What will organizations have to do to adapt to the influx of needed .pdf
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What will organizations have to do to adapt to the influx of needed Generation Y individuals? Solution simple thing is choose the right person for a work and let them free, they will create miracles. here generation Y people are those who can perform variaous tasks at a time and with more accuracy. these are multi talented, always carries courage, never hides their talent and ready to accept challenges. first to attract those kind of people organisations should change their existing rules, policies and regulations. all they were writen decades ago and still we are follwoing them. they need to redesign the work and the structure. if we are succede at recruiting the correct people for our work, then the remaining changes can automatically comes into the picture. the organisations should create the work environment to the Y generation people There are no precise dates when the generation starts and ends; most researchers and commentators use birth years ranging from the early 1980s to early 2000s. Generation Y people may be known as echo boomers because they are the children of baby boomers, or someone who was born during the period of increased birth rates that occurred between 1946 and 1964. The Generation Y cohort is also referred to as Millennials, the Internet generation, iGen and/or the Net generation.
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Solution
T is a bounded linear operator. Hence, T*T is also a bounded linear operator as composition of
two bounded linear operator is a bounded linear operator.
I is the identity operator. Hence, it is bounded linear operator. So, S = I + T*T is a bounded
linear operator as sum of two bounded linear operator is a bounded linear operator.
Now, S : H --> H is a bounded linear operator.
The identity operator I : H --> H can be written as S-1S : H --> H as I = S-1S.
Let, h H. Then I(h) = h
This can be written as S-1S(h) = h
Or, S-1(S(h)) = h
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Solution
Energy is of great importance in human life because of its benefits as the main resource for
human activity.
The application of nano technology or nano material in the field of energy, which involves
lithium-ion battery, fuel cell, light emitting diode (LED), ultra-capacitor, and solar cell
(including Grätzel cell).
Nanotechnology will bring significant benefits to the energy sector, especially to energy storage
and solar energy. Improved materials efficiency and reduced manufacturing costs are just two of
the real economic benefits that nanotechnology already brings these fields and that’s only the
beginning. Battery storage capacity could be extended, solar cells could be produced cheaper,
and the lifetime of solar cells or batteries for electric cars could be increased, all thanks to
continued development of nanotechnology.
The following nanomaterial technologies will be of particular importance: \"organic and printed
electronics\", \"nano-coatings\", \"nano-composites\", \"nano-fluids\", \"nano-catalysts\",
\"nanocarbons\" and \"nano-electrodes\".
Nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot
sensitized solar cells QDSSC) . Nanocomposites are also used for the capture of energy in
silicon-based solar cells, typically in the form of nanowires Nanocomposites can also be found in
antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O
fi lms via sol-gel-method) In the fi eld of energy storage the advantages of nanocomposites are
needed in thin, fl exible energy storage devices with nanocomposite units, e.g. supercapacitors,
Li-ion batteries and hybrid devices Nanotechnology in the sectors of solar energy and energy
storage Additionally, nanocomposites can be used in the process of semiconductor-assisted
photocatalysis for fuel production in order to obtain solar hydrogen or methanol For example, the
nanocomposites applied here are TiO2 -Gold composite nanoparticles or semiconductor-
semiconductor composites Another fi eld of application lies in the wind energy sector, where
lightweight nanocomposite materials are mainly used for rotor blades.
nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
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sensitized solar cells QDSSC) Nanocomposites are also used for the capture of energy in silicon-
based solar cells, typically in the form of nanowires . Nanocomposites can also be found in
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Differences
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activities create misunderstandings and frustrations
By the year 2014, 70 million Baby Boomer (including
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Harnessing the Potential of Today’s Multi Generational Workforce in Singapore...Centre for Executive Education
This document discusses managing a multi-generational workforce in Singapore. It identifies five generations currently in the workforce - Traditionalists, Baby Boomers, Generation X, Generation Y, and Generation Z. Each generation brings different experiences and expectations to work. Managing different generations effectively can create advantages for organizations. Leaders must understand generational characteristics to engage employees and leverage strengths. Highly engaged employees improve customer satisfaction and organizational success through leadership that creates a positive organizational climate.Harnessing The Potential of Today’s Multi-Gen Workforce - Today Manager Issue...
Harnessing The Potential of Today’s Multi-Gen Workforce - Today Manager Issue...Centre for Executive Education
This document discusses managing a multi-generational workforce in Singapore. It identifies five generations currently in the workforce - Traditionalists, Baby Boomers, Generation X, Generation Y, and Generation Z. Each generation brings different experiences and expectations to work. Managing different generations effectively can create advantages for organizations. Leaders must understand generational characteristics to engage employees and leverage strengths. This impacts organizational culture, employee engagement, customer satisfaction, and business success.HR in 2017 (Englisch version)
Engelstalige essay over HR in 2017.
Deze essay was deelname aan de NVP International Future HR-Award 2003 en bekroond met de eerste prijs!
1e plaats: HR in 2017, Espace-transformation-a future HR-essay
De bijdrage van Mark van den Broek, HR Development Directeur bij FCI, getuigt van een integrale zakelijke visie op de toekomst van de HR-professie. Zijn essay bevat een heldere probleemanalyse en toont een duidelijke link met de actuele en toekomstige HR beleidsdoelstellingen. Zijn bijdrage onderscheidt zich door een creatieve, soms gedurfde toepassing van huidige en/of nieuwe HRM-instrumenten, -methodieken en/of denkwijzen. Zijn bijdrage is verzorgd, prettig leesbaar, degelijk, compleet en zet de lezer aan tot denken.
NVTC Millennial Board Research v3
The document summarizes research conducted by a team of millennials on attracting and retaining millennials in Northern Virginia. The research included analyzing 28 sources, identifying 215 findings, and deriving 21 themes and 5 key stories. The stories covered millennial characteristics, living preferences, vital benefits, values, and how to manage millennials. Some of the main findings were that millennials are naturally disloyal but loyal to companies sharing their values; prefer cities and walkable, diverse communities; want career progression, flexibility, and benefits like tuition reimbursement; and highly value ethics, social responsibility, and diversity.
Understanding Generational Types
Understanding different generations is important for recruitment, retention, and training. As baby boomers retire, there will be labor shortages. Younger generations like millennials and Gen Z have different expectations of work-life balance, feedback, and career growth. Organizations need to understand generational traits and adapt practices to engage multi-generational employees. This includes knowledge transfer from retiring workers, clear career paths, frequent feedback, work-life balance, and leveraging each generation's strengths. Adapting to different needs is key to retaining talent amid competition for skilled workers.
Wellness Works_GEN_Y_WORKFORCE Edition 1
The document discusses generational differences in the workforce, focusing on Generation Y. It notes that Gen Y, consisting of 70 million people born between 1980-2000, will be increasingly important as Baby Boomers retire. Gen Y has different values and priorities than previous generations due to factors like technology and wants flexibility, fun, ethics, and mentoring from employers. Understanding Gen Y's perspectives is important for companies to attract and engage this upcoming segment of the workforce.
Generations
This document discusses generational differences in the workplace and provides strategies for managing a multigenerational workforce. It identifies the four main generations currently in the workforce - Veterans, Baby Boomers, Generation X, and Generation Y. Each generation has different values, work ethics, and preferences that can lead to conflicts if not properly managed. The document recommends identifying each employee's generation, drawing on the strengths of each, adapting management styles to different generations, accepting generational differences, and motivating employees in generation-specific ways. Effective multi-generational management can help reduce conflicts and maximize productivity.
Recruitment baby boomers, gen x & gen y
Difference between Baby Boomers, Generation X and Generation Y and their attitude towards work. How to manage Generation X & Y and what to expect from the three generation
Understanding Generation Y: The New generations at work. Attracting, Recruiti...
Our original 2006 paper on Generation Y. This has been one of the most internationally downloaded, shared, referenced and cited papers on Generation Y.
Preliminary report
This document discusses what skills employers want in young job applicants. It outlines that employers feel applicants lack "soft skills" needed to succeed in the workplace. These include skills like communication, decision making, commitment, flexibility, time management, and leadership. The document also examines how technology has impacted different generations and changed learning. It notes technology has created a gap in interpersonal skills vital for work. Employers want candidates who can combine digital skills with soft skills like problem solving, collaboration, and critical thinking.
Managing Millennial Generation Future Leaders
This document discusses strategies for managing the millennial generation in the workplace. It begins by outlining learning objectives around winning talent wars with millennials, adopting a new mindset, and creating a millennial-friendly culture. It then profiles different generations and their core values before focusing on millennials' characteristics like being confident, tech-savvy, and seeking work-life balance. The rest summarizes best practices for attracting, selecting, developing, and retaining millennial talent through flexible work, frequent feedback, mentorship, and an emotionally intelligent culture.
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Understanding Generational DifferencesUnderstand G.docx
Generational Differences In Workplace [Supervisory Training}
Generational Differences In Workplace [Supervisory Training}
Harnessing the potential of today’s multi gen workforce - today manager issue...
Harnessing the potential of today’s multi gen workforce - today manager issue...
Harnessing the Potential of Today’s Multi Generational Workforce in Singapore...
Harnessing the Potential of Today’s Multi Generational Workforce in Singapore...
Harnessing The Potential of Today’s Multi-Gen Workforce - Today Manager Issue...
Harnessing The Potential of Today’s Multi-Gen Workforce - Today Manager Issue...
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Understanding Generation Y: The New generations at work. Attracting, Recruiti...
More from arihantstoneart
Let S = I + TT H rightarrow H, where T is linear and bounded. Show.pdf
Let S = I + T*T: H rightarrow H, where T is linear and bounded. Show that S^-1: S(H)
rightarrow H exists.
Solution
T is a bounded linear operator. Hence, T*T is also a bounded linear operator as composition of
two bounded linear operator is a bounded linear operator.
I is the identity operator. Hence, it is bounded linear operator. So, S = I + T*T is a bounded
linear operator as sum of two bounded linear operator is a bounded linear operator.
Now, S : H --> H is a bounded linear operator.
The identity operator I : H --> H can be written as S-1S : H --> H as I = S-1S.
Let, h H. Then I(h) = h
This can be written as S-1S(h) = h
Or, S-1(S(h)) = h
As h H and S : H --> H is a bounded linear operator, S(h) S(H). Let S(h) = h1 S(H)
So, S-1(h1) = h where h H and h1 S(H)
As h is arbitrary, we can say that S-1 : S(H) --> H exists. Proved.
It is a nanotechnology question. In which you are able to pick one m.pdf
It is a nanotechnology question. In which you are able to pick one material for both questions or
one material for question one and another for question two.
Solution
Energy is of great importance in human life because of its benefits as the main resource for
human activity.
The application of nano technology or nano material in the field of energy, which involves
lithium-ion battery, fuel cell, light emitting diode (LED), ultra-capacitor, and solar cell
(including Grätzel cell).
Nanotechnology will bring significant benefits to the energy sector, especially to energy storage
and solar energy. Improved materials efficiency and reduced manufacturing costs are just two of
the real economic benefits that nanotechnology already brings these fields and that’s only the
beginning. Battery storage capacity could be extended, solar cells could be produced cheaper,
and the lifetime of solar cells or batteries for electric cars could be increased, all thanks to
continued development of nanotechnology.
The following nanomaterial technologies will be of particular importance: \"organic and printed
electronics\", \"nano-coatings\", \"nano-composites\", \"nano-fluids\", \"nano-catalysts\",
\"nanocarbons\" and \"nano-electrodes\".
Nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot
sensitized solar cells QDSSC) . Nanocomposites are also used for the capture of energy in
silicon-based solar cells, typically in the form of nanowires Nanocomposites can also be found in
antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O
fi lms via sol-gel-method) In the fi eld of energy storage the advantages of nanocomposites are
needed in thin, fl exible energy storage devices with nanocomposite units, e.g. supercapacitors,
Li-ion batteries and hybrid devices Nanotechnology in the sectors of solar energy and energy
storage Additionally, nanocomposites can be used in the process of semiconductor-assisted
photocatalysis for fuel production in order to obtain solar hydrogen or methanol For example, the
nanocomposites applied here are TiO2 -Gold composite nanoparticles or semiconductor-
semiconductor composites Another fi eld of application lies in the wind energy sector, where
lightweight nanocomposite materials are mainly used for rotor blades.
nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic
PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot
sensitized solar cells QDSSC) Nanocomposites are also used for the capture of energy in silicon-
based solar cells, typically in the form of nanowires . Nanocomposites can also be found in
antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O
fi lms via sol-gel-method) . In the fi eld of energy storage the advantages of nanocomposites .
I have this problem that I was given in class, but I cant for the .pdf
I have this problem that I was given in class, but I can\'t for the life of me figure out how to do
part (d). Could someone please do it for me if it\'s possible? If the problem is leaving out some
information, then let me know. I am having a problem finding a way to do this part with the
information given, so I will tell my teacher if this problem can\'t be answered with what is in the
problem statement. 1. During the Apollo program, the U.S. landed 12 men on the Moon and
returned them safely to Earth. It was an amazing feat and an exciting time to be an aerospace
engineer. The last Apollo mission was accomplished in December, 1972. The descent stage (DS)
separated from the command module (CM), which remained in lunar orbit with the CM pilot.
The DS carried two astronauts to a landing on the Moon. When the lunar expedition ended, the
ascent stage (AS) rendezvoused with the CM for the return trip to Earth. Here\'s your chance to
analyze the AS trajectory using actual values. 5.322 ft/sec (acceleration due to gravity on the
Moon) 5,187 lb. (Earth weightas 10,295 lb. (Earth weight including propellant); weight of AS
propellant- w weight). a) What is the mass of the AS including propellant? b) What does the AS
and propellant weigh on the Moon? c) What is the propellant mass? The thrust of the ascent
propulsion system (T-3,617 lb. The thrust duration is called action time (tama teton = 446 sec. Of
course, the mass of the AS varies as propellant is expended. d) Assuming the propellant mass
flow rate is constant, what is the mass flow rate during the motor burn (from t-0 to t-taction)?
Note that the AS mass, m, is a function of time such that m(t)=m() + m (t. t.) where m=-. Also let
t.-0, so that m(t)=m(0)+m t . Note that mc0 Now, the total force acting on the AS in the vertical
direction is T-mt)gThe vertical acceleration (up is positive) is T-m(t)g m(t) a(t) (From F ma). e)
What is the speed of the AS at the end of the motor burn, that is, when t? For a first-order
approximation, assume the AS trajectory is vertical and does not pitch over to rendezvous with
the CM. Further assumegarconstant. dt v()-sal)dt jdb.bm(a + bt) Some useful information: =-ln (a
+ bt
Solution
Autocorrect is almost as old as personal computers. Even some of the earliest word processors
had a spellcheck feature that suggested alternative words if a word you typed did not appear in
its internal dictionary. Certainly, today, autocorrect has come a long way since the Cupertino
effect. What’s the Cupertino effect? Well, in its early days, word processors would replace the
word “cooperation” with “Cupertino” (the city in Northern California where Apple and other
computercompanies are headquartered). This spelling suggestion made its way to many
documents published by the United Nations, NATO, and other official bodies, so the Cupertino
In the beginning autocorrect was primarily a smart spellchecker. If you typed “bfeore,” it
replaced it with “before.” Then in 2007, Microsoft int.
Implement this in Java Method to determine if a particular elemen.pdf
Implement this in Java
* Method to determine if a particular element is in this set.
* @param target * the element that needs to be found in this set
* @return * true if the target element is in this set, false otherwise
public boolean contains(int target)
{
}
Solution
public boolean contains(int target)
{
boolean flag=false;
for(Integer i:(this))
{
if(target==i.intValue())
{
flag=true;
break;
}
}
return flag;
}.
If an image is represented using 400 x 300 pixels and each pixel is .pdf
If an image is represented using 400 x 300 pixels and each pixel is represented using 8 bits, what
will be the exact amount of data needed in kilobytes to represent in the image?
Solution
there are 1200 pixels in total.
Each pixel is represented requires 8 bit for its representation which 1 byte of data.
Now for 1200 pixels we require 1200 bytes of data to represent a 400x300 pixel image.
1 byte = 1/1024kb
1200 byte = 1200*(1/1024) = 1.17 kb data is required to present an image..
Implement the unsorted single linked list as we did in the class and .pdf
Implement the unsorted single linked list as we did in the class and implement the following
operations: 1.DeletelastDuplicat(): For any element in the linked list, if there are multiple copies
(> = 2 copies), delete the last copy. DeleteSecondlastDuplicat(): For any element in the linked
list, if there are multiple copies (> = 2 copies), delete the second last copy Test your program
with the following operations: a)insert 5 b)Insert 7 c)Insert 11 d)Insert 5 e)Insert f) Insert 5 g)
Print out the list h) Delete the last duplicate of 5 i)Print out the list i) Delete the last duplicate
of 11 k) Print out the list I) Insert 11 m) Insert 7 n) Print out the list O) Delete the second last
duplicate of 5 p) Print out the list q) Delete the second last duplicate of 7 r) Print out the list
Solution
#include
#include
#include
using namespace std;
struct node
{
int data;
struct node *next;
}*start;
class LinkedList
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos(int);
void delete_lastdup();
void delete_seclast();
int search(int);
void display();
int get_count(int);
LinkedList()
{
start = NULL;
}
};
int main()
{
int choice, pos, value, res=0;
LinkedList sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<>pos;
sl.delete_pos(pos);
break;
case 5:
cout<<\"Search element in Link List: \"<>value;
res=sl.search(value);
cout<<\"Element \"<data = value;
temp->next = NULL;
return temp;
}
}
//function to insert element at beginning
void LinkedList::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<next != NULL)
{
if(s->data == value)
count++;
s = s->next;
}
return count;
}
//function to insert node at last
void LinkedList::insert_last()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<>value;
c=get_count(value);
if(c<2)
{
cout<<\"duplicate does not exist\"<>value;
c=get_count(value);
x=c-1;
if(c<2)
{
cout<<\"duplicate does not exist\"<data == value)
{
f++;
res=pos;
if(f==x)
break;
}
s = s->next;
}
delete_pos(res);
}
}
//delete element at p.
Hyposecretion of insulin is referred to as Type I diabetes insipidus.pdf
Hyposecretion of insulin is referred to as Type I diabetes insipidus. Type II diabetes mellitus
Type I diabetes mellitus sugarbeetes pancreatic dwarfism
Solution
33. Hyposecretion of insulin is referred to as -
C) Type I diabetes mellitus.
Type I diabetes mellitus : It is happened due to deficient amount of insulin production , by
pancreatic beta cells of islets of Langerhans..
How are the forces of diffusion and electrical force responsible for.pdf
How are the forces of diffusion and electrical force responsible for both the resting potential and
the action potential?
Solution
Ans). In a resting state, diffusion and electrical force balance each other to attain an equilibrium
for potassium and chloride ions. Large number of protein molecules cannot move through the
membrane due to their size because of their negative charge, they contribute to the relative
negativity of the intracellular fluid..
How does each of the differences between prokaryotes and eukaryotes .pdf
How does each of the differences between prokaryotes and eukaryotes influence bacterial
infection and treatment
Solution
The differences between prokaryotes and eukaryotes such as cell wall, nucleus, flagella,
membrane-bound organelles, genetic makeup, metabolic differences and antigenicity are
responsible for the infection and treatment of bacterial diseases. For example, the bacteria
contains cell wall, which is absence in animal cells. So, cell wall synthesis inhibition is a strategy
in the treatment of bacterial infections..
For MIMO system, (a) Please talk about the advantage and disadvantag.pdf
For MIMO system, (a) Please talk about the advantage and disadvantage of MIMO technique.
(b) Please explain why we need Cooperative Diversity to replace MIMO in cellular network.
Solution
A)MIMO advantages
• Capacity scales linearly with number of antennas
Channel knowledge/estimation at Rx needed
• MIMO offers potential for
larger data rate
larger spectral efficiency
larger number of users
improved range/coverage
better interference suppression
better quality of service (QoS), lower bit-error rate (BER)
lower Tx power
MIMO disavantages
• Hardware complexity:
Each antenna needs a radio-frequency (RF) unit
Powerful digital signal processing (DSP) unit required
• Software complexity:
Most signal processing algorithms are computationally intensive
Power consumption:
Battery lifetime of mobile devices
Thermal problems
Antennas:
Antenna spacing (electromagnetic mutual coupling-e.g. mobile handsets)
RF interference and antenna correlation
B)
Most of the present researches on cooperative
network in which the user nodes are equipped with a single
antenna or multiple, there have been some new results which
exploit the benefits of multiple antenna deployment.
Cooperative MIMO technology allows a wireless network
system to attain better performance gains than provided by
either usual MIMO or cooperative systems. It promised
significant improvement in spectral efficiency and network
coverage phenomena for different next generation wireless
communication systems. In wireless communication, the path
towards the various techniques that gives high service quality
and data rate has been through the use of the cooperative
network provided by the rich scattering wireless channels.
Due to their great aspects, MIMO and cooperative systems
have found their way into several standards for future
wireless communication systems, especially in cellular
networks and wireless local area networks (LAN) in this
review article we are presenting the comparative analysis..
Explain why and give examples. In OO programming Polymorphism is on.pdf
Explain why and give examples. \"In OO programming Polymorphism is only possible when
using Interface or Inheritance\".
Solution
Polymorphism is the ability of an object to take on many forms. The most common use of
polymorphism in OOP occurs when a parent class reference is used to refer to a child class
object.
Any Java object that can pass more than one IS-A test is considered to be polymorphic. In Java,
all Java objects are polymorphic since any object will pass the IS-A test for their own type and
for the class Object.
It is important to know that the only possible way to access an object is through a reference
variable. A reference variable can be of only one type. Once declared, the type of a reference
variable cannot be changed.
The reference variable can be reassigned to other objects provided that it is not declared final.
The type of the reference variable would determine the methods that it can invoke on the object.
A reference variable can refer to any object of its declared type or any subtype of its declared
type. A reference variable can be declared as a class or interface type.
Polymorphism using inheritance:
Polymorphism using interface:
class Main
{
}.
Eukaryotic cells modify RNA after transcription What critical RNA pr.pdf
Eukaryotic cells modify RNA after transcription What critical RNA processing events usually
happen to pre-mRNA\'s before they sent to the cytoplasm for translation? What is a 5\' cap?
What is a poly A tail? What do these end modifications do for the mRNA transcript? What is
RNA splicing? What are introns? What are exons? How are intros spliced out? What is a
spliceosome? What is a \"snurp\" (snRNA)?
Solution
After its synthesis, the eukaryotic mRNA will undergo extensive modification like capping,
polyadenylation and splicing to enter into the process of translation.
(B)
Capping: Here the 5\' end of the mRNA is modified by the addition of 7-methylguanosine (m7G)
and the main function of this cap is to protect the 5\' end of the primary RNA transcript from
attack by ribonucleases and this 5’ cap will be recognized by eukaryotic initiation factors, so that
it can assemble the mature mRNA with the ribosome to start the process of translation.
At the 3\' end of the RNA, we will have polyadenylation signal and during transcription itself
this sequence will be chopped by an enzyme and another enzyme will add about 100100100 -
200200200 adenine (A) nucleotides to the 3’ end and this will form the poly-A tail.
The main function of this poly A tail is proving stability to the transcript and also helping it to
get exported from the nucleus to the cytosol.RNA splicing
(C )Splicing is the third big RNA processing event and the pre-mRNA will have two sequences,
exons and introns.
Introns are the non-coding sequences and exons are the coding sequences, here in this step the
through splicing, the introns will be removed and exons will be attached together.
In RNA splicing, specific parts of the pre-mRNA (introns) will be recognized and removed by a
protein-and-RNA complex called the spliceosome. Mature mRNA will have only exons, but no
introns.
(D)The splicing signal exon/GU-intron-AG/exon will be present in nuclear mRNA precursors
and 5\' and 3\' splice sites always have consensus sequences extending beyond GU and AG
motifs. During splicing, the exon-intron boundaries will be recognized by snRNA and the
consensus sequences within introns will get hybridized and now the proteins other snRNAs will
assemble the spliceosome on the transcript, the unpaired A present at 3\' side of the introns will
attacks the 5\' exon -intron boundary with the help of 2\' OH and this will give rise to lariat
structure. The free 3\' OH of the upstream exon will displace the downstream junctional
nucleotide, like this introns will be removed and exons will be attached together.
(E)A spliceosome is a large complex formed due to the assembly of snRNAs and protein
complexes, and plays an important role in splicing of pre mRNA.
snRNPs (snurps) is the small nuclear ribonucleic proteins and this is a RNA-protein complexes
and they will form a larger complex with the unmodified pre-mRNA and various other proteins
to form the structure called spliceosome.(A)Critical RNA processing events.
Driverimport java.util.Scanner;A class that keeps a f.pdf
Driver:
import java.util.Scanner;
//A class that keeps a fleet of different types of vehicles
public class FleetOfVehicles {
//An array of vehicles
private Vehicle[] fleet;
public static final int MAX_VEHICLES = 100;
public FleetOfVehicles()
{
fleet = new Vehicle[MAX_VEHICLES];
}
public Vehicle[] getFleet()
{
return this.fleet;
}
//Adds a new vehicle to the first empty spot in the fleet array
public void addVehicle(Vehicle aV)
{
for(int i=0;i
Solution
import java.util.Scanner;
//A class that keeps a fleet of different types of vehicles
public class FleetOfVehicles {
//An array of vehicles
private Vehicle[] fleet;
public static final int MAX_VEHICLES = 100;
public FleetOfVehicles()
{
fleet = new Vehicle[MAX_VEHICLES];
}
public Vehicle[] getFleet()
{
return this.fleet;
}
//Adds a new vehicle to the first empty spot in the fleet array
public void addVehicle(Vehicle aV)
{
for(int i=0;i 0)
this.cylinders = cylinders;
else
this.cylinders = 0;
}
/**
* @return the ownersName
*/
public String getOwnersName() {
return ownersName;
}
/**
* @param ownersName
* the ownersName to set
*/
public void setOwnersName(String ownersName) {
this.ownersName = ownersName;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
if (obj instanceof Vehicle) {
Vehicle vehicle = (Vehicle) obj;
if ((this.getManuName().equals(vehicle.getManuName()))
&& (this.getOwnersName().equals(vehicle.getOwnersName()))
&& (this.getCylinders() == vehicle.getCylinders()))
return true;
else
return false;
} else
return false;
}
@Override
public String toString() {
// TODO Auto-generated method stub
return getManuName() + \" \" + getCylinders() + \" \" + getOwnersName();
}
}
public class Truck extends Vehicle {
double loadCap, towCap;
public Truck() {
// TODO Auto-generated constructor stub
super();
setLoadCap(0);
setTowCap(0);
}
/**
* @param manuName
* @param cylinders
* @param ownersName
* @param loadCap
* @param towCap
*/
public Truck(String manuName, int cylinders, String ownersName,
double loadCap, double towCap) {
super(manuName, cylinders, ownersName);
setLoadCap(loadCap);
setTowCap(towCap);
}
/**
* @return the loadCap
*/
public double getLoadCap() {
return loadCap;
}
/**
* @param loadCap
* the loadCap to set
*/
public void setLoadCap(double loadCap) {
if (loadCap > 0)
this.loadCap = loadCap;
else
this.loadCap = 0;
}
/**
* @return the towCap
*/
public double getTowCap() {
return towCap;
}
/**
* @param towCap
* the towCap to set
*/
public void setTowCap(double towCap) {
if (towCap > 0)
this.towCap = towCap;
else
this.towCap = 0;
}
@Override
public boolean equals(Object obj) {
// TODO Auto-generated method stub
if (obj instanceof Truck) {
if (super.equals(obj)) {
Truck vehicle = (Truck) obj;
if (this.getLoadCap() == vehicle.getLoadCap()
&& this.getTowCap() == vehicle.getTowCap())
return true;
else
return false;
} else
return false;
} else
return false;
}
@Override
public String toString() {
// TODO Auto-generated method stub
return super.toString() + \" \" + g.
Describe a data structure that supports both removeMin() and rem.pdf
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Solution
The idea here is to use the concept of two binary heaps along with the concept of list(doubly
linked list).
The doubly linked list contains all input items and indexes of corresponding min and max heap
nodes. The nodes of min and max heaps store addresses of nodes of doubly linked list. The root
node of min heap stores the address of minimum item in doubly linked list. Similarly, root of
max heap stores address of maximum item in doubly linked list. Following are the details of
operations.
1) findMax(): We get the address of maximum value node from root of Max Heap. So this is a
O(1) operation.
1) findMin(): We get the address of minimum value node from root of Min Heap. So this is a
O(1) operation.
3) removeMin(): We get the address of minimum value node from root of Min Heap. We use this
address to find the node in doubly linked list. From the doubly linked list, we get node of Max
Heap. We delete node from all three. We can delete a node from doubly linked list in O(1) time.
delete() operations for max and min heaps take O(Logn) time.
4) removeMax(): is similar to deleteMin()
5) Insert() We always insert at the beginning of linked list in O(1) time. Inserting the address in
Max and Min Heaps take O(Logn) time. So overall complexity is O(Logn)
#include
#include
#include
// A node of doubly linked list
struct LNode
{
int data;
int minHeapIndex;
int maxHeapIndex;
struct LNode *next, *prev;
};
// Structure for a doubly linked list
struct List
{
struct LNode *head;
};
// Structure for min heap
struct MinHeap
{
int size;
int capacity;
struct LNode* *array;
};
// Structure for max heap
struct MaxHeap
{
int size;
int capacity;
struct LNode* *array;
};
// The required data structure
struct CDS
{
struct MinHeap* minHeap;
struct MaxHeap* maxHeap;
struct List* list;
};
// A utility function to create a new List node
struct LNode* newLNode(int data)
{
struct LNode* node =
(struct LNode*) malloc(sizeof(struct LNode));
node->minHeapIndex = node->maxHeapIndex = -1;
node->data = data;
node->prev = node->next = NULL;
retu.
Columbus Incorporated just paid $4.3 per share dividend yesterday (i.pdf
Columbus Incorporated just paid $4.3 per share dividend yesterday (i.e.,D0). The dividend is
expected to grow at a constant rate of 6% a year. The required rate of return on the stock, r, is
11%. What is the value per share of the company\'s stock? Round your answer to two decimal
places.
Solution
Recent Dividend, D0 = $4.30
Growth Rate, g = 6%
Required Return, rs = 11%
D1 = D0 * (1 + g)
D1 = $4.30 * 1.06
D1 = $4.558
Value per share = D1 / (rs - g)
Value per share = $4.558 / (0.11 - 0.06)
Value per share = $4.558 / 0.05
Value per share = $91.16.
Anatomy Question Please Answer them all.A. Using a punnett square.pdf
Anatomy Question: Please Answer them all.
A. Using a punnett square, show the possible genotypes for John and Andrea\'s children? (John
has O blood type and Andrea has A blood type)
B. Based on the blood typing results, could Daryl be John and Andrea\'s Son? How Do you know
this? ( Daryl has O blood type)
Solution
A. Andrea :Mother has A blood type , her genotype will be either AO or AA
John :Father has O blood type ,his genotype will be OO.
O allele is masked by the presence of A allele or B allele.
OO blood group is expressed only in presence of OO alleles.
Father
Father
B. Daryl has O blood type.For O blood froup to be expressed in child, both parents should have
O alleles that together will contribute to make OO blood type of their son.So, yes Daryl is the
son of John and Andrea.MotherOOAAO A blood groupAO A blood groupOOO O blood
groupOO O blood group.
A person is lost in the desert. (S)he has no water and after a coupl.pdf
A person is lost in the desert. (S)he has no water and after a couple of days loses so much water
through perspiration that his/her body fluids become hypertonic. What happens to his cells?
What happens to his proteins? What happens to his carbohydrates?
Solution
Early sign of dehydration there is no symptom but later on there is dryness of mouth and tongue,
dark yellow urine,dryness of skin cramp in arm and leg. When fluid become hypertonic i.e
sodium is lost from cell this result in cell shrinkage.
After two days of starvation body will first utilize carbohydrate and then glycogen from liver and
muscle for energy. Protein from cell / muscle will be next to be used as source of energy for
body.
2. Describe an advantage and a limitation to tracing ancestry with b.pdf
Tracing ancestry with mitochondrial DNA and Y-DNA has the advantage of providing information about historical human evolution over thousands of years, but is limited in its ability to unravel full familial relationships. Autosomal DNA provides a more complete picture of ancestral inheritance patterns, but has ethical limitations when used for prenatal ancestral tracing.
Write your thought on the following1. petroski notes that most en.pdf
Write your thought on the following:
1. petroski notes that most engineering failures are much more involved than simple technical
mis-calculations and involve the failure of the design process or management culture.However,
not all engineering failures involve ethical issue.The infamous collapse of the first Tacoma
Narrow Brodge, and the losses of the Mars polar lander and Mars Climate Orbiter were technical
and design process failures.
Solution
In the face of new technology, how do we balance public welfare and progress? If Moisseiff had
designed a bridge similar to the ones which had already proven their stability, Tacoma Narrows
Bridge would never have collapsed costing thousands of dollars and endangering many lives. It
would also have been significantly more expensive. On the other hand, if engineers had never
tried innovative techniques, suspension bridges may never have been built at all. At the time of
their introduction, no one believed that a suspension bridge could safely accommodate trains.
Roebling, however, took a gamble, pushed the limits of the current technology, and built a
suspension bridge that he believed could safely support rail traffic. Luckily he was correct, and
suspension bridges soon became widely accepted (Petroski. Moisseiff also took a gamble, trying
to create a longer, sleeker, less expensive bridge, by pushing the limits of technology. He,
however, was not as lucky, and what could have been a breakthrough in technology turned into a
catastrophic failure. Every time engineers push the limits of technology they risk a similar loss
like losses of mas polar lander and mass climate orbiter in which there were technical error like
Mass polar lander
1. Inadequate specifications which includes what the software was supposed to do, but no
mention of what it must not to do.
2. Flawed review process: Violation of basic engineering safety pracitices in the digital part of
the system.
Mass Climate Orbiter
1. Units communication and attention to detail failure. The primary cause of this discrepancy
was that one piece of ground software supplied by Lockheed Martin produced results in a United
States customary unit, contrary to its Software Interface Specification (SIS), while a second
system, supplied by NASA, expected those results to be in SI units, in accordance with the SIS.
Specifically, software that calculated the total impulse produced by thruster firings calculated
results in pound-seconds. The trajectory calculation software then used these results - expected to
be in newton-seconds - to update the predicted position of the spacecraft.
sometimes even It can cause a loss of life. How much is too much? When is a possible advance
worth a risk to public safety? What can the engineering profession do to make the
implementation of new technology safer? Do our current peer review and building code
committee processes adequately protect public safety?.
You have two cell lines, a macrophage cell line and a CD4 T-cell lin.pdf
You have two cell lines, a macrophage cell line and a CD4 T-cell line. The macrophage cell line
is resistant to infection by HIV, whereas the T-cell line is fully susceptible and fully permissive.
Why are macrophage cell line resistant to infection by HIV? Why are T-cell fully susceptible and
fully permissive? Please explain.
Also, A heterokaryon cell made from the fusion of these two cells is able to be infected by HIV.
Explain why? Please explain in detail.
Solution
Macrophages are antigen- presenting cells so they process the viral peptides and present it to
uninfected CD4 T-Cell which is the important transmission process. Also HIV-infected
macrophages release soluble cytotoxic factors which induce apoptosis in CD4 T-Cells.
Macrophages produce certain host restriction factors which inhibits the release of viral progeny
from infected cells. So macrophages are resistant to lysis because of its genetic make up and its
phenotypes (restriction factors).
So when two cell lines are fused to form heterokaryon cell, the genes of macrophages are
suppressed or it is recessive or the T cell genes are dominant to make the heterokaryon
susceptible to infection..
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