Relying on the above pathway for biosynthesis of tryptophan, fill i.pdf
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Relying on the above pathway for biosynthesis of tryptophan, fill in the table below to predict growth (+) or failure to grow (-) when mutants defective in each of the trp genes are placed on media supplemented with the compounds listed. Solution Answer: A mutant for a particular gene in the tryptophan pathway will only grow when it is supplemented with the required nutrient so that the biosynthetic pathway for the tryptophan synthesis is resumed. The mutant strain (auxotroph) will proliferate only when the medium is supplemented with the required substance leading to the completion of the tryptophan biosynthetic pathway. + indicates growth when a particular supplement is given. Mutants which have dysfunctional trpB, trpC, trpD or trpE can still grow if supplemented with any of Indole, tryptophan or Indole+tryptophan because in all of these cases the strain will get the required tryptophan necessary for its survival.SupplementMutanttrpAtrpBtrpCtrpDtrpENone-----Anthranilic acid---- +Indole-++++Tryptophan+++++Typtophan and indole+++++.
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Write about Participatory Design Issues and Concerns (use your own.pdf
Write about Participatory Design: Issues and Concerns? (use your own word about 500 words
not copy paste from another website pl.)
Solution
Participatory design is also called as co-operative design or co-design.It can be said as an
approach to design which invloves active participation of all the stakeholders in the design
process to ensure the final result meets its requirement.It is rather a process and procedure of
design than a design style.According to market study and surveys designers do a better job in a
co-design environment than they do with there own ideas.
We discuss three main issues addressed by Participatory Design researchers -
1)politics of design : This is one of the main concern for many design researchers and
practitioners.For example, the introduction of new frameworks such as system design or
platforms related to the introduction of computer-based systems and power dynamics that
emerge within the workspace. The automation process of system platforms has created a huge
concern within the workers as it is threatening their ownership over the work and involvement.
2)nature of participation : One of the major international organization Project for Public Spaces
create opportunities for continuous participation in the design and creation of place which is
believed to be the essential ingredient for successful environments. Project for Public Spaces
creates a new platform for the community to participate and reflect their initimate knowledge.A
method called Place Performance Evaluation or Place Game is used where groups from the
community are taken on the site of proposed development.They use their knowledge to design
strategies from which society will be benefited.
3)method, tools and techniques for participation : Distributed participatory design is a design
approach that involves direct participation of users and other stakeholders in system analysis and
design work.At present design teams most often are distributed leads to a need for support and
knowledge gathered from design of distributed systems. Distributed Participatory design focuses
on developing an understanding between different stakeholders in distributed design teams by
giving each the opportunity to engage in hands-on activities.
Why do some vertebrates, like a hellbender salamander, exchange oxyg.pdf
Why do some vertebrates, like a hellbender salamander, exchange oxygen as well as carbon
dioxide through their skin but bony fishes, like a goldfish, only exchange oxygen? (Choose One)
A. Goldfish do not produce CO2 since they live completely underwater
B. Goldfish have gills that diffuse CO2 while hellbender salamanders do not
C. Only young goldfish diffuse just O2 through their skin but adult goldfish diffuse both gases
D. Hellbender salamanders have hair-like projections on their skin, enabling them to diffuse both
gases
Solution
Answer - Option D - Hellbender salamanders have hair like projections on their skin, enabling
them to diffuse both gases.
Because they have cutaneous respiration..
Which of the following is the most direct source of energy for cotran.pdf
Which of the following is the most direct source of energy for cotransport? The movement of
one of the transported substances up its concentration gradient The movement of one of the
transported substances down its concentration gradient ATP hydrolysis ATP formation
Cotransport requires no energy
Solution
ATP hydrolysis
The direct source of energy for Cotransport is ATP hydrolysis to ADP means it can be achieved
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Which minerals commonly show pleochroisma. Isometric minerals.pdf
Which minerals commonly show pleochroism?
a. Isometric minerals
b. Triclinic minerals
c. Framework silicates
d. Mafic minerals
e. Quartz
Solution
A substance which appears to be different colors when observed at different angles, with
polarized light is called Pleochroism. it is an optical phoenomenon.
Triclinic minerals show pleochroism.
Answer is (b)..
Two cards are drawn from a standard deck of cards. Determine the pro.pdf
Two cards are drawn from a standard deck of cards. Determine the probability that the first card
is a Heart and the second card is an Ace.
Solution
Here,
P(heart then ace) = P(heart) P(ace|first is heart of ace) + P(heart) P(ace|first is not heart of ace)
Thus,
P(heart then ace) = (1/4)(3/51) + (1/4)(4/51)
= 7/204 or 0.034313725 [ANSWER].
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The Northern Pacific Railroad actively lobbied for which of the foll.pdf
The Northern Pacific Railroad actively lobbied for which of the following?
Solution
The Northern Pacific Railroad did eventually lobby for the idea of a national park when in 1871
when Judge William D. Kelley sent a letter by way of Northern Pacific Railroad executive A. B.
Nettleton to Ferdinand Hayden to lobby in Congress for a national park..
The difference between vertical and horizontal transmission of plant .pdf
The difference between vertical and horizontal transmission of plant viruses is that vertical
transmission is transmission of a virus from a parent plant to its progeny, and horizontal
transmission is one plant spreading the virus to another plant the spread of viruses from trees
and tall plants to bushes and other smaller plants, and horizontal transmission is the spread of
viruses among plants of similar size the spread of viruses from upper leaves to lower leaves of
the plant, and horizontal transmission is the spread of a virus among leaves at the same general
level the transfer of DNA from a plant of one species to a plant of a different species, and
horizontal transmission is the spread of viruses among plants of the same species What is the
main structural difference between enveloped and nonenveloped viruses? Enveloped viruses
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a capsid and phospholipid membrane; but in the nonenveloped virus the genetic material is
between these two membranes, while in the enveloped virus the genetic material is inside both
membranes. Enveloped viruses have a phospholipid membrane outside their capsid, whereas
nonenveloped viruses do not have a phospholipid membrane. Nonenveloped viruses have only a
phospholipid membrane, while enveloped viruses have two membranes, the other one being a
protein capsid. If all prokaryotes on Earth suddenly vanished, which of the following would be
the most likely and most direct result? There would be no more pathogens on Earth. Human
populations would thrive in the absence of disease. The recycling of nutrients would be greatly
reduced, at least initially. Bacteriophage numbers would dramatically increase. Viruses are
referred to as obligate parasites because they cannot reproduce outside of a host cell viral DNA
always inserts itself into host DNA. they invariably kill any cell they infect. they can
iocorporate nucleic acids from other viruses they must use enzymes encoded by the virus itself
Bacteriophage DNA that have become integrated into the host cell chromosome are called
intemperate bacteriophages. traosposons. prophages. T-even phages. plasmids. Virulent
pahges undergo a(n) life cycles, whereas temperate phages are capable of undergoing a(n)
cycle. infective; retroviral lysogenic; lytic lytic; lysogenic retroviral; infective Infective;
benign
Solution
36.Answer a. In Horizontal transmission spread of virus takes place from one plant to other
while vertical transmission involves transfer of virus from plant to offspring
37.Answer c. Phospholipid enclosed virus they are enveloped virus while other lack
phospholipid membrane
38.Answer c.microbes help in recycling of material.
39.Answer a. Reproduction takes place inside host
40. Answer d. Known as phage
41.Answer c : Lyric and lysogenic phage.
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Write about Participatory Design Issues and Concerns (use your own.pdf
Write about Participatory Design: Issues and Concerns? (use your own word about 500 words
not copy paste from another website pl.)
Solution
Participatory design is also called as co-operative design or co-design.It can be said as an
approach to design which invloves active participation of all the stakeholders in the design
process to ensure the final result meets its requirement.It is rather a process and procedure of
design than a design style.According to market study and surveys designers do a better job in a
co-design environment than they do with there own ideas.
We discuss three main issues addressed by Participatory Design researchers -
1)politics of design : This is one of the main concern for many design researchers and
practitioners.For example, the introduction of new frameworks such as system design or
platforms related to the introduction of computer-based systems and power dynamics that
emerge within the workspace. The automation process of system platforms has created a huge
concern within the workers as it is threatening their ownership over the work and involvement.
2)nature of participation : One of the major international organization Project for Public Spaces
create opportunities for continuous participation in the design and creation of place which is
believed to be the essential ingredient for successful environments. Project for Public Spaces
creates a new platform for the community to participate and reflect their initimate knowledge.A
method called Place Performance Evaluation or Place Game is used where groups from the
community are taken on the site of proposed development.They use their knowledge to design
strategies from which society will be benefited.
3)method, tools and techniques for participation : Distributed participatory design is a design
approach that involves direct participation of users and other stakeholders in system analysis and
design work.At present design teams most often are distributed leads to a need for support and
knowledge gathered from design of distributed systems. Distributed Participatory design focuses
on developing an understanding between different stakeholders in distributed design teams by
giving each the opportunity to engage in hands-on activities.
Why do some vertebrates, like a hellbender salamander, exchange oxyg.pdf
Why do some vertebrates, like a hellbender salamander, exchange oxygen as well as carbon
dioxide through their skin but bony fishes, like a goldfish, only exchange oxygen? (Choose One)
A. Goldfish do not produce CO2 since they live completely underwater
B. Goldfish have gills that diffuse CO2 while hellbender salamanders do not
C. Only young goldfish diffuse just O2 through their skin but adult goldfish diffuse both gases
D. Hellbender salamanders have hair-like projections on their skin, enabling them to diffuse both
gases
Solution
Answer - Option D - Hellbender salamanders have hair like projections on their skin, enabling
them to diffuse both gases.
Because they have cutaneous respiration..
Which of the following is the most direct source of energy for cotran.pdf
Which of the following is the most direct source of energy for cotransport? The movement of
one of the transported substances up its concentration gradient The movement of one of the
transported substances down its concentration gradient ATP hydrolysis ATP formation
Cotransport requires no energy
Solution
ATP hydrolysis
The direct source of energy for Cotransport is ATP hydrolysis to ADP means it can be achieved
by active transport..
Which minerals commonly show pleochroisma. Isometric minerals.pdf
Which minerals commonly show pleochroism?
a. Isometric minerals
b. Triclinic minerals
c. Framework silicates
d. Mafic minerals
e. Quartz
Solution
A substance which appears to be different colors when observed at different angles, with
polarized light is called Pleochroism. it is an optical phoenomenon.
Triclinic minerals show pleochroism.
Answer is (b)..
Two cards are drawn from a standard deck of cards. Determine the pro.pdf
Two cards are drawn from a standard deck of cards. Determine the probability that the first card
is a Heart and the second card is an Ace.
Solution
Here,
P(heart then ace) = P(heart) P(ace|first is heart of ace) + P(heart) P(ace|first is not heart of ace)
Thus,
P(heart then ace) = (1/4)(3/51) + (1/4)(4/51)
= 7/204 or 0.034313725 [ANSWER].
The probability that a student will find a (desirable) job before gr.pdf
The probability that a student will find a (desirable) job before graduation is .47 and the
probability that a student will get married before graduation in .38. The probability that a student
will get a job and get married before graduation is .17. What is the probability that a student will
either find a job or get married before graduation?
Solution.
The Northern Pacific Railroad actively lobbied for which of the foll.pdf
The Northern Pacific Railroad actively lobbied for which of the following?
Solution
The Northern Pacific Railroad did eventually lobby for the idea of a national park when in 1871
when Judge William D. Kelley sent a letter by way of Northern Pacific Railroad executive A. B.
Nettleton to Ferdinand Hayden to lobby in Congress for a national park..
The difference between vertical and horizontal transmission of plant .pdf
The difference between vertical and horizontal transmission of plant viruses is that vertical
transmission is transmission of a virus from a parent plant to its progeny, and horizontal
transmission is one plant spreading the virus to another plant the spread of viruses from trees
and tall plants to bushes and other smaller plants, and horizontal transmission is the spread of
viruses among plants of similar size the spread of viruses from upper leaves to lower leaves of
the plant, and horizontal transmission is the spread of a virus among leaves at the same general
level the transfer of DNA from a plant of one species to a plant of a different species, and
horizontal transmission is the spread of viruses among plants of the same species What is the
main structural difference between enveloped and nonenveloped viruses? Enveloped viruses
have their genetic material enclosed by a layer made only of protein. Both types of viruses have
a capsid and phospholipid membrane; but in the nonenveloped virus the genetic material is
between these two membranes, while in the enveloped virus the genetic material is inside both
membranes. Enveloped viruses have a phospholipid membrane outside their capsid, whereas
nonenveloped viruses do not have a phospholipid membrane. Nonenveloped viruses have only a
phospholipid membrane, while enveloped viruses have two membranes, the other one being a
protein capsid. If all prokaryotes on Earth suddenly vanished, which of the following would be
the most likely and most direct result? There would be no more pathogens on Earth. Human
populations would thrive in the absence of disease. The recycling of nutrients would be greatly
reduced, at least initially. Bacteriophage numbers would dramatically increase. Viruses are
referred to as obligate parasites because they cannot reproduce outside of a host cell viral DNA
always inserts itself into host DNA. they invariably kill any cell they infect. they can
iocorporate nucleic acids from other viruses they must use enzymes encoded by the virus itself
Bacteriophage DNA that have become integrated into the host cell chromosome are called
intemperate bacteriophages. traosposons. prophages. T-even phages. plasmids. Virulent
pahges undergo a(n) life cycles, whereas temperate phages are capable of undergoing a(n)
cycle. infective; retroviral lysogenic; lytic lytic; lysogenic retroviral; infective Infective;
benign
Solution
36.Answer a. In Horizontal transmission spread of virus takes place from one plant to other
while vertical transmission involves transfer of virus from plant to offspring
37.Answer c. Phospholipid enclosed virus they are enveloped virus while other lack
phospholipid membrane
38.Answer c.microbes help in recycling of material.
39.Answer a. Reproduction takes place inside host
40. Answer d. Known as phage
41.Answer c : Lyric and lysogenic phage.
select all that apply to ground substanceSolutionLets us see w.pdf
select all that apply to ground substance
Solution
Lets us see what is ground substance.
It is an amorphous gel-like or a transparent flud substance surrounding the cells,present in the
connective tissue .
The ground substance and fibers between cells in a connective tissue (most abundant tissues in
the body-functions : binding and supporting; consists of relatively few cells in a generous matrix
) are surrounded and supported by matrix called as extracellular matrix.
The Ground substance includes all the other components of the extracellular matrix excluding
fibers (Eg: collagen and elastic fibers).
Depending upon the tissue, the components of the ground substances vary .
Ground substance is primarily composed of water, proteoglycans,glycosaminoglycans (most
notably hyaluronan-hyaluronic acid), calcium and glycoproteins.
In cytology, it may also refers to cytosol or protoplasm.
The stroma is the tissue that forms the ground substance, base or the foundation, or the
framework of an organ, as opposed to its functional parts (parenchyma)..
Question 7 (2 points)A researcher conducts a study on the realtion.pdf
Question 7 (2 points)
A researcher conducts a study on the realtionship between caffeine consumption and
performance on a math test. What kind of study is this?
Question 7 options:
There is not enough information here to tell.
Correlational
Quasi-experimental
Experimental
Save
Question 8 (2 points)
Which researcher is most likely to engage in content analysis?
Question 8 options:
Dr. Williams, who studies transcripts of police interviews with suspects.
Dr. McKenzie, who studies relationships between various personality traits.
Dr. Perryman, who studies the social skills of indigenous people.
Dr. Ramirez, who studies mathematical competence in adolescents.
Save
Question 9 (2 points)
Which of the following would be more likely to be found in an empirical research article on
bullying than in a popular magazine article on the same topic?
Question 9 options:
Opinions about why people bully from the point of view of the author of the article.
Excerpts from actual interviews with bullies and/or victims.
Summaries of the methods and results found by previous researchers on bullying accompanied
by reference citations.
Humerous anecdotes about how victims fantasize about getting back at their bullies.
Save
Question 10 (2 points)
Dr. Simmons is interested in studying attachment behavior in infants. She trained her research
assistants to recognize classic attachment behaviors such as fearful reactions to strangers and
proximity seeking with the attachment figure. The research assistants then learned to code these
behaivors while watching video tapes of infants and caregivers interacting while they waited in
the doctor\'s waiting room. What type of research is going on here?
Question 10 options:
Quasi-experimental research
Naturalistic observation
Archival data collection
Interrupted time-series designa)
There is not enough information here to tell.b)
Correlationalc)
Quasi-experimentald)
Experimental
Solution
b)
Correlational
a)
Dr. Williams, who studies transcripts of police interviews with suspects.
c)
Summaries of the methods and results found by previous researchers on bullying accompanied
by reference citations
c)
Archival data collection
b)
Correlational.
Primary difference between the income sections for form 1120 and .pdf
Primary difference between the income sections for form 1120 and 1120s
Primary difference between the income sections for form 1120 and 1120s
Solution
Answer :- The primary difference between form 1120 and 1120s is that the form 1120 is filed by
corporations on annual basis under internal revenue service (IRS) code whereas the form 1120s
is filed by S corporation (corporation who have opted for \"S\" status) on annual basis under
internal revenue service (IRS) code..
Once its readable i will be okay.SolutionI will be okay in.pdf
Once it\'s readable i will be okay.
Solution
\"I will be okay in the end. If it\'s not okay, it\'s not the end.\" What are some incidences when
this John Lennon quote was proved true in real life?
Many inspiring stories of people\'s lives show it...
Consider Abraham Lincoln. He met failure after failure throughout his life before he became the
President of USA at 52.
Thomas Edison... 10,000 failures before inventing the light bulb.
Henry Ford... was broke at the age of 40.
Conolel Sanders... broke at 65... knocked at 1000 doors before he could make his first sale of a
chicken dish...founder of KFC....
Each and everytime in life ,
On all the small tasks and bigger challenges...its proven true.
It may be preparing a cup of tea or coffee or
Working on a home . The tea or coffee after two boils are perfectly okay for me but you may
boil it more with extra milk.
I work on decor with few frames of modern art and exclusive vases but you may go for plain
walls.
A waiter takes the order and serves and is okay with it...but another one wears a smile and is
quite alert and won\'t let the glass of water to be empty because its his level of \"okay\".
So each of us working and completing our daily task work on this principle.
We try hard and reach our okay\'s.
There\'s a variation in \"okay\" of all of us...that\'s it..
Need the mutiple choice answered uslde .An)_can omunicate with a sys.pdf
Need the mutiple choice answered uslde .An)_can omunicate with a system inside a firewall
system to be used to filter out undesirable traffic and prevent employees from accessing
potentially hostile web sites Multiple-Choice Quiz e three types of event logs generated What are
the three t t bw Windows Which of the following is not a capability of 6. Windows 2003 and
Vista by A. Event, Process, and Security B. Application, User, and Security c. User, Event, and
Security network-based IDS A. Can detect denial-of-service attacks B. Can decrypt and read
encrypted traffic C. Can decode UDP and TCP packets D. Can be tuned to a particular network
Application, System, and Security What are the two main types of intrusion detection systems?
A. Network-based and host-based B. Signature-based and event-based C. Active and reactive D.
Intelligent and passive environment An active IDS can: A. Respond to attacks with TCP resets
B. Monitor for malicious activity C. A and B D. None of the above Honeypots are used to: , 7. s.
What was the first commercial, network-based 8. IDS product? A. Stalker B. NetRanger C. IDES
D. RealSecure What are the two main types of IDS signatures? 9. Egress filtering is used to
detect spam that is: A. Network-based and file-based B. Context-based and content-based C
Active and reactive A. Attract attackers by simulating systems with open network services B.
Monitor network usage by employees C. Process alarms from other IDSs D. Attract customers to
e-commerce sites A. Coming into an organization B. Sent from known spammers outside your
organization C. Leaving an organization D. Sent to mailing lists in your organization
Preventative intrusion detection systems: A. Are cheaper B. D. None of the above . Which of the
following describes a passive, host based IDS? 10. A. Runs on the local system Are designed to
stop malicious activity from occurring B. Does not C. Can l D. All of the above not interact with
the traffic around it C. Can only monitor activity D. Were the first types of IDS ook at system
event and error logs r 13:Indusion Detection Systems Detection Systems and Network Security
Solution
1. What are the three types of event logs generated by Windows 2003 and Vista systems?
D. Application, System, and Security
2. What are the two main types of intrusion detection systems?
A. Network-based and host-based
3. What was the first commercial, network-based IDS product?
B. NetRanger
4. What are the two main types of IDS signatures?
B. Context-based and content-based
5. Which of the following describes a passive, host-based IDS?
D. All of the above
6. Which of the following is not a capability of network-based IDS?
B. Can decrypt and read encrypted traffic
7. An active IDS can:
C. A and B
8. Honeypots are used to:
A. Attract attackers by simulating systems with open network services
9. Egress filtering is used to detect spam that is:
C. Leaving an organization
10. Preventative intrusion detection systems:
B. Are designed to stop m.
Need a strong biologistgeneticist to answer this P element question.pdf
Need a strong biologist/geneticist to answer this P element question for Drosophila... PLEASE
What do you know about P-elements in Drosophila? How might P-elements have entered
Drosophila melanogaster? What do you know about P-element structure? How big are the native
elements? How do P-elements move around the genome? Connect your ideas to the special P-
elements used in this lab. What is the connection between transposition of mobile DNA elements
and DNA replication? Why are P-elements important for biologists and Drosophila researchers
in particular? What do you know about the Gene Disruption Project (GDP) in Drosophila?
Solution
Ans.) P element is a type of transposon found particularly in Drosophila and is used for
mutagenesis and for the production of genetically modified flies that is extensively used in
various genetic and medical researches. It has ability to initiate a phenotype known as hybrid
dysgenesis. It encodes the protein P transposase and is flanked by terminal inverted repeats
which are important for its mobility.
The P element enters in Drosophila by the horizontal gene transfer method. It has a sequence of 4
exons with 3 introns. The absolute splicing of the intron produces the transposase enzyme,
whereas the alternative partial splicing of intron 1 and 2 produce the P element repressor.
The P element is a class II transposon and it is moved by DNA-based \"cut and paste\"
mechanism. The P element encodes the transposase enzyme which recognizes the 31 bp terminal
inverted repeats of the P element and catalyzes P element excision and re-insertion.
P element is important for biologist and genetics because nowadays it is widely used to solve
many puzzles of genetics through its use in fly transformation, insertional mutagenesis as well as
reverse genetics.
The Gene Disruption Project (GDP) is a public collection of mutant that is created by applying a
non-targeted transposon mutagenesis strategy. Its main objective is to acquire a transposable
element insertion in each gene and this collection is available online and new mutants may be
requested while they are being balanced. These balanced stocks are available at the Bloomington
Drosophila Stock Center..
Modify HuffmanTree.java and HuffmanNode.java to allow the user to se.pdf
Modify HuffmanTree.java and HuffmanNode.java to allow the user to select the value of n 9 to
construct an n ary Huffman Tree. Once finished, verify the accuracy of your tree by running it
for the English Alphabet for n = 3, 4, 5, 6, 7, 8, 9. For each value of n, compute the average
codeword length. Write these in a table (n vs. ACW L(Cn)).
/ HuffmanTree.java
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.ArrayList;
import java.util.Scanner;
public class HuffmanTree {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
System.out.print(\"Enter the file name: \");
String fileName = in.nextLine();
ArrayList nodes = new ArrayList<>();
ArrayList temp = new ArrayList<>();
try {
Scanner fin = new Scanner(new FileReader(fileName));
String symbol;
double prob;
while(fin.hasNext()){
symbol = fin.next();
prob = fin.nextDouble();
nodes.add(findInsertIndex(nodes, prob), new HuffmanNode(symbol, prob));
}
fin.close();
temp.addAll(nodes);
HuffmanNode root = createHuffmanTree(temp);
setCodeWords(root);
double avg = 0;
for (HuffmanNode node : nodes) {
System.out.println(node.getS() + \": \" + node.getC() + \": \" + node.getL());
avg += node.getL();
}
if(avg != 0)
avg /= nodes.size();
System.out.println(\"Average code length is: \" + avg);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
private static HuffmanNode createHuffmanTree(ArrayList nodes){
HuffmanNode root = null;
if(nodes.size() > 0) {
HuffmanNode node0, node1, node;
while (nodes.size() > 1) {
node0 = nodes.get(0);
node1 = nodes.get(1);
node = new HuffmanNode(null, node0.getP() + node1.getP());
node.setLeft(node0);
node.setRight(node1);
nodes.remove(0);
nodes.remove(0);
nodes.add(findInsertIndex(nodes, node.getP()), node);
}
root = nodes.get(0);
}
return root;
}
private static void setCodeWords(HuffmanNode root){
if(root != null){
setCodeWords(root, \"\");
}
}
private static void setCodeWords(HuffmanNode root, String codeWord){
if(root.getS() != null){
root.setC(codeWord);
root.setL(codeWord.length());
}
else{
setCodeWords(root.getLeft(), codeWord + \"0\");
setCodeWords(root.getRight(), codeWord + \"1\");
}
}
private static int findInsertIndex(ArrayList nodes, double prob){
for(int i = 0; i < nodes.size(); ++i){
if(prob < nodes.get(i).getP()){
return i;
}
}
return nodes.size();
}
}
// HuffmanNode.java
public class HuffmanNode {
private String S;
private double P;
private String C;
private int L;
private HuffmanNode left, right;
public HuffmanNode(String s, double p) {
S = s;
P = p;
}
public String getS() {
return S;
}
public void setS(String s) {
S = s;
}
public double getP() {
return P;
}
public void setP(double p) {
P = p;
}
public String getC() {
return C;
}
public void setC(String c) {
C = c;
}
public int getL() {
return L;
}
public void setL(int l) {
L = l;
}
public HuffmanNode getLeft() {
return left;
}
public void setLeft(HuffmanNode left) {
this.left = left;
}
public HuffmanNode getRight() {
return right;
}
public void se.
Listed are 2 of the core theories of evolution. For each, give a def.pdf
Listed are 2 of the core theories of evolution. For each, give a definition or brief explanation of
the theory and 2 examples of the type of evidence we have that supports the theory.
The theory of common descent –
2 examples of supporting evidence:
The theory of natural selection –
2 examples of supporting evidence:
Solution
Evolution:
Evolution is a process of change through time. It is the process by which modern organisms have
descended from ancient organisms.
In genetic terms, the change in the frequency of alleles in populations from one generation to
another generation. (change over time).
A. The theory of common descent:
According to the principle of common descent, all species (living and extinct) are descended
from ancient common ancestors.
In common descent theory darwin explains, that organisms are descended from common
ancestors that gradually change over time into entirely new organism.
To create new species, many generations must pass on traits with slight modifications.
Examples:
1. Archaeopteryx is between reptiles and birds.
2. Eustheopteron is an amphibious fish.
3. Seymouria is a reptile-like amphibian.
4. Therapsids were mammal-like reptiles.
B. Darwin\'s Natural selection:
According to the Darwin, It is the process that results in adaptation of an organism to its
environment which means of selectivity reproducing changes in the genotype.
In the process of evolution the natural selection is a basic mechanism, it is mainly happens due to
the mutations, migrations and genetic drift.
Examples:
1. The beaks of Darwin finches. The finches of the Galapagos islands were adopted to the island
in different ways, but are believed to share a common ancestor.
2. Peppered moths and industrial pollution: Natural selection favored dark colored moths in areas
of heavy pollution, while light colored moths are survived better in less polluted area.
3.The soapberry bugs have must mutated their beaks length genes, when they ran out of food.
Beak size variation:
In South Florida soapberry bugs feed on balloon vine tree with large fruits, longer beak present
in them.
In North Florida soapberry bugs feed on golden rain tree with small fruits, they have shorter
beaks..
Light is confined within the core of a simple clad optical fiber by.pdf
Light is confined within the core of a simple clad optical fiber by?
Solution
An optical fiber is a cylindrical dielec wave guide (nonconducting waveguide) that transmits
light along its axis, by the process of total internal reflection. To confine the optical signal in the
core, therefractive index of the core must be greater than that of the cladding..
Let L R^n rightarrow R^m be a linear transformation. Show that if n .pdf
Let L: R^n rightarrow R^m be a linear transformation. Show that if n > m, then L must have a
nontrivial kernel, i.e., dim(ker(L)) > 0. If n lessthanorequalto m does L have to be one-to-one?
Solution
For second proof,
If T is one-one, it can\'t map a non-zero vector into 0, since it already maps 0 into 0. This means
that its kernel can contain only 0.
Suppose the kernel contains just the zero vector. To check if T is one-one, check if it can map
two different vectors u and v in Rn into the same vector in Rm. If T(u) = T(v), then T(u – v) =
T(u) – T(v) = 0. This means that the non-zero vectoru – v is in the kernel of T, which is
impossible since we assumed the kernel contained just 0. Then T cannot map different vectors
into the same vector, and must be one-one..
IS path vector routing algorithm closer to the distance vector rou.pdf
: IS path vector routing algorithm closer to the distance vector routing algorithm or to the link
state routing algorithm? Explain.
Solution
The path-vector routing algorithm is truly distance-vector routing using the finest path instead of
the direct distance as the metric. Each node first create a forwarding table. Assuming it can only
arrive at instant neighbors . The forwarding table is slowly better as path vectors turn up from the
instant neighbors..
a growth medium that can determine if a bacterium contains several d.pdf
a growth medium that can determine if a bacterium contains several different types of enzymes
or characteristics.
a multitest
b selective
c differential
d enriched
e all of the above
Solution
b selective
Selective media allows the growth of certain type of organisms, it contains several different
types of enzymes or characteristics..
Identify the function of the circulatory system in fishes. Explain h.pdf
Identify the function of the circulatory system in fishes. Explain how it works.
Solution
The circulatory arrangement of fish is very straightforward. It comprises of a heart, blood, and
veins. The heart of a fish is a basic strong structure that is situated behind (and underneath) the
gills. It is encased by the pericardial layer or pericardium.
The heart comprises of a chamber, a ventricle, a thin-walled structure known as sinus venosus,
and a tube called bulbus arteriosus.
The blood contains plasma (the liquid part) and platelets. The red platelets or the erythrocytes
contain hemoglobin, a protein that conveys oxygen all through the body. The white platelets
contain a basic part of the invulnerable framework. The thrombocytes perform capacities that are
proportionate to the part of platelets in the human body, i.e. they help in blood coagulating.
Veins convey the blood all through the body. While corridors convey oxygenated blood from the
gills to whatever remains of the body, veins return deoxygenated blood from various parts of the
body to the heart.
The cardiovascular arrangement of a fish involve a heart, veins, corridors, blood, and fine
vessels. The vessels are minute vessels that shape a system called a fine bed, where the blood
vessel and venous blood get connected. Vessels have thin dividers that encourage dispersion, a
procedure through which oxygen and different supplements from the blood vessel blood are
moved into the cells. In the meantime, carbon dioxide and waste materials are moved into the
vessels.
Vessels with deoxygenated blood (contains carbon dioxide) deplete into little veins called
venules, which thusly deplete into bigger veins. The veins convey the deoxygenated blood into
the sinus venosus, which resembles a little gathering chamber. The sinus venosus has pacemaker
cells that are in charge of starting constrictions, so that the blood is moved into the thin-walled
chamber, which has not very many muscles.
The chamber produces feeble compressions to push blood into the ventricle. The ventricle is a
thick-walled structure with bunches of cardiovascular muscles. It produces enough weight to
pump the blood all through the body. The ventricle pumps blood inside it into bulbus arteriosus,
a little chamber with flexible segments.
While bulbus arteriosus is the name of the chamber in teleosts (rayfinned, hard fish), the
structure is known as conus arteriosus in elasmobranchs (angle with cartilaginous skeleton and
placoid scales). Conus arteriosus has numerous valves and muscles, while bulbus arteriosus has
no valves. The primary capacity of this structure is to diminish the beat weight created by the
ventricle, with a specific end goal to maintain a strategic distance from harm to the thin-walled
gills.
Gills are the essential respiratory organs of fish. They encourage trade of gasses, i.e. retention of
oxygen from water and end of carbon dioxide. Corridors convey the oxygenated blood (from the
gills) all through the body..
How did the discovery of three categories of petite mutations in yea.pdf
How did the discovery of three categories of petite mutations in yeast lead researchers to
postulate extranuclear inheritance of colony size?
Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive
petites followed non-Mendelian patterns that were consistent with the involvement of nuclear
inheritance.
Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive
petites followed non-Mendelian patterns that were consistent with the involvement of an
extranuclear agent.
Whereas suppressive petites exhibited Mendelian inheritance, both neutral and segregational
petites followed non-Mendelian patterns that were consistent with the involvement of an
extranuclear agent.
Whereas suppressive petites exhibited Mendelian inheritance, both neutral and segregational
petites followed non-Mendelian patterns that were consistent with the involvement of nuclear
inheritance.
Solution
Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive
petites followed non-Mendelian patterns that were consistent with the involvement of nuclear
inheritance.
---------
Extra nuclear inheritance means, inheritance patterns involving genetic material outside the
nucleus, here the genome is residing in mitochondria or chloroplast.
The segregation petite followed Mendelian type of inheritance (wild type and segregation when
mated, segregated in 1:1 ratio), and if any mutation has occurred in the cell nucleus, it affected
the protein production needed for mitochondrial function.
Where as neutral petites, they did not have most of their mt DNA and but then it was mated with
wild type resulted in all wild type 4: neutral 0. So, these are not following Mendelian type of
inheritance.
In case, of suppressive petites, when suppressive mutants mated with wild type resulted in All
wild type 4: 0 all petite. So they are also not following Mendelian inheritance..
A certain system can experience 2 difference types of defects. Let ,.pdf
A certain system can experience 2 difference types of defects. Let , Ai=1,2 denote the event that
the system has a defect of type i. Suppose that A1=0.15, A2=0.1, P(A1 and A2) = 0.08, and . If
the system has a type 1 defect, what is the probability that it has a type 2 defect? (Consider the
conditional probability).
a.) 0.47
b.) 0.33
c.) 0.252
d.) 0.53
Solution
The answer is (D).
Find to three decimal places, the coordinate of point P lying in the.pdf
Find to three decimal places, the coordinate of point P lying in the 4th quadrant on the Cartesian
plan such that the line tangent to the cosine function passes through the origin. X:[0,2Pi] Show
all work and describe all steps.
Solution
t = cos x
tangent passing through origin
t\' = -sin x
So,
tangent y = mx+c
y = -xsinx+c
here,
(0,0) is point on tangent,
0 = -0sin0 + c
c = 0
So,
y = -xsinx is tangent..
Find the mobility (DoF) and Grashof condition of the mechanism below..pdf
Find the mobility (DoF) and Grashof condition of the mechanism below. Also, which inversion
is it?
Solution
Mobility
n = 4
p = 4
DOF = 3*(n-1) - 2*p = 3*3 - 2*4 = 1
Grashof\'s Condition:
s + l = 0.68+1.62 = 2.3 in
p + q = 1.38 + 1.22 = 2.6
as (s+l)<(p+q) so it is Grashof linkage.
As the shortest part is the Frame (BasePart), so the linkage is a Double-Crank (or drag link)
mechanism..
5. Discuss the pros and cons of American business being excluded by .pdf
5. Discuss the pros and cons of American business being excluded by American law (Pres
Kennedy issued an Executive Order) from conducting any form of business activity on the island
of Cuba. Express your perspectives of the European Union and other first world nations (like
Canada) steadfastly generating billion of euros (and other currencies) from trade with Cuba.
Solution
There are significant pros and cons of prohibition of conducting trade with Cuba.
The pros of the prohibition are as follows
1. Cuba has not fulfilled the conditions of lifting the trade Ban which was set up by the US
government.
2. The US government made various attempts to reduce the restrictions of trade. However this
was met with aggression by the Cuban government.
3. There is no benefit to the US government from lifting the trade ban since there are hardly any
private businesses in Cuba.
4. Reversing the embargo could be ruinous to the US image.
The cons of the trade Ban are as follows
1. The rest of the world is in favour of reversing the trade ban and this ban reduces the US
goodwill in the world.
2. The Ban is considered to be ineffective since it was created to make Cuba promote capitalism
but has not made its objective.
3. The ban has an impact on the small and family Run enterprises in United States and is
expensive for the nation.
The main imports of Cuba are food, machinery and fuel. Its major exports are sugar,
pharmaceuticals and tobacco. The main trading partners of Cuba are Venezuela and European
Union. Tourism stemming from Canada has boosted the economy of Cuba. The European Union
is Cubas main trade partner and also the biggest foreign investor in Cuba. Cuba will benefit from
the European Union thematic program which covers issues such as support of
internationalisation of small and medium enterprises as well as increasing the technical
assistance in agriculture. The European Union is also exploring opportunities in fields of culture
and health..
Explain the chemical basis of why proteins have such a large variety .pdf
Explain the chemical basis of why proteins have such a large variety of structures but DNA has
so few.
Solution
The DNA is a molecule that carries the genetic information from one generation to another.
Structurally, DNA is made of 04 nitrogen containing nucleobases namely, Adenine (A), Guanine
(G), Cytosine (C) and Thymine (T). These chemical bases pairs with each other, like A pairs
with T with 02 Hydrogen Bonds and C pairs with G with 03 Hydrogen bonds to form units called
Base Pairs. Each base is then attached to a 5 carbon sugar molecule and a phosphate group.
Combined base, sugar and phosphate are called nucleotide. These nucleotides are arranged in
two long strands to form a spiral structure called double helix. The double helix structure looks
like ladder, the base pairs form the ‘ladders rugs’ and sugar with phosphate forms the vertical
sidepieces of ladder. Only four different bases in nucleic acid but it contains a long backbone
with million of bases. The appearance of nucleotides in nucleic acid codes for the information
that is being carried out in the molecule.
Protein is a very large biological molecule that is composed of one or more polypeptides that are
folded into specific structures. Structurally, proteins are the polymers of 20 different Amino
Acids (AA). The AA are the building block of the the proteins. Each AA consist of - carbon
atom bonded specifically with carboxyl group (COO-), amino group (NH3+) and distinctive side
chain. The specific chemical property of different AA side chains leads to the role of each AA in
protein structure and function.
The AA are joined together by peptide bonds between - amino group of 1 AA to - carboxyl
group of second. Each polypeptides has 2 distinct ends, one terminates in -amino group (N-
terminus) and other in -carboxyl group (C-terminus) attached to the same tetrahedron atom C.
The proteins have large variety of structure is because, proteins have 20 different Amino Acids
and each amino acid contains a vast range of functional groups including alcohol, thiols,
thioesters, carboxylic acid, carboxamides and variety of basic groups. Proteins are complex
structure of 20 AA, but there are only four bases in DNA (A,T, C,G). A minimum of three bases
are required to code for at least 20 AA . The AA can be distinguished on the basis of their
properties of side chains. Few contain non polar side chains that do not interact with water, other
have uncharged but polar side chains. Some with charged side chain with basic groups. Unlike
DNA which forms ladder like structure protein adopts three dimensional structures. The protein
structure has four different levels. The primary structure is depicted by sequence of Amino Acid
in polypeptide chain. The secondary structure depicts the transformation of AA sequence into
and chains. The third level depicts the completely folded 3 Dimensional structure of protein.
The quaternary structure consist of different interactions between different polypeptides.
Element# of protons# of neutrons# of electronsCarbon (atomic.pdf
Element
# of protons
# of neutrons
# of electrons
Carbon (atomic # 6)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Oxygen (atomic # 8)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Hydrogen (atomic # 1)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Nitrogen (atomic # 7)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
How does the number of electrons affect how many bonds are formed by these atoms?
Define the following (see boxes on pp 31, 33)
Polar covalent bond
Non-polar covalent bond
Ionic bond
Hydrogen bond
Hydrophobic
Hydrophilic
Amphipathic
Element
# of protons
# of neutrons
# of electrons
Carbon (atomic # 6)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Oxygen (atomic # 8)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Hydrogen (atomic # 1)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Nitrogen (atomic # 7)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Solution
Element
# of protons
# of electrons
# of neutrons
Ion
Isotopes
Carbon
6
6
6
C+4 or C-4
Carbon-12 which has 6 protons and 6 neutrons and 6 electrons.
Carbon-14 which has 6 protons, 8 neutrons and 6 electrons.
Oxygen
8
8
8
Oxygen generally forms O-2. It has 6 electrons in the outermost orbit.
Three stable isotopes of oxygen are found- O-16,O17 and O-18.
Hydrogen
1
1
1
Hydrogen has a charge of +1.
Hydrogen has 3 stable isotopes – protium (1 proton and 1 electron), deuterium (1 proton, 1
electron and 1 neutron) and tritium (1 proton, 1 electron and 2 neutrons)
Nitrogen
7
7
7
Usually has the charge -3.
N-14 has odd number of protons and neutrons (7 each). N-15 has protons and 8 neutrons.
Carbon does not usually form an ion by loss or gain of 4 electrons. It shares its 4 electrons with 4
others to complete its orbit and forms a covalent bond.
Oxygen has 6 valence lectrons in its outermost shell, therefore it can form 2 bonds
Hydrogen atom has one valence electron in its outermost orbit. The maximum capacity of its
orbital is 2 electrons.
Nonpolar covalent bond
Non-polar covalent bond is formed when the atoms (with similar electron affinity) share
electrons equally. The electronegativity difference between the two bonded atoms is usually less
than 0.5. Examples – diatomic molecules H2 and N2.
Polar covalent bond
Polar covalent bond is formed when the electrons are not equally shared i.e. the atom with higher
electronegativity (affinity) pulls the electrons towards itself. The compound which is formed has
electrostatic potential and has the ability to form weak bonds with other polar molecules. The
electronegativity difference between bonded atoms is 0.5 to 1.9. Examples include water, ozone.
Ionic bond – the valence electrons are completely transferred from one atom to another. The
metal lo.
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
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select all that apply to ground substance
Solution
Lets us see what is ground substance.
It is an amorphous gel-like or a transparent flud substance surrounding the cells,present in the
connective tissue .
The ground substance and fibers between cells in a connective tissue (most abundant tissues in
the body-functions : binding and supporting; consists of relatively few cells in a generous matrix
) are surrounded and supported by matrix called as extracellular matrix.
The Ground substance includes all the other components of the extracellular matrix excluding
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Question 7 (2 points)A researcher conducts a study on the realtion.pdf
Question 7 (2 points)
A researcher conducts a study on the realtionship between caffeine consumption and
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Question 7 options:
There is not enough information here to tell.
Correlational
Quasi-experimental
Experimental
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Question 8 (2 points)
Which researcher is most likely to engage in content analysis?
Question 8 options:
Dr. Williams, who studies transcripts of police interviews with suspects.
Dr. McKenzie, who studies relationships between various personality traits.
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Dr. Ramirez, who studies mathematical competence in adolescents.
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Question 9 (2 points)
Which of the following would be more likely to be found in an empirical research article on
bullying than in a popular magazine article on the same topic?
Question 9 options:
Opinions about why people bully from the point of view of the author of the article.
Excerpts from actual interviews with bullies and/or victims.
Summaries of the methods and results found by previous researchers on bullying accompanied
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Humerous anecdotes about how victims fantasize about getting back at their bullies.
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Question 10 (2 points)
Dr. Simmons is interested in studying attachment behavior in infants. She trained her research
assistants to recognize classic attachment behaviors such as fearful reactions to strangers and
proximity seeking with the attachment figure. The research assistants then learned to code these
behaivors while watching video tapes of infants and caregivers interacting while they waited in
the doctor\'s waiting room. What type of research is going on here?
Question 10 options:
Quasi-experimental research
Naturalistic observation
Archival data collection
Interrupted time-series designa)
There is not enough information here to tell.b)
Correlationalc)
Quasi-experimentald)
Experimental
Solution
b)
Correlational
a)
Dr. Williams, who studies transcripts of police interviews with suspects.
c)
Summaries of the methods and results found by previous researchers on bullying accompanied
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c)
Archival data collection
b)
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Primary difference between the income sections for form 1120 and .pdf
Primary difference between the income sections for form 1120 and 1120s
Primary difference between the income sections for form 1120 and 1120s
Solution
Answer :- The primary difference between form 1120 and 1120s is that the form 1120 is filed by
corporations on annual basis under internal revenue service (IRS) code whereas the form 1120s
is filed by S corporation (corporation who have opted for \"S\" status) on annual basis under
internal revenue service (IRS) code..
Once its readable i will be okay.SolutionI will be okay in.pdf
Once it\'s readable i will be okay.
Solution
\"I will be okay in the end. If it\'s not okay, it\'s not the end.\" What are some incidences when
this John Lennon quote was proved true in real life?
Many inspiring stories of people\'s lives show it...
Consider Abraham Lincoln. He met failure after failure throughout his life before he became the
President of USA at 52.
Thomas Edison... 10,000 failures before inventing the light bulb.
Henry Ford... was broke at the age of 40.
Conolel Sanders... broke at 65... knocked at 1000 doors before he could make his first sale of a
chicken dish...founder of KFC....
Each and everytime in life ,
On all the small tasks and bigger challenges...its proven true.
It may be preparing a cup of tea or coffee or
Working on a home . The tea or coffee after two boils are perfectly okay for me but you may
boil it more with extra milk.
I work on decor with few frames of modern art and exclusive vases but you may go for plain
walls.
A waiter takes the order and serves and is okay with it...but another one wears a smile and is
quite alert and won\'t let the glass of water to be empty because its his level of \"okay\".
So each of us working and completing our daily task work on this principle.
We try hard and reach our okay\'s.
There\'s a variation in \"okay\" of all of us...that\'s it..
Need the mutiple choice answered uslde .An)_can omunicate with a sys.pdf
Need the mutiple choice answered uslde .An)_can omunicate with a system inside a firewall
system to be used to filter out undesirable traffic and prevent employees from accessing
potentially hostile web sites Multiple-Choice Quiz e three types of event logs generated What are
the three t t bw Windows Which of the following is not a capability of 6. Windows 2003 and
Vista by A. Event, Process, and Security B. Application, User, and Security c. User, Event, and
Security network-based IDS A. Can detect denial-of-service attacks B. Can decrypt and read
encrypted traffic C. Can decode UDP and TCP packets D. Can be tuned to a particular network
Application, System, and Security What are the two main types of intrusion detection systems?
A. Network-based and host-based B. Signature-based and event-based C. Active and reactive D.
Intelligent and passive environment An active IDS can: A. Respond to attacks with TCP resets
B. Monitor for malicious activity C. A and B D. None of the above Honeypots are used to: , 7. s.
What was the first commercial, network-based 8. IDS product? A. Stalker B. NetRanger C. IDES
D. RealSecure What are the two main types of IDS signatures? 9. Egress filtering is used to
detect spam that is: A. Network-based and file-based B. Context-based and content-based C
Active and reactive A. Attract attackers by simulating systems with open network services B.
Monitor network usage by employees C. Process alarms from other IDSs D. Attract customers to
e-commerce sites A. Coming into an organization B. Sent from known spammers outside your
organization C. Leaving an organization D. Sent to mailing lists in your organization
Preventative intrusion detection systems: A. Are cheaper B. D. None of the above . Which of the
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stop malicious activity from occurring B. Does not C. Can l D. All of the above not interact with
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Solution
1. What are the three types of event logs generated by Windows 2003 and Vista systems?
D. Application, System, and Security
2. What are the two main types of intrusion detection systems?
A. Network-based and host-based
3. What was the first commercial, network-based IDS product?
B. NetRanger
4. What are the two main types of IDS signatures?
B. Context-based and content-based
5. Which of the following describes a passive, host-based IDS?
D. All of the above
6. Which of the following is not a capability of network-based IDS?
B. Can decrypt and read encrypted traffic
7. An active IDS can:
C. A and B
8. Honeypots are used to:
A. Attract attackers by simulating systems with open network services
9. Egress filtering is used to detect spam that is:
C. Leaving an organization
10. Preventative intrusion detection systems:
B. Are designed to stop m.
Need a strong biologistgeneticist to answer this P element question.pdf
Need a strong biologist/geneticist to answer this P element question for Drosophila... PLEASE
What do you know about P-elements in Drosophila? How might P-elements have entered
Drosophila melanogaster? What do you know about P-element structure? How big are the native
elements? How do P-elements move around the genome? Connect your ideas to the special P-
elements used in this lab. What is the connection between transposition of mobile DNA elements
and DNA replication? Why are P-elements important for biologists and Drosophila researchers
in particular? What do you know about the Gene Disruption Project (GDP) in Drosophila?
Solution
Ans.) P element is a type of transposon found particularly in Drosophila and is used for
mutagenesis and for the production of genetically modified flies that is extensively used in
various genetic and medical researches. It has ability to initiate a phenotype known as hybrid
dysgenesis. It encodes the protein P transposase and is flanked by terminal inverted repeats
which are important for its mobility.
The P element enters in Drosophila by the horizontal gene transfer method. It has a sequence of 4
exons with 3 introns. The absolute splicing of the intron produces the transposase enzyme,
whereas the alternative partial splicing of intron 1 and 2 produce the P element repressor.
The P element is a class II transposon and it is moved by DNA-based \"cut and paste\"
mechanism. The P element encodes the transposase enzyme which recognizes the 31 bp terminal
inverted repeats of the P element and catalyzes P element excision and re-insertion.
P element is important for biologist and genetics because nowadays it is widely used to solve
many puzzles of genetics through its use in fly transformation, insertional mutagenesis as well as
reverse genetics.
The Gene Disruption Project (GDP) is a public collection of mutant that is created by applying a
non-targeted transposon mutagenesis strategy. Its main objective is to acquire a transposable
element insertion in each gene and this collection is available online and new mutants may be
requested while they are being balanced. These balanced stocks are available at the Bloomington
Drosophila Stock Center..
Modify HuffmanTree.java and HuffmanNode.java to allow the user to se.pdf
Modify HuffmanTree.java and HuffmanNode.java to allow the user to select the value of n 9 to
construct an n ary Huffman Tree. Once finished, verify the accuracy of your tree by running it
for the English Alphabet for n = 3, 4, 5, 6, 7, 8, 9. For each value of n, compute the average
codeword length. Write these in a table (n vs. ACW L(Cn)).
/ HuffmanTree.java
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.ArrayList;
import java.util.Scanner;
public class HuffmanTree {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
System.out.print(\"Enter the file name: \");
String fileName = in.nextLine();
ArrayList nodes = new ArrayList<>();
ArrayList temp = new ArrayList<>();
try {
Scanner fin = new Scanner(new FileReader(fileName));
String symbol;
double prob;
while(fin.hasNext()){
symbol = fin.next();
prob = fin.nextDouble();
nodes.add(findInsertIndex(nodes, prob), new HuffmanNode(symbol, prob));
}
fin.close();
temp.addAll(nodes);
HuffmanNode root = createHuffmanTree(temp);
setCodeWords(root);
double avg = 0;
for (HuffmanNode node : nodes) {
System.out.println(node.getS() + \": \" + node.getC() + \": \" + node.getL());
avg += node.getL();
}
if(avg != 0)
avg /= nodes.size();
System.out.println(\"Average code length is: \" + avg);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
private static HuffmanNode createHuffmanTree(ArrayList nodes){
HuffmanNode root = null;
if(nodes.size() > 0) {
HuffmanNode node0, node1, node;
while (nodes.size() > 1) {
node0 = nodes.get(0);
node1 = nodes.get(1);
node = new HuffmanNode(null, node0.getP() + node1.getP());
node.setLeft(node0);
node.setRight(node1);
nodes.remove(0);
nodes.remove(0);
nodes.add(findInsertIndex(nodes, node.getP()), node);
}
root = nodes.get(0);
}
return root;
}
private static void setCodeWords(HuffmanNode root){
if(root != null){
setCodeWords(root, \"\");
}
}
private static void setCodeWords(HuffmanNode root, String codeWord){
if(root.getS() != null){
root.setC(codeWord);
root.setL(codeWord.length());
}
else{
setCodeWords(root.getLeft(), codeWord + \"0\");
setCodeWords(root.getRight(), codeWord + \"1\");
}
}
private static int findInsertIndex(ArrayList nodes, double prob){
for(int i = 0; i < nodes.size(); ++i){
if(prob < nodes.get(i).getP()){
return i;
}
}
return nodes.size();
}
}
// HuffmanNode.java
public class HuffmanNode {
private String S;
private double P;
private String C;
private int L;
private HuffmanNode left, right;
public HuffmanNode(String s, double p) {
S = s;
P = p;
}
public String getS() {
return S;
}
public void setS(String s) {
S = s;
}
public double getP() {
return P;
}
public void setP(double p) {
P = p;
}
public String getC() {
return C;
}
public void setC(String c) {
C = c;
}
public int getL() {
return L;
}
public void setL(int l) {
L = l;
}
public HuffmanNode getLeft() {
return left;
}
public void setLeft(HuffmanNode left) {
this.left = left;
}
public HuffmanNode getRight() {
return right;
}
public void se.
Listed are 2 of the core theories of evolution. For each, give a def.pdf
Listed are 2 of the core theories of evolution. For each, give a definition or brief explanation of
the theory and 2 examples of the type of evidence we have that supports the theory.
The theory of common descent –
2 examples of supporting evidence:
The theory of natural selection –
2 examples of supporting evidence:
Solution
Evolution:
Evolution is a process of change through time. It is the process by which modern organisms have
descended from ancient organisms.
In genetic terms, the change in the frequency of alleles in populations from one generation to
another generation. (change over time).
A. The theory of common descent:
According to the principle of common descent, all species (living and extinct) are descended
from ancient common ancestors.
In common descent theory darwin explains, that organisms are descended from common
ancestors that gradually change over time into entirely new organism.
To create new species, many generations must pass on traits with slight modifications.
Examples:
1. Archaeopteryx is between reptiles and birds.
2. Eustheopteron is an amphibious fish.
3. Seymouria is a reptile-like amphibian.
4. Therapsids were mammal-like reptiles.
B. Darwin\'s Natural selection:
According to the Darwin, It is the process that results in adaptation of an organism to its
environment which means of selectivity reproducing changes in the genotype.
In the process of evolution the natural selection is a basic mechanism, it is mainly happens due to
the mutations, migrations and genetic drift.
Examples:
1. The beaks of Darwin finches. The finches of the Galapagos islands were adopted to the island
in different ways, but are believed to share a common ancestor.
2. Peppered moths and industrial pollution: Natural selection favored dark colored moths in areas
of heavy pollution, while light colored moths are survived better in less polluted area.
3.The soapberry bugs have must mutated their beaks length genes, when they ran out of food.
Beak size variation:
In South Florida soapberry bugs feed on balloon vine tree with large fruits, longer beak present
in them.
In North Florida soapberry bugs feed on golden rain tree with small fruits, they have shorter
beaks..
Light is confined within the core of a simple clad optical fiber by.pdf
Light is confined within the core of a simple clad optical fiber by?
Solution
An optical fiber is a cylindrical dielec wave guide (nonconducting waveguide) that transmits
light along its axis, by the process of total internal reflection. To confine the optical signal in the
core, therefractive index of the core must be greater than that of the cladding..
Let L R^n rightarrow R^m be a linear transformation. Show that if n .pdf
Let L: R^n rightarrow R^m be a linear transformation. Show that if n > m, then L must have a
nontrivial kernel, i.e., dim(ker(L)) > 0. If n lessthanorequalto m does L have to be one-to-one?
Solution
For second proof,
If T is one-one, it can\'t map a non-zero vector into 0, since it already maps 0 into 0. This means
that its kernel can contain only 0.
Suppose the kernel contains just the zero vector. To check if T is one-one, check if it can map
two different vectors u and v in Rn into the same vector in Rm. If T(u) = T(v), then T(u – v) =
T(u) – T(v) = 0. This means that the non-zero vectoru – v is in the kernel of T, which is
impossible since we assumed the kernel contained just 0. Then T cannot map different vectors
into the same vector, and must be one-one..
IS path vector routing algorithm closer to the distance vector rou.pdf
: IS path vector routing algorithm closer to the distance vector routing algorithm or to the link
state routing algorithm? Explain.
Solution
The path-vector routing algorithm is truly distance-vector routing using the finest path instead of
the direct distance as the metric. Each node first create a forwarding table. Assuming it can only
arrive at instant neighbors . The forwarding table is slowly better as path vectors turn up from the
instant neighbors..
a growth medium that can determine if a bacterium contains several d.pdf
a growth medium that can determine if a bacterium contains several different types of enzymes
or characteristics.
a multitest
b selective
c differential
d enriched
e all of the above
Solution
b selective
Selective media allows the growth of certain type of organisms, it contains several different
types of enzymes or characteristics..
Identify the function of the circulatory system in fishes. Explain h.pdf
Identify the function of the circulatory system in fishes. Explain how it works.
Solution
The circulatory arrangement of fish is very straightforward. It comprises of a heart, blood, and
veins. The heart of a fish is a basic strong structure that is situated behind (and underneath) the
gills. It is encased by the pericardial layer or pericardium.
The heart comprises of a chamber, a ventricle, a thin-walled structure known as sinus venosus,
and a tube called bulbus arteriosus.
The blood contains plasma (the liquid part) and platelets. The red platelets or the erythrocytes
contain hemoglobin, a protein that conveys oxygen all through the body. The white platelets
contain a basic part of the invulnerable framework. The thrombocytes perform capacities that are
proportionate to the part of platelets in the human body, i.e. they help in blood coagulating.
Veins convey the blood all through the body. While corridors convey oxygenated blood from the
gills to whatever remains of the body, veins return deoxygenated blood from various parts of the
body to the heart.
The cardiovascular arrangement of a fish involve a heart, veins, corridors, blood, and fine
vessels. The vessels are minute vessels that shape a system called a fine bed, where the blood
vessel and venous blood get connected. Vessels have thin dividers that encourage dispersion, a
procedure through which oxygen and different supplements from the blood vessel blood are
moved into the cells. In the meantime, carbon dioxide and waste materials are moved into the
vessels.
Vessels with deoxygenated blood (contains carbon dioxide) deplete into little veins called
venules, which thusly deplete into bigger veins. The veins convey the deoxygenated blood into
the sinus venosus, which resembles a little gathering chamber. The sinus venosus has pacemaker
cells that are in charge of starting constrictions, so that the blood is moved into the thin-walled
chamber, which has not very many muscles.
The chamber produces feeble compressions to push blood into the ventricle. The ventricle is a
thick-walled structure with bunches of cardiovascular muscles. It produces enough weight to
pump the blood all through the body. The ventricle pumps blood inside it into bulbus arteriosus,
a little chamber with flexible segments.
While bulbus arteriosus is the name of the chamber in teleosts (rayfinned, hard fish), the
structure is known as conus arteriosus in elasmobranchs (angle with cartilaginous skeleton and
placoid scales). Conus arteriosus has numerous valves and muscles, while bulbus arteriosus has
no valves. The primary capacity of this structure is to diminish the beat weight created by the
ventricle, with a specific end goal to maintain a strategic distance from harm to the thin-walled
gills.
Gills are the essential respiratory organs of fish. They encourage trade of gasses, i.e. retention of
oxygen from water and end of carbon dioxide. Corridors convey the oxygenated blood (from the
gills) all through the body..
How did the discovery of three categories of petite mutations in yea.pdf
How did the discovery of three categories of petite mutations in yeast lead researchers to
postulate extranuclear inheritance of colony size?
Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive
petites followed non-Mendelian patterns that were consistent with the involvement of nuclear
inheritance.
Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive
petites followed non-Mendelian patterns that were consistent with the involvement of an
extranuclear agent.
Whereas suppressive petites exhibited Mendelian inheritance, both neutral and segregational
petites followed non-Mendelian patterns that were consistent with the involvement of an
extranuclear agent.
Whereas suppressive petites exhibited Mendelian inheritance, both neutral and segregational
petites followed non-Mendelian patterns that were consistent with the involvement of nuclear
inheritance.
Solution
Whereas segregational petites exhibited Mendelian inheritance, both neutral and suppressive
petites followed non-Mendelian patterns that were consistent with the involvement of nuclear
inheritance.
---------
Extra nuclear inheritance means, inheritance patterns involving genetic material outside the
nucleus, here the genome is residing in mitochondria or chloroplast.
The segregation petite followed Mendelian type of inheritance (wild type and segregation when
mated, segregated in 1:1 ratio), and if any mutation has occurred in the cell nucleus, it affected
the protein production needed for mitochondrial function.
Where as neutral petites, they did not have most of their mt DNA and but then it was mated with
wild type resulted in all wild type 4: neutral 0. So, these are not following Mendelian type of
inheritance.
In case, of suppressive petites, when suppressive mutants mated with wild type resulted in All
wild type 4: 0 all petite. So they are also not following Mendelian inheritance..
A certain system can experience 2 difference types of defects. Let ,.pdf
A certain system can experience 2 difference types of defects. Let , Ai=1,2 denote the event that
the system has a defect of type i. Suppose that A1=0.15, A2=0.1, P(A1 and A2) = 0.08, and . If
the system has a type 1 defect, what is the probability that it has a type 2 defect? (Consider the
conditional probability).
a.) 0.47
b.) 0.33
c.) 0.252
d.) 0.53
Solution
The answer is (D).
Find to three decimal places, the coordinate of point P lying in the.pdf
Find to three decimal places, the coordinate of point P lying in the 4th quadrant on the Cartesian
plan such that the line tangent to the cosine function passes through the origin. X:[0,2Pi] Show
all work and describe all steps.
Solution
t = cos x
tangent passing through origin
t\' = -sin x
So,
tangent y = mx+c
y = -xsinx+c
here,
(0,0) is point on tangent,
0 = -0sin0 + c
c = 0
So,
y = -xsinx is tangent..
Find the mobility (DoF) and Grashof condition of the mechanism below..pdf
Find the mobility (DoF) and Grashof condition of the mechanism below. Also, which inversion
is it?
Solution
Mobility
n = 4
p = 4
DOF = 3*(n-1) - 2*p = 3*3 - 2*4 = 1
Grashof\'s Condition:
s + l = 0.68+1.62 = 2.3 in
p + q = 1.38 + 1.22 = 2.6
as (s+l)<(p+q) so it is Grashof linkage.
As the shortest part is the Frame (BasePart), so the linkage is a Double-Crank (or drag link)
mechanism..
5. Discuss the pros and cons of American business being excluded by .pdf
5. Discuss the pros and cons of American business being excluded by American law (Pres
Kennedy issued an Executive Order) from conducting any form of business activity on the island
of Cuba. Express your perspectives of the European Union and other first world nations (like
Canada) steadfastly generating billion of euros (and other currencies) from trade with Cuba.
Solution
There are significant pros and cons of prohibition of conducting trade with Cuba.
The pros of the prohibition are as follows
1. Cuba has not fulfilled the conditions of lifting the trade Ban which was set up by the US
government.
2. The US government made various attempts to reduce the restrictions of trade. However this
was met with aggression by the Cuban government.
3. There is no benefit to the US government from lifting the trade ban since there are hardly any
private businesses in Cuba.
4. Reversing the embargo could be ruinous to the US image.
The cons of the trade Ban are as follows
1. The rest of the world is in favour of reversing the trade ban and this ban reduces the US
goodwill in the world.
2. The Ban is considered to be ineffective since it was created to make Cuba promote capitalism
but has not made its objective.
3. The ban has an impact on the small and family Run enterprises in United States and is
expensive for the nation.
The main imports of Cuba are food, machinery and fuel. Its major exports are sugar,
pharmaceuticals and tobacco. The main trading partners of Cuba are Venezuela and European
Union. Tourism stemming from Canada has boosted the economy of Cuba. The European Union
is Cubas main trade partner and also the biggest foreign investor in Cuba. Cuba will benefit from
the European Union thematic program which covers issues such as support of
internationalisation of small and medium enterprises as well as increasing the technical
assistance in agriculture. The European Union is also exploring opportunities in fields of culture
and health..
Explain the chemical basis of why proteins have such a large variety .pdf
Explain the chemical basis of why proteins have such a large variety of structures but DNA has
so few.
Solution
The DNA is a molecule that carries the genetic information from one generation to another.
Structurally, DNA is made of 04 nitrogen containing nucleobases namely, Adenine (A), Guanine
(G), Cytosine (C) and Thymine (T). These chemical bases pairs with each other, like A pairs
with T with 02 Hydrogen Bonds and C pairs with G with 03 Hydrogen bonds to form units called
Base Pairs. Each base is then attached to a 5 carbon sugar molecule and a phosphate group.
Combined base, sugar and phosphate are called nucleotide. These nucleotides are arranged in
two long strands to form a spiral structure called double helix. The double helix structure looks
like ladder, the base pairs form the ‘ladders rugs’ and sugar with phosphate forms the vertical
sidepieces of ladder. Only four different bases in nucleic acid but it contains a long backbone
with million of bases. The appearance of nucleotides in nucleic acid codes for the information
that is being carried out in the molecule.
Protein is a very large biological molecule that is composed of one or more polypeptides that are
folded into specific structures. Structurally, proteins are the polymers of 20 different Amino
Acids (AA). The AA are the building block of the the proteins. Each AA consist of - carbon
atom bonded specifically with carboxyl group (COO-), amino group (NH3+) and distinctive side
chain. The specific chemical property of different AA side chains leads to the role of each AA in
protein structure and function.
The AA are joined together by peptide bonds between - amino group of 1 AA to - carboxyl
group of second. Each polypeptides has 2 distinct ends, one terminates in -amino group (N-
terminus) and other in -carboxyl group (C-terminus) attached to the same tetrahedron atom C.
The proteins have large variety of structure is because, proteins have 20 different Amino Acids
and each amino acid contains a vast range of functional groups including alcohol, thiols,
thioesters, carboxylic acid, carboxamides and variety of basic groups. Proteins are complex
structure of 20 AA, but there are only four bases in DNA (A,T, C,G). A minimum of three bases
are required to code for at least 20 AA . The AA can be distinguished on the basis of their
properties of side chains. Few contain non polar side chains that do not interact with water, other
have uncharged but polar side chains. Some with charged side chain with basic groups. Unlike
DNA which forms ladder like structure protein adopts three dimensional structures. The protein
structure has four different levels. The primary structure is depicted by sequence of Amino Acid
in polypeptide chain. The secondary structure depicts the transformation of AA sequence into
and chains. The third level depicts the completely folded 3 Dimensional structure of protein.
The quaternary structure consist of different interactions between different polypeptides.
Element# of protons# of neutrons# of electronsCarbon (atomic.pdf
Element
# of protons
# of neutrons
# of electrons
Carbon (atomic # 6)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Oxygen (atomic # 8)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Hydrogen (atomic # 1)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Nitrogen (atomic # 7)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
How does the number of electrons affect how many bonds are formed by these atoms?
Define the following (see boxes on pp 31, 33)
Polar covalent bond
Non-polar covalent bond
Ionic bond
Hydrogen bond
Hydrophobic
Hydrophilic
Amphipathic
Element
# of protons
# of neutrons
# of electrons
Carbon (atomic # 6)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Oxygen (atomic # 8)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Hydrogen (atomic # 1)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Nitrogen (atomic # 7)
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Neutral atom:
Ion:
Isotope:
Solution
Element
# of protons
# of electrons
# of neutrons
Ion
Isotopes
Carbon
6
6
6
C+4 or C-4
Carbon-12 which has 6 protons and 6 neutrons and 6 electrons.
Carbon-14 which has 6 protons, 8 neutrons and 6 electrons.
Oxygen
8
8
8
Oxygen generally forms O-2. It has 6 electrons in the outermost orbit.
Three stable isotopes of oxygen are found- O-16,O17 and O-18.
Hydrogen
1
1
1
Hydrogen has a charge of +1.
Hydrogen has 3 stable isotopes – protium (1 proton and 1 electron), deuterium (1 proton, 1
electron and 1 neutron) and tritium (1 proton, 1 electron and 2 neutrons)
Nitrogen
7
7
7
Usually has the charge -3.
N-14 has odd number of protons and neutrons (7 each). N-15 has protons and 8 neutrons.
Carbon does not usually form an ion by loss or gain of 4 electrons. It shares its 4 electrons with 4
others to complete its orbit and forms a covalent bond.
Oxygen has 6 valence lectrons in its outermost shell, therefore it can form 2 bonds
Hydrogen atom has one valence electron in its outermost orbit. The maximum capacity of its
orbital is 2 electrons.
Nonpolar covalent bond
Non-polar covalent bond is formed when the atoms (with similar electron affinity) share
electrons equally. The electronegativity difference between the two bonded atoms is usually less
than 0.5. Examples – diatomic molecules H2 and N2.
Polar covalent bond
Polar covalent bond is formed when the electrons are not equally shared i.e. the atom with higher
electronegativity (affinity) pulls the electrons towards itself. The compound which is formed has
electrostatic potential and has the ability to form weak bonds with other polar molecules. The
electronegativity difference between bonded atoms is 0.5 to 1.9. Examples include water, ozone.
Ionic bond – the valence electrons are completely transferred from one atom to another. The
metal lo.
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Relying on the above pathway for biosynthesis of tryptophan, fill i.pdf
- 1. Relying on the above pathway for biosynthesis of tryptophan, fill in the table below to predict growth (+) or failure to grow (-) when mutants defective in each of the trp genes are placed on media supplemented with the compounds listed. Solution Answer: A mutant for a particular gene in the tryptophan pathway will only grow when it is supplemented with the required nutrient so that the biosynthetic pathway for the tryptophan synthesis is resumed. The mutant strain (auxotroph) will proliferate only when the medium is supplemented with the required substance leading to the completion of the tryptophan biosynthetic pathway. + indicates growth when a particular supplement is given. Mutants which have dysfunctional trpB, trpC, trpD or trpE can still grow if supplemented with any of Indole, tryptophan or Indole+tryptophan because in all of these cases the strain will get the required tryptophan necessary for its survival.SupplementMutanttrpAtrpBtrpCtrpDtrpENone-----Anthranilic acid---- +Indole-++++Tryptophan+++++Typtophan and indole+++++