A. Intergenic sequences make up 60 of the human genome. Where do t.pdf
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A. Intergenic sequences make up >60% of the human genome. Where do these intergenic sequences come from? B. What are some of their functions? a. May encode mRNAs b. May encode miRNAs c. May be nonfunctional sequences d. May serve as regulatory sequences for transcription Solution A)Intergenic regions are short -coding regions that are found with in genes iin eukaryotic organisms. It is one of the DNA sequences sometimes referred to as junk DNA, Intergenic regions are a subset of noncoding DNA. these intergenic dna consists of parasitic dna regions of transposable elements.An Intergenic sequence is a stretch of DNA sequences located between genes.Intergenic regions are a subset of noncoding DNA. some intergenicDNA acts to control genes. B) d)May serve as regulatory sequences for transcription.
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This document discusses RNA interference (RNAi), a biological mechanism that leads to post-transcriptional gene silencing triggered by double-stranded RNA molecules. It was discovered in 1998 by Fire and Mello, who received the Nobel Prize for this work. The mechanism involves double-stranded RNA being processed by the enzyme Dicer into small interfering RNAs that integrate into the RNA-induced silencing complex and guide mRNA degradation. RNAi can be induced by synthetic siRNAs or endogenous microRNAs and represents a powerful tool for studying gene function and developing therapies.
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RNA interference (RNAi) is a mechanism where double-stranded RNA inhibits gene expression. It was discovered in plants, fungi, and animals. The mechanism involves dicer enzymes cleaving long double-stranded RNA into short interfering RNAs (siRNAs). These siRNAs are incorporated into an RNA-induced silencing complex (RISC) which guides the complex to mRNAs with complementary sequences, resulting in their degradation. RNAi has applications in therapeutics for cancer, viruses, and genetic disorders, as well as research in gene function and pathways.
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transcribing in a gene at the right place on the DNA of a prokaryotic cell? What makes RNA
polymerase start transcribing in a gene at the right place on the DNA of a eukaryotic cell?
Solution
A)Transcription is the first step in gene expression. It involves copying a gene\'s DNA sequence
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Initiation
Elongation
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complementary nucleotides,making a chain that grows from 5\' to 3\'.The RNA transcript carries
the same information as the non-template (coding) strand of DNA,but it contains the base uracil
instead of thymine .
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are transcribed,they cause the transcript to be released from the RNA polymerase.
B)Prokaryotes do not have an organized nucleus,so the nuclear materials or DNA is in the
cytoplasm.Therefore,the transcription occurs in the cytoplasm and all the precursors needed for
the transcription are found in the cytoplasm.Prokaryotic transcription requires the RNA
polymerase enzyme in order for the transcription to be successfully completed.The enzyme binds
to the sigma factor and the promoter region,and then initiate the transcription by completing the
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directly.This could be advantageous when prokaryotes have overlapping genes.Transcription
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Gene silencing
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Gene silencing is a technique that reduces or eliminates the production of a protein from its corresponding gene through mechanisms other than genetic modification. There are two main types of gene silencing - transcriptional and post-transcriptional. Post-transcriptional gene silencing includes RNA interference (RNAi), which involves introducing double-stranded RNA that binds to mRNA and prevents replication, effectively silencing gene expression in a sequence-specific manner. RNAi is a powerful tool for gene function analysis and applications in biotechnology and agriculture.
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Non – coding dna presentation
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3. While the amount of non-coding DNA varies between species, it plays roles in chromosome structure, gene expression regulation, and protecting the genome from degradation.
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Gene silencing and editing in plant PPT
Gene silencing and editing techniques allow for precise regulation of gene expression. RNA interference uses small RNA molecules like siRNA and miRNA to degrade target mRNA, decreasing protein production. Genome editing tools like CRISPR/Cas9 create targeted DNA breaks that are repaired, resulting in mutations like insertions or deletions that can knock out gene function. These techniques have applications in research including studying gene function and biotechnology to modify traits in crops.
If messages are being sent from your email account without your know.pdf
If messages are being sent from your email account without your knowledge, this is a sign that
your computer is infected with
Solution
If messages are being sent from your email account without your knowledge, this is a sign that
your computer is infected with:
Identity Theft. In identity theft, some other machine is purported to be the other machine. Which
means, somebody else will steal your identity and will send an email to somebody else, as if they
were sending a mail from your email account..
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Solution
The female art movement originated in the late 1960s which sought to rewrite a falsely male-
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as follows; It sought to create an interaction between the artist and viewer by including women\'s
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If you were looking at an mRNA and saw the codon AUG, what would you conclude about it?
What does it mean to that the genetic code is redundant, but not ambiguous? The genetic code is
nearly universal, meaning the same RNA codon that designates tryptophan in humans, designates
tryptophan in bacteria. a. What has this knowledge allowed us to conclude about the code? b.
What has this knowledge allowed us to do with genes Transcription is the DNA-directed
synthesis of RNA: a closer look Describe the general model of transcription. Include in your
answer the steps of transcription and the key elements of each step. Compare and contrast
transcription between prokaryotes and eukaryotes What makes RNA polymerase start
transcribing in a gene at the right place on the DNA of a prokaryotic cell? What makes RNA
polymerase start transcribing in a gene at the right place on the DNA of a eukaryotic cell?
Solution
A)Transcription is the first step in gene expression. It involves copying a gene\'s DNA sequence
to make an RNA molecule.Transcription can be divided into four distinct stages:
Template recognition
Initiation
Elongation
Termination
Initiation:RNA polymerase binds to a sequence of DNA called the promoter,found near the
beginning of a gene.Each gene has its own promoter.Once bound,RNA polymerase separates the
DNA strands,providing the single-stranded template needed for transcription.
Elongation:One strand of DNA,the template strand,acts as a template for RNA polymerase.As it
reads this template one base at a time,the polymerase builds an RNA molecule out of
complementary nucleotides,making a chain that grows from 5\' to 3\'.The RNA transcript carries
the same information as the non-template (coding) strand of DNA,but it contains the base uracil
instead of thymine .
Termination:Sequences called terminators signal that the RNA transcript is complete.Once they
are transcribed,they cause the transcript to be released from the RNA polymerase.
B)Prokaryotes do not have an organized nucleus,so the nuclear materials or DNA is in the
cytoplasm.Therefore,the transcription occurs in the cytoplasm and all the precursors needed for
the transcription are found in the cytoplasm.Prokaryotic transcription requires the RNA
polymerase enzyme in order for the transcription to be successfully completed.The enzyme binds
to the sigma factor and the promoter region,and then initiate the transcription by completing the
holoenzyme.In prokaryotes,DNA is not bound to histones.Thus,the transcription initiates
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polymerase reads the termination signal.There are two types of termination signals,Rho-
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Gene silencing
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This presentation provides an overview of RNA interference (RNAi) including its history, components, mechanism, advantages, and applications. It discusses how RNAi involves long double-stranded RNAs being cut by the enzyme Dicer into short interfering RNAs (siRNAs) that then guide the RNA-induced silencing complex (RISC) to degrade messenger RNAs with complementary base sequences, preventing gene expression. The presentation also compares siRNAs and microRNAs (miRNAs), noting similarities in their biogenesis and roles in post-transcriptional gene silencing, while distinguishing their origins, sizes, targets, and effects on mRNA. Recent applications of RNAi modulation of viral replication and gene expression are highlighted.
RNA interference (RNAi)
RNA interference is a mechanism by which small RNA molecules like microRNA and siRNA regulate gene expression by binding to mRNA and preventing protein production. This mechanism is important for immunity and controlling gene expression. It was discovered in 1998 by Fire and Mello through experiments in C. elegans showing that double-stranded RNA, but not single stranded, could efficiently silence genes. RNAi has many applications in biotechnology, functional genomics, and medicine.
Gene silencing last
Gene silencing is a process by which gene expression is regulated without modifying the genetic sequence. It occurs through transcriptional or post-transcriptional mechanisms. RNA interference (RNAi) is a major method of post-transcriptional gene silencing where double-stranded RNA introduced into a cell leads to degradation of mRNA from complementary genes. RNAi has various applications including cancer treatment by targeting oncogenes, inhibiting HIV replication, and improving crops by reducing undesirable compounds.
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EPIGENETIC.pptx
Gene silencing is a technique that reduces or eliminates the production of a protein from its corresponding gene through mechanisms other than genetic modification. There are two main types of gene silencing - transcriptional and post-transcriptional. Post-transcriptional gene silencing includes RNA interference (RNAi), which involves introducing double-stranded RNA that binds to mRNA and prevents replication, effectively silencing gene expression in a sequence-specific manner. RNAi is a powerful tool for gene function analysis and applications in biotechnology and agriculture.
Micro rna
- MicroRNAs (miRNAs) are small non-coding RNAs that regulate gene expression through base pairing with messenger RNA (mRNA) molecules. They are encoded in the genome and are abundant in many human cell types.
- miRNAs play a vital role in genetic regulation and are involved in most biological processes. Aberrant miRNA expression has been implicated in many diseases.
- miRNAs are initially transcribed as long primary transcripts that are processed in the nucleus by the Drosha enzyme into hairpin-shaped precursor miRNAs. These are then exported into the cytoplasm and further processed by the Dicer enzyme into mature miRNAs that can regulate gene expression through pairing with mRNAs.
2 coding and non coding dna, genetic mapping
Coding DNA sequences that encode proteins make up less than 2% of the human genome. The remaining 98% is non-coding DNA, which includes regulatory sequences, introns, repetitive elements, and other sequences not represented in proteins. While protein-coding genes have been well-studied, non-coding DNA serves numerous important functions and makes up the vast majority of the genome. Gene mapping involves assigning DNA fragments to chromosomes to create a genetic map, with the distance between genes proportional to recombination frequency.
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The document summarizes the organization of the human genome. It describes that the genome includes nuclear DNA, mitochondrial DNA, and repetitive DNA sequences. Nuclear DNA contains the majority of the genome and is found within the cell nucleus. Mitochondrial DNA is found within mitochondria and accounts for a smaller portion of the genome. Repetitive DNA sequences include tandem repeats like satellite DNA and microsatellites, as well as interspersed repeats such as transposable elements.
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1. Non-coding DNA makes up the majority of DNA in eukaryotic organisms but does not encode for proteins. It was once referred to as "junk DNA" but is now known to have important functions.
2. There are several types of non-coding DNA, including regulatory elements, introns, pseudogenes, repeats, transposons, telomeres, and various non-coding RNAs.
3. While the amount of non-coding DNA varies between species, it plays roles in chromosome structure, gene expression regulation, and protecting the genome from degradation.
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Gene silencing and editing techniques allow for precise regulation of gene expression. RNA interference uses small RNA molecules like siRNA and miRNA to degrade target mRNA, decreasing protein production. Genome editing tools like CRISPR/Cas9 create targeted DNA breaks that are repaired, resulting in mutations like insertions or deletions that can knock out gene function. These techniques have applications in research including studying gene function and biotechnology to modify traits in crops.
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If messages are being sent from your email account without your know.pdf
If messages are being sent from your email account without your knowledge, this is a sign that
your computer is infected with
Solution
If messages are being sent from your email account without your knowledge, this is a sign that
your computer is infected with:
Identity Theft. In identity theft, some other machine is purported to be the other machine. Which
means, somebody else will steal your identity and will send an email to somebody else, as if they
were sending a mail from your email account..
Identify some of the postmodern trends movements that women artist .pdf
Identify some of the postmodern trends/ movements that women artist have particiapted in.
Solution
The female art movement originated in the late 1960s which sought to rewrite a falsely male-
dominated history and change the perception of the contemporary world. The main attributes are
as follows; It sought to create an interaction between the artist and viewer by including women\'s
perspective. Before the introduction of feminism, the presence of female artist was not
considered which was later changed by the female artists via establishing alternate venues to
increase their visibility in the market. They also embrace different materials which symbolized
feminine perspective..
i need the executable file of this program (.exe) This is Assembly.pdf
i need the executable file of this program (.exe)
This is Assembly code This program calculates the integer expression A=(A+B)-(C+D) using
registers INCLUDE Irvine32.inc. data A DWORD 150 B DWORD 100 C DWORD 50 D
DWORD 40 .code main PROC; save the integer values in registers mov eax, A; EAX=150
mov ebx.B; EBX=100 mov ecx.C; ECX=50 mov edx.D; EDX=40; calculates the integer
expression add eax.ebx; EAX: (A+B) add ecx.edx; ECX: (C+D) sub eax.ecx; EAX: (A+B)-
(C+D) mov A.eax; A=(A+B)-(C+D) exit main ENDP END main Here the name of program
file is intExp.asm; the program contains comments starting with a semicolon (;) symbol.
Solution
Ans: Its not possible to upload the file. The following steps you can get your intExp.exe from
intExp.asm.
Step 1:
a) Open a MS-DOS window
b) your program as intExp.asm
c) Assemble the program with
C:> MASM demo
MASM will create an object file intExp.obj
Set your path variable to MASM.EXE]
d) Link the program with
C:> LINK intExp ( + prog1 + prog2 ....)
This adds info to intExp.obj and creates
the executable intExp.exe.
e) Run the program with
C:> intExp
intExp expects the input of a character from KB: and sends it
to the screen.
2. Do Assembly and Linkage in one Step
From MASM 6.0 on MicroSoft delivered a program ML that allowed
to do assembly and linkage in one step.
C:> ML intExp.asm : creates intExp.exe
Useful switches : /Zi : add debug info ( codeview)
/Fl : add a list file demo.lst
/Fm : add a map file demo.map
A complete call looks like :
C:> ML /Zi /Fl /Fm intExp.asm
Have a look at intExp.lst, intExp.map.
The full list of options is available via :
C:> MASM /H
C:> LINK /HELP ; LINK /?
C:> ML /HELP ; ML /?
BORLAND version :
For Borland environment ( C:>\\BC5\\TASM....)
Prefer to use Borland assembler TASM and linker TLINK.
Assembly : C:> TASM /l /n /z intExp
creates intExp.obj
Switches : /l : create listing intExp.lst
/n : no symbol table
/z : display source lines with errors
Linkage : C:> TLINK /3 /m /v intExp
creates intExp.exe
Switches : /3 : use 32-bit registers
/m : create map file intExp.map
/v : add debug info for the debugger TD
List of all options : C:> TASM /? ; C:> TLINK /?
3. IDE : PWB
Start : Doubleclick the PWB icon or
use a DOS window and type C:> PWB
The edit-part of the functionality should be self.
File --> -New / -Open / - Save / -Save as / -Close / -Exit
Edit --> -Cut / -Copy / -Paste / -Delete ..
Before you can convert the asm source code into an exe file
you should tell PWB which kind of executable
you want to build. It can done by setting many options
in the OPTIONS menue :
Options --> Project Templates -->
activate : RunTimeSupport : Assembler
Project Template : DOS EXE ok
Options --> Build Options -->
activate : Use Debug Options ok
Options --> Language Options --> MASM-Options -->
activate : MASM 5.1 Compatibility
delete : Warnings treated as errors
go to --> Set Debug Options -->
activate : Generate Listing File
All the options are set you are able to build the
executable file.
Project.
Help me fix the error shown above in my code of image.cpp please~I.pdf
Help me fix the error shown above in my code of image.cpp please~
Image.cpp(the code i wrote):
#include
#include \"image.hpp\"
#include
using namespace std;
Image::Image() {
cols = 256;
rows = 256;
pixels = new uint8_t[rows * cols];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
pixels[i + i * j] = \'A\';
}
}
}
Image:: ~Image(){
delete[] pixels;
}
int Image::get_pixel(unsigned int x, unsigned int y, uint8_t* colorp) {
colorp= &pixels[x+x*y];
if (colorp) {
return 0;
}
return -1;
}
int Image::set_pixel(unsigned int x, unsigned int y, uint8_t color) {
pixels[x + x * y] = color;
if (color) {
return 0;
}
return -1;
}
int Image::resize(unsigned int width, unsigned int height, uint8_t fillcolor) {
pixels = new uint8_t[rows * cols];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
pixels[i + i * j] = fillcolor;
}
}
if (pixels) {
return 0;
}
return -1;
}
Image.hpp(resourse):
#include // for uint8_t
class Image {
public:
// made these public so I can peek at them
unsigned int cols;
unsigned int rows;
uint8_t* pixels;
/* Constructs an image of 0x0 pixels. */
Image();
/* Frees all memory allocated for img */
~Image();
/* Changes the size of an image, allocating memory as necessary, and
setting all pixels to fillcolor. Returns 0 on success, or a non-zero error code.*/
int resize( unsigned int width, unsigned int height, uint8_t fillcolor );
/* Sets the color of the pixel at (x,y) to color. Returns 0 on success, else a non-zero
error code. If (x,y) is not a valid pixel, the call fails and the image does not change.*/
int set_pixel( unsigned int x, unsigned int y, uint8_t color );
/* Gets the color of the pixel at (x,y) and stores at the address pointed to
by colorp. Returns 0 on success, else a non-zero error code. */
int get_pixel( unsigned int x, unsigned int y, uint8_t* colorp );
}; Running Program Done. Program returned 1 Error: Your program returned nonzero status
code 1 stdout Testing normal usage Constructing Image i Error: Empty image does not have zero
cols & rows i resize( 640, 480, 0 i resize 100, 100, 72 resize( 32, 32, 72) 0K Checking for correct
value in every pixel after resize after resize, pixel[257] does not have fillcolor value Error Error:
get pixel returned wrong pixel value (0 e) 0 Doing lots of get pixel and set pixel on bounds pixel
coords get pixel returned wrong pixel value (1 0) 0 Error: get pixel returned wrong pixel value
(20) 0 Error: get pixel returned wrong pixel value (3 0) 0 Error: get pixel returned wrong pixel
value (4 0) 0 Error: get pixel returned wrong pixel value (5 0) 0 Error: get pixel returned wrong
pixel value (6 0) 0 Error: get pixel returned wrong pixel value (7 0) 0 Error: get pixel returned
wrong pixel value (8 0) 0 Error: get pixel returned wrong pixel value (9 0) 0 Error: get pixel
returned wrong pixel value (10 0) 0 Error: get pixel returned wrong pixel value (11 0) 0 Error:
get pixel returned wrong pixel value (120) 0 Error: get pixel returned wrong pixel value (13 0) 0.
Explain how the format of the mark-up language differs from the docu.pdf
Explain how the format of the mark-up language differs from the document’s format,
Solution
A document format is a simple text without any differentiation between the elements in the text,
whereas mark- up language is a system for annotating a document in a way that is syntactically
distinguishable from the text. Instructions are expressed directly by tags or \"instruction text
encapsulated by tags.\" Examples include typesetting instructions such as those found in troff,
TeX and LaTeX, or structural markers such as XML tags. Markup instructs the software that
displays the text to carry out appropriate actions, but is omitted from the version of the text that
users see.Today, the most important markup languages are the Standard Generalized Markup
Language (SGML), the Hypertext Markup Language (HTML), and the Extensible Markup
Language (XML).
HTML
The Hypertext Markup Language is an SGML Document Type Definition (DTD) that was
designed specifically for the World Wide Web. In essence, HTML is a set of markup codes
inserted in a file that note logical structures and instruct a web browser how to display a web
page\'s words and images.
XML
The Extensible Markup Language (XML) is a subset of SGML, whose purpose is \"to enable
generic SGML to be served, received, and processed on the web\" through a system of notation
that is unlimited and self-defining. Although XML has been designed to be compatible with
HTML (as well as SGML), it is not another single, predefined markup language. It is a meta-
language—a language for describing other languages—for designing markup..
Here is the given code, and the things I need to be done. Ive post.pdf
The document provides code for a Java application with a menu bar. It asks the user to modify the code to:
1. Add a new menu item called "Open" that displays an image when clicked.
2. Modify the "New" menu item to display a quote and related image when clicked.
3. Test the modified program.
It then asks the user to further modify the code to add a second column to the menu bar with items like "Welcome", "About", and duplicate the existing menu options.
Each student in my class was given bodily fluids. Everyone was n.pdf
Each student in my class was given \"bodily fluids.\" Everyone was negative for \"HIV\" except
one. We have to figure out who was positive for HIV and passed it on to others. Sharing partners
are in order. All the informaton that is needed is there. There are three numbers in the \"sharing
partner\" column. The column next to it is blank. Student\'s name 13 Sharing partners or 4,10, i4
7,8, la 2,4 I l, 14) B 13 I, 123,10 t ll, 4,3
Solution
Answer:
Student name \"3\" is positive for HIV.
Explanation:
The sharing partners of 3 are 2,4,5, and 12. Student 12 is +ve, it shares room with 11, 14 and 3.
But from the data it was clear that 11 and 14 are -ve for HIV. So the only possibility to transfer
the HIV to 12 is from \"3\"..
Hello everyone,Im actually working on a fast food order program..pdf
Hello everyone,
I\'m actually working on a fast food order program. So far my program works good. Once the
user has finished the order the program displays a receipt with all the order\'s information.
What I want is to print the receipt information in a showMessagedialog box by making used of
JOptionPane. I have tryed but i has been impossible. Can you please guys give a hand with this.
Here is my program code:
import java.text.DecimalFormat;
import java.util.Date;
import java.util.Random;
import java.util.Scanner;
//Interface
interface Restaurant
{
//Method that adds items to order.
void addItem();
}
//class that contains item\'s menu and prices.
class ItemsMenu{
Restaurant[] Menu = {
new CheeseBurger(4.45),
new ChickenBurger(5.25),
new SuperHotDog(3.55),
new MiniHotDog(2.99),
new Fries(1.75),
new OnionRings(1.75),
new Soda(1.55),
new BottleWater(1.25)};
//Method that display menu and receipt.
void Menu()
{
Scanner CashierInput = new Scanner(System.in);
DecimalFormat df = new DecimalFormat(\"0.00\");
Date dateCreated = new Date();
//Declaring variables.
int Choice = 0;
int ItemNum = 0;
double Tax = 0;
double NYTax = 8.78;
double SubTotal = 0;
double Total = 0;
double TotalFinal = 0;
//Printing menu options.
System.out.println(\"\ /*/*/*/*/*/*/*/*/ Mac King Fast Food /*/*/*/*/*/*/*/*/\");
System.out.println(\" \ Burgers:\" + \" \" + \"Side Orders: \");
System.out.println(\" 1. CheeseBurger $\" + df.format(((CheeseBurger)
Menu[0]).cheeseBurger) + \" \" + \"5. Fries $\" + df.format(((Fries) Menu[4]).fries));
System.out.println(\" 2. ChickenBurger $\" + df.format(((ChickenBurger)
Menu[1]).chickenBurger) + \" \" + \"6. Onion Rings $\" + df.format(((OnionRings)
Menu[5]).onionRings));
System.out.println(\" \ Hot Dogs:\" + \" \" + \"Beverages: \");
System.out.println(\" 3. Super HotDog $\" + df.format(((SuperHotDog)
Menu[2]).superHotDog) + \" \" + \"7. Soda $\" + df.format(((Soda) Menu[6]).soda));
System.out.println(\" 4. Mini HotDog $\" + df.format(((MiniHotDog) Menu[3]).miniHotDog)
+ \" \" + \"8. Bottle of Water $\" + df.format(((BottleWater) Menu[7]).bottleWater));
//Loop to get order choice from user.
do {
if (Choice != 0)
{
System.out.print(\" !!Enter (0) to finish Order!!\ \");
}
System.out.print(\" \ Enter menu selection: \");
Choice = CashierInput.nextInt();
//Printing choices added.
//Counting items added.
//Counting number of items added.
//Calculating subtotal.
if (Choice == 1)
{
System.out.println(\"\ CheeseBurger added\");
SubTotal = ((CheeseBurger) Menu[0]).cheeseBurger;
Menu[0].addItem();
++CheeseBurger.cheeseBCount;
++ItemNum;
Total = Total + SubTotal;
System.out.println(\"Current Subtotal $\" + (df.format(Total)));
}
else if (Choice == 2)
{
System.out.println(\"\ ChickenBurger added\");
SubTotal = ((ChickenBurger) Menu[1]).chickenBurger;
Menu[1].addItem();
++ChickenBurger.chickenBCount;
++ItemNum;
Total = Total + SubTotal;
System.out.println(\"Current Subtotal $\" + (df.format(Total)));
}
else if (Choice == 3)
{
System.out.println(\.
Describe the attributes that make a vision meaningful to stakeholder.pdf
Describe the attributes that make a vision meaningful to stakeholders. Discuss why a vision may
need to change.
Solution
The very essence of leadership is that you have to have a vision. It’s good to be a vision you
articulate clearly and forcefully on every occasion. You can’t blow an uncertain trumpet. –
Theodore Hesburgh.
A vision provides direction and road map into the future, it describes the type of organisations
that you want to become and how it’s unique, it creates purpose and identity.
The following characteristics make a vision compelling and meaningful to it’s stakeholders:
Future focused: An effective vision describes the organisation’s desired future, its provides a
clear picture of what the organisation should look like in 5-10 years time. It sets the context for
action.
Directional- An effective vision provides direction as to where the organisation should be going.
It should be specific enough to shape decision making and allow innovative strategies to help
realise the vision.
Clear: An effective vision provides guidance for decision-making and independent action. This
requires the vision to be clearly articulated and easily understood. The vision must clarify focus,
direction and constraints, to ensure that scarce resources are focused on the most strategic
initiatives. Vision that is clear enables effective allocation of scare resources.
Relevant- An effective vision is a good fit with the organisation’s history, current reality, culture
and values. An effective vision connects what has happened in the past to the desired future this
gives the vision credibility.
Purpose-Driven: An effective vision provides a larger sense of purpose for the organisation and
it’s people. That purpose must be more meaningful than getting bigger or beating the
competition. Purpose is about why we exist and why anyone should care. Vision connects people
to a meaningful purpose, allowing them to feel that they are part of something bigger than
themselves. As Steve Jobs said, “We’re here to put a dent in the universe. Otherwise why else
even be here?”
Value based- an effective vision connects the people to an organization’s core values. Values are
ideas and beliefs about what’s considered good or bad. Values of an organisation percolates in
the behaviour and attitudes of the people.
Challenging: An effective vision challenges us, it’s an invitation to greatness. A vision is a goal
that should challenge us, stretch us and set a high standard for the organisation. Effective visions
represent a future that is beyond what is possible today or what we think possible tomorrow.
Unique: An effective vision reflects what’s unique about the organisation, it recognizes what
makes it different. A vision is unique when it declares what makes the organisation stand out and
why it matters.
Vivid: An effective vision provides a vivid mental image of what the organisation will be like in
the future. Well crafted visions describe the future in a way that is easy .
Create a C program which auto-completes suggestions when beginning t.pdf
Create a C program which auto-completes suggestions when beginning to type a command and
pressing tab. The program needs to read all the files names in a directory given by the user and
store them in a data structure made up of a 26-element array of linked lists. A linked list
implementation in C is required. Create a linked list for each letter of the alphabet in which
you\'ll store the filenames which begin with that letter in some order. After reading the
filenames, the user provides the beginning of a filename they\'ll like to search. The output should
be all the filenames which match that prefix. The program should continue to ask for beginnings
of filenames until an empty string is entered.
Solution
Code
-----------
#include
#include
#include
struct node
{
char data[10] ;
struct node *next;
}*head;
node* arrNode[26];
char lettersPosition[26] =
{\'a\',\'b\',\'c\',\'d\',\'e\',\'f\',\'g\',\'h\',\'i\',\'j\',\'k\',\'l\',\'m\',\'n\',\'o\',\'p\',\'q\',\'r\',\'s\',\'t\',\'u\',\'v\',\'w\',\'x
\',\'y\',\'z\'};
/*
* Creating Node
*/
node* create_node(char fileName[])
{
node* newnode = (node *)malloc(sizeof(node));
if (newnode == NULL)
{
printf(\"\ Memory was not allocated\");
return 0;
}
else
{
strcpy(newnode->data,fileName);
newnode->next = NULL;
return newnode;
}
}
void display(struct node *r)
{
r=head;
if(r==NULL)
{
return;
}
while(r!=NULL)
{
printf(\"%s \",r->data);
r=r->next;
}
printf(\"\ \");
}
void search(char beginningFilePrefix[])
{
node* newnode = NULL;
char str[10];
strcpy(str,beginningFilePrefix);
char dirPos = str[0];
node* r = arrNode[lettersPosition[dirPos]];
if(r==NULL){
printf( \"No file is avialable in directory\");
return;
}else{
while(r!=NULL){
char tempFile[10];
bool isMathched = true;
strcpy(tempFile,r->data);
for( int i = 0; i < 1 ; i++){
printf(\"%c %c\",str[i],tempFile[i]);
if(!(str[i] == tempFile[i])){
isMathched = false;
}
}
if(isMathched){
if(newnode == NULL){
newnode = (node *)malloc(sizeof(node));
strcpy(newnode->data,r->data);
}else{
struct node *temp,*right;
temp= (struct node *)malloc(sizeof(struct node));
strcpy(temp->data,r->data);
right=(struct node *)newnode;
while(right->next != NULL){
right=right->next;
}
right->next =temp;
right=temp;
right->next=NULL;
}
}
r=r->next;
}
}
if(newnode==NULL){
printf(\"File not found with given search critieria\");
}else{
printf(\" Below are the output or file list which are matched with search criteria \ \");
while(newnode!=NULL){
printf(\"%s \ \",newnode->data);
newnode=newnode->next;
}
}
printf(\"\ \");
}
void displayFromArray(int dirPosition)
{
node* r= arrNode[dirPosition];
if(r==NULL){
printf( \"r is null\");
return;
}
while(r!=NULL){
printf(\"%s \",r->data);
r=r->next;
}
printf(\"\ \");
}
void insertIntoArray(char fileName[],int dirPosition){
struct node *temp;
temp=(struct node *)malloc(sizeof(struct node));
strcpy(temp->data,fileName);
node* r = arrNode[dirPosition];
if(lettersPosition[dirPosition] == fileName[0]){
printf(\"directory starts with letter %c cannot hold fi.
Compare the DHCP and DHCPv6SolutionThe comparison between the .pdf
Compare the DHCP and DHCPv6
Solution
The comparison between the DHCP and DHCPv6 is highlighted as below
1) DHCP packets are send over IPv4 while as DHCPv6 packets are sent over IPv6.
2) DHCPV6 used IPV6 multicast messages while as DHCP used IPV4 broadcast messsages
over the LAN.
3) DHCP has a default IP gateway address for the client while as DHCPv6 Does not have it.
4) DHCP can be configures over a router while as DHCPv6 does not have this option.
5) Scopes and leases is the concept used by both protocols.
6) DHCP Client, Relay and Server is the concept used by both the protocols.
7) Both messages types are different but perform olmost the same funtion.
8) Both Protocols use the DHCP end node for the additional information..
Consider in a given economy For a given time period the following in.pdf
Consider in a given economy: For a given time period the following information indicated
numbers of goods and services needed (Items) Unit Price for these items for selected time are
shown in the table below. Assume base year is 1981: calculate CPI for 1971, 1981, 1992 and
2012, Make sure show all your calculations Consider the following information, Use the CPI
from part one, and assume base is 1992. Complete the table Calculate percentage change of
Nominal GDP from 1971 to 2012. Calculate percentage change of Real GDP from 19711 to
2012. Explain the difference between nominal percentage change and real percentage change.
Solution
CPI=current year expenditure * 100/base year expenditure.
Background Many cities such as Detroit, MI have been in the news re.pdf
Background: Many cities such as Detroit, MI have been in the news recently regarding
underfunded pension funds. This exercise addresses a simplified version of some of the issues
involved (doing this as a starting point is common: the key generally is to address the core work,
adding more complexity later to make the analysis more accurate).
Solution
We have to analysis the background by the background data.
Can someone put this code in a zip file. I tried running it last tim.pdf
Can someone put this code in a zip file. I tried running it last time it was answered but it did not
work.. Please and thanks in advance..
In my template for how to solve JH2, I assume you have seen the StringTokenizer which should
be covered in cps161. You can find it in the first 9 chapters of \"Absolute Java\" by Savitch. The
StringTokenizer is a simple class for parsing a String that has multiple words on it. Run the
following simple example to see what it does:
class participation: 5 points
file_operations(45 points)
Create a package named file_operations and a class named FileOperations which will receive
one or more command line pathnames to command files. Each command file will be opened and
processed line by line. Each line can contain one of the following commands:
? - This command will print out the legal commands available
createFile - The first string following \"createFile\" will be treated as a filename, and the
remaining strings will be written to the file separated by new lines.
printFile - The first string following \"printFile\" will be treated as a filename that will be opened
up and printed out to the screen.
lastModified - The first string following \"lastModified\" will be treated as a filename for which
we will print out the date when this file was last modified.
size - The first string following \"size\" will be treated as a filename for which we will print out
the number of bytes it contains.
rename - The first string following \"rename\" will be treated as the current filename and the
second string will be treated as the new filename we desire.
mkdir - The first string following \"mkdir\" will be treated as the name of a directory that should
be created.
delete - The first string following \"delete\" will be treated as the name of a file that should be
deleted.
list - The first string following \"list\" will be treated as the name of a directory for which we
want a list of the files it contains.
quit - exit program
Anything else is a bad command
For a non-existent command file, you would get something like:
For a good command file like cmd.txt
cmd.txt
Your output would look something like:
Here is a template that you can use.
Once you have your program written, I want you to add the following files to the top level of
your project and run the test that I request:
cmd1.txt
cmd2.txt
Run your program with the following command line:
cmd1.txt non-existent-file cmd2.txt
Insert the contents of your screen into the appropriate JH2 worksheet.
Please don\'t create a file with the name non-existent-file!!!
Solution
Hi,
PFB the class for the question. Please comment for any queries/feedbacks.
Thanks,
Anita
FileOperations.java
package file_operations;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.text.SimpleDat.
Assignment 1. List the three visual elements Brieny descrbe each. 2 W.pdf
Assignment 1. List the three visual elements Brieny descrbe each. 2 What is the PCL? Give an
example of how it is used 3. What is the file ending lor an Android project? Windows Project?
Windows Phone Project? Mac project? A. Explain where Form int can be found for Windows,
Windows phone, Android, and Mac applicatons. 5 Each platform must instantiate the APP class
(wihich is located in the PCL This occurs in a can to the oadApplication Explain where this call
occurs in the Windows, Windows Phone, Android, and Mac platforms. 6 What is the difference
between PCL and SAP? 7 Explain how to create a label 8 Explain how padding works Make sure
to include a differences that exist with damerent devices 9 How do you set a margin? 10 How do
you center a label? How do you center text within a label?
Solution
Form:
The form of a piece is its form, including its volume or perceived volume. A three-dimensional
artwork has depth as well as breadth and height. Three-dimensional form is the basis of
sculpture.However, two-dimensional artwork will win the illusion of type with the employment
of perspective and/or shading or modelling techniques. Formalism is the analysis of works by
their form or shapes in humanities or anthropology.
Line
Lines and curves are marks that span a distance between 2 points . As an component of visual
art, line is the use of assorted marks, outlines, and implied lines in design and style. A line has a
width, direction, and length. A line\'s width is typically referred to as its \"thickness\". Lines are
typically referred to as \"strokes\", especially once referring to lines in digital design.
Color
Color is the element of art that\'s made once light-weight, striking AN object, is reflected back
to the eye. There are 3 properties to color. The first is hue, which merely means that the name we
have a tendency to offer to a color . The second element is intensity, which refers to the
vividness of the color. A color\'s intensity is sometimes stated as its \"colorfulness\", its
\"saturation\", its \"purity\" or its \"strength\".The third and final property of color is its value,
meaning however light-weight or dark it is. The terms shade and tint refer to value changes in
colours. In painting, shades are created by adding black to a color, while tints ar created by
adding white to a color..
a) How can habitat selection and sexual selection drive sympatric sp.pdf
a) How can habitat selection and sexual selection drive sympatric speciation?
b) What are hybrid zones and why can they be viewed as “natural laboratories” in which to study
speciation?
Solution
a)sympatric speciation is when a small population becomes a new species without geographic
separation due to reduced gene flow by factors like polyploidy, habitat differentiation, and sexual
selection.This type of speciation takes place in geographically overlapping populations.There is
no physical barrier, but still interruption of gene flow is there.
For example in fruit flies, food source (habitat selection) and apple maget fruit flies
Ex. Cichlids(Differ in coloration of breeding males). Mate choice based on male breeding
coloration is the main reproductive barrier that normally keeps these gene pools of these two
species separate.
b)Hybrid zones are regions in which members of different species meet and mate, producing
some offspring of mixed ancestry. Such regions are \"natural laboratories\" in which to study
speciation because scientists can directly observe factors that cause (or fail to cause)
reproductive isolation..
A common test of balances in a revenue cycle is the inquiry of manage.pdf
A common test of balances in a revenue cycle is the inquiry of management about the likelihood
of delinquent accounts receivable. This test concerns primarily with which one of the 28.
following management assertions? a. Existence or occurrence assertion. c. Presentation and
disclosure assertion. b. Valuation assertion. d. Completeness assertion.
Solution
2A common test of balances in a revenue cycle is the inquiry of management about the
likelihood of delinquent account receivable. This test concerns primarily with which one of the
following management assertions?
a.Existence or occurence assertion.
7. Recently, Skooterville has experienced a large growth in populati.pdf
7. Recently, Skooterville has experienced a large growth in population. As a result, the market
demand curve for housing in Skooterville:has shifted to the right.
Solution
Answer
Recently, Skooterville has experienced a large growth in population. As a result, the market
demand curve for housing in Skooterville:
Has shifted to the left.
50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, Ka=1.91.pdf
50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, Ka=1.9*10^-5) is titrated with a
0.1000M solution of sodium Hydroxide. Calculate the pH at the following points a. Before
Titration b. After the addition of 5.00mL of NaOH c. After the addition of 25.00mL of NaOH d.
After the addition of 50.00mL of NaOH e. After the addition of 5.00mL of NaOH
Solution
HN3+ NaOH--> NaN3 +H2O
mini moles of HN3=5
ka=Ka=1.9*10^-5
pka=4.72
a) ph before addition
before addition only weak acid is present
alpha=sqrt(ka/c)
=1.37*10^-2
c.alpha= [H+]= 1.37*10^-3
ph=2.86
b) after addition of 5 ml
mini moles of naoh added=0.5
ph=4.72+log(0.5/(5-0.5))
ph=3.76
c) after addition of 25 ml Naoh
ph=4.72+log(2.5/(5-2.5))
ph=4.72
d)after addition of 50 ml
now this is equivalence point.
Here N1v1=N2v2
here salt hydrolysis will take place
N3- +H2O <=> HN3+OH-
pOH=1/2(pKw-pKa-logC)
C= 5/(50+50)
=1/20M
pOH=1/2(14-4.72+1.30)
=5.29
ph=8.71.
4. For each of the following electrochemical cells i. Identify the a.pdf
4. For each of the following electrochemical cells: i. Identify the anode, the cathode, the reduced
species and the oxidized species. ii. Write the half reactions for the anode and cathode, and the
net reaction. ii. Calculate the standard cell potential (E°). *Note- For learning purposes, the
cathodes are not always on the left! e. C)SO (),H\'(H2S03(a)Co\"(a)Co)
Solution
a.
Cathode : Cr2+/Cr
Anode : Sn/Sn2+
Oxidized : Sn
Reduced : Cr2+
half-reactions,
Sn --> Sn2+ + 2e-
Cr2+ + 2e- --> Cr
net reaction : Sn + Cr2+ --> Sn2+ + Cr
Eo = Ecathode - Eanode = -0.14 - (-0.91) = 0.77 V
b.
Cathode : Ni2+/Ni
Anode : Fe/Fe2+
Oxidized : Fe
Reduced : Ni2+
half-reactions,
Fe --> Fe2+ + 2e-
Ni2+ + 2e- --> Ni
net reaction : Fe + Ni2+ --> Fe2+ + Ni
Eo = Ecathode - Eanode = -0.25 - (-0.44) = 0.19 V
c.
Cathode : Cu2+/Cu
Anode : Pb/Pb2+
Oxidized : Pb
Reduced : Cu2+
half-reactions,
Pb --> Pb2+ + 2e-
Cu2+ + 2e- --> Cu
net reaction : Pb + Cu2+ --> Pb2+ + Cu
Eo = Ecathode - Eanode = 0.337 - 0.126 = 0.211 V
d.
Cathode : Cl2/Cl-
Anode : Zn/Zn2+
Oxidized : Zn
Reduced : Cl2
half-reactions,
Zn --> Zn2+ + 2e-
Cl2 + 2e- --> 2Cl-
net reaction : Zn + Cl2 --> Zn2+ + 2Cl-
Eo = Ecathode - Eanode = 1.358 - (-0.763) = 2.121 V
e.
Cathode : SO4^2-/H2SO3
Anode : Co/Co2+
Oxidized : Co
Reduced : SO4^2-
half-reactions,
Co --> Co2+ + 2e-
SO4^2- + 4H+ + 2e- --> H2SO3 + H2O
net reaction : 2Co + SO4^2- + 4H+ --> 2Co2+ + H2SO3 + H2O
Eo = Ecathode - Eanode = 0.17 - (-0.28) = 0.45 V
f.
Cathode : O2/H2O2
Anode : Cu2+/Cu
Oxidized : Cu
Reduced : O2
half-reactions,
Cu --> Cu2+ + 2e-
O2 + 2H+ + 2e- --> H2O2
net reaction : Cu + O2 + 2H+ --> Cu2+ + H2O2
Eo = Ecathode - Eanode = 0.682 - 0.337 = 0.345 V
g.
Cathode : Cr2O7^2-/Cr3+
Anode : Fe2+/Fe3+
Oxidized : Fe2+
Reduced : Cr2O7^2-
half-reactions,
Fe --> Fe2+ + 2e-
Cr2O7^2- + 14H+ 6e- --> 2Cr3+ + 7H2O
net reaction : 6Fe + Cr2O7^2- + 14H+ 6e- --> 6Fe2+ + 2Cr3+ + 7H2O
Eo = Ecathode - Eanode = 1.33 - 0.771 = 0.559 V.
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