1. Variables or subject characteristics that bias the results of res.pdf
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1. Variables or subject characteristics that bias the results of research are called _______________. 2. The independent variable (IV) can only be thought of as causing the dependent variable (DV) when you have eliminated threats to internal validity. Otherwise, you can only say that the two variables are ___________. Solution.
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h. What are epigenetic changes How may these contribute to the deve.pdf
h. What are epigenetic changes? How may these contribute to the development of cancer?
Describe the mechanism by which methylation of -G-C- sites may contribute to the formation of
cancerous cells. i. What are the six hallmarks of cancer, and what enabling trait allows cells to
acquire the six hallmarks?
Solution
Ques-1: What are epigenetic changes? How may these contribute to the development of cancer?
Epigenetic inheritance can be defined as the process in which complete heritable phenotypic
individual is resulted from chromosomal alterations but there will be no more changes in the
DNA sequence followed by differential gene expression of the phenotype. Recent research has
illustrated that epigenetic inheritance is involves both mitotically and meiotically. However,
mitotic or somatic inheritance is higher because of no reprogramming events of altered
chromatin. In meiosis for gametogenesis, a lot of reprogramming events to repair \"alterations in
chromatic\" has involved for altered cpG islands in both sperm cells and oocytes result in no
inheritance propagation from generation to generation. For instance, Epigenetic alterations can
be produced due to the exposure to the environment, in case of mice; dietary nutritional additives
can induce gene express of agouti gene, further result in the improper fur color and increased
level of cancer growth.
Ques-2: Describe the mechanism by which methylation of -G-C- sites may contribute to the
formation of cancerous cells.
Epigenetic changes in chromatin structure that affects gene expression. The epigenetic changes
involves in the viable gene expression inside the cells due to methylated -G-C- sites on DNA
with differential pattern finally silencing of large number of genes approximately 550 genes
(CpG dinucleotides) chemically methylated and modified to change the gene expression. These
events are leading to generation of abnormal phenotypes of cells finally become neoplastic &
undergoes metastasis
Que-3: What are the six hallmarks of cancer, and what enabling trait allows cells to acquire the
six hallmarks?
Invasion is mainly due to presence of cell adhesion molecules such as integrins on the cancer
cells finally promote the movement of cancer cells to another region. This process is going to
connect adjacent cells to tumor cells finally undergo epithelial-mesenchymal transition
Intravastation is the process in which cancer cells undergoes metastasis to enter into “blood or
lymph stream” and these cancer cells secrete metalloprotenases, proteases to degrade basement
membrane of the stromal surface for endothelial migration. Sometimes, these cancer cells are
going to promote angiogenesis (synthesis of blood vessels) towards them to obtain nutrients
Extravasation is the process in which cancer cells are going to undergoes metastasis and “escape
from lymphatic vessels or blood vessel” by detaching from endothelium
Six hallmarks of cancer:
Self sufficiency in generating growth signals, insensitivity .
in this assignment you are asked to write a simple driver program an.pdf
in this assignment you are asked to write a simple driver program and set of functions (maybein
a library) that can be performed on a binary search tree.
Your program should allow user to insert/delete integer values into the binary search tree along
with several other operations on the binary search tree. You can use the code given in slides. But
this time your key will be int! Specifically, your program will ask user to enter a command and
related parameters (if any) in a loop, and then perform the given commands. Here is the list of
commands that your program must implement:
* insert
*find\'
*delete
*list inorder
*list preorder
*list postorder
*list levelorder
* max
* min
* height
*count
* sum
*quit
As always, make sure you release (free) the dynamically allocated memories if you allocate any
memory in your programs. So, before submitting your program, run it with valgrind to see if
there is any memory leakage
//my proggram in C
struct tree_node {
int data;
struct tree_node *left, *right;
}
typedef struct nodeT {
int key;
struct nodeT *left, *right;
} nodeT, *treeT;
int main(){
while (TRUE) {
printf(\"> \");
line = GetLine();
ch = toupper(line[0]);
switch (ch) {
case \'I\': insert(); break;
case \'F\': find(); break;
case \'D\': delete(); break;
case \'LI\': listInorder; break;
case \'LPR\': listPreorder(); break;
case \'LPO\': listPostorder(); break;
case \'MAX\': max(); break;
case \'min\': min(); break;
case \'H\': height(); break;
case \'C\': count(); break;
case \'S\': sum(); break;
case \'Q\': exit(0);
default:printf(\"Illegal command\ \"); break;
}
}
}
nodeT *FindNode(nodeT *t, int key){
while(t !=NULL) {
if (key == t->key) return t;
if (key < t->key) {
t = t->left;
} else {
t = t->right;
}
return NULL;
}
void delete(nodeT **p){
nodeT
*target;
target=*p;
if (target->left==NULL && target->right==NULL) {
*p=NULL;
} else if (target->left == NULL) {
*p=target->right;
} else
if (target->right == NULL) {
*p=target->left;
} else {
/* target has two children, see next slide */
}
free(target);
}
void listInorder(nodeT *T){
if (t != NULL) {
DisplayTree(t->left);
printf(“%d “, t->key);
DisplayTree(t->right);
}
}
void listPreorder(nodeT *t) {
if (t != NULL) {
printf(“%d “, t->key);
DisplayTree(t->left);
DisplayTree(t->right);
}
}
void listPostOrder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
DisplayTree(t->right);
printf(“%d “, t->key);
}
}
void intsert(nodeT **tptr, int key){
nodeT*t, *tmp;
t=*tptr;
if (t == NULL) {
tmp=New(nodeT*);
tmp->key = key;
tmp->left=tmp->right=NULL;
*tptr=tmp;
return;
}
if (key < t->key) {
InsertNode
(&t->left, key);
} else {
InsertNode(&t->right, key);
}
}
int height(nodeT *t){
if (t == NULL)
return 0;
else
return (1 + maximumof(
height(t->left),
height(t->right)) );
}
int sum(struct tree_node *p){
if (p == NULL)
return 0;
else
return (p->data +
sum(p->left) +
sum(p->right) );
}
Solution
1. /*
2. * Java Program to Implement Binary Search Tree
3. */
4.
5. import java.util.Scanner;
6.
7. /* Class BSTNode */
8. cl.
How does alcohol affect IPSP and Potassium Why do the ions move.pdf
How does alcohol affect IPSP and Potassium? Why do the ions move?
Solution
ANS:
Entire body system is effected by alcohol consumption. alcohol is taken placed by every cell in
organs by through the blood and it results electrolytes imbalance. So potassium levels of the
body fluids is affected by electrically conductive ions.
The altered body fluid ionic concentration causes ionic imbalances.
Functions of Potassium:
The nerves function, muscle contraction, hormonal balance and water balance is mainly based on
Potassium levels.
The imbalance between sodium and potassium ions causes hypertension. It effects ADH
hormone levels.
The body dehydration is normally prevented by Anti-Diuretic Hormone. After alcohol
consumption alcohol suppresses ADH, upset the water balance and potassium ion concentration
in cells.
IPSP (Inhibitory postsynaptic potential):
It is a postsynaptic potential that hyperpolarized and inhibits the responses.
it decrease action potentials by moving membrane potential farther from threshold.
It is generally results from opening of Potassium or chloride channels.
Ions movement after alcohol consumption:
The potassium ion move through the membrane more easily than sodium. The ions of the
potassium is moved faster than sodium. Serum potassium levels were significantly lower in
alcoholic patients.
Ionic effect by Beer:
The water content in beer is high and the soluble nutrients concentration is low. Beer alcoholic
percent impairs normal ADH effected and the water content of beer is retained in body, so its
results over fluid load in blood stream.
The ion concentration of potassium is decreased and causes hormonal imbalance and thirst.
Ions effect by Hard liquor:
In Hard liquor the water concentration is low, so less water enters into body. Then the ADH is
suppressed by alcohol, and the kidney process high amount of water from blood stream into
urine. it leads to raise of potassium ion concentration in blood..
How has the social change driven the development of new information t.pdf
How has the social change driven the development of new information technologies, such as cell
phone, text messaging, movie recommendation software?
Solution
Information technological innovations (such as cell phone, text messaging, movie
recommendation softwarehave changed society in dramatic ways.Technology leads to a social
change and development process if used to articulate local needs and and contents.Social
changes like the industrial revolution leading to urbanization, the creation of a working class, and
so on) have driven the need and development of new technologies.
-> Radio and television are perceived as pivotal technologies for the shaping of public opinion.
-> Technology used to communicate (cell phone, telegraph) fuelled powerful new industries for
the exchange of news and other important information at the national and international level..
Describe the concept of web accessibility.Thank-you!Solution.pdf
Describe the concept of web accessibility.
Thank-you!
Solution
Web accessibility means that the people with disabilities can use the Web. More particularly,
Web accessibility means that people with disabilities can recognize, understand, find the way,
and communicate with the Web. This is also helpful to others, including old people with
changing abilities due to aging.
Web accessibility includes all disabilities that influence contact to the Web, including visual,
hearing, and physical and neurological disabilities.
Millions of people have disabilities that change their use of the Web. Presently most Web sites
and Web software have accessibility difficulties that make it impossible for many people with
disabilities to use the Web.
A key principle of Web accessibility is to design Web sites and software that are flexible to meet
different user needs, preferences, and conditions. This flexibility also helps people without
disabilities in certain situations, such as people using a slow Internet connection, people with
\"temporary disabilities\" such as a broken arm.
The Web is more and more important resource in many ways of life such as education,
employment, government, commerce, health care and more. It is necessary that the Web be
accessible in order to provide equal access and equal opportunity to people with disabilities and
help them to actively participate in society.
The Web offers the chance of unmatched access to information and communication for many
people with disabilities..
Good new technologies are a bit like good new roads Their social ben.pdf
Good new technologies are a bit like good new roads: Their social benefits far exceed what any
one person or company can get paid for creating them Michael Lewis, \" In defence of the
boom,\" New York times, October 27, 2002.
Solution
C. Tes. When marginal social benefits are positive, subsidies can increase output to achieve
efficient outcomes.
*****.
can you help me to solve this. Design a Person class similar to the .pdf
can you help me to solve this. Design a Person class similar to the Coin class defined in this
chapter. Then design and implement a driver class called SelectPerson whose main method
creates two Person objects, then randomly selects a gender in both to see in which object a
female gender is selected two times in a row. Consider the possibility that they might tie. Print
the results of each selection and then print the object name which comes up first with the female
gender being selected twice, consecutively
Solution
public class SelectPerson{
public static void main(String []args){
//create two person objects
Person p1=new Person();
Person p2=new Person();
//delcare variable to set intiial values
int count1=0,count2=0,counter=0;
//loop until two consequetive femailes aren\'t listed
while(count1<2 && count2<2)
{
counter++;
p1.setGender();
p2.setGender();
//display selection of eahc round
System.out.println(\"*******Round \"+counter+\"*****\");
System.out.println(\"Object1 = Gender :\"+p1);
System.out.println(\"Object2 = Gender :\"+p2);
//check for female occurance
if(p1.isFemale())
{
count1++;
}
if(p2.isFemale())
{
count2++;
}
}
//display the results
if(count1==count2)
System.out.println(\"Tie\");
else
if(count1==2)
System.out.println(\"p1\");
else if(count2==2)
System.out.println(\"p2\");
}
}
//definitin of Person class
class Person
{
//variable to hold gender
private int gender;
//dynamiclaly set gender
public void setGender()
{
gender=(int)(Math.random()*2);
}
//check if gender is female
public boolean isFemale()
{
return (gender==1);
}
//return gender of person object
public String toString()
{
String genderName;
if(gender==0)
genderName=\"Male\";
else
genderName=\"Female\";
return genderName;
}
}.
Cephalothorax, 6 pairs of jointed appendages. Aquatic, but comes to s.pdf
Cephalothorax, 6 pairs of jointed appendages. Aquatic, but comes to shore to mate and lay eggs
(spawn). Living fossil. Subphylum _ Class_ Many similar segments. Rounded body. Detritivore
and decomposer. Two pairs of jointed legs per segments. Subphylum _ Class _ Deuterostome
with 5 parts fused together. No arms. Spines. Elaborate jaws that eat kelp. Phylum _ Class _ Six
legs with 4 scaly wings. Undergoes complete metamorphosis. Wings often large and showy.
Pollinators. SUBPHYLUM _ ORDER _ Pentaradial symmetry. Arms with tube feet that have
SUCKERS. Members of this class eat mussels. Phylum _ Class _ Two papery wings. Includes
flies and mosquitos. ORDER_ First appendages are chelicerae, second are pedipalps. Terrestrial
animals that include spiders and scorpions. Subphylum _ Class _ Almost entirely aquatic.
Undergo ecdysiast. Jointed appendages. Includes Crab, lobsters, and pill bugs. Subphylum _
Class_
Solution
1. The organism here is the Horseshoe Crab. Subphylum: Chelicerata, Class: Merostomata
2. The organism here is millipede. Subphylum: Myriapoda, Class: Diplopoda
3. Sea urchins fit the description. Phylum: Echinodermata, Class: Echinoidea
4.The organism is butterfly. Subphylum: Hexapoda, Class: Insecta
5. Starfish is the organism. Phylum: Echinodermata, Class: Asteroidea
6. Order: Diptera
7. Subphylum: Chelicerata, Class: Arachnida
8. Subphylum: Crustacea, Class: Malacostraca.
Recommended
h. What are epigenetic changes How may these contribute to the deve.pdf
h. What are epigenetic changes? How may these contribute to the development of cancer?
Describe the mechanism by which methylation of -G-C- sites may contribute to the formation of
cancerous cells. i. What are the six hallmarks of cancer, and what enabling trait allows cells to
acquire the six hallmarks?
Solution
Ques-1: What are epigenetic changes? How may these contribute to the development of cancer?
Epigenetic inheritance can be defined as the process in which complete heritable phenotypic
individual is resulted from chromosomal alterations but there will be no more changes in the
DNA sequence followed by differential gene expression of the phenotype. Recent research has
illustrated that epigenetic inheritance is involves both mitotically and meiotically. However,
mitotic or somatic inheritance is higher because of no reprogramming events of altered
chromatin. In meiosis for gametogenesis, a lot of reprogramming events to repair \"alterations in
chromatic\" has involved for altered cpG islands in both sperm cells and oocytes result in no
inheritance propagation from generation to generation. For instance, Epigenetic alterations can
be produced due to the exposure to the environment, in case of mice; dietary nutritional additives
can induce gene express of agouti gene, further result in the improper fur color and increased
level of cancer growth.
Ques-2: Describe the mechanism by which methylation of -G-C- sites may contribute to the
formation of cancerous cells.
Epigenetic changes in chromatin structure that affects gene expression. The epigenetic changes
involves in the viable gene expression inside the cells due to methylated -G-C- sites on DNA
with differential pattern finally silencing of large number of genes approximately 550 genes
(CpG dinucleotides) chemically methylated and modified to change the gene expression. These
events are leading to generation of abnormal phenotypes of cells finally become neoplastic &
undergoes metastasis
Que-3: What are the six hallmarks of cancer, and what enabling trait allows cells to acquire the
six hallmarks?
Invasion is mainly due to presence of cell adhesion molecules such as integrins on the cancer
cells finally promote the movement of cancer cells to another region. This process is going to
connect adjacent cells to tumor cells finally undergo epithelial-mesenchymal transition
Intravastation is the process in which cancer cells undergoes metastasis to enter into “blood or
lymph stream” and these cancer cells secrete metalloprotenases, proteases to degrade basement
membrane of the stromal surface for endothelial migration. Sometimes, these cancer cells are
going to promote angiogenesis (synthesis of blood vessels) towards them to obtain nutrients
Extravasation is the process in which cancer cells are going to undergoes metastasis and “escape
from lymphatic vessels or blood vessel” by detaching from endothelium
Six hallmarks of cancer:
Self sufficiency in generating growth signals, insensitivity .
in this assignment you are asked to write a simple driver program an.pdf
in this assignment you are asked to write a simple driver program and set of functions (maybein
a library) that can be performed on a binary search tree.
Your program should allow user to insert/delete integer values into the binary search tree along
with several other operations on the binary search tree. You can use the code given in slides. But
this time your key will be int! Specifically, your program will ask user to enter a command and
related parameters (if any) in a loop, and then perform the given commands. Here is the list of
commands that your program must implement:
* insert
*find\'
*delete
*list inorder
*list preorder
*list postorder
*list levelorder
* max
* min
* height
*count
* sum
*quit
As always, make sure you release (free) the dynamically allocated memories if you allocate any
memory in your programs. So, before submitting your program, run it with valgrind to see if
there is any memory leakage
//my proggram in C
struct tree_node {
int data;
struct tree_node *left, *right;
}
typedef struct nodeT {
int key;
struct nodeT *left, *right;
} nodeT, *treeT;
int main(){
while (TRUE) {
printf(\"> \");
line = GetLine();
ch = toupper(line[0]);
switch (ch) {
case \'I\': insert(); break;
case \'F\': find(); break;
case \'D\': delete(); break;
case \'LI\': listInorder; break;
case \'LPR\': listPreorder(); break;
case \'LPO\': listPostorder(); break;
case \'MAX\': max(); break;
case \'min\': min(); break;
case \'H\': height(); break;
case \'C\': count(); break;
case \'S\': sum(); break;
case \'Q\': exit(0);
default:printf(\"Illegal command\ \"); break;
}
}
}
nodeT *FindNode(nodeT *t, int key){
while(t !=NULL) {
if (key == t->key) return t;
if (key < t->key) {
t = t->left;
} else {
t = t->right;
}
return NULL;
}
void delete(nodeT **p){
nodeT
*target;
target=*p;
if (target->left==NULL && target->right==NULL) {
*p=NULL;
} else if (target->left == NULL) {
*p=target->right;
} else
if (target->right == NULL) {
*p=target->left;
} else {
/* target has two children, see next slide */
}
free(target);
}
void listInorder(nodeT *T){
if (t != NULL) {
DisplayTree(t->left);
printf(“%d “, t->key);
DisplayTree(t->right);
}
}
void listPreorder(nodeT *t) {
if (t != NULL) {
printf(“%d “, t->key);
DisplayTree(t->left);
DisplayTree(t->right);
}
}
void listPostOrder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
DisplayTree(t->right);
printf(“%d “, t->key);
}
}
void intsert(nodeT **tptr, int key){
nodeT*t, *tmp;
t=*tptr;
if (t == NULL) {
tmp=New(nodeT*);
tmp->key = key;
tmp->left=tmp->right=NULL;
*tptr=tmp;
return;
}
if (key < t->key) {
InsertNode
(&t->left, key);
} else {
InsertNode(&t->right, key);
}
}
int height(nodeT *t){
if (t == NULL)
return 0;
else
return (1 + maximumof(
height(t->left),
height(t->right)) );
}
int sum(struct tree_node *p){
if (p == NULL)
return 0;
else
return (p->data +
sum(p->left) +
sum(p->right) );
}
Solution
1. /*
2. * Java Program to Implement Binary Search Tree
3. */
4.
5. import java.util.Scanner;
6.
7. /* Class BSTNode */
8. cl.
How does alcohol affect IPSP and Potassium Why do the ions move.pdf
How does alcohol affect IPSP and Potassium? Why do the ions move?
Solution
ANS:
Entire body system is effected by alcohol consumption. alcohol is taken placed by every cell in
organs by through the blood and it results electrolytes imbalance. So potassium levels of the
body fluids is affected by electrically conductive ions.
The altered body fluid ionic concentration causes ionic imbalances.
Functions of Potassium:
The nerves function, muscle contraction, hormonal balance and water balance is mainly based on
Potassium levels.
The imbalance between sodium and potassium ions causes hypertension. It effects ADH
hormone levels.
The body dehydration is normally prevented by Anti-Diuretic Hormone. After alcohol
consumption alcohol suppresses ADH, upset the water balance and potassium ion concentration
in cells.
IPSP (Inhibitory postsynaptic potential):
It is a postsynaptic potential that hyperpolarized and inhibits the responses.
it decrease action potentials by moving membrane potential farther from threshold.
It is generally results from opening of Potassium or chloride channels.
Ions movement after alcohol consumption:
The potassium ion move through the membrane more easily than sodium. The ions of the
potassium is moved faster than sodium. Serum potassium levels were significantly lower in
alcoholic patients.
Ionic effect by Beer:
The water content in beer is high and the soluble nutrients concentration is low. Beer alcoholic
percent impairs normal ADH effected and the water content of beer is retained in body, so its
results over fluid load in blood stream.
The ion concentration of potassium is decreased and causes hormonal imbalance and thirst.
Ions effect by Hard liquor:
In Hard liquor the water concentration is low, so less water enters into body. Then the ADH is
suppressed by alcohol, and the kidney process high amount of water from blood stream into
urine. it leads to raise of potassium ion concentration in blood..
How has the social change driven the development of new information t.pdf
How has the social change driven the development of new information technologies, such as cell
phone, text messaging, movie recommendation software?
Solution
Information technological innovations (such as cell phone, text messaging, movie
recommendation softwarehave changed society in dramatic ways.Technology leads to a social
change and development process if used to articulate local needs and and contents.Social
changes like the industrial revolution leading to urbanization, the creation of a working class, and
so on) have driven the need and development of new technologies.
-> Radio and television are perceived as pivotal technologies for the shaping of public opinion.
-> Technology used to communicate (cell phone, telegraph) fuelled powerful new industries for
the exchange of news and other important information at the national and international level..
Describe the concept of web accessibility.Thank-you!Solution.pdf
Describe the concept of web accessibility.
Thank-you!
Solution
Web accessibility means that the people with disabilities can use the Web. More particularly,
Web accessibility means that people with disabilities can recognize, understand, find the way,
and communicate with the Web. This is also helpful to others, including old people with
changing abilities due to aging.
Web accessibility includes all disabilities that influence contact to the Web, including visual,
hearing, and physical and neurological disabilities.
Millions of people have disabilities that change their use of the Web. Presently most Web sites
and Web software have accessibility difficulties that make it impossible for many people with
disabilities to use the Web.
A key principle of Web accessibility is to design Web sites and software that are flexible to meet
different user needs, preferences, and conditions. This flexibility also helps people without
disabilities in certain situations, such as people using a slow Internet connection, people with
\"temporary disabilities\" such as a broken arm.
The Web is more and more important resource in many ways of life such as education,
employment, government, commerce, health care and more. It is necessary that the Web be
accessible in order to provide equal access and equal opportunity to people with disabilities and
help them to actively participate in society.
The Web offers the chance of unmatched access to information and communication for many
people with disabilities..
Good new technologies are a bit like good new roads Their social ben.pdf
Good new technologies are a bit like good new roads: Their social benefits far exceed what any
one person or company can get paid for creating them Michael Lewis, \" In defence of the
boom,\" New York times, October 27, 2002.
Solution
C. Tes. When marginal social benefits are positive, subsidies can increase output to achieve
efficient outcomes.
*****.
can you help me to solve this. Design a Person class similar to the .pdf
can you help me to solve this. Design a Person class similar to the Coin class defined in this
chapter. Then design and implement a driver class called SelectPerson whose main method
creates two Person objects, then randomly selects a gender in both to see in which object a
female gender is selected two times in a row. Consider the possibility that they might tie. Print
the results of each selection and then print the object name which comes up first with the female
gender being selected twice, consecutively
Solution
public class SelectPerson{
public static void main(String []args){
//create two person objects
Person p1=new Person();
Person p2=new Person();
//delcare variable to set intiial values
int count1=0,count2=0,counter=0;
//loop until two consequetive femailes aren\'t listed
while(count1<2 && count2<2)
{
counter++;
p1.setGender();
p2.setGender();
//display selection of eahc round
System.out.println(\"*******Round \"+counter+\"*****\");
System.out.println(\"Object1 = Gender :\"+p1);
System.out.println(\"Object2 = Gender :\"+p2);
//check for female occurance
if(p1.isFemale())
{
count1++;
}
if(p2.isFemale())
{
count2++;
}
}
//display the results
if(count1==count2)
System.out.println(\"Tie\");
else
if(count1==2)
System.out.println(\"p1\");
else if(count2==2)
System.out.println(\"p2\");
}
}
//definitin of Person class
class Person
{
//variable to hold gender
private int gender;
//dynamiclaly set gender
public void setGender()
{
gender=(int)(Math.random()*2);
}
//check if gender is female
public boolean isFemale()
{
return (gender==1);
}
//return gender of person object
public String toString()
{
String genderName;
if(gender==0)
genderName=\"Male\";
else
genderName=\"Female\";
return genderName;
}
}.
Cephalothorax, 6 pairs of jointed appendages. Aquatic, but comes to s.pdf
Cephalothorax, 6 pairs of jointed appendages. Aquatic, but comes to shore to mate and lay eggs
(spawn). Living fossil. Subphylum _ Class_ Many similar segments. Rounded body. Detritivore
and decomposer. Two pairs of jointed legs per segments. Subphylum _ Class _ Deuterostome
with 5 parts fused together. No arms. Spines. Elaborate jaws that eat kelp. Phylum _ Class _ Six
legs with 4 scaly wings. Undergoes complete metamorphosis. Wings often large and showy.
Pollinators. SUBPHYLUM _ ORDER _ Pentaradial symmetry. Arms with tube feet that have
SUCKERS. Members of this class eat mussels. Phylum _ Class _ Two papery wings. Includes
flies and mosquitos. ORDER_ First appendages are chelicerae, second are pedipalps. Terrestrial
animals that include spiders and scorpions. Subphylum _ Class _ Almost entirely aquatic.
Undergo ecdysiast. Jointed appendages. Includes Crab, lobsters, and pill bugs. Subphylum _
Class_
Solution
1. The organism here is the Horseshoe Crab. Subphylum: Chelicerata, Class: Merostomata
2. The organism here is millipede. Subphylum: Myriapoda, Class: Diplopoda
3. Sea urchins fit the description. Phylum: Echinodermata, Class: Echinoidea
4.The organism is butterfly. Subphylum: Hexapoda, Class: Insecta
5. Starfish is the organism. Phylum: Echinodermata, Class: Asteroidea
6. Order: Diptera
7. Subphylum: Chelicerata, Class: Arachnida
8. Subphylum: Crustacea, Class: Malacostraca.
After reading the codes of ethics posted to the USAOnline site please.pdf
After reading the codes of ethics posted to the USAOnline site please do the following (when
you are asked to compare, consider the following: the length and details in each one, the topics
covered, what are your thoughts on what is included in each): 1) Compare the Alabama Board of
Licensure for Professional Engineers and Professional Land Surveyors (Alabama BELS) with
your primary engineering societies Code of Ethics (i.e. if you are a civil it is ASCE, if you are an
Electrical it is IEEE, etc.) 2) Compare your primary engineering societies Code of Ethics with
ASCE\'s Code of Ethics (if you are a civil compare ASCE with any of the other societies)
Solution
1) First topic in Alabama BELS says about conflict of interest and is as follows-
This topic is also present in ASCE ethics and is as follows -
Now, this is present (highlighted in bold below) in Engineering council of India - code of ethics
under article 3 titled Obligation to Maintain High Standard of Personal Behaviour in a
Responsible Manner which is as follows -
The second article in Alabama BELS is tiltled Qualified by Education and/or Experience. Details
as follows:
The engineer or land surveyor shall act competently and use proper care in performing
engineering or land surveying services for clients or employers and shall act only in fields in
which qualified by education or experience as follows:
This is present in ASCE under canon 2 tiltles service with competence which says that
Engineers shall perform services only in areas of their competence.
This is covered in Engineering Council of India ethics code under article 2 named Responsibility
to Maintain High Standards of Professional Quality which include the following-
Third part in Alabama Bels is related to safeguarding and preservation of the confidences and
private information of clients and employers which is not present in ASCE but present in
Engineeering council of india(ECI) ethics code.
Cannon iii of Alabama BELS is named Confidences of Clients and Employers and is as follows -
This is present in ECI code of ethics as under article 3 as 3.9 (Employers’ Business Secrets)
which says that Engineers shall not disclose by any means, confidential information of the
employer or client, unless otherwise authorized
Next in Alabama Bels is titled practice;
The engineer or land surveyor shall endeavor to build a practice and professional reputation on
the merit of his or her services as follows:
(a) The engineer or land surveyor shall not offer, or promise to pay or deliver, directly or
indirectly, any commission, political contribution, gift, favor, gratuity, benefit or reward as an
inducement to secure any specific professional engineering or professional land surveying work
or assignment; providing and excepting, however, that an engineer or land surveyor may pay a
duly licensed employment agency its fee or commission for securing engineering or land
surveying employment in a salaried position.
(b) The engineer or land surveyo.
A small boy is lost coming down Mount Washington. The leader of the .pdf
A small boy is lost coming down Mount Washington. The leader of the search team estimates
that there is a probability p that he came down on the east side and a probability 1-p that he came
down on the west side. He has n people in his search team who will search independently and, if
the boy is on the side being searched, each member will find the boy with probability u.
Determine how he should divide the n people into two groups to search the two sides of the
mountain so that he will have the highest probability of finding the boy. How does this depend
on u?
Solution.
Address how cooperative binding contributes to SSB function during D.pdf
Address how cooperative binding contributes to SSB function during DNA replication. What is
the advantage of such binding for SSB activity?
Solution
Single-strand DNA-binding protein (SSB) is 178 amino acids long protein, found in E. coli, that
binds to single-stranded poly nucleotide strand of double helix DNA and prevents reannealing of
poly nucleotide strands and allow various aspects of DNA metabolism: replication,
recombination, and repair. As in E. Coli, the DNA replication initiate at Ori site. During
initiation, the helicase opens the double stranded DNA into single poly nucleotide strands,
eventually the SSB proteins bind to the seperated poly nucleotide strands and prevents
reannealing and facilitates sliding of DNA polymerase and DNA replication. Here binding of
single SSB promotes massive binding of SSB proteins to the poly nucleotide strand, called
cooperative binding. Such cooperative binding facilitates the uninterupted synthesis of new DNA
molecules..
A perfect left-sided binary tree is a binary tree where every intern.pdf
A perfect left-sided binary tree is a binary tree where every internal node has two children, and
every left child is a leaf. Write an algorithm, based on DFS or BFS, that takes as input an
undirected graph G, and returns whether or not G is (or can be viewed as) a perfect left-sided
binary tree. Analyze the time complexity of your algorithm.
Solution
bfs(G)
{
list L = empty
tree T = empty
choose a starting vertex x
search(x)
while(L nonempty)
remove edge (v,w) from start of L
if w not yet visited
{
add (v,w) to T
search(w)
}
}
dfs(G)
{
list L = empty
tree T = empty
choose a starting vertex x
search(x)
while(L nonempty)
remove edge (v,w) from end of L
if w not yet visited
{
add (v,w) to T
search(w)
}
}
search(vertex v)
{
visit(v);
for each edge (v,w)
add edge (v,w) to end of L
}
mport java.util.*;
public class BST > implements Iterable
{
public static void main(String[] args)
{
Integer[] a = {1,5,2,7,4};
BST bst = new BST();
for(Integer n : a) bst.insert(n);
bst.preOrderTraversal();
System.out.println();
//testing comparator
//build a mirror BST with a rule: Left > Parent > Right
//code for the comparator at the bottom of the file
bst = new BST(new MyComp1());
for(Integer n : a) bst.insert(n);
bst.preOrderTraversal();
System.out.println();
bst.inOrderTraversal();
System.out.println();
for(Integer n : bst) System.out.print(n);
System.out.println();
System.out.println(bst);
//testing restoring a tree from two given traversals
bst.restore(new Integer[] {11,8,6,4,7,10,19,43,31,29,37,49},
new Integer[] {4,6,7,8,10,11,19,29,31,37,43,49});
bst.preOrderTraversal();
System.out.println();
bst.inOrderTraversal();
System.out.println();
//testing diameter
System.out.println(\"diameter = \" + bst.diameter());
//testing width
System.out.println(\"width = \" + bst.width());
}
private Node root;
private Comparator comparator;
public BST()
{
root = null;
comparator = null;
}
public BST(Comparator comp)
{
root = null;
comparator = comp;
}
private int compare(T x, T y)
{
if(comparator == null) return x.compareTo(y);
else
return comparator.compare(x,y);
}
/*****************************************************
*
* INSERT
*
******************************************************/
public void insert(T data)
{
root = insert(root, data);
}
private Node insert(Node p, T toInsert)
{
if (p == null)
return new Node(toInsert);
if (compare(toInsert, p.data) == 0)
return p;
if (compare(toInsert, p.data) < 0)
p.left = insert(p.left, toInsert);
else
p.right = insert(p.right, toInsert);
return p;
}
/*****************************************************
*
* SEARCH
*
******************************************************/
public boolean search(T toSearch)
{
return search(root, toSearch);
}
private boolean search(Node p, T toSearch)
{
if (p == null)
return false;
else
if (compare(toSearch, p.data) == 0)
return true;
else
if (compare(toSearch, p.data) < 0)
return search(p.left, toSearch);
else
return search(p.right, toSearch);
}
/*****************************************************.
Assembly Code Pep8 Question Write a Pep8 subroutine that is equiva.pdf
Assembly Code Pep/8 Question Write a Pep/8 subroutine that is equivalent to the following C
function int Q2 (int A, int B) {int M, C = 0; M = (A + B)/2; while (M > 0) {C += 3; M--;}
return C;}
Solution
Answer:
MIPS Assembly Language Code :
L0:
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-20], edi
mov DWORD PTR [rbp-24], esi
mov DWORD PTR [rbp-8], 0
mov edx, DWORD PTR [rbp-20]
mov eax, DWORD PTR [rbp-24]
add eax, edx
mov edx, eax
shr edx, 31
add eax, edx
sar eax
mov DWORD PTR [rbp-4], eax
.L3:
cmp DWORD PTR [rbp-4], 0
jle .L2
add DWORD PTR [rbp-8], 3
sub DWORD PTR [rbp-4], 1
jmp .L3
.L2:
mov eax, DWORD PTR [rbp-8]
pop rbp
ret.
why might bacteria need to alter gene expression when their environm.pdf
why might bacteria need to alter gene expression when their environment changes?
Solution
Bacteria need to alter the gene expression according to the changing environment. As bacterial
gene is expressed in the form of operons which function according to the regulatory element,
which can either be inducer or repressor.
If inducer is present in the environment gene expression is promoted as inducer binds with
regulator and RNA Polymerase can bind to DNA, while if repressor is present in the
environment, gene expression is suppressed.
therefore bacteria alter their gene expression according to the change in environment..
Which of the three principles from the Belmont Report were violated .pdf
Which of the three principles from the Belmont Report were violated in the Milgram
Experiment?
Solution
Answer:
Two concerns that have been raised against Milgram\'s obedience studies were:
1. Respect for persons: Deception
The study itself was extremely stressful to the teacher-participants.
2. Beneficence: Psychological harm
There were lasting effects of the study. Some participants were dramatically affected by learning
that they were
willing to harm another human being, simply because someone in authority told them to do it..
Where is the cell body of the neuron that directly causes SA node ce.pdf
Where is the cell body of the neuron that directly causes SA node cells to depolarize faster?
Solution
The cell body of the nerve fibres which directly causes the depolarization of Sino Atrial node or
SA node is located in the parasympathetic and sympathetic nerves of autonomous nervous
system.
This depolarization is spontaneous because the SA node is responsible for the cintraction of all
the other muscles of the heart. Therefore the action potential which is generated at SA node
passes down the other pacemeker cells like AV nodes and conduction of impuslse throughout the
heart muscles..
Which of the following is FALSE regarding phylogenetic inference met.pdf
Which of the following is FALSE regarding phylogenetic inference methods? a. Biologists
cannot use statistical analysis in inferring phylogeny. b. The logic of tree building is that species
with many characteristics in common are more likely to be closely related to one another than are
species with fewer characteristics in common. c. Biologists select a number of species (or other
taxa) and collect information about the characteristics, looking at which taxa have which
characteristics in common. d. Most phylogenetic methods today rely on molecular data for tree
building, but prior to molecular sequencing technology, most phylogenetic distances were
computed from differences in other types of characters.
Solution
a. Biologists cannot use statistical analysis in inferring phylogeny. ---> FALSE (computational
phylogenetics do use statistical analysis in interfering the data).
b. The logic of tree building is that species with many characteristics in common are more likely
to be closely related to one another than are species with fewer characteristics in common. --->
TRUE
c. Biologists select a number of species (or other taxa) and collect information about the
characteristics, looking at which taxa have which characteristics in common. ---> TRUE
d. Most phylogenetic methods today rely on molecular data for tree building, but prior to
molecular sequencing technology, most phylogenetic distances were computed from differences
in other types of characters. ---> TRUE.
What is the critical F value for a sample of four observations in th.pdf
What is the critical F value for a sample of four observations in the numerator and seven in the
denominator? Use a one-tailed test and the .01 significance level. can you show step by step of
how you got the answer.
Solution
For a a sample of four observations in the numerator and seven in the denominator using a one-
tailed test and the .01 significance level the critical F value is:
F(4-1,7-1,0.01)
=F(3,6,0.01)
=9.780.
what is the main function and sub-functions of a penSolutions.pdf
what is the main function and sub-functions of a pen?
Solution
solution
main function
- main function of pen is to write anything.
sub functions
- to stab the eye of the school bully
- pens are easily disassembled. Take the outer body and use it as a spitwad shooter
- Stylus
- Pointer
- Trache-tube (artificial airway used in tracheotomies)
- Emergency Axel for Bike
- Snorkle
- Straw
- Candle-Holder.
what are some barriers a company might need to overcome when enterin.pdf
what are some barriers a company might need to overcome when entering the global market?
and
what are some sources of financing for a new business? Limit your post to one source, and
include some pros and cons.
Solution
Barriers to entry into markets for firms include:
· Advertising - Incumbent firms can seek to make it difficult for new competitors by spending
heavily on advertising that new firms would find more difficult to afford. This is known as the
market power theory of advertising. Here, established firms\' use of advertising creates a
consumer perceived difference in its brand from other brands to a degree that consumers see its
brand as a slightly different product. Since the brand is seen as a slightly different product,
products from existing or potential competitors cannot be perfectly substituted in place of the
established firm\'s brand. This makes it hard for new competitors to gain consumer acceptance.
· Capital - need the capital to start up such as equipment, building, and raw materials
· Control of resources - If a single firm has control of a resource essential for a certain industry,
and then other firms are unable to compete in the industry.
· Cost advantages independent of scale - Proprietary technology, know-how, favorable access to
raw materials, favorable geographic locations, learning curve cost advantages.
· Customer loyalty - Large incumbent firms may have existing customers loyal to established
products. The presence of established strong brands within a market can be a barrier to entry in
this case.
· Distributor agreements - Exclusive agreements with key distributors or retailers can make it
difficult for other manufacturers to enter the industry. · Economy of scale - The increase in
efficiency of production as the number of goods being produced increases. Cost advantages can
sometimes be quickly reversed by advances in technology. For example, the development of
personal computers has allowed small companies to make use of database and communications
technology which was once extremely expensive and only available to large corporations.
· Government regulations - A rule of order having the force of law, prescribed by a superior or
competent authority, relating to the actions of those under the authority\'s control. Requirements
for licenses and permits may raise the investment needed to enter a market, creating an effective
barrier to entry.
· Inelastic demand - One strategy to penetrate a market is to sell at a lower price than the
incumbents. This is ineffective with price-insensitive consumers.
· Intellectual property - Potential entrant requires access to equally efficient production
technology as the combatant monopolist in order to freely enter a market. Patents give a firm the
legal right to stop other firms producing a product for a given period of time, and so restrict entry
into a market. Patents are intended to encourage invention and technological progress by
guaranteeing proceeds as an incentive.
True or False1 The highest abundance of microbial life in soils .pdf
True or False??
1 The highest abundance of microbial life in soils appears in the O horizon.
2 Typically, increased oxygenic photosynthetic activity results in a decrease in pH values during
the day and an increase in pH values at night.
3 Freshwater represents about 10% of the total water on Earth.
4 Oligotrophic organisms thrive in low nutrient conditions.
5 Establishment of a proton motive force (PMF) is required to generate ATP via substrate level
phosphorylation.
6 Siderphores are used to increase availability of manganese in aerobic environments.
Solution
1. True
2. False, increased oxygenic photosynthetic activity will increase pH (become more basic) during
the day and during the night pH will decrease (become more acidic).
3. False, fresh water accounts for 2.5% to 2.75% of total water on earth.
4. True
5. False, Generation of ATP via substrate level phosphorylation is independent of PMF. ATP is
formed by the direct transfer and donation of phosphoryl group. It takes place in erythrocytes as
they have no mitochondria.
6. False, Siderphores are iron chelating compounds secreted by microorganisms used in the
bioremediation of iron rich environments..
This pedigree illustrates a heritable disease named geneticitis.1..pdf
This pedigree illustrates a heritable disease named geneticitis.
1. Which mode of inheritance does the pedigree illustrate?
2. What are the genotypes of all of the numbered family members according to your chosen
mode of inheritance? :O 2 TO 4
Solution
The mode of inheritance pattern is X Linked recessieve inheritance.
The genotypes is
X - Normal X chromosome
x - Defective X chromosome
1 - XY
2 - xY
3 - Xx
4 - Xx.
The number of key comparisons in a sequential search depends on the .pdf
The number of key comparisons in a sequential search depends on the value of the search item.
True
False
QUESTION 2
The structured design approach is also known as ____.
bottom-up design
top-down design
object design
QUESTION 3
The expression vecList.front() moves an element to the front of the vector.
True
False
QUESTION 4
Containers are essentially used ____.
to update objects that are part of a given set of elements
to manage objects of a given type
to manipulate data
QUESTION 5
The operation ____ reinitializes the stack to an empty state.
alloc
init
initializeStack
QUESTION 6
In a doubly linked list, some of the operations require modification from how they were
implemented for a regular linked list, because of the ____ pointer(s) in each node.
null
two
three
QUESTION 7
We can traverse a singly linked list backward starting from the last node.
True
False
QUESTION 8
A precondition is a statement specifying the condition(s) that must be true before the function is
called.
True
False
QUESTION 9
To remove, or pop, an element from the stack ____.
invert stackTop
decrement stackTop by 1
increment stackTop by 1
QUESTION 10
To define new classes, you create new ____ files.
header
friend
placeholder
QUESTION 11
A technique in which one system models the behavior of another system is called ____.
bench testing
simulation
referencing
QUESTION 12
With the help of the ____, two values can be combined into a single unit and, therefore, can be
treated as one unit.
class unit
class tuple
class double
class pair
QUESTION 13
A ____ is a list in which each element contains a key, such that the key in the element at position
k in the list is at least as large as the key in the element at position 2k + 1 (if it exists) and 2k + 2
(if it exists).
heap
hemp
hash table
QUESTION 14
Default arguments can be used with an overloaded operator.
True
False
QUESTION 15
The syntax for accessing a class (struct) member using the operator -> is ____.
pointerVariableName.classMemberName
pointerVariableName&->classMemberName
pointerVariableName->classMemberName
QUESTION 16
The components of a class are called the ____ of the class.
objects
members
operators
QUESTION 17
Operations, such as finding the height, determining the number of nodes, checking whether the
tree is empty, tree traversal, and so on, on AVL trees cannot be implemented the same way they
are implemented on binary trees.
True
False
QUESTION 18
The analysis of algorithms enables programmers to decide which algorithm to use for a specific
application.
True
False
QUESTION 19
A linked list is a collection of ____.
addresses
classes
nodes
QUESTION 20
In a ____, customers or jobs with higher priority are pushed to the front of the queue.
false queue
priority queue
free queue
double queue
QUESTION 21
When the destructor ____ the queue, it deallocates the memory occupied by the elements of the
queue.
instantiates
allocates
destroys
QUESTION 22
In breadth first traversal, starting at the first vertex, the graph is tra.
The Pirerras are planning to go to Europe 2 years from now and have .pdf
The Pirerras are planning to go to Europe 2 years from now and have agreed to set aside
$170/month for their trip. If they deposit this money at the end of each month into a savings
account paying interest at the rate of 2%/year compounded monthly, how much money will be in
their travel fund at the end of the second year? (Round your answer to the nearest cent.)
Solution
FVoa = PMT [(((1 + i)^n) - 1) / i]
Where:
FVoa = Future Value of an Ordinary Annuity
PMT = Amount of each payment (170)
i = Interest Rate Per Period (0.02)
n = Number of Periods (2 x 12)=24
$5171.72.
51.Ingestion of material through pinocytosis can be found in (sele.pdf
51.
Ingestion of material through pinocytosis can be found in (select all that apply)
cells that line the kidney tubules
cells that line the intestinal cell walls
red blood cells
macrophages
52.
The sodium potassium pump
requires the expenditure of energy in the form of ATP
is an example of a facilitated diffusion carrier protien
Moves sodium against its concentration gradient
Moves potassium against its concentration gradient
53.
Facilitated diffusion and active transport processes are similar in that
both require the expenditure of cellular energy for the transport to occur
require the function of membrane transport proteins
both typically refer to the transport of material along its concentration gradient
both typically refer to the transport of material against its concentration gradient
57.
Bulk transport includes
endocytosis
exocytosis
the sodium potassium pump
both A and B
both B and CA.
cells that line the kidney tubulesB.
cells that line the intestinal cell wallsC.
red blood cellsD.
macrophages
Solution
51.A,B,D(Small proteins are reabsorbed by pinocytosis, digested, and the amino acids diffuse
into the blood by proximal convoluted tubule.Nutrientss are absorbed by microvilli of intestinal
wall by pinocytosis. Endocytosis is used by the macrophage to engulf a pathogen.)
52.A,B,C,D(done by active transport process involving the hydrolysis of ATP to provide the
necessary energy. The sodium potassium pump is an important and well understood example of
active transport in which move sodium ions out of the cell and potassium ions into the cell
against concentration gradient using specialised carrier protein.).
53.B(facilitated diffusion allows substances to follow the concentration gradient either way,
while active transport only has substances go one way, against their concentration
gradient.Active transport needs energy, while facilitated diffusion does not need energy.)
57.D(Both are two types of bulk transport;they transport ions and other substances in bulk).
The local city youth league needs a database system to help track ch.pdf
The local city youth league needs a database system to help track children that sign up to play
soccer. Data needs to be kept on each team and the children that will be playing on each team
and their parents. Also, data needs to be kept on the coaches for each team.
A. Entities required:
?- Team
?- Player
?- Coach
? -Parent
B. Attributes required:
? -Team: Team ID number, Team name, and Team colors.
?- Player: Player ID number, Player first name, Player last name, and Player age.
?- Coach: Coach ID number, Coach first name, Coach last name, and Coach home phone
number.
?- Parent: Parent ID number, Parent last name, Parent first name, Home phone number, and
Home Address (Street, City, State, and ZIP Code).
C. The following relationships must be defined:
? -Team is related to Player.
? -Team is related to Coach.
? -Player is related to Parent.
D. Connectivities (cardinalities) and participations are defined as follows:
?- A Team may or may not have a Player.
?- A Player must have a Team.
?- A Team may have many Players.
? -A Player has only one Team.
? -A Team may or may not have a Coach.
? -A Coach must have a Team.
?- A Team may have many Coaches.
?- A Coach has only one Team.
?- A Player must have a Parent.
?- A Parent must have a Player.
?- A Player may have many Parents.
?- A Parent may have many Players.
Solution
A. Entities required: Team
B. Attributes required: Team: Team ID number, Team name, and Team colors.
C. The following relationships must be defined: Team is related to Player.
D. Connectivities (cardinalities) and participations are defined as follows: A Team may or may
not have a Coach..
The idea that, when life originated on Earth, a macromolecule other .pdf
The idea that, when life originated on Earth, a macromolecule other than DNA served the role of
information storage and that this same macromolecule carried out catalysis is called the _____
hypothesis.
1) macromolecular world
2) water world
3) carbon world
4) RNA world
The idea that, when life originated on Earth, a macromolecule other than DNA served the role of
information storage and that this same macromolecule carried out catalysis is called the _____
hypothesis.
1) macromolecular world
2) water world
3) carbon world
4) RNA world
Solution
4) RNA world
In 1986, the RNA World Hypothesis was proposed by Walter Gilbert, on the possibility of RNA
self-replication.
However, the ability of the RNA molecules to catalyze the synthesis of another identical RNA is
questioned. Many of the ribozymes can catalyze either the formation or breakup of
phosphodiester bonds. In prokaryotes, the 23S and in eukaryotes, the 28S large subunits of rRNA
are ribozymes, which catalyze the formation of peptide bonds.
Many of the naturally occurring ribozymes can catalyze only one reaction, for example the self-
splicing introns. The properties these ribozymes are far away from the expectations of RNA
World Hypothesis. Perhaps, the self-replicating RNA was existed during the times of evolution
but was replaced by the more stable DNA molecules..
The amount of DNA per cell of a particular species is measured in cel.pdf
The amount of DNA per cell of a particular species is measured in cells found at various stages
of meiosis and the cell cycle, and the following amounts are obtained: 3.7 pg, 7.4 pg, and 14.6
pg. Cells that contain 3.7 pg of DNA are likely found in which of the following stages? G1 G2
Prophase I Following telophase II and cytokinesis Anaphase I The amount of DNA per cell of
a particular species is measured in cells found at various stages of meiosis and the cell cycle, and
the following amounts are obtained: 3.7 pg, 7.4 pg, and 14.8 pg. Cells that contain 7.4 pg of
DNA are likely found in which of the following stages? G1 G2 Prophase I Following
telophase II and cytokinesis Anaphase I
Solution
After the completion of replication and prior to cytokinesis, a cell will have double the amount of
DNA. Suppose a normal cell had X pg of DNA in its resting/G0 phase, then at the end of
replication and before cytokinesis it will have 2X pg of DNA. The contents of the cells are
halved at the end of meiosis I, therefore now, each cell would again have X pg of DNA (2X/2=
X). However at the end of meiosis II, a single cell would have divided to 4 cells, hence each cell
will contain X/2 pg of DNA.
At G1 phase, no replication of DNA occurs. Therefore the amount of DNA at this phase would
be normal. Hence the phase with 7.4 pg of DNA would be the G1 phase.
At G2 phase, the amount of DNA is double as it occurrs right after the S pahse. Hence, this phase
will have 14.8 pg of DNA.
At Prophase I, the amount of DNA is double the normal amount since cytokinesis has not yet
occurred. Therefore, this phase too will have 14.8 pg of DNA.
Following Telophase II and cytokinesis, each cell would have been divided into 4 cells. So each
cell will only contain X/2 pg of DNA. Hence, the cells at this phase will have 3.7 pg of DNA.
At Anaphase I, the amount of DNA is double the normal amount since cytokinesis has not yet
occurred. Therefore, this phase too will have 14.8 pg of DNA..
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Cambridge International AS A Level Biology Coursebook - EBook (MaryFosbery J...
for studentd in cabridge board
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After reading the codes of ethics posted to the USAOnline site please.pdf
After reading the codes of ethics posted to the USAOnline site please do the following (when
you are asked to compare, consider the following: the length and details in each one, the topics
covered, what are your thoughts on what is included in each): 1) Compare the Alabama Board of
Licensure for Professional Engineers and Professional Land Surveyors (Alabama BELS) with
your primary engineering societies Code of Ethics (i.e. if you are a civil it is ASCE, if you are an
Electrical it is IEEE, etc.) 2) Compare your primary engineering societies Code of Ethics with
ASCE\'s Code of Ethics (if you are a civil compare ASCE with any of the other societies)
Solution
1) First topic in Alabama BELS says about conflict of interest and is as follows-
This topic is also present in ASCE ethics and is as follows -
Now, this is present (highlighted in bold below) in Engineering council of India - code of ethics
under article 3 titled Obligation to Maintain High Standard of Personal Behaviour in a
Responsible Manner which is as follows -
The second article in Alabama BELS is tiltled Qualified by Education and/or Experience. Details
as follows:
The engineer or land surveyor shall act competently and use proper care in performing
engineering or land surveying services for clients or employers and shall act only in fields in
which qualified by education or experience as follows:
This is present in ASCE under canon 2 tiltles service with competence which says that
Engineers shall perform services only in areas of their competence.
This is covered in Engineering Council of India ethics code under article 2 named Responsibility
to Maintain High Standards of Professional Quality which include the following-
Third part in Alabama Bels is related to safeguarding and preservation of the confidences and
private information of clients and employers which is not present in ASCE but present in
Engineeering council of india(ECI) ethics code.
Cannon iii of Alabama BELS is named Confidences of Clients and Employers and is as follows -
This is present in ECI code of ethics as under article 3 as 3.9 (Employers’ Business Secrets)
which says that Engineers shall not disclose by any means, confidential information of the
employer or client, unless otherwise authorized
Next in Alabama Bels is titled practice;
The engineer or land surveyor shall endeavor to build a practice and professional reputation on
the merit of his or her services as follows:
(a) The engineer or land surveyor shall not offer, or promise to pay or deliver, directly or
indirectly, any commission, political contribution, gift, favor, gratuity, benefit or reward as an
inducement to secure any specific professional engineering or professional land surveying work
or assignment; providing and excepting, however, that an engineer or land surveyor may pay a
duly licensed employment agency its fee or commission for securing engineering or land
surveying employment in a salaried position.
(b) The engineer or land surveyo.
A small boy is lost coming down Mount Washington. The leader of the .pdf
A small boy is lost coming down Mount Washington. The leader of the search team estimates
that there is a probability p that he came down on the east side and a probability 1-p that he came
down on the west side. He has n people in his search team who will search independently and, if
the boy is on the side being searched, each member will find the boy with probability u.
Determine how he should divide the n people into two groups to search the two sides of the
mountain so that he will have the highest probability of finding the boy. How does this depend
on u?
Solution.
Address how cooperative binding contributes to SSB function during D.pdf
Address how cooperative binding contributes to SSB function during DNA replication. What is
the advantage of such binding for SSB activity?
Solution
Single-strand DNA-binding protein (SSB) is 178 amino acids long protein, found in E. coli, that
binds to single-stranded poly nucleotide strand of double helix DNA and prevents reannealing of
poly nucleotide strands and allow various aspects of DNA metabolism: replication,
recombination, and repair. As in E. Coli, the DNA replication initiate at Ori site. During
initiation, the helicase opens the double stranded DNA into single poly nucleotide strands,
eventually the SSB proteins bind to the seperated poly nucleotide strands and prevents
reannealing and facilitates sliding of DNA polymerase and DNA replication. Here binding of
single SSB promotes massive binding of SSB proteins to the poly nucleotide strand, called
cooperative binding. Such cooperative binding facilitates the uninterupted synthesis of new DNA
molecules..
A perfect left-sided binary tree is a binary tree where every intern.pdf
A perfect left-sided binary tree is a binary tree where every internal node has two children, and
every left child is a leaf. Write an algorithm, based on DFS or BFS, that takes as input an
undirected graph G, and returns whether or not G is (or can be viewed as) a perfect left-sided
binary tree. Analyze the time complexity of your algorithm.
Solution
bfs(G)
{
list L = empty
tree T = empty
choose a starting vertex x
search(x)
while(L nonempty)
remove edge (v,w) from start of L
if w not yet visited
{
add (v,w) to T
search(w)
}
}
dfs(G)
{
list L = empty
tree T = empty
choose a starting vertex x
search(x)
while(L nonempty)
remove edge (v,w) from end of L
if w not yet visited
{
add (v,w) to T
search(w)
}
}
search(vertex v)
{
visit(v);
for each edge (v,w)
add edge (v,w) to end of L
}
mport java.util.*;
public class BST > implements Iterable
{
public static void main(String[] args)
{
Integer[] a = {1,5,2,7,4};
BST bst = new BST();
for(Integer n : a) bst.insert(n);
bst.preOrderTraversal();
System.out.println();
//testing comparator
//build a mirror BST with a rule: Left > Parent > Right
//code for the comparator at the bottom of the file
bst = new BST(new MyComp1());
for(Integer n : a) bst.insert(n);
bst.preOrderTraversal();
System.out.println();
bst.inOrderTraversal();
System.out.println();
for(Integer n : bst) System.out.print(n);
System.out.println();
System.out.println(bst);
//testing restoring a tree from two given traversals
bst.restore(new Integer[] {11,8,6,4,7,10,19,43,31,29,37,49},
new Integer[] {4,6,7,8,10,11,19,29,31,37,43,49});
bst.preOrderTraversal();
System.out.println();
bst.inOrderTraversal();
System.out.println();
//testing diameter
System.out.println(\"diameter = \" + bst.diameter());
//testing width
System.out.println(\"width = \" + bst.width());
}
private Node root;
private Comparator comparator;
public BST()
{
root = null;
comparator = null;
}
public BST(Comparator comp)
{
root = null;
comparator = comp;
}
private int compare(T x, T y)
{
if(comparator == null) return x.compareTo(y);
else
return comparator.compare(x,y);
}
/*****************************************************
*
* INSERT
*
******************************************************/
public void insert(T data)
{
root = insert(root, data);
}
private Node insert(Node p, T toInsert)
{
if (p == null)
return new Node(toInsert);
if (compare(toInsert, p.data) == 0)
return p;
if (compare(toInsert, p.data) < 0)
p.left = insert(p.left, toInsert);
else
p.right = insert(p.right, toInsert);
return p;
}
/*****************************************************
*
* SEARCH
*
******************************************************/
public boolean search(T toSearch)
{
return search(root, toSearch);
}
private boolean search(Node p, T toSearch)
{
if (p == null)
return false;
else
if (compare(toSearch, p.data) == 0)
return true;
else
if (compare(toSearch, p.data) < 0)
return search(p.left, toSearch);
else
return search(p.right, toSearch);
}
/*****************************************************.
Assembly Code Pep8 Question Write a Pep8 subroutine that is equiva.pdf
Assembly Code Pep/8 Question Write a Pep/8 subroutine that is equivalent to the following C
function int Q2 (int A, int B) {int M, C = 0; M = (A + B)/2; while (M > 0) {C += 3; M--;}
return C;}
Solution
Answer:
MIPS Assembly Language Code :
L0:
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-20], edi
mov DWORD PTR [rbp-24], esi
mov DWORD PTR [rbp-8], 0
mov edx, DWORD PTR [rbp-20]
mov eax, DWORD PTR [rbp-24]
add eax, edx
mov edx, eax
shr edx, 31
add eax, edx
sar eax
mov DWORD PTR [rbp-4], eax
.L3:
cmp DWORD PTR [rbp-4], 0
jle .L2
add DWORD PTR [rbp-8], 3
sub DWORD PTR [rbp-4], 1
jmp .L3
.L2:
mov eax, DWORD PTR [rbp-8]
pop rbp
ret.
why might bacteria need to alter gene expression when their environm.pdf
why might bacteria need to alter gene expression when their environment changes?
Solution
Bacteria need to alter the gene expression according to the changing environment. As bacterial
gene is expressed in the form of operons which function according to the regulatory element,
which can either be inducer or repressor.
If inducer is present in the environment gene expression is promoted as inducer binds with
regulator and RNA Polymerase can bind to DNA, while if repressor is present in the
environment, gene expression is suppressed.
therefore bacteria alter their gene expression according to the change in environment..
Which of the three principles from the Belmont Report were violated .pdf
Which of the three principles from the Belmont Report were violated in the Milgram
Experiment?
Solution
Answer:
Two concerns that have been raised against Milgram\'s obedience studies were:
1. Respect for persons: Deception
The study itself was extremely stressful to the teacher-participants.
2. Beneficence: Psychological harm
There were lasting effects of the study. Some participants were dramatically affected by learning
that they were
willing to harm another human being, simply because someone in authority told them to do it..
Where is the cell body of the neuron that directly causes SA node ce.pdf
Where is the cell body of the neuron that directly causes SA node cells to depolarize faster?
Solution
The cell body of the nerve fibres which directly causes the depolarization of Sino Atrial node or
SA node is located in the parasympathetic and sympathetic nerves of autonomous nervous
system.
This depolarization is spontaneous because the SA node is responsible for the cintraction of all
the other muscles of the heart. Therefore the action potential which is generated at SA node
passes down the other pacemeker cells like AV nodes and conduction of impuslse throughout the
heart muscles..
Which of the following is FALSE regarding phylogenetic inference met.pdf
Which of the following is FALSE regarding phylogenetic inference methods? a. Biologists
cannot use statistical analysis in inferring phylogeny. b. The logic of tree building is that species
with many characteristics in common are more likely to be closely related to one another than are
species with fewer characteristics in common. c. Biologists select a number of species (or other
taxa) and collect information about the characteristics, looking at which taxa have which
characteristics in common. d. Most phylogenetic methods today rely on molecular data for tree
building, but prior to molecular sequencing technology, most phylogenetic distances were
computed from differences in other types of characters.
Solution
a. Biologists cannot use statistical analysis in inferring phylogeny. ---> FALSE (computational
phylogenetics do use statistical analysis in interfering the data).
b. The logic of tree building is that species with many characteristics in common are more likely
to be closely related to one another than are species with fewer characteristics in common. --->
TRUE
c. Biologists select a number of species (or other taxa) and collect information about the
characteristics, looking at which taxa have which characteristics in common. ---> TRUE
d. Most phylogenetic methods today rely on molecular data for tree building, but prior to
molecular sequencing technology, most phylogenetic distances were computed from differences
in other types of characters. ---> TRUE.
What is the critical F value for a sample of four observations in th.pdf
What is the critical F value for a sample of four observations in the numerator and seven in the
denominator? Use a one-tailed test and the .01 significance level. can you show step by step of
how you got the answer.
Solution
For a a sample of four observations in the numerator and seven in the denominator using a one-
tailed test and the .01 significance level the critical F value is:
F(4-1,7-1,0.01)
=F(3,6,0.01)
=9.780.
what is the main function and sub-functions of a penSolutions.pdf
what is the main function and sub-functions of a pen?
Solution
solution
main function
- main function of pen is to write anything.
sub functions
- to stab the eye of the school bully
- pens are easily disassembled. Take the outer body and use it as a spitwad shooter
- Stylus
- Pointer
- Trache-tube (artificial airway used in tracheotomies)
- Emergency Axel for Bike
- Snorkle
- Straw
- Candle-Holder.
what are some barriers a company might need to overcome when enterin.pdf
what are some barriers a company might need to overcome when entering the global market?
and
what are some sources of financing for a new business? Limit your post to one source, and
include some pros and cons.
Solution
Barriers to entry into markets for firms include:
· Advertising - Incumbent firms can seek to make it difficult for new competitors by spending
heavily on advertising that new firms would find more difficult to afford. This is known as the
market power theory of advertising. Here, established firms\' use of advertising creates a
consumer perceived difference in its brand from other brands to a degree that consumers see its
brand as a slightly different product. Since the brand is seen as a slightly different product,
products from existing or potential competitors cannot be perfectly substituted in place of the
established firm\'s brand. This makes it hard for new competitors to gain consumer acceptance.
· Capital - need the capital to start up such as equipment, building, and raw materials
· Control of resources - If a single firm has control of a resource essential for a certain industry,
and then other firms are unable to compete in the industry.
· Cost advantages independent of scale - Proprietary technology, know-how, favorable access to
raw materials, favorable geographic locations, learning curve cost advantages.
· Customer loyalty - Large incumbent firms may have existing customers loyal to established
products. The presence of established strong brands within a market can be a barrier to entry in
this case.
· Distributor agreements - Exclusive agreements with key distributors or retailers can make it
difficult for other manufacturers to enter the industry. · Economy of scale - The increase in
efficiency of production as the number of goods being produced increases. Cost advantages can
sometimes be quickly reversed by advances in technology. For example, the development of
personal computers has allowed small companies to make use of database and communications
technology which was once extremely expensive and only available to large corporations.
· Government regulations - A rule of order having the force of law, prescribed by a superior or
competent authority, relating to the actions of those under the authority\'s control. Requirements
for licenses and permits may raise the investment needed to enter a market, creating an effective
barrier to entry.
· Inelastic demand - One strategy to penetrate a market is to sell at a lower price than the
incumbents. This is ineffective with price-insensitive consumers.
· Intellectual property - Potential entrant requires access to equally efficient production
technology as the combatant monopolist in order to freely enter a market. Patents give a firm the
legal right to stop other firms producing a product for a given period of time, and so restrict entry
into a market. Patents are intended to encourage invention and technological progress by
guaranteeing proceeds as an incentive.
True or False1 The highest abundance of microbial life in soils .pdf
True or False??
1 The highest abundance of microbial life in soils appears in the O horizon.
2 Typically, increased oxygenic photosynthetic activity results in a decrease in pH values during
the day and an increase in pH values at night.
3 Freshwater represents about 10% of the total water on Earth.
4 Oligotrophic organisms thrive in low nutrient conditions.
5 Establishment of a proton motive force (PMF) is required to generate ATP via substrate level
phosphorylation.
6 Siderphores are used to increase availability of manganese in aerobic environments.
Solution
1. True
2. False, increased oxygenic photosynthetic activity will increase pH (become more basic) during
the day and during the night pH will decrease (become more acidic).
3. False, fresh water accounts for 2.5% to 2.75% of total water on earth.
4. True
5. False, Generation of ATP via substrate level phosphorylation is independent of PMF. ATP is
formed by the direct transfer and donation of phosphoryl group. It takes place in erythrocytes as
they have no mitochondria.
6. False, Siderphores are iron chelating compounds secreted by microorganisms used in the
bioremediation of iron rich environments..
This pedigree illustrates a heritable disease named geneticitis.1..pdf
This pedigree illustrates a heritable disease named geneticitis.
1. Which mode of inheritance does the pedigree illustrate?
2. What are the genotypes of all of the numbered family members according to your chosen
mode of inheritance? :O 2 TO 4
Solution
The mode of inheritance pattern is X Linked recessieve inheritance.
The genotypes is
X - Normal X chromosome
x - Defective X chromosome
1 - XY
2 - xY
3 - Xx
4 - Xx.
The number of key comparisons in a sequential search depends on the .pdf
The number of key comparisons in a sequential search depends on the value of the search item.
True
False
QUESTION 2
The structured design approach is also known as ____.
bottom-up design
top-down design
object design
QUESTION 3
The expression vecList.front() moves an element to the front of the vector.
True
False
QUESTION 4
Containers are essentially used ____.
to update objects that are part of a given set of elements
to manage objects of a given type
to manipulate data
QUESTION 5
The operation ____ reinitializes the stack to an empty state.
alloc
init
initializeStack
QUESTION 6
In a doubly linked list, some of the operations require modification from how they were
implemented for a regular linked list, because of the ____ pointer(s) in each node.
null
two
three
QUESTION 7
We can traverse a singly linked list backward starting from the last node.
True
False
QUESTION 8
A precondition is a statement specifying the condition(s) that must be true before the function is
called.
True
False
QUESTION 9
To remove, or pop, an element from the stack ____.
invert stackTop
decrement stackTop by 1
increment stackTop by 1
QUESTION 10
To define new classes, you create new ____ files.
header
friend
placeholder
QUESTION 11
A technique in which one system models the behavior of another system is called ____.
bench testing
simulation
referencing
QUESTION 12
With the help of the ____, two values can be combined into a single unit and, therefore, can be
treated as one unit.
class unit
class tuple
class double
class pair
QUESTION 13
A ____ is a list in which each element contains a key, such that the key in the element at position
k in the list is at least as large as the key in the element at position 2k + 1 (if it exists) and 2k + 2
(if it exists).
heap
hemp
hash table
QUESTION 14
Default arguments can be used with an overloaded operator.
True
False
QUESTION 15
The syntax for accessing a class (struct) member using the operator -> is ____.
pointerVariableName.classMemberName
pointerVariableName&->classMemberName
pointerVariableName->classMemberName
QUESTION 16
The components of a class are called the ____ of the class.
objects
members
operators
QUESTION 17
Operations, such as finding the height, determining the number of nodes, checking whether the
tree is empty, tree traversal, and so on, on AVL trees cannot be implemented the same way they
are implemented on binary trees.
True
False
QUESTION 18
The analysis of algorithms enables programmers to decide which algorithm to use for a specific
application.
True
False
QUESTION 19
A linked list is a collection of ____.
addresses
classes
nodes
QUESTION 20
In a ____, customers or jobs with higher priority are pushed to the front of the queue.
false queue
priority queue
free queue
double queue
QUESTION 21
When the destructor ____ the queue, it deallocates the memory occupied by the elements of the
queue.
instantiates
allocates
destroys
QUESTION 22
In breadth first traversal, starting at the first vertex, the graph is tra.
The Pirerras are planning to go to Europe 2 years from now and have .pdf
The Pirerras are planning to go to Europe 2 years from now and have agreed to set aside
$170/month for their trip. If they deposit this money at the end of each month into a savings
account paying interest at the rate of 2%/year compounded monthly, how much money will be in
their travel fund at the end of the second year? (Round your answer to the nearest cent.)
Solution
FVoa = PMT [(((1 + i)^n) - 1) / i]
Where:
FVoa = Future Value of an Ordinary Annuity
PMT = Amount of each payment (170)
i = Interest Rate Per Period (0.02)
n = Number of Periods (2 x 12)=24
$5171.72.
51.Ingestion of material through pinocytosis can be found in (sele.pdf
51.
Ingestion of material through pinocytosis can be found in (select all that apply)
cells that line the kidney tubules
cells that line the intestinal cell walls
red blood cells
macrophages
52.
The sodium potassium pump
requires the expenditure of energy in the form of ATP
is an example of a facilitated diffusion carrier protien
Moves sodium against its concentration gradient
Moves potassium against its concentration gradient
53.
Facilitated diffusion and active transport processes are similar in that
both require the expenditure of cellular energy for the transport to occur
require the function of membrane transport proteins
both typically refer to the transport of material along its concentration gradient
both typically refer to the transport of material against its concentration gradient
57.
Bulk transport includes
endocytosis
exocytosis
the sodium potassium pump
both A and B
both B and CA.
cells that line the kidney tubulesB.
cells that line the intestinal cell wallsC.
red blood cellsD.
macrophages
Solution
51.A,B,D(Small proteins are reabsorbed by pinocytosis, digested, and the amino acids diffuse
into the blood by proximal convoluted tubule.Nutrientss are absorbed by microvilli of intestinal
wall by pinocytosis. Endocytosis is used by the macrophage to engulf a pathogen.)
52.A,B,C,D(done by active transport process involving the hydrolysis of ATP to provide the
necessary energy. The sodium potassium pump is an important and well understood example of
active transport in which move sodium ions out of the cell and potassium ions into the cell
against concentration gradient using specialised carrier protein.).
53.B(facilitated diffusion allows substances to follow the concentration gradient either way,
while active transport only has substances go one way, against their concentration
gradient.Active transport needs energy, while facilitated diffusion does not need energy.)
57.D(Both are two types of bulk transport;they transport ions and other substances in bulk).
The local city youth league needs a database system to help track ch.pdf
The local city youth league needs a database system to help track children that sign up to play
soccer. Data needs to be kept on each team and the children that will be playing on each team
and their parents. Also, data needs to be kept on the coaches for each team.
A. Entities required:
?- Team
?- Player
?- Coach
? -Parent
B. Attributes required:
? -Team: Team ID number, Team name, and Team colors.
?- Player: Player ID number, Player first name, Player last name, and Player age.
?- Coach: Coach ID number, Coach first name, Coach last name, and Coach home phone
number.
?- Parent: Parent ID number, Parent last name, Parent first name, Home phone number, and
Home Address (Street, City, State, and ZIP Code).
C. The following relationships must be defined:
? -Team is related to Player.
? -Team is related to Coach.
? -Player is related to Parent.
D. Connectivities (cardinalities) and participations are defined as follows:
?- A Team may or may not have a Player.
?- A Player must have a Team.
?- A Team may have many Players.
? -A Player has only one Team.
? -A Team may or may not have a Coach.
? -A Coach must have a Team.
?- A Team may have many Coaches.
?- A Coach has only one Team.
?- A Player must have a Parent.
?- A Parent must have a Player.
?- A Player may have many Parents.
?- A Parent may have many Players.
Solution
A. Entities required: Team
B. Attributes required: Team: Team ID number, Team name, and Team colors.
C. The following relationships must be defined: Team is related to Player.
D. Connectivities (cardinalities) and participations are defined as follows: A Team may or may
not have a Coach..
The idea that, when life originated on Earth, a macromolecule other .pdf
The idea that, when life originated on Earth, a macromolecule other than DNA served the role of
information storage and that this same macromolecule carried out catalysis is called the _____
hypothesis.
1) macromolecular world
2) water world
3) carbon world
4) RNA world
The idea that, when life originated on Earth, a macromolecule other than DNA served the role of
information storage and that this same macromolecule carried out catalysis is called the _____
hypothesis.
1) macromolecular world
2) water world
3) carbon world
4) RNA world
Solution
4) RNA world
In 1986, the RNA World Hypothesis was proposed by Walter Gilbert, on the possibility of RNA
self-replication.
However, the ability of the RNA molecules to catalyze the synthesis of another identical RNA is
questioned. Many of the ribozymes can catalyze either the formation or breakup of
phosphodiester bonds. In prokaryotes, the 23S and in eukaryotes, the 28S large subunits of rRNA
are ribozymes, which catalyze the formation of peptide bonds.
Many of the naturally occurring ribozymes can catalyze only one reaction, for example the self-
splicing introns. The properties these ribozymes are far away from the expectations of RNA
World Hypothesis. Perhaps, the self-replicating RNA was existed during the times of evolution
but was replaced by the more stable DNA molecules..
The amount of DNA per cell of a particular species is measured in cel.pdf
The amount of DNA per cell of a particular species is measured in cells found at various stages
of meiosis and the cell cycle, and the following amounts are obtained: 3.7 pg, 7.4 pg, and 14.6
pg. Cells that contain 3.7 pg of DNA are likely found in which of the following stages? G1 G2
Prophase I Following telophase II and cytokinesis Anaphase I The amount of DNA per cell of
a particular species is measured in cells found at various stages of meiosis and the cell cycle, and
the following amounts are obtained: 3.7 pg, 7.4 pg, and 14.8 pg. Cells that contain 7.4 pg of
DNA are likely found in which of the following stages? G1 G2 Prophase I Following
telophase II and cytokinesis Anaphase I
Solution
After the completion of replication and prior to cytokinesis, a cell will have double the amount of
DNA. Suppose a normal cell had X pg of DNA in its resting/G0 phase, then at the end of
replication and before cytokinesis it will have 2X pg of DNA. The contents of the cells are
halved at the end of meiosis I, therefore now, each cell would again have X pg of DNA (2X/2=
X). However at the end of meiosis II, a single cell would have divided to 4 cells, hence each cell
will contain X/2 pg of DNA.
At G1 phase, no replication of DNA occurs. Therefore the amount of DNA at this phase would
be normal. Hence the phase with 7.4 pg of DNA would be the G1 phase.
At G2 phase, the amount of DNA is double as it occurrs right after the S pahse. Hence, this phase
will have 14.8 pg of DNA.
At Prophase I, the amount of DNA is double the normal amount since cytokinesis has not yet
occurred. Therefore, this phase too will have 14.8 pg of DNA.
Following Telophase II and cytokinesis, each cell would have been divided into 4 cells. So each
cell will only contain X/2 pg of DNA. Hence, the cells at this phase will have 3.7 pg of DNA.
At Anaphase I, the amount of DNA is double the normal amount since cytokinesis has not yet
occurred. Therefore, this phase too will have 14.8 pg of DNA..
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