Progeny of triploid tomato plants often contain parts of an extra ch.pdf
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Progeny of triploid tomato plants often contain parts of an extra chromosome, in addition to the normal complement of 24 chromosomes. Mutants with a part of an extra chromosome are referred to as secondaries. James and Margaret Lesley observed that secondaries arise from triploid (3n), trisomic (3n 1), and double trisomic (3n 1 1) parents, but never from diploids (2n). (J. W. Lesley and M. M. Lesley. 1929. Genetics 14:321–336). Triploid and trisomic plants are vulnerable to infection by the tomato mosaic O virus, so they acquire extra chromosomal material by the integration of viral DNA, whereas diploid plants are not infected by the virus. The trivalent chromosomes would still be able to associate through c prior to segregation during meiosis. asma Triploid and trisomic conditions in tomato plants are unnatural, and this genomic imbalance leads to higher levels of mitotic recombination. Due to some unknown mechanism, triploid and trisomic tomato plants O occasionally over-replicate part of a chromosome, which leads to the generation of mutant tomato plants. Solution The second option is correct. The secondary chromosomes are mainly due to abnormalities in Anaphase segregatio, Which is popularly called Non Disjunction. So answer is \"THE TRIVIAL CHROMOSOMES WOULD STILL BE ABLE TO ASSOCIATES THROUGH CHIASMA PRIOR TO SEGREGATION DURING MEIOSIS.
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What is the possible reason that secondaries arise from parents that.pdf
What is the possible reason that secondaries arise from parents that have unpaired chromosomes
but not from parents that are normal diploids? Map Genetics: A Conceptual Approach 6th
Edition Pierce MHE/Freeman presented by Sapling Learning Progeny of triploid tomato plants
often contain parts of an extra chromosome, in addition to the normal complement of 24
chromosomes. Mutants with a part of an extra chromosome are referred to as secondaries. James
and Margaret Lesley observed that secondaries arise from triploid (3n), trisomic (3n 1), and
double trisomic (3n 1 1) parents, but never from diploids (2n). (J. W. Lesley and M. M. Lesley.
1929. Genetics 14:321-336) What is a possible reason that secondaries arise from parents that
have unpaired chromosomes but not from parents that are normal diploids? Triploid and trisomic
conditions in tomato plants are unnatural, and this genomic imbalance leads to higher levels of
mitotic recombination. The trivalent chromosomes would still be able to associate through
chiasma prior to segregation during meiosis. Due to some unknown mechanism, triploid and
trisom c tomato plants O occasionally over-replicate part of a chromosome, which leads to the
generation of mutant tomato plants. Triploid and trisomic plants are vulnerable to infection by
the tomato mosaic O virus, so they acquire extra chromosomal material by the integration of
viral DNA, whereas diploid plants are not infected by the virus.
Solution
Ans: B
2n+1 condition is called trisomic in which an organism contains two complete genome + 1 extra
chromosome. During the process of meiosis 1 chromosomal page separate so that 1 members of
each state is distributed to one daughter nucleus and other members goes to other daughter
nucleus but rarely one pair of chromosomes fail to separate and moves to the pole giving half of
the daughter cells extra chromosome and remaining half lost one chromosome. As a result n - 1
and n+ 1 chromosomes appear. When the gamut with n+ 1 chromosome fuses with normal
gamut the resulting zygote will have 2n + 1 chromosomes producing trisomics organism.
It was found that trisomies or white study nature and studied in plants like tomatoes tobacco
wheat datura and so on..
You are presented with a patient, named Gerry, who has Down syndrome .pdf
You are presented with a patient, named Gerry, who has Down syndrome (trisomy 21). You
determine that the nondisjunction event occurred in his mother, named Mary, during meiosis I.
You are interested in the TPTE gene, which is near the centromere of chromosome 21. Virtually
no recombination occurs between the centromere and the TPTE gene. Mary has two different
alleles of the TPTE gene. Would Gerry\'s two chromosome 21s from his mother have the same
or different alleles of the TPTE gene? Or do you not have enough information to decide?
Explain your answer by drawing meiosis I and II for Mary\'s chromosome 21. Use \"A\" for one
allele and \"a\" for the other allele of the TPTE gene. Only draw the chromosomes relevant to
this question (Do not draw all 46 chromosomes).
Solution
a. Nondisjunction produces abnormal gametes. Nondisjunction in meiosis I occurs during
anaphase I when one pair of homologous chromosomes fails to separate.Trisomy is the condition
of having 3 copies of one chromosome type. It is designated as 2n + 1 because the cell has the
normal two sets of each 23 types of chromosomes plus an extra copy of one chromosome.
I think Gerrys two chromosome 21s from his mother have different alleles (A and a) of the TPTE
gene. But more information could be needed to explain and understand this in a much better way..
Chromosomal abberations.pptx
chromosomekjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjkkkkkkkkkkkkkkkkkk,;lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll
in C++ , Design a linked list class named IntegerList to hold a seri.pdf
in C++ , Design a linked list class named IntegerList to hold a series of integers. The class must
have public member functions for appending , inserting, deleting and displaying nodes. The
append function must append a node to the end of the list. The insert function must insert node in
ascending order. The delete function must delete the node with the given value. The display
function should display the list from head to tail. The IntegerList class should also have a
destructor that destroys the list and release memory. The list node must be declared as a structure
using keyword struct. This structure is a private member of IntegerList. You should also store the
head of the list as a private member.
Write a test program to test the implementation your IntegerList class works properly.
Solution
Here is the code for the above scenario:
#include
using namespace std;
struct ListNode
{
float value;
ListNode *next;
};
ListNode *head;
class LinkedList
{
public:
int insertNode(float num);
int appendNode(float num);
void deleteNode(float num);
void destroyList();
void displayList();
LinkedList(void) {head = NULL;}
~LinkedList(void) {destroyList();}
};
int LinkedList::appendNode(float num)
{
ListNode *newNode, *nodePtr = head;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head == NULL)
{
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while(nodePtr->next != NULL)
nodePtr = nodePtr->next;
nodePtr->next = newNode;
}
return 0;
}
int LinkedList::insertNode(float num)
{
struct ListNode *newNode, *nodePtr = head, *prevNodePtr = NULL;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head==NULL) {
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while((nodePtr != NULL) && (nodePtr->value < num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(prevNodePtr==NULL)
newNode->next = head;
else
newNode->next = nodePtr; prevNodePtr->next = newNode;
}
return 0;
}
void LinkedList::deleteNode(float num)
{
struct ListNode *nodePtr = head, *prevNodePtr = NULL;
if(head==NULL) {
cout << \"The list was empty!\ \";
return;
}
if(head->value == num) {
head = nodePtr->next;
delete [] nodePtr;
}
else
{
while((nodePtr!= NULL)&&(nodePtr->value != num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(nodePtr==NULL)
cout << \"The value \" << num << \" is not in this list!\ \";
else {
prevNodePtr->next = nodePtr->next;
delete [] nodePtr;
}
}
}
void LinkedList::destroyList()
{
struct ListNode *nodePtr = head, *nextNode = nodePtr;
if(head==NULL) {
cout << \"The list is empty!\ \";
return;
}
while (nodePtr != NULL) {
nextNode = nodePtr->next;
delete [] nodePtr;
nodePtr = nextNode;
}
}
void LinkedList.
Recommended
What is the possible reason that secondaries arise from parents that.pdf
What is the possible reason that secondaries arise from parents that have unpaired chromosomes
but not from parents that are normal diploids? Map Genetics: A Conceptual Approach 6th
Edition Pierce MHE/Freeman presented by Sapling Learning Progeny of triploid tomato plants
often contain parts of an extra chromosome, in addition to the normal complement of 24
chromosomes. Mutants with a part of an extra chromosome are referred to as secondaries. James
and Margaret Lesley observed that secondaries arise from triploid (3n), trisomic (3n 1), and
double trisomic (3n 1 1) parents, but never from diploids (2n). (J. W. Lesley and M. M. Lesley.
1929. Genetics 14:321-336) What is a possible reason that secondaries arise from parents that
have unpaired chromosomes but not from parents that are normal diploids? Triploid and trisomic
conditions in tomato plants are unnatural, and this genomic imbalance leads to higher levels of
mitotic recombination. The trivalent chromosomes would still be able to associate through
chiasma prior to segregation during meiosis. Due to some unknown mechanism, triploid and
trisom c tomato plants O occasionally over-replicate part of a chromosome, which leads to the
generation of mutant tomato plants. Triploid and trisomic plants are vulnerable to infection by
the tomato mosaic O virus, so they acquire extra chromosomal material by the integration of
viral DNA, whereas diploid plants are not infected by the virus.
Solution
Ans: B
2n+1 condition is called trisomic in which an organism contains two complete genome + 1 extra
chromosome. During the process of meiosis 1 chromosomal page separate so that 1 members of
each state is distributed to one daughter nucleus and other members goes to other daughter
nucleus but rarely one pair of chromosomes fail to separate and moves to the pole giving half of
the daughter cells extra chromosome and remaining half lost one chromosome. As a result n - 1
and n+ 1 chromosomes appear. When the gamut with n+ 1 chromosome fuses with normal
gamut the resulting zygote will have 2n + 1 chromosomes producing trisomics organism.
It was found that trisomies or white study nature and studied in plants like tomatoes tobacco
wheat datura and so on..
You are presented with a patient, named Gerry, who has Down syndrome .pdf
You are presented with a patient, named Gerry, who has Down syndrome (trisomy 21). You
determine that the nondisjunction event occurred in his mother, named Mary, during meiosis I.
You are interested in the TPTE gene, which is near the centromere of chromosome 21. Virtually
no recombination occurs between the centromere and the TPTE gene. Mary has two different
alleles of the TPTE gene. Would Gerry\'s two chromosome 21s from his mother have the same
or different alleles of the TPTE gene? Or do you not have enough information to decide?
Explain your answer by drawing meiosis I and II for Mary\'s chromosome 21. Use \"A\" for one
allele and \"a\" for the other allele of the TPTE gene. Only draw the chromosomes relevant to
this question (Do not draw all 46 chromosomes).
Solution
a. Nondisjunction produces abnormal gametes. Nondisjunction in meiosis I occurs during
anaphase I when one pair of homologous chromosomes fails to separate.Trisomy is the condition
of having 3 copies of one chromosome type. It is designated as 2n + 1 because the cell has the
normal two sets of each 23 types of chromosomes plus an extra copy of one chromosome.
I think Gerrys two chromosome 21s from his mother have different alleles (A and a) of the TPTE
gene. But more information could be needed to explain and understand this in a much better way..
Chromosomal abberations.pptx
chromosomekjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjkkkkkkkkkkkkkkkkkk,;lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll
in C++ , Design a linked list class named IntegerList to hold a seri.pdf
in C++ , Design a linked list class named IntegerList to hold a series of integers. The class must
have public member functions for appending , inserting, deleting and displaying nodes. The
append function must append a node to the end of the list. The insert function must insert node in
ascending order. The delete function must delete the node with the given value. The display
function should display the list from head to tail. The IntegerList class should also have a
destructor that destroys the list and release memory. The list node must be declared as a structure
using keyword struct. This structure is a private member of IntegerList. You should also store the
head of the list as a private member.
Write a test program to test the implementation your IntegerList class works properly.
Solution
Here is the code for the above scenario:
#include
using namespace std;
struct ListNode
{
float value;
ListNode *next;
};
ListNode *head;
class LinkedList
{
public:
int insertNode(float num);
int appendNode(float num);
void deleteNode(float num);
void destroyList();
void displayList();
LinkedList(void) {head = NULL;}
~LinkedList(void) {destroyList();}
};
int LinkedList::appendNode(float num)
{
ListNode *newNode, *nodePtr = head;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head == NULL)
{
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while(nodePtr->next != NULL)
nodePtr = nodePtr->next;
nodePtr->next = newNode;
}
return 0;
}
int LinkedList::insertNode(float num)
{
struct ListNode *newNode, *nodePtr = head, *prevNodePtr = NULL;
newNode = new ListNode;
if(newNode == NULL)
{
cout << \"Error allocating memory for new list member!\ \";
return 1;
}
newNode->value = num;
newNode->next = NULL;
if(head==NULL) {
cout << \"List was empty - \" << newNode->value;
cout << \" is part of list\'s first node.\ \";
head = newNode;
}
else
{
while((nodePtr != NULL) && (nodePtr->value < num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(prevNodePtr==NULL)
newNode->next = head;
else
newNode->next = nodePtr; prevNodePtr->next = newNode;
}
return 0;
}
void LinkedList::deleteNode(float num)
{
struct ListNode *nodePtr = head, *prevNodePtr = NULL;
if(head==NULL) {
cout << \"The list was empty!\ \";
return;
}
if(head->value == num) {
head = nodePtr->next;
delete [] nodePtr;
}
else
{
while((nodePtr!= NULL)&&(nodePtr->value != num)) {
prevNodePtr = nodePtr;
nodePtr = nodePtr->next;
}
if(nodePtr==NULL)
cout << \"The value \" << num << \" is not in this list!\ \";
else {
prevNodePtr->next = nodePtr->next;
delete [] nodePtr;
}
}
}
void LinkedList::destroyList()
{
struct ListNode *nodePtr = head, *nextNode = nodePtr;
if(head==NULL) {
cout << \"The list is empty!\ \";
return;
}
while (nodePtr != NULL) {
nextNode = nodePtr->next;
delete [] nodePtr;
nodePtr = nextNode;
}
}
void LinkedList.
In addition to the fossil record, what other kinds of evidence reinfo.pdf
In addition to the fossil record, what other kinds of evidence reinforce the evolutionary of life?
You may want to use your textbook as a reference.
Solution
In addition to fossil record, application of Molecular Systematics is very much helpful for easy
identification of fossilized organism. Molecular systematics involves comparisons of nucleic
acids or other molecules to presume the degree of relatedness which can disclose phylogenetic
relationships which usually cannot be determined by non-molecular methods like comparative
anatomy. For example, molecular systematics helps us to uncover evolutionary relationships
between groups and allow us to reconstruct phylogenic trees among groups of present-day
organisms for which the fossil record is poor or lacking entirely. For example, the DNA that
encodes for ribosomal RNA (rRNA) changes comparatively slowly. Therefore, comparisons of
DNA sequences in these genes are valuable for investigating relationships between taxa that
diverged hundreds of millions of years ago. Other molecular methods useful for determining
phylogeny from fossil record include Nucleic acid sequencing, Genomic fingerprinting (RFLP)
and Amino acid sequencing..
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.pdf
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.51 M ascorbic acid
(H2C6H6O6) to a pH of 11.60?
Solution
pH of acidic buffer = pka2 + log(salt/acid)
No of mol of ascorbic acid = 0.161*4.51 = 0.726 mol
No of mol of salt = x mol
pka2 = 11.6, pH = 11.6
11.6 = 11.6 + log(x/(0.726-x))
x = 0.363
NO of mol of NaOH required = 2x = 2*0.363 = 0.726 mol
volume of NaOH must take = 0.726/0.521 = 1.4 L
answer: 1.4 L.
I am writing a program that places bolcks on a grid the the sape of .pdf
I am writing a program that places bolcks on a grid the the sape of different letters. It can also
move the blocks and checks if they overlap. I have most of the test cases figured out except for
four. Can someone help me out?
Thanks in advance.
output:
input:
output:
input:
output:
I CAN ONLY MODIFY ONE FILE SHAPE.CPP
#ifndef SHAPE_CPP
#define SHAPE_CPP
#include
#include
#include \"Shape.h\"
using namespace std;
// Constructor for O
O::O(int posx, int posy)
{
x = new int[1];
y = new int[1];
x[0] = posx;
y[0] = posy;
}
char O::name() const
{
return \'O\';
}
int O::size() const
{
return 1;
}
// Constructor for I
I::I(int posx, int posy)
{
x = new int[2];
y = new int[2];
x[0] = x[1] = posx;
y[0] = posy;
y[1] = posy+1;
}
char I::name() const
{
return \'I\';
}
int I::size() const
{
return 2;
}
// Constructor for L
L::L(int posx, int posy)
{
x = new int[3];
y = new int[3];
x[0] = x[2] = posx;
y[0] = y[1] = posy;
x[1] = posx+1;
y[2] = posy+1;
}
char L::name() const
{
return \'L\';
}
int L::size() const
{
return 3;
}
// Constructor for S
S::S(int posx, int posy)
{
x = new int[4];
y = new int[4];
x[0] = posx;
x[1] = posx+1;
x[2] = posx+1;
x[3] = posx+2;
y[0] = y[1] = posy;
y[2] = y[3] = posy+1;
}
char S::name() const
{
return \'S\';
}
int S::size() const
{
return 4;
}
/*
Constructor for X
the constructor initialises the cell co-ordinates
X is spread across 5 cells
*
* * *
*
The Numbering of cells is done from bottom-top and left to right
(x[0], y[0]) is the position of bottom most cell
(x[1], y[1]) is the position of left most cell of the 2nd row from bottom
(x[2], y[2]) is the position of the middle cell in the 2nd row from bottom
....
*/
X::X(int posx, int posy)
{
x = new int[5];
y = new int[5];
x[0] = x[2] = x[4] =posx;
x[1] = posx-1;
x[3] = posx+1;
y[0] = posy;
y[1] = y[2] = y[3] = posy+1;
y[4] = posy+2;
}
/*
This function returns the name of type of object i.e it is a X, or L, or U ...
*/
char X::name() const
{
return \'X\';
}
int X::size() const
{
return 5;
}
// Constructor for U
U::U(int posx, int posy)
{
x = new int[7];
y = new int[7];
x[0] = x[3] = x[5] = posx;
x[1] = posx+1;
x[2] = x[4] = x[6] = posx+2;
y[0] = y[1] = y[2] = posy;
y[3] = y[4] = posy+1;
y[5] = y[6] = posy+2;
}
char U::name() const
{
return \'U\';
}
int U::size() const
{
return 7;
}
void Shape::print() const
{
//get the size of the object i.e no of cells across which this is spread
int sz = size();
// get the type of object i.e O, L, X...
char n = name();
cout< > st;
int sz1 = t.size();
int sz2 = size();
for(int i = 0 ; i < sz1 ; i++)
{
st.insert(make_pair(t.x[i], t.y[i]));
}
for(int i = 0 ; i < sz2 ; i++)
{
if(st.find(make_pair(x[i], y[i]))!= st.end())
return true;
}
return false;
}
/*
This function creates a new object of type ch
*/
Shape * Shape::makeShape(char ch, int posx, int posy)
{
if(ch == \'O\')
return new O(posx, posy);
else if(ch == \'I\')
return new I(posx, posy);
else if(ch == \'L\')
return new L(posx, posy);
else if(ch == \'S\')
return new S(posx, posy);
else if.
How do teams bring value to an organizationHow are high performin.pdf
How do teams bring value to an organization?
How are high performing teams built?
When a team is working well together and the project is proceeding as planned what can happen
to derail even the high performing teams?
Solution
Teams are of important value to any organization as they determine not only the success of an
organization but also the level with which the organization grow.
1. Teams bring about exchange of ideas which benefit the organization in terms of innovation
as well as different ways of approaching difficulties.
2. Teams use the skill sets that come varied personalities to form a comprehensive part of the
work force.
3. Teams help an organization to develop good communication and commitment values a
factor that strengthens an organization both internally and externally.
4. Teams help organizations realize goals set through combination of effort.
High performing teams are build through choosing of a visionary, enthusiastic and committed
team leader. The leader then sets to choose a group of equally visionary, skilled and hard
working individuals who are united and willing to work towards a common goal a process that
entails forging of trust between the leader and the team. The team is then put through various
projects in which it is expected to complete with utmost success a process which defines and
makes the team better with team to a level of high performance.
1. Calamities such as sickness or death or a team member may derail the high performance.
2. Lack of motivation especially by the team leader may derail the high performance.
3. Laziness may be also stand as a factor especially when the teams assume that the set
objectives have been achieved
4. Conflict amongst the team members or the team leader and members may derail the teams\'
performance..
how can we use empricism to figure out if a fact is trueSolutio.pdf
how can we use empricism to figure out if a fact is true?
Solution
Empiricism is nothing but Knowledge by observation. Nothing is accepted without touching it,
Experiencing it or sensing the things directly.
So, empiricism helps a person to accept that fact by working on it or experimenting on it.
Empiricism demands logic to believe any particular fact to accept.
By using these many ways one can believe the fact is true from Empiricism..
Gene RegulationFunction of transcription factor domains, i.e. DNA.pdf
Gene Regulation:
Function of transcription factor domains, i.e. DNA binding vs. activation domain and co-
activators. Explain how they affect gene regulation in DNA.
Solution
Transcription is the process of synthesis of RNA from DNA template a- a fist step in gene
expression.The enzyme RNA polymerase copies a particular segment of DNA into RNa and it
represents the primary level at which gene expression is regulated in prokaryotes and eukaryotic
systems .the mechanism is similar in both prokayotes and eukaryotes .
In eukaryotic systems distinct factors known as the transcription initiation factors are required
and have a promoter region upstream from the gene or an enhancer region either up or
downstream from the gene and with specific motifs that are recognized by various types of
transcription factors .
The transcription factors bind and attract other transcripton factors to create a complex thus
facilitating the binding of RNA polymerase ii and hence the process of transcription. These
transcription factors are classified into genreral transcription factors which represent the basic
transcription machinery and gene specific transcription factor.
The transcrption factors are specific proteins that are required for RNA polymerase II to initiate
transcrption and contain one or more DNA binding domains (DBD) which attach to specific
DNA sequence adjacent to the gene that they regulate.
DNA binding domains is an independently folded domain which contains one motif that can
recognize a specific DNA sequence or general affinity to DNA. There are several different type
of DNA binding domains like the Helix turn Helix domain, Zinc finger domain ,Dimerization
domain , Leucine zipper etc.
Helix turn Helix domain: is characteristic of DNA binding proteins, composed of two -helices
joined by a short strand of amino acids and is found commonly in several proteins that regulate
gene expression. helix-turn-helix proteins include the homeodomain proteins, which play critical
roles in the regulation of gene expression during embryonic development
Zinc finger domain :They contain repeats of cysteine and histidine residues that bind zinc ions
and fold into looped structures that bind DNA first identified in the polymerase III transcription
factor TFIIA but are also found in transcription factors that regulate polymerse II promoters.
,Dimerization domain :,These domains bring about dimerization of two polypeptide chains as
seen in helix turn helkix domians , leucine zipper domain.
The transcrptional activator domains or the TAD are the regions of transcription factors which in
conjunction with DNA binding domain can activate transcrption from promoter by contacting
transcriptional machinery either directly or through other protein coactivators.
The transcriptional repressor domain TRDS are the regions of the transcription factors which in
conjuation with the DNA binding domain can repress transcirption from a promoter by
contacting transcrptional machinery eit.
For each step of DNA replication, predict the outcome ifconcentra.pdf
For each step of DNA replication, predict the outcome if:
concentration of nitrogenous bases is low
primase is inhibited
DNA polymerase were inhibited
DNA ligase were inhibited
Helicase were inhibited
AND
Identify cellular process(es) that require DNA replication
THANKS
Solution
concentration of nitrogenous bases is low- DNA molecule is made up of nitrogenous bases
adenine, guanine, cytosine, and thymine. These nitrogenous bases form the template and the
sequence of DNA in replication. They specify the order of these bases in the complementary
template strand being formed.
primase is inhibited- Enzyme primase is of key importance in DNA replication because it is the
only enzyme that can initiate DNA synthesis strand without an initial PNA or CNA primer.
Hence primase is essential as it catalysis the synthesis of a primer complementary to ssDNA
template aiding in the initiation of DNA replication to take place.
DNA polymerase were inhibited- DNA polymerase is an essential enzyme involved in the repair
of any damage to DNA molecule and it also plays a vital role in DNA replication; it connects the
Okazaki fragments (complementary fragements to the lagging template strand ) by deleting RNA
primers and replacing it with DNA.
DNA ligase were inhibited- Ligase enzyme is useful in DNA repair (single-strand or double-
strand breaks of DNA) and in replication by joining the double DNA strands through
phosphodiester bond formation (3\'OH end and 5\'P end). DNA double helicular structure is thus
formed through ligation of both strands on replication.
Helicase were inhibited- Helicases help in the seperation of nucleic acid strands required during
the cellular processes like DNA replication, transcription, repair. This seperation is done by
Helicase enzyme by moving through a nucleic acid phosphodiester backbone that exists in DNA,
RNA, or DNA-RNA hybrid.
cellular process that require DNA replication- In mitosis DNA replication is essential (S stage of
interphase) which gets distributed to the resulting daughter cells at the end of cytokinesis. This
ensures two similar daughter cells having common genetic material..
Explain THREE unique characteristics of fungi.SolutionThree un.pdf
Explain THREE unique characteristics of fungi.
Solution
Three unique characteristics of fungi are-
1)Fungi are filamentous and not truly multicellular. They are composed of unique thread-like
filamentous structures called hyphae. In higher fungi, these hyphae anastomose to form a
network called mycelium.The cell wall of fungi is composed of chitin or cellulose. Fungi can
build mutual relationships with other organisms in order to get nutritional benefits from such
association. For example: The symbiotic relationship between funfi and algae is called lichen.
2) They need preformed food, optimum temperature, pH and moisture to grow properly. In order
to avoid dessication, they also require some amount of water to grow. They are resistant to
osmotic pressure and hence, can grow in relatively high salt and sugar concentrations.
3)Fungi are heterotrophic and they lack plastids.They collect nutrients from the environment,
from dead or living organisms. Fungi are capable of metabolising complex carbohydrates (like
lignin)..
Explain the basic relationship between humans and microorganisms.pdf
Explain the basic relationship between humans and microorganisms
Solution
Answer=
The basic relationship between humans & microorganisms are as follows
The surfaces of human body inside & outside ,for example skin ,mouth & intestines are covered
in millions of individual microbes
The microbes that do not harm in fact they helps to protect us becoming infected with harmful
microbes they are known as normal body flora
There are 3 types of symbiotic relationships between humans & bacteria
1] Commensalism = Is a relationship that is beneficial to the bacteria ,but does not help or
harmthe host
most are reside on epithelial surfaces that come in contact with the external environment
commonly found on skin ,respiratory tract & gastrointestinal tract
They aquire nutrients & place to live & grow from their host
some are pathogenic & cause disease may provide benefit to host
2] Mutuliastic relationship-=
Both bacteria & host are beneficial
for example there are several kinds of bacteria that live on skin & inside the mouth ,nose,throat
& intestines of humans
These bacteria received a place to live & feed while keeping other harmful microbes from taking
up residence
bacteria in digestive system assist in nutrient metabolism ,vitamin production & waste
processing
3] Parasitic relatioship= is one in which the bacteria benefit while the host is harmed
pathogenic parasites ,which cause disease,do so by resisting the hosts defences & growing at the
expense of host
These bacteria produces poisonous substances called endotoxins & exotoxins are responsible for
the symptoms to occure with an illness
Disease causing bacteria are responsible for a number of diseases including meningitis
,pneumonia .tuberculosis& several types of food borne diseases.
discuss the rationale for the global harmonization of financial repo.pdf
discuss the rationale for the global harmonization of financial reporting and disclosure
requirements
Solution
The focus and rationale of global harmonization of financial reporting and disclosure
requirements is for achieving a level of compatibility for the statements. There is less
complication for the companies and individuals to present the financial statements and its
interpretations. The differences between financial reporting standards are erased with
international harmonization of financial reporting.
When there is achievement of international harmonization in finial reporting then it benefits
countries and improves the access effectively for the financial market. The confidence and
investor\'s K owned he is increased. According to Wiley (2000) even though the rationale behind
harmonization is smooth preparation and presentation of financial statements; there may be
differences based on the factors like taxation and political aspects.
It focuses on creation of similar set of procedures with the help of boundary establishment and
how they can differ globally.
It also benefits accounting standards on national and international levels.
The disclosure requirements include the information related to nature and the extent of risks
which arises from financial instruments. The importance of financial instutments for the purpose
of finial position and performance of entity. It includes financial assets, held to maturity
investments, loans and receivables, sale assets etc..
Describe the need to multitask in BBC (behavior-based control) syste.pdf
Describe the need to multitask in BBC (behavior-based control) systems?
Solution
Behavior-based control employs a set of distributed, in-teracting modules, called behaviors that
collectively achieve the desired system-level behavior. To an ex-ternal observer, behaviors are
patterns of the robot’s activity emerging from interactions between the robot and its
environment. To a programmer, behaviors are control modules that cluster sets of constraints in
order to achieve and maintain a goal. Each behavior receives inputs from sensors and/or other
behaviors in the system, and provides outputs to the robot’s actuators or to other behaviors. Thus,
a behavior-based controller is a structured network of interacting behaviors, with no centralized
world representation or focus of control. In-stead, individual behaviors and networks of
behaviors maintain any state information and models.
The basic principles of behavior-based control can be summarized briefly as follows:
• Behaviors are implemented as control laws (some-times similar to those used in control
theory), either in software or hardware, as a processing element or as a procedure.
• Each behavior can take inputs from the robot’s sen-sors (e.g., proximity sensors, range
detectors, contact sensors, camera) and/or from other modules in the
system, and send outputs to the robot’s effectors (e.g., wheels, grippers, arm, speech) and/or to
other modules.
• Many different behaviors may independently re- ceive input from the same sensors and output
action commands to the same actuators.
• Behaviors are encoded to be relatively simple, and are added to the system incrementally.
• Behaviors (or subsets thereof) are executed con- currently, not sequentially, in order to exploit
parallelism and speed of computation, as well as the interaction dynamics among behaviors and
between behaviors and the environment.
The ability to improve performance over time and to reason about the world, in the context of a
chang-ing and dynamic environment, is an important area of research in situated robotics. Unlike
in classical ma-chine learning, where the goal is typically to optimize performance over a long
period of time, in situated learning the aim is to adapt relatively quickly, toward
attaining efficiency in the light of uncertainty. Models from biology are often considered, given
its proper- ties of learning directly from environmental feedback. Variations and adaptations of
machine learning, and in particular reinforcement learning, have been effectively applied to
behavior-based robots, which have demon- strated learning to walk [38.
, navigate and create topological maps, di-vide tasks, behave socially , and even identify
opponents and score goals in robot soc-
cer. Methods from artificial life, evolutionary computation/genetic algorithms, fuzzy logic,
vision and learning, multi-agent systems, and many other research areas continue to be actively
explored and applied to behavior-based robots as their role in ani.
A spinner with 7 equally sized slices is shown below. The dial is spu.pdf
A spinner with 7 equally sized slices is shown below. The dial is spun and stops on a slice at
random. What are the odds in favor of landing on a black slice?
Solution
4 slice grey
1 black slice
2 white slice
odds are ratio of favorable outcomes to non favorable outcomes
Odds in favour of landing on a black slice = 1:6.
ble below to calculate the Gross Domestic Product uning the ral Gro.pdf
ble below to calculate the Gross Domestic Product uning the ral Gross Domestic Product? Fully
explain your anower. (4 mak) or rea overnment expenditures on goods and service Wages,
salaries and supplementary labour income 2,000 reciation nvestment 400 sumption expenditure
et exports direct taxes 150
Solution
Gross domestic product be expenditure method= consumption expenditure+ government
expenditure+ investment+ net export
=2200+500+400+(-50)
=3050
This is the nominal GDP as well as real GDP because base year price and current year price is
not given.
Hence real GDP and nominal GDP will be same..
1.The limbic systemA. matures before the prefrontal cortexB. is.pdf
1.The limbic system:
A. matures before the prefrontal cortex
B. is another name for the prefrontal cortex
C. is responsible for emotional regulation and impulse control
D. matures before the amygdala
Solution
The limbic system is responsible for emotional regulationand impules control.
Limbic system is that portion of the brain which deals with the emotions, arousal, learning aw
well as formation of memories. It has a complicated structure that lies on the both side of the
thalamus just beneath the cerebrum. It includes amygdala, hippocampus, thalamus,
hypothalamus, basal ganglia and cingulate gyrus.
Amygdala also known as the emotional centre of the brain. Hippocampus helps in the formation
of new memories from our past experinces. While thalamus and hypothalamus are governs the
changes occuring to the emotional activities. Basal ganglia involved in motor control and habbit
learning. And cingulate gyrus manage smell and sight and regulate our behaviour..
Why is leadership a critical factor in implementing change and qua.pdf
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Solution
Leadership is required for inspiring and motivating followers to work towards common
objectives. The leader employs enthusiasm and charisma to motivate the employees to think out
of the box.Leader lays emphasis on the values, morals, ideas and needs of the organization and
the followers.
Leadership is critical for transformation and change in the firm in a turbulent and challenging
environment. By concentrating on innovation and creation rather than just daily planning and
execution, the leader ensures that the changes are implemented with minimum resistance.
Below is how the leaders can implement change and quality improvement initiatives while
overcoming resistance-
1. Communicating the objective of the change and how it would positively affect the
productivity of the employees and the organization
2. Developing training and mentoring programs to train the employees to adopt to the new
process or system
3. Removing fears from the minds of employees through openhouse discussions etc.
4. Explaining the employees how the change would help in the development of the individuals..
Why are there bubbles in beer and champagneSolution Bubbles of .pdf
Why are there bubbles in beer and champagne?
Solution
Bubbles of gas, typically carbon dioxide are produced from beer and champagne , rising to the
surface. Because . yeasts ferment sugars and form carbon dioxide gas during fermentation. In the
bottle, where the pressure is high, the carbon dioxide disappears. However, when you open the
bottle and lower the pressure, the carbon dioxide escapes in the form of bubbles. In Champagne
and other sparkling wines, the bubbles pop, while in beer, the bubbles remain and form the
beer\'shead..
Which of the following is true of a traditional master budget a. It.pdf
Which of the following is true of a traditional master budget? a. It uses the zero-base budgeting
approach. b. It is department oriented. c. It recognizes interdependencies. d. It is dynamic in
nature.
Solution
Correct Answer is b. it is department Oriented
One of the critisism of Traditional Master Budget is that it is Department oriented and does not
recognise independencies among departments, further it is static not dynamic in nature..
Which ethnic groups were a part of the former YugoslaviaCroats, S.pdf
Which ethnic groups were a part of the former Yugoslavia?
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Solution
The ethnic group can be defined as a social category based on common culture or nationality.
In other words, it can be said that an ethnic group is a category of people who recognize with
each other based on similarities for example as a common ancestry, language, society, culture or
nation.
First Yugoslavia contains people of the Kingdom of Serbs, Croats and Slovenes (1918–29).
Hence it can be said that the ethnic groups which were a part of the former Yugoslavia were
Croats, Serbs, Bosnian Muslims, Macedonians.
Hence option first is the correct answer.
Option first; Croats, Serbs, Bosnian Muslims, Macedonians.
What would the components of the development be in an early learning.pdf
What would the components of the development be in an early learning(1st – 3rd) website
environment be?
Solution
For a website,major components include Front end,images,contents navigation,page structure.
For early learning website.As it for the children,content along with effective graphic design is
most preferable.
Graphic design:
Many of the elements described, such as the logo, the navigation menus, the layout, images, etc.,
fall under the general category of graphic design. But graphic design is more than these. It is the
look and feel the website will have as a result of proper use and integration of all these elements.
Content management is considered.
The website is not recommended to be a static as it should be more creative and changes is
necessary.
Content management system:
This is the ability to update your website without having to directly edit the html. Simple
systems create areas on a web page that can be easily changed on a regular basis.
Include multiedia and databases.
Multi-media.
It is for than just text and images. Photo-tours, video-clips, sound-clips all can add to the
experience if they are well matched to the type of site and profile of the target audience..
What is true of all fungi They area. eukaryotic, heterotrophic pl.pdf
What is true of all fungi? They are
a. eukaryotic, heterotrophic plants.
b. eukaryotic, parasitic plants.
c. saprobic/decomposing plants.
d. eukaryotic heterotrophs with cell walls.
e. saprobic heterotrophs.
Solution
Answer: a. eukaryotic, heterotrophic plants.
Fungi are non-chlorophyllous(absence of chloroplasts) plants. They are eukaryotes due to the
presence of true nucleus. Fungi live as heterotrophic, parasitic, saprophytic, etc..
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
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In addition to the fossil record, what other kinds of evidence reinfo.pdf
In addition to the fossil record, what other kinds of evidence reinforce the evolutionary of life?
You may want to use your textbook as a reference.
Solution
In addition to fossil record, application of Molecular Systematics is very much helpful for easy
identification of fossilized organism. Molecular systematics involves comparisons of nucleic
acids or other molecules to presume the degree of relatedness which can disclose phylogenetic
relationships which usually cannot be determined by non-molecular methods like comparative
anatomy. For example, molecular systematics helps us to uncover evolutionary relationships
between groups and allow us to reconstruct phylogenic trees among groups of present-day
organisms for which the fossil record is poor or lacking entirely. For example, the DNA that
encodes for ribosomal RNA (rRNA) changes comparatively slowly. Therefore, comparisons of
DNA sequences in these genes are valuable for investigating relationships between taxa that
diverged hundreds of millions of years ago. Other molecular methods useful for determining
phylogeny from fossil record include Nucleic acid sequencing, Genomic fingerprinting (RFLP)
and Amino acid sequencing..
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.pdf
How much 0.521 M NaOH will be needed to raise the pH of 0.161 L of 4.51 M ascorbic acid
(H2C6H6O6) to a pH of 11.60?
Solution
pH of acidic buffer = pka2 + log(salt/acid)
No of mol of ascorbic acid = 0.161*4.51 = 0.726 mol
No of mol of salt = x mol
pka2 = 11.6, pH = 11.6
11.6 = 11.6 + log(x/(0.726-x))
x = 0.363
NO of mol of NaOH required = 2x = 2*0.363 = 0.726 mol
volume of NaOH must take = 0.726/0.521 = 1.4 L
answer: 1.4 L.
I am writing a program that places bolcks on a grid the the sape of .pdf
I am writing a program that places bolcks on a grid the the sape of different letters. It can also
move the blocks and checks if they overlap. I have most of the test cases figured out except for
four. Can someone help me out?
Thanks in advance.
output:
input:
output:
input:
output:
I CAN ONLY MODIFY ONE FILE SHAPE.CPP
#ifndef SHAPE_CPP
#define SHAPE_CPP
#include
#include
#include \"Shape.h\"
using namespace std;
// Constructor for O
O::O(int posx, int posy)
{
x = new int[1];
y = new int[1];
x[0] = posx;
y[0] = posy;
}
char O::name() const
{
return \'O\';
}
int O::size() const
{
return 1;
}
// Constructor for I
I::I(int posx, int posy)
{
x = new int[2];
y = new int[2];
x[0] = x[1] = posx;
y[0] = posy;
y[1] = posy+1;
}
char I::name() const
{
return \'I\';
}
int I::size() const
{
return 2;
}
// Constructor for L
L::L(int posx, int posy)
{
x = new int[3];
y = new int[3];
x[0] = x[2] = posx;
y[0] = y[1] = posy;
x[1] = posx+1;
y[2] = posy+1;
}
char L::name() const
{
return \'L\';
}
int L::size() const
{
return 3;
}
// Constructor for S
S::S(int posx, int posy)
{
x = new int[4];
y = new int[4];
x[0] = posx;
x[1] = posx+1;
x[2] = posx+1;
x[3] = posx+2;
y[0] = y[1] = posy;
y[2] = y[3] = posy+1;
}
char S::name() const
{
return \'S\';
}
int S::size() const
{
return 4;
}
/*
Constructor for X
the constructor initialises the cell co-ordinates
X is spread across 5 cells
*
* * *
*
The Numbering of cells is done from bottom-top and left to right
(x[0], y[0]) is the position of bottom most cell
(x[1], y[1]) is the position of left most cell of the 2nd row from bottom
(x[2], y[2]) is the position of the middle cell in the 2nd row from bottom
....
*/
X::X(int posx, int posy)
{
x = new int[5];
y = new int[5];
x[0] = x[2] = x[4] =posx;
x[1] = posx-1;
x[3] = posx+1;
y[0] = posy;
y[1] = y[2] = y[3] = posy+1;
y[4] = posy+2;
}
/*
This function returns the name of type of object i.e it is a X, or L, or U ...
*/
char X::name() const
{
return \'X\';
}
int X::size() const
{
return 5;
}
// Constructor for U
U::U(int posx, int posy)
{
x = new int[7];
y = new int[7];
x[0] = x[3] = x[5] = posx;
x[1] = posx+1;
x[2] = x[4] = x[6] = posx+2;
y[0] = y[1] = y[2] = posy;
y[3] = y[4] = posy+1;
y[5] = y[6] = posy+2;
}
char U::name() const
{
return \'U\';
}
int U::size() const
{
return 7;
}
void Shape::print() const
{
//get the size of the object i.e no of cells across which this is spread
int sz = size();
// get the type of object i.e O, L, X...
char n = name();
cout< > st;
int sz1 = t.size();
int sz2 = size();
for(int i = 0 ; i < sz1 ; i++)
{
st.insert(make_pair(t.x[i], t.y[i]));
}
for(int i = 0 ; i < sz2 ; i++)
{
if(st.find(make_pair(x[i], y[i]))!= st.end())
return true;
}
return false;
}
/*
This function creates a new object of type ch
*/
Shape * Shape::makeShape(char ch, int posx, int posy)
{
if(ch == \'O\')
return new O(posx, posy);
else if(ch == \'I\')
return new I(posx, posy);
else if(ch == \'L\')
return new L(posx, posy);
else if(ch == \'S\')
return new S(posx, posy);
else if.
How do teams bring value to an organizationHow are high performin.pdf
How do teams bring value to an organization?
How are high performing teams built?
When a team is working well together and the project is proceeding as planned what can happen
to derail even the high performing teams?
Solution
Teams are of important value to any organization as they determine not only the success of an
organization but also the level with which the organization grow.
1. Teams bring about exchange of ideas which benefit the organization in terms of innovation
as well as different ways of approaching difficulties.
2. Teams use the skill sets that come varied personalities to form a comprehensive part of the
work force.
3. Teams help an organization to develop good communication and commitment values a
factor that strengthens an organization both internally and externally.
4. Teams help organizations realize goals set through combination of effort.
High performing teams are build through choosing of a visionary, enthusiastic and committed
team leader. The leader then sets to choose a group of equally visionary, skilled and hard
working individuals who are united and willing to work towards a common goal a process that
entails forging of trust between the leader and the team. The team is then put through various
projects in which it is expected to complete with utmost success a process which defines and
makes the team better with team to a level of high performance.
1. Calamities such as sickness or death or a team member may derail the high performance.
2. Lack of motivation especially by the team leader may derail the high performance.
3. Laziness may be also stand as a factor especially when the teams assume that the set
objectives have been achieved
4. Conflict amongst the team members or the team leader and members may derail the teams\'
performance..
how can we use empricism to figure out if a fact is trueSolutio.pdf
how can we use empricism to figure out if a fact is true?
Solution
Empiricism is nothing but Knowledge by observation. Nothing is accepted without touching it,
Experiencing it or sensing the things directly.
So, empiricism helps a person to accept that fact by working on it or experimenting on it.
Empiricism demands logic to believe any particular fact to accept.
By using these many ways one can believe the fact is true from Empiricism..
Gene RegulationFunction of transcription factor domains, i.e. DNA.pdf
Gene Regulation:
Function of transcription factor domains, i.e. DNA binding vs. activation domain and co-
activators. Explain how they affect gene regulation in DNA.
Solution
Transcription is the process of synthesis of RNA from DNA template a- a fist step in gene
expression.The enzyme RNA polymerase copies a particular segment of DNA into RNa and it
represents the primary level at which gene expression is regulated in prokaryotes and eukaryotic
systems .the mechanism is similar in both prokayotes and eukaryotes .
In eukaryotic systems distinct factors known as the transcription initiation factors are required
and have a promoter region upstream from the gene or an enhancer region either up or
downstream from the gene and with specific motifs that are recognized by various types of
transcription factors .
The transcription factors bind and attract other transcripton factors to create a complex thus
facilitating the binding of RNA polymerase ii and hence the process of transcription. These
transcription factors are classified into genreral transcription factors which represent the basic
transcription machinery and gene specific transcription factor.
The transcrption factors are specific proteins that are required for RNA polymerase II to initiate
transcrption and contain one or more DNA binding domains (DBD) which attach to specific
DNA sequence adjacent to the gene that they regulate.
DNA binding domains is an independently folded domain which contains one motif that can
recognize a specific DNA sequence or general affinity to DNA. There are several different type
of DNA binding domains like the Helix turn Helix domain, Zinc finger domain ,Dimerization
domain , Leucine zipper etc.
Helix turn Helix domain: is characteristic of DNA binding proteins, composed of two -helices
joined by a short strand of amino acids and is found commonly in several proteins that regulate
gene expression. helix-turn-helix proteins include the homeodomain proteins, which play critical
roles in the regulation of gene expression during embryonic development
Zinc finger domain :They contain repeats of cysteine and histidine residues that bind zinc ions
and fold into looped structures that bind DNA first identified in the polymerase III transcription
factor TFIIA but are also found in transcription factors that regulate polymerse II promoters.
,Dimerization domain :,These domains bring about dimerization of two polypeptide chains as
seen in helix turn helkix domians , leucine zipper domain.
The transcrptional activator domains or the TAD are the regions of transcription factors which in
conjunction with DNA binding domain can activate transcrption from promoter by contacting
transcriptional machinery either directly or through other protein coactivators.
The transcriptional repressor domain TRDS are the regions of the transcription factors which in
conjuation with the DNA binding domain can repress transcirption from a promoter by
contacting transcrptional machinery eit.
For each step of DNA replication, predict the outcome ifconcentra.pdf
For each step of DNA replication, predict the outcome if:
concentration of nitrogenous bases is low
primase is inhibited
DNA polymerase were inhibited
DNA ligase were inhibited
Helicase were inhibited
AND
Identify cellular process(es) that require DNA replication
THANKS
Solution
concentration of nitrogenous bases is low- DNA molecule is made up of nitrogenous bases
adenine, guanine, cytosine, and thymine. These nitrogenous bases form the template and the
sequence of DNA in replication. They specify the order of these bases in the complementary
template strand being formed.
primase is inhibited- Enzyme primase is of key importance in DNA replication because it is the
only enzyme that can initiate DNA synthesis strand without an initial PNA or CNA primer.
Hence primase is essential as it catalysis the synthesis of a primer complementary to ssDNA
template aiding in the initiation of DNA replication to take place.
DNA polymerase were inhibited- DNA polymerase is an essential enzyme involved in the repair
of any damage to DNA molecule and it also plays a vital role in DNA replication; it connects the
Okazaki fragments (complementary fragements to the lagging template strand ) by deleting RNA
primers and replacing it with DNA.
DNA ligase were inhibited- Ligase enzyme is useful in DNA repair (single-strand or double-
strand breaks of DNA) and in replication by joining the double DNA strands through
phosphodiester bond formation (3\'OH end and 5\'P end). DNA double helicular structure is thus
formed through ligation of both strands on replication.
Helicase were inhibited- Helicases help in the seperation of nucleic acid strands required during
the cellular processes like DNA replication, transcription, repair. This seperation is done by
Helicase enzyme by moving through a nucleic acid phosphodiester backbone that exists in DNA,
RNA, or DNA-RNA hybrid.
cellular process that require DNA replication- In mitosis DNA replication is essential (S stage of
interphase) which gets distributed to the resulting daughter cells at the end of cytokinesis. This
ensures two similar daughter cells having common genetic material..
Explain THREE unique characteristics of fungi.SolutionThree un.pdf
Explain THREE unique characteristics of fungi.
Solution
Three unique characteristics of fungi are-
1)Fungi are filamentous and not truly multicellular. They are composed of unique thread-like
filamentous structures called hyphae. In higher fungi, these hyphae anastomose to form a
network called mycelium.The cell wall of fungi is composed of chitin or cellulose. Fungi can
build mutual relationships with other organisms in order to get nutritional benefits from such
association. For example: The symbiotic relationship between funfi and algae is called lichen.
2) They need preformed food, optimum temperature, pH and moisture to grow properly. In order
to avoid dessication, they also require some amount of water to grow. They are resistant to
osmotic pressure and hence, can grow in relatively high salt and sugar concentrations.
3)Fungi are heterotrophic and they lack plastids.They collect nutrients from the environment,
from dead or living organisms. Fungi are capable of metabolising complex carbohydrates (like
lignin)..
Explain the basic relationship between humans and microorganisms.pdf
Explain the basic relationship between humans and microorganisms
Solution
Answer=
The basic relationship between humans & microorganisms are as follows
The surfaces of human body inside & outside ,for example skin ,mouth & intestines are covered
in millions of individual microbes
The microbes that do not harm in fact they helps to protect us becoming infected with harmful
microbes they are known as normal body flora
There are 3 types of symbiotic relationships between humans & bacteria
1] Commensalism = Is a relationship that is beneficial to the bacteria ,but does not help or
harmthe host
most are reside on epithelial surfaces that come in contact with the external environment
commonly found on skin ,respiratory tract & gastrointestinal tract
They aquire nutrients & place to live & grow from their host
some are pathogenic & cause disease may provide benefit to host
2] Mutuliastic relationship-=
Both bacteria & host are beneficial
for example there are several kinds of bacteria that live on skin & inside the mouth ,nose,throat
& intestines of humans
These bacteria received a place to live & feed while keeping other harmful microbes from taking
up residence
bacteria in digestive system assist in nutrient metabolism ,vitamin production & waste
processing
3] Parasitic relatioship= is one in which the bacteria benefit while the host is harmed
pathogenic parasites ,which cause disease,do so by resisting the hosts defences & growing at the
expense of host
These bacteria produces poisonous substances called endotoxins & exotoxins are responsible for
the symptoms to occure with an illness
Disease causing bacteria are responsible for a number of diseases including meningitis
,pneumonia .tuberculosis& several types of food borne diseases.
discuss the rationale for the global harmonization of financial repo.pdf
discuss the rationale for the global harmonization of financial reporting and disclosure
requirements
Solution
The focus and rationale of global harmonization of financial reporting and disclosure
requirements is for achieving a level of compatibility for the statements. There is less
complication for the companies and individuals to present the financial statements and its
interpretations. The differences between financial reporting standards are erased with
international harmonization of financial reporting.
When there is achievement of international harmonization in finial reporting then it benefits
countries and improves the access effectively for the financial market. The confidence and
investor\'s K owned he is increased. According to Wiley (2000) even though the rationale behind
harmonization is smooth preparation and presentation of financial statements; there may be
differences based on the factors like taxation and political aspects.
It focuses on creation of similar set of procedures with the help of boundary establishment and
how they can differ globally.
It also benefits accounting standards on national and international levels.
The disclosure requirements include the information related to nature and the extent of risks
which arises from financial instruments. The importance of financial instutments for the purpose
of finial position and performance of entity. It includes financial assets, held to maturity
investments, loans and receivables, sale assets etc..
Describe the need to multitask in BBC (behavior-based control) syste.pdf
Describe the need to multitask in BBC (behavior-based control) systems?
Solution
Behavior-based control employs a set of distributed, in-teracting modules, called behaviors that
collectively achieve the desired system-level behavior. To an ex-ternal observer, behaviors are
patterns of the robot’s activity emerging from interactions between the robot and its
environment. To a programmer, behaviors are control modules that cluster sets of constraints in
order to achieve and maintain a goal. Each behavior receives inputs from sensors and/or other
behaviors in the system, and provides outputs to the robot’s actuators or to other behaviors. Thus,
a behavior-based controller is a structured network of interacting behaviors, with no centralized
world representation or focus of control. In-stead, individual behaviors and networks of
behaviors maintain any state information and models.
The basic principles of behavior-based control can be summarized briefly as follows:
• Behaviors are implemented as control laws (some-times similar to those used in control
theory), either in software or hardware, as a processing element or as a procedure.
• Each behavior can take inputs from the robot’s sen-sors (e.g., proximity sensors, range
detectors, contact sensors, camera) and/or from other modules in the
system, and send outputs to the robot’s effectors (e.g., wheels, grippers, arm, speech) and/or to
other modules.
• Many different behaviors may independently re- ceive input from the same sensors and output
action commands to the same actuators.
• Behaviors are encoded to be relatively simple, and are added to the system incrementally.
• Behaviors (or subsets thereof) are executed con- currently, not sequentially, in order to exploit
parallelism and speed of computation, as well as the interaction dynamics among behaviors and
between behaviors and the environment.
The ability to improve performance over time and to reason about the world, in the context of a
chang-ing and dynamic environment, is an important area of research in situated robotics. Unlike
in classical ma-chine learning, where the goal is typically to optimize performance over a long
period of time, in situated learning the aim is to adapt relatively quickly, toward
attaining efficiency in the light of uncertainty. Models from biology are often considered, given
its proper- ties of learning directly from environmental feedback. Variations and adaptations of
machine learning, and in particular reinforcement learning, have been effectively applied to
behavior-based robots, which have demon- strated learning to walk [38.
, navigate and create topological maps, di-vide tasks, behave socially , and even identify
opponents and score goals in robot soc-
cer. Methods from artificial life, evolutionary computation/genetic algorithms, fuzzy logic,
vision and learning, multi-agent systems, and many other research areas continue to be actively
explored and applied to behavior-based robots as their role in ani.
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A spinner with 7 equally sized slices is shown below. The dial is spun and stops on a slice at
random. What are the odds in favor of landing on a black slice?
Solution
4 slice grey
1 black slice
2 white slice
odds are ratio of favorable outcomes to non favorable outcomes
Odds in favour of landing on a black slice = 1:6.
ble below to calculate the Gross Domestic Product uning the ral Gro.pdf
ble below to calculate the Gross Domestic Product uning the ral Gross Domestic Product? Fully
explain your anower. (4 mak) or rea overnment expenditures on goods and service Wages,
salaries and supplementary labour income 2,000 reciation nvestment 400 sumption expenditure
et exports direct taxes 150
Solution
Gross domestic product be expenditure method= consumption expenditure+ government
expenditure+ investment+ net export
=2200+500+400+(-50)
=3050
This is the nominal GDP as well as real GDP because base year price and current year price is
not given.
Hence real GDP and nominal GDP will be same..
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1.The limbic system:
A. matures before the prefrontal cortex
B. is another name for the prefrontal cortex
C. is responsible for emotional regulation and impulse control
D. matures before the amygdala
Solution
The limbic system is responsible for emotional regulationand impules control.
Limbic system is that portion of the brain which deals with the emotions, arousal, learning aw
well as formation of memories. It has a complicated structure that lies on the both side of the
thalamus just beneath the cerebrum. It includes amygdala, hippocampus, thalamus,
hypothalamus, basal ganglia and cingulate gyrus.
Amygdala also known as the emotional centre of the brain. Hippocampus helps in the formation
of new memories from our past experinces. While thalamus and hypothalamus are governs the
changes occuring to the emotional activities. Basal ganglia involved in motor control and habbit
learning. And cingulate gyrus manage smell and sight and regulate our behaviour..
Why is leadership a critical factor in implementing change and qua.pdf
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Why is leadership a critical factor in implementing change and quality improvement initiatives?
Solution
Leadership is required for inspiring and motivating followers to work towards common
objectives. The leader employs enthusiasm and charisma to motivate the employees to think out
of the box.Leader lays emphasis on the values, morals, ideas and needs of the organization and
the followers.
Leadership is critical for transformation and change in the firm in a turbulent and challenging
environment. By concentrating on innovation and creation rather than just daily planning and
execution, the leader ensures that the changes are implemented with minimum resistance.
Below is how the leaders can implement change and quality improvement initiatives while
overcoming resistance-
1. Communicating the objective of the change and how it would positively affect the
productivity of the employees and the organization
2. Developing training and mentoring programs to train the employees to adopt to the new
process or system
3. Removing fears from the minds of employees through openhouse discussions etc.
4. Explaining the employees how the change would help in the development of the individuals..
Why are there bubbles in beer and champagneSolution Bubbles of .pdf
Why are there bubbles in beer and champagne?
Solution
Bubbles of gas, typically carbon dioxide are produced from beer and champagne , rising to the
surface. Because . yeasts ferment sugars and form carbon dioxide gas during fermentation. In the
bottle, where the pressure is high, the carbon dioxide disappears. However, when you open the
bottle and lower the pressure, the carbon dioxide escapes in the form of bubbles. In Champagne
and other sparkling wines, the bubbles pop, while in beer, the bubbles remain and form the
beer\'shead..
Which of the following is true of a traditional master budget a. It.pdf
Which of the following is true of a traditional master budget? a. It uses the zero-base budgeting
approach. b. It is department oriented. c. It recognizes interdependencies. d. It is dynamic in
nature.
Solution
Correct Answer is b. it is department Oriented
One of the critisism of Traditional Master Budget is that it is Department oriented and does not
recognise independencies among departments, further it is static not dynamic in nature..
Which ethnic groups were a part of the former YugoslaviaCroats, S.pdf
Which ethnic groups were a part of the former Yugoslavia?
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Croats, Serbs, Bosnian Muslims, Macedonians
Janjaweed
Israelis, Palestinians, Arabs
Hutu & Tutsis
Solution
The ethnic group can be defined as a social category based on common culture or nationality.
In other words, it can be said that an ethnic group is a category of people who recognize with
each other based on similarities for example as a common ancestry, language, society, culture or
nation.
First Yugoslavia contains people of the Kingdom of Serbs, Croats and Slovenes (1918–29).
Hence it can be said that the ethnic groups which were a part of the former Yugoslavia were
Croats, Serbs, Bosnian Muslims, Macedonians.
Hence option first is the correct answer.
Option first; Croats, Serbs, Bosnian Muslims, Macedonians.
What would the components of the development be in an early learning.pdf
What would the components of the development be in an early learning(1st – 3rd) website
environment be?
Solution
For a website,major components include Front end,images,contents navigation,page structure.
For early learning website.As it for the children,content along with effective graphic design is
most preferable.
Graphic design:
Many of the elements described, such as the logo, the navigation menus, the layout, images, etc.,
fall under the general category of graphic design. But graphic design is more than these. It is the
look and feel the website will have as a result of proper use and integration of all these elements.
Content management is considered.
The website is not recommended to be a static as it should be more creative and changes is
necessary.
Content management system:
This is the ability to update your website without having to directly edit the html. Simple
systems create areas on a web page that can be easily changed on a regular basis.
Include multiedia and databases.
Multi-media.
It is for than just text and images. Photo-tours, video-clips, sound-clips all can add to the
experience if they are well matched to the type of site and profile of the target audience..
What is true of all fungi They area. eukaryotic, heterotrophic pl.pdf
What is true of all fungi? They are
a. eukaryotic, heterotrophic plants.
b. eukaryotic, parasitic plants.
c. saprobic/decomposing plants.
d. eukaryotic heterotrophs with cell walls.
e. saprobic heterotrophs.
Solution
Answer: a. eukaryotic, heterotrophic plants.
Fungi are non-chlorophyllous(absence of chloroplasts) plants. They are eukaryotes due to the
presence of true nucleus. Fungi live as heterotrophic, parasitic, saprophytic, etc..
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