Signal transduction cascades are produced by molecular assemblies of .pdf
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Signal transduction cascades are produced by molecular assemblies of which of the following components? Enzymes Regulatory proteins Receptors Transmembrane channels Nuclear pores Solution Answer : The correct answer is C (Receptors) Explanation : Receptors are found on the plasmamembrane of the target cells which are bound by primary signal molecules called ligands/hormones. This inturn brings about the synthesis of secondary messengers such as cAMP/cGMP etc., that brings about the changes in cellular physiology through cascade of events..
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In what ways does interferon eliminate pathogensSolutionInter.pdf
In what ways does interferon eliminate pathogens?
Solution
Interferons are a group of proteins called cytokines produced by white blood cells, fibroblasts, or
T-cells as part of an immune response to a viral infection or other immune trigger. IFNs are
released by host cells in response to the presence of pathogens. Interferons do not directly kill
viral or cancerous cells; they boost the immune system response and reduce the growth of cancer
cells by regulating the action of several genes that control the secretion of numerous cellular
proteins that affect growth. The type I IFNs (IFN- and ) instruct infected cells to die by
apoptosis. During apoptosis the virus is prevented from replicating and apoptosis also halts the
spread of the infection..
In 2006 you joined a group of 20 Richland College graduates to start.pdf
In 2006 you joined a group of 20 Richland College graduates to start a company that provides
innovative tutoring and educational services You are responsible for all accounting activities.
After applying to the Secretary of State in Austin, Shannon’s Tutoring Corporation
(STC)received its corporate charter and began business as a Texas Corporation on December 1,
2006. During the first month of business the following Transactions occurred. You must analyze
each transaction’s effect on the Accounting Equation and Prepare the Income Statement,
Statement of Retained Earnings and the Balance Sheet for the Year ended 12/31/2006.
12/01/06
STC began business by investing $50,000 in exchange for 10,000 shares of Common Stock.
12/1/06
STC Paid $12,000 for 1 year of rent for the 12 month period beginning December 1, 2006
through November 30, 2007. (Hint: Prepaying rent in advance creates an Asset called Prepaid
Rent )
12/1/06
STC received $18,000 cash in advance from the Texas School System and signed a contract to
provide monthly tutoring services to students from December 2006 through May of 2007.
12/02/06
STC purchased a Computer System for $35,000 cash.
12/09/06
STC earned & received $1,000 in cash by providing tutoring services.
12/12/06
STC Billed the RISD $2,500 for tutoring services performed at their school. Payment has not
been received.
12/12/06
STC Purchased $500 of Supplies on Account.
12/21/06
STC Paid $400 Wages to employees.
12/23/06
STC Received $1,500 Cash from RISD for services billed on 12/12/06(#6).
12/29/06
STC Paid a $300 Utility Bill.
12/01/06
STC began business by investing $50,000 in exchange for 10,000 shares of Common Stock.
12/1/06
STC Paid $12,000 for 1 year of rent for the 12 month period beginning December 1, 2006
through November 30, 2007. (Hint: Prepaying rent in advance creates an Asset called Prepaid
Rent )
12/1/06
STC received $18,000 cash in advance from the Texas School System and signed a contract to
provide monthly tutoring services to students from December 2006 through May of 2007.
12/02/06
STC purchased a Computer System for $35,000 cash.
12/09/06
STC earned & received $1,000 in cash by providing tutoring services.
12/12/06
STC Billed the RISD $2,500 for tutoring services performed at their school. Payment has not
been received.
12/12/06
STC Purchased $500 of Supplies on Account.
12/21/06
STC Paid $400 Wages to employees.
12/23/06
STC Received $1,500 Cash from RISD for services billed on 12/12/06(#6).
12/29/06
STC Paid a $300 Utility Bill.
Solution
Answer
Accounting Equation: Total Assets = Capital/Equity + Liabilities
Note: After every transaction, A column = B column
Date
Tran saction no
Total Assets
=
Equity
+
Liabilities
TOTAL (B)
Cash
Fixed Assets
Supplies
Accounts Receivables
Prepaid Rent
TOTAL(A)
=
Common Stock
+
Advance from Customer
12-01-2006
1
50000
50000
=
50000
+
50000
TOTAL
50000
50000
=
50000
+
50000
12-01-2006
2
-12000
11000
-1000
=
-1000
+
-1000
TOTAL
38000
11000
49000
=
49000
+
49000
12-01-20.
If the leader has been dealing with the follower appropriately by us.pdf
If the leader has been dealing with the follower appropriately by using an S4 leadership style and
the follower suddenly begins performing at Performance Readiness® level R3, what should the
leader do? (10 points)
Solution
the leader has to approach the follower and ask him to do it in a right way only. otherwise the
total efforts are going to be spoil. even the other people will suffer with one mistake by one
person in the team..
How does the structure of vertebrae aid in their functionSoluti.pdf
How does the structure of vertebrae aid in their function?
Solution
The vertebral column or the back bone is the main part of the axial skeleton performing the
function of support. It comprises of 33 vertebrae,each of them having specific function and
structure.
Vertebrae are made up of building blocks of spines,stacked one upon the other with a disc
between them.the disc provides cushion and help in the absorption of shock /load and absorbs
energy.
The vertebrae are further classified into the following types based on their loction :
Cervical spine: it is the uppermost part of the verebral column made up of 7 vertebrae labelled as
c1 to c7 from the top. Rotation of the neck is brought about by the two vertebrae. The c1 or the
atlas connects the head and the rest of the spine.
C2 or the axis has a bony process called the odontoid process which fits within a hole in the atlas
to allow the rotation of the neck.The first spinal curve is located at the cervical spine , bent
slightly inward and called the lordotic curve.The cervical spine has opening in each vertebral
body that allows the blood vessels to carry blood the blood to the brain.
Thoraicic spine : It is made up of 12 veretbrae labelled as T1 -T12 in the thoraicic region or the
chest region.The vertebral curve is bent outwards like backward C and is called the kyphotic
curve.This region shows less motion and hence subjected to less wear and tear.
Lumbar spine: Has 5 vertebrae L1-L5 connects the spine and the pelvis. .Bear maximum weight
of the body and have the largest vertebrae.The cuve is bent inward . The paired facets , joints on
the c\\back are aligned in such a way that they allow the flexion and extension but not rotation.
Sacrum lies below the lumbar and is made up of many number of fused small bone and forms the
base of the spine and the back of pelvis. This is followed by the coccyx or the tail bone made up
of sevral small bones.
All the vertebrae are made up of :
vertebral body: which is cylindrical,bony with discs to absorb shock.Carries most of the weight.
spinous processes: Posterior or the rear portions of the vertebrae.
Laminae : two small plates that join in the back of the vertebrae.
Pedicels:short thick bones on either side of the vertebrae which help in the attachment of the
muscles and tendons.
pedicels :Short bumpy backward projecting structures from the upper part.
transverse processes: bony projections on either side of the vertebra where the laminae join the
pedicles which help in the Muscles and ligaments attachment of Muscles and ligament to the
spine— These are the spinal joints, the areas on the spine where one vertebra comes into contact
with another.
facet joints— these are the spinal joints, the areas on the spine where one vertebra comes into
contact with another.allow vertebrae to move against one another allow felxibility and
movement.help us in bending forward and backward and also side to side.the spinal canal allows
the spinal cord and the nerves to .
Homework W1 Homew Score 0 of 1 pt 9.4.17 X Determine the vo.pdf
Homework: W1 Homew Score: 0 of 1 pt 9.4.17 X Determine the volume of the triangular prism.
The volume of the triangular prism is m or Mype an in decima
Solution
The Volume of a prism = Base Area * Height (of prism)
Here the base is a triangle and its area = 1/2 * 9 * 9 = 81/2 m2
The height of the prism is given as 21 m.
Therefore the volume = 81/2 * 21 = 850.5 m3.
Explain the security implications of HIPPA requirements for hospital.pdf
Explain the security implications of HIPPA requirements for hospital networks.
your responce should be 300 words
Solution
HIPAAstands for Health Insurance Portability and Accountability Act.
Passed in 1996 HIPAA is a federal law that sets a national standard to protect medical records
and other personal health information. The rule defines \"protected health information\" as health
information that:
1. Identifies an individual and
2. Is maintained or exchanged electronically or in hard copy.
If the information has any components that could be used to identify a person, it would be
protected. The protection would stay with the information as long as the information is in the
hands of a covered entity or a business associate.
HIPAA Security Rules
The portion of the HIPAA law that most impacts technology interests is the section on
Administrative Simplification (Title II, Subtitle F). Administrative Simplification seeks to force
uniform standards in the electronic interchange of health information (through the Transaction
Rule) and also mandates guidelines for the security (Security rules) and privacy (Privacy rules)
of that information whether in transit or stored. The HIPAA Security regulations apply to that
protected health information that is electronically maintained or used in an electronic
transmission1
. Administrative Simplification is divided in to Transaction, Security and Privacy Rules.
The HIPAA Security rules are divided into four sections:
· Administrative Safeguards
· Physical Safeguards
· Security Services
· Security Mechanisms
Administrative safeguards deal with those administrative policies, procedures and practices that
are used by a covered entity to handle protected health information. These generally take the
form of written policies and procedures that are practiced in normal day-to-day operations.
Physical safeguards deal with physical access to data and facilities within that contain protected
health information. Security services and security mechanisms specifically address technical
systems, networks and applications that possess or transmit protected
health information.
The HIPAA Security rules mandate that if healthcare information (also referred to in the HIPAA
text as protected health information) is stored or processed electronically, then the security rule
applies to that covered entity. This would seem to exempt pure paper-based operations from the
Security rules, but even
these organizations likely use fax technology, which is covered by the HIPAA security rule.
Accordingly, there are very few healthcare organizations that will escape the grasp of the HIPAA
regulations as very few are entirely paper-based.
HIPAA Security rules essentially resemble a collection of the recommended best practices for
security management and operations. For this reason, if the healthcare organization has already
adopted sound security practices, the HIPAA-compliance effort should be minimal. Given that
Security is not a prime conc.
Explain briefly about the major enhancements in ARM processor archite.pdf
Explain briefly about the major enhancements in ARM processor architecture compared to
conventional RSIC computational model.
Solution
ARM:
• ARM stands for Advanced RISC Machine based on the RISC.
• It has high code density, low power consumption & low silicon area.
• It is a load-store architecture, data processing through registers and does not involve changes
directly within memory and it gives good speed vs power consumption ratio.
RISC features
Instructions:
Lower number of instructions compared to CISC. The compiler or programmer synthesizes
complicated operations by combining several simple instructions. Each instruction is a fixed
length to allow the pipeline to fetch future instructions before decoding the current instruction.
Pipeline:
The processing of instructions is broken down into smaller units that can be executed in parallel
by pipelines. Ideally the pipeline advances by one step on each cycle for maximum throughput.
Instructions can be decoded in one pipeline stage. There is no need for an instruction to be
executed by a mini-program called microcode as on CISC processors.
Fixed number of instruction cycles: Most instructions single cycle.
Registers:
RISC have a large number of general purpose registers while CISC have special purpose
registers. In RISC any register can contain either data or an address. Registers act as the fast
local memory store for all data processing operations.
Load-store architecture -The processor operates on data held in registers. Separate load and store
instructions transfer data between the register bank and external memory. Memory accesses are
costly, so separating memory accesses from data processing pro-vides an advantage because you
can use data items held in the register bank multiple times without needing multiple memory
accesses. In contrast, with a CISC design the data processing operations can act on memory
directly.
ARM feature improvements over RISC
Variable cycle execution for certain instructions-Not every ARM instruction executes in a single
cycle. For example, load-store-multiple instructions vary in the number of execution cycles
depending upon the number of registers being transferred. The transfer can occur on sequential
memory addresses, which increases performance since sequential memory accesses are often
faster than random accesses.
Inline barrel shifter leading to more complex instructions-The inline barrel shifter is a hardware
component that preprocesses one of the input registers before it is used by an instruction. This
expands the capability of many instructions to improve core performance and code density.
ARM has enhanced the processor core by adding a second 16 bit instruction set called Thumb.
This thumb instruction permits the ARM core to execute either 16 bit or 32 bit instructions. The
16 bit instructions improve code density by about 30 percent compare to 32 bit instructions of
fixed length.
Conditional execution-An instruction is only executed when a specific.
Chapter Cellular Structure and Function Wor which consist of fatty ac.pdf
Chapter Cellular Structure and Function Wor which consist of fatty acids and alcohol. The phos-
The plasma membrane composed mainly of phospholipids, d bilayer As shown in the figure and
the tails pholipids in the plasma membrane are tails. bead \"loves\" water the head and two the
membrane. whereas below, each a on the interior of ne cell because water aydophobic). The
water-hating tails are tars are hand, point outwards, either the cytoplusm or dhe nuid surrounds if
they the other of heads pass the plasma hydrophilic, on exterior Mtolecules that are b can easily
membrane. Molecule that are water-loving like the hating like the interior the cause they are
through de plasma least rot without help pass Phospholipid molecule Phospholipld bilayer
Hydrophilic head Hydrophobic tails hydrophobic, or waterhating, interior The phospholipid
bilayer consists of two layers of phospholipids 0eft), with a depicted on the and a hydrophilic, or
water-loving, esterior, Asingle phospholipid molecule is Other Molecules in the Plasma
Membrane molecules in The plasma membrane also contains other molecules primarily other
lipids and proteins. The green the figure above, for example, are the lipid Molecules of
cholesterol belp the plasma membrane keep its shape. Many of the proteins in the plasma
membrane assist other substances in crossing the membrane. Euensions of the Plasma Membrane
plasma membrane may haveextensions, such as whip like nagelt or ke cilia. In single-celled
organisms. the membrane estensions may belp organisms move. In multicellular organisms, the
extensions have other For example, the cilia the particles the mouth and Dose. Mung cells sweep
foreign and mucustoward What is the plasma membrane? 2. What is the meaning of semipen
nneabiliry?
Solution
1). What is the plasma membrane?
The plasma membrane is a membrane that encloses the cell organelles of a cell. It separates the
cell interior and the outside environment. The plasma membrane is semi-permeable membrane
that allows selective entry and exit to charged ions and organic molecules. It provides protection
to the cell.
2). What is the meaning of semi-permeability?
Semi-permeability is the quality of a membrane to allow passage of some particles through it
while restricting the flow of others. The plasma membrane is composed of two layers of
phospholipid molecules which have polar phosphate head on the either sides and two non-polar
lipid tails embedded inside. Due to this structure, the hydrophilic molecules can not pass through
the membrane. Besides, the membrane has transmembrane proteins that may act as channels to
allow the passage of specific molecules through them with or without any expense of energy..
Recommended
In what ways does interferon eliminate pathogensSolutionInter.pdf
In what ways does interferon eliminate pathogens?
Solution
Interferons are a group of proteins called cytokines produced by white blood cells, fibroblasts, or
T-cells as part of an immune response to a viral infection or other immune trigger. IFNs are
released by host cells in response to the presence of pathogens. Interferons do not directly kill
viral or cancerous cells; they boost the immune system response and reduce the growth of cancer
cells by regulating the action of several genes that control the secretion of numerous cellular
proteins that affect growth. The type I IFNs (IFN- and ) instruct infected cells to die by
apoptosis. During apoptosis the virus is prevented from replicating and apoptosis also halts the
spread of the infection..
In 2006 you joined a group of 20 Richland College graduates to start.pdf
In 2006 you joined a group of 20 Richland College graduates to start a company that provides
innovative tutoring and educational services You are responsible for all accounting activities.
After applying to the Secretary of State in Austin, Shannon’s Tutoring Corporation
(STC)received its corporate charter and began business as a Texas Corporation on December 1,
2006. During the first month of business the following Transactions occurred. You must analyze
each transaction’s effect on the Accounting Equation and Prepare the Income Statement,
Statement of Retained Earnings and the Balance Sheet for the Year ended 12/31/2006.
12/01/06
STC began business by investing $50,000 in exchange for 10,000 shares of Common Stock.
12/1/06
STC Paid $12,000 for 1 year of rent for the 12 month period beginning December 1, 2006
through November 30, 2007. (Hint: Prepaying rent in advance creates an Asset called Prepaid
Rent )
12/1/06
STC received $18,000 cash in advance from the Texas School System and signed a contract to
provide monthly tutoring services to students from December 2006 through May of 2007.
12/02/06
STC purchased a Computer System for $35,000 cash.
12/09/06
STC earned & received $1,000 in cash by providing tutoring services.
12/12/06
STC Billed the RISD $2,500 for tutoring services performed at their school. Payment has not
been received.
12/12/06
STC Purchased $500 of Supplies on Account.
12/21/06
STC Paid $400 Wages to employees.
12/23/06
STC Received $1,500 Cash from RISD for services billed on 12/12/06(#6).
12/29/06
STC Paid a $300 Utility Bill.
12/01/06
STC began business by investing $50,000 in exchange for 10,000 shares of Common Stock.
12/1/06
STC Paid $12,000 for 1 year of rent for the 12 month period beginning December 1, 2006
through November 30, 2007. (Hint: Prepaying rent in advance creates an Asset called Prepaid
Rent )
12/1/06
STC received $18,000 cash in advance from the Texas School System and signed a contract to
provide monthly tutoring services to students from December 2006 through May of 2007.
12/02/06
STC purchased a Computer System for $35,000 cash.
12/09/06
STC earned & received $1,000 in cash by providing tutoring services.
12/12/06
STC Billed the RISD $2,500 for tutoring services performed at their school. Payment has not
been received.
12/12/06
STC Purchased $500 of Supplies on Account.
12/21/06
STC Paid $400 Wages to employees.
12/23/06
STC Received $1,500 Cash from RISD for services billed on 12/12/06(#6).
12/29/06
STC Paid a $300 Utility Bill.
Solution
Answer
Accounting Equation: Total Assets = Capital/Equity + Liabilities
Note: After every transaction, A column = B column
Date
Tran saction no
Total Assets
=
Equity
+
Liabilities
TOTAL (B)
Cash
Fixed Assets
Supplies
Accounts Receivables
Prepaid Rent
TOTAL(A)
=
Common Stock
+
Advance from Customer
12-01-2006
1
50000
50000
=
50000
+
50000
TOTAL
50000
50000
=
50000
+
50000
12-01-2006
2
-12000
11000
-1000
=
-1000
+
-1000
TOTAL
38000
11000
49000
=
49000
+
49000
12-01-20.
If the leader has been dealing with the follower appropriately by us.pdf
If the leader has been dealing with the follower appropriately by using an S4 leadership style and
the follower suddenly begins performing at Performance Readiness® level R3, what should the
leader do? (10 points)
Solution
the leader has to approach the follower and ask him to do it in a right way only. otherwise the
total efforts are going to be spoil. even the other people will suffer with one mistake by one
person in the team..
How does the structure of vertebrae aid in their functionSoluti.pdf
How does the structure of vertebrae aid in their function?
Solution
The vertebral column or the back bone is the main part of the axial skeleton performing the
function of support. It comprises of 33 vertebrae,each of them having specific function and
structure.
Vertebrae are made up of building blocks of spines,stacked one upon the other with a disc
between them.the disc provides cushion and help in the absorption of shock /load and absorbs
energy.
The vertebrae are further classified into the following types based on their loction :
Cervical spine: it is the uppermost part of the verebral column made up of 7 vertebrae labelled as
c1 to c7 from the top. Rotation of the neck is brought about by the two vertebrae. The c1 or the
atlas connects the head and the rest of the spine.
C2 or the axis has a bony process called the odontoid process which fits within a hole in the atlas
to allow the rotation of the neck.The first spinal curve is located at the cervical spine , bent
slightly inward and called the lordotic curve.The cervical spine has opening in each vertebral
body that allows the blood vessels to carry blood the blood to the brain.
Thoraicic spine : It is made up of 12 veretbrae labelled as T1 -T12 in the thoraicic region or the
chest region.The vertebral curve is bent outwards like backward C and is called the kyphotic
curve.This region shows less motion and hence subjected to less wear and tear.
Lumbar spine: Has 5 vertebrae L1-L5 connects the spine and the pelvis. .Bear maximum weight
of the body and have the largest vertebrae.The cuve is bent inward . The paired facets , joints on
the c\\back are aligned in such a way that they allow the flexion and extension but not rotation.
Sacrum lies below the lumbar and is made up of many number of fused small bone and forms the
base of the spine and the back of pelvis. This is followed by the coccyx or the tail bone made up
of sevral small bones.
All the vertebrae are made up of :
vertebral body: which is cylindrical,bony with discs to absorb shock.Carries most of the weight.
spinous processes: Posterior or the rear portions of the vertebrae.
Laminae : two small plates that join in the back of the vertebrae.
Pedicels:short thick bones on either side of the vertebrae which help in the attachment of the
muscles and tendons.
pedicels :Short bumpy backward projecting structures from the upper part.
transverse processes: bony projections on either side of the vertebra where the laminae join the
pedicles which help in the Muscles and ligaments attachment of Muscles and ligament to the
spine— These are the spinal joints, the areas on the spine where one vertebra comes into contact
with another.
facet joints— these are the spinal joints, the areas on the spine where one vertebra comes into
contact with another.allow vertebrae to move against one another allow felxibility and
movement.help us in bending forward and backward and also side to side.the spinal canal allows
the spinal cord and the nerves to .
Homework W1 Homew Score 0 of 1 pt 9.4.17 X Determine the vo.pdf
Homework: W1 Homew Score: 0 of 1 pt 9.4.17 X Determine the volume of the triangular prism.
The volume of the triangular prism is m or Mype an in decima
Solution
The Volume of a prism = Base Area * Height (of prism)
Here the base is a triangle and its area = 1/2 * 9 * 9 = 81/2 m2
The height of the prism is given as 21 m.
Therefore the volume = 81/2 * 21 = 850.5 m3.
Explain the security implications of HIPPA requirements for hospital.pdf
Explain the security implications of HIPPA requirements for hospital networks.
your responce should be 300 words
Solution
HIPAAstands for Health Insurance Portability and Accountability Act.
Passed in 1996 HIPAA is a federal law that sets a national standard to protect medical records
and other personal health information. The rule defines \"protected health information\" as health
information that:
1. Identifies an individual and
2. Is maintained or exchanged electronically or in hard copy.
If the information has any components that could be used to identify a person, it would be
protected. The protection would stay with the information as long as the information is in the
hands of a covered entity or a business associate.
HIPAA Security Rules
The portion of the HIPAA law that most impacts technology interests is the section on
Administrative Simplification (Title II, Subtitle F). Administrative Simplification seeks to force
uniform standards in the electronic interchange of health information (through the Transaction
Rule) and also mandates guidelines for the security (Security rules) and privacy (Privacy rules)
of that information whether in transit or stored. The HIPAA Security regulations apply to that
protected health information that is electronically maintained or used in an electronic
transmission1
. Administrative Simplification is divided in to Transaction, Security and Privacy Rules.
The HIPAA Security rules are divided into four sections:
· Administrative Safeguards
· Physical Safeguards
· Security Services
· Security Mechanisms
Administrative safeguards deal with those administrative policies, procedures and practices that
are used by a covered entity to handle protected health information. These generally take the
form of written policies and procedures that are practiced in normal day-to-day operations.
Physical safeguards deal with physical access to data and facilities within that contain protected
health information. Security services and security mechanisms specifically address technical
systems, networks and applications that possess or transmit protected
health information.
The HIPAA Security rules mandate that if healthcare information (also referred to in the HIPAA
text as protected health information) is stored or processed electronically, then the security rule
applies to that covered entity. This would seem to exempt pure paper-based operations from the
Security rules, but even
these organizations likely use fax technology, which is covered by the HIPAA security rule.
Accordingly, there are very few healthcare organizations that will escape the grasp of the HIPAA
regulations as very few are entirely paper-based.
HIPAA Security rules essentially resemble a collection of the recommended best practices for
security management and operations. For this reason, if the healthcare organization has already
adopted sound security practices, the HIPAA-compliance effort should be minimal. Given that
Security is not a prime conc.
Explain briefly about the major enhancements in ARM processor archite.pdf
Explain briefly about the major enhancements in ARM processor architecture compared to
conventional RSIC computational model.
Solution
ARM:
• ARM stands for Advanced RISC Machine based on the RISC.
• It has high code density, low power consumption & low silicon area.
• It is a load-store architecture, data processing through registers and does not involve changes
directly within memory and it gives good speed vs power consumption ratio.
RISC features
Instructions:
Lower number of instructions compared to CISC. The compiler or programmer synthesizes
complicated operations by combining several simple instructions. Each instruction is a fixed
length to allow the pipeline to fetch future instructions before decoding the current instruction.
Pipeline:
The processing of instructions is broken down into smaller units that can be executed in parallel
by pipelines. Ideally the pipeline advances by one step on each cycle for maximum throughput.
Instructions can be decoded in one pipeline stage. There is no need for an instruction to be
executed by a mini-program called microcode as on CISC processors.
Fixed number of instruction cycles: Most instructions single cycle.
Registers:
RISC have a large number of general purpose registers while CISC have special purpose
registers. In RISC any register can contain either data or an address. Registers act as the fast
local memory store for all data processing operations.
Load-store architecture -The processor operates on data held in registers. Separate load and store
instructions transfer data between the register bank and external memory. Memory accesses are
costly, so separating memory accesses from data processing pro-vides an advantage because you
can use data items held in the register bank multiple times without needing multiple memory
accesses. In contrast, with a CISC design the data processing operations can act on memory
directly.
ARM feature improvements over RISC
Variable cycle execution for certain instructions-Not every ARM instruction executes in a single
cycle. For example, load-store-multiple instructions vary in the number of execution cycles
depending upon the number of registers being transferred. The transfer can occur on sequential
memory addresses, which increases performance since sequential memory accesses are often
faster than random accesses.
Inline barrel shifter leading to more complex instructions-The inline barrel shifter is a hardware
component that preprocesses one of the input registers before it is used by an instruction. This
expands the capability of many instructions to improve core performance and code density.
ARM has enhanced the processor core by adding a second 16 bit instruction set called Thumb.
This thumb instruction permits the ARM core to execute either 16 bit or 32 bit instructions. The
16 bit instructions improve code density by about 30 percent compare to 32 bit instructions of
fixed length.
Conditional execution-An instruction is only executed when a specific.
Chapter Cellular Structure and Function Wor which consist of fatty ac.pdf
Chapter Cellular Structure and Function Wor which consist of fatty acids and alcohol. The phos-
The plasma membrane composed mainly of phospholipids, d bilayer As shown in the figure and
the tails pholipids in the plasma membrane are tails. bead \"loves\" water the head and two the
membrane. whereas below, each a on the interior of ne cell because water aydophobic). The
water-hating tails are tars are hand, point outwards, either the cytoplusm or dhe nuid surrounds if
they the other of heads pass the plasma hydrophilic, on exterior Mtolecules that are b can easily
membrane. Molecule that are water-loving like the hating like the interior the cause they are
through de plasma least rot without help pass Phospholipid molecule Phospholipld bilayer
Hydrophilic head Hydrophobic tails hydrophobic, or waterhating, interior The phospholipid
bilayer consists of two layers of phospholipids 0eft), with a depicted on the and a hydrophilic, or
water-loving, esterior, Asingle phospholipid molecule is Other Molecules in the Plasma
Membrane molecules in The plasma membrane also contains other molecules primarily other
lipids and proteins. The green the figure above, for example, are the lipid Molecules of
cholesterol belp the plasma membrane keep its shape. Many of the proteins in the plasma
membrane assist other substances in crossing the membrane. Euensions of the Plasma Membrane
plasma membrane may haveextensions, such as whip like nagelt or ke cilia. In single-celled
organisms. the membrane estensions may belp organisms move. In multicellular organisms, the
extensions have other For example, the cilia the particles the mouth and Dose. Mung cells sweep
foreign and mucustoward What is the plasma membrane? 2. What is the meaning of semipen
nneabiliry?
Solution
1). What is the plasma membrane?
The plasma membrane is a membrane that encloses the cell organelles of a cell. It separates the
cell interior and the outside environment. The plasma membrane is semi-permeable membrane
that allows selective entry and exit to charged ions and organic molecules. It provides protection
to the cell.
2). What is the meaning of semi-permeability?
Semi-permeability is the quality of a membrane to allow passage of some particles through it
while restricting the flow of others. The plasma membrane is composed of two layers of
phospholipid molecules which have polar phosphate head on the either sides and two non-polar
lipid tails embedded inside. Due to this structure, the hydrophilic molecules can not pass through
the membrane. Besides, the membrane has transmembrane proteins that may act as channels to
allow the passage of specific molecules through them with or without any expense of energy..
Ellen died with the following property house in joint tenancy wi.pdf
Ellen died with the following property: house in joint tenancy with her husband, stocks owned
individually, a bank account held individually, and household goods. What schedules must her
executor complete?
Ellen died with the following property: house in joint tenancy with her husband, stocks owned
individually, a bank account held individually, and household goods. What schedules must her
executor complete?
Solution
Schedules B, C, E, F of form 706
Schedule B - Stocks and bonds
Schedule C - Mortgages, notes and cash (bank account)
Schedule E - Jointly owned Property (house in joint tenancy with her husband)
Schedule F - Miscellaneous Property (Household Goods).
essages in the process of communication. Active listening is very im.pdf
essages in the process of communication. Active listening is very important. There are so many
listening habits. People have both good and bad listening habits. I also have some good and bad
listening habits. My good listening habits are like; I pay full
Solution
I pay full attention while listening
I concentrate on the subject while listening
I won\'t intervene in between.
Ed Norton, a city maintenance worker, is inspecting a sewer line. To.pdf
Ed Norton, a city maintenance worker, is inspecting a sewer line. To access the sewer line he
removes a manhole cover and climbs into the sewer line. City policy requires that protective
barriers be placed around any open manholes but Ed is in a hurry so instead he sets out a warning
sign and leaves the flashing yellow light on over his truck. Betsy is walking along the street
texting with her boyfriend. Not watching where she is going Betsy falls into the manhole and
suffers serious injuries. Betsy sues Ed in negligence.
6.Assume that Betsy suffered $50,000 in damages. Discuss the contributory fault defenses that
Ed might assert and how the suit would be resolved under the different approaches.
Solution
In such a case, the fault is with both the parties. Ed should highlight the contributory negligence.
Contributory Negligence addresses a way to allocate fault between parties when it is not clear
that whose fault is it. If you are negligent, and your negligence causes another person to become
injured, then you are legally responsible for paying damages. The concept of contributory
negligence is used to characterize conduct that creates an unreasonable risk to one\'s self. The
idea is that an individual has a duty to act as a reasonable person. When a person does not act
this way and injury occurs, that person may be held entirely or partially responsible for the
resulting injury, even though another party was involved in the accident.
Ed was at fault because he did not put a protective barrier around the open manholes. But if
Betsy had been paying attention to the open manhole, she would not have sufferred injuries.
Betsy was not cautious adn is at fault. So Ed definitely does not owe $50,000 in damanges. He
owes less than $50,000 (may be 50% or more)..
Can -1^ (14) be expressed in terms of iSolutionTo express (-1.pdf
Can -1^ (1/4) be expressed in terms of i
Solution
To express (-1)^1/4 and (-1)^(1/6) interms of i.
We know that i= (-1)^(1/2) or i^2 = -1.
-1 = cos pi +sinpi = cos(2n+1)pi +isin(2n+1)pi
Therefore (-1)^1/4 = Cos(2n+1)pi/4+sin(2n+1)pi)/4 = cos(2n+1)pi/2+isin(2n+1)pi/2 , for n =
0,1,2...
1/2^(1/2) +i/sqrt2 , -1/2^(1/2) +i/2^1/2) , -1/2^1/2) -i/2^(1/2) and (1/2^(1/2) +i/2^(1/2).
(-1)^1/6 = {cos(2npi+1) +isin(2npi+1)}^(1/6)
= cos(2n+1)pi/6 +isin(2n+1)pi/6. for n = 0,1,2,3,..
= (1/2)*3^(1/2) +i(1/2) , i , -(1/2)3^(1/2 ) +i/2 , (1/2)*3^(1/2)-i/2, -i , (1/2)3^(1/2) - i ,.
digestive disorders of the GI tractSolutionThe two common dige.pdf
digestive disorders of the GI tract
Solution
The two common digestive disorders of the GI tract are,
1).
Peptic ulcer disease (PUD) is a lesion of the mucosal layer n the stomach or duodenum that may
be caused by pathogenic bacteria or pepsin and other gastric acids.
The most common causes of peptic ulcers include the infection caused by the bacterium
Helicobacter pylori or H. Pylori, medications that disturb the gastric mucosa, psychological
pressures and irregular food habits that result in improper acid secretions that damages gastric
mucosa.
2). Gastro esophageal reflux disorder (GERD) is a physiological abnormality in which the acidic
contents of stomach reflex into esophagus.
The causes of GERD include weakened lower esophageal sphincter, impaired acid clearance in
the stomach due to unhealthy food habits and pregnancy, infection with H.pylori and stress.
Heart burn and acid indigestion are the common symptoms of GERD..
Define ionophore and describe how the compound A23187 affects intrac.pdf
Define ionophore and describe how the compound A23187 affects intracellular Ca2+ levels in
sperm and the capacitation process.
Solution
Ionophore are small hydrophobic lipid soluble molecule that disolve in lipid biolayer and
increase their permeability to specific inorganic ions . Ionophores shield the charge of the ions to
be transported, enabling it to penetrate the hydrophobic interior of the lipid bilayer. there are two
kind of ionophores.
1. Mobile ion channels It transports the K+ ions down its electrochemical gradient
eg.(Valinomycin )
2. channel formers - In membrane it form channel. the channel permits the passage of water and
univalent cations but not anions eg. (gramisidin).
The molecule metioned in the question is a Protein-Tyrosine Kinase (HCK) which involves in
various signaling events capacitation is also one of them.The capacitation depends on PKA
activation mediated by the Ca2+ and HCO3-dependent soluble adenylyl cyclase. It has been
suggested that Ca2+ is transported into the cell by the sperm-specific Ca2+ channel (CatSper)
and HCO3 by the Na+/HCO3 cotransporter..
Describe how a prioritized set of system requirements is developed i.pdf
Describe how a prioritized set of system requirements is developed in the traditional system
development life cycle (SDLC).
Solution
in a tradition sdlc life cycle the system requirements is developed in the following way
step1:it first check or evaluate the system and identify the deficiencies.so to evaluate the system
they will interview the user of the respected system and discussion will be done with the
supporting
person
step2:afer the evaluation is done then the new system requirements are defined.
Consider the equation 1 2x = sin x. (a) Use the Intermediate Val.pdf
Consider the equation 1 ? 2x = sin x.
(a) Use the Intermediate Value Theorem to prove that this equation has a
root in the interval (0, pie/4). Check all the criterions in the IVT.
(b) Use a graphing calculator to find an interval of length 0.01 that contains
this root. Explain how to obtain this interval.
Solution
(a)f(x)=sinx+2x-1 Notice that, f(0) < 0 and f(pi/4) > 0 Therefore, by the
Intermediate Value theorem there exists a zero in the interval [0, pi/4]. Now, is this the only
root? f\'(x) = cos x+2 for all x. Therefore, f(x) is strictly increasing and therefore, the root we
found above is the ONLY root..
A binary encoded signal has a bit-rate of Rb = 28 kbps. An Alternate .pdf
A binary encoded signal has a bit-rate of Rb = 28 kbps. An Alternate Mark Inversion line-code
is used to encode these bits for baseband transmission over copper wire. What would be the
transmission bandwidth required to transmit this signal? B = 112 kHz B = infinity B = 28 kHz
B = 56 kHz B = 14 kHz B = 7 kHz
Solution
Speed or bit rate = nfs = Rb
Bandwidth minimum is Rb/2
Bandwidth maximum is Rb
Normally bandwidth consider to be half of speed
Bandwidth = 28/2
Bandwidth = 14 kbps.
15.) What kind of reaction occurs between Sudan III and hydrophobic .pdf
Sudan III forms hydrophobic interactions with the hydrocarbon chains of lipids. The Sudan III/IV test uses the red, non-polar Sudan III dye to detect lipid hydrocarbon chains through hydrophobic interactions. Alternatively, the Brown Bag test can identify lipids based on the oily nature of their hydrocarbon chains.
1.) a.What was the original definition of a virus before it could be.pdf
1.) a.What was the original definition of a virus before it could be seen with electron
microscopy? Why is that definition no longer functionally useful? What is the current definition
of a virus?
b.What are the differences between cellular organisms (i.e., eukaryotes & prokaryotes) and
viruses? Can antibiotics be used to treat viral infections? Interferons? Why or why not? Why are
some more recently discovered viruses challenging some of these paradigms, especially the one
that classifies viruses as non-living? Explain.
Solution
Ques-1: 1.) What was the original definition of a virus before it could be seen with electron
microscopy? Why is that definition no longer functionally useful? What is the current definition
of a virus?
Answer:
According to degeneracy hypothesis, viruses were considered as small cells that can live as
parasites in other cells. The other definition of viruses before they were seen under electron
microscopy is that the viruses are evolved from macromolecules such as protein and nucleic acid
at the time of cells initially appeared on Earth. at the same time as cells first appeared on earth.
These definitions are not worked because these viruses are not complete cells because they do
not have conspicuous nucleus or cell membrane or any morphological cellular features when
observed under electron microscope. Therefore, the new definition of virus has proposed that
these are technically living species and it is a small in size composed of virion and a segment of
nucleic acid covered by a protein coat referred as capsid often infectious in host cell.
1.Is the following discrete or continuous, number of brothers a pers.pdf
1.Is the following discrete or continuous, number of brothers a person has?
2.Is the following a statistic or parameter? I asked 4 of the 12 people in our family, if this
weekend would be a good time for the birthday party?
Solution
1) number of brothers is clearly discrete as it only takes point values (integers) rather than a
continous range.
2) This is a statistic because it is description of a sample.
1. What two types of ICMP messages are received at the sending hos.pdf
1. What two types of ICMP messages are received at the sending host executing the traceroute
program?
2. True or false: When an OSPF route sends its link state information, it is sent only to those
nodes directly attached neighbors. Explain.
1. What two types of ICMP messages are received at the sending host executing the traceroute
program?
2. True or false: When an OSPF route sends its link state information, it is sent only to those
nodes directly attached neighbors. Explain.
2. True or false: When an OSPF route sends its link state information, it is sent only to those
nodes directly attached neighbors. Explain.
Solution
(1).The two types of messages are received at the srending host executing the traceroute
programm is as follows:
1.Port unreachable
2.Time exceeded i.e TTL=0
(2).Answer:True
Reason:it utilizes the IP multicast to send link state updates .it confirms little processing on
routers which are not concentrating on OSF packets..
C++Write a method Node Nodereverse() which reverses a list..pdf
C++
Write a method Node *Node::reverse() which reverses a list.
Solution
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to reverse the linked list */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* Function to push a node */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
printf(\"Given linked list\ \");
printList(head);
reverse(&head);
printf(\"\ Reversed Linked list \ \");
printList(head);
getchar();
}.
Briefly describe the pathogenesis of typhoid fever when the bacteria.pdf
Briefly describe the pathogenesis of typhoid fever when the bacteria are ingested. Include what
happens to an infected individual systemically and how the person succumbs to the infection.
Solution
Typhoid fever :
Typhoid fever, otherwise called enteric fever, is a possibly lethal multisystemic ailment brought
about essentially by Salmonella enterica, subspecies enterica serovar typhi and, to a lesser
degree, related serovars paratyphi A, B, and C.
The mutable appearances of typhoid fever make this malady a genuine symptomatic test. The
exemplary presentation incorporates fever, disquietude, diffuse stomach torment, and clogging.
Untreated, typhoid fever is an overwhelming sickness that may advance to daze, obtundation,
intestinal discharge, inside puncturing, and passing inside 1 month of onset. Survivors might be
left with long haul or changeless neuropsychiatric confusions.
All pathogenic Salmonella species, when present in the gut are overwhelmed by phagocytic
cells, which then go them through the mucosa and present them to the macrophages in the lamina
propria. Nontyphoidal salmonellae are phagocytized all through the distal ileum and colon. With
toll-like receptor (TLR)–5 and TLR-4/MD2/CD-14 complex, macrophages perceive pathogen-
related atomic examples (PAMPs, for example, flagella and lipopolysaccharides. Macrophages
and intestinal epithelial cells then pull in T cells and neutrophils with interleukin 8 (IL-8),
bringing about irritation and smothering the contamination.
As opposed to the nontyphoidal salmonellae, S typhi and paratyphi enter the host\'s framework
fundamentally through the distal ileum. They have particular fimbriae that stick to the epithelium
over groups of lymphoid tissue in the ileum (Peyer patches), the primary transfer point for
macrophages going from the gut into the lymphatic framework. The microscopic organisms then
actuate their host macrophages to pull in more macrophages.
S typhi has a Vi capsular antigen that veils PAMPs, keeping away from neutrophil-based
irritation, while the most well-known paratyphi serovar, paratyphi A, does not. This may clarify
the more prominent infectivity of typhi contrasted and the majority of its cousins.
Typhoidal salmonella co-pick the macrophages\' cell apparatus for their own particular
proliferation as they are helped through the mesenteric lymph hubs to the thoracic conduit and
the lymphatics and after that through to the reticuloendothelial tissues of the liver, spleen, bone
marrow, and lymph hubs. Once there, they interruption and keep on multiplying until some basic
thickness is come to. A while later, the microorganisms incite macrophage apoptosis, breaking
out into the circulatory system to attack whatever remains of the body. The microorganisms then
taint the gallbladder by means of either bacteremia or direct expansion of contaminated bile. The
outcome is that the living being re-enters the gastrointestinal tract in the bile and reinfects Peyer
patches.
Constant tr.
You are working In a lab with several different macromolecules from F.pdf
You are working In a lab with several different macromolecules from Francisella tulameria a
bacterial species that causes an illness similar to bubonic plague. You drop the lest tube rack
containing purified metabolic enzymes, plasma membranes, chromosomes, and cell walls
Unfortunately your labels also fell off when the tubes hit the floor. Your lab can detect pentose,
hexose, adenine, cytosine, cysteine. and glycerol (AH your other testing equipment is beyond
your training and you don\'t want your boss to know you dropped test tubes In lab (bad call,
especially r a lab. while working with a biohazardous organism)) What test would you use to
differentiate between the tube with DMA and the tube with proteins? What test would you use
to differentiate between tubes with carbohydrates and those with DNA? What test would you
use to identify the test tube with the plasma membranes\' Fill in the cell structures with an A-D
Metabolic enzymes Plasma membranes Chromosomes Cell Walls In the last 20 years, there
has been a dramatic increase in the incidence of fluoroquinolone resistant gram negative bacteria
in patient samples. Fluoroquinolones are antibiotics (like Cipro and Levaquin) that bind to and
inactivate gyrase. Gyrase is an enzyme that is needed to relieve the tension formed by opening a
replication fork on both sides of a replication bubble DNA replication of circular DNA. Without
functional gyrase. DNA cannot be copied in the affected cells. Researchers have isolated and
sequenced the DNA coding part of the gyrase enzyme (gyrA) and found that resistant strains of
bacteria typically have mutations that cause them to substitute one amino acid for another. Based
on your reading of section A1.3 and review of Module 3 s notes and study guide, predict which
alterations would lead to resistance and which would lead to cell death. Explain.
Solution
We use adenine +cytosine and cystiene detection test perform for DNA and protein. Adenine and
cytosine are found in DNA and cystiene is found in protein as making disulphide bridges. So we
take 10-15 ul of each sample in different vials and add detection test enzymes of both on each
vial. Then we can detect via observing optical density.
For carbohydrate and DNA we take hexose (for detecting carbohydrate) and adenine +cytosine
for DNA.
For plasma membrane, glycerol detection test is required. Glycerol is a key constitute of
membrane..
Write a Regular Expression, requirements1. They contain letters, .pdf
Write a Regular Expression, requirements:
1. They contain letters, digits, and punctuation symbols
2. They must contain at least a letter and a number, and at least one but at most two punctuation
symbols
3. They cannot start with a punctuation symbol
Using exclusively the regular expresssion operators epsilon(empty), * (zero or more copies), and
+ (concatenation), write a regular expression denoting the set of all passwords that satisfy these
requirements. For brevity, use: L, D, and P to denote Letter, Digit, and Punctuation, respectively.
My answer: ( LD ) ( P + PP ) ( L + D )* + ( DL ) ( P + PP ) ( L + D )* ... but I don\'t think it is
correct.
Solution
So the required expression can be written as follows:
^[ LDP ]* + [ LD ] [ LDP ]*.
Write a program Grader that that will be used by a program driver to.pdf
Write a program Grader that that will be used by a program driver to enter in grades that a
student has earned and the prints out their final grade. In this particular scenario there are nine
(9) grades for a student throughout the course of the semester. These assignments are
1. HW 01
2. HW 02
3. Lab 01
4. Lab 02
5. Lab 03
6. Lab 04
7. Midterm 01
8. Midterm 02
9. Final Exam
When computing the final numerical grade, use the following weights • Homework - 10% • Labs
- 40% • Midterms - 25% • Final Exam - 25%. Once you have the numerical result for the student,
the Alphabetical grade can be computed using
A - total 90 B+ - 90 > total 85 B - 85 > total 80 C+ - 80 > total 75 C - 75 > total 70 D+ - 70 >
total 65 D - 65 > total 60 F - 60 > total
You have been supplied with a “Driver class” called TestGrader which creates an instance of a
Grader object and runs each method from the Grader class. The methods that you will need to
create are:
The default constructor Grader()- You should create class variables for each assignment, the
numerical total, and (for brevity) a scanner object for your grader class. Initialize all of the
numerical variables to zero (0), and create your Scanner object for user input.
A private method called checkRange that takes a double as an argument and checks to see if the
supplied double is within the acceptable range (0-100). If not, print “Invalid Grade” and exit.
Public “setter” methods (also known as modifiers) for the class variables corresponding to each
assignment (Ex. setHW01(), setLab02(), etc)
A method called calculateFinalGrade that calculates the final grade using the weights specified
above.
A method called printFinalGrade that prints the final alphabetical grade that corresponds to the
numerical total. A method called printGrades() that prints the assignment data
This is what is given by the \"Driver Class\"
/*
*
*/
import java.util.Scanner;
import java.util.InputMismatchException;
public class TestGrader {
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
Grader myGrader = new Grader();
while (true) {
prompt(myGrader);
try {
// Get the user\'s answer
int userinput = scan.nextInt();
// Catch the eol
scan.nextLine();
switch (userinput) {
case 1:
myGrader.setHW01();
myGrader.calculateFinalGrade();
break;
case 2:
myGrader.setHW02();
myGrader.calculateFinalGrade();
break;
case 3:
myGrader.setLab01();
myGrader.calculateFinalGrade();
break;
case 4:
myGrader.setLab02();
myGrader.calculateFinalGrade();
break;
case 5:
myGrader.setLab03();
myGrader.calculateFinalGrade();
break;
case 6:
myGrader.setLab04();
myGrader.calculateFinalGrade();
break;
case 7:
myGrader.setMidterm01();
myGrader.calculateFinalGrade();
break;
case 8:
myGrader.setMidterm02();
myGrader.calculateFinalGrade();
break;
case 9:
myGrader.setFinalExam();
myGrader.calculateFinalGrade();
break;
default:
System.exit(1);
}
} catch (Exception e) {
System.out.println(\"You did not enter a valid input. Exiting.\");
System.exit(1);
}.
Why can Benedicts solution be used to distinguish between glucose .pdf
Benedict's solution can distinguish between glucose and sucrose because it will release carbon dioxide bubbles from sucrose but not from glucose, it turns orange in the presence of sucrose but remains blue in the presence of glucose, and glucose is a reducing sugar while sucrose is not a reducing sugar.
Which of the following is TRUE about alternation of generationsIt.pdf
Which of the following is TRUE about alternation of generations?
It occurs in haploid organisms
Which of the following is TRUE about alternation of generations?
It occurs in haploid organisms
Solution
Answer: It occurs in some multicellular organisms
Reason:
Alternation of generations lifecycle of the land plants: It occurs in some eukaryotes, in which
sporophytic & gametophytic stages both are exhibited by alternating generations (haploid to
diploid & diploid to haploid) of plant species and in fungal species. This phenomenon does not
occur in all sexually reproducing species
In case of land plants, decreased significance of haploid stages of gametophytes and more
advanced and developed large pollen and egg cells (megakaryophytes) are produced due to
production of flowers with inflorescence, stigma, style, anthers, ovules etc.
More dominant- stage of the land plant life cycle (gymnosperm or angiosperm) is the
development of sporophytes from a zygote due to fusion of haploid gametes of pollen with
female ovules finally to form \"a diploid zygote\" with diploid stages, production of endosperms
with cotyledons, seeds, inflorescence, flowers and fruits..
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Ellen died with the following property house in joint tenancy wi.pdf
Ellen died with the following property: house in joint tenancy with her husband, stocks owned
individually, a bank account held individually, and household goods. What schedules must her
executor complete?
Ellen died with the following property: house in joint tenancy with her husband, stocks owned
individually, a bank account held individually, and household goods. What schedules must her
executor complete?
Solution
Schedules B, C, E, F of form 706
Schedule B - Stocks and bonds
Schedule C - Mortgages, notes and cash (bank account)
Schedule E - Jointly owned Property (house in joint tenancy with her husband)
Schedule F - Miscellaneous Property (Household Goods).
essages in the process of communication. Active listening is very im.pdf
essages in the process of communication. Active listening is very important. There are so many
listening habits. People have both good and bad listening habits. I also have some good and bad
listening habits. My good listening habits are like; I pay full
Solution
I pay full attention while listening
I concentrate on the subject while listening
I won\'t intervene in between.
Ed Norton, a city maintenance worker, is inspecting a sewer line. To.pdf
Ed Norton, a city maintenance worker, is inspecting a sewer line. To access the sewer line he
removes a manhole cover and climbs into the sewer line. City policy requires that protective
barriers be placed around any open manholes but Ed is in a hurry so instead he sets out a warning
sign and leaves the flashing yellow light on over his truck. Betsy is walking along the street
texting with her boyfriend. Not watching where she is going Betsy falls into the manhole and
suffers serious injuries. Betsy sues Ed in negligence.
6.Assume that Betsy suffered $50,000 in damages. Discuss the contributory fault defenses that
Ed might assert and how the suit would be resolved under the different approaches.
Solution
In such a case, the fault is with both the parties. Ed should highlight the contributory negligence.
Contributory Negligence addresses a way to allocate fault between parties when it is not clear
that whose fault is it. If you are negligent, and your negligence causes another person to become
injured, then you are legally responsible for paying damages. The concept of contributory
negligence is used to characterize conduct that creates an unreasonable risk to one\'s self. The
idea is that an individual has a duty to act as a reasonable person. When a person does not act
this way and injury occurs, that person may be held entirely or partially responsible for the
resulting injury, even though another party was involved in the accident.
Ed was at fault because he did not put a protective barrier around the open manholes. But if
Betsy had been paying attention to the open manhole, she would not have sufferred injuries.
Betsy was not cautious adn is at fault. So Ed definitely does not owe $50,000 in damanges. He
owes less than $50,000 (may be 50% or more)..
Can -1^ (14) be expressed in terms of iSolutionTo express (-1.pdf
Can -1^ (1/4) be expressed in terms of i
Solution
To express (-1)^1/4 and (-1)^(1/6) interms of i.
We know that i= (-1)^(1/2) or i^2 = -1.
-1 = cos pi +sinpi = cos(2n+1)pi +isin(2n+1)pi
Therefore (-1)^1/4 = Cos(2n+1)pi/4+sin(2n+1)pi)/4 = cos(2n+1)pi/2+isin(2n+1)pi/2 , for n =
0,1,2...
1/2^(1/2) +i/sqrt2 , -1/2^(1/2) +i/2^1/2) , -1/2^1/2) -i/2^(1/2) and (1/2^(1/2) +i/2^(1/2).
(-1)^1/6 = {cos(2npi+1) +isin(2npi+1)}^(1/6)
= cos(2n+1)pi/6 +isin(2n+1)pi/6. for n = 0,1,2,3,..
= (1/2)*3^(1/2) +i(1/2) , i , -(1/2)3^(1/2 ) +i/2 , (1/2)*3^(1/2)-i/2, -i , (1/2)3^(1/2) - i ,.
digestive disorders of the GI tractSolutionThe two common dige.pdf
digestive disorders of the GI tract
Solution
The two common digestive disorders of the GI tract are,
1).
Peptic ulcer disease (PUD) is a lesion of the mucosal layer n the stomach or duodenum that may
be caused by pathogenic bacteria or pepsin and other gastric acids.
The most common causes of peptic ulcers include the infection caused by the bacterium
Helicobacter pylori or H. Pylori, medications that disturb the gastric mucosa, psychological
pressures and irregular food habits that result in improper acid secretions that damages gastric
mucosa.
2). Gastro esophageal reflux disorder (GERD) is a physiological abnormality in which the acidic
contents of stomach reflex into esophagus.
The causes of GERD include weakened lower esophageal sphincter, impaired acid clearance in
the stomach due to unhealthy food habits and pregnancy, infection with H.pylori and stress.
Heart burn and acid indigestion are the common symptoms of GERD..
Define ionophore and describe how the compound A23187 affects intrac.pdf
Define ionophore and describe how the compound A23187 affects intracellular Ca2+ levels in
sperm and the capacitation process.
Solution
Ionophore are small hydrophobic lipid soluble molecule that disolve in lipid biolayer and
increase their permeability to specific inorganic ions . Ionophores shield the charge of the ions to
be transported, enabling it to penetrate the hydrophobic interior of the lipid bilayer. there are two
kind of ionophores.
1. Mobile ion channels It transports the K+ ions down its electrochemical gradient
eg.(Valinomycin )
2. channel formers - In membrane it form channel. the channel permits the passage of water and
univalent cations but not anions eg. (gramisidin).
The molecule metioned in the question is a Protein-Tyrosine Kinase (HCK) which involves in
various signaling events capacitation is also one of them.The capacitation depends on PKA
activation mediated by the Ca2+ and HCO3-dependent soluble adenylyl cyclase. It has been
suggested that Ca2+ is transported into the cell by the sperm-specific Ca2+ channel (CatSper)
and HCO3 by the Na+/HCO3 cotransporter..
Describe how a prioritized set of system requirements is developed i.pdf
Describe how a prioritized set of system requirements is developed in the traditional system
development life cycle (SDLC).
Solution
in a tradition sdlc life cycle the system requirements is developed in the following way
step1:it first check or evaluate the system and identify the deficiencies.so to evaluate the system
they will interview the user of the respected system and discussion will be done with the
supporting
person
step2:afer the evaluation is done then the new system requirements are defined.
Consider the equation 1 2x = sin x. (a) Use the Intermediate Val.pdf
Consider the equation 1 ? 2x = sin x.
(a) Use the Intermediate Value Theorem to prove that this equation has a
root in the interval (0, pie/4). Check all the criterions in the IVT.
(b) Use a graphing calculator to find an interval of length 0.01 that contains
this root. Explain how to obtain this interval.
Solution
(a)f(x)=sinx+2x-1 Notice that, f(0) < 0 and f(pi/4) > 0 Therefore, by the
Intermediate Value theorem there exists a zero in the interval [0, pi/4]. Now, is this the only
root? f\'(x) = cos x+2 for all x. Therefore, f(x) is strictly increasing and therefore, the root we
found above is the ONLY root..
A binary encoded signal has a bit-rate of Rb = 28 kbps. An Alternate .pdf
A binary encoded signal has a bit-rate of Rb = 28 kbps. An Alternate Mark Inversion line-code
is used to encode these bits for baseband transmission over copper wire. What would be the
transmission bandwidth required to transmit this signal? B = 112 kHz B = infinity B = 28 kHz
B = 56 kHz B = 14 kHz B = 7 kHz
Solution
Speed or bit rate = nfs = Rb
Bandwidth minimum is Rb/2
Bandwidth maximum is Rb
Normally bandwidth consider to be half of speed
Bandwidth = 28/2
Bandwidth = 14 kbps.
15.) What kind of reaction occurs between Sudan III and hydrophobic .pdf
Sudan III forms hydrophobic interactions with the hydrocarbon chains of lipids. The Sudan III/IV test uses the red, non-polar Sudan III dye to detect lipid hydrocarbon chains through hydrophobic interactions. Alternatively, the Brown Bag test can identify lipids based on the oily nature of their hydrocarbon chains.
1.) a.What was the original definition of a virus before it could be.pdf
1.) a.What was the original definition of a virus before it could be seen with electron
microscopy? Why is that definition no longer functionally useful? What is the current definition
of a virus?
b.What are the differences between cellular organisms (i.e., eukaryotes & prokaryotes) and
viruses? Can antibiotics be used to treat viral infections? Interferons? Why or why not? Why are
some more recently discovered viruses challenging some of these paradigms, especially the one
that classifies viruses as non-living? Explain.
Solution
Ques-1: 1.) What was the original definition of a virus before it could be seen with electron
microscopy? Why is that definition no longer functionally useful? What is the current definition
of a virus?
Answer:
According to degeneracy hypothesis, viruses were considered as small cells that can live as
parasites in other cells. The other definition of viruses before they were seen under electron
microscopy is that the viruses are evolved from macromolecules such as protein and nucleic acid
at the time of cells initially appeared on Earth. at the same time as cells first appeared on earth.
These definitions are not worked because these viruses are not complete cells because they do
not have conspicuous nucleus or cell membrane or any morphological cellular features when
observed under electron microscope. Therefore, the new definition of virus has proposed that
these are technically living species and it is a small in size composed of virion and a segment of
nucleic acid covered by a protein coat referred as capsid often infectious in host cell.
1.Is the following discrete or continuous, number of brothers a pers.pdf
1.Is the following discrete or continuous, number of brothers a person has?
2.Is the following a statistic or parameter? I asked 4 of the 12 people in our family, if this
weekend would be a good time for the birthday party?
Solution
1) number of brothers is clearly discrete as it only takes point values (integers) rather than a
continous range.
2) This is a statistic because it is description of a sample.
1. What two types of ICMP messages are received at the sending hos.pdf
1. What two types of ICMP messages are received at the sending host executing the traceroute
program?
2. True or false: When an OSPF route sends its link state information, it is sent only to those
nodes directly attached neighbors. Explain.
1. What two types of ICMP messages are received at the sending host executing the traceroute
program?
2. True or false: When an OSPF route sends its link state information, it is sent only to those
nodes directly attached neighbors. Explain.
2. True or false: When an OSPF route sends its link state information, it is sent only to those
nodes directly attached neighbors. Explain.
Solution
(1).The two types of messages are received at the srending host executing the traceroute
programm is as follows:
1.Port unreachable
2.Time exceeded i.e TTL=0
(2).Answer:True
Reason:it utilizes the IP multicast to send link state updates .it confirms little processing on
routers which are not concentrating on OSF packets..
C++Write a method Node Nodereverse() which reverses a list..pdf
C++
Write a method Node *Node::reverse() which reverses a list.
Solution
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to reverse the linked list */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* Function to push a node */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
printf(\"Given linked list\ \");
printList(head);
reverse(&head);
printf(\"\ Reversed Linked list \ \");
printList(head);
getchar();
}.
Briefly describe the pathogenesis of typhoid fever when the bacteria.pdf
Briefly describe the pathogenesis of typhoid fever when the bacteria are ingested. Include what
happens to an infected individual systemically and how the person succumbs to the infection.
Solution
Typhoid fever :
Typhoid fever, otherwise called enteric fever, is a possibly lethal multisystemic ailment brought
about essentially by Salmonella enterica, subspecies enterica serovar typhi and, to a lesser
degree, related serovars paratyphi A, B, and C.
The mutable appearances of typhoid fever make this malady a genuine symptomatic test. The
exemplary presentation incorporates fever, disquietude, diffuse stomach torment, and clogging.
Untreated, typhoid fever is an overwhelming sickness that may advance to daze, obtundation,
intestinal discharge, inside puncturing, and passing inside 1 month of onset. Survivors might be
left with long haul or changeless neuropsychiatric confusions.
All pathogenic Salmonella species, when present in the gut are overwhelmed by phagocytic
cells, which then go them through the mucosa and present them to the macrophages in the lamina
propria. Nontyphoidal salmonellae are phagocytized all through the distal ileum and colon. With
toll-like receptor (TLR)–5 and TLR-4/MD2/CD-14 complex, macrophages perceive pathogen-
related atomic examples (PAMPs, for example, flagella and lipopolysaccharides. Macrophages
and intestinal epithelial cells then pull in T cells and neutrophils with interleukin 8 (IL-8),
bringing about irritation and smothering the contamination.
As opposed to the nontyphoidal salmonellae, S typhi and paratyphi enter the host\'s framework
fundamentally through the distal ileum. They have particular fimbriae that stick to the epithelium
over groups of lymphoid tissue in the ileum (Peyer patches), the primary transfer point for
macrophages going from the gut into the lymphatic framework. The microscopic organisms then
actuate their host macrophages to pull in more macrophages.
S typhi has a Vi capsular antigen that veils PAMPs, keeping away from neutrophil-based
irritation, while the most well-known paratyphi serovar, paratyphi A, does not. This may clarify
the more prominent infectivity of typhi contrasted and the majority of its cousins.
Typhoidal salmonella co-pick the macrophages\' cell apparatus for their own particular
proliferation as they are helped through the mesenteric lymph hubs to the thoracic conduit and
the lymphatics and after that through to the reticuloendothelial tissues of the liver, spleen, bone
marrow, and lymph hubs. Once there, they interruption and keep on multiplying until some basic
thickness is come to. A while later, the microorganisms incite macrophage apoptosis, breaking
out into the circulatory system to attack whatever remains of the body. The microorganisms then
taint the gallbladder by means of either bacteremia or direct expansion of contaminated bile. The
outcome is that the living being re-enters the gastrointestinal tract in the bile and reinfects Peyer
patches.
Constant tr.
You are working In a lab with several different macromolecules from F.pdf
You are working In a lab with several different macromolecules from Francisella tulameria a
bacterial species that causes an illness similar to bubonic plague. You drop the lest tube rack
containing purified metabolic enzymes, plasma membranes, chromosomes, and cell walls
Unfortunately your labels also fell off when the tubes hit the floor. Your lab can detect pentose,
hexose, adenine, cytosine, cysteine. and glycerol (AH your other testing equipment is beyond
your training and you don\'t want your boss to know you dropped test tubes In lab (bad call,
especially r a lab. while working with a biohazardous organism)) What test would you use to
differentiate between the tube with DMA and the tube with proteins? What test would you use
to differentiate between tubes with carbohydrates and those with DNA? What test would you
use to identify the test tube with the plasma membranes\' Fill in the cell structures with an A-D
Metabolic enzymes Plasma membranes Chromosomes Cell Walls In the last 20 years, there
has been a dramatic increase in the incidence of fluoroquinolone resistant gram negative bacteria
in patient samples. Fluoroquinolones are antibiotics (like Cipro and Levaquin) that bind to and
inactivate gyrase. Gyrase is an enzyme that is needed to relieve the tension formed by opening a
replication fork on both sides of a replication bubble DNA replication of circular DNA. Without
functional gyrase. DNA cannot be copied in the affected cells. Researchers have isolated and
sequenced the DNA coding part of the gyrase enzyme (gyrA) and found that resistant strains of
bacteria typically have mutations that cause them to substitute one amino acid for another. Based
on your reading of section A1.3 and review of Module 3 s notes and study guide, predict which
alterations would lead to resistance and which would lead to cell death. Explain.
Solution
We use adenine +cytosine and cystiene detection test perform for DNA and protein. Adenine and
cytosine are found in DNA and cystiene is found in protein as making disulphide bridges. So we
take 10-15 ul of each sample in different vials and add detection test enzymes of both on each
vial. Then we can detect via observing optical density.
For carbohydrate and DNA we take hexose (for detecting carbohydrate) and adenine +cytosine
for DNA.
For plasma membrane, glycerol detection test is required. Glycerol is a key constitute of
membrane..
Write a Regular Expression, requirements1. They contain letters, .pdf
Write a Regular Expression, requirements:
1. They contain letters, digits, and punctuation symbols
2. They must contain at least a letter and a number, and at least one but at most two punctuation
symbols
3. They cannot start with a punctuation symbol
Using exclusively the regular expresssion operators epsilon(empty), * (zero or more copies), and
+ (concatenation), write a regular expression denoting the set of all passwords that satisfy these
requirements. For brevity, use: L, D, and P to denote Letter, Digit, and Punctuation, respectively.
My answer: ( LD ) ( P + PP ) ( L + D )* + ( DL ) ( P + PP ) ( L + D )* ... but I don\'t think it is
correct.
Solution
So the required expression can be written as follows:
^[ LDP ]* + [ LD ] [ LDP ]*.
Write a program Grader that that will be used by a program driver to.pdf
Write a program Grader that that will be used by a program driver to enter in grades that a
student has earned and the prints out their final grade. In this particular scenario there are nine
(9) grades for a student throughout the course of the semester. These assignments are
1. HW 01
2. HW 02
3. Lab 01
4. Lab 02
5. Lab 03
6. Lab 04
7. Midterm 01
8. Midterm 02
9. Final Exam
When computing the final numerical grade, use the following weights • Homework - 10% • Labs
- 40% • Midterms - 25% • Final Exam - 25%. Once you have the numerical result for the student,
the Alphabetical grade can be computed using
A - total 90 B+ - 90 > total 85 B - 85 > total 80 C+ - 80 > total 75 C - 75 > total 70 D+ - 70 >
total 65 D - 65 > total 60 F - 60 > total
You have been supplied with a “Driver class” called TestGrader which creates an instance of a
Grader object and runs each method from the Grader class. The methods that you will need to
create are:
The default constructor Grader()- You should create class variables for each assignment, the
numerical total, and (for brevity) a scanner object for your grader class. Initialize all of the
numerical variables to zero (0), and create your Scanner object for user input.
A private method called checkRange that takes a double as an argument and checks to see if the
supplied double is within the acceptable range (0-100). If not, print “Invalid Grade” and exit.
Public “setter” methods (also known as modifiers) for the class variables corresponding to each
assignment (Ex. setHW01(), setLab02(), etc)
A method called calculateFinalGrade that calculates the final grade using the weights specified
above.
A method called printFinalGrade that prints the final alphabetical grade that corresponds to the
numerical total. A method called printGrades() that prints the assignment data
This is what is given by the \"Driver Class\"
/*
*
*/
import java.util.Scanner;
import java.util.InputMismatchException;
public class TestGrader {
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
Grader myGrader = new Grader();
while (true) {
prompt(myGrader);
try {
// Get the user\'s answer
int userinput = scan.nextInt();
// Catch the eol
scan.nextLine();
switch (userinput) {
case 1:
myGrader.setHW01();
myGrader.calculateFinalGrade();
break;
case 2:
myGrader.setHW02();
myGrader.calculateFinalGrade();
break;
case 3:
myGrader.setLab01();
myGrader.calculateFinalGrade();
break;
case 4:
myGrader.setLab02();
myGrader.calculateFinalGrade();
break;
case 5:
myGrader.setLab03();
myGrader.calculateFinalGrade();
break;
case 6:
myGrader.setLab04();
myGrader.calculateFinalGrade();
break;
case 7:
myGrader.setMidterm01();
myGrader.calculateFinalGrade();
break;
case 8:
myGrader.setMidterm02();
myGrader.calculateFinalGrade();
break;
case 9:
myGrader.setFinalExam();
myGrader.calculateFinalGrade();
break;
default:
System.exit(1);
}
} catch (Exception e) {
System.out.println(\"You did not enter a valid input. Exiting.\");
System.exit(1);
}.
Why can Benedicts solution be used to distinguish between glucose .pdf
Benedict's solution can distinguish between glucose and sucrose because it will release carbon dioxide bubbles from sucrose but not from glucose, it turns orange in the presence of sucrose but remains blue in the presence of glucose, and glucose is a reducing sugar while sucrose is not a reducing sugar.
Which of the following is TRUE about alternation of generationsIt.pdf
Which of the following is TRUE about alternation of generations?
It occurs in haploid organisms
Which of the following is TRUE about alternation of generations?
It occurs in haploid organisms
Solution
Answer: It occurs in some multicellular organisms
Reason:
Alternation of generations lifecycle of the land plants: It occurs in some eukaryotes, in which
sporophytic & gametophytic stages both are exhibited by alternating generations (haploid to
diploid & diploid to haploid) of plant species and in fungal species. This phenomenon does not
occur in all sexually reproducing species
In case of land plants, decreased significance of haploid stages of gametophytes and more
advanced and developed large pollen and egg cells (megakaryophytes) are produced due to
production of flowers with inflorescence, stigma, style, anthers, ovules etc.
More dominant- stage of the land plant life cycle (gymnosperm or angiosperm) is the
development of sporophytes from a zygote due to fusion of haploid gametes of pollen with
female ovules finally to form \"a diploid zygote\" with diploid stages, production of endosperms
with cotyledons, seeds, inflorescence, flowers and fruits..
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Briefly describe the pathogenesis of typhoid fever when the bacteria.pdf
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- 1. Signal transduction cascades are produced by molecular assemblies of which of the following components? Enzymes Regulatory proteins Receptors Transmembrane channels Nuclear pores Solution Answer : The correct answer is C (Receptors) Explanation : Receptors are found on the plasmamembrane of the target cells which are bound by primary signal molecules called ligands/hormones. This inturn brings about the synthesis of secondary messengers such as cAMP/cGMP etc., that brings about the changes in cellular physiology through cascade of events.