why do people eating outside rather than home cook SolutionFi.pdf
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why do people eating outside rather than home cook ? Solution First : They dont have enough time to cook. And financially they are strong so that they can afford to eat outside. Second: They dont like home cook food. Third: May be the person as very wide social circle due that he/she has to attend many parties or social gathering. That leads to eat outside..
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I need help with this two methods in java. Here are the guidelines. .pdf
I need help with this two methods in java. Here are the guidelines. The methods private int
stringHash(String s) and private int RabinKarpHashes(String s, int[] hashes, int pos, int length).
EmployeeManager
EmployeeManager
- employees : Employee[]
- employeeMax : final int = 10
-currentEmployees : int
<> EmployeeManager
+ addEmployee( type : int, fn : String, ln : String, m : char, g : char, en : int, ft : boolean, amount
: double)
+ removeEmployee( index : int)
+ listAll()
+ listHourly()
+ listSalary()
+ listCommision()
+ resetWeek()
+ calculatePayout() : double
+ getIndex( empNum : int ) : int
+ annualRaises()
+ holidayBonuses() : double
+ increaseHours( index : int, amount : double)
+ increaseSales( index : int, amount : double)
+ findAllBySubstring(find : String) : Employee[]
- RabinKarp(name : String, find : String) : int
- stringHash(s : String) : int
- charNumericValue(c : char) : int
- RabinKarpHashes(s : String, hashes : int[], pos : int, length : int) : int
- linearSearchRecursive(nameHashes : int[], findHash : int, pos : int) : int
public EmployeeManager()
Constructor, creates the Employee array, sets currentEmployees to 0.
public void addEmployee(int, String, String, char, char, int, Boolean, double)
Takes an int representing the type of Employee to be added (1 – Hourly, 2 – Salary, 3 –
Commission) as well as the required data to create that Employee. If one of these values is not
passed output the line, “Invalid Employee Type, None Added”, and exit the method. If an
Employee with the given Employee Number already exists do not add the Employee and output
the line, “Duplicate Not Added”, and exit the method. If the array is at maximum capacity do not
add the new Employee, and output the line, \"Cannot add more Employees\".
public void removeEmployee(int)
Removes an Employee located at the given index from the Employee array.
public void listAll()
Lists all the current Employees. Outputs there are none if there are none.
public void listHourly()
Lists all the current HourlyEmployees. Outputs there are none if there are none.
public void listSalary()
Lists all the current SalaryEmployees. Outputs there are none if there are none.
public void listCommission()
Lists all the current CommissionEmployees. Outputs there are none if there are none.
public void resetWeek()
Resets the week for all Employees.
public double calculatePayout()
Returns the total weekly payout for all Employees.
public int getIndex(int)
Given an Employee Number, returns the index of that Employee in the array, if the Employee
doesn’t exist retuns -1.
public void annualRaises()
Applies annual raise to all current Employees.
public double holidayBonuses()
Outputs and returns the total holiday bonus of all Employees.
public void increaseHours(int, double)
Increase the hours worked of the Employee at the given index by the given double amount.
public void increaseSales(int, double)
Increase the sales of the Employee at the given index by the given double amount.
publi.
How would you design an experiment to test whether M-CSF growth fact.pdf
How would you design an experiment to test whether M-CSF growth factor receptor signaling is
dependent on dimerization of the receptor?
Solution
Answer:
M-CSF growth factor receptor is a member of a family of homodimeric cytokines that share
some sequence and structural similarity. Binding of its cognate bivalent ligands by this class of
receptors stabilizes their non-covalent dimerization, permitting receptor activation and trans-
tyrosine phosphorylation of one ICD (Intra-cellular domain) by the other.
Once the growth factor has bound to and activated the receptor, newly phosphorylated tyrosine
residues on the intracellular carboxy terminus of the receptor serve as binding sites for cytosolic
or membrane-associated proteins that contain phosphotyrosine-binding domains. The best
characterized of these domains are the rc-omology 2 (SH2) domains that share characteristic
features first described in the phosphotyrosine-binding region of the cytosolic tyrosine kinase.
Mona/Gads is a Grb2-related, Src homology 3 (SH3) and SH2 domain-containing adapter
protein whose expression is restricted to cells of hematopoietic lineage. It is known to interact
with the phosphorylated Y697 site of the M-CSFR.
A simple experment to test that M-CSF growth factor receptor signaling is dependent on
dimerization of the receptor is as follows:
1. Preparation of a WT (which can dimerize) and mutant version of M-CSF receptor such that it
cannot dimerize. This can be done through over expression of the protein in insect cells (Sf9
cells) with FLAG-tag and judging for the oligomeric state through size exclusion
chromatography (FPLC). (The insect cells sholud be co-infected with the kinase baculovirus as
well, if the insect cells doesn\'t contain the same. This would facilitate phosphorylation of ICD)
2. Both the receptors can be purified and a Co-Immunoprecipitation (Co-IP) experiment can be
performed with purified Mona/Gads or with cell extracts expressing the same.
3. After elution, a western blot can be performed and probed for the FLAG-tagged receptors. The
dimerized WT receptor should bind to Mona/Gads and appear as bands in the western blot after
developing, whereas the monomeric form should not show up in the blot.
This would prove that Mana/Gads binds to dimeric receptor as compared to the monomeric form
and facilitate the downstream signaling cascades. Similar experiments can be performed in-vivo,
such as in HEK293 or HeLa cells by overexpressing tagged receptor and Mona/Gads and
probing for the binding through western blot..
Explain what is meant by a positive relationship between two variabl.pdf
Explain what is meant by a positive relationship between two variables and a negative
relationship between two variables.
Solution
Positive relationship means that the two variable tend to grow together, i.e, that when one
variable increases, the other tends to increase as well. NEgative relationship is the opposite of
positive, i.e, when one variable increases, the other tends to decrease..
Ethidium homodimer is responsible for the red color observed. How is.pdf
Ethidium homodimer is responsible for the red color observed but cannot cross intact plasma membranes. Calcein is a green dye that can cross membranes but is then trapped inside cells by esterase, which makes it negatively charged. Ethidium homodimer, a positively charged stain, also becomes trapped inside cells with calcein.
BMP signaling can be fine-tuned at the level of Smad158 linker phos.pdf
BMP signaling can be fine-tuned through phosphorylation of Smad1/5/8 linker regions. Smad1 linker phosphorylation integrates diverse signals to control BMP signaling through interplay between phosphorylation, ubiquitination, and exclusion from the nucleus. The level of BMP activity in developing tissues is also fine-tuned by secreted proteins that modulate BMP interaction with receptors and proteases that regulate BMP trap proteins. The 1942 Spemann Organizer graft experiment demonstrated that transplanting the dorsal lip of a salamander embryo induced neighboring cells to adopt more dorsal fates, organizing nervous system, somite and kidney tissues and establishing patterning of the embryo.
Describe the specific brain region(s) associated with language, spee.pdf
Describe the specific brain region(s) associated with language, speech, emotions, memory, and
learning.
Solution
The different brain region(s) associated with following functions are described as below:
Thus, it seems that specific parts of brain are associated with deciphering information of its own
kind but a comprehensive reaction or response is produced only when a combined effort of inter-
related neurons takes place by relaying the information from one part of brain to another. This
makes the brain a highly complex and networked centre which controls all the functions of the
body..
Collected from flower of mountain laurel (Kalmia latifolia) plant. T.pdf
Collected from flower of mountain laurel (Kalmia latifolia) plant. The local media features the
work of your team on their nightly news. During a live interview the reporter asks you to explain
this organism\'s development. How would you respond?
Solution
Kalmialatifolia, commonly known as Mountain laurels\', a broad leaved evergreen shrub found
in the eastern united states.
Field Classification of Kalmia latifolia:
Domain: Eukaryote
Kingdom: Plantae (Plants)
Phylum: Anthophyta
Class: Magnoliopsida (Dicotyledons).
Based on the number of Barr bodies seen in the somatic cells of the f.pdf
Based on the number of Barr bodies seen in the somatic cells of the following individuals
assume their most probable phenotypes and karyotypes: a. For example, John has no Barr body:
i. Phenotype: Normal man ii. Karyotype: 46,XY b. Sue has no Barr body: i. Phenotype: i.
Karyotype:. c. Julie has three Barr bodies: Phenotype: ii. Karyotype:. i. d. Margie is clinically
diagnosed as a case of Turner syndrome. She has 46 chromosomes and an unusually very small
Barr body. In spite of having genotype Aa, she does not show phenotypic mosaicism. Why: Janet
is a carrier for Huntington disease and has decided not to have any child. For this and other
reasons she married Jeff who was diagnosed as a case of klinefleter, But, surprisingly, Janet is
pregnant, and fetal paternity test shows that Jeff is the father of the fetus. Based on Jeff\'s
chromosomal study, the couple have been informed that Jeff still might father another child.
Jeff\'s chromosomal constitution could be e. f. Jim is nfertile, has 46 chromosomes and one Barr
body i. Phenotype could be Karyotype could be
Solution
B. Sue.
Phenotype: Turner syndrome Female.
Karyotype: 45, XO
C.Julie:
Phenotype: Poly X female.
Karyotype: 48, XXXX.
D. In the case of Margie, she has 46 chromosomes and a single Barr body. It is the karyotype of
normal female and not Turner syndrome female. In case of Turner syndrome, one X
chromosome is missing in females and resulting karyotype is 45, XO.
E.
F. Jim being infertile with one Barr body
Phenotype: klinefelter syndrome male.
Karyotype: 47, XXY..
Recommended
I need help with this two methods in java. Here are the guidelines. .pdf
I need help with this two methods in java. Here are the guidelines. The methods private int
stringHash(String s) and private int RabinKarpHashes(String s, int[] hashes, int pos, int length).
EmployeeManager
EmployeeManager
- employees : Employee[]
- employeeMax : final int = 10
-currentEmployees : int
<> EmployeeManager
+ addEmployee( type : int, fn : String, ln : String, m : char, g : char, en : int, ft : boolean, amount
: double)
+ removeEmployee( index : int)
+ listAll()
+ listHourly()
+ listSalary()
+ listCommision()
+ resetWeek()
+ calculatePayout() : double
+ getIndex( empNum : int ) : int
+ annualRaises()
+ holidayBonuses() : double
+ increaseHours( index : int, amount : double)
+ increaseSales( index : int, amount : double)
+ findAllBySubstring(find : String) : Employee[]
- RabinKarp(name : String, find : String) : int
- stringHash(s : String) : int
- charNumericValue(c : char) : int
- RabinKarpHashes(s : String, hashes : int[], pos : int, length : int) : int
- linearSearchRecursive(nameHashes : int[], findHash : int, pos : int) : int
public EmployeeManager()
Constructor, creates the Employee array, sets currentEmployees to 0.
public void addEmployee(int, String, String, char, char, int, Boolean, double)
Takes an int representing the type of Employee to be added (1 – Hourly, 2 – Salary, 3 –
Commission) as well as the required data to create that Employee. If one of these values is not
passed output the line, “Invalid Employee Type, None Added”, and exit the method. If an
Employee with the given Employee Number already exists do not add the Employee and output
the line, “Duplicate Not Added”, and exit the method. If the array is at maximum capacity do not
add the new Employee, and output the line, \"Cannot add more Employees\".
public void removeEmployee(int)
Removes an Employee located at the given index from the Employee array.
public void listAll()
Lists all the current Employees. Outputs there are none if there are none.
public void listHourly()
Lists all the current HourlyEmployees. Outputs there are none if there are none.
public void listSalary()
Lists all the current SalaryEmployees. Outputs there are none if there are none.
public void listCommission()
Lists all the current CommissionEmployees. Outputs there are none if there are none.
public void resetWeek()
Resets the week for all Employees.
public double calculatePayout()
Returns the total weekly payout for all Employees.
public int getIndex(int)
Given an Employee Number, returns the index of that Employee in the array, if the Employee
doesn’t exist retuns -1.
public void annualRaises()
Applies annual raise to all current Employees.
public double holidayBonuses()
Outputs and returns the total holiday bonus of all Employees.
public void increaseHours(int, double)
Increase the hours worked of the Employee at the given index by the given double amount.
public void increaseSales(int, double)
Increase the sales of the Employee at the given index by the given double amount.
publi.
How would you design an experiment to test whether M-CSF growth fact.pdf
How would you design an experiment to test whether M-CSF growth factor receptor signaling is
dependent on dimerization of the receptor?
Solution
Answer:
M-CSF growth factor receptor is a member of a family of homodimeric cytokines that share
some sequence and structural similarity. Binding of its cognate bivalent ligands by this class of
receptors stabilizes their non-covalent dimerization, permitting receptor activation and trans-
tyrosine phosphorylation of one ICD (Intra-cellular domain) by the other.
Once the growth factor has bound to and activated the receptor, newly phosphorylated tyrosine
residues on the intracellular carboxy terminus of the receptor serve as binding sites for cytosolic
or membrane-associated proteins that contain phosphotyrosine-binding domains. The best
characterized of these domains are the rc-omology 2 (SH2) domains that share characteristic
features first described in the phosphotyrosine-binding region of the cytosolic tyrosine kinase.
Mona/Gads is a Grb2-related, Src homology 3 (SH3) and SH2 domain-containing adapter
protein whose expression is restricted to cells of hematopoietic lineage. It is known to interact
with the phosphorylated Y697 site of the M-CSFR.
A simple experment to test that M-CSF growth factor receptor signaling is dependent on
dimerization of the receptor is as follows:
1. Preparation of a WT (which can dimerize) and mutant version of M-CSF receptor such that it
cannot dimerize. This can be done through over expression of the protein in insect cells (Sf9
cells) with FLAG-tag and judging for the oligomeric state through size exclusion
chromatography (FPLC). (The insect cells sholud be co-infected with the kinase baculovirus as
well, if the insect cells doesn\'t contain the same. This would facilitate phosphorylation of ICD)
2. Both the receptors can be purified and a Co-Immunoprecipitation (Co-IP) experiment can be
performed with purified Mona/Gads or with cell extracts expressing the same.
3. After elution, a western blot can be performed and probed for the FLAG-tagged receptors. The
dimerized WT receptor should bind to Mona/Gads and appear as bands in the western blot after
developing, whereas the monomeric form should not show up in the blot.
This would prove that Mana/Gads binds to dimeric receptor as compared to the monomeric form
and facilitate the downstream signaling cascades. Similar experiments can be performed in-vivo,
such as in HEK293 or HeLa cells by overexpressing tagged receptor and Mona/Gads and
probing for the binding through western blot..
Explain what is meant by a positive relationship between two variabl.pdf
Explain what is meant by a positive relationship between two variables and a negative
relationship between two variables.
Solution
Positive relationship means that the two variable tend to grow together, i.e, that when one
variable increases, the other tends to increase as well. NEgative relationship is the opposite of
positive, i.e, when one variable increases, the other tends to decrease..
Ethidium homodimer is responsible for the red color observed. How is.pdf
Ethidium homodimer is responsible for the red color observed but cannot cross intact plasma membranes. Calcein is a green dye that can cross membranes but is then trapped inside cells by esterase, which makes it negatively charged. Ethidium homodimer, a positively charged stain, also becomes trapped inside cells with calcein.
BMP signaling can be fine-tuned at the level of Smad158 linker phos.pdf
BMP signaling can be fine-tuned through phosphorylation of Smad1/5/8 linker regions. Smad1 linker phosphorylation integrates diverse signals to control BMP signaling through interplay between phosphorylation, ubiquitination, and exclusion from the nucleus. The level of BMP activity in developing tissues is also fine-tuned by secreted proteins that modulate BMP interaction with receptors and proteases that regulate BMP trap proteins. The 1942 Spemann Organizer graft experiment demonstrated that transplanting the dorsal lip of a salamander embryo induced neighboring cells to adopt more dorsal fates, organizing nervous system, somite and kidney tissues and establishing patterning of the embryo.
Describe the specific brain region(s) associated with language, spee.pdf
Describe the specific brain region(s) associated with language, speech, emotions, memory, and
learning.
Solution
The different brain region(s) associated with following functions are described as below:
Thus, it seems that specific parts of brain are associated with deciphering information of its own
kind but a comprehensive reaction or response is produced only when a combined effort of inter-
related neurons takes place by relaying the information from one part of brain to another. This
makes the brain a highly complex and networked centre which controls all the functions of the
body..
Collected from flower of mountain laurel (Kalmia latifolia) plant. T.pdf
Collected from flower of mountain laurel (Kalmia latifolia) plant. The local media features the
work of your team on their nightly news. During a live interview the reporter asks you to explain
this organism\'s development. How would you respond?
Solution
Kalmialatifolia, commonly known as Mountain laurels\', a broad leaved evergreen shrub found
in the eastern united states.
Field Classification of Kalmia latifolia:
Domain: Eukaryote
Kingdom: Plantae (Plants)
Phylum: Anthophyta
Class: Magnoliopsida (Dicotyledons).
Based on the number of Barr bodies seen in the somatic cells of the f.pdf
Based on the number of Barr bodies seen in the somatic cells of the following individuals
assume their most probable phenotypes and karyotypes: a. For example, John has no Barr body:
i. Phenotype: Normal man ii. Karyotype: 46,XY b. Sue has no Barr body: i. Phenotype: i.
Karyotype:. c. Julie has three Barr bodies: Phenotype: ii. Karyotype:. i. d. Margie is clinically
diagnosed as a case of Turner syndrome. She has 46 chromosomes and an unusually very small
Barr body. In spite of having genotype Aa, she does not show phenotypic mosaicism. Why: Janet
is a carrier for Huntington disease and has decided not to have any child. For this and other
reasons she married Jeff who was diagnosed as a case of klinefleter, But, surprisingly, Janet is
pregnant, and fetal paternity test shows that Jeff is the father of the fetus. Based on Jeff\'s
chromosomal study, the couple have been informed that Jeff still might father another child.
Jeff\'s chromosomal constitution could be e. f. Jim is nfertile, has 46 chromosomes and one Barr
body i. Phenotype could be Karyotype could be
Solution
B. Sue.
Phenotype: Turner syndrome Female.
Karyotype: 45, XO
C.Julie:
Phenotype: Poly X female.
Karyotype: 48, XXXX.
D. In the case of Margie, she has 46 chromosomes and a single Barr body. It is the karyotype of
normal female and not Turner syndrome female. In case of Turner syndrome, one X
chromosome is missing in females and resulting karyotype is 45, XO.
E.
F. Jim being infertile with one Barr body
Phenotype: klinefelter syndrome male.
Karyotype: 47, XXY..
About 13 of the population is nervous around strangers. If two peop.pdf
About 13% of the population is nervous around strangers. If two people are randomly selected,
what is the probability that both are nervous around strangers? What is the probability at least
one is nervous around strangers?
Solution
P(nervous around stranger)=0.07
(a)P(Both will be nervous around stranger)=P(nervous around stranger)*P(nervous around
stranger)
=0.07*0.07=0.0049
(B) P(atleast one will be nervous)=1-P(none will be nercvous)
=1-0.93*0.93=0.1351.
Assume that a cross is made between tall and dwarf Alsatian red-ribb.pdf
Assume that a cross is made between tall and dwarf Alsatian red-ribbed spider plants. The F1
generation showed intermediate height, while the F2 generation showed a distribution of height
ranging from tall to dwarf, like the original parents, with many heights between the extremes.
These data are consistent with the following mode of inheritance:
A. artificial selection
B. codominance
C. multifactorial inheritance
D. incomplete penetrance
E. alternation of generations
F. hemizygosity
Solution
D. Incomplete penetrance.
Penetrance is the frequency with which individuals of a guven genotype manifest at least some
degree of specific mutant phenotype associated with a triat.In incomplete or reduced penetrance,
some individuals will not express the trait even though they carry the allele..
An example would be height in humans Traits have a wide range of valu.pdf
An example would be height in humans Traits have a wide range of values Can be the result of a
multiple allelic series Usually the result of one to a few genes Traits are discrete and can be
classified into one to a few categories An example would be flower color in Mendel s
experiments Controlled by several to many genes and can be influenced by environmental
conditions Continuous Traits Discontinuous Traits
Solution
Continuous trait : are often measured and given quantitative value these are controlled by many
genes.height etc.here individual differ by small degree.
Following comes under continuous trait:
Discontinuous trait: trait occur in distinct categories .Triat is there or not.
A group of BIOL 1114 students were able to measure various metabolic.pdf
A group of BIOL 1114 students were able to measure various metabolic functions of a white
mouse before and after they let it run on an exercise wheel for 10 minutes. You would correctly
predict that the mouse would_______than before it ran on the exercise wheel
a) move protons (H+) across the inner mitochondrial membrane faster
b) make less metabolic H2O because fewer electrons will reach oxygen
c) show a decreased rate of CO2 production
d) pump more glucose out of the mitochondria
e) pump more oxygen out of the mitochondria
Solution
a) move protons (H+) across the inner mitochondrial membrane faster
During exercise, the metabolic rate increases, which results in more oxygen and glucose
consumption. So, the metabolic water production and carbon dioxide production also increases.
The protons move at a faster rate because the rate of metabolism increases during exercise..
2) By the mid-1950s, there was an increases study inFixed asset m.pdf
2) By the mid-1950s, there was an increases study in:
Fixed asset management: capital budgeting.
Efficient utilization of current assets.
Capital structure composition.
Dividend policy
How was this related to the study of finance?
Solution
Finance is a link between accounting, economics, and other related fields ofstudy.A financial
manager must understand the Federal Reserve System, thecommercial banking system, and the
interrelationships between various sectorsof the economy.BThe financial manager must know
how to interpret and use financial statementsin assessing the firm’s performance and allocating
the firm’s financial resourcesto generate the best return possible in the long run.C.The demand
for financial management skills exists in many sectors of oursociety including corporate
management, financial institutions, and consulting..Evolution of the Field of FinanceA.The field
of finance is closely tied to economics and accounting but achievedrecognition as a separate field
of study in response to the mergers, acquisitionsand growth of corporations in the early
1900s.B.As a result of the Great Depression, emphasis shifted from raising capital, tocapital
preservation, maintenance of liquidity, and increased governmentintervention in the financial
aspects of businesses.C.By the mid-1950s the field of finance had become more decision
oriented withan emphasis on the analysis of resource utilization within the firm. Thisdecision-
making focus was manifested in the increased study of:1.Fixed asset management:capital
budgeting.2.Efficient utilization of current assets.3.Capital structure composition..
6. Which of the following has the lowest orbit a. weather satel.pdf
6. Which of the following has the lowest orbit? a. weather satellite b. spy satellite c. GPS d. TV
satellite
6. Which of the following has the lowest orbit? a. weather satellite b. spy satellite c. GPS d. TV
satellite
6. Which of the following has the lowest orbit? a. weather satellite b. spy satellite c. GPS d. TV
satellite
Solution
a. weather satellite
Weather satellites are in a nearly circular, low Earth orbit. The lowest orbit would be just above
the surface of the earth, A satellite with a low eccentricity orbit moves in a near circle around the
Earth.
You can fill the cell background with color or an image A.color b.pdf
You can fill the cell background with color or an image? A.color b.image c. Nothing d. Color or
an image.
You can fill the cell background with color or an image? A.color b.image c. Nothing d. Color or
an image.
Solution
Ans: D. Color or an Image
You can add background images, solid color, or a color gradient to individual table cells or to an
entire table, depending on what you have selected when you add the background..
When a judge sends an innocent person to jail he is making aa. Ty.pdf
When a judge sends an innocent person to jail he is making a:
a. Type I error
b. Type II error
c. Type III error
d. Type IV error
e. none of the above
Solution
in court verdicts, null hypothesis is: the person is innocent (untill proven guilty)
detecting an innocent person as guilty is equal to rejecting true null hypothesis, i.e,
a. Type I error.
Using an inoculating loop, demonstrate how to aseptically remove some.pdf
Using an inoculating loop, demonstrate how to aseptically remove some inoculum from either a
broth tube, slant tube, or petri plate and inoculate a sterile broth tube, slant tube or petri plate
without introducing outside contamination.
Solution
Answer:
1) Before inserting loop into bacterial culture (broth tube, slant tube and petriplate), loop should
be kept in flame until it appears red, after that wait for few seconds before picking a bacterial
culture (as loop with more heat may kill bacterial culture) and touch from one end culture and
transfer into to new nutrient medium.
All this procedure should be done in sterile cabin called as laminar air flow which blows the air
continuosly..
What are the common DSS analysis techniquesSolutionDSS analys.pdf
What are the common DSS analysis techniques?
Solution
DSS analysis techniques are computer based information systems. These techniques supports an
organization in their decision making process.
Some techniques of DSS analysis are communication driven DSS (used for collaborating,
conducting meetings etc.), data driven DSS (used to query a data base, and deployed via maon
frame), document driven DSS (used to search web pages), knowledge driven DSS (used for
providing management advice) and model driven DSS (helps in selecting different options).
Sensitivity analysis, solver in the excel, what if analysis are some of the DSS methods..
What are the vulnerabilities in the boot process What can an attack.pdf
What are the vulnerabilities in the boot process? What can an attacker exploit?
Solution
attacker can read or write into sdram arbitarily by bypassing firmware write restrictions or
protections
2) A local authenticated attacker may be able to bypass Secure Boot and/or perform an arbitrary
reflash of the platform firmware despite the presence of signed firmware update enforcement.
3)the attacker could corrupt the platform firmware and cause the system to become unusable
4)because of race condition the authenticated attacker can bypass the BIOS write protection
mechanism and write malicious code to the platform firmware.
5).
True or False Chromosome replication occurs during meiosis and not .pdf
True or False: Chromosome replication occurs during meiosis and not during mitosis.
Solution
Chromosome replication occurs during meiosis and not during mitosis --> FALSE
The replication of chromosomes (or DNA replication) occurs both during mitosis and meiosis.
This occurs during the S phase (synthesis phase) of the cell cycle..
What does thyroid-stimulating hormone (TSH), follicle-stimulating ho.pdf
What does thyroid-stimulating hormone (TSH), follicle-stimulating hormone (FSH), and
luteinizing hormone (LH) all have in common?
Solution
Tropic hormones have thier target as other endocrine glands hence these hormones control the
release of other hormones from thier target endocrine glands all these hormones:
thyroid-stimulating hormone (TSH)
follicle-stimulating hormone (FSH)
luteinizing hormone (LH)
are tropic hormones and these hormones are inturn controlled by the secretion of releasing
hormones of the hypothalamus..
To which kind of system can the phasor technique be used Linear sys.pdf
To which kind of system can the phasor technique be used? Linear systems that are periodic in
time. Linear systems that are periodic in space. Linear or nonlinear systems that are periodic in
time. Linear or nonlinear systems that are periodic in space.
Solution
The answer of the above question is ,phasar technique mostly used linear systems that are
periodic in time.
The term Memory-mapped IO means that devices, such as ports and spe.pdf
The term Memory-mapped I/O means that devices, such as ports and special registers, as well as
memory locations, are all accessed or referred to by a unique memory address.
Solution
No in memory mapping I/o scheme same address bus is used to address both memory and I/o
ports or devices so it is not unique.
8.) What is the role of FACT in eukaryotic transcription9.) The c.pdf
8.) What is the role of FACT in eukaryotic transcription?
9.) The chicken ovalbumin gene has five introns and interrupting the coding sequences or exons.
How many R-loops will be observed if DNA of the gene and cytoplasmic mRNA are mixed
under conditions that allow hybridization? How many R-loops will be observed if nuclear pre-
mRNA is used?
10.) What is the function of RNA editing? How does it occur?
11.) Compare and contrast Rho dependent and Rho independent termination in prokaryotes.
12.) Describes the characteristics of promoters and enhancers? What is the function of each?
Solution
8) FACT, the abbreviated form of Facilitates chromatin transcription, is a heterodimeric protein
which helps in the process of elongation in eukaryotic DNA transcription.
The chromatin-coated DNA, which is wound along with histone proteins in nucleosomes, are
difficult to be transcribed because the RNA Polymerase II cannot access the DNA. FACT binds
to the H2A-H2B dimer of histone proteins upstream and removes it from the DNA, allowing
RNA PolI to carry out elongation. Behind the RNA Pol II, FACT reassembles the H2A-H2B
dimer into nucleosomes.
9) R-loop structure is formed by 3-nucleic acids, wherein a double-stranded DNA is hybridized
to an RNA component, leaving behind a looped out single-stranded DNA.
In the laboratory, an R-loop can be formed by hybridization of a double-stranded DNA
(containing introns) to a mature mRNA. The non-complementary region of each intron will form
a single stranded R-loop. Therefore, the chicken ovalbumin gene having 5 introns when
hybridized to cytoplasmic mRNA, will produce 5 R-loops. This is because, the mRNA have
spliced out the intronic regions.
In a pre-mRNA, splicing of introns has not yet taken place. Therefore when a nuclear pre-mRNA
is hybridized to the given DNA, no R-loops will be observed.
10) RNA editing is a process by which transcribed mRNA undergo changes in few or more bases
throughout the polynucleotide chain. These changes are different from the RNA processing
events such as capping, polyadenylation of splicing. In RNA editing, the changes comprise of
base insertion, deletion or substitution events and mostly requires a guide RNA (gRNA) to
accomplish these editing.
11) In prokaryotes, the termination of transcription occurs by either two ways: Rho-independent
or Rho-dependent termination.
Rho-dependent termination: The RNA being transcribed forms a hairpin loop structure which
causes the RNA polymerase to dissociate and causes termination. This is also known an intrinsic
termination.
Rho-dependent termination: A protein called Rho-factor, attaches itself at the termination site of
RNA. When it encounters the RNA Polymerase, it binds and removes the polymerase from the
template, causing termination.
12) Promoters are regions of the DNA located immediately upstream of the transcription start
site. These regions are 200 bp or more in length and usually, have few dispersed conserved
sequences.
Enhancers are reg.
Thank you for the helpSolutionHave a look at the number that i.pdf
Thank you for the help
Solution
Have a look at the number that is multiplying x in your function, i.e., ;
is given by /|b|. Since: period = 1
Use the form a tan(bxc)+d
to find the amplitude, period, phase shift, and verticalshift.
Vertical Asymptotes: x=(1/2)+ n, where n is an integer
Amplitude: None
Period: 1
Phase Shift: 1 (1 to the right)
Vertical Shift:
x =1,5/4,7/4,2,9/4
y=0,1/5,-1/5,0,1/5
graph is last one i.e d).
So this is Bacillus subtillis. Please tell me the cell shape, cell a.pdf
So this is Bacillus subtillis. Please tell me the cell shape, cell arrangements, and the color please.
Thanks.
Solution
Bacillus subtilis as the name indicates is rod shaped bacteria. It is gram positive, catalase positive
bacteria. It is purple when stained with Gram staining. They form small clumps, short chains, or
single cells.
It also forms endospores to tolerate extreme environment..
QUESTION 4 Critical Thinking Scenario The Vice President for Complian.pdf
QUESTION 4 Critical Thinking Scenario The Vice President for Compliance Management,
Alexander Goodenuff, supports the position taken by Henry Wilburn, the plant manager.
Alexander directs the Chief Engineer, Cathy Martin, to prepare a confidential daily
contamination discharge report for him. Cathy is directed to keep this report strictly confidential
and to share it with no one other than Henry Wilburn. Alexander informs both Cathy and Henry
that no new contamination control equipment will be purchased and installed unless or until there
is data indicating a serious violation of environmental regulations is continuing. True False
QUESTION 5 Critical Thinking Scenario Cathy Martin, PE, the Chief Engineer for Metals
Extraction Corporation, is assigned the responsibility for reporting contamination in plant water
discharges to regulatory agencies. In a situation where she is directed by upper management to
make \"adjusted\" or inaccurate reports she is still professionally responsible for her reports to
regulatory agencies True False
Solution
Q4)True-without the any proper and correct report the cheif Engineer can\'t allow the or approve
the purchase of the new material.
Q5)True-even though she is responsible for the improvement or the betterment of the company
but there is a professional work ethics in which she is required to report th correct report..
Problem 4.8 Refer to problem 4.. It is estimated that 0.16 of the .pdf
Problem 4.8: Refer to problem 4.. It is estimated that 0.16% of the adult population in this
country without tuberculosis is infected with HIV. a) Compute the probability that a randomly
selected adult is infected with HIV. b) Given that a randomly selected adult is infected with HIV,
compute the probability that this adult is co-infected with tuberculosis
Solution.
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About 13% of the population is nervous around strangers. If two people are randomly selected,
what is the probability that both are nervous around strangers? What is the probability at least
one is nervous around strangers?
Solution
P(nervous around stranger)=0.07
(a)P(Both will be nervous around stranger)=P(nervous around stranger)*P(nervous around
stranger)
=0.07*0.07=0.0049
(B) P(atleast one will be nervous)=1-P(none will be nercvous)
=1-0.93*0.93=0.1351.
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Assume that a cross is made between tall and dwarf Alsatian red-ribbed spider plants. The F1
generation showed intermediate height, while the F2 generation showed a distribution of height
ranging from tall to dwarf, like the original parents, with many heights between the extremes.
These data are consistent with the following mode of inheritance:
A. artificial selection
B. codominance
C. multifactorial inheritance
D. incomplete penetrance
E. alternation of generations
F. hemizygosity
Solution
D. Incomplete penetrance.
Penetrance is the frequency with which individuals of a guven genotype manifest at least some
degree of specific mutant phenotype associated with a triat.In incomplete or reduced penetrance,
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An example would be height in humans Traits have a wide range of values Can be the result of a
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classified into one to a few categories An example would be flower color in Mendel s
experiments Controlled by several to many genes and can be influenced by environmental
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Solution
Continuous trait : are often measured and given quantitative value these are controlled by many
genes.height etc.here individual differ by small degree.
Following comes under continuous trait:
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predict that the mouse would_______than before it ran on the exercise wheel
a) move protons (H+) across the inner mitochondrial membrane faster
b) make less metabolic H2O because fewer electrons will reach oxygen
c) show a decreased rate of CO2 production
d) pump more glucose out of the mitochondria
e) pump more oxygen out of the mitochondria
Solution
a) move protons (H+) across the inner mitochondrial membrane faster
During exercise, the metabolic rate increases, which results in more oxygen and glucose
consumption. So, the metabolic water production and carbon dioxide production also increases.
The protons move at a faster rate because the rate of metabolism increases during exercise..
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2) By the mid-1950s, there was an increases study in:
Fixed asset management: capital budgeting.
Efficient utilization of current assets.
Capital structure composition.
Dividend policy
How was this related to the study of finance?
Solution
Finance is a link between accounting, economics, and other related fields ofstudy.A financial
manager must understand the Federal Reserve System, thecommercial banking system, and the
interrelationships between various sectorsof the economy.BThe financial manager must know
how to interpret and use financial statementsin assessing the firm’s performance and allocating
the firm’s financial resourcesto generate the best return possible in the long run.C.The demand
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management, financial institutions, and consulting..Evolution of the Field of FinanceA.The field
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6. Which of the following has the lowest orbit? a. weather satellite b. spy satellite c. GPS d. TV
satellite
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satellite
Solution
a. weather satellite
Weather satellites are in a nearly circular, low Earth orbit. The lowest orbit would be just above
the surface of the earth, A satellite with a low eccentricity orbit moves in a near circle around the
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Ans: D. Color or an Image
You can add background images, solid color, or a color gradient to individual table cells or to an
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When a judge sends an innocent person to jail he is making a:
a. Type I error
b. Type II error
c. Type III error
d. Type IV error
e. none of the above
Solution
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Solution
Answer:
1) Before inserting loop into bacterial culture (broth tube, slant tube and petriplate), loop should
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Solution
DSS analysis techniques are computer based information systems. These techniques supports an
organization in their decision making process.
Some techniques of DSS analysis are communication driven DSS (used for collaborating,
conducting meetings etc.), data driven DSS (used to query a data base, and deployed via maon
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reflash of the platform firmware despite the presence of signed firmware update enforcement.
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5).
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follicle-stimulating hormone (FSH)
luteinizing hormone (LH)
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8.) What is the role of FACT in eukaryotic transcription?
9.) The chicken ovalbumin gene has five introns and interrupting the coding sequences or exons.
How many R-loops will be observed if DNA of the gene and cytoplasmic mRNA are mixed
under conditions that allow hybridization? How many R-loops will be observed if nuclear pre-
mRNA is used?
10.) What is the function of RNA editing? How does it occur?
11.) Compare and contrast Rho dependent and Rho independent termination in prokaryotes.
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Solution
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which helps in the process of elongation in eukaryotic DNA transcription.
The chromatin-coated DNA, which is wound along with histone proteins in nucleosomes, are
difficult to be transcribed because the RNA Polymerase II cannot access the DNA. FACT binds
to the H2A-H2B dimer of histone proteins upstream and removes it from the DNA, allowing
RNA PolI to carry out elongation. Behind the RNA Pol II, FACT reassembles the H2A-H2B
dimer into nucleosomes.
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to an RNA component, leaving behind a looped out single-stranded DNA.
In the laboratory, an R-loop can be formed by hybridization of a double-stranded DNA
(containing introns) to a mature mRNA. The non-complementary region of each intron will form
a single stranded R-loop. Therefore, the chicken ovalbumin gene having 5 introns when
hybridized to cytoplasmic mRNA, will produce 5 R-loops. This is because, the mRNA have
spliced out the intronic regions.
In a pre-mRNA, splicing of introns has not yet taken place. Therefore when a nuclear pre-mRNA
is hybridized to the given DNA, no R-loops will be observed.
10) RNA editing is a process by which transcribed mRNA undergo changes in few or more bases
throughout the polynucleotide chain. These changes are different from the RNA processing
events such as capping, polyadenylation of splicing. In RNA editing, the changes comprise of
base insertion, deletion or substitution events and mostly requires a guide RNA (gRNA) to
accomplish these editing.
11) In prokaryotes, the termination of transcription occurs by either two ways: Rho-independent
or Rho-dependent termination.
Rho-dependent termination: The RNA being transcribed forms a hairpin loop structure which
causes the RNA polymerase to dissociate and causes termination. This is also known an intrinsic
termination.
Rho-dependent termination: A protein called Rho-factor, attaches itself at the termination site of
RNA. When it encounters the RNA Polymerase, it binds and removes the polymerase from the
template, causing termination.
12) Promoters are regions of the DNA located immediately upstream of the transcription start
site. These regions are 200 bp or more in length and usually, have few dispersed conserved
sequences.
Enhancers are reg.
Thank you for the helpSolutionHave a look at the number that i.pdf
Thank you for the help
Solution
Have a look at the number that is multiplying x in your function, i.e., ;
is given by /|b|. Since: period = 1
Use the form a tan(bxc)+d
to find the amplitude, period, phase shift, and verticalshift.
Vertical Asymptotes: x=(1/2)+ n, where n is an integer
Amplitude: None
Period: 1
Phase Shift: 1 (1 to the right)
Vertical Shift:
x =1,5/4,7/4,2,9/4
y=0,1/5,-1/5,0,1/5
graph is last one i.e d).
So this is Bacillus subtillis. Please tell me the cell shape, cell a.pdf
So this is Bacillus subtillis. Please tell me the cell shape, cell arrangements, and the color please.
Thanks.
Solution
Bacillus subtilis as the name indicates is rod shaped bacteria. It is gram positive, catalase positive
bacteria. It is purple when stained with Gram staining. They form small clumps, short chains, or
single cells.
It also forms endospores to tolerate extreme environment..
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Alexander Goodenuff, supports the position taken by Henry Wilburn, the plant manager.
Alexander directs the Chief Engineer, Cathy Martin, to prepare a confidential daily
contamination discharge report for him. Cathy is directed to keep this report strictly confidential
and to share it with no one other than Henry Wilburn. Alexander informs both Cathy and Henry
that no new contamination control equipment will be purchased and installed unless or until there
is data indicating a serious violation of environmental regulations is continuing. True False
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Extraction Corporation, is assigned the responsibility for reporting contamination in plant water
discharges to regulatory agencies. In a situation where she is directed by upper management to
make \"adjusted\" or inaccurate reports she is still professionally responsible for her reports to
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Solution
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Event Link:- https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-mule-event-processing-models/
Agenda
● What is event processing in MuleSoft?
● Types of event processing models in Mule 4
● Distinction between the reactive, parallel, blocking & non-blocking processing
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shivani Yasaswi - https://www.linkedin.com/in/shivaniyasaswi/
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
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