the reaction between SnCl4 and pyridine conducted.pdfanonaeon
the reaction between SnCl4 and pyridine conducted in EtOH gives another
hexacoordinated compound [(py)4 SnCl¢] C12
Solution
the reaction between SnCl4 and pyridine conducted in EtOH gives another
hexacoordinated compound [(py)4 SnCl¢] C12.
The number of mols of HBr initially add to the so.pdfanonaeon
The number of mols of HBr initially add to the solution of Zn(OH)2 = 0.350 L *0.5
M = 0.175 mol Even after addition of HBr to the Zn(OH)2 solution, the solution is acidic, The
number of mols of H+ present after the reaction completed is calculated by using the titration
with NaOH. the number of mols of NaOH consumed = (0.5 M*88.5 mL*1.0 L) /1000 mL =
0.04425 mol therefore the number of mols of HBr remained = 0.04425 mol Therefore number of
mols of HBr consumed with the Zn(OH)2 = 0.175 -0.04425 = 0.1307 mol The balnced equation
between then Zn(OH)2 and HBr is - Zn(OH)2 (Aq) + 2 HBr (aq ) -------------> ZnBr2 (Aq) + 2
H2O(l) 0.1307 mol Therfore - number of moles of Zn(OH)2 reacted with HBr = 0.1307 /2 =
0.06535 mol therfore mass of Zn(OH)2 added to the solution = 99.38 g/mol*0.06535 = 6.5 g
Solution
The number of mols of HBr initially add to the solution of Zn(OH)2 = 0.350 L *0.5
M = 0.175 mol Even after addition of HBr to the Zn(OH)2 solution, the solution is acidic, The
number of mols of H+ present after the reaction completed is calculated by using the titration
with NaOH. the number of mols of NaOH consumed = (0.5 M*88.5 mL*1.0 L) /1000 mL =
0.04425 mol therefore the number of mols of HBr remained = 0.04425 mol Therefore number of
mols of HBr consumed with the Zn(OH)2 = 0.175 -0.04425 = 0.1307 mol The balnced equation
between then Zn(OH)2 and HBr is - Zn(OH)2 (Aq) + 2 HBr (aq ) -------------> ZnBr2 (Aq) + 2
H2O(l) 0.1307 mol Therfore - number of moles of Zn(OH)2 reacted with HBr = 0.1307 /2 =
0.06535 mol therfore mass of Zn(OH)2 added to the solution = 99.38 g/mol*0.06535 = 6.5 g.
One or more hydrogens can be replaced by halogens.pdfanonaeon
One or more hydrogens can be replaced by halogens. ... geometric or cis
Solution
One or more hydrogens can be replaced by halogens. ... geometric or cis.
linear This is Because here Double Bonds in a st.pdfanonaeon
linear This is Because here Double Bonds in a straight line are present
Solution
linear This is Because here Double Bonds in a straight line are present.
Eugenol, as a phenol, is a weak acid, whereas ace.pdfanonaeon
Eugenol, as a phenol, is a weak acid, whereas acetyleugenol is an ester. When the
crude mixture of them was treated with a weak base (NaOH), Eugenol was converted ionic
sodium salt of eugenol, which is very soluble in water. while, acetyleugenol is insouble in water.
This allows seperate eugenol from acetyleugenol using extraction.
Solution
Eugenol, as a phenol, is a weak acid, whereas acetyleugenol is an ester. When the
crude mixture of them was treated with a weak base (NaOH), Eugenol was converted ionic
sodium salt of eugenol, which is very soluble in water. while, acetyleugenol is insouble in water.
This allows seperate eugenol from acetyleugenol using extraction..
Hg(NO3)2 is completely ionic HgCl2 is partially i.pdfanonaeon
Hg(NO3)2 is completely ionic HgCl2 is partially ionic with more covalent character
Solution
Hg(NO3)2 is completely ionic HgCl2 is partially ionic with more covalent character.
the reaction between SnCl4 and pyridine conducted.pdfanonaeon
the reaction between SnCl4 and pyridine conducted in EtOH gives another
hexacoordinated compound [(py)4 SnCl¢] C12
Solution
the reaction between SnCl4 and pyridine conducted in EtOH gives another
hexacoordinated compound [(py)4 SnCl¢] C12.
The number of mols of HBr initially add to the so.pdfanonaeon
The number of mols of HBr initially add to the solution of Zn(OH)2 = 0.350 L *0.5
M = 0.175 mol Even after addition of HBr to the Zn(OH)2 solution, the solution is acidic, The
number of mols of H+ present after the reaction completed is calculated by using the titration
with NaOH. the number of mols of NaOH consumed = (0.5 M*88.5 mL*1.0 L) /1000 mL =
0.04425 mol therefore the number of mols of HBr remained = 0.04425 mol Therefore number of
mols of HBr consumed with the Zn(OH)2 = 0.175 -0.04425 = 0.1307 mol The balnced equation
between then Zn(OH)2 and HBr is - Zn(OH)2 (Aq) + 2 HBr (aq ) -------------> ZnBr2 (Aq) + 2
H2O(l) 0.1307 mol Therfore - number of moles of Zn(OH)2 reacted with HBr = 0.1307 /2 =
0.06535 mol therfore mass of Zn(OH)2 added to the solution = 99.38 g/mol*0.06535 = 6.5 g
Solution
The number of mols of HBr initially add to the solution of Zn(OH)2 = 0.350 L *0.5
M = 0.175 mol Even after addition of HBr to the Zn(OH)2 solution, the solution is acidic, The
number of mols of H+ present after the reaction completed is calculated by using the titration
with NaOH. the number of mols of NaOH consumed = (0.5 M*88.5 mL*1.0 L) /1000 mL =
0.04425 mol therefore the number of mols of HBr remained = 0.04425 mol Therefore number of
mols of HBr consumed with the Zn(OH)2 = 0.175 -0.04425 = 0.1307 mol The balnced equation
between then Zn(OH)2 and HBr is - Zn(OH)2 (Aq) + 2 HBr (aq ) -------------> ZnBr2 (Aq) + 2
H2O(l) 0.1307 mol Therfore - number of moles of Zn(OH)2 reacted with HBr = 0.1307 /2 =
0.06535 mol therfore mass of Zn(OH)2 added to the solution = 99.38 g/mol*0.06535 = 6.5 g.
One or more hydrogens can be replaced by halogens.pdfanonaeon
One or more hydrogens can be replaced by halogens. ... geometric or cis
Solution
One or more hydrogens can be replaced by halogens. ... geometric or cis.
linear This is Because here Double Bonds in a st.pdfanonaeon
linear This is Because here Double Bonds in a straight line are present
Solution
linear This is Because here Double Bonds in a straight line are present.
Eugenol, as a phenol, is a weak acid, whereas ace.pdfanonaeon
Eugenol, as a phenol, is a weak acid, whereas acetyleugenol is an ester. When the
crude mixture of them was treated with a weak base (NaOH), Eugenol was converted ionic
sodium salt of eugenol, which is very soluble in water. while, acetyleugenol is insouble in water.
This allows seperate eugenol from acetyleugenol using extraction.
Solution
Eugenol, as a phenol, is a weak acid, whereas acetyleugenol is an ester. When the
crude mixture of them was treated with a weak base (NaOH), Eugenol was converted ionic
sodium salt of eugenol, which is very soluble in water. while, acetyleugenol is insouble in water.
This allows seperate eugenol from acetyleugenol using extraction..
Hg(NO3)2 is completely ionic HgCl2 is partially i.pdfanonaeon
Hg(NO3)2 is completely ionic HgCl2 is partially ionic with more covalent character
Solution
Hg(NO3)2 is completely ionic HgCl2 is partially ionic with more covalent character.
y+y=x It is a linear ordinary differential eqn of the Bernouille.pdfanonaeon
y\'+y=x
It is a linear ordinary differential eqn of the Bernouille form.
So integrating factor, IF=exp[integral (dx)]=e^x
so y\'e^x+ye^x=xe^x
=>d(ye^x)/dx=xe^x
=>d(ye^x)=xe^xdx
Integrating we have
ye^x=integral[xe^xdx]
=xintegral[e^xdx]-integral[d(x)/dx{integral(e^xdx)}dx] [integrating by parts]
=xe^x-integral[1.e^xdx]
=xe^x-integral[e^xdx]
=xe^x-e^x+c
=>y=x-1+ce^(-x)
y(0)=2
=>-1+c=2
=>c=3
so y=x+3e^(-x)-1
y\'+y=e^(-x)
IF=exp[integral(dx)]=e^x
so y\'e^x+ye^x=1
=>d(ye^x)/dx=1
=>d(ye^x)=dx
=>ye^x=x+c
=>y=xe^(-x)+ce^(-x)
y(0)=1
=>c=1
so y=xe^(-x)+e^(-x)
=>y=(x+1)e^(-x)
xy\'+2y=10x^2
=>y\'+2y/x=10x
IF=exp[integral{(2/x)dx}]=exp[2lnx]=x^2
so y\'x^2+2xy=10x^3
=>d(x^2y)/dx=10x^3
=>d(x^2y)=10x^3dx
=>yx^2=integral[10x^3dx]=10x^4/4+c
=>y=5x^2/2+cx^(-2)
y(1)=3
=>5/2+c=3
=>c=3-5/2=1/2
so y=5x^2/2+1/(2x^2)
Solution
y\'+y=x
It is a linear ordinary differential eqn of the Bernouille form.
So integrating factor, IF=exp[integral (dx)]=e^x
so y\'e^x+ye^x=xe^x
=>d(ye^x)/dx=xe^x
=>d(ye^x)=xe^xdx
Integrating we have
ye^x=integral[xe^xdx]
=xintegral[e^xdx]-integral[d(x)/dx{integral(e^xdx)}dx] [integrating by parts]
=xe^x-integral[1.e^xdx]
=xe^x-integral[e^xdx]
=xe^x-e^x+c
=>y=x-1+ce^(-x)
y(0)=2
=>-1+c=2
=>c=3
so y=x+3e^(-x)-1
y\'+y=e^(-x)
IF=exp[integral(dx)]=e^x
so y\'e^x+ye^x=1
=>d(ye^x)/dx=1
=>d(ye^x)=dx
=>ye^x=x+c
=>y=xe^(-x)+ce^(-x)
y(0)=1
=>c=1
so y=xe^(-x)+e^(-x)
=>y=(x+1)e^(-x)
xy\'+2y=10x^2
=>y\'+2y/x=10x
IF=exp[integral{(2/x)dx}]=exp[2lnx]=x^2
so y\'x^2+2xy=10x^3
=>d(x^2y)/dx=10x^3
=>d(x^2y)=10x^3dx
=>yx^2=integral[10x^3dx]=10x^4/4+c
=>y=5x^2/2+cx^(-2)
y(1)=3
=>5/2+c=3
=>c=3-5/2=1/2
so y=5x^2/2+1/(2x^2).
True Since E union Ec is the whole sample space which has probabilit.pdfanonaeon
True Since E union Ec is the whole sample space which has probability 1
Solution
True Since E union Ec is the whole sample space which has probability 1.
There are many ethical and legal issues surrounding the US human ser.pdfanonaeon
There are many ethical and legal issues surrounding the US human service system. There are
many ethical issues faced by social workers which involves legal matters. They have to disclose
confidential information about a person according to legal guideline without client\'s consent.
Secondly humanitarian ground sometime become above Legal matters it is difficult for them to
abide both simultaneously.
Solution
There are many ethical and legal issues surrounding the US human service system. There are
many ethical issues faced by social workers which involves legal matters. They have to disclose
confidential information about a person according to legal guideline without client\'s consent.
Secondly humanitarian ground sometime become above Legal matters it is difficult for them to
abide both simultaneously..
The rituals and ceremony that follows the death of a person in order.pdfanonaeon
The rituals and ceremony that follows the death of a person in order to dispose off the body
along with religious fervor is called a funeral. Normally family and friends are the attendees who
follow the funerary customs and beliefs based on complex religious practices to remember and
respect the dead. Grieving and death rituals vary across cultures and are often heavily influenced
by religion (Chachkes & Jennings, 1994; Younoszai, 1993). How and when rituals are practiced
vary depending on the country of origin and level of acculturation into the mainstream society.
The duration, frequency, and intensity of the grief process may also vary based on the manner of
death and the individual family and cultural beliefs (Clements et al., 2003). These rituals give the
families and friends a chance to give up the deceased and properly mourn the loss. Common
secular motivations for funerals include mourning the deceased, celebrating their life, and
offering support and sympathy to the bereaved.
Types of funeral rituals based on religions:
Assignment topic - Organ donation
Organ donation is considered as an honorary way of disposing off a dead body. On deceased
donor can save up to eight lives and can save and enhance more than 100 lives through tissue
donation. Organs that can be donated after death are the heart, liver, kidneys, lungs, pancreas and
small intestines. Tissues include corneas, skin, veins, heart valves, tendons, ligaments and bones.
The cornea is the most commonly transplanted tissue. More than 40,000 corneal transplants take
place each year in the United States. Most major religions support organ and tissue donation.
Typically, religions view organ and tissue donation as acts of charity and goodwill. Costs
associated with recovering and processing organs and tissues for transplant are never passed on
to the donor family. The family may be expected to pay for medical expenses incurred before
death is declared and for expenses involving funeral arrangements. the average North American
traditional funeral costs between $7,000 and $10,000. This price range includes the services at
the funeral home, burial in a cemetery, and the installation of a headstone.
While the cost of organ donation is a widely accepted part of the procedure today, some scholars
have analysed the ethical incentives involved in this act rather than its monetary counterpart.
There have been instances in the past where the act itself was performed with an altruist motive
and the donors were left uncompensated, whether dead or alive. Despite that, there are other
ethical incentives apart from a sense of magnanimity that a person feels from organ donation.
Board (2002) talks about various forms of such incentives that influence people’s behaviour. A
donor medal of honour can be thought of as similar to an award for employee of the month in
terms of the satisfaction that it provides to people. Apart from that he also talks about a medical
leave for organ donation in order t.
The primary objectivegoal of this paper is to study the effect of w.pdfanonaeon
The primary objective/goal of this paper is to study the effect of wallowing behavior of bisons on
the above-ground net primary production (ANPP), plant species richness and diversity and plant
life from richness and diversity.
Solution
The primary objective/goal of this paper is to study the effect of wallowing behavior of bisons on
the above-ground net primary production (ANPP), plant species richness and diversity and plant
life from richness and diversity..
Symmetry It is the property where parts of things resembles each o.pdfanonaeon
Symmetry : It is the property where parts of things resembles each other. Bilateral symmetry and
radial symmetry are found mostly in animals.
Radial symmetry: There is presence of similar body parts that are distributed in circular manner
around a central axis e.g. in case of echinoderms and coelenterates .Animals having radial
symmetry possess two dorsal and ventral sides instead of right and left sides.
Bilateral symmetry: In this case body is divided into two equal parts through central plane. The
two equal halves are called right and left e.g. human body which can be divided into two equal
parts through sagittal plane.
Differences between radial and bilateral symmetry:
1. Radial symmetry possess symmetric axis while as bilateral symmetry has a symmetric plane.
2. In radial symmetry more than two similar portions of the body can be identified while as in
bilateral symmetry only two similar portions can be identified.
3. Most of the radially symmetric animals have aquatic habitat while as bilaterally symmetric
animals are found in both land and water.
4. Among animals bilateral symmetry is more common as compared to radial symmetry.
5.There is special type of symmetry found in echinoderms known as penta radial symmetry
while as bilaterally symmetric animals can be divided in to two parts only.
6. Cnidarians, Echinoderms and unicellular organisms are radially symmetric while as all other
phyla in animal kingdom exhibit bilateral symmetry.
Solution
Symmetry : It is the property where parts of things resembles each other. Bilateral symmetry and
radial symmetry are found mostly in animals.
Radial symmetry: There is presence of similar body parts that are distributed in circular manner
around a central axis e.g. in case of echinoderms and coelenterates .Animals having radial
symmetry possess two dorsal and ventral sides instead of right and left sides.
Bilateral symmetry: In this case body is divided into two equal parts through central plane. The
two equal halves are called right and left e.g. human body which can be divided into two equal
parts through sagittal plane.
Differences between radial and bilateral symmetry:
1. Radial symmetry possess symmetric axis while as bilateral symmetry has a symmetric plane.
2. In radial symmetry more than two similar portions of the body can be identified while as in
bilateral symmetry only two similar portions can be identified.
3. Most of the radially symmetric animals have aquatic habitat while as bilaterally symmetric
animals are found in both land and water.
4. Among animals bilateral symmetry is more common as compared to radial symmetry.
5.There is special type of symmetry found in echinoderms known as penta radial symmetry
while as bilaterally symmetric animals can be divided in to two parts only.
6. Cnidarians, Echinoderms and unicellular organisms are radially symmetric while as all other
phyla in animal kingdom exhibit bilateral symmetry..
struct procedure { Date dateOfProcedure; int procedureID.pdfanonaeon
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
Slide to depict the basic layout of the pages in the website. Usiing.pdfanonaeon
Slide to depict the basic layout of the pages in the website. Usiing place holders images or shapes
for depiction.
Ok
Solution
Slide to depict the basic layout of the pages in the website. Usiing place holders images or shapes
for depiction.
Ok.
Part A-iv) Parietal CellsExplanation - The lining of the stoma.pdfanonaeon
Part A:-
iv) Parietal Cells
Explanation :- The lining of the stomach is consists of tough, Hydrochloric acid producing cells,
which are termed as Parietal cells.
Part B :-
iv) Colon
Explanation :- The large part of Intestine is the colon, which absorbs excess water, salts from the
digested food.
Part C
iv) Bile salts : Bile salts along with monoglycerides of the fats assist the diffusion of fat globules
into the intestinal epithelium of the small intestine, so that it forms chylomicrons and transported
through ER by the formation of Micelle into Lacteal membrane.
Part D
iii) Liver
This is largest gland in the body, which conisist of storage of majority of nutrients, which
supplied into blood.
Part E
i) Pancreas : The majority of the digestive enzymes like Trypsin, Chymotrypsin, Pancreatic
lipase are produced from Pancrease, so this is termed so.
Part F
iv) D increase surface area and enchance digestion and absorption of nutrients in small intestine.
The small intestine consists of many 1,00,000 of small finger like projections called Villi, which
increase area of absorption and digestion of the digested food material into blood stream.
Solution
Part A:-
iv) Parietal Cells
Explanation :- The lining of the stomach is consists of tough, Hydrochloric acid producing cells,
which are termed as Parietal cells.
Part B :-
iv) Colon
Explanation :- The large part of Intestine is the colon, which absorbs excess water, salts from the
digested food.
Part C
iv) Bile salts : Bile salts along with monoglycerides of the fats assist the diffusion of fat globules
into the intestinal epithelium of the small intestine, so that it forms chylomicrons and transported
through ER by the formation of Micelle into Lacteal membrane.
Part D
iii) Liver
This is largest gland in the body, which conisist of storage of majority of nutrients, which
supplied into blood.
Part E
i) Pancreas : The majority of the digestive enzymes like Trypsin, Chymotrypsin, Pancreatic
lipase are produced from Pancrease, so this is termed so.
Part F
iv) D increase surface area and enchance digestion and absorption of nutrients in small intestine.
The small intestine consists of many 1,00,000 of small finger like projections called Villi, which
increase area of absorption and digestion of the digested food material into blood stream..
Microplate capture and detection assay is used to detect the specifi.pdfanonaeon
Microplate capture and detection assay is used to detect the specific protein-DNA interactions
and identifies the proteins. The method includes immobilization of DNA probes on the surface of
microplates (having wells) coated with streptavidin. The cellular extract, prepared in binding
buffer, is added for enough time to facilitate binding protein to bind to the oligonucleotide
followed by washing of wells several times to remove nonspecifically bound proteins. A protein
specific antibody is used for detection of protein.
Comparison of advantages and disadvantages of microplate capture and detection assay and
EMSA:
A: Advantages
1: Microplate capture and detection assay is highly sensitive, fast and throughput technique that
is based on ELISA while EMSA is based on comparative velocity of migration of DNA-protein
complexes through non denaturing gel electrophoresis.
2: Both techniques can skip the use of radioacitve probes by using biotinylated or fluorescently
labeled DNA probes.
3: EMSA uses DNA probe mutational analysis for testing the binding affinity of proteins.
4: Though EMSA can detect the presence of low quantities of DNA binding proteins from
lysates, Microplate capture and detection assay can detect even lesser abundance of protein in the
well by using with enzyme-labeled antibodies and a chemiluminescent substrate.
Disadvantages:
1: Both techniques require supershift antibodies i.e. antibodies which can selectively bind to
DNA bound native proteins.
2: The technique gives little information about corresponding changes in transcription factor-
DNA.
3: The techniques have been used for a few target proteins so far while EMSA is applicable
under in vitro conditions only.
Solution
Microplate capture and detection assay is used to detect the specific protein-DNA interactions
and identifies the proteins. The method includes immobilization of DNA probes on the surface of
microplates (having wells) coated with streptavidin. The cellular extract, prepared in binding
buffer, is added for enough time to facilitate binding protein to bind to the oligonucleotide
followed by washing of wells several times to remove nonspecifically bound proteins. A protein
specific antibody is used for detection of protein.
Comparison of advantages and disadvantages of microplate capture and detection assay and
EMSA:
A: Advantages
1: Microplate capture and detection assay is highly sensitive, fast and throughput technique that
is based on ELISA while EMSA is based on comparative velocity of migration of DNA-protein
complexes through non denaturing gel electrophoresis.
2: Both techniques can skip the use of radioacitve probes by using biotinylated or fluorescently
labeled DNA probes.
3: EMSA uses DNA probe mutational analysis for testing the binding affinity of proteins.
4: Though EMSA can detect the presence of low quantities of DNA binding proteins from
lysates, Microplate capture and detection assay can detect even lesser abundance of protein in the
well by us.
C. Amide - Organic functional group found in Nylo.pdfanonaeon
C. Amide - Organic functional group found in Nylon D. Amine - Organic functional
group found in peptide bonds A. Polyethylene terephtalate - Used to make soda bottles B. High
density polyethylene - Used to make milk jugs
Solution
C. Amide - Organic functional group found in Nylon D. Amine - Organic functional
group found in peptide bonds A. Polyethylene terephtalate - Used to make soda bottles B. High
density polyethylene - Used to make milk jugs.
H2PO4^-ThereforeP has Oxidation state = +7H has oxidation stat.pdfanonaeon
H2PO4^-
Therefore
P has Oxidation state = +7
H has oxidation state = + 1
O has Oxidation state = - 2
Solution
H2PO4^-
Therefore
P has Oxidation state = +7
H has oxidation state = + 1
O has Oxidation state = - 2.
First of all this program has compilation errors.Line 14 WilmaL.pdfanonaeon
First of all this program has compilation errors.
Line 14: Wilma
Line 15:Fred, Pebbles, Barney, Bam-Bam, Betty
____________________________________
The Index of the ArrayList Starts from 0
Till line 11, the elements in the Array list are:
[Fred, Barney, Wilma, Betty]
In line 12,we are adding \"BaM-Bam\" at index 3
[Fred, Barney, Wilma, Bam-Bam, Betty]
In Line 13, we are adding \"Pebbles\" at index 1
[Fred, Pebbles, Barney, Wilma, Bam-Bam, Betty]
When this statement in the line 14 is executed
System.out.println(peopleList.remove(3));
This line will removes And displays Wilma
The line 15 will displays the elements in the array list:
Fred, Pebbles, Barney, Bam-Bam, Betty
In line 17 it is wrong atement.The program wont compile.
suppose if it is collections.sort(null) we will get null pointer exception.
Then line 19 will not get executed.
Solution
First of all this program has compilation errors.
Line 14: Wilma
Line 15:Fred, Pebbles, Barney, Bam-Bam, Betty
____________________________________
The Index of the ArrayList Starts from 0
Till line 11, the elements in the Array list are:
[Fred, Barney, Wilma, Betty]
In line 12,we are adding \"BaM-Bam\" at index 3
[Fred, Barney, Wilma, Bam-Bam, Betty]
In Line 13, we are adding \"Pebbles\" at index 1
[Fred, Pebbles, Barney, Wilma, Bam-Bam, Betty]
When this statement in the line 14 is executed
System.out.println(peopleList.remove(3));
This line will removes And displays Wilma
The line 15 will displays the elements in the array list:
Fred, Pebbles, Barney, Bam-Bam, Betty
In line 17 it is wrong atement.The program wont compile.
suppose if it is collections.sort(null) we will get null pointer exception.
Then line 19 will not get executed..
C and D. TCPIP, or Transmission Control ProtocolInternet Protocol,.pdfanonaeon
C and D. TCP/IP, or Transmission Control Protocol/Internet Protocol, is the basic
communication protocol for the Internet and is also used on private networks such as intranets
(web environments within a company) or extranets (web environments between two or more
business partners or between a vendor and customers). NetBIOS, or Network Basic Input/Output
System, is a program that allows computers to communicate on local area networks, or LANs,
but does not support transport into wide area networks, or WANs, such as the Internet.
Solution
C and D. TCP/IP, or Transmission Control Protocol/Internet Protocol, is the basic
communication protocol for the Internet and is also used on private networks such as intranets
(web environments within a company) or extranets (web environments between two or more
business partners or between a vendor and customers). NetBIOS, or Network Basic Input/Output
System, is a program that allows computers to communicate on local area networks, or LANs,
but does not support transport into wide area networks, or WANs, such as the Internet..
Answer is,D. Greater activation of left hemisphere blood flow.Op.pdfanonaeon
Answer is,
D. Greater activation of left hemisphere blood flow.
Options A , B and C are related to the evolutionary origin of the hemisphere asymmetry.
Solution
Answer is,
D. Greater activation of left hemisphere blood flow.
Options A , B and C are related to the evolutionary origin of the hemisphere asymmetry..
Answer Real Estate Life Cycle is having 3 stages at Universal leve.pdfanonaeon
Answer : Real Estate Life Cycle is having 3 stages at Universal level i.e. Pre- development stage,
Development Stage and Developed stage.
1) Pre development stage - It is stage when the land is raw and there is no structure on it, betting
at such location would be risky as defining the value in future is difficult. since the area may be
devoid of people but would come in future based on government policies or any other attraction
for people to come and live.
2) Development stage - It will come after the area or land has been defined for development, that
will attract developers to create buildings or structure and suddenly will shoot the prices of land.
3) Developed stage - When the inventory is ready and it needs to be sold to the buyers, it is the
end satge of cycle and everyone can come to get benefit from it. But the pricing are speculative
at this stage and based on the supply and demand of the property.
Why these are important for investment and finance purpose
Pre development stage, it is difficult to predict time taken for its conversion and hence most
illiquid stage, buying at this point would be a risk.
Development Stage, prediction is easy and developer would gonna sell it may be after
development or during construction . So easy to predict but reward may not be that much.
Developed stage can be risky as it is based more n speculation, if demand is more investment can
be beneficial but if not then it will be a risky investment.
Solution
Answer : Real Estate Life Cycle is having 3 stages at Universal level i.e. Pre- development stage,
Development Stage and Developed stage.
1) Pre development stage - It is stage when the land is raw and there is no structure on it, betting
at such location would be risky as defining the value in future is difficult. since the area may be
devoid of people but would come in future based on government policies or any other attraction
for people to come and live.
2) Development stage - It will come after the area or land has been defined for development, that
will attract developers to create buildings or structure and suddenly will shoot the prices of land.
3) Developed stage - When the inventory is ready and it needs to be sold to the buyers, it is the
end satge of cycle and everyone can come to get benefit from it. But the pricing are speculative
at this stage and based on the supply and demand of the property.
Why these are important for investment and finance purpose
Pre development stage, it is difficult to predict time taken for its conversion and hence most
illiquid stage, buying at this point would be a risk.
Development Stage, prediction is easy and developer would gonna sell it may be after
development or during construction . So easy to predict but reward may not be that much.
Developed stage can be risky as it is based more n speculation, if demand is more investment can
be beneficial but if not then it will be a risky investment..
1.static int HORIZONTAL_WRAP2. Container view controllers are most.pdfanonaeon
1.static int HORIZONTAL_WRAP
2. Container view controllers are most often used to facilitate navigation and to create new user
interface types based on existing content. Examples of container view controllers in UIKit
includeUINavigationController, UITabBarController, and UISplitViewController, all of which
facilitate navigation between different parts of your user interface.
3. Andoid supports .png(preferred),.jpg(accepteble),.gif(discouraged)
4.Each element in an array is called an Element
5.For Eclipse, typically all you need to do is set up a folder somewhere within your source code
directory. For instance, if the directory containing your source is /src then you can create a
/src/resources folder to place your images/files in. Then, within your class you do a
getResource(\"/resources/image.png\") to retrieve it.
6.Native apps have the ability to use device-specific hardware and software, meaning that native
apps can take advantage of the latest technology available on mobile devices such as a global
positioning system (GPS) and camera.
The main similarity is that both emulated mobile apps and Web apps run on servers external to
the mobile device and require the use of a browser on the mobile device to display and then use
the app user interface (UI).
7.Marshmallow 6.0-6.0.1
8.This ColorModel class supports two representations of pixel values. A pixel value can be a
single 32-bit int or an array of primitive types.
9.By changing the android:label field in your application node in AndroidManifest.xml.
10.\"android:layout_span” to span view in 2 cells, and “android:layout_column” to display the
view in specified column.
11.1.ListActivity.java
2.activity__twopane.xml
3.activity__list.xml
4.ListFragment.java
5.DetailActivity.java
13.
14.A dp is a density-independent pixel that corresponds to the physical size of a pixel at 160 dpi.
An sp is the same base unit, but is scaled by the user\'s preferred text size (it\'s a scale-
independent pixel), so you should use this measurement unit when defining text size (but never
for layout sizes).
15.Tween animation calculates the animation with information such as the start point, end point,
size, rotation, and other common aspects of an animation.
16.Setting a View\'s size to wrap_content will force it to expand only far enough to contain the
values (or child controls) it contains. For controls -- like text boxes (TextView) or images
(ImageView) -- this will wrap the text or image being shown. For layout elements it will resize
the layout to fit the controls / layouts added as its children.
17.An animation defined in XML that modifies properties of the target object, such as
background color or alpha value, over a set amount of time.
18.9-patch image
Solution
1.static int HORIZONTAL_WRAP
2. Container view controllers are most often used to facilitate navigation and to create new user
interface types based on existing content. Examples of container view controllers in UIKit
includeUINavigationController.
y+y=x It is a linear ordinary differential eqn of the Bernouille.pdfanonaeon
y\'+y=x
It is a linear ordinary differential eqn of the Bernouille form.
So integrating factor, IF=exp[integral (dx)]=e^x
so y\'e^x+ye^x=xe^x
=>d(ye^x)/dx=xe^x
=>d(ye^x)=xe^xdx
Integrating we have
ye^x=integral[xe^xdx]
=xintegral[e^xdx]-integral[d(x)/dx{integral(e^xdx)}dx] [integrating by parts]
=xe^x-integral[1.e^xdx]
=xe^x-integral[e^xdx]
=xe^x-e^x+c
=>y=x-1+ce^(-x)
y(0)=2
=>-1+c=2
=>c=3
so y=x+3e^(-x)-1
y\'+y=e^(-x)
IF=exp[integral(dx)]=e^x
so y\'e^x+ye^x=1
=>d(ye^x)/dx=1
=>d(ye^x)=dx
=>ye^x=x+c
=>y=xe^(-x)+ce^(-x)
y(0)=1
=>c=1
so y=xe^(-x)+e^(-x)
=>y=(x+1)e^(-x)
xy\'+2y=10x^2
=>y\'+2y/x=10x
IF=exp[integral{(2/x)dx}]=exp[2lnx]=x^2
so y\'x^2+2xy=10x^3
=>d(x^2y)/dx=10x^3
=>d(x^2y)=10x^3dx
=>yx^2=integral[10x^3dx]=10x^4/4+c
=>y=5x^2/2+cx^(-2)
y(1)=3
=>5/2+c=3
=>c=3-5/2=1/2
so y=5x^2/2+1/(2x^2)
Solution
y\'+y=x
It is a linear ordinary differential eqn of the Bernouille form.
So integrating factor, IF=exp[integral (dx)]=e^x
so y\'e^x+ye^x=xe^x
=>d(ye^x)/dx=xe^x
=>d(ye^x)=xe^xdx
Integrating we have
ye^x=integral[xe^xdx]
=xintegral[e^xdx]-integral[d(x)/dx{integral(e^xdx)}dx] [integrating by parts]
=xe^x-integral[1.e^xdx]
=xe^x-integral[e^xdx]
=xe^x-e^x+c
=>y=x-1+ce^(-x)
y(0)=2
=>-1+c=2
=>c=3
so y=x+3e^(-x)-1
y\'+y=e^(-x)
IF=exp[integral(dx)]=e^x
so y\'e^x+ye^x=1
=>d(ye^x)/dx=1
=>d(ye^x)=dx
=>ye^x=x+c
=>y=xe^(-x)+ce^(-x)
y(0)=1
=>c=1
so y=xe^(-x)+e^(-x)
=>y=(x+1)e^(-x)
xy\'+2y=10x^2
=>y\'+2y/x=10x
IF=exp[integral{(2/x)dx}]=exp[2lnx]=x^2
so y\'x^2+2xy=10x^3
=>d(x^2y)/dx=10x^3
=>d(x^2y)=10x^3dx
=>yx^2=integral[10x^3dx]=10x^4/4+c
=>y=5x^2/2+cx^(-2)
y(1)=3
=>5/2+c=3
=>c=3-5/2=1/2
so y=5x^2/2+1/(2x^2).
True Since E union Ec is the whole sample space which has probabilit.pdfanonaeon
True Since E union Ec is the whole sample space which has probability 1
Solution
True Since E union Ec is the whole sample space which has probability 1.
There are many ethical and legal issues surrounding the US human ser.pdfanonaeon
There are many ethical and legal issues surrounding the US human service system. There are
many ethical issues faced by social workers which involves legal matters. They have to disclose
confidential information about a person according to legal guideline without client\'s consent.
Secondly humanitarian ground sometime become above Legal matters it is difficult for them to
abide both simultaneously.
Solution
There are many ethical and legal issues surrounding the US human service system. There are
many ethical issues faced by social workers which involves legal matters. They have to disclose
confidential information about a person according to legal guideline without client\'s consent.
Secondly humanitarian ground sometime become above Legal matters it is difficult for them to
abide both simultaneously..
The rituals and ceremony that follows the death of a person in order.pdfanonaeon
The rituals and ceremony that follows the death of a person in order to dispose off the body
along with religious fervor is called a funeral. Normally family and friends are the attendees who
follow the funerary customs and beliefs based on complex religious practices to remember and
respect the dead. Grieving and death rituals vary across cultures and are often heavily influenced
by religion (Chachkes & Jennings, 1994; Younoszai, 1993). How and when rituals are practiced
vary depending on the country of origin and level of acculturation into the mainstream society.
The duration, frequency, and intensity of the grief process may also vary based on the manner of
death and the individual family and cultural beliefs (Clements et al., 2003). These rituals give the
families and friends a chance to give up the deceased and properly mourn the loss. Common
secular motivations for funerals include mourning the deceased, celebrating their life, and
offering support and sympathy to the bereaved.
Types of funeral rituals based on religions:
Assignment topic - Organ donation
Organ donation is considered as an honorary way of disposing off a dead body. On deceased
donor can save up to eight lives and can save and enhance more than 100 lives through tissue
donation. Organs that can be donated after death are the heart, liver, kidneys, lungs, pancreas and
small intestines. Tissues include corneas, skin, veins, heart valves, tendons, ligaments and bones.
The cornea is the most commonly transplanted tissue. More than 40,000 corneal transplants take
place each year in the United States. Most major religions support organ and tissue donation.
Typically, religions view organ and tissue donation as acts of charity and goodwill. Costs
associated with recovering and processing organs and tissues for transplant are never passed on
to the donor family. The family may be expected to pay for medical expenses incurred before
death is declared and for expenses involving funeral arrangements. the average North American
traditional funeral costs between $7,000 and $10,000. This price range includes the services at
the funeral home, burial in a cemetery, and the installation of a headstone.
While the cost of organ donation is a widely accepted part of the procedure today, some scholars
have analysed the ethical incentives involved in this act rather than its monetary counterpart.
There have been instances in the past where the act itself was performed with an altruist motive
and the donors were left uncompensated, whether dead or alive. Despite that, there are other
ethical incentives apart from a sense of magnanimity that a person feels from organ donation.
Board (2002) talks about various forms of such incentives that influence people’s behaviour. A
donor medal of honour can be thought of as similar to an award for employee of the month in
terms of the satisfaction that it provides to people. Apart from that he also talks about a medical
leave for organ donation in order t.
The primary objectivegoal of this paper is to study the effect of w.pdfanonaeon
The primary objective/goal of this paper is to study the effect of wallowing behavior of bisons on
the above-ground net primary production (ANPP), plant species richness and diversity and plant
life from richness and diversity.
Solution
The primary objective/goal of this paper is to study the effect of wallowing behavior of bisons on
the above-ground net primary production (ANPP), plant species richness and diversity and plant
life from richness and diversity..
Symmetry It is the property where parts of things resembles each o.pdfanonaeon
Symmetry : It is the property where parts of things resembles each other. Bilateral symmetry and
radial symmetry are found mostly in animals.
Radial symmetry: There is presence of similar body parts that are distributed in circular manner
around a central axis e.g. in case of echinoderms and coelenterates .Animals having radial
symmetry possess two dorsal and ventral sides instead of right and left sides.
Bilateral symmetry: In this case body is divided into two equal parts through central plane. The
two equal halves are called right and left e.g. human body which can be divided into two equal
parts through sagittal plane.
Differences between radial and bilateral symmetry:
1. Radial symmetry possess symmetric axis while as bilateral symmetry has a symmetric plane.
2. In radial symmetry more than two similar portions of the body can be identified while as in
bilateral symmetry only two similar portions can be identified.
3. Most of the radially symmetric animals have aquatic habitat while as bilaterally symmetric
animals are found in both land and water.
4. Among animals bilateral symmetry is more common as compared to radial symmetry.
5.There is special type of symmetry found in echinoderms known as penta radial symmetry
while as bilaterally symmetric animals can be divided in to two parts only.
6. Cnidarians, Echinoderms and unicellular organisms are radially symmetric while as all other
phyla in animal kingdom exhibit bilateral symmetry.
Solution
Symmetry : It is the property where parts of things resembles each other. Bilateral symmetry and
radial symmetry are found mostly in animals.
Radial symmetry: There is presence of similar body parts that are distributed in circular manner
around a central axis e.g. in case of echinoderms and coelenterates .Animals having radial
symmetry possess two dorsal and ventral sides instead of right and left sides.
Bilateral symmetry: In this case body is divided into two equal parts through central plane. The
two equal halves are called right and left e.g. human body which can be divided into two equal
parts through sagittal plane.
Differences between radial and bilateral symmetry:
1. Radial symmetry possess symmetric axis while as bilateral symmetry has a symmetric plane.
2. In radial symmetry more than two similar portions of the body can be identified while as in
bilateral symmetry only two similar portions can be identified.
3. Most of the radially symmetric animals have aquatic habitat while as bilaterally symmetric
animals are found in both land and water.
4. Among animals bilateral symmetry is more common as compared to radial symmetry.
5.There is special type of symmetry found in echinoderms known as penta radial symmetry
while as bilaterally symmetric animals can be divided in to two parts only.
6. Cnidarians, Echinoderms and unicellular organisms are radially symmetric while as all other
phyla in animal kingdom exhibit bilateral symmetry..
struct procedure { Date dateOfProcedure; int procedureID.pdfanonaeon
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
Slide to depict the basic layout of the pages in the website. Usiing.pdfanonaeon
Slide to depict the basic layout of the pages in the website. Usiing place holders images or shapes
for depiction.
Ok
Solution
Slide to depict the basic layout of the pages in the website. Usiing place holders images or shapes
for depiction.
Ok.
Part A-iv) Parietal CellsExplanation - The lining of the stoma.pdfanonaeon
Part A:-
iv) Parietal Cells
Explanation :- The lining of the stomach is consists of tough, Hydrochloric acid producing cells,
which are termed as Parietal cells.
Part B :-
iv) Colon
Explanation :- The large part of Intestine is the colon, which absorbs excess water, salts from the
digested food.
Part C
iv) Bile salts : Bile salts along with monoglycerides of the fats assist the diffusion of fat globules
into the intestinal epithelium of the small intestine, so that it forms chylomicrons and transported
through ER by the formation of Micelle into Lacteal membrane.
Part D
iii) Liver
This is largest gland in the body, which conisist of storage of majority of nutrients, which
supplied into blood.
Part E
i) Pancreas : The majority of the digestive enzymes like Trypsin, Chymotrypsin, Pancreatic
lipase are produced from Pancrease, so this is termed so.
Part F
iv) D increase surface area and enchance digestion and absorption of nutrients in small intestine.
The small intestine consists of many 1,00,000 of small finger like projections called Villi, which
increase area of absorption and digestion of the digested food material into blood stream.
Solution
Part A:-
iv) Parietal Cells
Explanation :- The lining of the stomach is consists of tough, Hydrochloric acid producing cells,
which are termed as Parietal cells.
Part B :-
iv) Colon
Explanation :- The large part of Intestine is the colon, which absorbs excess water, salts from the
digested food.
Part C
iv) Bile salts : Bile salts along with monoglycerides of the fats assist the diffusion of fat globules
into the intestinal epithelium of the small intestine, so that it forms chylomicrons and transported
through ER by the formation of Micelle into Lacteal membrane.
Part D
iii) Liver
This is largest gland in the body, which conisist of storage of majority of nutrients, which
supplied into blood.
Part E
i) Pancreas : The majority of the digestive enzymes like Trypsin, Chymotrypsin, Pancreatic
lipase are produced from Pancrease, so this is termed so.
Part F
iv) D increase surface area and enchance digestion and absorption of nutrients in small intestine.
The small intestine consists of many 1,00,000 of small finger like projections called Villi, which
increase area of absorption and digestion of the digested food material into blood stream..
Microplate capture and detection assay is used to detect the specifi.pdfanonaeon
Microplate capture and detection assay is used to detect the specific protein-DNA interactions
and identifies the proteins. The method includes immobilization of DNA probes on the surface of
microplates (having wells) coated with streptavidin. The cellular extract, prepared in binding
buffer, is added for enough time to facilitate binding protein to bind to the oligonucleotide
followed by washing of wells several times to remove nonspecifically bound proteins. A protein
specific antibody is used for detection of protein.
Comparison of advantages and disadvantages of microplate capture and detection assay and
EMSA:
A: Advantages
1: Microplate capture and detection assay is highly sensitive, fast and throughput technique that
is based on ELISA while EMSA is based on comparative velocity of migration of DNA-protein
complexes through non denaturing gel electrophoresis.
2: Both techniques can skip the use of radioacitve probes by using biotinylated or fluorescently
labeled DNA probes.
3: EMSA uses DNA probe mutational analysis for testing the binding affinity of proteins.
4: Though EMSA can detect the presence of low quantities of DNA binding proteins from
lysates, Microplate capture and detection assay can detect even lesser abundance of protein in the
well by using with enzyme-labeled antibodies and a chemiluminescent substrate.
Disadvantages:
1: Both techniques require supershift antibodies i.e. antibodies which can selectively bind to
DNA bound native proteins.
2: The technique gives little information about corresponding changes in transcription factor-
DNA.
3: The techniques have been used for a few target proteins so far while EMSA is applicable
under in vitro conditions only.
Solution
Microplate capture and detection assay is used to detect the specific protein-DNA interactions
and identifies the proteins. The method includes immobilization of DNA probes on the surface of
microplates (having wells) coated with streptavidin. The cellular extract, prepared in binding
buffer, is added for enough time to facilitate binding protein to bind to the oligonucleotide
followed by washing of wells several times to remove nonspecifically bound proteins. A protein
specific antibody is used for detection of protein.
Comparison of advantages and disadvantages of microplate capture and detection assay and
EMSA:
A: Advantages
1: Microplate capture and detection assay is highly sensitive, fast and throughput technique that
is based on ELISA while EMSA is based on comparative velocity of migration of DNA-protein
complexes through non denaturing gel electrophoresis.
2: Both techniques can skip the use of radioacitve probes by using biotinylated or fluorescently
labeled DNA probes.
3: EMSA uses DNA probe mutational analysis for testing the binding affinity of proteins.
4: Though EMSA can detect the presence of low quantities of DNA binding proteins from
lysates, Microplate capture and detection assay can detect even lesser abundance of protein in the
well by us.
C. Amide - Organic functional group found in Nylo.pdfanonaeon
C. Amide - Organic functional group found in Nylon D. Amine - Organic functional
group found in peptide bonds A. Polyethylene terephtalate - Used to make soda bottles B. High
density polyethylene - Used to make milk jugs
Solution
C. Amide - Organic functional group found in Nylon D. Amine - Organic functional
group found in peptide bonds A. Polyethylene terephtalate - Used to make soda bottles B. High
density polyethylene - Used to make milk jugs.
H2PO4^-ThereforeP has Oxidation state = +7H has oxidation stat.pdfanonaeon
H2PO4^-
Therefore
P has Oxidation state = +7
H has oxidation state = + 1
O has Oxidation state = - 2
Solution
H2PO4^-
Therefore
P has Oxidation state = +7
H has oxidation state = + 1
O has Oxidation state = - 2.
First of all this program has compilation errors.Line 14 WilmaL.pdfanonaeon
First of all this program has compilation errors.
Line 14: Wilma
Line 15:Fred, Pebbles, Barney, Bam-Bam, Betty
____________________________________
The Index of the ArrayList Starts from 0
Till line 11, the elements in the Array list are:
[Fred, Barney, Wilma, Betty]
In line 12,we are adding \"BaM-Bam\" at index 3
[Fred, Barney, Wilma, Bam-Bam, Betty]
In Line 13, we are adding \"Pebbles\" at index 1
[Fred, Pebbles, Barney, Wilma, Bam-Bam, Betty]
When this statement in the line 14 is executed
System.out.println(peopleList.remove(3));
This line will removes And displays Wilma
The line 15 will displays the elements in the array list:
Fred, Pebbles, Barney, Bam-Bam, Betty
In line 17 it is wrong atement.The program wont compile.
suppose if it is collections.sort(null) we will get null pointer exception.
Then line 19 will not get executed.
Solution
First of all this program has compilation errors.
Line 14: Wilma
Line 15:Fred, Pebbles, Barney, Bam-Bam, Betty
____________________________________
The Index of the ArrayList Starts from 0
Till line 11, the elements in the Array list are:
[Fred, Barney, Wilma, Betty]
In line 12,we are adding \"BaM-Bam\" at index 3
[Fred, Barney, Wilma, Bam-Bam, Betty]
In Line 13, we are adding \"Pebbles\" at index 1
[Fred, Pebbles, Barney, Wilma, Bam-Bam, Betty]
When this statement in the line 14 is executed
System.out.println(peopleList.remove(3));
This line will removes And displays Wilma
The line 15 will displays the elements in the array list:
Fred, Pebbles, Barney, Bam-Bam, Betty
In line 17 it is wrong atement.The program wont compile.
suppose if it is collections.sort(null) we will get null pointer exception.
Then line 19 will not get executed..
C and D. TCPIP, or Transmission Control ProtocolInternet Protocol,.pdfanonaeon
C and D. TCP/IP, or Transmission Control Protocol/Internet Protocol, is the basic
communication protocol for the Internet and is also used on private networks such as intranets
(web environments within a company) or extranets (web environments between two or more
business partners or between a vendor and customers). NetBIOS, or Network Basic Input/Output
System, is a program that allows computers to communicate on local area networks, or LANs,
but does not support transport into wide area networks, or WANs, such as the Internet.
Solution
C and D. TCP/IP, or Transmission Control Protocol/Internet Protocol, is the basic
communication protocol for the Internet and is also used on private networks such as intranets
(web environments within a company) or extranets (web environments between two or more
business partners or between a vendor and customers). NetBIOS, or Network Basic Input/Output
System, is a program that allows computers to communicate on local area networks, or LANs,
but does not support transport into wide area networks, or WANs, such as the Internet..
Answer is,D. Greater activation of left hemisphere blood flow.Op.pdfanonaeon
Answer is,
D. Greater activation of left hemisphere blood flow.
Options A , B and C are related to the evolutionary origin of the hemisphere asymmetry.
Solution
Answer is,
D. Greater activation of left hemisphere blood flow.
Options A , B and C are related to the evolutionary origin of the hemisphere asymmetry..
Answer Real Estate Life Cycle is having 3 stages at Universal leve.pdfanonaeon
Answer : Real Estate Life Cycle is having 3 stages at Universal level i.e. Pre- development stage,
Development Stage and Developed stage.
1) Pre development stage - It is stage when the land is raw and there is no structure on it, betting
at such location would be risky as defining the value in future is difficult. since the area may be
devoid of people but would come in future based on government policies or any other attraction
for people to come and live.
2) Development stage - It will come after the area or land has been defined for development, that
will attract developers to create buildings or structure and suddenly will shoot the prices of land.
3) Developed stage - When the inventory is ready and it needs to be sold to the buyers, it is the
end satge of cycle and everyone can come to get benefit from it. But the pricing are speculative
at this stage and based on the supply and demand of the property.
Why these are important for investment and finance purpose
Pre development stage, it is difficult to predict time taken for its conversion and hence most
illiquid stage, buying at this point would be a risk.
Development Stage, prediction is easy and developer would gonna sell it may be after
development or during construction . So easy to predict but reward may not be that much.
Developed stage can be risky as it is based more n speculation, if demand is more investment can
be beneficial but if not then it will be a risky investment.
Solution
Answer : Real Estate Life Cycle is having 3 stages at Universal level i.e. Pre- development stage,
Development Stage and Developed stage.
1) Pre development stage - It is stage when the land is raw and there is no structure on it, betting
at such location would be risky as defining the value in future is difficult. since the area may be
devoid of people but would come in future based on government policies or any other attraction
for people to come and live.
2) Development stage - It will come after the area or land has been defined for development, that
will attract developers to create buildings or structure and suddenly will shoot the prices of land.
3) Developed stage - When the inventory is ready and it needs to be sold to the buyers, it is the
end satge of cycle and everyone can come to get benefit from it. But the pricing are speculative
at this stage and based on the supply and demand of the property.
Why these are important for investment and finance purpose
Pre development stage, it is difficult to predict time taken for its conversion and hence most
illiquid stage, buying at this point would be a risk.
Development Stage, prediction is easy and developer would gonna sell it may be after
development or during construction . So easy to predict but reward may not be that much.
Developed stage can be risky as it is based more n speculation, if demand is more investment can
be beneficial but if not then it will be a risky investment..
1.static int HORIZONTAL_WRAP2. Container view controllers are most.pdfanonaeon
1.static int HORIZONTAL_WRAP
2. Container view controllers are most often used to facilitate navigation and to create new user
interface types based on existing content. Examples of container view controllers in UIKit
includeUINavigationController, UITabBarController, and UISplitViewController, all of which
facilitate navigation between different parts of your user interface.
3. Andoid supports .png(preferred),.jpg(accepteble),.gif(discouraged)
4.Each element in an array is called an Element
5.For Eclipse, typically all you need to do is set up a folder somewhere within your source code
directory. For instance, if the directory containing your source is /src then you can create a
/src/resources folder to place your images/files in. Then, within your class you do a
getResource(\"/resources/image.png\") to retrieve it.
6.Native apps have the ability to use device-specific hardware and software, meaning that native
apps can take advantage of the latest technology available on mobile devices such as a global
positioning system (GPS) and camera.
The main similarity is that both emulated mobile apps and Web apps run on servers external to
the mobile device and require the use of a browser on the mobile device to display and then use
the app user interface (UI).
7.Marshmallow 6.0-6.0.1
8.This ColorModel class supports two representations of pixel values. A pixel value can be a
single 32-bit int or an array of primitive types.
9.By changing the android:label field in your application node in AndroidManifest.xml.
10.\"android:layout_span” to span view in 2 cells, and “android:layout_column” to display the
view in specified column.
11.1.ListActivity.java
2.activity__twopane.xml
3.activity__list.xml
4.ListFragment.java
5.DetailActivity.java
13.
14.A dp is a density-independent pixel that corresponds to the physical size of a pixel at 160 dpi.
An sp is the same base unit, but is scaled by the user\'s preferred text size (it\'s a scale-
independent pixel), so you should use this measurement unit when defining text size (but never
for layout sizes).
15.Tween animation calculates the animation with information such as the start point, end point,
size, rotation, and other common aspects of an animation.
16.Setting a View\'s size to wrap_content will force it to expand only far enough to contain the
values (or child controls) it contains. For controls -- like text boxes (TextView) or images
(ImageView) -- this will wrap the text or image being shown. For layout elements it will resize
the layout to fit the controls / layouts added as its children.
17.An animation defined in XML that modifies properties of the target object, such as
background color or alpha value, over a set amount of time.
18.9-patch image
Solution
1.static int HORIZONTAL_WRAP
2. Container view controllers are most often used to facilitate navigation and to create new user
interface types based on existing content. Examples of container view controllers in UIKit
includeUINavigationController.
1. Code :
import java.awt.*;
import javax.swing.*;
import java.awt.event.*;
/*Designed and programmed by Srujana */
/*Designed and programmed by Srujana */
public class Calculator implements ActionListener{
char o;
int ctr=0;
String value="", cv="", oBtn;
Double answer, v1, v2;
Double NumberConverted;
Frame f;
Panel p1, p2, p3, p4, p5, p6;
private TextField tField;
private Menu EditMenu;
private MenuBar menuBar;
private MenuItem fmi1, fmi2, fmi3;
private Button num0, num1, num2, num3, num4, num5, num6, num7, num8,
num9;
private Button bAdd, bSub, bMul, bDiv, bPer, bSqrt, bFrac, bInt, bDot,
bCE, equals, backspace, clear;
Calculator(){
f = new Frame("Calculator");
menuBar = new MenuBar();
EditMenu = new Menu ("Edit");
fmi1 = new MenuItem(" Copy ");
fmi2 = new MenuItem(" Paste ");
fmi3 = new MenuItem(" Quit ");
EditMenu.add(fmi1);
EditMenu.add(fmi2);
EditMenu.addSeparator();
EditMenu.add(fmi3);
2. p1 = new Panel();
p2 = new Panel();
p3 = new Panel();
p4 = new Panel();
p5 = new Panel();
p6 = new Panel();
tField = new TextField(35);
num0 = new Button ("0");
num1 = new Button ("1");
num2 = new Button ("2");
num3 = new Button ("3");
num4 = new Button ("4");
num5 = new Button ("5");
num6 = new Button ("6");
num7 = new Button ("7");
num8 = new Button ("8");
num9 = new Button ("9");
bAdd = new Button ("+");
bSub = new Button ("-");
bMul = new Button ("x");
bDiv = new Button ("/");
bPer = new Button ("%");
bSqrt = new Button ("sqrt");
bFrac = new Button ("1/x");
bInt = new Button ("+/-");
bDot = new Button (".");
bCE = new Button ("CE");
equals = new Button ("=");
backspace = new Button ("Backspace");
clear = new Button ("C");
}
public void launchFrame(){
tField.setText("0.");
tField.setEnabled(false);
menuBar.add(EditMenu);
p2.add(backspace);
