. In few
cases, insect stings can cause shock.
4) Diagnosis includes, 1) Allergy testing (skin allergy test), 2)Differential diagnosis.
5) Skin test is available only to test effect of penicillin drug allergy.so, it is risk for allergies..
As shown in the pedigree below, a pair of first cousins (I and K) ma.pdf
As shown in the pedigree below, a pair of first cousins (I and K) marry an unrelated pair of first
cousins (J and L). Their children M and N produce child O.
1. Calculate the inbreeding coefficient (f) of O (express answers to 1 and 3 to three decimal
places). Assume that f= 0 for all ancestors. (Hint: To help you out, I have shaded individuals
who cannot contribute to autozygosity.) To show your work, use the letter names to indicate each
path/loop that contributes to autozygosity. For example: path 1 = M-I-E…
2. Interpret your result. What does this value of f indicate?
3. Recalculate f for child O when f=1/2 for great-great grandmother D (f=0 for all other
ancestors). Hint: After you have calculated all the loops above, you’ll adjust just the loop that
includes ancester D to account for her ancestral inbreeding using the formula presented in class
(and text). As shown in the pedigree below, a pair of first cousins (I and K marry a unrelated pair
of first cousin s and L). Their children M and N produce child O. 1. Calculate the inbreeding
coefficient (f) of O (express answers to 1 and 3 to three decimal places). Assume that f 0 for all
ancestors. (Hint: To help you out, I have shaded individuals who cannot contribute to
autozyBosity.) o show your work, use the letter names to indicate each path/loop that contributes
to autoz osit For example: path 1 M-1-E 2. Interpret your result. What does this value of
findicate? 3. Recalculate f for child o when f-1/2 for great-great grandmother D (f 0 for all other
ancestors). Hint: After you have calculated all the loops above, you\'ll adjust just the loop that
includes ancester D to account for her ancestral breeding using the formula presented in class
(and text)
Solution
1. Path 1:
So, the probability that a copy of gene in A makes the way down to O via E and a copy of gene
in B makes the way down to O via F =( ½)n = (½)7=0.00781. Here n= node=7
Path 2:
Similarly like the first one the total probability in path 2 will be same that is 0.00781.
Thus inbreeding co-efficient = 0.00781+0.00781=0.01562.
2. Generally, the higher the level of inbreeding closer the coefficient of relationship approaches a
value of 1 and approaches a value of 0 for individuals with remote common ancestors.
As here the value of inbreeding co-efficient is aproximately 0 so we can conclude that
3. When a common ancestor is iself inbreed the inbreeding coefficient is = (1/2)n (1+ inbreeding
co-efficient of common ancestor) ,
The probability that a copy of gene in D makes the way to O = (1/2)7 = 0.00781
Therefore inbreeding co-efficient is = 0.00781 * (1 + 1/2 )=0.01172.
An individual with genes A, H, Se and lele has which of the followin.pdf
An individual with genes A, H, Se and lele has which of the following phenotypes?
AH, Le(a-b-)
ABH, Le(a+b-)
ABH, Le(a-b-)
AH, Le(a+b-)
The Rh phenotype results for a patient are: anti-C = 4+, anti-c = 3+, anti-D = NEG, anti-E =
NEG, anti-e = 4+. The patients most probable genotype using the Wiener nomenclature is?
r\'r
R1r
r\"r
rr
With respect to the Lewis blood group system, which statement is false.
Lewis antibodies occur frequently in the sera of pregnant women.
Lewis blood group antigens are made by tissue cells, secreted into body fluids and plasma, and
adsorbed onto the red cell membrane.
The Le gene codes for L-fucosyltransferase which adds L-fucose to the type 1 chain.
Lewis antigens are well developed at birth.
Lewis antibodies are generally IgM, and are therefore not a cause of HDN.
The major immunoglobulin class of anti-B in a group A individual is:
IgM.
None of the alternatives
IgA.
IgG.
Antibody screening cells may not detect antibodies against low frequency antigens. Which of the
following is most likely to go undetected?
K
Kpa
M
k
e
Rh antigens are inherited as?
Codominant alleles.
X linked.
Sex-linked genes.
Autosomal recessive alleles.
AH, Le(a-b-)
ABH, Le(a+b-)
ABH, Le(a-b-)
AH, Le(a+b-)
Solution
An individual with genes A, H, Se and lele has which of the following phenotypes?
answer: AH, Le(a-b-) . if B antigen is not found.
if B antigen is present then ABH, Le(a-b-)
The Rh phenotype results for a patient are: anti-C = 4+, anti-c = 3+, anti-D = NEG, anti-E =
NEG, anti-e = 4+. The patients most probable genotype using the Wiener nomenclature is?
answer is R1r
positive sign shows that antigens are present. so we can see here that D,d, C,c,e are present on
RBC. D and C are homo and heterozygous whereas e is only homozygous.
combinations are DCe/dce, Dce/dCe, Dce/DCe. when they are converted into fischer race chart.
among these phenotypes,R1r is the most probable one.
With respect to the Lewis blood group system, which statement is false.
answer: Lewis antigens are well developed at birth. they cannot be detected until 2 years of birth.
The major immunoglobulin class of anti-B in a group A individual is
answer: IgM. Major immunoglobulin class for anti-b in group A individual or anti-a in group B
individual is IgM.
Antibody screening cells may not detect antibodies against low frequency antigens. Which of the
following is most likely to go undetected?
answer: Kp-a, low frequency antigens are K,Kp-a, Le-a and Js-a
Rh antigens are inherited as?
answer: codominant alleles which are present on autosomes.
8.8 Program Playlist (Java)You will be building a linked list. Ma.pdf
8.8 Program: Playlist (Java)
You will be building a linked list. Make sure to keep track of both the head and tail nodes.
(1) Create two files to submit.
SongEntry.java - Class declaration
Playlist.java - Contains main() method
Build the SongEntry class per the following specifications. Note: Some methods can initially be
method stubs (empty methods), to be completed in later steps.
Private fields
String uniqueID - Initialized to \"none\" in default constructor
string songName - Initialized to \"none\" in default constructor
string artistName - Initialized to \"none\" in default constructor
int songLength - Initialized to 0 in default constructor
SongEntry nextNode - Initialized to null in default constructor
Default constructor (1 pt)
Parameterized constructor (1 pt)
Public member methods
insertAfter() (1 pt)
setNext() - Mutator (1 pt)
getID() - Accessor
getSongName() - Accessor
getArtistName() - Accessor
getSongLength() - Accessor
getNext() - Accessor
printPlaylistSongs()
Ex. of printPlaylistSongs output:
(2) In main(), prompt the user for the title of the playlist. (1 pt)
Ex:
(3) Implement the printMenu() method. printMenu() takes the playlist title as a parameter and
outputs a menu of options to manipulate the playlist. Each option is represented by a single
character. Build and output the menu within the method.
If an invalid character is entered, continue to prompt for a valid choice. Hint: Implement Quit
before implementing other options. Call printMenu() in the main() method. Continue to execute
the menu until the user enters q to Quit. (3 pts)
Ex:
(4) Implement \"Output full playlist\" menu option. If the list is empty, output: Playlist is empty
(3 pts)
Ex:
(5) Implement the \"Add song\" menu item. New additions are added to the end of the list. (2
pts)
Ex:
(6) Implement the \"Remove song\" method. Prompt the user for the unique ID of the song to be
removed.(4 pts)
Ex:
(7) Implement the \"Change position of song\" menu option. Prompt the user for the current
position of the song and the desired new position. Valid new positions are 1 - n (the number of
nodes). If the user enters a new position that is less than 1, move the node to the position 1 (the
head). If the user enters a new position greater than n, move the node to position n (the tail). 6
cases will be tested:
Moving the head node (1 pt)
Moving the tail node (1 pt)
Moving a node to the head (1 pt)
Moving a node to the tail (1 pt)
Moving a node up the list (1 pt)
Moving a node down the list (1 pt)
Ex:
(8) Implement the \"Output songs by specific artist\" menu option. Prompt the user for the
artist\'s name, and output the node\'s information, starting with the node\'s current position. (2 pt)
Ex:
(9) Implement the \"Output total time of playlist\" menu option. Output the sum of the time of
the playlist\'s songs (in seconds). (2 pts)
Ex:
Solution
SongEntry.java
package playlist;
public class SongEntry {
private String uniqueID;
private String songName;
private String artistName.
5. Cells with an intracellular osmolarity of 300mosmL are suspended.pdf
5. Cells with an intracellular osmolarity of 300mosm/L are suspended in the following solutions.
Describe the sequence of events that take place.
(a)0.3M urea (membrane permeable)
(b)0.3M glucose (membrane impermeable)
(c)0.6M urea
(d)0.6M glucose
Solution
Osmolarity of a solution refers to the concentration of the solution in terms of how strongly it
attracts water molecules across a semipermeable membrane. It is given the intracellular
osmolarity of the cell is 300 mosm/L. It can also be expressed as 0.3 osm/L or 0.3 M.
(a) 0.3 M Urea (membrane permeable)
This solution is isosmotic to the intracellular osmotic pressure. But, the cell membrane being
selectively permeable, allows the entry of urea until both the solutions become isotonic. This
leads to bursting ofcell as along with urea water enters the cell.
(b) 0.3 M Glucose (membrane impermeable)
This solution is also isosmotic to the intracellular osmotic pressure. But given that the membrane
is impermeable, glucose cannot enter to cell and an isotonic condition cannot be reached. Hence
the cell would be unstable and would distort due to the uneven tonicity.
(c) 0.6 M urea
This solution is hypertonic and hyperosmotic as compared to the cell\'s intracellular fluid. Thus,
water will move out through the pores of the cell membrane causing the cell to shrink or become
flaccid.
(d) 0.6 M glucose
This solution is hypertonic and hyperosmotic as compared to the cell\'s intracellular fluid. Thus,
water will move out through the pores of the cell membrane causing the cell to shrink or become
flaccid..
Which symptom of inflammation is NOT paired with its sourceA) swell.pdf
Which symptom of inflammation is NOT paired with its source?A) swelling: accumulation of
fluid in connective tissueB) heat: due to increased blood flowC) redness: blood leaking into
surrounding tissuesD) pain: pressure of fluid on nerve endingsE) pain: release of inflammatory
chemicals affect nerve endings directlyF) All of the listed responses are correctly paired.
Solution
Option C is wrong. Because both redness and heat is mainly due to increased blood flow to the
inflamed area..
Which is true of the data shown in the histogram I. The distribution.pdf
Which is true of the data shown in the histogram? I. The distribution is approximately
symmetric. II. The mean and median are approximately equal. III. The median and IQR
summarize the data better than the mean and standard deviation. Click the icon to view the
histogram. I and III I, II and III I only I and II III only
Solution
(I) The data shows that the shape of this histogram is bell-shaped. Also,if a line is drawn down
the middle of the graph, the two sides will mirror each other. So,the distribution is approximately
symmetric.
(II) In the given histogram where the values of data occur at regular frequencies. So, the mean,
median and mode occur at the same point.
(III) The IQR is the difference between the third and the first quartile. IQR makes the values by
estimate of outliers where mean and standard deviation are not effected against outliers. So, the
median and IQR is not summarize the data well.
Answer is (I) and (II).
Which of the following is a member of the domain ArchaeaA. The ce.pdf
Which of the following is a member of the domain Archaea?
A. The cell is eukaryotic
B. The cell has a nucleus
C. The cell is prokaryotic
D. The cell lack phospholipid
Solution
C. The cell is prokaryotic
Archea are the unicellular prokaryotes that can inhabit an extreme range of environments due to
their cellular adaptations. Their cell wall does not contain peptidoglycans but contains, S-layers..
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Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
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● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
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