15. Let’s say you are systems administrator for a hospital’s network.pdf
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15. Let’s say you are systems administrator for a hospital’s network. Keep in mind that as system administrator, your job is to provide the correct level of accessibility to authorized users while denying access to those who lack authorization. What are the top five policies that you would implement? How would you enforce compliance with these policies? How would you educate your users to convince them to adhere to your policies? Solution The policies that system administrator use to provide the correct level of accessibility to authorized users while denying access to unauthorized users are- 1. Using antivirus software and running them regularly to protect systems from attack by malicious softwares. 2. Using firewalls to maintain access control based on sender\'s and reciever\'s ip addresses. 3. By providing Login system policy to ensure authorized access to system by individuals (using login ID and passwords). 4. Using high level encryption techniques to secure access of information on transit between two users. 5. By monitoring network communication. These policies can be enforced by system administrator by using distributed operating system. Because DO/S works on the property of global control of all assets. Users can be provided with individual login and password as unauthorized access may lead to breach in security of hospital\'s information. As security breach in Hospital\'s information may lead to adverse effects like- breach in information of patients, their diseases, medicines prescribed and many more..
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A robust network security strategy is one that actively involves the entire stakeholders of the system. The network administrator has the responsibility of ensuring that best practices in information security management are implemented throughout the entirety of the system they oversee. Threats to a system exist both within and outside an organization. This necessitates the need for a comprehensive security strategy that can cover all those potential threats. Information security threats are of a dynamic nature and the network administrator should take this consideration to ensure that they are always on top of any emerging threats. System vulnerabilities should be sought and effectively sealed and this should be a regular task.
End User Security Recommendations
Best practice in network security will require that the users and the firm abide by the following:
• Training and awareness – all employees of the company should have a firm grasp of matters pertaining network security. This will come through the training that should be offered by the company. The training should involve how to spot and identify threats, how to combat them, and how to handle them should they occur. As new threats emerge, the firm will need to create a continuous awareness program to inform its employees on them.
• Effective monitoring program – even after training has been done, this is not reason enough to believe employees will adhere to the lessons learnt. As such, the IT personnel should be empowered to conduct random checks on the security behavior of the firm’s employees. This will help in identifying potential weak spots.
• Unique user credentials – each and every employee that has been granted use of computer resources should do so with their own unique username and a password that should not be shared with any other user. The password should be complex enough that no one could possibly guess. The user should avoid using passwords from familiar objects or people. A strong password should have a mix of alphanumeric and special characters. For every activity a user does on any computer, they will be required to use their own unique credentials. This should leave an audit that can be followed should there be an incident.
• Automatic logoff – it is possible that a user might leave a computer without logging out from their session. This opens the possibility that another user might access resources using the logged in credentials. This could be devastating should the unauthorized have malicious intent and the logged on credentials have advanced permissions. Automatic logoff should be set to happen after a given period of time. This should especially happen after the end of prescribed business hours.
• Regular event log audits – event logs are very important when it comes to monitoring the performance of a given system. They can also be used to spot any anomalies within the system. Event log.
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Untitled document.pdf
Outline:
I. Introduction
Definition of horizontal privilege escalation
Importance of understanding the risks
II. Common Vulnerabilities and Exploits
Misconfigured access controls
Weak authentication mechanisms
Software vulnerabilities
Social engineering attacks
III. Impact of Horizontal Privilege Escalation
Unauthorized access to sensitive information
Data breaches and privacy violations
Financial losses and legal consequences
Reputational damage
IV. Examples of Horizontal Privilege Escalation
Case study 1: Exploiting a misconfigured access control
Case study 2: Leveraging weak authentication
Case study 3: Exploiting software vulnerabilities
V. Mitigation Strategies
Implementing strong access controls
Regularly updating and patching software
Conducting security audits and penetration testing
Educating employees about social engineering attacks
VI. Best Practices for Prevention
Principle of least privilege
Implementing multi-factor authentication
Regularly monitoring and logging system activities
Implementing intrusion detection and prevention systems
VII. Conclusion
VIII. FAQs
What is horizontal privilege escalation?
How can misconfigured access controls lead to horizontal privilege escalation?
What are some examples of software vulnerabilities that can be exploited for horizontal privilege escalation?
How can organizations prevent horizontal privilege escalation?
What are the potential consequences of horizontal privilege escalation?
The Risks of Horizontal Privilege Escalation
Horizontal privilege escalation refers to a critical security vulnerability that can have severe consequences for organizations and individuals alike. It occurs when an unauthorized user gains access to resources, data, or privileges that they should not have within the same level of authorization. In this article, we will delve into the risks associated with horizontal privilege escalation and explore mitigation strategies to protect against this type of attack.
Introduction
Horizontal privilege escalation poses a significant threat to the security of computer systems, networks, and sensitive data. It occurs when an attacker exploits vulnerabilities or weaknesses within a system to gain unauthorized access to resources or privileges. Understanding the risks associated with this type of attack is crucial for organizations to implement effective security measures.
Common Vulnerabilities and Exploits
Misconfigured access controls: Improperly configured access controls can allow unauthorized users to gain access to sensitive information or perform actions beyond their authorized privileges. Attackers can exploit these misconfigurations to elevate their privileges and access critical resources.
Weak authentication mechanisms: Weak passwords, default credentials, or insufficient authentication processes provide opportunities for attackers to gain unauthorized access to user accounts and escalate their privileges within a system.
Software vulnerabilities:
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The Risks of Horizontal Privilege Escalation.pdf
I. Introduction
Definition of horizontal privilege escalation
Importance of understanding the risks
II. Common Vulnerabilities and Exploits
Misconfigured access controls
Weak authentication mechanisms
Software vulnerabilities
Social engineering attacks
III. Impact of Horizontal Privilege Escalation
Unauthorized access to sensitive information
Data breaches and privacy violations
Financial losses and legal consequences
Reputational damage
IV. Examples of Horizontal Privilege Escalation
Case study 1: Exploiting a misconfigured access control
Case study 2: Leveraging weak authentication
Case study 3: Exploiting software vulnerabilities
V. Mitigation Strategies
Implementing strong access controls
Regularly updating and patching software
Conducting security audits and penetration testing
Educating employees about social engineering attacks
VI. Best Practices for Prevention
Principle of least privilege
Implementing multi-factor authentication
Regularly monitoring and logging system activities
Implementing intrusion detection and prevention systems
VII. Conclusion
VIII. FAQs
What is horizontal privilege escalation?
How can misconfigured access controls lead to horizontal privilege escalation?
What are some examples of software vulnerabilities that can be exploited for horizontal privilege escalation?
How can organizations prevent horizontal privilege escalation?
What are the potential consequences of horizontal privilege escalation?
The Risks of Horizontal Privilege Escalation
Horizontal privilege escalation refers to a critical security vulnerability that can have severe consequences for organizations and individuals alike. It occurs when an unauthorized user gains access to resources, data, or privileges that they should not have within the same level of authorization. In this article, we will delve into the risks associated with horizontal privilege escalation and explore mitigation strategies to protect against this type of attack.
Introduction
Horizontal privilege escalation poses a significant threat to the security of computer systems, networks, and sensitive data. It occurs when an attacker exploits vulnerabilities or weaknesses within a system to gain unauthorized access to resources or privileges. Understanding the risks associated with this type of attack is crucial for organizations to implement effective security measures.
Common Vulnerabilities and Exploits
Misconfigured access controls: Improperly configured access controls can allow unauthorized users to gain access to sensitive information or perform actions beyond their authorized privileges. Attackers can exploit these misconfigurations to elevate their privileges and access critical resources.
Weak authentication mechanisms: Weak passwords, default credentials, or insufficient authentication processes provide opportunities for attackers to gain unauthorized access to user accounts and escalate their privileges within
system.
Software vulnerabilities: Unpatched software or applicatio
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All fungi have tiny tube-like structures call hyphae. One is labeled .pdf
All fungi have tiny tube-like structures call hyphae. One is labeled within the collection. What
do you think these structures do for the fungi?
Solution
Answer:
Hyphae is a long, branching, filamentous structure of a fungus, it contain the cytoplasm or cell
sap, including the nuclei containing genetic material.
-absorb nutrients from the environment and transport them to other parts of the thallus
- hyphae represents the vegetative stage of fungi..
What would the predicted result be if Griffith injected mice with a .pdf
What would the predicted result be if Griffith injected mice with a mixture of heat-killed R
bacteria and living S bacteria?
a. All mice would live, but only if DNAse was added to the mixture prior to injection
b. All mice would die.
c. All mice would live.
d. All mice would die, but only if DNAse was added to the mixture prior to injection.
Solution
Griffith’s Experiment-A Quick Review: Fredrick Griffith, an English physician, first reported
transformation of one strain of Streptococcus pneumoniae into other strain in 1928. The bacteria
cause pneumonia. A strain of this bacterium produces capsule and gives a smooth, shiny colony
on culture media, thus called S-strain (smooth strain). The S-strain exhibits pathogenicity in
laboratory mice. Another strain without capsule produces rough and coarse colony on culture
medium, thus called R-strain (rough strain). The R-strain is avirulent (non-pathogenic).
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead
When injected with live S-strain bacteria, mice developed pneumonia and died. When injected
with live R-strain bacteria, mice did not develop pneumonia and survived. When heat killed S-
strain injected to mice, they remain unaffected. However, when heat killed S-strain bacteria
mixed with live R-strain injected to mice, they developed pneumonia and died. He was also able
to isolate and culture S-strain bacteria from affected mice. Since heat killed bacteria could have
no way been revived, he concluded that some molecules from heat killed S-strain had
transformed the R-strain into S-train.
Now,
The Current Perspective: Presence of living S-strain causes pneumonia development in mice
killing them at last. It is noteworthy that transformation of R-strain into S-strain occurs only
when there is alive R-strain and fragments of DNA from heat-killed S-strain.
Option A: Incorrect. DNAse is a large protein incapable of entering into bacterial cytoplasm
through bacterial membrane of living S-strain. So, this treatment further cleaves DNA fragments
of heat killed R-strain BUT leaves the S-strain unaffected. So, injecting the mixture to mice kills
them as live S-strain causes pneumonia.
Option B: Correct. Presence of living S-strain causes pneumonia development in mice killing
them at last.
Option C: Incorrect. All mice die because S-strain cause pneumonia to them.
Option D: Incorrect. All mice would die, BUT NOT because of addition of DNAse to the
mixture. As mentioned in option A explanation, DNAse does not penetrate into live S-strain,
thus leave them unaffected. Therefore, mice die irrespective of DNAse treatment.
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead.
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Explanation of the most common types of technical risks
Check out "Explanation of the Most Common Types of Technical Risks" from Netspective Communications| https://www.netspective.com/opsfolio/
Cyber Security Measures for LMS Platforms
Cyber Security is a critical concern for LMS platforms, as they store a plethora of sensitive information and are an attractive target for cybercriminals. Implementing robust cybersecurity measures, including encryption, authentication, access control, regular updates and patching, monitoring and detection, backup and recovery, training and awareness, and third-party security, is essential to safeguard against unauthorized access, data breaches, and other cyber threats.
Untitled document.pdf
Outline:
I. Introduction
Definition of horizontal privilege escalation
Importance of understanding the risks
II. Common Vulnerabilities and Exploits
Misconfigured access controls
Weak authentication mechanisms
Software vulnerabilities
Social engineering attacks
III. Impact of Horizontal Privilege Escalation
Unauthorized access to sensitive information
Data breaches and privacy violations
Financial losses and legal consequences
Reputational damage
IV. Examples of Horizontal Privilege Escalation
Case study 1: Exploiting a misconfigured access control
Case study 2: Leveraging weak authentication
Case study 3: Exploiting software vulnerabilities
V. Mitigation Strategies
Implementing strong access controls
Regularly updating and patching software
Conducting security audits and penetration testing
Educating employees about social engineering attacks
VI. Best Practices for Prevention
Principle of least privilege
Implementing multi-factor authentication
Regularly monitoring and logging system activities
Implementing intrusion detection and prevention systems
VII. Conclusion
VIII. FAQs
What is horizontal privilege escalation?
How can misconfigured access controls lead to horizontal privilege escalation?
What are some examples of software vulnerabilities that can be exploited for horizontal privilege escalation?
How can organizations prevent horizontal privilege escalation?
What are the potential consequences of horizontal privilege escalation?
The Risks of Horizontal Privilege Escalation
Horizontal privilege escalation refers to a critical security vulnerability that can have severe consequences for organizations and individuals alike. It occurs when an unauthorized user gains access to resources, data, or privileges that they should not have within the same level of authorization. In this article, we will delve into the risks associated with horizontal privilege escalation and explore mitigation strategies to protect against this type of attack.
Introduction
Horizontal privilege escalation poses a significant threat to the security of computer systems, networks, and sensitive data. It occurs when an attacker exploits vulnerabilities or weaknesses within a system to gain unauthorized access to resources or privileges. Understanding the risks associated with this type of attack is crucial for organizations to implement effective security measures.
Common Vulnerabilities and Exploits
Misconfigured access controls: Improperly configured access controls can allow unauthorized users to gain access to sensitive information or perform actions beyond their authorized privileges. Attackers can exploit these misconfigurations to elevate their privileges and access critical resources.
Weak authentication mechanisms: Weak passwords, default credentials, or insufficient authentication processes provide opportunities for attackers to gain unauthorized access to user accounts and escalate their privileges within a system.
Software vulnerabilities:
Computer Security Primer - Eric Vanderburg - JURINNOV
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The Risks of Horizontal Privilege Escalation.pdf
I. Introduction
Definition of horizontal privilege escalation
Importance of understanding the risks
II. Common Vulnerabilities and Exploits
Misconfigured access controls
Weak authentication mechanisms
Software vulnerabilities
Social engineering attacks
III. Impact of Horizontal Privilege Escalation
Unauthorized access to sensitive information
Data breaches and privacy violations
Financial losses and legal consequences
Reputational damage
IV. Examples of Horizontal Privilege Escalation
Case study 1: Exploiting a misconfigured access control
Case study 2: Leveraging weak authentication
Case study 3: Exploiting software vulnerabilities
V. Mitigation Strategies
Implementing strong access controls
Regularly updating and patching software
Conducting security audits and penetration testing
Educating employees about social engineering attacks
VI. Best Practices for Prevention
Principle of least privilege
Implementing multi-factor authentication
Regularly monitoring and logging system activities
Implementing intrusion detection and prevention systems
VII. Conclusion
VIII. FAQs
What is horizontal privilege escalation?
How can misconfigured access controls lead to horizontal privilege escalation?
What are some examples of software vulnerabilities that can be exploited for horizontal privilege escalation?
How can organizations prevent horizontal privilege escalation?
What are the potential consequences of horizontal privilege escalation?
The Risks of Horizontal Privilege Escalation
Horizontal privilege escalation refers to a critical security vulnerability that can have severe consequences for organizations and individuals alike. It occurs when an unauthorized user gains access to resources, data, or privileges that they should not have within the same level of authorization. In this article, we will delve into the risks associated with horizontal privilege escalation and explore mitigation strategies to protect against this type of attack.
Introduction
Horizontal privilege escalation poses a significant threat to the security of computer systems, networks, and sensitive data. It occurs when an attacker exploits vulnerabilities or weaknesses within a system to gain unauthorized access to resources or privileges. Understanding the risks associated with this type of attack is crucial for organizations to implement effective security measures.
Common Vulnerabilities and Exploits
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Weak authentication mechanisms: Weak passwords, default credentials, or insufficient authentication processes provide opportunities for attackers to gain unauthorized access to user accounts and escalate their privileges within
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All fungi have tiny tube-like structures call hyphae. One is labeled .pdf
All fungi have tiny tube-like structures call hyphae. One is labeled within the collection. What
do you think these structures do for the fungi?
Solution
Answer:
Hyphae is a long, branching, filamentous structure of a fungus, it contain the cytoplasm or cell
sap, including the nuclei containing genetic material.
-absorb nutrients from the environment and transport them to other parts of the thallus
- hyphae represents the vegetative stage of fungi..
What would the predicted result be if Griffith injected mice with a .pdf
What would the predicted result be if Griffith injected mice with a mixture of heat-killed R
bacteria and living S bacteria?
a. All mice would live, but only if DNAse was added to the mixture prior to injection
b. All mice would die.
c. All mice would live.
d. All mice would die, but only if DNAse was added to the mixture prior to injection.
Solution
Griffith’s Experiment-A Quick Review: Fredrick Griffith, an English physician, first reported
transformation of one strain of Streptococcus pneumoniae into other strain in 1928. The bacteria
cause pneumonia. A strain of this bacterium produces capsule and gives a smooth, shiny colony
on culture media, thus called S-strain (smooth strain). The S-strain exhibits pathogenicity in
laboratory mice. Another strain without capsule produces rough and coarse colony on culture
medium, thus called R-strain (rough strain). The R-strain is avirulent (non-pathogenic).
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead
When injected with live S-strain bacteria, mice developed pneumonia and died. When injected
with live R-strain bacteria, mice did not develop pneumonia and survived. When heat killed S-
strain injected to mice, they remain unaffected. However, when heat killed S-strain bacteria
mixed with live R-strain injected to mice, they developed pneumonia and died. He was also able
to isolate and culture S-strain bacteria from affected mice. Since heat killed bacteria could have
no way been revived, he concluded that some molecules from heat killed S-strain had
transformed the R-strain into S-train.
Now,
The Current Perspective: Presence of living S-strain causes pneumonia development in mice
killing them at last. It is noteworthy that transformation of R-strain into S-strain occurs only
when there is alive R-strain and fragments of DNA from heat-killed S-strain.
Option A: Incorrect. DNAse is a large protein incapable of entering into bacterial cytoplasm
through bacterial membrane of living S-strain. So, this treatment further cleaves DNA fragments
of heat killed R-strain BUT leaves the S-strain unaffected. So, injecting the mixture to mice kills
them as live S-strain causes pneumonia.
Option B: Correct. Presence of living S-strain causes pneumonia development in mice killing
them at last.
Option C: Incorrect. All mice die because S-strain cause pneumonia to them.
Option D: Incorrect. All mice would die, BUT NOT because of addition of DNAse to the
mixture. As mentioned in option A explanation, DNAse does not penetrate into live S-strain,
thus leave them unaffected. Therefore, mice die irrespective of DNAse treatment.
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
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Write a function called myfind that will search for a key in a vecto.pdf
Write a function called myfind that will search for a key in a vector and return the indices of all
occurrences of the key, like the built-in find function. It will receive two arguments— the vector
and the key—and will return a vector of indices or the empty vector [ ] if the key is not found.
The function will be invoked as follows:
idx = myfind(vec, key);
Following is an example of correct function behavior:
>> vec = randi([5, 10], 1, 10)
vec = 5 10 10 7 9 5 7 10 9 10
>> key = 10;
>> idx = myfind(vec, key)
idx = 2 3 8 10
>> key = 2;
>> idx = myfind(vec, key)
idx = Empty matrix: 1-by-0
Do not use the in-built find function.
Solution
% matlab code find all index of key in vector
function idx = myfind(vec, key)
index = 1;
for i=1:length(vec)
if vec(i) == key
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index = index + 1;
end
end
end
vec = randi([5, 10], 1, 10);
disp(\"Vector: \");
disp(vec);
key = 10;
fprintf(\"\ Key: %d\ \",key);
idx = myfind(vec, key);
disp(\"\ Index: \");
disp(idx);
%{
sample output:
Vector:
10 10 8 10 5 10 5 8 9 5
Key: 10
Index:
1 2 4 6
%}.
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Without the membrane proteins, cell to cell interaction and transport of molecules across the
membrane are not possible. The membrane receptor proteins help cell communication with the
use of hormones, neurotransmitters, and other signalling molecules. Transport proteins through
facilitated diffusion transport of molecules across the membrane. Structural proteins help to give
the cell support and shape..
What is the state of water at 200 kPa and 250oCQuestion 7 options.pdf
What is the state of water at 200 kPa and 250oC?
Question 7 options:
Saturated vapor
Superheated vapor
Compressed liquid
Saturated liquid
Saturated mixturA)
Saturated vaporB)
Superheated vaporC)
Compressed liquidD)
Saturated liquidE)
Saturated mixtur
Solution
at 200 kPa
saturation temperature is 120.21 C
but the temperature is 250 C
so, it is a superheated vapour.
What is ransomware How can these cyber-attacks result in data loss.pdf
What is ransomware? How can these cyber-attacks result in data loss? In your experience, do
you believe ransomware attacks should be treated as a cybercrime? What are the possible
consequences of ransomware? Be sure to cite examples from credible sources to support your
answers.
Solution
Ransomware is a virus or malware of the computer which installs secretly on computer without
user authorization and encrypts the data that is present on the victims computer and in order to
decrypt that message it asks for payment for decryption. Yes this ransomware cyber attacks
result in data loss because the attacker demands the victim for money in order to give the decrypt
key whether if he has not paid the decrypt key will be lost and the data is also lost forever and
results in data loss.This ransomware is a cybercrime because so much of data is lost and they are
thefting the data and demands money from the victims in order to give the data it is like
kidnapping the data. The possible consequences of ransomware malware are dataloss and
datatheft. This malware are sent through the Email attachments and messages in social media
and whenever we click that it installs without authorization..
What are the 3 ways a Java method must match with its definition and.pdf
What are the 3 ways a Java method must match with its definition and call?
Solution
The three ways are:
1. Name of the method must be exactly same in definition and call.
2. The number of variables mentioned in the function definition must be same as function call.
2.The type of variables mentioned in the function definition must be same as function call..
Using the Internet, compare and contrast two commercially available .pdf
Using the Internet, compare and contrast two commercially available Vulnerability Assessment
solutions.
What are the differences between the two? How are they similar?
Make a purchasing recommendation and defend your suggestion. Be detailed and specific.
(Ensure you provide links to each product offering.)
Solution
An interruption detection system is a software application that monitors network behavior for
cruel activities or policy violation and produce information to a running station. A few systems
may attempt to prevent an intrusion attempt monitoring system. IDPSes have turn into a
necessary addition to the security communications of almost every organization.
IDPSes normally record information related to pragmatic events notify safety administrator of
significant observed events and create reports. A lot of IDPSes can also react to a detected threat
by attempt to stop it from succeeding.
WAFs are intended to protect web applications from web based attacks that IPSs cannot put off.
In the same regard as an IPS, WAFs can be network based. They sit in order and monitor traffic
to and from web applications.
WAFs are normally deployed in some sort of proxy manner just in front of the web application.
CONCLUSION:
Web Applications Firewalls are a particular type of product used to detect attacks next to web
applications in extra depth than an Intrusion anticipation System. WAFs can be used in our
environment to provide better protection to web applications..
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Saplin.pdf
uetator Periodic Table Question 6 of 30 Incorrect Incorrect DA Sapling Learning sponsible
syndrome is anX-inked, recessive condition caused by mutation of the HPRT1 muscle
movements, The syndrome is characterized byjuvenile-onset gout, in and behavioral disorders. A
mother and father have a young daughter who was diagnosed with Lesch-Nyhan syndrome. No
one in the with family was ever diagnosed the The maternal grandfather was diagnosed
syndrome, but neither one of her parents was diagnosed with condition. Select all the
descriptions that could explain how their daughter inherited Lesch-Nyhan syndrome. The
daughter could not have Nyhan because her maternal grandparents did not have disease. The girl
must have acquired the disease through a random mutation. The daughter inherited two
xchromosomes with the of the mothers X mutation. Nondisjunction chromosome chromosomes
occurred during meiosis, and the daughter inherited from her father. The father did not contribute
a sex chromosome to his daughter due to nondisjunction of the sex chromosomes. The daughter
is xo and her only X chromosome came from her mother, who was a O The d from her father,
who was a carrier. camies the Lesch-Nyhan mutation and was Her inherited There is a hint
available! bottom divider bar. Click on the divider bar again to hide the hint. Previous ® Give
Up & View solution Check Answer 0 Next HExi Hint
Solution
1) the hint here is that the mother\'s maternal was diagnosed with Lesch-Nyhan syndrome. Since
it is an X-chromosome related disease, it is very rare to have females affected by it. If a girl has
LNS, she must have inherited two effected X chromosomes from each parent. Also, one third of
cases arise from new mutations even when there is no family history of any. So the first two
points are the possible reasons for the girls to have LNS..
The starship Epidemiology Team has cone many missions exploring the u.pdf
The starship Epidemiology Team has cone many missions exploring the universe. The captain
has noticed that many of the crew members who worked in the weapons bay returned home with
low back pain (LBP). It is important to note that people who worked in the Weapon\'s Bay only
worked in that area of the ship and are the only people who are required to lift heavy equipment.
After their last voyage, the captain decided to sample the crew members who had L8P and ones
who didn\'t. Note that the captain only recorded male crew members that were of the same age.
Based on the captain\'s recordings below, was working in the weapons bay area associated with
LBP? What are some possible interventions the captain could use to reduce the occurrence of
LBP on his ship?
Solution
a) Based on the captains recordings, working in weapons bay is not associated with LBP.
b) Captain can shift the persons with LBP to other areas of ship which does not have any
equipment lifting.
Captain would suggest the person\'s with LBP to go for a medical checkup..
Question 3)Perform a regression analysis using compa as the depend.pdf
Question 3)
Perform a regression analysis using compa as the dependent variable and the same independent
variables as used in question 2. (Question #2 found below the table). Show the result, and
interpret your findings by answering the same questions.
Note: be sure to include the appropriate hypothesis statements.
A ) Regression hypotheses
Ho:
Ha:
Coefficient hyhpotheses (one to stand for all the separate variables)
Ho:
Ha:
B )Interpretation:
For the Regression as a whole:
What is the value of the F statistic:
What is the p-value associated with this value:
Is the p-value < 0.05?
Do you reject or not reject the null hypothesis:
What does this decision mean for our equal pay question (Do M and F get paid equaly?):
C )For each of the coefficients: Intercept, Midpoint, Age, Perf. rating, Service, Gender, Degree
What is the coefficient\'s p-value for each of the variables:
Is the p-value < 0.05?
Do you reject or not reject each null hypothesis:
What are the coefficients for the significant variables?
Using only the significant variables, what is the equation? Compa =
Is gender a significant factor in compa:
If so, who gets paid more with all other things being equal?
How do we know?
ID
Salary
Compa
Midpoint
Age
Performance Rating
Service
Gender
Raise
Degree
Gender1
Gr
1
66.1
1.159
57
34
85
8
0
5.7
0
M
E
2
25.9
0.834
31
52
80
7
0
3.9
0
M
B
3
35.2
1.135
31
30
75
5
1
3.6
1
F
B
4
55.3
0.971
57
42
100
16
0
5.5
1
M
E
5
49.6
1.033
48
36
90
16
0
5.7
1
M
D
6
78.3
1.168
67
36
70
12
0
4.5
1
M
F
7
42.3
1.058
40
32
100
8
1
5.7
1
F
C
8
22.8
0.990
23
32
90
9
1
5.8
1
F
A
9
78
1.164
67
49
100
10
0
4
1
M
F
10
23.3
1.014
23
30
80
7
1
4.7
1
F
A
11
23.6
1.025
23
41
100
19
1
4.8
1
F
A
12
60.8
1.067
57
52
95
22
0
4.5
0
M
E
13
40.6
1.014
40
30
100
2
1
4.7
0
F
C
14
21.7
0.943
23
32
90
12
1
6
1
F
A
15
21.8
0.949
23
32
80
8
1
4.9
1
F
A
16
37.4
0.934
40
44
90
4
0
5.7
0
M
C
17
57
1.000
57
27
55
3
1
3
1
F
E
18
33.5
1.081
31
31
80
11
1
5.6
0
F
B
19
23
1.000
23
32
85
1
0
4.6
1
M
A
20
36
1.162
31
44
70
16
1
4.8
0
F
B
21
76
1.135
67
43
95
13
0
6.3
1
M
F
22
43.7
0.911
48
48
65
6
1
3.8
1
F
D
23
25.3
1.098
23
36
65
6
1
3.3
0
F
A
24
48.9
1.019
48
30
75
9
1
3.8
0
F
D
25
25.8
1.122
23
41
70
4
0
4
0
M
A
26
23.3
1.013
23
22
95
2
1
6.2
0
F
A
27
42.3
1.057
40
35
80
7
0
3.9
1
M
C
28
75.2
1.122
67
44
95
9
1
4.4
0
F
F
29
80.9
1.208
67
52
95
5
0
5.4
0
M
F
30
49
1.020
48
45
90
18
0
4.3
0
M
D
31
24.2
1.054
23
29
60
4
1
3.9
1
F
A
32
27.5
0.886
31
25
95
4
0
5.6
0
M
B
33
63.6
1.115
57
35
90
9
0
5.5
1
M
E
34
28.6
0.922
31
26
80
2
0
4.9
1
M
B
35
22.4
0.976
23
23
90
4
1
5.3
0
F
A
36
23.6
1.026
23
27
75
3
1
4.3
0
F
A
37
24.3
1.057
23
22
95
2
1
6.2
0
F
A
38
63
1.105
57
45
95
11
0
4.5
0
M
E
39
34.8
1.123
31
27
90
6
1
5.5
0
F
B
40
24.3
1.057
23
24
90
2
0
6.3
0
M
A
41
42.8
1.071
40
25
80
5
0
4.3
0
M
C
42
23
0.998
23
32
100
8
1
5.7
1
F
A
43
75.4
1.125
67
42
95
20
1
5.5
0
F
F
44
60.7
1.065
57
45
90
16
0
5.2
1
M
E
45
57.9
1.206
48
36
95
8
1
5.2
1
F
D
46
62.2
1.091
57
39
75
20
0
3.9
1
M
E
47
62.2
1.091
57
37
.
Problems in PYTHON3.5 1. Define a function sumOfDigits that has one .pdf
Problems in PYTHON3.5 1. Define a function sumOfDigits that has one parameter of type
string. The function should return the sum of the digits in the string. Do not treat multiple digit
string as one number – “2014” should be treated as 4 different digits of 2, 0, 1, 4. The function
should return 17 for the string “Today’s date is 09/01/2014”. Then write a main program that
asks a user to enter a sentence and then displays the sum of digits that are in the sentence by
using the function, sumOfDigits. The function sumOfDigits should only do what’s asked to do. It
should not do any additional tasks such as print or input!!!
Solution
Before implementing the function, we need to check whether the integer comes into the string or
not.
To do that there is an in-built function in Python, isdigit() which checks whether the input is the
integer
or some other character.
def sumOfDigits(input_str):
summation = 0
for number in list(input_str):
if number.isdigit():
summation += int(number)
return summation
Hence, this will loop through the entire string and check for each of them whether it is integer or
some other character.For all the integer it is summing up the values and returning it at the end of
the function definition..
Only a very small percentage of organisms actually fossilize. Based o.pdf
Only a very small percentage of organisms actually fossilize. Based on the information in the
table below, rank the likely abundance of fossils of organisms A-E. Rank the fossils of
organisms A-E from most common to least common. If two fossils have the same commonness,
overlap them.
Solution
Answer:
Ranking based on the abundance of fossils of organisms, A-E:
1. Organism E is most likely to fossilize
This organism has a hard shell and lives with sediments near the shore and has a large population
with long existence.
2. Organism A
It has the larger population and longer existence time. This can lead to A having a better chance
at fossilizing than B.
3. Organism B
Organism B has a small population with a short existence time.
4. Organism C
It lives with no sediment, has a large population with a long existence thus leaving it more
susceptible to becoming fossilize than D.
5. Organism D (being the least common fossil)
It has a soft body, lives with no sediment deposits, has a short life and a small population..
package lab7; import java.util.Scanner; Sammy Student, Program.pdf
package lab7;
import java.util.Scanner;
//Sammy Student, Programmer
public class Maze
{
static Scanner sc = new Scanner(System.in);
// maze movements
static char myMove = \'\\0\';
// cell position
static int currentCell = 0;
static int score = 0;
static boolean advance = true;
static boolean checkThis = false;
public static void main(String args[])
{
// the local variables declared and initialized
char answer = \'Y\';
displayMenu();
while(answer == \'Y\' || answer == \'y\')
{
displayMovement();
makeYourMove();
checkThis = checkYourMove();
mazeStatus();
System.out.println(\"move again(Y or N)?\");
answer = sc.next().charAt(0);
}
System.out.println(\"***************************\");
}// end main() method
static void displayMenu()
{
System.out.println(\"\");
System.out.println(\"***************************\");
System.out.println(\"----The Maze Strategy---\");
System.out.println(\"\");
}// end method
static void displayMovement()
{
if(currentCell == 0)
{
System.out.println(\"You have entered the maze!!\");
System.out.println(\"There is no turning back!!\");
currentCell = 1;
mazeStatus();
advance = true;
}
System.out.println(\"make your move (W, A, S, D)\");
System.out.println(\"W = up, A = left, S = down, D = right)\");
}// end method
static void makeYourMove()
{
myMove = sc.next().charAt(0);
switch(myMove)
{
case \'W\': { MoveUp(); break; }
case \'A\': { MoveLeft(); break; }
case \'S\': { MoveDown(); break; }
case \'D\': { MoveRight(); break; }
}
// end program menu
}// end method
static boolean checkYourMove()
{
if(currentCell == 1 && advance == true)
{
if (myMove == \'W\')
{
advance = false;
System.out.println(\"try again\");
return advance;
}
if (myMove == \'A\')
{
advance = false;
System.out.println(\"SORRY, there is no return\");
return advance;
}
if (myMove == \'D\')
{
currentCell = 2;
advance = true;
System.out.println(\"continue through the maze\");
return advance;
}
if (myMove == \'S\')
{
advance = false;
System.out.println(\"continue through the maze\");
return advance;
}
}
if(currentCell == 2 && advance == true)
{
if (myMove == \'W\')
{
advance = false;
System.out.println(\"try again\");
return advance;
}
if (myMove == \'A\')
{
advance = false;
System.out.println(\"SORRY, there is no return\");
return advance;
}
if (myMove == \'D\')
{
advance = false;
System.out.println(\"continue through the maze\");
return advance;
}
if (myMove == \'S\')
{
currentCell = 5;
advance = true;
System.out.println(\"continue through the maze\");
return advance;
}
}
return advance;
// end program menu
}// end method
static void MoveLeft()
{
System.out.println(\"you moved to the left\");
}//end method
static void MoveRight()
{
System.out.println(\"you moved to the right\");
}//end method
static void MoveUp()
{
System.out.println(\"you moved up (forward)\");
}//end method
static void MoveDown()
{
System.out.println(\"you moved down (downward)\");
}//end method
static void mazeStatus()
{
System.out.println(\"current position: cell \" + currentCell);
}//end meth.
Odin and Zeus travel to the outermost reaches of our galaxy using wa.pdf
Odin and Zeus travel to the outermost reaches of our galaxy using warp drive spaceship
searching for other forms of life. They make contact with highly advanced civilization on planet
X329. However, while trying to land at a guided landing site, they are exposed to high energy
gamma rays which produces defects in their classical complement and some cytokine genes, and
obliterates monocyte mactrophages, and granulocytes. Their mission mandates staying with their
friends for three days and then journey back to earth. The trip back to Earth will take between 6-
7 days. Will Odin and Zeus survive meeting their new friends on planet X329? If so what
immune mechanisms will help their survival?
Solution
In the present case, Odin and Zeus are exposed to high energy gamma radiations. Gamma rays
are highly penetrating, and can cause genetic damage leading to detrimental effects. Its ionizing
energy breaks up bonds of genetci material. In the case presented here, gamma rays have caused
defects in classical complement system and also the cytokine genes. Classical complement is the
term given group of more than 30 serum proteins that enhance the ability to produce antibodies
and phagocytic cells, that are critically important in our defense against infections. Similarly,
cytokines which include chemokines, interferons, interleukins, lymphokines and TNFs etc are
small proteins involved in autocrine signalling, paracrine signalling and endocrine signalling,
acting as immunomodulating agents.Thus defects produced by gammma radiations in phagocytic
cells, classical complement, cytokines, cytokine receptors or in molecules that mediate effector
responses might lead to various immunodeficiency diseases. Also, the individuals have suffered
from obliteration of monocyte, macrophages, and granulocytes due to gamma rays exposure.
Thus they have practically no defence against (no cells to eat up) cellular debris, foreign
substances, microbes, cancer cells.
In such cases of severe exposure and damage, immediate medical aid is required. Delay of 6-7
days (as per the given mission mandate) could be fatal for the individuals exposed, especially in
case of acute exposure to ioninzing radiations. Odin and Zeus may not survive for meeting their
new friends on planet X329 without emergency medical help, change of clothes and
radioprotecive drugs..
Many new species of desert have arisen from a population which was .pdf
Many new species of desert have arisen from a population which was introduced into the
American southwest. This example best describes: allopatric speciation. sympatric speciation.
microevolution. adaptive radiation. the effect of climatic change. What is the only source of all
new ? genetic drift gene flew genetic polymorphism mutation natural selection Extensive
data from hospitals have shown that born with intermediate weights are the ones most likely to
survive. This is a classic example of which type of selection? stabilizing selection directional
selection no selection disruptive selection natural selection What are considered to be the two
most important factors that have been involved in past mass ? glaciers and earthquakes disease
and parasites plate tectonics and climatic factors fire and weather events environmental and
biological factors Which was a possible source of high energy no early. Earth that likely
contributed to the origin of
Solution
11. Adaptive radiation: Adaptive radiation is when a species diversify to many different species
from one ancestral species when there is a change in the environment. Change in environment
would be when they are exposed to a different environment like in this case, to a new land.
Change in the environment presents new challenges and advantages in term of resources,
climate, predators etc. Such changes force a species to evolve to adapt.
12. Mutation: Mutation is change in the genetic code which can occur due to many reasons
which change the basic DNA code as compared to what the parent cell has. Mutation can give
the cell new characteristics or even hamper basic functions of a cell or an organism. Hence,
mutation is the only source of change in alleles
Genetic drift can happen only with mutation. Gene flow is basically the inheritance of genes
from the parents to offspring. Genetic polymorphism does not guarantee change in phenotype,
hence not change in alleles. Natural selection is basically survival of the fittest.
13. Stabilizing selection : It is when for a particular trait, the genetic diversity decreases and the
population mean stabilizes. It is assumed to be the most common type of natural selection. When
it comes to phenotypes it supports the intermediate variations.
Human babies who are underweight are susceptible to diseases and lose heat more quickly and
have high mortality rate and similarly for the babies with more than normal weight, which are
difficult to deliver..
List and analyze some of the most salient social factors of group un.pdf
List and analyze some of the most salient social factors of group unrest AND deviance in cities.
Solution
Some of the important social factors form group unrest are listed below.
1. Social inequality
2. Corruption
3. Inflation
4. Government apathy
If the people of the group feels that all members of the group are not being treated equal, sense of
inferiority arises and the member of the group starts feeling harassed. If the same amount of
resources are not available to all for exploitation, there will always be group unrest. When people
are not treated equal among different sections of the society unrest happens. They can be
harassed on the basis of colour, caste, ethnicity and creed.
Corrpution is also an important factor for group unrest. When there is corruption in the society
people of the group feels being cheated. There is impression that they are paying more than what
they should have been. The exploited lots will cause social unrest and raise awareness.
Inflation affects whole section of the society and all members of the group in a society. The rise
in the expenditure on the same amount of commodities bought in a month will provoke the
buyers to take action, sometimes result in social unrest. Usually when the authority is unable to
control inflation and when there is apathy of the authorities towards the members of the group,
the social unrest happens..
J. A. Moore investigated the inheritance of spotting patterns in leop.pdf
J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore. 1943.
Journal of Heredity 34:3-7). The pipiens phenotype had the normal spots that give leopard frogs
their name. In contrast, the burnsi phenotype lacked spots on its back. Moore earned out the
following crosses, producing the progeny indicated. Parent phenotypes Progeny phenotypes
burnsi X burnsi 42 burnsi, 8 pipiens burnsi X pipiens 25 burnsi, 34 pipiens burnsi X pipiens 194
bumsi, 208 pipiens On the basis of these results, what is the most likely mode of inheritance of
the burnsi phenotype? Give the most likely genotypes of the parent in each cross. Use a chi-
square test to evaluate the fit of the observed numbers of progeny to the number expected on the
basis of your proposed genotypes for each of the three matings shown.
Solution
a. On the basis of the above crosses, the mode of inheritance of burnsi phenotype is
DOMINANT PHENOTYPE.
b. The Genotypes of the parents
Burnsi X Burnsi = Bb X Bb
Burnsi X pipiens = Bb X bb
Burnsi X pipiens = Bb X bb
C.
1.
Null hypothesis: The observed values are not deviating from 3:1 ratio.
Chisquare test:
Category
Burnsi
pipiens
Total
Observed values
42
8
50
Exptected Ratio
3
1
4
Exprected Values
37.5
12.5
Deviation
4.5
-4.5
D^2
20.25
20.25
D^2/E
0.54
1.62
2.16
X^2
2.16
Degrees of freedom
-
1
Inference: The calculated chisquare values is less than the table i.e. 3.84 at 1 DF and 0.05
probability, hence null hypothesis accepted.
2.
Null hypothesis: The observed values are not deviating from 1:1 ratio.
Chisquare test:
Category
Burnsi
pipiens
Total
Observed values
28
34
62
Exptected Ratio
1
1
2
Exprected Values
31
31
Deviation
-3
3
D^2
9
9
D^2/E
0.290323
0.290323
0.580645
X^2
0.580645
Degrees of freedom
-
1
Inference: The calculated chisquare values is less than the table i.e. 3.84 at 1 DF and 0.05
probability, hence null hypothesis accepted.
3.
Null hypothesis: The observed values are not deviating from 1:1 ratio.
Chisquare test:
Category
Burnsi
pipiens
Total
Observed values
194
208
402
Exptected Ratio
1
1
2
Exprected Values
201
201
Deviation
-7
7
D^2
49
49
D^2/E
0.243781
0.243781
0.487562
X^2
0.487562
Degrees of freedom
-
1
Inference: The calculated chisquare values is less than the table i.e. 3.84 at 1 DF and 0.05
probability, hence null hypothesis accepted.
Category
Burnsi
pipiens
Total
Observed values
42
8
50
Exptected Ratio
3
1
4
Exprected Values
37.5
12.5
Deviation
4.5
-4.5
D^2
20.25
20.25
D^2/E
0.54
1.62
2.16
X^2
2.16
Degrees of freedom
-
1.
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