Do The statement of stockholders equity explains changes in equity.pdf
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Do The statement of stockholder\'s equity explains changes in equity from net income? Solution Statement of equity is a summary of changes in share holders equity during the reporting period. The statemnt includes all changes including issue of new stocks, payment of dividend, issue of bonus shares, buy back of shares. It also includes changes to retained earnings, reserves and stock reserve. Distribution of dividend from net income and transfer of remaining balance to reserves are shown at the bottom of P&L account. The remaining balance of profits transferred to reserves are reflected in Statement of changes in equity as addition to retained earnings. Hope I have answered the question, for any further clarification please comment..
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JAVALAB #8 - ARRAY BASED LISTSThe next exercise is based on this.pdf
JAVA
LAB #8 - ARRAY BASED LISTS
The next exercise is based on this implemetation for an UnorderedArrayList of integers:
//Interface: ArrayListADT
//works for int
public interface ArrayListADT {
public boolean isEmpty(); //Method to determine whether the list is empty.
public boolean isFull(); //Method to determine whether the list is full.
public int listSize(); //Method to return the number of elements in the list.
public int maxListSize(); //Method to return the maximum size of the list.
public void print(); //Method to output the elements of the list.
public boolean isItemAtEqual(int location, int item); //Method to determine whether item is
the same as the item in the list at location.
public void insertAt(int location, int insertItem); //Method to insert insertItem in the list at
the position
public void insertEnd(int insertItem); //Method to insert insertItem at the end of the list.
public void removeAt(int location); //Method to remove the item from the list at location.
public int retrieveAt(int location); //Method to retrieve the element from the list at location.
public void replaceAt(int location, int repItem); //Method to replace the element in the list at
location with repItem.
public void clearList(); //Method to remove all the elements from the list.
public int search(int searchItem); //Method to determine whether searchItem is in the list.
public void remove(int removeItem); //Method to remove an item from the list.
}
//Class: ArrayListClass implements
//Interface: ArrayListADT
public abstract class ArrayListClass implements ArrayListADT {
protected int length; //to store the length of the list
protected int maxSize; //to store the maximum size of the list
protected int[] list; //array to hold the list elements
//Default constructor
public ArrayListClass() {
maxSize = 100;
length = 0;
list = new int[maxSize];
}
//Alternate Constructor
public ArrayListClass(int size) {
if(size <= 0) {
System.err.println(\"The array size must be positive. Creating an array of size 100.\");
maxSize = 100;
}
else
maxSize = size;
length = 0;
list = new int[maxSize];
}
public boolean isEmpty() {
return (length == 0);
}
public boolean isFull() {
return (length == maxSize);
}
public int listSize() {
return length;
}
public int maxListSize() {
return maxSize;
}
public void print() {
for (int i = 0; i < length; i++)
System.out.print(list[i] + \" \");
System.out.println();
}
public boolean isItemAtEqual(int location, int item) {
if (location < 0 || location >= length) {
System.err.println(\"The location of the item to be compared is out of range.\");
return false;
}
return list[location]== item;
}
public void clearList() {
for (int i = 0; i < length; i++)
list[i] = 0;
length = 0;
System.gc(); //invoke the Java garbage collector
}
public void removeAt(int location) {
if (location < 0 || location >= length)
System.err.println(\"The location of the item to be removed is out of range.\");
else {
for(int i = location; i < length - 1; i++)
list[i] = list[i + 1.
In the year 2000, a halophile was isolated from a 250 million year o.pdf
In the year 2000, a halophile was isolated from a 250 million year old salt crystal and grown up
in a university laboratory in Westchester PA. What did the scientists isolate and why do you
think it survived 250 million years?
Solution
Halophile is salt loving, They are extremophiles which live in high saline environment. They
adapt different strategies to survive in the extreme salt conditions. These adaptations are-
Hence the halophile survived 250 million years.
In in the image seen here, in which direction would you expect water .pdf
In in the image seen here, in which direction would you expect water to move if the membrane
is not permeable to the solute? It would move equally between both sides. Right to left Left to
right
Solution
The process shown here is called as Osmosis.. According to which the movement of water
molecules take place from the region higher concentration to a region of lower water
concentration when separated by the semi - permeable Membrane..
In this the left side contains more water molecules concentration then right side as the right side
contains more solute molecules...
Soo, the movement will be from Left Side to Right side...!!!!.
How does the movement of DNA fragments on the gel and the movemen.pdf
How does the movement of DNA fragments on the gel and the movement of the ethilisim
troemide in the eel affect the appearance of the DNA fragments?
Solution
Answer:
the movement of DNA happens from a negative charged electrode to positively charged
electrode as the DNA is negatively charged.
By applying the electric field through the agarose gel matrix DNA is seperated on the basis of
size.
Ethidium bromide is a flourescent dye which intercalates into circular DNA and can
sunsequently change the charge, length, as well as the superhelicity of the DNA molecule,
EtBr presence in gel during electrophoresis can affect movement of DNA. Agarose gel
electrophoresis can be used to resolve circular DNA with different supercoiling topology.
as more EtBr binds between the DNA bases, the molecule stiffens and hence mobility is
decreased as compared to same frangment being run without EtBr.
Ethedium bromide is positively charged and runs from the positive electrode towards the
negative electrode. i.e opposite direction of movement of DNA.
hence when the two encounter, the intercalaton happens, leading to flourescence of orange red
colour under UV light..
Genetics Homework Problem 1 You are mapping genes in a species of f.pdf
Genetics Homework Problem 1: You are mapping genes in a species of fruit fly. You cross a
wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w),
and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs).
You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a.
What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many
progeny of each phenotype would be expected if the genes were assorting independently? c. Can
you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d.
Assuming these genes are all linked on the same chromosome, draw a genetic map showing the
order and map distances between the three genes.
Solution
Answer:
a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X gg swsw hlhl
b). If the genes were assorting independently, 8 types of phenotypes are expected in equl
proportions
1:1:1:1:1:1:1:1
Green, short, hair: 1
Red, long, smooth: 1
Green, long, smooth: 1
Green, short, smooth: 1
Red, short, hairy: 1
Red, long, hairy: 1
Green, long, hairy: 1
Red, short, hairy: 1
c).
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
19
361.00
14.44
Red, long, smooth
42
25
17
289.00
11.56
Green, long, smooth
20
25
-5
25.00
1.00
Green, short, smooth
5
25
-20
400.00
16.00
Red, short, hairy
20
25
-5
25.00
1.00
Red, long, hairy
7
25
-18
324.00
12.96
Green, long, hairy
30
25
5
25.00
1.00
Red, short, hairy
32
25
7
49.00
1.96
200
200
59.92
Chi-square value = 59.92
The independent assortment at P=0.05 is rejected.
d).
Explanation
Shorthand
Number of progeny
g sw hl
44
G Sw Hl
42
g Sw Hl
20
g sw Hl
5
G sw hl
20
G Sw hl
7
g Sw hl
30
G sw Hl
32
Imagine the order is
G SW Hl / g sw hl
1. If single cross over (SCO) occurs between G & Sw
Normal order= G---------Sw & g-----sw
After cross over= G-----sw & g------Sw
G-----sw recombinants are 20+ 32 = 52
g----- Sw recombinants are 20+ 30 = 50
Total recombinants = 102
RF = (102/200.)*100 =51%
2. If single cross over (SCO) occurs between Sw & Hl
Normal order= Sw---------Hl & sw----hl
After cross over= Sw-----hl & sw------Hl
Sw-----hl recombinants are 30 + 7 = 37
sw------Hl recombinants are 32+5 = 37
Total recombinants = 74
RF = (74/200)*100 = 37%
3. If single cross over (SCO) occurs between G & Hl
Normal order= G---------Hl & g------hl
After cross over= G-----hl & g------Hl
G-----hl recombinants are 20 + 7=27
g------Hl recombinants are 20+5=25
Total recombinants = 52
RF = (52/200)*100 = 26%
% RF = Map unit distance
The order of genes is -----
G------26 m.u.--------Hl------37 m.u.---------Sw
Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene
for wings & gene for eye.
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
19
361.00
14.44
Red, long, smooth
42
25
17
289.00
11.56
Green, long, smooth
20
25
-5
25.00
1.00
Green, sho.
For f(x) =2x, find a formula for the Riemann sum obtained by dividin.pdf
For f(x) =2x, find a formula for the Riemann sum obtained by dividing the interval [2.5]
subintervals and using the right hand endpoint for each ck. Simplify the sum and take the limit as
n--> infinity to calculate the area under the curve over [2,5]
please show all of your work as be as descriptive as you can I appreciate your help thank you!
Solution
a=2, b =5, x =(b-a)/n =(5-2)/n =3/n
f(x)=2x
xi=a+ ix =2+i(3/n)
f(xi)=2(2+3(i/n))
the area under the curve over [2,5] is =limn->[i =1 to n]f(xi)x
=limn->[i =1 to n]2(2+3(i/n))(3/n)
=limn->[i =1 to n]6((2/n)+3(i/n2))
=limn-> 6((2n/n)+3(n(n+1)/2n2))
=limn-> 6(2 +3((1+(1/n))/2))
= 6(2 +3((1+0)/2))
=6(2 +3(1/2))
=6(2 +(3/2))
=6(7/2)
=3*7
=21
the area under the curve over [2,5] is 21.
Fibroblasts are the typical cells of simple squamous epithelium. Dens.pdf
Fibroblasts are the typical cells of simple squamous epithelium. Dense connective tissue, cardiac
muscle tissue. skeletal muscle tissue. nervous tissue.
Solution
Answer :- option B
Explanation:- Fibroblast are the type of cells that synthesise extracellular matrix and collagen
and are part of connective tissue..
Explain the difference between an observational study and a designed.pdf
Explain the difference between an observational study and a designed experiment?
Solution
An observational study simply take down the properties of its subjects, without manipulating
them or interfering with their activities.
In a designed experiment, the researcher may change the environment or circumstances of some
of the sample in order to compare, say, a control and a treatment group..
Recommended
JAVALAB #8 - ARRAY BASED LISTSThe next exercise is based on this.pdf
JAVA
LAB #8 - ARRAY BASED LISTS
The next exercise is based on this implemetation for an UnorderedArrayList of integers:
//Interface: ArrayListADT
//works for int
public interface ArrayListADT {
public boolean isEmpty(); //Method to determine whether the list is empty.
public boolean isFull(); //Method to determine whether the list is full.
public int listSize(); //Method to return the number of elements in the list.
public int maxListSize(); //Method to return the maximum size of the list.
public void print(); //Method to output the elements of the list.
public boolean isItemAtEqual(int location, int item); //Method to determine whether item is
the same as the item in the list at location.
public void insertAt(int location, int insertItem); //Method to insert insertItem in the list at
the position
public void insertEnd(int insertItem); //Method to insert insertItem at the end of the list.
public void removeAt(int location); //Method to remove the item from the list at location.
public int retrieveAt(int location); //Method to retrieve the element from the list at location.
public void replaceAt(int location, int repItem); //Method to replace the element in the list at
location with repItem.
public void clearList(); //Method to remove all the elements from the list.
public int search(int searchItem); //Method to determine whether searchItem is in the list.
public void remove(int removeItem); //Method to remove an item from the list.
}
//Class: ArrayListClass implements
//Interface: ArrayListADT
public abstract class ArrayListClass implements ArrayListADT {
protected int length; //to store the length of the list
protected int maxSize; //to store the maximum size of the list
protected int[] list; //array to hold the list elements
//Default constructor
public ArrayListClass() {
maxSize = 100;
length = 0;
list = new int[maxSize];
}
//Alternate Constructor
public ArrayListClass(int size) {
if(size <= 0) {
System.err.println(\"The array size must be positive. Creating an array of size 100.\");
maxSize = 100;
}
else
maxSize = size;
length = 0;
list = new int[maxSize];
}
public boolean isEmpty() {
return (length == 0);
}
public boolean isFull() {
return (length == maxSize);
}
public int listSize() {
return length;
}
public int maxListSize() {
return maxSize;
}
public void print() {
for (int i = 0; i < length; i++)
System.out.print(list[i] + \" \");
System.out.println();
}
public boolean isItemAtEqual(int location, int item) {
if (location < 0 || location >= length) {
System.err.println(\"The location of the item to be compared is out of range.\");
return false;
}
return list[location]== item;
}
public void clearList() {
for (int i = 0; i < length; i++)
list[i] = 0;
length = 0;
System.gc(); //invoke the Java garbage collector
}
public void removeAt(int location) {
if (location < 0 || location >= length)
System.err.println(\"The location of the item to be removed is out of range.\");
else {
for(int i = location; i < length - 1; i++)
list[i] = list[i + 1.
In the year 2000, a halophile was isolated from a 250 million year o.pdf
In the year 2000, a halophile was isolated from a 250 million year old salt crystal and grown up
in a university laboratory in Westchester PA. What did the scientists isolate and why do you
think it survived 250 million years?
Solution
Halophile is salt loving, They are extremophiles which live in high saline environment. They
adapt different strategies to survive in the extreme salt conditions. These adaptations are-
Hence the halophile survived 250 million years.
In in the image seen here, in which direction would you expect water .pdf
In in the image seen here, in which direction would you expect water to move if the membrane
is not permeable to the solute? It would move equally between both sides. Right to left Left to
right
Solution
The process shown here is called as Osmosis.. According to which the movement of water
molecules take place from the region higher concentration to a region of lower water
concentration when separated by the semi - permeable Membrane..
In this the left side contains more water molecules concentration then right side as the right side
contains more solute molecules...
Soo, the movement will be from Left Side to Right side...!!!!.
How does the movement of DNA fragments on the gel and the movemen.pdf
How does the movement of DNA fragments on the gel and the movement of the ethilisim
troemide in the eel affect the appearance of the DNA fragments?
Solution
Answer:
the movement of DNA happens from a negative charged electrode to positively charged
electrode as the DNA is negatively charged.
By applying the electric field through the agarose gel matrix DNA is seperated on the basis of
size.
Ethidium bromide is a flourescent dye which intercalates into circular DNA and can
sunsequently change the charge, length, as well as the superhelicity of the DNA molecule,
EtBr presence in gel during electrophoresis can affect movement of DNA. Agarose gel
electrophoresis can be used to resolve circular DNA with different supercoiling topology.
as more EtBr binds between the DNA bases, the molecule stiffens and hence mobility is
decreased as compared to same frangment being run without EtBr.
Ethedium bromide is positively charged and runs from the positive electrode towards the
negative electrode. i.e opposite direction of movement of DNA.
hence when the two encounter, the intercalaton happens, leading to flourescence of orange red
colour under UV light..
Genetics Homework Problem 1 You are mapping genes in a species of f.pdf
Genetics Homework Problem 1: You are mapping genes in a species of fruit fly. You cross a
wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w),
and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs).
You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a.
What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many
progeny of each phenotype would be expected if the genes were assorting independently? c. Can
you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d.
Assuming these genes are all linked on the same chromosome, draw a genetic map showing the
order and map distances between the three genes.
Solution
Answer:
a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X gg swsw hlhl
b). If the genes were assorting independently, 8 types of phenotypes are expected in equl
proportions
1:1:1:1:1:1:1:1
Green, short, hair: 1
Red, long, smooth: 1
Green, long, smooth: 1
Green, short, smooth: 1
Red, short, hairy: 1
Red, long, hairy: 1
Green, long, hairy: 1
Red, short, hairy: 1
c).
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
19
361.00
14.44
Red, long, smooth
42
25
17
289.00
11.56
Green, long, smooth
20
25
-5
25.00
1.00
Green, short, smooth
5
25
-20
400.00
16.00
Red, short, hairy
20
25
-5
25.00
1.00
Red, long, hairy
7
25
-18
324.00
12.96
Green, long, hairy
30
25
5
25.00
1.00
Red, short, hairy
32
25
7
49.00
1.96
200
200
59.92
Chi-square value = 59.92
The independent assortment at P=0.05 is rejected.
d).
Explanation
Shorthand
Number of progeny
g sw hl
44
G Sw Hl
42
g Sw Hl
20
g sw Hl
5
G sw hl
20
G Sw hl
7
g Sw hl
30
G sw Hl
32
Imagine the order is
G SW Hl / g sw hl
1. If single cross over (SCO) occurs between G & Sw
Normal order= G---------Sw & g-----sw
After cross over= G-----sw & g------Sw
G-----sw recombinants are 20+ 32 = 52
g----- Sw recombinants are 20+ 30 = 50
Total recombinants = 102
RF = (102/200.)*100 =51%
2. If single cross over (SCO) occurs between Sw & Hl
Normal order= Sw---------Hl & sw----hl
After cross over= Sw-----hl & sw------Hl
Sw-----hl recombinants are 30 + 7 = 37
sw------Hl recombinants are 32+5 = 37
Total recombinants = 74
RF = (74/200)*100 = 37%
3. If single cross over (SCO) occurs between G & Hl
Normal order= G---------Hl & g------hl
After cross over= G-----hl & g------Hl
G-----hl recombinants are 20 + 7=27
g------Hl recombinants are 20+5=25
Total recombinants = 52
RF = (52/200)*100 = 26%
% RF = Map unit distance
The order of genes is -----
G------26 m.u.--------Hl------37 m.u.---------Sw
Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene
for wings & gene for eye.
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Green, short, hair:
44
25
19
361.00
14.44
Red, long, smooth
42
25
17
289.00
11.56
Green, long, smooth
20
25
-5
25.00
1.00
Green, sho.
For f(x) =2x, find a formula for the Riemann sum obtained by dividin.pdf
For f(x) =2x, find a formula for the Riemann sum obtained by dividing the interval [2.5]
subintervals and using the right hand endpoint for each ck. Simplify the sum and take the limit as
n--> infinity to calculate the area under the curve over [2,5]
please show all of your work as be as descriptive as you can I appreciate your help thank you!
Solution
a=2, b =5, x =(b-a)/n =(5-2)/n =3/n
f(x)=2x
xi=a+ ix =2+i(3/n)
f(xi)=2(2+3(i/n))
the area under the curve over [2,5] is =limn->[i =1 to n]f(xi)x
=limn->[i =1 to n]2(2+3(i/n))(3/n)
=limn->[i =1 to n]6((2/n)+3(i/n2))
=limn-> 6((2n/n)+3(n(n+1)/2n2))
=limn-> 6(2 +3((1+(1/n))/2))
= 6(2 +3((1+0)/2))
=6(2 +3(1/2))
=6(2 +(3/2))
=6(7/2)
=3*7
=21
the area under the curve over [2,5] is 21.
Fibroblasts are the typical cells of simple squamous epithelium. Dens.pdf
Fibroblasts are the typical cells of simple squamous epithelium. Dense connective tissue, cardiac
muscle tissue. skeletal muscle tissue. nervous tissue.
Solution
Answer :- option B
Explanation:- Fibroblast are the type of cells that synthesise extracellular matrix and collagen
and are part of connective tissue..
Explain the difference between an observational study and a designed.pdf
Explain the difference between an observational study and a designed experiment?
Solution
An observational study simply take down the properties of its subjects, without manipulating
them or interfering with their activities.
In a designed experiment, the researcher may change the environment or circumstances of some
of the sample in order to compare, say, a control and a treatment group..
Discuss the recruitment and selection procedure done by human resour.pdf
Discuss the recruitment and selection procedure done by human resource management?
Solution
The procedure of recruitment and selection done by the human resource management involves
five distinct and basic aspects which are as follows:
1. CRITERIA DEVELOPMENT: Those individuals which are involved in the process of hiring
should be trained enough to develop criteria , to conduct interview , to review the resumes , to
develop the questions for interview , to weigh the candidates.
In this step the basic thing is to plan the interview and develop the criteria as which of the
sources are to be used during interview and how to use these sources to score better in the
interview.The criteria should be job focussed and also the focus should be on the personality.
In this process the involvement of discussion of the skills, abilities and some personal
characteristics takes place that are required to have to the job.Before reviewing the resumes by
developing this criteria can be sure that he is fair in selecting the applicants to interview.The
application should include the details of the candidate , previous job experience certificates and
educational qualification certificates.
2. APPLICATION AND REVIEWING OF RESUME : After the completion of step 1 the
applications can now be reviewed. There are different ways to go through this process but there
are various computer based programs that can search for the keywords in the resumes and
shortlist the number of resumes that can b looked into.
3. INTERVIEW PROCESS : After shortlisting those applicants who have met the minimum
criteria.The HR manager selects these applicants for the interview.Due to lack of time sometimes
interview is being done on phone.
4. TEST ADMINISTRATION : Before the hiring decision is made any number of test are
administered. The tests may be physical tests ,drug tests, response test, credit report checks,
personality tests, etc.
5. OFFER MAKING PROCESS : This is the final step in which the choosen candidate is offered
the job position. The offer can b developed through email or through the job letter.All the
benefits and compensation are discussed in the offer letter..
Describe what is meant by “radical cure” and identify the drug that .pdf
Describe what is meant by “radical cure” and identify the drug that is produces a radical cure for
malaria.
Solution
Radical cure refers to the elimination of malaria parasites completely from the body.It
specifically refers to the elimination of hypnozoites of P.vivax and P. ovale which are the
parasitic stages in the liver and could result in relapse of the infection. A combination of two
drugs i.e chloroquine and primaquine produces a radical cure for malaria..
Describe the nutritional modes of organisms based upon where they obt.pdf
Describe the nutritional modes of organisms based upon where they obtain carbon energy.
Include an example of each.
Solution
Describe the nutritional modes of organisms based upon where they obtain carbon and energy.
Almost all organisms on the earth need energy to do their work, and carbon to use in building
their chemical structures. And these two (energy and carbon) has to be obtained from external
sources, based on the way they obtain energy and carbon, organisms are broadly classified into
four different groups.
Classification
Obtaining energy and carbon
Examples
Autotrophs
Photoautotrophs
They get energy from sunlight and will not depend on other organisms for carbon. Use CO2 as a
carbon source.
All photosynthetic prokaryotes
Plants (eukaryotes)
Chemoautotrophs
These make use of inorganic chemicals as an energy source. For example, H2S, Fe2+ etc and
CO2 will serve as a carbon source.
Archaea living in hydrothermal vents.
Heterotrophs
Photoheterotrophs
Use light as an energy source and will use organic molecules as a carbon source. These type of
organisms are rare.
Heliobacteria, non-sulfur bacteria, green non-sulfur bacteria etc
Chemoheterotrophs
Use organic molecules like sugars as an energy source.
Use organic molecules like proteins, sugars, fats, nucleic acids, etc. as a carbon source. This type
of nutrition is most common form of nutrition on Earth
Most of the prokaryotes, protists, fungi, and animals
Classification
Obtaining energy and carbon
Examples
Autotrophs
Photoautotrophs
They get energy from sunlight and will not depend on other organisms for carbon. Use CO2 as a
carbon source.
All photosynthetic prokaryotes
Plants (eukaryotes)
Chemoautotrophs
These make use of inorganic chemicals as an energy source. For example, H2S, Fe2+ etc and
CO2 will serve as a carbon source.
Archaea living in hydrothermal vents.
Heterotrophs
Photoheterotrophs
Use light as an energy source and will use organic molecules as a carbon source. These type of
organisms are rare.
Heliobacteria, non-sulfur bacteria, green non-sulfur bacteria etc
Chemoheterotrophs
Use organic molecules like sugars as an energy source.
Use organic molecules like proteins, sugars, fats, nucleic acids, etc. as a carbon source. This
type of nutrition is most common form of nutrition on Earth
Most of the prokaryotes, protists, fungi, and animals.
Decasper and Spence (1986) measured babies sucking rates as they lis.pdf
Decasper and Spence (1986) measured babies sucking rates as they liscenced to a story they had
heard in the womb (the cat in the hat) and to a story they had not heard (the king, the mice, and
the cheese) the study concluded that:
a.) amniotic fluid blocks sound waves from reaching the fetus
b) increased heart rate showed that an infant can learn even before it is born
c) different sucking styles and rates showed that an infant can learn even before it is born
d) the cadence of the reading material not previous experience altered sucking style and rate
Solution
different sucking styles and rates showed that an infant can learn even before it is born.
Corhin Eckles and Connor Hayes have recieved walkie talkies for Chri.pdf
Corhin Eckles and Connor Hayes have recieved walkie talkies for Christmas. If they leave from
the same point at the same time, Corhin walking north at 2.5 mph and Connor walking east at 3
mph, how long will they be able to talk to each other if the range of the walkie-talkies is 4 mi?
Solution
Corhin DATA:
distance= x miles; rate=2.5 mph; time=d/r= x/2.5
Connor DATA:
distance= y miles; rate=3 mph; time=y/4
EQUATIONS:
1) x^2+y^2=4^2
2) x/2.5 = y/4
Using 2) you get y=4x/2.5
Substitute into 1) to get x^2+(4x/2.5)^2=16
x^2+(16x^2/6.25)=16
6.25x^2+16x^2=16(6.25)
22.25x^2=100
x^2=100/22.25=4.49...
x=2.12 hrs.
x/2.5= 0.85 hrs. = 50.87 minutes
This is Corhin\'s time, but it is the same for Connor..
Consider three genes in a hypothetical bird. One controls the color o.pdf
Consider three genes in a hypothetical bird. One controls the color of feathers on the top of the
head (red or brown), one controls the color of feathers on the chest (black or brown) and one
controls the number of syllables in the song (3 or 4). You notice that birds frequently have either
red heads and black chests or brown heads and brown chests, but very few have red heads and
brown chests or brown heads and black chests. However, there doesn\'t seem to be any
association between the syllables in the bird song and color. In other words, red heads are
equally likely to sing songs with 3 or 4 syllables, and the same goes for the brown heads. Based
on what you know about how the genome is organized and passed down to the next generation,
explain this observation.
Solution
The observation can be attributed to the mechanism of genetic linkage.
It is a well known fact that genes get passed from one generation to another. Mendel\'s second
law of inheritance states that during this process, the inheritance of one gene is independent of
the inheritance of another gene. Howeve, this law holds true only when the different genes are
sufficiently spaced apart from each other. Inother words, if the genes are close to each other, it is
difficult for them to be inherited in an independent manner. If one of the genes is inherited, then
the other gene (by virtue of its proximity to the first gene) also tends to get inherited. This
phenomenon is called as genetic linkage, and the genes involved are said to be Linked Genes.
The same mechanism is in action in the above case.
The genes for Red Head and Black Crest are linked, as are the genes for Brown Head and Brown
Crest. Therefore, these two genes are inherited in a linked manner, and not in an independent
manner. The two genes therefore behave as One Single Gene.
The gene for song syllable is at a sufficient distance from the linked genes, and follows
Mendel\'s law of independent assortment and is not dependent on the presence or absence of the
former two traits..
Answer the following questions What is the domain of logarithmic fun.pdf
Answer the following questions: What is the domain of logarithmic function What is the range
of Logarithmic function What is the value of e(approximate to 4 decimal places)
Solution
SOLUTION
===========
the domain of any logarithmic function is positive real numbers
1) Domain =(0,inf)
2) Range =(-inf,+inf)
3) e=2.7183.
are organic compounds that bind very tightly to iron ions for transp.pdf
are organic compounds that bind very tightly to iron ions for transport into cells.
Solution
Organisms like bacteria secrerte substances called siderophores which bind to iron for transport
inside the cell. Iron is one of the important micronutrients for the bacterial metabolism.
Ferrichrome,Desferrioxamine B and Desferrioxamine E are some of the important siderophores
secreted by various bacteria..
Book Business Its Legal, Ethical, Global Environment 10th Edition.pdf
Book: Business It\'s Legal, Ethical, Global Environment 10th Edition
Page 204-205
Question and Problems
# 2
#5
Solution
Ethical issues connect intimately with economic issues. Take the economic practice of doing a
cost-benefit analysis. You could spend one hundred dollars for a night on the town, or you could
donate that one hundred dollars to the reelection campaign of your favorite politician. Which
option is better? The night on the town increases pleasure. A politician’s successful campaign
may lead to more liberty in the long term. We regularly make decisions like this, weighing our
options by measuring their likely costs and likely benefits against each other.
This connects economics directly to a major issue in ethics: By what standard do we determine
what counts as a benefit or a cost? A list of competing candidates for the status of ultimate value
standard includes happiness, satisfying the will of God, long-term survival, liberty, duty, and
equality.
Economists implicitly adopt a value framework when beginning a cost-benefit analysis. Different
value commitments can lead to the same item being considered a cost from one perspective and a
benefit from another. For example, those whose standard of value is increasing human happiness
would count a new road to a scenic mountain vista as a benefit, while those whose standard is
maintaining an unchanged natural environment would count it as a cost.
The results of economic analysis also lead directly to ethical issues. For example, one result of
the nineteenth- and twentieth-century debate over capitalism and socialism is a general
consensus that capitalism is effective at producing wealth and socialism is effective at keeping
people poor. Advocates of capitalism use these results to argue that capitalism is good; others
might respond that “socialism is good in theory, but unfortunately it is not practical.” Implicit in
the capitalist position is the view that practical consequences determine goodness. By contrast,
implicit in the position of those who believe socialism to be an impractical moral ideal is the
view that goodness is distinct from practical consequences.
This connects economics to a second major issue in ethics: Is goodness or badness determined by
real-world practical consequences or by some other means, such as revelations from God, faith in
authorities or authoritative institutions, appeals to rational consistency, felt senses of empathy, or
an innate conscience? The point for economic analysis, most of which is a matter of
understanding and predicting the consequences of various actions, is that the relevance of
economic analysis to policymaking depends, in part, on what one believes is the final source of
value standards..
According the United States (U.S.) Centers for Disease Control and P.pdf
According the United States (U.S.) Centers for Disease Control and Prevention, in 2015
influenza and pneumonia combined were the 8th leading causes of death (55,227 deaths in total)
and the only communicable diseases ranked among the top ten killers in the U.S. What were the
10 causes of mortality in the U.S. in 1900 and 2015? Since communicable disease caused a
relatively few deaths in the U.S. in 2015, why should public and public health professionals be
concerned about them anymore and devote resources in their prevention and control?
Solution
Top 10 leading causes of deaths in US: Hearts diseases Cancer Airway diseases which are non
infectious Cardio vascular disorders Accidents Alzheimer\'s disease Diabetes Pneumonia or
influenza Nephropathys Suicide. The period of communicability is less for the communicable
diseases, hence they are prevention of disease in early stages is possible. Prevention and control
of influenza: Good ventilation of public buildings Encouraging sufferers to cover their faces
with a handkerchief when coughing and sneezing Stay in home at the first sign of influenza
Vaccination against the influenza, may be killed or level attenuated vaccines Prevention of
exposure to influenza strains by hygiene practice Antiviral drugs like amantadine or rimantidine
used Prevention and control of pneumonia: Pneumococcal pneumonia vaccination, PPV 23
AND PCV7. Education of mother to observe the sings of pneumonia. History taking and clinical
examination are very important for management Physical examination for, breath count, chest
indrawing, listen for stridor, wheezing and for severe malnutrition, cyanosis Any signs detected,
they are treated early by antibiotics and other measures Improving the primary medical care
services and development..
9. For some non-Mendelian inheritance patterns, the genotype of an i.pdf
9. For some non-Mendelian inheritance patterns, the genotype of an individual is not sufficient to
predict the phenotype. For the following traits, what else do you need to know to predict the
phenotype?
a) Snail coiling direction
b) Angelman syndrome due to deleted UBE3A gene
Solution
The inheritance mechanisms of both snail coiling and Angelman syndrome involve strong
maternal influences. Snail coiling direction is governed by the genotype of the oocyte, not of the
zygote, so much so that the snail coiling direction can be predicted from the maternal genotype
alone. Similarly, Angelman syndrome (due to deletions in ube3A) is governed by the maternal
copy of the gene alone. If the copy of the gene derived from the mother is functional, the
individuals are healthy. If the maternal copy is defective or deficient, the individual suffers from
the Angelman syndrome..
You have mated a phenotypically wild-type female of the genotype Xy+.pdf
You have mated a phenotypically wild-type female of the genotype Xy+w+Xyw to a male with
the genotype XywY. You counted 2205 total F1 progeny and found 17 flies with gray bodies and
white eyes and 12 flies that had yellow body and red eyes. What is the distance between these
genes controlling body color and eye color on the X chromosome?
rev: 10_27_2012
A. 0.78 map units
B. 1.32 map units
C. 2.63 map units
D. 98.7 map units
Solution
In this problem total progeny is 2205. Among them recombinant individuals are 17 flies with
gray body and white eyes; 12 flies with yellow body and red eyes.
Then % of Recombination = No. Of Recombinant Individuals x 100 / Total progeny
= (17 + 12) x 100 / 2205
= 29 x 100 / 2205
= 1.32 % or 1.32 mU.
You are an oncologist specializing in colon cancer. A patient presen.pdf
You are an oncologist specializing in colon cancer. A patient presents to you with disease that
has spread to his bones and liver and there is no treatment available. What stage cancer is
this?2345
Solution
Answer is Stage IV (4)
Stages of colon cancer are:
Stage 0
It is very early cancer and it will be affecting the innermost layer of the intestine
Stage I
In stage I, cancer is seen in the inner layers of the colon
Stage II
In stage II, cancer is seen spreading through the muscle wall of the colon
Stage III
In stage III, cancer is seen spreading to the lymph nodes
Stage IV
In stage IV, cancer is seen spreading to other organs outside the colon
In case, of colon cancer, at stage IV, it will be spreading to the distant location in the body (liver,
lungs, bones, distant lymph nodes or other sites). At this stage there is no cure for this cancer or
patient will be left with very few options.
Stage 0
It is very early cancer and it will be affecting the innermost layer of the intestine
Stage I
In stage I, cancer is seen in the inner layers of the colon
Stage II
In stage II, cancer is seen spreading through the muscle wall of the colon
Stage III
In stage III, cancer is seen spreading to the lymph nodes
Stage IV
In stage IV, cancer is seen spreading to other organs outside the colon.
You are required to open an input file (shown below) then read one c.pdf
You are required to open an input file (shown below) then read one character at a time until it is
finished. While reading the content of the file, keep a count of the following:
1) Count of each digit read. The counts should be kept in an int array of size 10.
2) Count of each alphabetic character read irrespective if it is an upper case or lower case. The
counts should be kept in an int array of size 26.
3) Count of other characters (Not a digit and not an alphabet). Whitespace character such as
newline \ , blank or tab \\t should NOT be counted here. Hence, if you find ! ; @ the count will
be 3; you will ignore spaces. This will just be an int value.
4) Total count. This will be the count for everything read, including blanks, new line, etc. Again,
this will be an int value.
After reading and processing the file, and get each count correctly, you will need to sort the
count values in numerical order (lowest to highest) before displaying the result.
Your solution must use FUNCTIONAL DECOMPOSITION.
Hence, if the input file was the one given below, your unsorted output is shown below. Please
note that your output must also has the values sorted in numerical order from lowest to highest.
Here is the content of your input file.
Here will be the output generated, minus the gap. That\'s just the end of the page.
Solution
int main(int argc, char *argv[]) {
#include
FILE *ptr_file;
char buf[1000];
int digit[10];
int alpha[26];
int other=0;
char alp=65;
int i;
for(i=0;i<=9;i++)
digit[i]=0;
for(i=0;i<=25;i++)
alpha[i]=0;
ptr_file =fopen(\"D:/input.txt\",\"r\");
if (!ptr_file)
return 1;
while (fgets(buf,1000, ptr_file)!=NULL)
printf(\"\ Original File:=>\ \");
printf(\"%s\ \",buf);
fclose(ptr_file);
for(i=0;buf[i]!=NULL;i++)
{
//Digits Count
if(buf[i]==\'0\')
digit[0]=digit[0]+1;
if(buf[i]==\'1\')
digit[1]=digit[1]+1;
if(buf[i]==\'2\')
digit[2]=digit[2]+1;
if(buf[i]==\'3\')
digit[3]=digit[3]+1;
if(buf[i]==\'4\')
digit[4]=digit[4]+1;
if(buf[i]==\'5\')
digit[5]=digit[0]+1;
if(buf[i]==\'6\')
digit[6]=digit[6]+1;
if(buf[i]==\'7\')
digit[7]=digit[7]+1;
if(buf[i]==\'8\')
digit[8]=digit[8]+1;
if(buf[i]==\'9\')
digit[9]=digit[9]+1;
//Alphabatic count
if(buf[i]==\'a\'||buf[i]==\'A\')
alpha[0]=alpha[0]+1;
if(buf[i]==\'b\'||buf[i]==\'B\')
alpha[1]=alpha[1]+1;
if(buf[i]==\'c\'||buf[i]==\'C\')
alpha[2]=alpha[2]+1;
if(buf[i]==\'d\'||buf[i]==\'D\')
alpha[3]=alpha[3]+1;
if(buf[i]==\'e\'||buf[i]==\'E\')
alpha[4]=alpha[4]+1;
if(buf[i]==\'f\'||buf[i]==\'F\')
alpha[5]=alpha[5]+1;
if(buf[i]==\'g\'||buf[i]==\'G\')
alpha[6]=alpha[6]+1;
if(buf[i]==\'h\'||buf[i]==\'H\')
alpha[7]=alpha[7]+1;
if(buf[i]==\'i\'||buf[i]==\'I\')
alpha[8]=alpha[8]+1;
if(buf[i]==\'j\'||buf[i]==\'J\')
alpha[9]=alpha[9]+1;
if(buf[i]==\'k\'||buf[i]==\'K\')
alpha[10]=alpha[10]+1;
if(buf[i]==\'l\'||buf[i]==\'L\')
alpha[11]=alpha[11]+1;
if(buf[i]==\'m\'||buf[i]==\'M\')
alpha[12]=alpha[12]+1;
if(buf[i]==\'n\'||buf[i]==\'N\')
alpha[13]=alpha[13]+1;
if(buf[i]==\'o\'||buf[i]==\'O\')
alpha[14]=alpha[14]+1;
if(bu.
1.Which of Marys cranial nerves is affected in this episode( Mary.pdf
1.Which of Mary\'s cranial nerves is affected in this episode?( Mary felt a stabbing pain in her
upper jaw and teeth that radiated out to the side of her nose. Over the next several days, she
experienced several more episodes of this intense pain. A visit to the dentist revealed no
abnormalities and she was referred to her physician for an evaluation. Prior to her appointment,
she noticed that the symptoms were subsiding as they had previously. Her physician scheduled
an appointment for a complete neurological exam the following week.)
2. In the MRI images shown in the case, you can see the lesions as bright \"white spots\" on the
brain. Using what you know about the structure of a neuron, explain what is causing this spot to
appear in the MRI?( The results of the MRI, shown below, were consistent with a diagnosis of
relapsing-remitting multiple sclerosis (RRMS). Relapsing-remitting multiple sclerosis is a form
of MS in which symptoms randomly flare up (Mary\'s episodes) and then resolve on their own.
The lesions seen on the MRI on the left were associated with another episode in which Mary
experienced sensory and motor disjunction in her left lower extremity. A subsequent MRI (image
on the right) appeared to show improvement after three months)
3. Three months later, you can see that the spots in the MRI appear to be smaller. Using what
you know about the structure of a neuron, explain what is happening to the neurons in the area
where the lesions are disappearing? Signa 1.0T srsanIRMtoce Go 3/3 28/40 A S35.8 TR 4000
TE:BSVEf 5.0 hk/2.0 04:16 A112 P108 Mary Rier Shutterstock gna 1.0T TR 4000 TEE85VEf
EC 2/2 15.6kHz 5.0thi/2. A112 P108 Mary Rice
Solution
1. The cranial nerves that are affected in this case are the olfactory nerves and the maxillary
nerves , which are a part of the trigeminal nerves.
2. The spot in the MRI is occurring as a result of demyelination of the neurons in that particular
area, which is resulting in the appearance of the white lesion. The condition of multiple sclerosis
is an auto immune condition of the CNS, and the lesions seen are called as MS lesions or
plaques.
3. After a period of three months, when the MRI was conducted again, it was observed that the
plaques found in the areas earlier had now disappeared. This implies that the myelin sheath that
were demyelinated have now recovered , as shown with a complete recovery of the affected
nerves, in this case. MS is also referred to be a relapsing remitting disease..
Which phyla have stomata and guard cells Bryophyta Monilophyta Co.pdf
Which phyla have stomata and guard cells? Bryophyta Monilophyta Coniferophyta
Anthophyta Which phyla have xylem cells? Bryophyta Monilophyta Coniferophyta
Anthophyta Which phyla have xylem vessels? Bryophyta Monilophyta Coniferophyta
Anthophyta
Solution
Answer:
34). Anthophyta
These are flowering plants and the possess stomata that are surrounded by guard cells which
helps in gaseous exchange.
35). Coniferophyta
36). Anthophyta
Vessel elements are typically found in flowering plants (anthophyta) but absent from most
gymnosperms such as conifers..
Which of the follwing traits would you expect to find in the transit.pdf
Which of the follwing traits would you expect to find in the transitional organisms linking flying
birds and their dinosaur ancestors?
A. Feathers
B. Claws on its fingers
C. Amniotic egg
D. Only A and B
E. A, B, and C
Solution
Answer: D)
The transitional organisms, during evolution shows some of the traits of both, but not linked by
direct descent. As birds evolved from dinosaurs, the follwing traits would be expected to find in
the transitional organisms linking flying birds and their dinosaur ancestors:
A. Feathers
B. Claws on its fingers.
What is DNA fingerprintingPlease, no wikipedia or google definiti.pdf
What is DNA fingerprinting?
Please, no wikipedia or google definitions.
Solution
DNA fingerprinting is applied to the scientific process whereby samples of DNA are
collected,collated and used to match other samples of DNA.DNA fingerprinting works on the
principle that each individuals DNA structure or genetic makeup is unique and therefore cannot
be forged,faked or altered in any way.No two individuals share the same genetic code except for
identical twins..
What are parametric and nonparametric inferential testsSolution.pdf
What are parametric and nonparametric inferential tests?
Solution
Inferential statistics suggest statements about a population based on a sample from that
population. Parametric inferential tests are carried out on data that follow certain parameters: the
data will be normal (i.e. the distribution parallels the bell curve); numbers can be added,
subtracted, multiplied and divided; variances are equal when comparing two or more groups; and
the sample should be large and randomly selected.
There are generally more statistical technique options for the analysis of parametric than non-
parametric data, and parametric statistics are considered to be the more powerful. Common
examples of parametric tests are: correlated t-tests and the Pearson r correlation coefficient.
Inferential statistics suggest statements or make predictions about a population based on a
sample from that population. Non-parametric tests relate to data that are flexible and do not
follow a normal distribution.
They are also known as.
1. How does body complexity differ across the 3 phyla (Platyhelmin.pdf
1. How does body complexity differ across the 3 phyla (Platyhelminthes, Nematoda, and
Annelida)? What contributes to this increased complexity? Give 2 examples.
1. How does body complexity differ across the 3 phyla (Platyhelminthes, Nematoda, and
Annelida)? What contributes to this increased complexity? Give 2 examples.
Solution
Answer:
The variable body complexity is differ across the 3 phyla such as platyhelminthes, nematoda, and
annelida because these are multicellular eukaryotes evolved with morphological innovations
including development of locomotors organs and development of various adaptations to survive
in the nature for predation. These are phyla such as platyhelimthes and nematyhelmithes. The
helminthes parasites and species have developed various morphological adaptations for their
survival in the host along with strong scolex to attach to the host internal surfaces. Nematodes
locomotion (round worms, hook worms) and body complexity is variable because the
locomotion of nematodes involves the somatic muscles mainly located under the hypodermis and
these muscle contractions promoted to apply sufficient high hydrostatic pressure laterally to the
cuticle with dorso-ventral bending to move in a sinusoidal manner (C. elegans) as an
evolutionary adaptations.
Annelids have circulatory system, primitive neural systems, and primitive evolutionary features
of invertebrates. Annelid\'s body complexity has increased from lower phyla, platyhelminthes
and nematoda. The body complexity has symmetrical and coelomate & they possess specific
“parapodia” for locomotion. In some species such as earthworms, they have specialized a
hydrostatic skeleton surrounded by several layers of longitudinal and circular muscles to promote
locomotion into the deeper layers of soil.
High metaolic activity, reproductive activity and increased body size have led to this increased
complexity, as part of evolutionary aspects
Platyhelmethis (tape worms): These are mostly parasitic in internal cavities of vertebrate
digestive system with complex asymmetrical body- complexity. The most prominent
segmentation phenomenon has promoted them with sexual reproduction and they use circular
and longitudinal muscle fibers of somatic region located within proglattids; used in locomotion.
For example, Hymenolepis diminuta is also called as rat tapeworm, belongs to cestoda and
possess scolex and proglattids as structural adaptations..
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Discuss the recruitment and selection procedure done by human resour.pdf
Discuss the recruitment and selection procedure done by human resource management?
Solution
The procedure of recruitment and selection done by the human resource management involves
five distinct and basic aspects which are as follows:
1. CRITERIA DEVELOPMENT: Those individuals which are involved in the process of hiring
should be trained enough to develop criteria , to conduct interview , to review the resumes , to
develop the questions for interview , to weigh the candidates.
In this step the basic thing is to plan the interview and develop the criteria as which of the
sources are to be used during interview and how to use these sources to score better in the
interview.The criteria should be job focussed and also the focus should be on the personality.
In this process the involvement of discussion of the skills, abilities and some personal
characteristics takes place that are required to have to the job.Before reviewing the resumes by
developing this criteria can be sure that he is fair in selecting the applicants to interview.The
application should include the details of the candidate , previous job experience certificates and
educational qualification certificates.
2. APPLICATION AND REVIEWING OF RESUME : After the completion of step 1 the
applications can now be reviewed. There are different ways to go through this process but there
are various computer based programs that can search for the keywords in the resumes and
shortlist the number of resumes that can b looked into.
3. INTERVIEW PROCESS : After shortlisting those applicants who have met the minimum
criteria.The HR manager selects these applicants for the interview.Due to lack of time sometimes
interview is being done on phone.
4. TEST ADMINISTRATION : Before the hiring decision is made any number of test are
administered. The tests may be physical tests ,drug tests, response test, credit report checks,
personality tests, etc.
5. OFFER MAKING PROCESS : This is the final step in which the choosen candidate is offered
the job position. The offer can b developed through email or through the job letter.All the
benefits and compensation are discussed in the offer letter..
Describe what is meant by “radical cure” and identify the drug that .pdf
Describe what is meant by “radical cure” and identify the drug that is produces a radical cure for
malaria.
Solution
Radical cure refers to the elimination of malaria parasites completely from the body.It
specifically refers to the elimination of hypnozoites of P.vivax and P. ovale which are the
parasitic stages in the liver and could result in relapse of the infection. A combination of two
drugs i.e chloroquine and primaquine produces a radical cure for malaria..
Describe the nutritional modes of organisms based upon where they obt.pdf
Describe the nutritional modes of organisms based upon where they obtain carbon energy.
Include an example of each.
Solution
Describe the nutritional modes of organisms based upon where they obtain carbon and energy.
Almost all organisms on the earth need energy to do their work, and carbon to use in building
their chemical structures. And these two (energy and carbon) has to be obtained from external
sources, based on the way they obtain energy and carbon, organisms are broadly classified into
four different groups.
Classification
Obtaining energy and carbon
Examples
Autotrophs
Photoautotrophs
They get energy from sunlight and will not depend on other organisms for carbon. Use CO2 as a
carbon source.
All photosynthetic prokaryotes
Plants (eukaryotes)
Chemoautotrophs
These make use of inorganic chemicals as an energy source. For example, H2S, Fe2+ etc and
CO2 will serve as a carbon source.
Archaea living in hydrothermal vents.
Heterotrophs
Photoheterotrophs
Use light as an energy source and will use organic molecules as a carbon source. These type of
organisms are rare.
Heliobacteria, non-sulfur bacteria, green non-sulfur bacteria etc
Chemoheterotrophs
Use organic molecules like sugars as an energy source.
Use organic molecules like proteins, sugars, fats, nucleic acids, etc. as a carbon source. This type
of nutrition is most common form of nutrition on Earth
Most of the prokaryotes, protists, fungi, and animals
Classification
Obtaining energy and carbon
Examples
Autotrophs
Photoautotrophs
They get energy from sunlight and will not depend on other organisms for carbon. Use CO2 as a
carbon source.
All photosynthetic prokaryotes
Plants (eukaryotes)
Chemoautotrophs
These make use of inorganic chemicals as an energy source. For example, H2S, Fe2+ etc and
CO2 will serve as a carbon source.
Archaea living in hydrothermal vents.
Heterotrophs
Photoheterotrophs
Use light as an energy source and will use organic molecules as a carbon source. These type of
organisms are rare.
Heliobacteria, non-sulfur bacteria, green non-sulfur bacteria etc
Chemoheterotrophs
Use organic molecules like sugars as an energy source.
Use organic molecules like proteins, sugars, fats, nucleic acids, etc. as a carbon source. This
type of nutrition is most common form of nutrition on Earth
Most of the prokaryotes, protists, fungi, and animals.
Decasper and Spence (1986) measured babies sucking rates as they lis.pdf
Decasper and Spence (1986) measured babies sucking rates as they liscenced to a story they had
heard in the womb (the cat in the hat) and to a story they had not heard (the king, the mice, and
the cheese) the study concluded that:
a.) amniotic fluid blocks sound waves from reaching the fetus
b) increased heart rate showed that an infant can learn even before it is born
c) different sucking styles and rates showed that an infant can learn even before it is born
d) the cadence of the reading material not previous experience altered sucking style and rate
Solution
different sucking styles and rates showed that an infant can learn even before it is born.
Corhin Eckles and Connor Hayes have recieved walkie talkies for Chri.pdf
Corhin Eckles and Connor Hayes have recieved walkie talkies for Christmas. If they leave from
the same point at the same time, Corhin walking north at 2.5 mph and Connor walking east at 3
mph, how long will they be able to talk to each other if the range of the walkie-talkies is 4 mi?
Solution
Corhin DATA:
distance= x miles; rate=2.5 mph; time=d/r= x/2.5
Connor DATA:
distance= y miles; rate=3 mph; time=y/4
EQUATIONS:
1) x^2+y^2=4^2
2) x/2.5 = y/4
Using 2) you get y=4x/2.5
Substitute into 1) to get x^2+(4x/2.5)^2=16
x^2+(16x^2/6.25)=16
6.25x^2+16x^2=16(6.25)
22.25x^2=100
x^2=100/22.25=4.49...
x=2.12 hrs.
x/2.5= 0.85 hrs. = 50.87 minutes
This is Corhin\'s time, but it is the same for Connor..
Consider three genes in a hypothetical bird. One controls the color o.pdf
Consider three genes in a hypothetical bird. One controls the color of feathers on the top of the
head (red or brown), one controls the color of feathers on the chest (black or brown) and one
controls the number of syllables in the song (3 or 4). You notice that birds frequently have either
red heads and black chests or brown heads and brown chests, but very few have red heads and
brown chests or brown heads and black chests. However, there doesn\'t seem to be any
association between the syllables in the bird song and color. In other words, red heads are
equally likely to sing songs with 3 or 4 syllables, and the same goes for the brown heads. Based
on what you know about how the genome is organized and passed down to the next generation,
explain this observation.
Solution
The observation can be attributed to the mechanism of genetic linkage.
It is a well known fact that genes get passed from one generation to another. Mendel\'s second
law of inheritance states that during this process, the inheritance of one gene is independent of
the inheritance of another gene. Howeve, this law holds true only when the different genes are
sufficiently spaced apart from each other. Inother words, if the genes are close to each other, it is
difficult for them to be inherited in an independent manner. If one of the genes is inherited, then
the other gene (by virtue of its proximity to the first gene) also tends to get inherited. This
phenomenon is called as genetic linkage, and the genes involved are said to be Linked Genes.
The same mechanism is in action in the above case.
The genes for Red Head and Black Crest are linked, as are the genes for Brown Head and Brown
Crest. Therefore, these two genes are inherited in a linked manner, and not in an independent
manner. The two genes therefore behave as One Single Gene.
The gene for song syllable is at a sufficient distance from the linked genes, and follows
Mendel\'s law of independent assortment and is not dependent on the presence or absence of the
former two traits..
Answer the following questions What is the domain of logarithmic fun.pdf
Answer the following questions: What is the domain of logarithmic function What is the range
of Logarithmic function What is the value of e(approximate to 4 decimal places)
Solution
SOLUTION
===========
the domain of any logarithmic function is positive real numbers
1) Domain =(0,inf)
2) Range =(-inf,+inf)
3) e=2.7183.
are organic compounds that bind very tightly to iron ions for transp.pdf
are organic compounds that bind very tightly to iron ions for transport into cells.
Solution
Organisms like bacteria secrerte substances called siderophores which bind to iron for transport
inside the cell. Iron is one of the important micronutrients for the bacterial metabolism.
Ferrichrome,Desferrioxamine B and Desferrioxamine E are some of the important siderophores
secreted by various bacteria..
Book Business Its Legal, Ethical, Global Environment 10th Edition.pdf
Book: Business It\'s Legal, Ethical, Global Environment 10th Edition
Page 204-205
Question and Problems
# 2
#5
Solution
Ethical issues connect intimately with economic issues. Take the economic practice of doing a
cost-benefit analysis. You could spend one hundred dollars for a night on the town, or you could
donate that one hundred dollars to the reelection campaign of your favorite politician. Which
option is better? The night on the town increases pleasure. A politician’s successful campaign
may lead to more liberty in the long term. We regularly make decisions like this, weighing our
options by measuring their likely costs and likely benefits against each other.
This connects economics directly to a major issue in ethics: By what standard do we determine
what counts as a benefit or a cost? A list of competing candidates for the status of ultimate value
standard includes happiness, satisfying the will of God, long-term survival, liberty, duty, and
equality.
Economists implicitly adopt a value framework when beginning a cost-benefit analysis. Different
value commitments can lead to the same item being considered a cost from one perspective and a
benefit from another. For example, those whose standard of value is increasing human happiness
would count a new road to a scenic mountain vista as a benefit, while those whose standard is
maintaining an unchanged natural environment would count it as a cost.
The results of economic analysis also lead directly to ethical issues. For example, one result of
the nineteenth- and twentieth-century debate over capitalism and socialism is a general
consensus that capitalism is effective at producing wealth and socialism is effective at keeping
people poor. Advocates of capitalism use these results to argue that capitalism is good; others
might respond that “socialism is good in theory, but unfortunately it is not practical.” Implicit in
the capitalist position is the view that practical consequences determine goodness. By contrast,
implicit in the position of those who believe socialism to be an impractical moral ideal is the
view that goodness is distinct from practical consequences.
This connects economics to a second major issue in ethics: Is goodness or badness determined by
real-world practical consequences or by some other means, such as revelations from God, faith in
authorities or authoritative institutions, appeals to rational consistency, felt senses of empathy, or
an innate conscience? The point for economic analysis, most of which is a matter of
understanding and predicting the consequences of various actions, is that the relevance of
economic analysis to policymaking depends, in part, on what one believes is the final source of
value standards..
According the United States (U.S.) Centers for Disease Control and P.pdf
According the United States (U.S.) Centers for Disease Control and Prevention, in 2015
influenza and pneumonia combined were the 8th leading causes of death (55,227 deaths in total)
and the only communicable diseases ranked among the top ten killers in the U.S. What were the
10 causes of mortality in the U.S. in 1900 and 2015? Since communicable disease caused a
relatively few deaths in the U.S. in 2015, why should public and public health professionals be
concerned about them anymore and devote resources in their prevention and control?
Solution
Top 10 leading causes of deaths in US: Hearts diseases Cancer Airway diseases which are non
infectious Cardio vascular disorders Accidents Alzheimer\'s disease Diabetes Pneumonia or
influenza Nephropathys Suicide. The period of communicability is less for the communicable
diseases, hence they are prevention of disease in early stages is possible. Prevention and control
of influenza: Good ventilation of public buildings Encouraging sufferers to cover their faces
with a handkerchief when coughing and sneezing Stay in home at the first sign of influenza
Vaccination against the influenza, may be killed or level attenuated vaccines Prevention of
exposure to influenza strains by hygiene practice Antiviral drugs like amantadine or rimantidine
used Prevention and control of pneumonia: Pneumococcal pneumonia vaccination, PPV 23
AND PCV7. Education of mother to observe the sings of pneumonia. History taking and clinical
examination are very important for management Physical examination for, breath count, chest
indrawing, listen for stridor, wheezing and for severe malnutrition, cyanosis Any signs detected,
they are treated early by antibiotics and other measures Improving the primary medical care
services and development..
9. For some non-Mendelian inheritance patterns, the genotype of an i.pdf
9. For some non-Mendelian inheritance patterns, the genotype of an individual is not sufficient to
predict the phenotype. For the following traits, what else do you need to know to predict the
phenotype?
a) Snail coiling direction
b) Angelman syndrome due to deleted UBE3A gene
Solution
The inheritance mechanisms of both snail coiling and Angelman syndrome involve strong
maternal influences. Snail coiling direction is governed by the genotype of the oocyte, not of the
zygote, so much so that the snail coiling direction can be predicted from the maternal genotype
alone. Similarly, Angelman syndrome (due to deletions in ube3A) is governed by the maternal
copy of the gene alone. If the copy of the gene derived from the mother is functional, the
individuals are healthy. If the maternal copy is defective or deficient, the individual suffers from
the Angelman syndrome..
You have mated a phenotypically wild-type female of the genotype Xy+.pdf
You have mated a phenotypically wild-type female of the genotype Xy+w+Xyw to a male with
the genotype XywY. You counted 2205 total F1 progeny and found 17 flies with gray bodies and
white eyes and 12 flies that had yellow body and red eyes. What is the distance between these
genes controlling body color and eye color on the X chromosome?
rev: 10_27_2012
A. 0.78 map units
B. 1.32 map units
C. 2.63 map units
D. 98.7 map units
Solution
In this problem total progeny is 2205. Among them recombinant individuals are 17 flies with
gray body and white eyes; 12 flies with yellow body and red eyes.
Then % of Recombination = No. Of Recombinant Individuals x 100 / Total progeny
= (17 + 12) x 100 / 2205
= 29 x 100 / 2205
= 1.32 % or 1.32 mU.
You are an oncologist specializing in colon cancer. A patient presen.pdf
You are an oncologist specializing in colon cancer. A patient presents to you with disease that
has spread to his bones and liver and there is no treatment available. What stage cancer is
this?2345
Solution
Answer is Stage IV (4)
Stages of colon cancer are:
Stage 0
It is very early cancer and it will be affecting the innermost layer of the intestine
Stage I
In stage I, cancer is seen in the inner layers of the colon
Stage II
In stage II, cancer is seen spreading through the muscle wall of the colon
Stage III
In stage III, cancer is seen spreading to the lymph nodes
Stage IV
In stage IV, cancer is seen spreading to other organs outside the colon
In case, of colon cancer, at stage IV, it will be spreading to the distant location in the body (liver,
lungs, bones, distant lymph nodes or other sites). At this stage there is no cure for this cancer or
patient will be left with very few options.
Stage 0
It is very early cancer and it will be affecting the innermost layer of the intestine
Stage I
In stage I, cancer is seen in the inner layers of the colon
Stage II
In stage II, cancer is seen spreading through the muscle wall of the colon
Stage III
In stage III, cancer is seen spreading to the lymph nodes
Stage IV
In stage IV, cancer is seen spreading to other organs outside the colon.
You are required to open an input file (shown below) then read one c.pdf
You are required to open an input file (shown below) then read one character at a time until it is
finished. While reading the content of the file, keep a count of the following:
1) Count of each digit read. The counts should be kept in an int array of size 10.
2) Count of each alphabetic character read irrespective if it is an upper case or lower case. The
counts should be kept in an int array of size 26.
3) Count of other characters (Not a digit and not an alphabet). Whitespace character such as
newline \ , blank or tab \\t should NOT be counted here. Hence, if you find ! ; @ the count will
be 3; you will ignore spaces. This will just be an int value.
4) Total count. This will be the count for everything read, including blanks, new line, etc. Again,
this will be an int value.
After reading and processing the file, and get each count correctly, you will need to sort the
count values in numerical order (lowest to highest) before displaying the result.
Your solution must use FUNCTIONAL DECOMPOSITION.
Hence, if the input file was the one given below, your unsorted output is shown below. Please
note that your output must also has the values sorted in numerical order from lowest to highest.
Here is the content of your input file.
Here will be the output generated, minus the gap. That\'s just the end of the page.
Solution
int main(int argc, char *argv[]) {
#include
FILE *ptr_file;
char buf[1000];
int digit[10];
int alpha[26];
int other=0;
char alp=65;
int i;
for(i=0;i<=9;i++)
digit[i]=0;
for(i=0;i<=25;i++)
alpha[i]=0;
ptr_file =fopen(\"D:/input.txt\",\"r\");
if (!ptr_file)
return 1;
while (fgets(buf,1000, ptr_file)!=NULL)
printf(\"\ Original File:=>\ \");
printf(\"%s\ \",buf);
fclose(ptr_file);
for(i=0;buf[i]!=NULL;i++)
{
//Digits Count
if(buf[i]==\'0\')
digit[0]=digit[0]+1;
if(buf[i]==\'1\')
digit[1]=digit[1]+1;
if(buf[i]==\'2\')
digit[2]=digit[2]+1;
if(buf[i]==\'3\')
digit[3]=digit[3]+1;
if(buf[i]==\'4\')
digit[4]=digit[4]+1;
if(buf[i]==\'5\')
digit[5]=digit[0]+1;
if(buf[i]==\'6\')
digit[6]=digit[6]+1;
if(buf[i]==\'7\')
digit[7]=digit[7]+1;
if(buf[i]==\'8\')
digit[8]=digit[8]+1;
if(buf[i]==\'9\')
digit[9]=digit[9]+1;
//Alphabatic count
if(buf[i]==\'a\'||buf[i]==\'A\')
alpha[0]=alpha[0]+1;
if(buf[i]==\'b\'||buf[i]==\'B\')
alpha[1]=alpha[1]+1;
if(buf[i]==\'c\'||buf[i]==\'C\')
alpha[2]=alpha[2]+1;
if(buf[i]==\'d\'||buf[i]==\'D\')
alpha[3]=alpha[3]+1;
if(buf[i]==\'e\'||buf[i]==\'E\')
alpha[4]=alpha[4]+1;
if(buf[i]==\'f\'||buf[i]==\'F\')
alpha[5]=alpha[5]+1;
if(buf[i]==\'g\'||buf[i]==\'G\')
alpha[6]=alpha[6]+1;
if(buf[i]==\'h\'||buf[i]==\'H\')
alpha[7]=alpha[7]+1;
if(buf[i]==\'i\'||buf[i]==\'I\')
alpha[8]=alpha[8]+1;
if(buf[i]==\'j\'||buf[i]==\'J\')
alpha[9]=alpha[9]+1;
if(buf[i]==\'k\'||buf[i]==\'K\')
alpha[10]=alpha[10]+1;
if(buf[i]==\'l\'||buf[i]==\'L\')
alpha[11]=alpha[11]+1;
if(buf[i]==\'m\'||buf[i]==\'M\')
alpha[12]=alpha[12]+1;
if(buf[i]==\'n\'||buf[i]==\'N\')
alpha[13]=alpha[13]+1;
if(buf[i]==\'o\'||buf[i]==\'O\')
alpha[14]=alpha[14]+1;
if(bu.
1.Which of Marys cranial nerves is affected in this episode( Mary.pdf
1.Which of Mary\'s cranial nerves is affected in this episode?( Mary felt a stabbing pain in her
upper jaw and teeth that radiated out to the side of her nose. Over the next several days, she
experienced several more episodes of this intense pain. A visit to the dentist revealed no
abnormalities and she was referred to her physician for an evaluation. Prior to her appointment,
she noticed that the symptoms were subsiding as they had previously. Her physician scheduled
an appointment for a complete neurological exam the following week.)
2. In the MRI images shown in the case, you can see the lesions as bright \"white spots\" on the
brain. Using what you know about the structure of a neuron, explain what is causing this spot to
appear in the MRI?( The results of the MRI, shown below, were consistent with a diagnosis of
relapsing-remitting multiple sclerosis (RRMS). Relapsing-remitting multiple sclerosis is a form
of MS in which symptoms randomly flare up (Mary\'s episodes) and then resolve on their own.
The lesions seen on the MRI on the left were associated with another episode in which Mary
experienced sensory and motor disjunction in her left lower extremity. A subsequent MRI (image
on the right) appeared to show improvement after three months)
3. Three months later, you can see that the spots in the MRI appear to be smaller. Using what
you know about the structure of a neuron, explain what is happening to the neurons in the area
where the lesions are disappearing? Signa 1.0T srsanIRMtoce Go 3/3 28/40 A S35.8 TR 4000
TE:BSVEf 5.0 hk/2.0 04:16 A112 P108 Mary Rier Shutterstock gna 1.0T TR 4000 TEE85VEf
EC 2/2 15.6kHz 5.0thi/2. A112 P108 Mary Rice
Solution
1. The cranial nerves that are affected in this case are the olfactory nerves and the maxillary
nerves , which are a part of the trigeminal nerves.
2. The spot in the MRI is occurring as a result of demyelination of the neurons in that particular
area, which is resulting in the appearance of the white lesion. The condition of multiple sclerosis
is an auto immune condition of the CNS, and the lesions seen are called as MS lesions or
plaques.
3. After a period of three months, when the MRI was conducted again, it was observed that the
plaques found in the areas earlier had now disappeared. This implies that the myelin sheath that
were demyelinated have now recovered , as shown with a complete recovery of the affected
nerves, in this case. MS is also referred to be a relapsing remitting disease..
Which phyla have stomata and guard cells Bryophyta Monilophyta Co.pdf
Which phyla have stomata and guard cells? Bryophyta Monilophyta Coniferophyta
Anthophyta Which phyla have xylem cells? Bryophyta Monilophyta Coniferophyta
Anthophyta Which phyla have xylem vessels? Bryophyta Monilophyta Coniferophyta
Anthophyta
Solution
Answer:
34). Anthophyta
These are flowering plants and the possess stomata that are surrounded by guard cells which
helps in gaseous exchange.
35). Coniferophyta
36). Anthophyta
Vessel elements are typically found in flowering plants (anthophyta) but absent from most
gymnosperms such as conifers..
Which of the follwing traits would you expect to find in the transit.pdf
Which of the follwing traits would you expect to find in the transitional organisms linking flying
birds and their dinosaur ancestors?
A. Feathers
B. Claws on its fingers
C. Amniotic egg
D. Only A and B
E. A, B, and C
Solution
Answer: D)
The transitional organisms, during evolution shows some of the traits of both, but not linked by
direct descent. As birds evolved from dinosaurs, the follwing traits would be expected to find in
the transitional organisms linking flying birds and their dinosaur ancestors:
A. Feathers
B. Claws on its fingers.
What is DNA fingerprintingPlease, no wikipedia or google definiti.pdf
What is DNA fingerprinting?
Please, no wikipedia or google definitions.
Solution
DNA fingerprinting is applied to the scientific process whereby samples of DNA are
collected,collated and used to match other samples of DNA.DNA fingerprinting works on the
principle that each individuals DNA structure or genetic makeup is unique and therefore cannot
be forged,faked or altered in any way.No two individuals share the same genetic code except for
identical twins..
What are parametric and nonparametric inferential testsSolution.pdf
What are parametric and nonparametric inferential tests?
Solution
Inferential statistics suggest statements about a population based on a sample from that
population. Parametric inferential tests are carried out on data that follow certain parameters: the
data will be normal (i.e. the distribution parallels the bell curve); numbers can be added,
subtracted, multiplied and divided; variances are equal when comparing two or more groups; and
the sample should be large and randomly selected.
There are generally more statistical technique options for the analysis of parametric than non-
parametric data, and parametric statistics are considered to be the more powerful. Common
examples of parametric tests are: correlated t-tests and the Pearson r correlation coefficient.
Inferential statistics suggest statements or make predictions about a population based on a
sample from that population. Non-parametric tests relate to data that are flexible and do not
follow a normal distribution.
They are also known as.
1. How does body complexity differ across the 3 phyla (Platyhelmin.pdf
1. How does body complexity differ across the 3 phyla (Platyhelminthes, Nematoda, and
Annelida)? What contributes to this increased complexity? Give 2 examples.
1. How does body complexity differ across the 3 phyla (Platyhelminthes, Nematoda, and
Annelida)? What contributes to this increased complexity? Give 2 examples.
Solution
Answer:
The variable body complexity is differ across the 3 phyla such as platyhelminthes, nematoda, and
annelida because these are multicellular eukaryotes evolved with morphological innovations
including development of locomotors organs and development of various adaptations to survive
in the nature for predation. These are phyla such as platyhelimthes and nematyhelmithes. The
helminthes parasites and species have developed various morphological adaptations for their
survival in the host along with strong scolex to attach to the host internal surfaces. Nematodes
locomotion (round worms, hook worms) and body complexity is variable because the
locomotion of nematodes involves the somatic muscles mainly located under the hypodermis and
these muscle contractions promoted to apply sufficient high hydrostatic pressure laterally to the
cuticle with dorso-ventral bending to move in a sinusoidal manner (C. elegans) as an
evolutionary adaptations.
Annelids have circulatory system, primitive neural systems, and primitive evolutionary features
of invertebrates. Annelid\'s body complexity has increased from lower phyla, platyhelminthes
and nematoda. The body complexity has symmetrical and coelomate & they possess specific
“parapodia” for locomotion. In some species such as earthworms, they have specialized a
hydrostatic skeleton surrounded by several layers of longitudinal and circular muscles to promote
locomotion into the deeper layers of soil.
High metaolic activity, reproductive activity and increased body size have led to this increased
complexity, as part of evolutionary aspects
Platyhelmethis (tape worms): These are mostly parasitic in internal cavities of vertebrate
digestive system with complex asymmetrical body- complexity. The most prominent
segmentation phenomenon has promoted them with sexual reproduction and they use circular
and longitudinal muscle fibers of somatic region located within proglattids; used in locomotion.
For example, Hymenolepis diminuta is also called as rat tapeworm, belongs to cestoda and
possess scolex and proglattids as structural adaptations..
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Wednesday 22 May 2024, 14:00-15:00.
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Embracing GenAI - A Strategic Imperative
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● 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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