Which of the following protocols is used for router multicastingA.pdf
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Which of the following protocols is used for router multicasting? A. ICMP B. RIPv1 C. 224.0.0.1 D. IGMP Solution D. Internet Group Management Protocol (IGMP) is used by hosts to report multicast group memberships to neighboring multicast routers. Security problems exist with IGMP because anyone can start a multicast group or join an existing one. Answer A is incorrect because ICMP is used for logical errors and diagnostics. Answer B is incorrect because the Routing Information Protocol (RIP) is a broadcast-based routing protocol. Answer C is incorrect because although 224.0.0.1 is a multicast address, it is not a protocol used for multicast management..
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Mitosis promoting factor is associated with cyclin b. Which class of .pdf
Mitosis promoting factor is associated with cyclin b. Which class of enzymes are activated by
cyclin b: Hydrolases Kinases Lipases Cytochalasin B disrupts: microtubules microfilaments
intermediate fibers tubulin Which of the species has a egg that is not classified as isolecithal.
amphioxus molluscs mammals amphibians nematodes A spore is a haploid gamete True
False D
Solution
(C) Kinases are activated by cyclin b. (B) Cytochalasin B disrupts microfilament formation. (D)
Eggs of amphibians are telolecithal.It is not classified as isolecithal (A) A spore is a haploid
gamete..
LTE uses an all IP data network, with voice calls reduced to a multim.pdf
LTE uses an all IP data network, with voice calls reduced to a multimedia application. Describe
the benefits to the network operator of this approach, including the relationship between the core
network and the radio access network, as well as other benefits from using a full IP network
rather than the GPRS approach.
Solution
Voice calls over LTE is commonly known as VoLTE (Voice over LTE) which delivers voice
service as data flows over LTE channels. VoLTE service contains an IP multimedia subsystem
network, known as multimedia telephony service (MMTel). This is a global standard for real
time communication using voice, video text and files. This serves to be the core network for
VoLTE and eliminates the need for a legacy circuit switched network. The main advantages for a
network operator in this case are:
The radio access network is a part of the remote device (e.g. Mobile) that provides access to the
operator’s core network. In case of VoLTE, it is termed as Evolved UMTS Terrestrial Radio
Access Network (E-UTRAN). This radio access network is different and incompatible with any
earlier technology like HSPA/HSUPA. It is more optimized for packet data and provides much
higher data rates and lower latency. EUTRAN uses orthogonal frequency division multiple
access (OFDM) for the downlink and single carrier frequency division multiple access (SC-
FDMA) for the uplink. EUTRAN consists of eNodeB (enhanced Node B) which transmit and
receive signals from the mobile devices. They interface with the core network using S1-AP
protocol S1 mobile management entity (MME) interface for the control plane traffic. It uses the
GTP-U protocol on the S1-U interface with the service gateway for user (data) plane traffic.
Collectively the S1-MME and S1-U interfaces are known as the S1 interface, which represents
the interface from the radio access network to the core network for VoLTE..
I Need Serious Help with this assignment. Plenty Points to be earned.pdf
I Need Serious Help with this assignment. Plenty Points to be earned here
Objective: To implement Thread scheduling on the OSP2 simulator using a modified version of
the Linux Completely Fair Scheduler strategy. You will implement the module ThreadCB.java to
further your understanding of CPU scheduling. Required to turn in: Follow all directions
regarding hand-in procedures. Points will be deducted if you do not follow directions precisely.
You must submit an electronic copy of your Linux Completely Fair Scheduler solution (or your
best try) via dropbox. Late assignments will be assessed a 10% penalty per day. You must
document your code and provide a one-page explanation of how you accomplished the
assignment (or what you have currently and why you could not complete). You should describe
your use, creation, and manipulation of data structures to accomplish the assignment. Building
and executing the simulation As in the previous assignment, I suggest that you create a new
directory. Copy your files from the previous cpu scheduling project and starting with that version
of ThreadCB.java modify it to create the Linux Completely Fair Scheduler solution. The changes
you need to make to your code from the previous project are: 1) You will need to add three
instance variables and three class variables to the ThreadCB class: Private long lastDispatch; an
instance variable that keeps track of when a thread is dispatched. private int lastCpuBurst; an
instance variable that you use to update vRunTime. private long vRunTime; an instance variable
that keeps track of the virtual run time of each thread, and private static long numThreads; a
static class variable that keeps track of how many threads there are. private static ThreadCB[]
heap; a static class variable that is used as a heap structure. private static int heapSize; a static
class variable that is used keep track of how many threads are in the heap. Be sure to initialize it
to 0. 2) Each time you change the status of a thread from ThreadRunning to any other state, you
will update its vRunTime value. You will start by determining its current burst time. You
calculate the value using the following equation: thread.lastCpuBurst = HClock.get() -
thread.lastDispatch; As you can see, you will need the value of lastDispatch in order to do this.
Be sure to set this variable each time you dispatch a new thread, i.e., thread.lastDispatch =
HClock.get() just before returning from do_dispatch(). Next you update vRunTime using the
following expression: thread.vRunTime = ((thread.vRunTime+thread.lastCpuBurst); 3) Instead
of using a GenericList readyQueue, you will use a priority queue implemented as a heap. You
will code the heap yourself. This is much simpler than the red black tree that Linux actually uses.
While you could create a separate heap class, I suggest that you simply declare the heap and the
two methods indicated below. My suggestion is that you look at the implementation of the heap
from Mark W.
I have code that reads and writes sstrings to file but I am having t.pdf
I have code that reads and writes sstrings to file but I am having trouble meeting these
requirements. My program writes the string names into the file and not the number of bags. I
think a for loop is needed to keep track of how many times their name is written and then check
their on board status but I\'m not sure. This is the code I have
#include
#include
#include
using namespace std;
int openOutputFile();
int openInputFile();
void readInputRecord(ifstream& inpt);
void writeOutputRecord(ofstream& opt);
int count_b = 0;
int count_p = 0;
string PassengerArray[100];
string BaggageArray[300];
int main()
{
openInputFile();
openOutputFile();
system(\"pause\");
}
void readInputRecord(ifstream& inpt)
{
string line;
int i = 0;
//read passanger from file
while (getline(inpt, line))
{
PassengerArray[i] = line;
i++;
}
cout << \"Passanger Array \" << endl;
for (int k = 0; k
Solution
while (getline(inpt, line) && line != \"\")
{
BaggageArray[i] = line;
i++;
}
count_b = i;
cout << \"Baggage Array\" << endl;
for (int k = 0; k.
Genes A and B are on different chromosomes. A dihybrid with the geno.pdf
Genes A and B are on different chromosomes. A dihybrid with the genotype AaBb undergoes
meiosis. Which of the following depicts a possible genotype in a gamete? Genes A and B are on
different chromosomes. A dihybrid with the genotype AaBb undergoes meiosis. Which of the
following depicts a possible genotype in a gamete?
A) Aa
B) A
C) Bb
D) B
E) AB
Solution
Since the genotype of the dihybrid is AaBb, the four possible gametes it can form are Ab, AB, ab
and aB.
Therefore from the choices given above, E is the correct answer..
Do the following program in C++- Create a item class... with and i.pdf
Do the following program in C++
- Create a item class... with and integer key and value
- Create a Hash class, that has an array of can input a item and uses a hash algorithm to place it
in a array of item nodes.
Include the necessary functions...
-Use CHAINING with pointers, when there is a collision.
-Have a function that prints out the array to verify the hashing and collision code works...
Explain each entry you insert into the hash table...
Solution
#include
using namespace std;
const int TABLE_SIZE = 128;
/*
* HashNode Class Declaration
*/
class HashNode
{
public:
int key;
int value;
HashNode* next;
HashNode(int key, int value)
{
this->key = key;
this->value = value;
this->next = NULL;
}
};
/*
* HashMap Class Declaration
*/
class HashMap
{
private:
HashNode** htable;
public:
HashMap()
{
htable = new HashNode*[TABLE_SIZE];
for (int i = 0; i < TABLE_SIZE; i++)
htable[i] = NULL;
}
~HashMap()
{
for (int i = 0; i < TABLE_SIZE; ++i)
{
HashNode* entry = htable[i];
while (entry != NULL)
{
HashNode* prev = entry;
entry = entry->next;
delete prev;
}
}
delete[] htable;
}
/*
* Hash Function
*/
void Display()
{
for (int i = 0; i < TABLE_SIZE; ++i)
{
HashNode* entry = htable[i];
while (entry != NULL)
{
cout<<\"Key \"<key<<\" Value \"<value<next;
//delete prev;
}
}
}
int HashFunc(int key)
{
return key % TABLE_SIZE;
}
/*
* Insert Element at a key
*/
void Insert(int key, int value)
{
int hash_val = HashFunc(key);
HashNode* prev = NULL;
HashNode* entry = htable[hash_val];
while (entry != NULL)
{
prev = entry;
entry = entry->next;
}
if (entry == NULL)
{
entry = new HashNode(key, value);
if (prev == NULL)
{
htable[hash_val] = entry;
}
else
{
prev->next = entry;
}
}
else
{
entry->value = value;
}
}
/*
* Remove Element at a key
*/
void Remove(int key)
{
int hash_val = HashFunc(key);
HashNode* entry = htable[hash_val];
HashNode* prev = NULL;
if (entry == NULL || entry->key != key)
{
cout<<\"No Element found at key \"<next != NULL)
{
prev = entry;
entry = entry->next;
}
if (prev != NULL)
{
prev->next = entry->next;
}
delete entry;
cout<<\"Element Deleted\"<key == key)
{
cout<value<<\" \";
flag = true;
}
entry = entry->next;
}
if (!flag)
return -1;
}
};
/*
* Main Contains Menu
*/
int main()
{
HashMap hash;
int key, value;
int choice;
while (1)
{
cout<<\"\ ----------------------\"<>choice;
switch(choice)
{
case 1:
cout<<\"Enter element to be inserted: \";
cin>>value;
cout<<\"Enter key at which element to be inserted: \";
cin>>key;
hash.Insert(key, value);
break;
case 2:
cout<<\"Enter key of the element to be searched: \";
cin>>key;
cout<<\"Element at key \"<>key;
hash.Remove(key);
break;
case 4:
hash.Display();break;
case 5:
return 0;
default:
cout<<\"\ Enter correct option\ \";
}
}
return 0;
}
==================================================================
Output:
akshay@akshay-Inspiron-3537:~/Chegg$ g++ hash.cpp
akshay@akshay-Inspiron-3537:~/Chegg$ ./a.out
----------------------
Operations on Hash Table
----------------------
1.Insert element into the table.
Explain the process of RNAi and how it contributes to differential .pdf
Explain the process of RNAi and how it contributes to differential gene expression
Solution
RNA interference is a biological process in which RNA molecules inhibit gene expression of
targeted mRNA. RNAi is mediated by two small RNA: miRNA and siRNA. these small RNA
moleules bind to target mRNA and either increase or decrease their activity by inhibiting
translation.
This process is initiated by dicer enzyme that cleave long RNA strands into small double
stranded siRNAs. so RNA unwound to yield two single stranded RNA molecules, of which one
is called passenger strand and the other is guide RNA. Guide RNA is incorporated into RNA
induced silencing complex (RISC). Guide RNA complementarily bind to mRNA moleule and
then cleaved by argonaute-2, the catalytic component of RISC.
Gene expression is maintained mainly by miRNA. miRNAs repress translation, involved in
regulation of development, timing of morphogenesis, maintenance of stem cells..
Consider the following scenario A population of moderately pigmented.pdf
Consider the following scenario: A population of moderately pigmented snakes from Indiana
migrate to northern Ohio. Several individuals from this population then make it out to some
islands in Lake Erie. Eventually we start to see that snakes on the islands become lighter in color
over time, whereas the snakes on the mainland become darker. Treating all the snakes as a single
population in which all individuals were originally banded, graph how selection is affecting
phenotype over time. For a few more points: Also indicate the relationship between fitness and
phenotype as a result of the given mode of selection on a separate graph. Write down which
mode of selection is operating and label all your axes PROPERLY.
Solution
Thanks for the question.
Some important points here are
1. since the nature alllowed these two snake into different geographical locations, because of the
change in environmental condition and climatic conditions could have resulted in the change of
color.
2. The dietery changes, the type and kind of food available may also have led to the change in
the fitneess and skin color.
3. The food catching methods and processess, maybe different for the 2 types of populations and
this could also have led to the changes in color and fitness.
4. the availability of additional necessary requirements such as availability of different types of
water and extra nutrition in the form of substances available only in certain geographical
locations may have resulted in changes in color and fitness.
The type of mode that may be active here is stabilizing selection as the trait in one location is
favored more than the other.
Regards.
Recommended
Mitosis promoting factor is associated with cyclin b. Which class of .pdf
Mitosis promoting factor is associated with cyclin b. Which class of enzymes are activated by
cyclin b: Hydrolases Kinases Lipases Cytochalasin B disrupts: microtubules microfilaments
intermediate fibers tubulin Which of the species has a egg that is not classified as isolecithal.
amphioxus molluscs mammals amphibians nematodes A spore is a haploid gamete True
False D
Solution
(C) Kinases are activated by cyclin b. (B) Cytochalasin B disrupts microfilament formation. (D)
Eggs of amphibians are telolecithal.It is not classified as isolecithal (A) A spore is a haploid
gamete..
LTE uses an all IP data network, with voice calls reduced to a multim.pdf
LTE uses an all IP data network, with voice calls reduced to a multimedia application. Describe
the benefits to the network operator of this approach, including the relationship between the core
network and the radio access network, as well as other benefits from using a full IP network
rather than the GPRS approach.
Solution
Voice calls over LTE is commonly known as VoLTE (Voice over LTE) which delivers voice
service as data flows over LTE channels. VoLTE service contains an IP multimedia subsystem
network, known as multimedia telephony service (MMTel). This is a global standard for real
time communication using voice, video text and files. This serves to be the core network for
VoLTE and eliminates the need for a legacy circuit switched network. The main advantages for a
network operator in this case are:
The radio access network is a part of the remote device (e.g. Mobile) that provides access to the
operator’s core network. In case of VoLTE, it is termed as Evolved UMTS Terrestrial Radio
Access Network (E-UTRAN). This radio access network is different and incompatible with any
earlier technology like HSPA/HSUPA. It is more optimized for packet data and provides much
higher data rates and lower latency. EUTRAN uses orthogonal frequency division multiple
access (OFDM) for the downlink and single carrier frequency division multiple access (SC-
FDMA) for the uplink. EUTRAN consists of eNodeB (enhanced Node B) which transmit and
receive signals from the mobile devices. They interface with the core network using S1-AP
protocol S1 mobile management entity (MME) interface for the control plane traffic. It uses the
GTP-U protocol on the S1-U interface with the service gateway for user (data) plane traffic.
Collectively the S1-MME and S1-U interfaces are known as the S1 interface, which represents
the interface from the radio access network to the core network for VoLTE..
I Need Serious Help with this assignment. Plenty Points to be earned.pdf
I Need Serious Help with this assignment. Plenty Points to be earned here
Objective: To implement Thread scheduling on the OSP2 simulator using a modified version of
the Linux Completely Fair Scheduler strategy. You will implement the module ThreadCB.java to
further your understanding of CPU scheduling. Required to turn in: Follow all directions
regarding hand-in procedures. Points will be deducted if you do not follow directions precisely.
You must submit an electronic copy of your Linux Completely Fair Scheduler solution (or your
best try) via dropbox. Late assignments will be assessed a 10% penalty per day. You must
document your code and provide a one-page explanation of how you accomplished the
assignment (or what you have currently and why you could not complete). You should describe
your use, creation, and manipulation of data structures to accomplish the assignment. Building
and executing the simulation As in the previous assignment, I suggest that you create a new
directory. Copy your files from the previous cpu scheduling project and starting with that version
of ThreadCB.java modify it to create the Linux Completely Fair Scheduler solution. The changes
you need to make to your code from the previous project are: 1) You will need to add three
instance variables and three class variables to the ThreadCB class: Private long lastDispatch; an
instance variable that keeps track of when a thread is dispatched. private int lastCpuBurst; an
instance variable that you use to update vRunTime. private long vRunTime; an instance variable
that keeps track of the virtual run time of each thread, and private static long numThreads; a
static class variable that keeps track of how many threads there are. private static ThreadCB[]
heap; a static class variable that is used as a heap structure. private static int heapSize; a static
class variable that is used keep track of how many threads are in the heap. Be sure to initialize it
to 0. 2) Each time you change the status of a thread from ThreadRunning to any other state, you
will update its vRunTime value. You will start by determining its current burst time. You
calculate the value using the following equation: thread.lastCpuBurst = HClock.get() -
thread.lastDispatch; As you can see, you will need the value of lastDispatch in order to do this.
Be sure to set this variable each time you dispatch a new thread, i.e., thread.lastDispatch =
HClock.get() just before returning from do_dispatch(). Next you update vRunTime using the
following expression: thread.vRunTime = ((thread.vRunTime+thread.lastCpuBurst); 3) Instead
of using a GenericList readyQueue, you will use a priority queue implemented as a heap. You
will code the heap yourself. This is much simpler than the red black tree that Linux actually uses.
While you could create a separate heap class, I suggest that you simply declare the heap and the
two methods indicated below. My suggestion is that you look at the implementation of the heap
from Mark W.
I have code that reads and writes sstrings to file but I am having t.pdf
I have code that reads and writes sstrings to file but I am having trouble meeting these
requirements. My program writes the string names into the file and not the number of bags. I
think a for loop is needed to keep track of how many times their name is written and then check
their on board status but I\'m not sure. This is the code I have
#include
#include
#include
using namespace std;
int openOutputFile();
int openInputFile();
void readInputRecord(ifstream& inpt);
void writeOutputRecord(ofstream& opt);
int count_b = 0;
int count_p = 0;
string PassengerArray[100];
string BaggageArray[300];
int main()
{
openInputFile();
openOutputFile();
system(\"pause\");
}
void readInputRecord(ifstream& inpt)
{
string line;
int i = 0;
//read passanger from file
while (getline(inpt, line))
{
PassengerArray[i] = line;
i++;
}
cout << \"Passanger Array \" << endl;
for (int k = 0; k
Solution
while (getline(inpt, line) && line != \"\")
{
BaggageArray[i] = line;
i++;
}
count_b = i;
cout << \"Baggage Array\" << endl;
for (int k = 0; k.
Genes A and B are on different chromosomes. A dihybrid with the geno.pdf
Genes A and B are on different chromosomes. A dihybrid with the genotype AaBb undergoes
meiosis. Which of the following depicts a possible genotype in a gamete? Genes A and B are on
different chromosomes. A dihybrid with the genotype AaBb undergoes meiosis. Which of the
following depicts a possible genotype in a gamete?
A) Aa
B) A
C) Bb
D) B
E) AB
Solution
Since the genotype of the dihybrid is AaBb, the four possible gametes it can form are Ab, AB, ab
and aB.
Therefore from the choices given above, E is the correct answer..
Do the following program in C++- Create a item class... with and i.pdf
Do the following program in C++
- Create a item class... with and integer key and value
- Create a Hash class, that has an array of can input a item and uses a hash algorithm to place it
in a array of item nodes.
Include the necessary functions...
-Use CHAINING with pointers, when there is a collision.
-Have a function that prints out the array to verify the hashing and collision code works...
Explain each entry you insert into the hash table...
Solution
#include
using namespace std;
const int TABLE_SIZE = 128;
/*
* HashNode Class Declaration
*/
class HashNode
{
public:
int key;
int value;
HashNode* next;
HashNode(int key, int value)
{
this->key = key;
this->value = value;
this->next = NULL;
}
};
/*
* HashMap Class Declaration
*/
class HashMap
{
private:
HashNode** htable;
public:
HashMap()
{
htable = new HashNode*[TABLE_SIZE];
for (int i = 0; i < TABLE_SIZE; i++)
htable[i] = NULL;
}
~HashMap()
{
for (int i = 0; i < TABLE_SIZE; ++i)
{
HashNode* entry = htable[i];
while (entry != NULL)
{
HashNode* prev = entry;
entry = entry->next;
delete prev;
}
}
delete[] htable;
}
/*
* Hash Function
*/
void Display()
{
for (int i = 0; i < TABLE_SIZE; ++i)
{
HashNode* entry = htable[i];
while (entry != NULL)
{
cout<<\"Key \"<key<<\" Value \"<value<next;
//delete prev;
}
}
}
int HashFunc(int key)
{
return key % TABLE_SIZE;
}
/*
* Insert Element at a key
*/
void Insert(int key, int value)
{
int hash_val = HashFunc(key);
HashNode* prev = NULL;
HashNode* entry = htable[hash_val];
while (entry != NULL)
{
prev = entry;
entry = entry->next;
}
if (entry == NULL)
{
entry = new HashNode(key, value);
if (prev == NULL)
{
htable[hash_val] = entry;
}
else
{
prev->next = entry;
}
}
else
{
entry->value = value;
}
}
/*
* Remove Element at a key
*/
void Remove(int key)
{
int hash_val = HashFunc(key);
HashNode* entry = htable[hash_val];
HashNode* prev = NULL;
if (entry == NULL || entry->key != key)
{
cout<<\"No Element found at key \"<next != NULL)
{
prev = entry;
entry = entry->next;
}
if (prev != NULL)
{
prev->next = entry->next;
}
delete entry;
cout<<\"Element Deleted\"<key == key)
{
cout<value<<\" \";
flag = true;
}
entry = entry->next;
}
if (!flag)
return -1;
}
};
/*
* Main Contains Menu
*/
int main()
{
HashMap hash;
int key, value;
int choice;
while (1)
{
cout<<\"\ ----------------------\"<>choice;
switch(choice)
{
case 1:
cout<<\"Enter element to be inserted: \";
cin>>value;
cout<<\"Enter key at which element to be inserted: \";
cin>>key;
hash.Insert(key, value);
break;
case 2:
cout<<\"Enter key of the element to be searched: \";
cin>>key;
cout<<\"Element at key \"<>key;
hash.Remove(key);
break;
case 4:
hash.Display();break;
case 5:
return 0;
default:
cout<<\"\ Enter correct option\ \";
}
}
return 0;
}
==================================================================
Output:
akshay@akshay-Inspiron-3537:~/Chegg$ g++ hash.cpp
akshay@akshay-Inspiron-3537:~/Chegg$ ./a.out
----------------------
Operations on Hash Table
----------------------
1.Insert element into the table.
Explain the process of RNAi and how it contributes to differential .pdf
Explain the process of RNAi and how it contributes to differential gene expression
Solution
RNA interference is a biological process in which RNA molecules inhibit gene expression of
targeted mRNA. RNAi is mediated by two small RNA: miRNA and siRNA. these small RNA
moleules bind to target mRNA and either increase or decrease their activity by inhibiting
translation.
This process is initiated by dicer enzyme that cleave long RNA strands into small double
stranded siRNAs. so RNA unwound to yield two single stranded RNA molecules, of which one
is called passenger strand and the other is guide RNA. Guide RNA is incorporated into RNA
induced silencing complex (RISC). Guide RNA complementarily bind to mRNA moleule and
then cleaved by argonaute-2, the catalytic component of RISC.
Gene expression is maintained mainly by miRNA. miRNAs repress translation, involved in
regulation of development, timing of morphogenesis, maintenance of stem cells..
Consider the following scenario A population of moderately pigmented.pdf
Consider the following scenario: A population of moderately pigmented snakes from Indiana
migrate to northern Ohio. Several individuals from this population then make it out to some
islands in Lake Erie. Eventually we start to see that snakes on the islands become lighter in color
over time, whereas the snakes on the mainland become darker. Treating all the snakes as a single
population in which all individuals were originally banded, graph how selection is affecting
phenotype over time. For a few more points: Also indicate the relationship between fitness and
phenotype as a result of the given mode of selection on a separate graph. Write down which
mode of selection is operating and label all your axes PROPERLY.
Solution
Thanks for the question.
Some important points here are
1. since the nature alllowed these two snake into different geographical locations, because of the
change in environmental condition and climatic conditions could have resulted in the change of
color.
2. The dietery changes, the type and kind of food available may also have led to the change in
the fitneess and skin color.
3. The food catching methods and processess, maybe different for the 2 types of populations and
this could also have led to the changes in color and fitness.
4. the availability of additional necessary requirements such as availability of different types of
water and extra nutrition in the form of substances available only in certain geographical
locations may have resulted in changes in color and fitness.
The type of mode that may be active here is stabilizing selection as the trait in one location is
favored more than the other.
Regards.
Consider the simple analog signal defined by x(t) = sin(2700t) shown.pdf
Consider the simple analog signal defined by x(t) = sin(2700t) shown in Figure P2–13. Draw the
spectrum of x(n) showing all spectral components, labeling their frequency locations, in the
frequency range 2fs to +2fs.
Figure P2–13
Solution
close all;
clear all;
clc;
n=-2:0.002:2;
x=sin(2*pi*700*n);
plot(n,x).
Cell adhesion, both cell-cell and cell-ECM, can influence whether a c.pdf
Cell adhesion, both cell-cell and cell-ECM, can influence whether a cell will proceed through
cell division or die by apoptosis. Compare/contrast the influence on cell survival and cell
division between cell-cell and cell-ECM adhesion. What signal in behavior is an integrin sending
to a cell when the integrin is binding its ligand? Explain your answer. What signal in behavior is
a cadherin sending to a cell when the cadherin is binding to its ligand? Explain your answer.
Compare and contrast how normal cells and cancer cells interpret cell adhesion (either cell-cell
and cell-ECM) to a change in cell behavior.
Solution
Answer B
Integrins and cadherins are adhesive molecules that are also called as cell adhesion molecules or
CAM. Cell matrix adhesion is brought over by integrins. Integrins are the receptors that bring
about ECM adhesion. Integrins activate signal transduction pathways, once they are activated by
ligand binding. These signal transduction pathways mediate cell signals for cell cycle
regulations, intracellular cytoskeleton organisation and new receptor movements to the cell
membranes.
AnswerC
Integrins and cadherins are adhesive molecules that are also called as cell adhesion molecules or
CAM. Junction adherin and cell cell interactions are generally carried out by cadherins. The
cadherins release a beta catenin that plays a role in cell cell adhesion and carries out cadherins
mediated cell division at the plasma membrane. Their name cadherin comes from their
dependency on Ca2+ ions required to function. There are different classes of cadherins. It is
generally seen that that cells containing a specific cadherin subtype tend to cluster together to the
exclusion of other types, both in cell culture and during development. When E cadherin is
expressed there is a release signal for the epithelial cadherin, followed by the expression of N
cadherin which is present for neural development. When the N cadherin is expressed E cadherin
is reduced , finally when there is development of notochord and so mites, E-P and N , cadherins
are expressed.
Answer D
In cancer cells , many of the stem cells show loosened attachment and higher motility. This is
generally seen in the cases where the cancer is highly invasive or malignant, affecting more than
one part of the body. The molecular adhesion fingerprint is altered in cases where the tumour
calls show high mobility showing a high need to dissociate.
Note- could not answer the first part due to lack of appropriate descriptive evidences. Thank you..
A(n) is a method of interacting with a PC by manipulating visual ele.pdf
A(n) is a method of interacting with a PC by manipulating visual elements such as icons and
windows. The is the narrow box across the bottom of the screen The bars across the top and
bottom of the Fedora 13 desktop are called A(n) is a small, single-function utility. is a quick
and fun way to resize open windows simply by dragging them to the edges of your screen. A(n)
appears to give you the power of X-ray vision - you can peer past all your open windows straight
to the Windows 7 desktop. Using a(n) list, you can go directly to the documents, pictures,
songs, and Web sites you use each day. makes it easy to read, organize, and compare windows.
are navigation aids used in an address bar and other parts of the GUI to help keep track of your
locations. are like folders but they can display the contents of other folders from all over your
PC. As the fills with material you have deleted, older items are removed automatically to make
room for newer items. (True or False) Items that you delete from external drives, such as I \'SB
drives, are permanently deleted and are not sent to the Recycle Bin.
Solution
1. A GUI(Graphical User Interface) is a method of interacting with a PC by manipulating visual
elements such as icons and windows.
2. The taskbar is the narrow box accross the bottom of the screen.
3. The bars across the top and bottom of the Fedora 13 desktop are called panels.
4. An applet is a small single function utility.
1. Snap is a quick and fun way to resize and open windows simply by dragging them to the edges
of the screen.
2. A peek appears to give you the power of X-ray vision - you can peer past all your open
windows straight to the windows 7 desktop..
You are about to perform a clean install of Windows 7 on a brandnew .pdf
You are about to perform a clean install of Windows 7 on a brandnew PC. Are there tasks you
must perform before installing the operating system?
A. You must manually format the disk first.
B. You must manually partition the disk first.
C. Windows 7 will let you perform partitioning and formatting from Setup.
D. The hard disk is already partitioned and formatted at the factory.
Solution
B. WET, or Windows Easy Transfer, is a utility used to perform a single transfer from a
Windows XP machine to a Windows Vista or Windows 7 machine. It is not suitable to use to
transfer settings from a single machine to multiple other computers..
Write the equation of a 4th degree polynomial function with the follo.pdf
Write the equation of a 4th degree polynomial function with the following roots: x=1 with
multiplicity 2, and x i )
Solution
if one root is i other will be -i as conjugate pair
(x-i)(x+i) will be a factor
(x^2+1) will be a factor as i^2 =-1
another root is 1 with multiplicity 2
so the polynomial will be
(x-1)^2(x^2+1)
Thank you..
Write a program which repeatedly calls a function named replaceIt() a.pdf
Write a program which repeatedly calls a function named replaceIt() as long as the user wishes
to. replaceIt() will have 3 parameters: originalList, old, and new. The function replaces all
occurrences of old with new in list originalList. You may NOT use the built-in function
replace()!!! Write your own.
Solution
ReplaceIt(originalList,old,new){
}.
Which, of the below choices of inference rules, is used to show that.pdf
Which, of the below choices of inference rules, is used to show that from the hypotheses “Every
kid in school learns arithmetic” and “Antonio is a kid in school”, we can conclude that Antonio
learns arithmetic? Explain...
Modus Tollens
Modus Ponens
Hypothetical Syllogism
Universal Instantiation
Solution
Universal Instantiation
Reason: Universal instantiation is a valid rule of inference from a truth about each member of a
class of individuals to the truth about a particular individual of that class.
Example: No humans can fly. John Doe is human. Therefore John Doe can not fly.
Which of the following isare not produced by homosporous plants [se.pdf
Which of the following is/are not produced by homosporous plants? [select all that apply]
Spores Megasporangia Embryos Microsporangia Eggs and sperm
Solution
Answer:
Megasporangi, microsporangia, Eggs and sperms
Explanation:
Homosporous plants produce spores of one kind only that are not differentiated by sex. The
spores develop into bisexual gametophytes.
When Gregor Mendel was developing his ideas of inheritance of cha.pdf
When Gregor Mendel was developing his ideas of inheritance of characteristics, molecular
genetics was not yet discovered. Explain in your own terms, why his two most important ant
laws have stood the test of time. What assumptions about them were correct? Describe a scenario
in which patterns of inheritance did NOT adhere to Mendelian genetics.
When Gregor Mendel was developing his ideas of inheritance of characteristics, molecular
genetics was not yet discovered. Explain in your own terms, why his two most important ant
laws have stood the test of time. What assumptions about them were correct? Describe a scenario
in which patterns of inheritance did NOT adhere to Mendelian genetics.
Solution
At the time of Gregor Mendel, the molecular genetics was not yet discovered but he was a
statistician by training and fortunately started his study on pea plant growing in his church lawn.
By careful observation and applying statistical inference, he came up with his two laws i.e. law
of segregation and law of independent assortment. Law of segregation states that the factors do
not mix with each other but maintain their identity and segregates (separate from each other) at
the time of meiosis when gametes form (Meiosis was not discovered at that time but people
knew about gametes). The other law was law of independent assortment which stated that the
factors responsible for different traits assort independent of each other into gametes giving rise to
all types of genotypes in F2 generation. His careful observation and mathematical background
enabled him to make robust inferences which stood the test of time. The assumptions that were
correct were that every factor has two alternative forms. These forms do not mix with each other
and segregate in each generation when gametes are formed.
There was another law that is law of dominance which states that one form of the factor is
dominant over the other due to which the effect of latter is suppressed. But because only 2 laws
were asked so we will put the best laws.
A scenario in which pattern of inheritence did not adhere to Mendelian genetics is linkage. As
many genes are present on same chromosomes, they can not assort independent of each other
resulting in linkage. The amount of recombination will depend upon the distance between two
genes but in general, these genes are not assorting independently..
what kind of short tandem repeats DNA markers can be used for tracin.pdf
what kind of short tandem repeats DNA markers can be used for tracing paternal lineage in DNA
ancestry testing?
Solution
DNA Ancestry testing can help you to trace your paternal and maternal lineages. There are two
types of DNA Ancestry tests available: Y-DNA STR marker testing for tracing paternal lineage
and mtDNA sequencing for tracing maternal lineage.
Y-DNA test method: PCR analysis of short tandem repeat (STR) markers in the Y-DNA. 20 to
101 Y-DNA STR markers can be tested..
What is the job of a layout managerTo render user interface compo.pdf
What is the job of a layout manager?
To render user interface components in a container.
To render a drawing on a canvas.
To position user interface components in a container.
To position a frame on a desktop.
Solution
Answer:To position user interface components in a container.
The job of a layout manager is to position user interface components in a container. These
layout managers are designed for displaying multiple components at once in a container.
There are five Layout managers available in Swings/AWT.
BorderLayout
BoxLayout
FlowLayout
GridBagLayout
and GridLayout.
Using the Bernoulli method, solve the differential equation t^2 dyd.pdf
Using the Bernoulli method, solve the differential equation: t^2 dy/dt + y^2 = ty.
Solution
t^2y\'+y^2=ty
t^2y\'-ty=-y^2
n=2 is highest power of y
So substitution is
u=y^{1-n}=y^{1-2}=1/y
y=1/u
y\'=-u\'/u^2
SUbstituting gives
t^2y\'-ty=-y^2
-t^2u\'/u^2-t/u=-1/u^2
t^2u\'+tu=1
tu\'+u=1/t
(tu)\'=1/t
Integrating gives
tu=ln(t)+C
u=(ln(t)+C)/t
Hence,
y=t/(C+ln(t)).
What are the implications of Darwins ideas for the traditional .pdf
What are the implications of Darwin\'s ideas for the traditional view of human nature (both
rationalistic and Judeo-Christian versions)?
What are the implications of Darwin\'s ideas for the traditional view of human nature (both
rationalistic and Judeo-Christian versions)?
Solution
Traditional Western views of human nature assume that all humans have the same human nature:
They are conscious, rational selves who have a purpose. Some Traditional views refer to this
immaterially defined self as the \"soul\". One important version of the Traditional Western view
of human nature is the ancient Greek view that sees humans as uniquely rational beings with a
purpose. This view contends that reason can and should rule over human desire and
aggressiveness. The Judeo-Christian religious view claims that humans are made in the image of
God, who has endowed them with rational self-consciousness and an ability to love. In the
Christian version, the self is immaterial and distinct from the body.. The Traditional view of
human nature and our ordinary thinking assume that humans have a self that endures through
time. Descartes claims that the enduring self is a soul. Locke argues that memory produces the
enduring self. Buddhism and Hume suggest that there is no enduring self.
Some scientific views challenge the Traditional view of human nature. Darwin argued that
humans evolved from earlier animal species through random variations and a natural selection
that is the result of a struggle for existence. Darwin\'s view has been taken to imply that human
nature has no purpose and is not unique..
What are the factors that underlie the observation that EBV infectio.pdf
What are the factors that underlie the observation that EBV infection give rise to so many tumor
types?
a. Different virus latency programs combined with the state of B cell differentiation
b. A single virus latency program acting on B cell mutations during maturation
c. Different virus latency programs in B blast cells cause different tumor types
d. A single virus latency program acting on different states of B cell differentiation
Solution
c. Different virus latency programs in B blast cells cause different tumor types
Some of the EBV infected lymphocytes remain in the circulation, the virus remains in a latent
state. When activated, each of these cells give rise to a spontaneous outgrowth of EBV
transformed, immortalized cell lines..
What is an advantage of the hanging drop method verses simple sta.pdf
What is an advantage of the hanging drop method verses simple staining? What is a
disadvantage?
What is an advantage of the hanging drop method verses simple staining? What is a
disadvantage?
Solution
Advantages of Hanging drop method over simple staining:
Simple staining includes staining and viewing microorganisms using basic dyes which are
attracted to negatively charged microbial cytoplasm materials. However, if microbes are large
and if motility is to be studied; staining becomes unnecessary. In these cases, wet mount is
prepared i.e. a drop of microbes is directly put on slide and observed. Hanging drop method
includes placing a drop of sample below a cover-slip which is when inverted moves through
hollowed-out slide. This method is advantageous because it keeps the sample hydrated i.e
doesn’t dry out. This is a rapid method and makes it easy to observe especially larger microbes
for examining motility.
Disadvantages:.
What are key factors to consider when selecting a flow sensorSo.pdf
What are key factors to consider when selecting a flow sensor?
Solution
1)Viscosity and Transparency
viscosity plays an important role as a sensor will not work with liquids that are too dense
2) Flow Sensor Bodies and Hazardous Chemicals
Hazardous chemicals can pose a problem to the flow sensor as the chemicals can damage
sensitive parts of the device such as the fittings or O-rings.
Flow Range Needed
Flow range sizing is important because a single flow meter can only accurately measure typically
10:1, so a unit rated for 10 L/minute will only measure as low as 1 L/minute accurately
Some other options and considerations may be.
The process of DNA replication creates a particular problem for repl.pdf
The process of DNA replication creates a particular problem for replicating the ends of linear
chromosomes.
1. Describe why the DNA replication machinery has difficulty replicating DNA ends.
2. Telomerase has been identified as an enzyme that can reverse the outcome of end replication.
Describe the near-universal mammalian chromosome telomere sequence and how telomerase
solves the problem of end replication.
Solution
Answer:
1. Describe why DNA replication machinery has difficulty replicating DNA ends:
Eukaryotic chromosomes are linear and not circular like bacterial chromosomes. The end of
eukaryotic chromosomes contains repetitive sequences (telomere) and cannot be fully copied
during each round of replication. DNA replication begins by unwinding of DNA double helix
and the synthesis of new strand by DNA polymerase. Since DNA polymerase can only work in a
unidirectional way (5’ to 3’) therefore one of the two new strands of DNA is synthesized
continuously (leading strand) and the other strand is synthesized in small stretches called as
Okazaki fragments (lagging strand). At the end of the lagging strand, since it is not possible to
attach an RNA primer, thererfore, the terminal region of chromosomes does not get covered by
Okazaki fragments. As a result, with each round of replication, the terminal chromosome
sequence cannot be copied leaving single stranded overhangs and gradual shortening of
chromosomes. This is called as “end replication problem”.
2. Describe Near Universal mammalian chromosome telomere sequence and how telomerase
solves the problem of end replication:
The termini of eukaryotic chromosome have specialized DNA caps called telomere which
comprise of short repetitive DNA sequence. Each time a cell divides some portion of the
telomere is lost.
Mammalian Telomere contains hundreds or thousands of tandem repeats of the same short DNA
sequence 5’-TTAGGG-3’. With each cell division telomeric sequence are lost thereby
protecting the critical chromosomal regions that contains gene.Telomerase (telomere terminal
transferase) is an RNA dependent DNA polymerase made up of RNA subunits and protein and is
involved in elongating the length of chromosome by extending the telomeres. Telomerase
enzyme binds to a special RNA molecule that contains a sequence complementary to the
telomeric repeat. By using the complementary RNA as a template, it extends the overhanging
strand of the telomere DNA. Since DNA polymerase cannot replicate the DNA sequence at the
end of the chromosomes therefore telomerase enzyme aids in terminal replication process by
attaching to the end of the chromosome. Absence of telomerase activity would result in
progressive shortening of chromosome with each cell division..
The lifetime T of a certain device (in 1000 hour units) has the expo.pdf
The lifetime T of a certain device (in 1000 hour units) has the exponential distribution with
parameter ½. Find P(T > 2).
Solution
Given that T is the lifetime of a certain device.
T has Exponential distribution with paramter with parameter 1/2.
We have to find the P(T > 2).
The p.d.f. of T is,
f(T) = e-t t > 0 and > 0.
given that = 1 / 2 = 0.5
P(T>2) = f(T) dt (from 2 to )
= e-t dt
= e-t dt
= [ e-t / - ]
= - / [ e- - e-0.5*2 ]
= - [ 0 - 0.3679 ]
P(T >2) = 0.3679.
The Federal Reserve controls the nations money supply by using var.pdf
The Federal Reserve controls the nation\'s money supply by using various tools that it has at its
disposal. Which of the following is not a tool used by the Fed to change the money supply?
Discount window lending.
The federal funds rate.
Open-market operations.
Changing reserve requirements.
Discount window lending.
The federal funds rate.
Open-market operations.
Changing reserve requirements.
Solution
Fed doesnt use DISCOUNT WINDOW LENDING as one of the tools to change money supply..
Suppose S is a set people, and R is the following relation on S aRb.pdf
Suppose S is a set people, and R is the following relation on S: aRb if and only if a has chatted
with b on the internet. Explain your answers to the following questions and use common sense.
(a) Is R reflexive? (b) Is R symmetric? (c) Is R transitive?
Solution
a.it is not reflexive because a has not chatted with himself.
b. Yes it is symmetric beacause a and b belong to set of people. As a chatted with b same can be
said for b that b also chatted with a.
c. There is no clear solution for transitive as it has not mentioned involvement of any third
person apart from a and b..
Introduction to AI for Nonprofits with Tapp Network
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
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Consider the simple analog signal defined by x(t) = sin(2700t) shown.pdf
Consider the simple analog signal defined by x(t) = sin(2700t) shown in Figure P2–13. Draw the
spectrum of x(n) showing all spectral components, labeling their frequency locations, in the
frequency range 2fs to +2fs.
Figure P2–13
Solution
close all;
clear all;
clc;
n=-2:0.002:2;
x=sin(2*pi*700*n);
plot(n,x).
Cell adhesion, both cell-cell and cell-ECM, can influence whether a c.pdf
Cell adhesion, both cell-cell and cell-ECM, can influence whether a cell will proceed through
cell division or die by apoptosis. Compare/contrast the influence on cell survival and cell
division between cell-cell and cell-ECM adhesion. What signal in behavior is an integrin sending
to a cell when the integrin is binding its ligand? Explain your answer. What signal in behavior is
a cadherin sending to a cell when the cadherin is binding to its ligand? Explain your answer.
Compare and contrast how normal cells and cancer cells interpret cell adhesion (either cell-cell
and cell-ECM) to a change in cell behavior.
Solution
Answer B
Integrins and cadherins are adhesive molecules that are also called as cell adhesion molecules or
CAM. Cell matrix adhesion is brought over by integrins. Integrins are the receptors that bring
about ECM adhesion. Integrins activate signal transduction pathways, once they are activated by
ligand binding. These signal transduction pathways mediate cell signals for cell cycle
regulations, intracellular cytoskeleton organisation and new receptor movements to the cell
membranes.
AnswerC
Integrins and cadherins are adhesive molecules that are also called as cell adhesion molecules or
CAM. Junction adherin and cell cell interactions are generally carried out by cadherins. The
cadherins release a beta catenin that plays a role in cell cell adhesion and carries out cadherins
mediated cell division at the plasma membrane. Their name cadherin comes from their
dependency on Ca2+ ions required to function. There are different classes of cadherins. It is
generally seen that that cells containing a specific cadherin subtype tend to cluster together to the
exclusion of other types, both in cell culture and during development. When E cadherin is
expressed there is a release signal for the epithelial cadherin, followed by the expression of N
cadherin which is present for neural development. When the N cadherin is expressed E cadherin
is reduced , finally when there is development of notochord and so mites, E-P and N , cadherins
are expressed.
Answer D
In cancer cells , many of the stem cells show loosened attachment and higher motility. This is
generally seen in the cases where the cancer is highly invasive or malignant, affecting more than
one part of the body. The molecular adhesion fingerprint is altered in cases where the tumour
calls show high mobility showing a high need to dissociate.
Note- could not answer the first part due to lack of appropriate descriptive evidences. Thank you..
A(n) is a method of interacting with a PC by manipulating visual ele.pdf
A(n) is a method of interacting with a PC by manipulating visual elements such as icons and
windows. The is the narrow box across the bottom of the screen The bars across the top and
bottom of the Fedora 13 desktop are called A(n) is a small, single-function utility. is a quick
and fun way to resize open windows simply by dragging them to the edges of your screen. A(n)
appears to give you the power of X-ray vision - you can peer past all your open windows straight
to the Windows 7 desktop. Using a(n) list, you can go directly to the documents, pictures,
songs, and Web sites you use each day. makes it easy to read, organize, and compare windows.
are navigation aids used in an address bar and other parts of the GUI to help keep track of your
locations. are like folders but they can display the contents of other folders from all over your
PC. As the fills with material you have deleted, older items are removed automatically to make
room for newer items. (True or False) Items that you delete from external drives, such as I \'SB
drives, are permanently deleted and are not sent to the Recycle Bin.
Solution
1. A GUI(Graphical User Interface) is a method of interacting with a PC by manipulating visual
elements such as icons and windows.
2. The taskbar is the narrow box accross the bottom of the screen.
3. The bars across the top and bottom of the Fedora 13 desktop are called panels.
4. An applet is a small single function utility.
1. Snap is a quick and fun way to resize and open windows simply by dragging them to the edges
of the screen.
2. A peek appears to give you the power of X-ray vision - you can peer past all your open
windows straight to the windows 7 desktop..
You are about to perform a clean install of Windows 7 on a brandnew .pdf
You are about to perform a clean install of Windows 7 on a brandnew PC. Are there tasks you
must perform before installing the operating system?
A. You must manually format the disk first.
B. You must manually partition the disk first.
C. Windows 7 will let you perform partitioning and formatting from Setup.
D. The hard disk is already partitioned and formatted at the factory.
Solution
B. WET, or Windows Easy Transfer, is a utility used to perform a single transfer from a
Windows XP machine to a Windows Vista or Windows 7 machine. It is not suitable to use to
transfer settings from a single machine to multiple other computers..
Write the equation of a 4th degree polynomial function with the follo.pdf
Write the equation of a 4th degree polynomial function with the following roots: x=1 with
multiplicity 2, and x i )
Solution
if one root is i other will be -i as conjugate pair
(x-i)(x+i) will be a factor
(x^2+1) will be a factor as i^2 =-1
another root is 1 with multiplicity 2
so the polynomial will be
(x-1)^2(x^2+1)
Thank you..
Write a program which repeatedly calls a function named replaceIt() a.pdf
Write a program which repeatedly calls a function named replaceIt() as long as the user wishes
to. replaceIt() will have 3 parameters: originalList, old, and new. The function replaces all
occurrences of old with new in list originalList. You may NOT use the built-in function
replace()!!! Write your own.
Solution
ReplaceIt(originalList,old,new){
}.
Which, of the below choices of inference rules, is used to show that.pdf
Which, of the below choices of inference rules, is used to show that from the hypotheses “Every
kid in school learns arithmetic” and “Antonio is a kid in school”, we can conclude that Antonio
learns arithmetic? Explain...
Modus Tollens
Modus Ponens
Hypothetical Syllogism
Universal Instantiation
Solution
Universal Instantiation
Reason: Universal instantiation is a valid rule of inference from a truth about each member of a
class of individuals to the truth about a particular individual of that class.
Example: No humans can fly. John Doe is human. Therefore John Doe can not fly.
Which of the following isare not produced by homosporous plants [se.pdf
Which of the following is/are not produced by homosporous plants? [select all that apply]
Spores Megasporangia Embryos Microsporangia Eggs and sperm
Solution
Answer:
Megasporangi, microsporangia, Eggs and sperms
Explanation:
Homosporous plants produce spores of one kind only that are not differentiated by sex. The
spores develop into bisexual gametophytes.
When Gregor Mendel was developing his ideas of inheritance of cha.pdf
When Gregor Mendel was developing his ideas of inheritance of characteristics, molecular
genetics was not yet discovered. Explain in your own terms, why his two most important ant
laws have stood the test of time. What assumptions about them were correct? Describe a scenario
in which patterns of inheritance did NOT adhere to Mendelian genetics.
When Gregor Mendel was developing his ideas of inheritance of characteristics, molecular
genetics was not yet discovered. Explain in your own terms, why his two most important ant
laws have stood the test of time. What assumptions about them were correct? Describe a scenario
in which patterns of inheritance did NOT adhere to Mendelian genetics.
Solution
At the time of Gregor Mendel, the molecular genetics was not yet discovered but he was a
statistician by training and fortunately started his study on pea plant growing in his church lawn.
By careful observation and applying statistical inference, he came up with his two laws i.e. law
of segregation and law of independent assortment. Law of segregation states that the factors do
not mix with each other but maintain their identity and segregates (separate from each other) at
the time of meiosis when gametes form (Meiosis was not discovered at that time but people
knew about gametes). The other law was law of independent assortment which stated that the
factors responsible for different traits assort independent of each other into gametes giving rise to
all types of genotypes in F2 generation. His careful observation and mathematical background
enabled him to make robust inferences which stood the test of time. The assumptions that were
correct were that every factor has two alternative forms. These forms do not mix with each other
and segregate in each generation when gametes are formed.
There was another law that is law of dominance which states that one form of the factor is
dominant over the other due to which the effect of latter is suppressed. But because only 2 laws
were asked so we will put the best laws.
A scenario in which pattern of inheritence did not adhere to Mendelian genetics is linkage. As
many genes are present on same chromosomes, they can not assort independent of each other
resulting in linkage. The amount of recombination will depend upon the distance between two
genes but in general, these genes are not assorting independently..
what kind of short tandem repeats DNA markers can be used for tracin.pdf
what kind of short tandem repeats DNA markers can be used for tracing paternal lineage in DNA
ancestry testing?
Solution
DNA Ancestry testing can help you to trace your paternal and maternal lineages. There are two
types of DNA Ancestry tests available: Y-DNA STR marker testing for tracing paternal lineage
and mtDNA sequencing for tracing maternal lineage.
Y-DNA test method: PCR analysis of short tandem repeat (STR) markers in the Y-DNA. 20 to
101 Y-DNA STR markers can be tested..
What is the job of a layout managerTo render user interface compo.pdf
What is the job of a layout manager?
To render user interface components in a container.
To render a drawing on a canvas.
To position user interface components in a container.
To position a frame on a desktop.
Solution
Answer:To position user interface components in a container.
The job of a layout manager is to position user interface components in a container. These
layout managers are designed for displaying multiple components at once in a container.
There are five Layout managers available in Swings/AWT.
BorderLayout
BoxLayout
FlowLayout
GridBagLayout
and GridLayout.
Using the Bernoulli method, solve the differential equation t^2 dyd.pdf
Using the Bernoulli method, solve the differential equation: t^2 dy/dt + y^2 = ty.
Solution
t^2y\'+y^2=ty
t^2y\'-ty=-y^2
n=2 is highest power of y
So substitution is
u=y^{1-n}=y^{1-2}=1/y
y=1/u
y\'=-u\'/u^2
SUbstituting gives
t^2y\'-ty=-y^2
-t^2u\'/u^2-t/u=-1/u^2
t^2u\'+tu=1
tu\'+u=1/t
(tu)\'=1/t
Integrating gives
tu=ln(t)+C
u=(ln(t)+C)/t
Hence,
y=t/(C+ln(t)).
What are the implications of Darwins ideas for the traditional .pdf
What are the implications of Darwin\'s ideas for the traditional view of human nature (both
rationalistic and Judeo-Christian versions)?
What are the implications of Darwin\'s ideas for the traditional view of human nature (both
rationalistic and Judeo-Christian versions)?
Solution
Traditional Western views of human nature assume that all humans have the same human nature:
They are conscious, rational selves who have a purpose. Some Traditional views refer to this
immaterially defined self as the \"soul\". One important version of the Traditional Western view
of human nature is the ancient Greek view that sees humans as uniquely rational beings with a
purpose. This view contends that reason can and should rule over human desire and
aggressiveness. The Judeo-Christian religious view claims that humans are made in the image of
God, who has endowed them with rational self-consciousness and an ability to love. In the
Christian version, the self is immaterial and distinct from the body.. The Traditional view of
human nature and our ordinary thinking assume that humans have a self that endures through
time. Descartes claims that the enduring self is a soul. Locke argues that memory produces the
enduring self. Buddhism and Hume suggest that there is no enduring self.
Some scientific views challenge the Traditional view of human nature. Darwin argued that
humans evolved from earlier animal species through random variations and a natural selection
that is the result of a struggle for existence. Darwin\'s view has been taken to imply that human
nature has no purpose and is not unique..
What are the factors that underlie the observation that EBV infectio.pdf
What are the factors that underlie the observation that EBV infection give rise to so many tumor
types?
a. Different virus latency programs combined with the state of B cell differentiation
b. A single virus latency program acting on B cell mutations during maturation
c. Different virus latency programs in B blast cells cause different tumor types
d. A single virus latency program acting on different states of B cell differentiation
Solution
c. Different virus latency programs in B blast cells cause different tumor types
Some of the EBV infected lymphocytes remain in the circulation, the virus remains in a latent
state. When activated, each of these cells give rise to a spontaneous outgrowth of EBV
transformed, immortalized cell lines..
What is an advantage of the hanging drop method verses simple sta.pdf
What is an advantage of the hanging drop method verses simple staining? What is a
disadvantage?
What is an advantage of the hanging drop method verses simple staining? What is a
disadvantage?
Solution
Advantages of Hanging drop method over simple staining:
Simple staining includes staining and viewing microorganisms using basic dyes which are
attracted to negatively charged microbial cytoplasm materials. However, if microbes are large
and if motility is to be studied; staining becomes unnecessary. In these cases, wet mount is
prepared i.e. a drop of microbes is directly put on slide and observed. Hanging drop method
includes placing a drop of sample below a cover-slip which is when inverted moves through
hollowed-out slide. This method is advantageous because it keeps the sample hydrated i.e
doesn’t dry out. This is a rapid method and makes it easy to observe especially larger microbes
for examining motility.
Disadvantages:.
What are key factors to consider when selecting a flow sensorSo.pdf
What are key factors to consider when selecting a flow sensor?
Solution
1)Viscosity and Transparency
viscosity plays an important role as a sensor will not work with liquids that are too dense
2) Flow Sensor Bodies and Hazardous Chemicals
Hazardous chemicals can pose a problem to the flow sensor as the chemicals can damage
sensitive parts of the device such as the fittings or O-rings.
Flow Range Needed
Flow range sizing is important because a single flow meter can only accurately measure typically
10:1, so a unit rated for 10 L/minute will only measure as low as 1 L/minute accurately
Some other options and considerations may be.
The process of DNA replication creates a particular problem for repl.pdf
The process of DNA replication creates a particular problem for replicating the ends of linear
chromosomes.
1. Describe why the DNA replication machinery has difficulty replicating DNA ends.
2. Telomerase has been identified as an enzyme that can reverse the outcome of end replication.
Describe the near-universal mammalian chromosome telomere sequence and how telomerase
solves the problem of end replication.
Solution
Answer:
1. Describe why DNA replication machinery has difficulty replicating DNA ends:
Eukaryotic chromosomes are linear and not circular like bacterial chromosomes. The end of
eukaryotic chromosomes contains repetitive sequences (telomere) and cannot be fully copied
during each round of replication. DNA replication begins by unwinding of DNA double helix
and the synthesis of new strand by DNA polymerase. Since DNA polymerase can only work in a
unidirectional way (5’ to 3’) therefore one of the two new strands of DNA is synthesized
continuously (leading strand) and the other strand is synthesized in small stretches called as
Okazaki fragments (lagging strand). At the end of the lagging strand, since it is not possible to
attach an RNA primer, thererfore, the terminal region of chromosomes does not get covered by
Okazaki fragments. As a result, with each round of replication, the terminal chromosome
sequence cannot be copied leaving single stranded overhangs and gradual shortening of
chromosomes. This is called as “end replication problem”.
2. Describe Near Universal mammalian chromosome telomere sequence and how telomerase
solves the problem of end replication:
The termini of eukaryotic chromosome have specialized DNA caps called telomere which
comprise of short repetitive DNA sequence. Each time a cell divides some portion of the
telomere is lost.
Mammalian Telomere contains hundreds or thousands of tandem repeats of the same short DNA
sequence 5’-TTAGGG-3’. With each cell division telomeric sequence are lost thereby
protecting the critical chromosomal regions that contains gene.Telomerase (telomere terminal
transferase) is an RNA dependent DNA polymerase made up of RNA subunits and protein and is
involved in elongating the length of chromosome by extending the telomeres. Telomerase
enzyme binds to a special RNA molecule that contains a sequence complementary to the
telomeric repeat. By using the complementary RNA as a template, it extends the overhanging
strand of the telomere DNA. Since DNA polymerase cannot replicate the DNA sequence at the
end of the chromosomes therefore telomerase enzyme aids in terminal replication process by
attaching to the end of the chromosome. Absence of telomerase activity would result in
progressive shortening of chromosome with each cell division..
The lifetime T of a certain device (in 1000 hour units) has the expo.pdf
The lifetime T of a certain device (in 1000 hour units) has the exponential distribution with
parameter ½. Find P(T > 2).
Solution
Given that T is the lifetime of a certain device.
T has Exponential distribution with paramter with parameter 1/2.
We have to find the P(T > 2).
The p.d.f. of T is,
f(T) = e-t t > 0 and > 0.
given that = 1 / 2 = 0.5
P(T>2) = f(T) dt (from 2 to )
= e-t dt
= e-t dt
= [ e-t / - ]
= - / [ e- - e-0.5*2 ]
= - [ 0 - 0.3679 ]
P(T >2) = 0.3679.
The Federal Reserve controls the nations money supply by using var.pdf
The Federal Reserve controls the nation\'s money supply by using various tools that it has at its
disposal. Which of the following is not a tool used by the Fed to change the money supply?
Discount window lending.
The federal funds rate.
Open-market operations.
Changing reserve requirements.
Discount window lending.
The federal funds rate.
Open-market operations.
Changing reserve requirements.
Solution
Fed doesnt use DISCOUNT WINDOW LENDING as one of the tools to change money supply..
Suppose S is a set people, and R is the following relation on S aRb.pdf
Suppose S is a set people, and R is the following relation on S: aRb if and only if a has chatted
with b on the internet. Explain your answers to the following questions and use common sense.
(a) Is R reflexive? (b) Is R symmetric? (c) Is R transitive?
Solution
a.it is not reflexive because a has not chatted with himself.
b. Yes it is symmetric beacause a and b belong to set of people. As a chatted with b same can be
said for b that b also chatted with a.
c. There is no clear solution for transitive as it has not mentioned involvement of any third
person apart from a and b..
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